![Theorem 2.3. [Huffman’s algorithm] The following algorithm solves Prob-
lem 2.2 with n messages. Order the messages so that p1 ≥ p2 ≥ · · · ≥ pn.
Solve the problem with n − 1 messages M′
1, M′
2, . . . , M′
n−1 such that M′
j has
probability pj for 1 ≤ j ≤ n − 2, but M′
n−1 has probability pn−1 + pn. If
C′
j is the code word corresponding to M′
j, the original problem is solved by
assigning Mj the code word C′
j for 1 ≤ j ≤ n − 2 and Mn−1 the code word
consisting of C′
n−1 followed by 0 and Mn the code word consisting of C′
n−1
followed by 1.
Since the problem is trivial when n = 2 (give M1 the code word 0 and
M2 the code word 1) this gives us what computer programmers and logicians
call a recursive solution.
Recursive programs are often better adapted to machines than human
beings, but it is very easy to follow the steps of Huffman’s algorithm ‘by
hand’.
Example 2.4. Suppose n = 4, Mj has probability j/10 for 1 ≤ j ≤ 4. Apply
Huffman’s algorithm.
Solution. (Note that we do not bother to reorder messages.) Combining
messages in the suggested way, we get
1, 2, 3, 4
[1, 2], 3, 4
[[1, 2], 3], 4.
Working backwards, we get
C[[1,2],3] = 0 . . . , C4 = 1
C[1,2] = 01 . . . , C3 = 00
C1 = 011, C2 = 010.
The reader is strongly advised to do a slightly more complicated example
like the next.
Exercise 2.5. Suppose Mj has probability j/45 for 1 ≤ j ≤ 9. Apply Huff-
man’s algorithm.
As we indicated earlier, the effects of Huffman’s algorithm will be most
marked when a few messages are highly probable.
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![In order to use the Huffman algorithm we need to know the probabilities
of the n possible messages. Suppose we do not. After we have sent k messages
we will know that message Mj has been sent kj times and so will the recipient
of the message. If we decide to use a Huffman code for the next message, it
is not unreasonable (lifting our hat in the direction of the Reverend Thomas
Bayes) to take
pj =
kj + 1
k + n
.
Provided the recipient knows the exact version of the Huffman algorithm
that we use, she can reconstruct our Huffman code and decode our next
message. Variants of this idea are known as ‘Huffman-on-the-fly’ and form
the basis of the kind of compression programs used in your computer. Notice
however, that whilst Theorem 2.3 is an examinable theorem, the contents of
this paragraph form a non-examinable plausible statement.
3 More on prefix-free codes
It might be thought that Huffman’s algorithm says all that is to be said on
the problem it addresses. However, there are two important points that need
to be considered. The first is whether we could get better results by using
codes which are not prefix-free. The object of this section is to show that
this is not the case.
As in section 1, we consider two alphabets A and B and a coding function
c : A → B∗
(where, as we said earlier, B∗
consists of all finite sequences of
elements of B). For most of this course B = {0, 1}, but in this section we
allow B to have D elements. The elements of B∗
are called words.
Lemma 3.1. [Kraft’s inequality 1] If a prefix-free code C consists of n
words Cj of length lj, then
n
X
j=1
D−lj
≤ 1.
Lemma 3.2. [Kraft’s inequality 2] Given strictly positive integers lj sat-
isfying
n
X
j=1
D−lj
≤ 1,
we can find a prefix-free code C consisting of n words Cj of length lj.
Proof. Take l1 ≤ l2 ≤ · · · ≤ ln. We give an inductive construction for an
appropriate prefix-free code. Start by choosing C1 to be any code word of
length l1.
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![Suppose that we have found a collection of r prefix-free code words Ck
of length lk [1 ≤ k ≤ r]. If r = n we are done. If not, consider all possible
code words of length lr+1. Of these Dlr+1−lk
will have prefix Ck so at most
(in fact, exactly)
r
X
k=1
Dlr+1−lk
will have one of the code words already selected as prefix. By hypothesis
r
X
k=1
Dlr+1−lk
= Dlr+1
r
X
k=1
D−lk
< Dlr+1
.
Since there are Dlr+1
possible code words of length lr+1 there is at least
one ‘good code word’ which does not have one of the code words already
selected as prefix. Choose one of the good code words as Cr+1 and restart
the induction.
The method used in the proof is called a ‘greedy algorithm’ because we
just try to do the best we can at each stage without considering future
consequences.
Lemma 3.1 is pretty but not deep. MacMillan showed that the same
inequality applies to all decodable codes. The proof is extremely elegant and
(after one has thought about it long enough) natural.
Theorem 3.3. [The MacMillan inequality] If a decodable code C consists
of n words Cj of length lj, then
n
X
j=1
D−lj
≤ 1.
Using Lemma 3.2 we get the immediate corollary.
Lemma 3.4. If there exists a decodable code C consisting of n words Cj of
length lj, then there exists a prefix-free code C′
consisting of n words C′
j of
length lj.
Thus if we are only concerned with the length of code words we need only
consider prefix-free codes.
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![The Lagrangian is
L(x, λ) =
n
X
j=1
pjxj − λ
n
X
j=1
D−xj
.
Since
∂L
∂xj
= pj + (λ log D)D−xj
we know that, at any stationary point,
D−xj
= K0(λ)pj
for some K0(λ) > 0. Since
Pn
j=1 D−xj
= 1, our original problem will have a
stationarising solution when
D−xj
= pj, that is to say xj = −
log pj
log D
and n
X
j=1
pjxj = −
n
X
j=1
pj
log pj
log D
.
It is not hard to convince oneself that the stationarising solution just
found is, in fact, maximising, but it is an unfortunate fact that IB variational
calculus is suggestive rather than conclusive.
The next two exercises (which will be done in lectures and form part of
the course) provide a rigorous proof.
Exercise 4.4. (i) Show that
log t ≤ t − 1
for t > 0 with equality if and only if t = 1.
(ii) [Gibbs’ inequality] Suppose that pj, qj > 0 and
n
X
j=1
pj =
n
X
j=1
qj = 1.
By applying (i) with t = qj/pj, show that
n
X
j=1
pj log pj ≥
n
X
j=1
pj log qj
with equality if and only if pj = qj.
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![Here |c(A)| denotes the length of c(A). Notice that the result takes a
particularly simple form when D = 2.
In Problem 4.3 the xj are just positive real numbers but in Problem 4.2
the dj are integers. Choosing dj as close as possible to the best xj may not
give the best dj, but it is certainly worth a try.
Theorem 4.9. [Shannon–Fano encoding] Let A and B be finite alphabets
and let B have D symbols. If A is an A-valued random variable, then there
exists a prefix-free (so decodable) code c : A → B which satisfies
E|c(A)| ≤ 1 +
H(A)
log2 D
.
Proof. By Lemma 3.2 (which states that given lengths satisfying Kraft’s
inequality we can construct an associated prefix-free code), it suffices to find
strictly positive integers la such that
X
a∈A
D−la
≤ 1, but
X
a∈A
pala ≤ 1 +
H(A)
log2 D
.
If we take
la = ⌈− logD pa⌉,
that is to say, we take la to be the smallest integer no smaller than − logD pa,
then these conditions are satisfied and we are done.
It is very easy to use the method just indicated to find an appropriate
code. (Such codes are called Shannon–Fano codes5
. Fano was the professor
who set the homework for Huffman. The point of view adopted here means
that for some problems there may be more than one Shannon–Fano code.)
Exercise 4.10. (i) Let A = {1, 2, 3, 4}. Suppose that the probability that
letter k is chosen is k/10. Observing that log 2−n
= −n write down an
appropriate Shannon–Fano code c.
(ii) We found a Huffman code ch for the system in Example 2.4. Show6
that the entropy is approximately 1.85, that E|c(A)| = 2.4 and that E|ch(A)| =
1.9. Check that these results are consistent with our previous theorems.
Putting Theorems 4.8 and Theorem 4.9 together, we get the following
remarkable result.
5
Wikipedia and several other sources give a definition of Shannon–Fano codes which
is definitely inconsistent with that given here. Within a Cambridge examination context
you may assume that Shannon–Fano codes are those considered here.
6
Unless you are on a desert island in which case the calculations are rather tedious.
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![Theorem 4.11. [Shannon’s noiseless coding theorem] Let A and B
be finite alphabets and let B have D symbols. If A is a A-valued random
variable, then any decodable code c which minimises E|c(A)| satisfies
H(A)
log2 D
≤ E|c(A)| ≤ 1 +
H(A)
log2 D
.
In particular, Huffman’s code ch for two symbols satisfies
H(A) ≤ E|ch(A)| ≤ 1 + H(A).
Exercise 4.12. (i) Sketch h(t) = −t log t for 0 ≤ t ≤ 1. (We define h(0) =
0.)
(ii) Let
Γ =
(
p ∈ Rn
: pj ≥ 0,
n
X
j=1
pj = 1
)
and let H : Γ → R be defined by
H(p) =
n
X
j=1
h(pj).
Find the maximum and minimum of H and describe the points where these
values are attained.
(iii) If n = 2r
+s with 0 ≤ s < 2r
and pj = 1/n, describe the Huffman code
ch for two symbols and verify directly that (with notation of Theorem 4.11)
H(A) ≤ E|ch(A)| ≤ 1 + H(A).
Waving our hands about wildly, we may say that ‘A system with low
Shannon entropy is highly organised and, knowing the system, it is usually
quite easy to identify an individual from the system’.
Exercise 4.13. The notorious Trinity gang has just been rounded up and
Trubshaw of the Yard wishes to identify the leader (or Master, as he is called).
Sam the Snitch makes the following offer. Presented with any collection of
members of the gang he will (by a slight twitch of his left ear) indicate if the
Master is among them. However, in view of the danger involved, he demands
ten pounds for each such encounter. Trubshaw believes that the probability of
the jth member of the gang being the Master is pj [1 ≤ j ≤ n] and wishes to
minimise the expected drain on the public purse. Advise him.
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![to the expected pictures, but then changes from what is expected can be dis-
astrous. At present, digital TV encoders merely expect the picture to consist
of blocks which move at nearly constant velocity remaining more or less un-
changed from frame to frame7
. In this, as in other applications, we know
that after compression the signal still has non-trivial statistical properties,
but we do not know enough about them to exploit them.
6 What is an error correcting code?
In the introductory Section 1, we discussed ‘telegraph codes’ in which one
five letter combination ‘QWADR’, say, meant ‘please book quiet room for
two’ and another ‘QWNDR’ meant ‘please book cheapest room for one’.
Obviously, also, an error of one letter in this code could have unpleasant
consequences8
.
Today, we transmit and store long strings of binary sequences, but face the
same problem that some digits may not be transmitted or stored correctly.
We suppose that the string is the result of data compression and so, as we
said at the end of the last section, although the string may have non-trivial
statistical properties, we do not know enough to exploit this fact. (If we knew
how to exploit any statistical regularity, we could build a prediction device
and compress the data still further.) Because of this, we shall assume that
we are asked to consider a collection of m messages each of which is equally
likely.
Our model is the following. When the ‘source’ produces one of the m
possible messages µi say, it is fed into a ‘coder’ which outputs a string ci of n
binary digits. The string is then transmitted one digit at a time along a ‘com-
munication channel’. Each digit has probability p of being mistransmitted
(so that 0 becomes 1 or 1 becomes 0) independently of what happens to the
other digits [0 ≤ p < 1/2]. The transmitted message is then passed through
a ‘decoder’ which either produces a message µj (where we hope that j = i)
or an error message and passes it on to the ‘receiver’. The technical term
for our model is the binary symmetric channel (binary because we use two
symbols, symmetric because the probability of error is the same whichever
symbol we use).
Exercise 6.1. Why do we not consider the case 1 ≥ p > 1/2? What if
p = 1/2?
7
Watch what happens when things go wrong.
8
This is a made up example, since compilers of such codes understood the problem.
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![distance between distinct code words.
We call a code of length n, size m and distance d an [n, m, d] code. Less
briefly, a set C ⊆ Fn
2 , with |C| = m and
min{d(x, y) : x, y ∈ C, x 6= y} = d
is called an [n, m, d] code. By an [n, m] code we shall simply mean a code of
length n and size m.
Lemma 8.2. A code of minimum distance d can detect d − 1 errors18
and
correct ⌊d−1
2
⌋ errors. It cannot detect all sets of d errors and cannot correct
all sets of ⌊d−1
2
⌋ + 1 errors.
It is natural, here and elsewhere, to make use of the geometrical insight
provided by the (closed) Hamming ball
B(x, r) = {y : d(x, y) ≤ r}.
Observe that
|B(x, r)| = |B(0, r)|
for all x and so, writing
V (n, r) = |B(0, r)|,
we know that V (n, r) is the number of points in any Hamming ball of radius
r. A simple counting argument shows that
V (n, r) =
r
X
j=0
n
j
.
Theorem 8.3. [Hamming’s bound] If a code C is e error correcting, then
|C| ≤
2n
V (n, e)
.
There is an obvious fascination (if not utility) in the search for codes
which attain the exact Hamming bound.
Definition 8.4. A code C of length n and size m which can correct e errors
is called perfect if
m =
2n
V (n, e)
.
18
This is not as useful as it looks when d is large. If we know that our message is likely
to contain many errors, all that an error detecting code can do is confirm our expectations.
Error detection is only useful when errors are unlikely.
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![Lemma 8.5. Hamming’s original code is a [7, 16, 3] code. It is perfect.
It may be worth remarking in this context that, if a code which can correct
e errors is perfect (i.e. has a perfect packing of Hamming balls of radius e),
then the decoder must invariably give the wrong answer when presented with
e + 1 errors. We note also that, if (as will usually be the case) 2n
/V (n, e) is
not an integer, no perfect e error correcting code can exist.
Exercise 8.6. Even if 2n
/V (n, e) is an integer, no perfect code may exist.
(i) Verify that
290
V (90, 2)
= 278
.
(ii) Suppose that C is a perfect 2 error correcting code of length 90 and
size 278
. Explain why we may suppose without loss of generality that 0 ∈ C.
(iii) Let C be as in (ii) with 0 ∈ C. Consider the set
X = {x ∈ F90
2 : x1 = 1, x2 = 1, d(0, x) = 3}.
Show that, corresponding to each x ∈ X, we can find a unique c(x) ∈ C such
that d(c(x), x) = 2.
(iv) Continuing with the argument of (iii), show that
d(c(x), 0) = 5
and that ci(x) = 1 whenever xi = 1. If y ∈ X, find the number of solutions
to the equation c(x) = c(y) with x ∈ X and, by considering the number of
elements of X, obtain a contradiction.
(v) Conclude that there is no perfect [90, 278
] code.
The result of Exercise 8.6 was obtained by Golay. Far more importantly,
he found another case when 2n
/V (n, e) is an integer and there does exist an
associated perfect code (the Golay code).
Exercise 8.7. Show that V (23, 3) is a power of 2.
Unfortunately the proof that the Golay code is perfect is too long to be
given in the course,
We obtained the Hamming bound, which places an upper bound on how
good a code can be, by a packing argument. A covering argument gives us
the GSV (Gilbert, Shannon, Varshamov) bound in the opposite direction.
Let us write A(n, d) for the size of the largest code with minimum distance
d.
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![Theorem 8.8. [Gilbert, Shannon, Varshamov] We have
A(n, d) ≥
2n
V (n, d − 1)
.
Until recently there were no general explicit constructions for codes which
achieved the GSV bound (i.e. codes whose minimum distance d satisfied the
inequality A(n, d)V (n, d − 1) ≥ 2n
). Such a construction was finally found
by Garcia and Stichtenoth by using ‘Goppa’ codes.
9 Some elementary probability
Engineers are, of course, interested in ‘best codes’ of length n for reasonably
small values of n, but mathematicians are particularly interested in what
happens as n → ∞.
We recall some elementary probability.
Lemma 9.1. [Tchebychev’s inequality] If X is a bounded real valued
random variable and a 0, then
Pr(|X − EX| ≥ a) ≤
var X
a2
.
Theorem 9.2. [Weak law of large numbers] If X1, X2, . . . is a sequence
of independent identically distributed real valued bounded random variables
and a 0, then
Pr n−1
n
X
j=1
Xj − EX ≥ a
!
→ 0
as N → ∞.
Applying the weak law of large numbers, we obtain the following impor-
tant result.
Lemma 9.3. Consider the model of a noisy transmission channel used in
this course in which each digit has probability p of being wrongly transmitted
independently of what happens to the other digits. If ǫ 0, then
Pr number of errors in transmission for message of n digits ≥ (1 + ǫ)pn
→ 0
as n → ∞.
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![By Lemma 8.2, a code of minimum distance d can correct ⌊d−1
2
⌋ errors.
Thus, if we have an error rate p and ǫ 0, we know that the probability
that a code of length n with error correcting capacity ⌈(1 + ǫ)pn⌉ will fail
to correct a transmitted message falls to zero as n → ∞. By definition, the
biggest code with minimum distance ⌈2(1 + ǫ)pn⌉ has size A(n, ⌈2(1 + ǫ)pn⌉)
and so has information rate log2 A(n, ⌈2(1+ǫ)pn⌉)/n. Study of the behaviour
of log2 A(n, nδ)/n will thus tell us how large an information rate is possible
in the presence of a given error rate.
Definition 9.4. If 0 δ 1/2 we write
α(δ) = lim sup
n→∞
log2 A(n, nδ)
n
.
Definition 9.5. We define the entropy function H : [0, 1] → R by H(0) =
H(1) = 0 and
H(t) = −t log2(t) − (1 − t) log2(1 − t).
Exercise 9.6. (i) We have already met Shannon entropy in Definition 4.7.
Give a simple system such that, using the notation of that definition,
H(A) = H(t).
(ii) Sketch H. What is the value of H(1/2)?
Theorem 9.7. With the definitions just given,
1 − H(δ) ≤ α(δ) ≤ 1 − H(δ/2)
for all 0 ≤ δ 1/2.
Using the Hamming bound (Theorem 8.3) and the GSV bound (Theo-
rem 8.8), we see that Theorem 9.7 follows at once from the following result.
Theorem 9.8. We have
log2 V (n, nδ)
n
→ H(δ)
as n → ∞.
Our proof of Theorem 9.8 depends, as one might expect, on a version of
Stirling’s formula. We only need the very simplest version proved in IA.
Lemma 9.9 (Stirling). We have
loge n! = n loge n − n + O(log2 n).
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![always possible to have an information rate greater than 1 − H(2δ) for an
error rate δ 1/4.
Although we can use repetition codes to get a positive information rate
when 1/4 ≤ δ 1/2 it looks very hard at first (and indeed second) glance to
improve these results.
However, Shannon realised that we do not care whether errors arise be-
cause of noise in transmission or imperfections in our coding scheme. By
allowing our coding scheme to be less than perfect (in this connection, see
Question 25.13) we can actually improve the information rate whilst still
keeping the error rate low.
Theorem 10.1. [Shannon’s noisy coding theorem] Suppose 0 p 1/2
and η 0. Then there exists an n0(p, η) such that, for any n n0, we can
find codes of length n which have the property that (under our standard model
of a symmetric binary channel with probability of error p) the probability
that any codeword is mistaken is less than η and still have information rate
1 − H(p) − η.
Shannon’s theorem is a masterly display of the power of elementary prob-
abilistic arguments to overcome problems which appear insuperable by other
means22
.
However, it merely asserts that good codes exist and gives no means of
finding them apart from exhaustive search. More seriously, random codes
will have no useful structure and the only way to use them is to ‘search
through a large dictionary’ at the coding end and ‘search through an enor-
mous dictionary’ at the decoding end. It should also be noted that n0(p, η)
will be very large when p is close to 1/2.
Exercise 10.2. Why, in the absence of suitable structure, is the dictionary
at the decoding end much larger than the dictionary at the coding end?
It is relatively simple to obtain a converse to Shannon’s theorem.
Theorem 10.3. Suppose 0 p 1/2 and η 0. Then there exists an
n0(p, η) such that, for any n n0, it is impossible to find codes of length
n which have the property that (under our standard model of a symmetric
binary channel with probability of error p) the probability that any codeword
is mistaken is less than 1/2 and the code has information rate 1 − H(p) + η.
22
Conway says that in order to achieve success in a mathematical field you must either
be first or be clever. However, as in the case of Shannon, most of those who are first to
recognise a new mathematical field are also clever.
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![where Yn+1 = uw + (1 − w) if the n + 1st throw is heads and Yn+1 = 1 − w
if it is tails.
Using the weak law of large numbers, we have the following result.
Lemma 11.1. Suppose Y , Y1, Y2, . . . are identically distributed independent
random variables taking values in [a, b] with 0 a b. If we write Zn =
Y1 . . . Yn, then
Pr(|n−1
log Zn − E log Y | ǫ) → 0
as n → 0.
Thus you should choose w to maximise
E log Yn = p log uw + (1 − w)
+ (1 − p) log(1 − w).
Exercise 11.2. (i) Show that, for the situation described, you should not bet
if up ≤ 1 and should take
w =
up − 1
u − 1
if up 1.
(ii) We write q = 1−p. Show that, if up 1 and we choose the optimum
w,
E log Yn = p log p + q log q + log u − q log(u − 1).
We have seen the expression −(p log p + q log q) before as (a multiple
of) the Shannon information entropy of a simple probabilistic system. In
a paper entitled A New Interpretation of Information Rate24
Kelly showed
how to interpret this and similar situations using communication theory. In
his model a gambler receives information over a noisy channel about which
horse is going to win. Just as Shannon’s theorem shows that information
can be transmitted over such a channel at a rate close to channel capacity
with negligible risk of error (provided the messages are long enough), so that
the gambler can (with arbitrarily high probability) increase her fortune at a
certain optimum rate provided that she can continue to bet long enough.
Although the analogy between betting and communication channels is
very pretty, it was the suggestion that those making a long sequence of bets
should aim to maximise the expectation of the logarithm (now called Kelly’s
criterion) which made the paper famous. Although Kelly seems never to
have used his idea in practice, mathematicians like Thorp, Berlekamp and
24
Available on the web. The exposition is slightly opaque because the Bell company
which employed Kelly was anxious not draw attention to the use of telephones for betting
fraud.
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![Shannon himself have made substantial fortunes in the stock market and
claim to have used Kelly’s ideas25
.
Kelly is also famous for an early demonstration of speech synthesis in
which a computer sang ‘Daisy Bell’. This inspired the corresponding scene
in the film 2001.
Before rushing out to the race track or stock exchange26
, the reader is
invited to run computer simulations of the result of Kelly gambling for various
values of u and p. She will observe that although, in the very long run, the
system works, the short run can be be very unpleasant indeed.
Exercise 11.3. Returning to our original problem, show that, if you bet
less than the optimal proportion, your fortune will still tend to increase but
more slowly, but, if you bet more than some proportion w1, your fortune will
decrease. Write down the equation for w1.
[Moral: If you use the Kelly criterion veer on the side under-betting.]
12 Linear codes
The next few sections involve no probability at all. We shall only be inter-
ested in constructing codes which are easy to handle and have all their code
words at least a certain Hamming distance apart.
Just as Rn
is a vector space over R and Cn
is a vector space over C, so
Fn
2 is a vector space over F2. (If you know about vector spaces over fields,
so much the better, if not, just follow the obvious paths.) A linear code is a
subspace of Fn
2 . More formally, we have the following definition.
Definition 12.1. A linear code is a subset of Fn
2 such that
(i) 0 ∈ C,
(ii) if x, y ∈ C, then x + y ∈ C.
Note that, if λ ∈ F, then λ = 0 or λ = 1, so that condition (i) of the
definition just given guarantees that λx ∈ C whenever x ∈ C. We shall see
that linear codes have many useful properties.
Example 12.2. (i) The repetition code with
C = {x : x = (x, x, . . . x)}
is a linear code.
25
However, we hear more about mathematicians who win on the stock market than
those who lose.
26
A sprat which thinks it’s a shark will have a very short life.
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![Lemma 12.20. Under the assumptions and with the notation of Defini-
tion 12.19, the following results are true.
(i) WC is a homogeneous polynomial of degree n.
(ii) If C has rank r, then WC(1, 1) = 2r
.
(iii) WC(0, 1) = 1.
(iv) WC(1, 0) takes the value 0 or 1.
(v) WC(s, t) = WC(t, s) for all s and t if and only if WC(1, 0) = 1.
Lemma 12.21. For our standard model of communication along an error
prone channel with independent errors of probability p and a linear code C
of length n,
WC(p, 1 − p) = Pr(receive a code word | code word transmitted)
and
Pr(receive incorrect code word | code word transmitted) = WC(p, 1−p)−(1−p)n
.
Example 12.22. (i) If C is the repetition code, WC(s, t) = sn
+ tn
.
(ii) If C is the paper tape code of length n, WC(s, t) = 1
2
((s+t)n
+(t−s)n
).
Example 12.22 is a special case of the MacWilliams identity.
Theorem 12.23. [MacWilliams identity] If C is a linear code
WC⊥ (s, t) = 2− dim C
WC(t − s, t + s).
We give a proof as Exercise 26.9. (The result is thus not bookwork though
it could be set as a problem with appropriate hints.)
13 Some general constructions
However interesting the theoretical study of codes may be to a pure mathe-
matician, the engineer would prefer to have an arsenal of practical codes so
that she can select the one most suitable for the job in hand. In this section
we discuss the general Hamming codes and the Reed-Muller codes as well as
some simple methods of obtaining new codes from old.
Definition 13.1. Let d be a strictly positive integer and let n = 2d
− 1.
Consider the (column) vector space D = Fd
2. Write down a d × n matrix H
whose columns are the 2d
− 1 distinct non-zero vectors of D. The Hamming
(n, n − d) code is the linear code of length n with HT
as parity check matrix.
38](https://image.slidesharecdn.com/shan-250416103225-19ae7e70/85/Shan-pdfFully-Homomorphic-Encryption-FHE-38-320.jpg)
![Of course the Hamming (n, n−d) code is only defined up to permutation
of coordinates. We note that H has rank d, so a simple use of the rank nullity
theorem shows that our notation is consistent.
Lemma 13.2. The Hamming (n, n−d) code is a linear code of length n and
rank n − d [n = 2d
− 1].
Example 13.3. The Hamming (7, 4) code is the original Hamming code.
The fact that any two rows of H are linearly independent and a look at the
appropriate syndromes gives us the main property of the general Hamming
code.
Lemma 13.4. The Hamming (n, n − d) code has minimum weight 3 and is
a perfect 1 error correcting code [n = 2d
− 1].
Hamming codes are ideal in situations where very long strings of binary
digits must be transmitted but the chance of an error in any individual
digit is very small. (Look at Exercise 7.5.) Although the search for perfect
codes other than the Hamming codes produced the Golay code (not discussed
here) and much interesting combinatorics, the reader is warned that, from a
practical point of view, it represents a dead end27
.
Here are a number of simple tricks for creating new codes from old.
Definition 13.5. If C is a code of length n, the parity check extension C+
of C is the code of length n + 1 given by
C+
=
(
x ∈ Fn+1
2 : (x1, x2, . . . , xn) ∈ C,
n+1
X
j=1
xj = 0
)
.
Definition 13.6. If C is a code of length n, the truncation C−
of C is the
code of length n − 1 given by
C−
= {(x1, x2, . . . , xn−1) : (x1, x2, . . . , xn) ∈ C for some xn ∈ F2}.
27
If we confine ourselves to the binary codes discussed in this course, it is known that
perfect codes of length n with Hamming spheres of radius ρ exist for ρ = 0, ρ = n,
ρ = (n− 1)/2, with n odd (the three codes just mentioned are easy to identify), ρ = 3 and
n = 23 (the Golay code, found by direct search) and ρ = 1 and n = 2m
− 1. There are
known to be non-Hamming codes with ρ = 1 and n = 2m
− 1, it is suspected that there
are many of them and they are the subject of much research, but, of course they present
no practical advantages. The only linear perfect codes with ρ = 1 and n = 2m
− 1 are the
Hamming codes.
39](https://image.slidesharecdn.com/shan-250416103225-19ae7e70/85/Shan-pdfFully-Homomorphic-Encryption-FHE-39-320.jpg)
![probes (a task which has been compared to signalling across the Atlantic with
a child’s torch28
).
We start by considering the 2d
points P0, P1, . . . , P2d−1 of the space
X = Fd
2. Our code words will be of length n = 2d
and will correspond to the
indicator functions IA on X. More specifically, the possible code word cA
is
given by
cA
i = 1 if Pi ∈ A
cA
i = 0 otherwise.
for some A ⊆ X.
In addition to the usual vector space structure on Fn
2 , we define a new
operation
cA
∧ cB
= cA∩B
.
Thus, if x, y ∈ Fn
2 ,
(x0, x1, . . . , xn−1) ∧ (y0, y1, . . . , yn−1) = (x0y0, x1y1, . . . , xn−1yn−1).
Finally we consider the collection of d hyperplanes
πj = {p ∈ X : pj = 0} [1 ≤ j ≤ d]
in Fn
2 and the corresponding indicator functions
hj
= cπj
,
together with the special vector
h0
= cX
= (1, 1, . . . , 1).
Exercise 13.12. Suppose that x, y, z ∈ Fn
2 and A, B ⊆ X.
(i) Show that x ∧ y = y ∧ x.
(ii) Show that (x + y) ∧ z = x ∧ z + y ∧ z.
(iii) Show that h0
∧ x = x.
(iv) If cA
+ cB
= cE
, find E in terms of A and B.
(v) If h0
+ cA
= cE
, find E in terms of A.
We refer to A0 = {h0
} as the set of terms of order zero. If Ak is the set
of terms of order at most k, then the set Ak+1 of terms of order at most k +1
is defined by
Ak+1 = {a ∧ hj
: a ∈ Ak, 1 ≤ j ≤ d}.
Less formally, but more clearly, the elements of order 1 are the hi
, the ele-
ments of order 2 are the hi
∧ hj
with i j, the elements of order 3 are the
hi
∧ hj
∧ hk
with i j k and so on.
28
Strictly speaking, the comparison is meaningless. However, it sounds impressive and
that is the main thing.
41](https://image.slidesharecdn.com/shan-250416103225-19ae7e70/85/Shan-pdfFully-Homomorphic-Encryption-FHE-41-320.jpg)
![14 Polynomials and fields
This section is starred. Its object is to make plausible the few facts from
modern29
algebra that we shall need. They were covered, along with much
else, in various post-IA algebra courses, but attendance at those courses is
no more required for this course than is reading Joyce’s Ulysses before going
for a night out at an Irish pub. Anyone capable of criticising the imprecision
and general slackness of the account that follows obviously can do better
themselves and should rewrite this section in an appropriate manner.
A field K is an object equipped with addition and multiplication which
follow the same rules as do addition and multiplication in R. The only rule
which will cause us trouble is
If x ∈ K and x 6= 0, then we can find y ∈ K such that xy = 1. ⋆
Obvious examples of fields include R, C and F2.
We are particularly interested in polynomials over fields, but here an
interesting difficulty arises.
Example 14.1. We have t2
+ t = 0 for all t ∈ F2.
To get round this, we distinguish between the polynomial in the ‘indeter-
minate’ X
P(X) =
n
X
j=0
ajXj
with coefficients aj ∈ K and its evaluation P(t) =
Pn
j=0 ajtj
for some t ∈
K. We manipulate polynomials in X according to the standard rules for
polynomials, but say that
n
X
j=0
ajXj
= 0
if and only if aj = 0 for all j. Thus X2
+ X is a non-zero polynomial over
F2 all of whose values are zero.
The following result is familiar, in essence, from school mathematics.
Lemma 14.2. [Remainder theorem] (i) If P is a polynomial over a field
K and a ∈ K, then we can find a polynomial Q and an r ∈ K such that
P(X) = (X − a)Q(X) + r.
(ii) If P is a polynomial over a field K and a ∈ K is such that P(a) = 0,
then we can find a polynomial Q such that
P(X) = (X − a)Q(X).
29
Modern, that is, in 1920.
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![The key to much of the elementary theory of polynomials lies in the fact
that we can apply Euclid’s algorithm to obtain results like the following.
Theorem 14.3. Suppose that P is a set of polynomials which contains at
least one non-zero polynomial and has the following properties.
(i) If Q is any polynomial and P ∈ P, then the product PQ ∈ P.
(ii) If P1, P2 ∈ P, then P1 + P2 ∈ P.
Then we can find a non-zero P0 ∈ P which divides every P ∈ P.
Proof. Consider a non-zero polynomial P0 of smallest degree in P.
Recall that the polynomial P(X) = X2
+ 1 has no roots in R (that is
P(t) 6= 0 for all t ∈ R). However, by considering the collection of formal
expressions a + bi [a, b ∈ R] with the obvious formal definitions of addition
and multiplication and subject to the further condition i2
+ 1 = 0, we obtain
a field C ⊇ R in which P has a root (since P(i) = 0). We can perform a
similar trick with other fields.
Example 14.4. If P(X) = X2
+X +1, then P has no roots in F2. However,
if we consider
F2[ω] = {0, 1, ω, 1 + ω}
with obvious formal definitions of addition and multiplication and subject to
the further condition ω2
+ ω + 1 = 0, then F2[ω] is a field containing F2 in
which P has a root (since P(ω) = 0).
Proof. The only thing we really need prove is that F2[ω] is a field and to do
that the only thing we need to prove is that ⋆ holds. Since
(1 + ω)ω = 1
this is easy.
In order to state a correct generalisation of the ideas of the previous
paragraph we need a preliminary definition.
Definition 14.5. If P is a polynomial over a field K, we say that P is
reducible if there exists a non-constant polynomial Q of degree strictly less
than P which divides P. If P is a non-constant polynomial which is not
reducible, then P is irreducible.
Theorem 14.6. If P is an irreducible polynomial of degree n ≥ 2 over a
field K, then P has no roots in K. However, if we consider
K[ω] =
(n−1
X
j=0
ajωj
: aj ∈ K
)
44](https://image.slidesharecdn.com/shan-250416103225-19ae7e70/85/Shan-pdfFully-Homomorphic-Encryption-FHE-44-320.jpg)
![with the obvious formal definitions of addition and multiplication and subject
to the further condition P(ω) = 0, then K[ω] is a field containing K in which
P has a root.
Proof. The only thing we really need prove is that K[ω] is a field and to do
that the only thing we need to prove is that ⋆ holds. Let Q be a non-zero
polynomial of degree at most n − 1. Since P is irreducible, the polynomials
P and Q have no common factor of degree 1 or more. Hence, by Euclid’s
algorithm, we can find polynomials R and S such that
R(X)Q(X) + S(X)P(X) = 1
and so R(ω)Q(ω) + S(ω)P(ω) = 1. But P(ω) = 0, so R(ω)Q(ω) = 1 and we
have proved ⋆.
In a proper algebra course we would simply define
K[ω] = K[X]/(P(X))
where (P(X)) is the ideal generated by P(X). This is a cleaner procedure
which avoids the use of such phrases as ‘the obvious formal definitions of
addition and multiplication’ but the underlying idea remains the same.
Lemma 14.7. If P is a polynomial over a field K which does not factorise
completely into linear factors, then we can find a field L ⊇ K in which P has
more linear factors.
Proof. Factor P into irreducible factors and choose a factor Q which is not
linear. By Theorem 14.6, we can find a field L ⊇ K in which Q has a root α
say and so, by Lemma 14.2, a linear factor X − α. Since any linear factor of
P in K remains a factor in the bigger field L, we are done.
Theorem 14.8. If P is a polynomial over a field K, then we can find a field
L ⊇ K in which P factorises completely into linear factors.
We shall be interested in finite fields (that is fields K with only a finite
number of elements). A glance at our method of proving Theorem 14.8 shows
that the following result holds.
Lemma 14.9. If P is a polynomial over a finite field K, then we can find a
finite field L ⊇ K in which P factorises completely.
In this context, we note yet another useful simple consequence of Euclid’s
algorithm.
45](https://image.slidesharecdn.com/shan-250416103225-19ae7e70/85/Shan-pdfFully-Homomorphic-Encryption-FHE-45-320.jpg)
![Let us establish a correspondence between Fn
2 and the polynomials on F2
modulo Xn
− 1 by setting
Pa =
n−1
X
j=0
ajXj
whenever a ∈ Fn
2 . (Of course, Xn
− 1 = Xn
+ 1 but in this context the first
expression seems more natural.)
Exercise 15.2. With the notation just established, show that
(i) Pa + Pb = Pa+b,
(ii) Pa = 0 if and only if a = 0.
Lemma 15.3. A code C in Fn
2 is cyclic if and only if PC = {Pa : a ∈ C}
satisfies the following two conditions (working modulo Xn
− 1).
(i) If f, g ∈ PC, then f + g ∈ PC.
(ii) If f ∈ PC and g is any polynomial, then the product fg ∈ PC.
(In the language of abstract algebra, C is cyclic if and only if PC is an ideal
of the quotient ring F2[X]/(Xn
− 1).)
From now on we shall talk of the code word f(X) when we mean the code
word a with Pa(X) = f(X). An application of Euclid’s algorithm gives the
following useful result.
Lemma 15.4. A code C of length n is cyclic if and only if (working modulo
Xn
−1, and using the conventions established above) there exists a polynomial
g such that
C = {f(X)g(X) : f a polynomial}
(In the language of abstract algebra, F2[X] is a Euclidean domain and so a
principal ideal domain. Thus the quotient F2[X]/(Xn
−1) is a principal ideal
domain.) We call g(X) a generator polynomial for C.
Lemma 15.5. A polynomial g is a generator for a cyclic code of length n if
and only if it divides Xn
− 1.
Thus we must seek generators among the factors of Xn
− 1 = Xn
+ 1. If
there are no conditions on n, the result can be rather disappointing.
Exercise 15.6. If we work with polynomials over F2, then
X2r
+ 1 = (X + 1)2r
.
In order to avoid this problem and to be able to make use of Lemma 14.12,
we shall take n odd from now on. (In this case, the cyclic codes are said to be
separable.) Notice that the task of finding irreducible factors (that is factors
with no further factorisation) is a finite one.
48](https://image.slidesharecdn.com/shan-250416103225-19ae7e70/85/Shan-pdfFully-Homomorphic-Encryption-FHE-48-320.jpg)
![Lemma 15.7. Consider codes of length n. Suppose that g(X)h(X) = Xn
−1.
Then g is a generator of a cyclic code C and h is a generator for a cyclic
code which is the reverse of C⊥
.
As an immediate corollary, we have the following remark.
Lemma 15.8. The dual of a cyclic code is itself cyclic.
Lemma 15.9. If a cyclic code C of length n has generator g of degree n − r
then g(X), Xg(X), . . . , Xr−1
g(X) form a basis for C.
Cyclic codes are thus easy to specify (we just need to write down the
generator polynomial g) and to encode.
We know that Xn
+ 1 factorises completely over some larger finite field
and, since n is odd, we know, by Lemma 14.12, that it has no repeated
factors. The same is therefore true for any polynomial dividing it.
Lemma 15.10. Suppose that g is a generator of a cyclic code C of odd length
n. Suppose further that g factorises completely into linear factors in some
field K containing F2. If g = g1g2 . . . gk with each gj irreducible over F2 and
A is a subset of the set of all the roots of all the gj and containing at least
one root of each gj [1 ≤ j ≤ k], then
C = {f ∈ F2[X] : f(α) = 0 for all α ∈ A}.
Definition 15.11. A defining set for a cyclic code C is a set A of elements
in some field K containing F2 such that f ∈ F2[X] belongs to C if and only
if f(α) = 0 for all α ∈ A.
(Note that, if C has length n, A must be a set of zeros of Xn
− 1.)
Lemma 15.12. Suppose that
A = {α1, α2, . . . , αr}
is a defining set for a cyclic code C in some field K containing F2. Let B be
the n × r matrix over K whose jth column is
(1, αj, α2
j , . . . , αn−1
j )T
Then a vector a ∈ Fn
2 is a code word in C if and only if
aB = 0
in K.
49](https://image.slidesharecdn.com/shan-250416103225-19ae7e70/85/Shan-pdfFully-Homomorphic-Encryption-FHE-49-320.jpg)
![The columns in B are not parity checks in the usual sense since the code
entries lie in F2 and the computations take place in the larger field K.
With this background we can discuss a famous family of codes known
as the BCH (Bose, Ray-Chaudhuri, Hocquenghem) codes. Recall that a
primitive nth root of unity is an root α of Xn
− 1 = 0 such that every root
is a power of α.
Definition 15.13. Suppose that n is odd and K is a field containing F2 in
which Xn
−1 factorises into linear factors. Suppose that α ∈ K is a primitive
nth root of unity. A cyclic code C with defining set
A = {α, α2
, . . . , αδ−1
}
is a BCH code of design distance δ.
Note that the rank of C will be n − k, where k is the degree of the product
of those irreducible factors of Xn
− 1 over F2 which have a zero in A. Notice
also that k may be very much larger than δ.
Example 15.14. (i) If K is a field containing F2, then (a + b)2
= a2
+ b2
for all a, b ∈ K.
(ii) If P ∈ F2[X] and K is a field containing F2, then P(a)2
= P(a2
) for
all a ∈ K.
(iii) Let K be a field containing F2 in which X7
− 1 factorises into linear
factors. If β is a root of X3
+X +1 in K, then β is a primitive root of unity
and β2
is also a root of X3
+ X + 1.
(iv) We continue with the notation (iii). The BCH code with {β, β2
} as
defining set is Hamming’s original (7,4) code.
The next theorem contains the key fact about BCH codes.
Theorem 15.15. The minimum distance for a BCH code is at least as great
as the design distance.
Our proof of Theorem 15.15 relies on showing that the matrix B of
Lemma 15.12 is of full rank for a BCH. To do this we use a result which
every undergraduate knew in 1950.
Lemma 15.16. [The van der Monde determinant] We work over a field
K. The determinant
1 1 1 . . . 1
x1 x2 x3 . . . xn
x2
1 x2
2 x2
3 . . . x2
n
.
.
.
.
.
.
.
.
.
...
.
.
.
xn−1
1 xn−1
2 xn−1
3 . . . xn−1
n
=
Y
1≤ji≤n
(xi − xj).
50](https://image.slidesharecdn.com/shan-250416103225-19ae7e70/85/Shan-pdfFully-Homomorphic-Encryption-FHE-50-320.jpg)
![Lemma 15.18. If the error locator polynomial is given the value of e and
so of c can be obtained directly.
We wish to make use of relations of the form
1
1 − αjX
=
∞
X
r=0
(αj
X)r
.
Unfortunately, it is not clear what meaning to assign to such a relation. One
way round is to work modulo Z2t
(more formally, to work in K[Z]/(Z2t
)).
We then have Zu
≡ 0 for all integers u ≥ 2t.
Lemma 15.19. If we work modulo Z2t
then
(1 − αj
Z)
2t−1
X
m=0
(αj
Z)m
≡ 1.
Thus, if we work modulo Z2t
, as we shall from now on, we may define
1
1 − αjZ
=
2t−1
X
m=0
(αj
Z)m
.
Lemma 15.20. With the conventions already introduced.
(i)
ω(Z)
σ(Z)
≡
2t−1
X
m=0
Zm
e(αm+1
).
(ii) e(αm
) = r(αm
) for all 0 ≤ m ≤ 2t − 1.
(iii)
ω(Z)
σ(Z)
≡
2t−1
X
m=0
Zm
r(αm+1
).
(iv) ω(Z) ≡
P2t−1
m=0 Zm
r(αm+1
)σ(Z).
(v) ωj =
X
u+v=j
r(αu+1
)σv for all 0 ≤ j ≤ t − 1.
(vi) 0 =
X
u+v=j
r(αu+1
)σv for all t ≤ j ≤ 2t − 1.
(vii) The conditions in (vi) determine σ completely.
Part (vi) of Lemma 15.20 completes our search for a decoding method, since
σ determines E, E determines e and e determines c. It is worth noting that
the system of equations in part (v) suffice to determine the pair σ and ω
directly.
Compact disc players use BCH codes. Of course, errors are likely to
occur in bursts (corresponding to scratches etc) and this is dealt with by
52](https://image.slidesharecdn.com/shan-250416103225-19ae7e70/85/Shan-pdfFully-Homomorphic-Encryption-FHE-52-320.jpg)
![distributing the bits (digits) in a single codeword over a much longer stretch
of track. The code used can correct a burst of 4000 consecutive errors (2.5
mm of track).
Unfortunately, none of the codes we have considered work anywhere near
the Shannon bound (see Theorem 10.1). We might suspect that this is be-
cause they are linear, but Elias has shown that this is not the case. (We just
state the result without proof.)
Theorem 15.21. In Theorem 10.1 we can replace ‘code’ by ‘linear code’.
The advance of computational power and the ingenuity of the discover-
ers30
have lead to new codes which appear to come close to the Shannon
bounds. But that is another story.
Just as pure algebra has contributed greatly to the study of error correct-
ing codes, so the study of error correcting codes has contributed greatly to
the study of pure algebra. The story of one such contribution is set out in
T. M. Thompson’s From Error-correcting Codes through Sphere Packings to
Simple Groups [9] — a good, not too mathematical, account of the discovery
of the last sporadic simple groups by Conway and others.
16 Shift registers
In this section we move towards cryptography, but the topic discussed will
turn out to have connections with the decoding of BCH codes as well.
Definition 16.1. A general feedback shift register is a map f : Fd
2 → Fd
2
given by
f(x0, x1, . . . , xd−2, xd−1) = (x1, x2, . . . , xd−1, C(x0, x1, . . . , xd−2, xd−1))
with C a map C : Fd
2 → F2. The stream associated to an initial fill
(y0, y1, . . . , yd−1) is the sequence
y0, y1, . . . , yj, yj+1, . . . with yn = C(yn−d, yn−d+1, . . . , yn−1) for all n ≥ d.
Example 16.2. If the general feedback shift f given in Definition 16.1 is a
permutation, then C is linear in the first variable, i.e.
C(x0, x1, . . . , xd−2, xd−1) = x0 + C′
(x1, x2, . . . , xd−2, xd−1).
30
People like David MacKay, now better known for his superb ‘Sustainable Energy
Without the Hot Air’ — rush out and read it.
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![are linearly independent in the sense that
m
X
j=0
bjfj(n) = 0
for all n implies bj = 0 for 0 ≤ j ≤ m.
(iv) Suppose K is a field containing F2 such that the auxiliary polynomial
C factorises completely into linear factors. If the root αu has multiplicity
m(u), [1 ≤ u ≤ q], show that the general solution of ⋆ in K is
xn =
q
X
u=1
m(u)−1
X
v=0
bu,v
n
v
αn
u
for some bu,v ∈ K. If x0, x1, . . . , xd−1 ∈ F2, show that xn ∈ F2 for all n.
A strong link with the problem of BCH decoding is provided by Theo-
rem 16.7 below.
Definition 16.6. If we have a sequence (or stream) x0, x1, x2, . . . of elements
of F2 then its generating function G is given by
G(Z) =
∞
X
n=0
xjZj
.
If the recurrence relation for a linear feedback generator is
d
X
j=0
cjxn−j = 0
for n ≥ d with c0, cd 6= 0 we call
C(z) =
d
X
j=0
cjZj
the auxiliary polynomial of the generator.
Theorem 16.7. The stream (xn) comes from a linear feedback generator
with auxiliary polynomial C if and only if the generating function for the
stream is (formally) of the form
G(Z) =
B(Z)
C(Z)
with B a polynomial of degree strictly smaller than that of C.
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![from everybody else) and tells everybody the value of p. She then burns her
calculations.
Suppose that k members of the Faculty Board with shadow pairs yj, Q(j)
=
xrj
, P(rj)
[1 ≤ j ≤ k] are together. By the properties of the Van der Monde
determinant (see Lemma 15.16)
1 y1 y2
1 . . . yk−1
1
1 y2 y2
2 . . . yk−1
2
1 y3 y2
3 . . . yk−1
3
.
.
.
.
.
.
.
.
.
...
.
.
.
1 yk y2
k . . . yk−1
k
≡
1 1 1 . . . 1
y1 y2 y3 . . . yk
y2
1 y2
2 y2
3 . . . y2
k
.
.
.
.
.
.
.
.
.
...
.
.
.
yk−1
1 yk−1
2 yk−1
3 . . . yk−1
k
≡
Y
1≤ji≤k−1
(yi − yj) 6≡ 0 mod p.
Thus the system of equations
z0 + y1z1 + y2
1z2 + . . . + yk−1
1 zk−1 ≡ Q1
z0 + y2z1 + y2
2z2 + . . . + yk−1
2 zk−1 ≡ Q2
z0 + y3z1 + y2
3z2 + . . . + yk−1
3 zk−1 ≡ Q3
.
.
.
z0 + ykz1 + y2
kz2 + . . . + yk−1
k zk−1 ≡ Qk
has a unique solution z. But we know that a is a solution, so z = a and the
secret S = z0.
On the other hand,
y1 y2
1 . . . yk−1
1
y2 y2
2 . . . yk−1
2
y3 y2
3 . . . yk−1
3
.
.
.
.
.
.
...
.
.
.
yk−1 y2
k−1 . . . yk−1
k−1
≡ y1y2 . . . yk−1
Y
1≤ji≤k−1
(yi − yj) 6≡ 0 mod p,
so the system of equations
z0 + y1z1 + y2
1z2 + . . . + yk−1
1 zk−2 ≡ Q1
z0 + y2z1 + y2
2z2 + . . . + yk−1
2 zk−2 ≡ Q2
z0 + y3z1 + y2
3z2 + . . . + yk−1
3 zk−2 ≡ Q3
.
.
.
z0 + yk−1z1 + y2
k−1z2 + . . . + yk−1
k−1zk−2 ≡ Qk−1
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![has a solution, whatever value of z0 we take, so k −1 members of the Faculty
Board have no way of saying that any possible values of S is more likely than
any other.
One way of looking at this method of ‘secret sharing’ is to note that a
polynomial of degree k − 1 can be recovered from its value at k points but
not from its value at k −1 points. However, the proof that the method works
needs to be substantially more careful.
Exercise 16.10. Is the secret compromised if the values of the xj become
known?
17 A short homily on cryptography
Cryptography is the science of code making. Cryptanalysis is the art of code
breaking.
Two thousand years ago, Lucretius wrote that ‘Only recently has the
true nature of things been discovered’. In the same way, mathematicians
are apt to feel that ‘Only recently has the true nature of cryptography been
discovered’. The new mathematical science of cryptography with its promise
of codes which are ‘provably hard to break’ seems to make everything that
has gone before irrelevant.
It should, however, be observed that the best cryptographic systems of our
ancestors (such as diplomatic ‘book codes’) served their purpose of ensuring
secrecy for a relatively small number of messages between a relatively small
number of people extremely well. It is the modern requirement for secrecy
on an industrial scale to cover endless streams of messages between many
centres which has made necessary the modern science of cryptography.
More pertinently, it should be remembered that the German Naval Enigma
codes not only appeared to be ‘provably hard to break’ (though not against
the modern criteria of what this should mean) but, considered in isolation,
probably were unbreakable in practice33
. Fortunately the Submarine codes
formed part of an ‘Enigma system’ with certain exploitable weaknesses. (For
an account of how these weaknesses arose and how they were exploited see
Kahn’s Seizing the Enigma [4].)
Even the best codes are like the lock on a safe. However good the lock
is, the safe may be broken open by brute force, or stolen together with its
contents, or a key holder may be persuaded by fraud or force to open the
lock, or the presumed contents of the safe may have been tampered with
before they go into the safe, or . . . . The coding schemes we shall consider,
33
Some versions remained unbroken until the end of the war.
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![sequence k. If a random sequence is reused it ceases to be random (it becomes
‘the same code as last Wednesday’ or the ‘the same code as Paris uses’) so,
when there is a great deal of code traffic35
, new one-time pads must be sent
out. If random bits can be safely communicated, so can ordinary messages
and the exercise becomes pointless.
In practice, we would like to start from a short shared secret ‘seed’ and
generate a ciphering string k that ‘behaves like a random sequence’. This
leads us straight into deep philosophical waters36
. As might be expected,
there is an illuminating discussion in Chapter III of Knuth’s marvellous The
Art of Computing Programming [7]. Note, in particular, his warning:
. . . random numbers should not be generated with a method chosen
at random. Some theory should be used.
One way that we might try to generate our ciphering string is to use a gen-
eral feedback shift register f of length d with the initial fill (k0, k1, . . . , kd−1)
as the secret seed.
Lemma 18.1. If f is a general feedback shift register of length d, then, given
any initial fill (k0, k1, . . . , kd−1), there will exist N, M ≤ 2d
such that the
output stream k satisfies kr+N = kr for all r ≥ M.
Exercise 18.2. Show that the decimal expansion of a rational number must
be a recurrent expansion. Give a bound for the period in terms of the quo-
tient. Conversely, by considering geometric series, or otherwise, show that a
recurrent decimal represents a rational number.
Lemma 18.3. Suppose that f is a linear feedback register of length d.
(i) f(x0, x1, . . . , xd−1) = (x0, x1, . . . , xd−1) if (x0, x1, . . . , xd−1) = (0, 0, . . . , 0).
(ii) Given any initial fill (k0, k1, . . . , kd−1), there will exist N, M ≤ 2d
− 1
such that the output stream k satisfies kr+N = kr for all r ≥ M.
We can complement Lemma 18.3 by using Lemma 14.16 and the associ-
ated discussion.
35
In 1941, the Soviet Union’s need for one-time pads suddenly increased and it appears
that pages were reused in different pads. If the reader reflects, she will see that, though
this is a mistake, it is one which it is very difficult to exploit. However, under the pressure
of the cold war, US code-breakers managed to decode messages which, although several
years old, still provided useful information. After 1944, the Soviet Union’s one-time pads
became genuinely one-time again and the coded messages became indecipherable.
36
Where we drown at once, since the best (at least, in my opinion) modern view is that
any sequence that can be generated by a program of reasonable length from a ‘seed’ of
reasonable size is automatically non-random.
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![profits of a telephone company are essentially unaffected by the .1% which
‘break the system’.
What happens if we try some simple tricks to increase the complexity of
the cipher text stream?
Lemma 18.6. If xn is a stream produced by a linear feedback system of
length N with auxiliary polynomial P and yn is a stream produced by a linear
feedback system of length M with auxiliary polynomial Q, then xn + yn is a
stream produced by a linear feedback system of length N + M with auxiliary
polynomial P(X)Q(X).
Note that this means that adding streams from two linear feedback system
is no more economical than producing the same effect with one. Indeed the
situation may be worse since a stream produced by linear feedback system of
given length may, possibly, also be produced by another linear feedback system
of shorter length.
Lemma 18.7. Suppose that xn is a stream produced by a linear feedback
system of length N with auxiliary polynomial P and yn is a stream produced
by a linear feedback system of length M with auxiliary polynomial Q. Let P
have roots α1, α2, . . . αN and Q have roots β1, β2, . . . βM over some field
K ⊇ F2. Then xnyn is a stream produced by a linear feedback system of
length NM with auxiliary polynomial
Y
1≤i≤N
Y
1≤i≤M
(X − αiβj).
We shall probably only prove Lemmas 18.6 and 18.7 in the case when all
roots are distinct, leaving the more general case as an easy exercise. We
shall also not prove that the polynomial
Q
1≤i≤N
Q
1≤i≤M (X −αiβj) obtained
in Lemma 18.7 actually lies in F2[X] but (for those who are familiar with the
phrase in quotes) this is an easy exercise in ‘symmetric functions of roots’.
Here is an even easier remark.
Lemma 18.8. Suppose that xn is a stream which is periodic with period N
and yn is a stream which is periodic with period M. Then the streams xn +yn
and xnyn are periodic with periods dividing the lowest common multiple of N
and M.
Exercise 18.9. One of the most confidential German codes (called FISH by
the British) involved a complex mechanism which the British found could be
simulated by two loops of paper tape of length 1501 and 1497. If kn = xn +yn
where xn is a stream of period 1501 and yn is a stream of period 1497, what
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![Another approach is to observe that, so far as the potential code breaker
is concerned, the cipher stream method only combines the ‘unknown secret’
(here the feedback generator f together with the seed (k0, k1, . . . , kd−1)) with
the unknown message p in a rather simple way. It might be better to consider
a system with two functions F : Fm
2 × Fn
2 → Fq
2 and G : Fm
2 × Fq
2 → Fn
2 such
that
G(k, F(k, p)) = p.
Here k will be the shared secret, p the message, and z = F(k, p) the encoded
message which can be decoded by using the fact that G(k, z) = p.
In the next section we shall see that an even better arrangement is possi-
ble. However, arrangements like this have the disadvantage that the message
p must be entirely known before it is transmitted and the encoded message
z must have been entirely received before it can be decoded. Stream ciphers
have the advantage that they can be decoded ‘on the fly’. They are also much
more error tolerant. A mistake in the coding, transmission or decoding of a
single element only produces an error in a single place of the sequence. There
will continue to be circumstances where stream ciphers are appropriate.
There is one further remark to be made. Suppose, as is often the case,
that we know F, that n = q and we know the ‘encoded message’ z. Suppose
also that we know that the ‘unknown secret’ or ‘key’ k ∈ K ⊆ Fm
2 and the
‘unknown message’ p ∈ P ⊆ Fn
2 . We are then faced with the problem:- Solve
the system
z = F(k, p) where k ∈ K, p ∈ P. ⋆
Speaking roughly, the task is hopeless unless ⋆ has a unique solution38
Speaking even more roughly, this is unlikely to happen if |K||P| 2n
and is
likely to happen if 2n
is substantially greater than |K||P|. (Here, as usual,
|B| denotes the number of elements of B.)
Now recall the definition of the information rate given in Definition 6.2.
If the message set M has information rate µ and the key set (that is the
shared secret set) K has information rate κ, then, taking logarithms, we see
that, if
n − mκ − nµ
38
‘According to some, the primordial Torah was inscribed in black flames on white fire.
At the moment of its creation, it appeared as a series of letters not yet joined up in
the form of words. For this reason, in the Torah rolls there appear neither vowels nor
punctuation, nor accents; for the original Torah was nothing but a disordered heap of
letters. Furthermore, had it not been for Adam’s sin, these letters might have been joined
differently to form another story. For the kabalist, God will abolish the present ordering
of the letters, or else will teach us how to read them according to a new disposition only
after the coming of the Messiah.’ ([1], Chapter 2.) A reader of this footnote has directed
me to the International Torah Codes Society.
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![work back from H(M) to M. The second requirement is that S should be
large so that a forger can not (sensibly) hope to hit on H(M) by luck.
Obviously we should aim at the same kind of security as that offered by
our ‘level 2’ for codes:-
Prospective opponents should find it hard to find H(M) given M
even if they are in possession of a plentiful supply of message–
signature pairs (Mi, H(Mi)) of messages Mi together with their
encodings Ci.
I leave it to the reader to think about level 3 security (or to look at section
12.6 of [10]).
Here is a signature scheme due to Elgamal46
. The message sender A
chooses a very large prime p, some integer 1 g p and some other integer
u with 1 u p (as usual, some randomisation scheme should be used). A
then releases the values of p, g and y = gu
(modulo p) but keeps the value of
u secret. Whenever he sends a message m (some positive integer), he chooses
another integer k with 1 ≤ k ≤ p − 2 at random and computes r and s with
1 ≤ r ≤ p − 1 and 0 ≤ s ≤ p − 2 by the rules47
r ≡ gk
(mod p), (*)
m ≡ ur + ks (mod p − 1). (**)
Lemma 21.4. If conditions (*) and (**) are satisfied, then
gm
≡ yr
rs
(mod p).
If A sends the message m followed by the signature (r, s), the recipient need
only verify the relation gm
≡ yr
rs
(mod p) to check that the message is
authentic48
.
Since k is random, it is believed that the only way to forge signatures is
to find u from gu
(or k from gk
) and it is believed that this problem, which
is known as the discrete logarithm problem, is very hard.
Needless to say, even if it is impossible to tamper with a message–signature
pair it is always possible to copy one. Every message should thus contain a
unique identifier such as a time stamp.
46
This is Dr Elgamal’s own choice of spelling according to Wikipedia.
47
There is a small point which I have glossed over here and elsewhere. Unless k and
and p − 1 are coprime the equation (**) may not be soluble. However the quickest way to
solve (**), if it is soluble, is Euclid’s algorithm which will also reveal if (**) is insoluble.
If (**) is insoluble, we simply choose another k at random and try again.
48
Sometimes, m is replaced by some hash function H(m) of m so (∗∗) becomes H(m) ≡
ur + ks (mod p − 1). In this case the recipient checks that gH(m)
≡ yr
rs
(mod p).
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![There are known unknowns.
That is to say
We know there are some things
We do not know.
But there are also unknown unknowns,
The ones we don’t know
We don’t know52
.
23 Further reading
For many students this will be one of the last university mathematics course
they will take. Although the twin subjects of error-correcting codes and
cryptography occupy a small place in the grand panorama of modern math-
ematics, it seems to me that they form a very suitable topic for such a final
course.
Outsiders often think of mathematicians as guardians of abstruse but set-
tled knowledge. Even those who understand that there are still problems un-
settled, ask what mathematicians will do when they run out of problems. At
a more subtle level, Kline’s magnificent Mathematical Thought from Ancient
to Modern Times [5] is pervaded by the melancholy thought that, though the
problems will not run out, they may become more and more baroque and
inbred. ‘You are not the mathematicians your parents were’ whispers Kline
‘and your problems are not the problems your parents’ were.’
However, when we look at this course, we see that the idea of error-
correcting codes did not exist before 1940. The best designs of such codes
depend on the kind of ‘abstract algebra’ that historians like Kline and Bell
consider a dead end, and lie behind the superior performance of CD players
and similar artifacts.
In order to go further into the study of codes, whether secret or error
correcting, we need to go into the question of how the information content of
a message is to be measured. ‘Information theory’ has its roots in the code
breaking of World War II (though technological needs would doubtless have
led to the same ideas shortly thereafter anyway). Its development required a
level of sophistication in treating probability which was simply not available
in the 19th century. (Even the Markov chain is essentially 20th century53
.)
The question of what makes a calculation difficult could not even have
52
Rumsfeld
53
We are now in the 21st century, but I suspect that we are still part of the mathematical
‘long 20th century’ which started in the 1880s with the work of Cantor and like minded
contemporaries.
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![been thought about until Gödel’s theorem (itself a product of the great ‘foun-
dations crisis’ at the beginning of the 20th century). Developments by Turing
and Church of Gödel’s theorem gave us a theory of computational complex-
ity which is still under development today. The question of whether there
exist ‘provably hard’ public codes is intertwined with still unanswered ques-
tions in complexity theory. There are links with the profound (and very 20th
century) question of what constitutes a random number.
Finally, the invention of the electronic computer has produced a cultural
change in the attitude of mathematicians towards algorithms. Before 1950,
the construction of algorithms was a minor interest of a few mathematicians.
(Gauss and Jacobi were considered unusual in the amount of thought they
gave to actual computation.) Today, we would consider a mathematician as
much as a maker of algorithms as a prover of theorems. The notion of the
probabilistic algorithm which hovered over much of our discussion of secret
codes is a typical invention of the last decades of the 20th century.
Although both the subjects of error correcting and secret codes are now
‘mature’ in the sense that they provide usable and well tested tools for prac-
tical application, they still contain deep unanswered questions. For example
How close to the Shannon bound can a ‘computationally easy’ error cor-
recting code get?
Do provably hard public codes exist?
Even if these questions are too hard, there must surely exist error cor-
recting and public codes based on new ideas54
. Such ideas would be most
welcome and, although they are most likely to come from the professionals,
they might come from outside the usual charmed circles.
Those who wish to learn about error correction from the horse’s mouth
will consult Hamming’s own book on the matter [2]. For the present course,
the best book I know for further reading is Welsh [10]. After this, the book
of Goldie and Pinch [8] provides a deeper idea of the meaning of information
and its connection with the topic. The book by Koblitz [6] develops the
number theoretic background. The economic and practical importance of
transmitting, storing and processing data far outweighs the importance of
hiding it. However, hiding data is more romantic. For budding cryptologists
and cryptographers (as well as those who want a good read), Kahn’s The
Codebreakers [3] has the same role as is taken by Bell’s Men of Mathematics
for budding mathematicians.
I conclude with a quotation from Galbraith (referring to his time as am-
bassador to India) taken from Koblitz’s entertaining text [6].
I had asked that a cable from Washington to New Delhi . . . be
54
Just as quantum cryptography was.
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![reported to me through the Toronto consulate. It arrived in code;
no facilities existed for decoding. They brought it to me at the
airport — a mass of numbers. I asked if they assumed I could
read it. They said no. I asked how they managed. They said
that when something arrived in code, they phoned Washington
and had the original read to them.
References
[1] U. Eco The Search for the Perfect Language (English translation), Black-
well, Oxford 1995.
[2] R. W. Hamming Coding and Information Theory (2nd edition) Prentice
Hall, 1986.
[3] D. Kahn The Codebreakers: The Story of Secret Writing MacMillan,
New York, 1967. (A lightly revised edition has recently appeared.)
[4] D. Kahn Seizing the Enigma Houghton Mifflin, Boston, 1991.
[5] M. Kline Mathematical Thought from Ancient to Modern Times OUP,
1972.
[6] N. Koblitz A Course in Number Theory and Cryptography Springer,
1987.
[7] D. E. Knuth The Art of Computing Programming Addison-Wesley. The
third edition of Volumes I to III is appearing during this year and the
next (1998–9).
[8] G. M. Goldie and R. G. E. Pinch Communication Theory CUP, 1991.
[9] T. M. Thompson From Error-correcting Codes through Sphere Packings
to Simple Groups Carus Mathematical Monographs 21, MAA, Wash-
ington DC, 1983.
[10] D. Welsh Codes and Cryptography OUP, 1988.
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![(i) Purely as a matter of interest, we consider the following question. If
the contents of a hand are conveyed by one player to their partner by a
series of nods and shakes of the head how many movements of the head are
required? Show that at least 40 movements are required. Give a simple code
requiring 52 movements.
[You may assume for simplicity that the player to whom the information
is being communicated does not look at her own cards. (In fact this does not
make a difference since the two players do not acquire any shared information
by looking at their own cards.)]
(ii) If instead the player uses the initial letters of words (say using the 16
most common letters), how many words will you need to utter55
?
Q 24.9. (i) In a comma code, like Morse code, one symbol from an alphabet of
m letters is reserved to end each code word. Show that this code is prefix-free
and give a direct argument to show that it must satisfy Kraft’s inequality.
(ii) Give an example of a code satisfying Kraft’s inequality which is not
decodable.
Q 24.10. Show that if an optimal binary code has word lengths s1, s2, . . . sm
then
m log2 m ≤ s1 + s2 + · · · + sm ≤ (m2
+ m − 2)/2.
Q 24.11. (i) It is known that exactly one member of the starship Emphasise
has contracted the Macguffin virus. A test is available that will detect the
virus at any dilution. However, the power required is such that the ship’s
force shields must be switched off56
for a minute during each test. Blood
samples are taken from all crew members. The ship’s computer has worked
out that the probability of crew member number i harbouring the virus is pi.
(Thus the probability that the captain, who is, of course, number 1, has the
disease is p1.) Explain how, by testing pooled samples, the expected number
of tests can be minimised. Write down the exact form of the test when there
are 2n
crew members and pi = 2−n
.
(ii) Questions like (i) are rather artificial, since they require that exactly
one person carries the virus. Suppose that the probability that any member
of a population of 2n
has a certain disease is p (and that the probability
55
‘Marked cards, M. l’Anglais?’ I said, with a chilling sneer. ’They are used, I am told,
to trap players–not unbirched schoolboys.’
’Yet I say that they are marked!’ he replied hotly, in his queer foreign jargon. ’In my
last hand I had nothing. You doubled the stakes. Bah, sir, you knew! You have swindled
me!’
’Monsieur is easy to swindle – when he plays with a mirror behind him,’ I answered
tartly. Under the Red Robe S. J. Weyman
56
‘Captain, ye canna be serious.’
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![and show that you do indeed have a solution. Show that your solution also
minimises the maximum number of weighings that you might have to do.
(ii) Suppose the problem is as before but m = 12 and the odd coin may
be heavier or lighter. Show that you need at least 3 weighings.
[In fact you can always do it in 3 weighings, but the problem of showing
this ‘is said to have been planted during the war . . . by enemy agents since
Operational Research spent so many man-hours on its solution.’57
]
Q 24.14. Extend the definition of entropy to a random variable X taking
values in the non-negative integers. (You must allow for the possibility of
infinite entropy.)
Compute the expected value EY and entropy H(Y ) in the case when
Y has the geometric distribution, that is to say Pr(Y = k) = pk
(1 − p)
[0 p 1]. Show that, amongst all random variables X taking values in
the non-negative integers with the same expected value µ [0 µ ∞], the
geometric distribution maximises the entropy.
Q 24.15. A source produces a set A of messages M1, M2, . . . , Mn with
non-zero probabilities p1, p2, . . . pn. Let S be the codeword length when the
message is encoded by a decodable code c : A → B∗
where B is an alphabet
of k letters.
(i) Show that
n
X
i=1
√
pi
!2
≤ E(kS
)
[Hint: Cauchy–Schwarz, p
1/2
i = p
1/2
i ksi/2
k−si/2
.]
(ii) Show that
min E(kS
) ≤ k
n
X
i=1
√
pi
!2
.
where the minimum is taken over all decodable codes.
[Hint: Look for a code with codeword lengths si = ⌈− logk p
1/2
i /λ⌉ for an
appropriate λ.]
57
The quotation comes from Pedoe’s The Gentle Art of Mathematics which also gives a
very pretty solution. As might be expected, there are many accounts of this problem on
the web.
89](https://image.slidesharecdn.com/shan-250416103225-19ae7e70/85/Shan-pdfFully-Homomorphic-Encryption-FHE-89-320.jpg)
![(ii) Three code words are chosen at random from Fn
2 . If kn is the expected
value of the distance between the closest two, show that n−1
kn → 1/2 as
n → ∞.
[There are many ways to do (ii). One way is to consider Tchebychev’s in-
equality.]
Q 25.8. (Exercises 11.2 and 11.3.) Consider the situation described in the
first paragraph of Section 11.
(i) Show that for the situation described you should not bet if up ≤ 1
and should take
w =
up − 1
u − 1
if up 1.
(ii) Let us write q = 1 − p. Show that, if up 1 and we choose the
optimum w,
E log Yn = p log p + q log q + log u − q log(u − 1).
(iii) Show that, if you bet less than the optimal proportion, your fortune
will still tend to increase but more slowly, but, if you bet more than some
proportion w1, your fortune will decrease. Write down the equation for w1.
[Moral: If you use the Kelly criterion veer on the side under-betting.]
Q 25.9. Your employer announces that he is abandoning the old-fashioned
paternalistic scheme under which he guarantees you a fixed sum Kx (where,
of course, K, x 0) when you retire. Instead, he will empower you by giving
you a fixed sum x now, to invest as you wish. In order to help you and the
rest of the staff, your employer arranges that you should obtain advice from
a financial whizkid with a top degree from Cambridge. After a long lecture
in which the whizkid manages to be simultaneously condescending, boring
and incomprehensible, you come away with the following information.
When you retire, the world will be in exactly one of n states. By means
of a piece of financial wizardry called ditching (or something like that) the
whizkid can offer you a pension plan which for the cost of xi will return
Kxiq−1
i if the world is in state i, but nothing otherwise. (Here qi 0 and
Pn
i=1 qi = 1.) The probability that the world will be in state i is pi. You
must invest the entire fixed sum. (Formally,
Pn
i=1 xi = x. You must also
take xi ≥ 0.) On philosophical grounds you decide to maximise the expected
value S of the logarithm of the sum received on retirement. Assuming that
you will have to live off this sum for the rest of your life, explain, in your
opinion, why this choice is reasonable or explain why it is unreasonable.
Find the appropriate choices of xi. Do they depend on the qi?
92](https://image.slidesharecdn.com/shan-250416103225-19ae7e70/85/Shan-pdfFully-Homomorphic-Encryption-FHE-92-320.jpg)
![26 Exercise Sheet 3
Q 26.1. A message passes through a binary symmetric channel with prob-
ability p of error for each bit and the resulting message is passed through
a second binary symmetric channel which is identical except that there is
probability q of error [0 p, q 1/2]. Show that the result behaves as if
it had been passed through a binary symmetric channel with probability of
error to be determined. Show that the probability of error is less than 1/2.
Can we improve the rate at which messages are transmitted (with low error)
by coding, sending through the first channel, decoding with error correction
and then recoding, sending through the second channel and decoding with
error correction again or will this produce no improvement on treating the
whole thing as a single channel and coding and decoding only once?
Q 26.2. Write down the weight enumerators of the trivial code (that is to
say, Fn
2 ), the zero code (that is to say, {0}), the repetition code and the
simple parity code.
Q 26.3. List the codewords of the Hamming (7,4) code and its dual. Write
down the weight enumerators and verify that they satisfy the MacWilliams
identity.
Q 26.4. (a) Show that if C is linear, then so are its extension C+
, truncation
C−
and puncturing C′
, provided the symbol chosen to puncture by is 0. Give
an example to show that C′
may not be linear if we puncture by 1.
(b) Show that extension followed by truncation does not change a code.
Is this true if we replace ‘truncation’ by ‘puncturing’?
(c) Give an example where puncturing reduces the information rate and
an example where puncturing increases the information rate.
(d) Show that the minimum distance of the parity extension C+
is the
least even integer n with n ≥ d(C).
(e) Show that the minimum distance of the truncation C−
is d(C) or
d(C) − 1 and that both cases can occur.
(f) Show that puncturing cannot decrease the minimum distance, but give
examples to show that the minimum distance can stay the same or increase.
Q 26.5. If C1 and C2 are linear codes of appropriate type with generator
matrices G1 and G2, write down a generator matrix for C1|C2.
Q 26.6. Show that the weight enumerator of RM(d, 1) is
y2d
+ (2d+1
− 2)x2d−1
y2d−1
+ x2d
.
95](https://image.slidesharecdn.com/shan-250416103225-19ae7e70/85/Shan-pdfFully-Homomorphic-Encryption-FHE-95-320.jpg)
![Q 26.7. (i) Show that every codeword in RM(d, d − 1) has even weight.
(ii) Show that RM(m, m − r − 1) ⊆ RM(m, r)⊥
.
(iii) By considering dimension, or otherwise, show that RM(m, r) has
dual code RM(m, m − r − 1).
Q 26.8. (Exercises 8.6 and 8.7.) We show that, even if 2n
/V (n, e) is an
integer, no perfect code may exist.
(i) Verify that
290
V (90, 2)
= 278
.
(ii) Suppose that C is a perfect 2 error correcting code of length 90 and
size 278
. Explain why we may suppose, without loss of generality, that 0 ∈ C.
(iii) Let C be as in (ii) with 0 ∈ C. Consider the set
X = {x ∈ F90
2 : x1 = 1, x2 = 1, d(0, x) = 3}.
Show that, corresponding to each x ∈ X, we can find a unique c(x) ∈ C
such that d(c(x), x) = 2.
(iv) Continuing with the argument of (iii), show that
d(c(x), 0) = 5
and that ci(x) = 1 whenever xi = 1. If y ∈ X, find the number of solutions
to the equation c(x) = c(y) with x ∈ X and, by considering the number of
elements of X, obtain a contradiction.
(v) Conclude that there is no perfect [90, 278
] code.
(vi) Show that V (3, 23) is a power of 2. (In this case a perfect code exists
called the binary Golay code.)
Q 26.9. [The MacWilliams identity for binary codes] Let C ⊆ Fn
2 be
a linear code of dimension k.
(i) Show that
X
x∈C
(−1)x.y
=
(
2k
if y ∈ C⊥
0 if y /
∈ C⊥
.
(ii) If t ∈ R, show that
X
y∈Fn
2
tw(y)
(−1)x.y
= (1 − t)w(x)
(1 + t)n−w(x)
.
(iii) By using parts (i) and (ii) to evaluate
X
x∈C
X
y∈Fn
2
(−1)x.y
s
t
w(y)
96](https://image.slidesharecdn.com/shan-250416103225-19ae7e70/85/Shan-pdfFully-Homomorphic-Encryption-FHE-96-320.jpg)
![in two different ways, obtain the MacWilliams identity
WC⊥ (s, t) = 2− dim C
WC(t − s, t + s).
Q 26.10. An erasure is a digit which has been made unreadable in trans-
mission. Why are they easier to deal with than errors? Find a necessary and
sufficient condition on the parity check matrix for it to be always possible
to correct t erasures. Find a necessary and sufficient condition on the parity
check matrix for it never to be possible to correct t erasures (ie whatever
message you choose and whatever t erasures are made the recipient cannot
tell what you sent).
Q 26.11. Consider the collection K of polynomials
a0 + a1ω
with aj ∈ F2 manipulated subject to the usual rules of polynomial arithmetic
and to the further condition
1 + ω + ω2
= 0.
Show by finding a generator and writing out its powers that K∗
= K {0}
is a cyclic group under multiplication and deduce that K is a finite field.
[Of course, this follows directly from general theory but direct calculation is
not uninstructive.]
Q 26.12. (i) Identify the cyclic codes of length n corresponding to each of
the polynomials 1, X − 1 and Xn−1
+ Xn−2
+ · · · + X + 1.
(ii) Show that there are three cyclic codes of length 7 corresponding to
irreducible polynomials of which two are versions of Hamming’s original code.
What are the other cyclic codes?
(iii) Identify the dual codes for each of the codes in (ii).
Q 26.13. (Example 15.14.) Prove the following results.
(i) If K is a field containing F2, then (a + b)2
= a2
+ b2
for all a, b ∈ K.
(ii) If P ∈ F2[X] and K is a field containing F2, then P(a)2
= P(a2
) for
all a ∈ K.
(iii) Let K be a field containing F2 in which X7
− 1 factorises into linear
factors. If β is a root of X3
+ X + 1 in K, then β is a primitive root of unity
and β2
is also a root of X3
+ X + 1.
(iv) We continue with the notation of (iii). The BCH code with {β, β2
}
as defining set is Hamming’s original (7,4) code.
97](https://image.slidesharecdn.com/shan-250416103225-19ae7e70/85/Shan-pdfFully-Homomorphic-Encryption-FHE-97-320.jpg)
![Q 26.14. Let C be a binary linear code of length n, rank k and distance d.
(i) Show that C contains a codeword x with exactly d non-zero digits.
(ii) Show that n ≥ d + k − 1.
(iii) Prove that truncating C on the non-zero digits of x produces a code
C′
of length n − d, rank k − 1 and distance d′
≥ ⌈d
2
⌉.
[Hint: To show d′
≥ ⌈d
2
⌉, consider, for y ∈ C, the coordinates where xj = yj
and the coordinates where xj 6= yj.]
(iv) Show that
n ≥ d +
k−1
X
u=1
⌈ d
2u ⌉.
Why does (iv) imply (ii)? Give an example where n d + k − 1.
Q 26.15. Implement the secret sharing method of page 57 with k = 2, n = 3,
xj = j + 1 p = 7, a0 = S = 2, a1 = 3. Check directly that any two people
can find S but no single individual can.
If we take k = 3, n = 4, p = 6, xj = j+1 show that the first two members
and the fourth member of the Faculty Board will be unable to determine S
uniquely. Why does this not invalidate our method?
98](https://image.slidesharecdn.com/shan-250416103225-19ae7e70/85/Shan-pdfFully-Homomorphic-Encryption-FHE-98-320.jpg)
![27 Exercise Sheet 4
Q 27.1. (Exercise 18.2.) Show that the decimal expansion of a rational
number must be a recurrent expansion. Give a bound for the period in terms
of the quotient. Conversely, by considering geometric series, or otherwise,
show that a recurrent decimal represents a rational number.
Q 27.2. A binary non-linear feedback register of length 4 has defining rela-
tion
xn+1 = xnxn−1 + xn−3.
Show that the state space contains 4 cycles of lengths 1, 2, 4 and 9
Q 27.3. A binary LFR was used to generate the following stream
110001110001 . . .
Recover the feedback polynomial by the Berlekamp–Massey method. [The
LFR has length 4 but you should work through the trials for length r for
1 ≤ r ≤ 4.]
Q 27.4. (Exercise 16.5.) Consider the linear recurrence
xn = a0xn−d + a1xn−d+1 + . . . + ad−1xn−1 ⋆
with aj ∈ F2 and a0 6= 0.
(i) Suppose K is a field containing F2 such that the auxiliary polynomial
C has a root α in K. Show that xn = αn
is a solution of ⋆ in K.
(ii) Suppose K is a field containing F2 such that the auxiliary polynomial
C has d distinct roots α1, α2, . . . , αd in K. Show that the general solution
of ⋆ in K is
xn =
d
X
j=1
bjαn
j
for some bj ∈ K. If x0, x1, . . . , xd−1 ∈ F2, show that xn ∈ F2 for all n.
(iii) Work out the first few lines of Pascal’s triangle modulo 2. Show that
the functions fj : Z → F2
fj(n) =
n
j
are linearly independent in the sense that
m
X
j=0
bjfj(n) = 0
99](https://image.slidesharecdn.com/shan-250416103225-19ae7e70/85/Shan-pdfFully-Homomorphic-Encryption-FHE-99-320.jpg)
![for all n implies bj = 0 for 1 ≤ j ≤ m.
(iv) Suppose K is a field containing F2 such that the auxiliary polynomial
C factorises completely into linear factors. If the root αu has multiplicity
m(u) [1 ≤ u ≤ q], show that the general solution of ⋆ in K is
xn =
q
X
u=1
m(u)−1
X
v=0
bu,v
n
v
αn
u
for some bu,v ∈ K. If x0, x1, . . . , xd−1 ∈ F2, show that xn ∈ F2 for all n.
Q 27.5. Consider the recurrence relation
un+p +
n−1
X
j=0
cjuj+p = 0
over a field (if you wish, you may take the field to be R but the algebra is
the same for all fields.) We suppose c0 6= 0. Write down an n × n matrix M
such that
u1
u2
.
.
.
un
= M
u0
u1
.
.
.
un−1
.
Find the characteristic and minimal polynomials for M. Would your
answers be the same if c0 = 0?
Q 27.6. (Exercise 18.9.) One of the most confidential German codes (called
FISH by the British) involved a complex mechanism which the British found
could be simulated by two loops of paper tape of length 1501 and 1497. If
kn = xn +yn where xn is a stream of period 1501 and yn is a stream of period
1497, what is the longest possible period of kn? How many consecutive values
of kn would you need to to find the underlying linear feedback register using
the Berlekamp–Massey method if you did not have the information given in
the question? If you had all the information given in the question how many
values of kn would you need? (Hint, look at xn+1497 − xn.)
You have shown that, given kn for sufficiently many consecutive n we can
find kn for all n. Can you find xn for all n?
Q 27.7. We work in F2. I have a secret sequence k1, k2, . . . and a message
p1, p2, . . . , pN . I transmit p1 + k1, p2 + k2, . . . pN + kN and then, by error,
transmit p1 + k2, p2 + k3, . . . pN + kN+1. Assuming that you know this and
that my message makes sense, how would you go about finding my message?
Can you now decipher other messages sent using the same part of my secret
sequence?
100](https://image.slidesharecdn.com/shan-250416103225-19ae7e70/85/Shan-pdfFully-Homomorphic-Encryption-FHE-100-320.jpg)
![Q 27.8. Give an example of a homomorphism attack on an RSA code. Show
in reasonable detail that the Elgamal signature scheme defeats it.
Q 27.9. I announce that I shall be using the Rabin–Williams scheme with
modulus N. My agent in X’Dofdro sends me a message m (with 1 ≤ m ≤
N − 1) encoded in the requisite form. Unfortunately, my cat eats the piece
of paper on which the prime factors of N are recorded, so I am unable to
decipher it. I therefore find a new pair of primes and announce that I shall
be using the Rabin–Williams scheme with modulus N′
N. My agent now
recodes the message and sends it to me again.
The dreaded SNDO of X’Dofdro intercept both code messages. Show that
they can find m. Can they decipher any other messages sent to me using
only one of the coding schemes?
Q 27.10. Extend the Diffie–Hellman key exchange system to cover three
participants in a way that is likely to be as secure as the two party scheme.
Extend the system to n parties in such a way that they can compute their
common secret key by at most n2
−n communications of ‘Diffie–Hellman type
numbers’. (The numbers p and g of our original Diffie-Hellman system are
known by everybody in advance.) Show that this can be done using at most
2n − 2 communications by including several ‘Diffie–Hellman type numbers’
in one message.
Q 27.11. St Abacus, who established written Abacan, was led, on theological
grounds, to use an alphabet containing only three letters A, B and C and to
avoid the use of spaces. (Thus an Abacan book consists of single word.) In
modern Abacan, the letter A has frequency .5 and the letters B and C both
have frequency .25. In order to disguise this, the Abacan Navy uses codes in
which the 3r + ith number is x3r+i + yi modulo 3 [0 ≤ i ≤ 2] where xj = 0
if the jth letter of the message is A, xj = 1 if the jth letter of the message
is B, xj = 2 if the jth letter of the message is C and y0, y1 and y2 are the
numbers 0, 1, 2 in some order.
Radio interception has picked up the following message.
120022010211121001001021002021
Although nobody in Naval Intelligence reads Abacan, it is believed that
the last letter of the message will be B if the Abacan fleet is at sea. The
Admiralty are desperate to know the last letter and send a representative to
your rooms in Baker Street to ask your advice. Give it.
101](https://image.slidesharecdn.com/shan-250416103225-19ae7e70/85/Shan-pdfFully-Homomorphic-Encryption-FHE-101-320.jpg)
Shan.pdfFully Homomorphic Encryption (FHE)
- 1. Coding and Cryptography T. W. Körner December 20, 2018 Transmitting messages is an important practical problem. Coding theory includes the study of compression codes which enable us to send messages cheaply and error correcting codes which ensure that messages remain legible even in the presence of errors. Cryptography on the other hand, makes sure that messages remain unreadable — except to the intended recipient. These techniques turn out to have much in common. Many Part II courses go deeply into one topic so that you need to un- derstand the whole course before you understand any part of it. They often require a firm grasp of some preceding course. Although this course has an underlying theme, it splits into parts which can be understood separately and, although it does require knowledge from various earlier courses, it does not require mastery of that knowledge. All that is needed is a little probability, a little algebra and a fair amount of common sense. On the other hand, the variety of techniques and ideas probably makes it harder to understand everything in the course than in a more monolithic course. Small print The syllabus for the course is defined by the Faculty Board Schedules (which are minimal for lecturing and maximal for examining). I should very much appreciate being told of any corrections or possible improvements however minor. This document is written in L A TEX2e and should be available from my home page http://www.dpmms.cam.ac.uk/˜twk in latex, dvi, ps and pdf formats. Supervisors can obtain comments on the exercises at the end of these notes from the secretaries in DPMMS or by e-mail from me. My e-mail address is twk@dpmms. These notes are based on notes taken in the course of a previous lecturer Dr Pinch, on the excellent set of notes available from Dr Carne’s home page and on Dr Fisher’s collection of examples. Dr Parker and Dr Lawther produced two very useful list of corrections. Any credit for these notes belongs to them, any discredit to me. This is a course outline. A few proofs are included or sketched in these notes but most are omitted. Please note that vectors are row vectors unless otherwise stated. 1
- 2. Contents 1 Codes and alphabets 3 2 Huffman’s algorithm 5 3 More on prefix-free codes 9 4 Shannon’s noiseless coding theorem 11 5 Non-independence 16 6 What is an error correcting code? 18 7 Hamming’s breakthrough 20 8 General considerations 24 9 Some elementary probability 27 10 Shannon’s noisy coding theorem 29 11 A holiday at the race track 31 12 Linear codes 33 13 Some general constructions 38 14 Polynomials and fields 43 15 Cyclic codes 47 16 Shift registers 53 17 A short homily on cryptography 59 18 Stream ciphers 61 19 Asymmetric systems 68 20 Commutative public key systems 71 21 Trapdoors and signatures 76 22 Quantum cryptography 78 2
- 3. 23 Further reading 82 24 Exercise Sheet 1 85 25 Exercise Sheet 2 90 26 Exercise Sheet 3 95 27 Exercise Sheet 4 99 1 Codes and alphabets Originally, a code was a device for making messages hard to read. The study of such codes and their successors is called cryptography and will form the subject of the last quarter of these notes. However, in the 19th century the optical1 and then the electrical telegraph made it possible to send mes- sages speedily, but only after they had been translated from ordinary written English or French into a string of symbols. The best known of the early codes is the Morse code used in electronic telegraphy. We think of it as consisting of dots and dashes but, in fact, it had three symbols dot, dash and pause which we write as •, − and ∗. Morse assigned a code word consisting of a sequence of symbols to each of the letters of the alphabet and each digit. Here are some typical examples. A 7→ • − ∗ B 7→ − • • • ∗ C 7→ − • − • ∗ D 7→ − • •∗ E 7→ •∗ F 7→ • • − • ∗ O 7→ − − −∗ S 7→ • • •∗ 7 7→ − − • • •∗ The symbols of the original message would be encoded and the code words sent in sequence, as in SOS 7→ • • • ∗ − − − ∗ • • •∗, and then decoded in sequence at the other end to recreate the original mes- sage. Exercise 1.1. Decode − • − • ∗ − − − ∗ − • • ∗ • ∗. 1 See The Count of Monte Cristo and various Napoleonic sea stories. A statue to the inventor of the optical telegraph (semaphore) was put up in Paris in 1893 but melted down during World War II and not replaced (http://hamradio.nikhef.nl/tech/rtty/chappe/). In the parallel universe of Disc World the clacks is one of the wonders of the Century of the Anchovy. 3
- 4. Morse’s system was intended for human beings. Once machines took over the business of encoding, other systems developed. A very influential one called ASCII was developed in the 1960s. This uses two symbols 0 and 1 and all code words have seven symbols. In principle, this would give 128 possibilities, but 0000000 and 1111111 are not used, so there are 126 code words allowing the original message to contain a greater variety of symbols than Morse code. Here are some typical examples A 7→ 1000001 B 7→ 1000010 C 7→ 1000011 a 7→ 1100001 b 7→ 1100010 c 7→ 1100011 + 7→ 0101011 ! 7→ 0100001 7 7→ 0110111 Exercise 1.2. Encode b7!. Decode 110001111000011100010. More generally, we have two alphabets A and B and a coding function c : A → B∗ where B∗ consists of all finite sequences of elements of B. If A∗ consists of all finite sequences of elements of A, then the encoding function c∗ : A∗ → B∗ is given by c∗ (a1a2 . . . an) = c(a1)c(a2) . . . c(an). We demand that c∗ is injective, since otherwise it is possible to produce two messages which become indistinguishable once encoded. We call codes for which c∗ is injective decodable. For many purposes, we are more interested in the collection of code words C = c(A) than the coding function c. If we look at the code words of Morse code and the ASCII code, we observe a very important difference. All the code words in ASCII have the same length (so we have a fixed length code), but this is not true for the Morse code (so we have a variable length code). Exercise 1.3. Explain why (if c is injective) any fixed length code is decod- able. A variable length code need not be decodable even if c is injective. Exercise 1.4. (i) Let A = B = {0, 1}. If c(0) = 0, c(1) = 00 show that c is injective but c∗ is not. (ii) Let A = {1, 2, 3, 4, 5, 6} and B = {0, 1}. Show that there is a variable length coding c such that c is injective and all code words have length 2 or less. Show that there is no decodable coding c such that all code words have length 2 or less. However, there is a family of variable length codes which are decodable in a natural way. 4
- 5. Definition 1.5. Let B be an alphabet. We say that a finite subset C of B∗ is prefix-free if, whenever w ∈ C is an initial sequence of w′ ∈ C, then w = w′ . If c : A → B∗ is a coding function, we say that c is prefix-free if c is injective and c(A) is prefix-free. If c is prefix-free, then, not only is c∗ injective, but we can decode messages on the fly. Suppose that we receive a sequence b1, b2, . . . . The moment we have received some c(a1), we know that the first message was a1 and we can proceed to look for the second message. (For this reason prefix-free codes are sometimes called instantaneous codes or self punctuation codes.) Exercise 1.6. Let A = {0, 1, 2, 3}, B = {0, 1}. If c, c̃ : A → B∗ are given by c(0) = 0 c̃(0) = 0 c(1) = 10 c̃(1) = 01 c(2) = 110 c̃(2) = 011 c(3) = 111 c̃(3) = 111 show that c is prefix-free, but c̃ is not. By thinking about the way c̃ is obtained from c, or otherwise, show that c̃∗ is injective. Exercise 1.7. Why is every injective fixed length code automatically prefix- free? From now on, unless explicitly stated otherwise, c will be injective and the codes used will be prefix-free. In section 3 we show that we lose nothing by confining ourselves to prefix-free codes. 2 Huffman’s algorithm An electric telegraph is expensive to build and maintain. However good a telegraphist was, he could only send or receive a limited number of dots and dashes each minute. (This is why Morse chose a variable length code. The telegraphist would need to send the letter E far more often than the letter Q so Morse gave E the short code •∗ and Q the long code − − • − ∗.) It is possible to increase the rate at which symbols are sent and received by using machines, but the laws of physics (backed up by results in Fourier analysis) place limits on the number of symbols that can be correctly transmitted over a given line. (The slowest rates were associated with undersea cables.) Customers were therefore charged so much a letter or, more usually, so much a word2 (with a limit on the permitted word length). Obviously it made 2 Leading to a prose style known as telegraphese. ‘Arrived Venice. Streets flooded. Advise.’ 5
- 6. sense to have books of ‘telegraph codes’ in which one five letter combination, say, ‘FTCGI’ meant ‘are you willing to split the difference?’ and another ‘FTCSU’ meant ‘cannot see any difference’3 . Today messages are usually sent in as binary sequences like 01110010 . . ., but the transmission of each digit still costs money. If we know that there are n possible messages that can be sent and that n ≤ 2m , then we can assign each message a different string of m zeros and ones (usually called bits) and each message will cost mK cents where K is the cost of sending one bit. However, this may not be the best way of saving money. If, as often happens, one message (such as ‘nothing to report’) is much more frequent than any other then it may be cheaper on average to assign it a shorter code word even at the cost of lengthening the other code words. Problem 2.1. Given n messages M1, M2, . . . , Mn such that the probability that Mj will be chosen is pj, find distinct code words Cj consisting of lj bits so that the expected cost K n X j=1 pjlj of sending the code word corresponding to the chosen message is minimised. Of course, we suppose K > 0. The problem is interesting as it stands, but we have not taken into account the fact that a variable length code may not be decodable. To deal with this problem we add an extra constraint. Problem 2.2. Given n messages M1, M2, . . . , Mn such that the probability that Mj will be chosen is pj, find a prefix-free collection of code words Cj consisting of lj bits so that the expected cost K n X j=1 pjlj of sending the code word corresponding to the chosen message is minimised. In 1951 Huffman was asked to write an essay on this problem as an end of term university exam. Instead of writing about the problem, he solved it completely. 3 If the telegraph company insisted on ordinary words you got codes like ‘FLIRT’ for ‘quality of crop good’. Google ‘telegraphic codes and message practice, 1870-1945’ for lots of examples. 6
- 7. Theorem 2.3. [Huffman’s algorithm] The following algorithm solves Prob- lem 2.2 with n messages. Order the messages so that p1 ≥ p2 ≥ · · · ≥ pn. Solve the problem with n − 1 messages M′ 1, M′ 2, . . . , M′ n−1 such that M′ j has probability pj for 1 ≤ j ≤ n − 2, but M′ n−1 has probability pn−1 + pn. If C′ j is the code word corresponding to M′ j, the original problem is solved by assigning Mj the code word C′ j for 1 ≤ j ≤ n − 2 and Mn−1 the code word consisting of C′ n−1 followed by 0 and Mn the code word consisting of C′ n−1 followed by 1. Since the problem is trivial when n = 2 (give M1 the code word 0 and M2 the code word 1) this gives us what computer programmers and logicians call a recursive solution. Recursive programs are often better adapted to machines than human beings, but it is very easy to follow the steps of Huffman’s algorithm ‘by hand’. Example 2.4. Suppose n = 4, Mj has probability j/10 for 1 ≤ j ≤ 4. Apply Huffman’s algorithm. Solution. (Note that we do not bother to reorder messages.) Combining messages in the suggested way, we get 1, 2, 3, 4 [1, 2], 3, 4 [[1, 2], 3], 4. Working backwards, we get C[[1,2],3] = 0 . . . , C4 = 1 C[1,2] = 01 . . . , C3 = 00 C1 = 011, C2 = 010. The reader is strongly advised to do a slightly more complicated example like the next. Exercise 2.5. Suppose Mj has probability j/45 for 1 ≤ j ≤ 9. Apply Huff- man’s algorithm. As we indicated earlier, the effects of Huffman’s algorithm will be most marked when a few messages are highly probable. 7
- 8. Exercise 2.6. Suppose n = 64, M1 has probability 1/2, M2 has probability 1/4 and Mj has probability 1/248 for 3 ≤ j ≤ 64. Explain why, if we use code words of equal length then the length of a code word must be at least 6. By using the ideas of Huffman’s algorithm (you should not need to go through all the steps) obtain a set of code words such that the expected length of a code word sent is not more than 3. Whilst doing the exercises the reader must already have been struck by the fact that minor variations in the algorithm produce different codes. (Note, for example that, if we have a Huffman code, then interchanging the role of 0 and 1 will produce another Huffman type code.) In fact, although the Huffman algorithm will always produce a best code (in the sense of Prob- lem 2.2), there may be other equally good codes which could not be obtained in this manner. Exercise 2.7. Suppose n = 4, M1 has probability .23, M2 has probability .24, M3 has probability .26 and M4 has probability .27. Show that any assignment of the code words 00, 01, 10 and 11 produces a best code in the sense of Problem 2.2. The fact that the Huffman code may not be the unique best solution means that we need to approach the proof of Theorem 2.3 with caution. We observe that reading a code word from a prefix-free code is like climbing a tree with 0 telling us to take the left branch and 1 the right branch. The fact that the code is prefix-free tells us that each code word may be represented by a leaf at the end of a final branch. Thus, for example, the code word 00101 is represented by the leaf found by following left branch, left branch, right branch, left branch, right branch. The next lemma contains the essence of our proof of Theorem 2.3. Lemma 2.8. (i) If we have a best code then it will split into a left branch and right branch at every stage. (ii) If we label every branch by the sum of the probabilities of all the leaves that spring from it then, if we have a best code, every branch belonging to a particular stage of growth will have at least as large a number associated with it as any branch belonging to a later stage. (iii) If we have a best code then interchanging the probabilities of leaves belonging to the last stage (ie the longest code words) still gives a best code. (iv) If we have a best code then two of the leaves with the lowest probabil- ities will appear at the last stage. (v) There is a best code in which two of the leaves with the lowest proba- bilities are neighbours (have code words differing only in the last place). 8
- 9. In order to use the Huffman algorithm we need to know the probabilities of the n possible messages. Suppose we do not. After we have sent k messages we will know that message Mj has been sent kj times and so will the recipient of the message. If we decide to use a Huffman code for the next message, it is not unreasonable (lifting our hat in the direction of the Reverend Thomas Bayes) to take pj = kj + 1 k + n . Provided the recipient knows the exact version of the Huffman algorithm that we use, she can reconstruct our Huffman code and decode our next message. Variants of this idea are known as ‘Huffman-on-the-fly’ and form the basis of the kind of compression programs used in your computer. Notice however, that whilst Theorem 2.3 is an examinable theorem, the contents of this paragraph form a non-examinable plausible statement. 3 More on prefix-free codes It might be thought that Huffman’s algorithm says all that is to be said on the problem it addresses. However, there are two important points that need to be considered. The first is whether we could get better results by using codes which are not prefix-free. The object of this section is to show that this is not the case. As in section 1, we consider two alphabets A and B and a coding function c : A → B∗ (where, as we said earlier, B∗ consists of all finite sequences of elements of B). For most of this course B = {0, 1}, but in this section we allow B to have D elements. The elements of B∗ are called words. Lemma 3.1. [Kraft’s inequality 1] If a prefix-free code C consists of n words Cj of length lj, then n X j=1 D−lj ≤ 1. Lemma 3.2. [Kraft’s inequality 2] Given strictly positive integers lj sat- isfying n X j=1 D−lj ≤ 1, we can find a prefix-free code C consisting of n words Cj of length lj. Proof. Take l1 ≤ l2 ≤ · · · ≤ ln. We give an inductive construction for an appropriate prefix-free code. Start by choosing C1 to be any code word of length l1. 9
- 10. Suppose that we have found a collection of r prefix-free code words Ck of length lk [1 ≤ k ≤ r]. If r = n we are done. If not, consider all possible code words of length lr+1. Of these Dlr+1−lk will have prefix Ck so at most (in fact, exactly) r X k=1 Dlr+1−lk will have one of the code words already selected as prefix. By hypothesis r X k=1 Dlr+1−lk = Dlr+1 r X k=1 D−lk < Dlr+1 . Since there are Dlr+1 possible code words of length lr+1 there is at least one ‘good code word’ which does not have one of the code words already selected as prefix. Choose one of the good code words as Cr+1 and restart the induction. The method used in the proof is called a ‘greedy algorithm’ because we just try to do the best we can at each stage without considering future consequences. Lemma 3.1 is pretty but not deep. MacMillan showed that the same inequality applies to all decodable codes. The proof is extremely elegant and (after one has thought about it long enough) natural. Theorem 3.3. [The MacMillan inequality] If a decodable code C consists of n words Cj of length lj, then n X j=1 D−lj ≤ 1. Using Lemma 3.2 we get the immediate corollary. Lemma 3.4. If there exists a decodable code C consisting of n words Cj of length lj, then there exists a prefix-free code C′ consisting of n words C′ j of length lj. Thus if we are only concerned with the length of code words we need only consider prefix-free codes. 10
- 11. 4 Shannon’s noiseless coding theorem In the previous section we indicated that there was a second question we should ask about Huffman’s algorithm. We know that Huffman’s algorithm is best possible, but we have not discussed how good the best possible should be. Let us restate our problem. (In this section we allow the coding alphabet B to have D elements.) Problem 4.1. Given n messages M1, M2, . . . , Mn such that the probability that Mj will be chosen is pj, find a decodable code C whose code words Cj consist of lj bits so that the expected cost K n X j=1 pjlj of sending the code word corresponding to the chosen message is minimised. In view of Lemma 3.2 (any system of lengths satisfying Kraft’s inequality is associated with a prefix-free and so decodable code) and Theorem 3.3 (any decodable code satisfies Kraft’s inequality), Problem 4.1 reduces to an abstract minimising problem. Problem 4.2. Suppose pj ≥ 0 for 1 ≤ j ≤ n and Pn j=1 pj = 1. Find strictly positive integers lj minimising n X j=1 pjlj subject to n X j=1 D−lj ≤ 1. Problem 4.2 is hard because we restrict the lj to be integers. If we drop the restriction we end up with a problem in Part IB variational calculus. Problem 4.3. Suppose pj ≥ 0 for 1 ≤ j ≤ n and Pn j=1 pj = 1. Find strictly positive real numbers xj minimising n X j=1 pjxj subject to n X j=1 D−xj ≤ 1. Calculus solution. Observe that decreasing any xk decreases Pn j=1 pjxj and increases Pn j=1 D−xj . Thus we may demand n X j=1 D−xj = 1. 11
- 12. The Lagrangian is L(x, λ) = n X j=1 pjxj − λ n X j=1 D−xj . Since ∂L ∂xj = pj + (λ log D)D−xj we know that, at any stationary point, D−xj = K0(λ)pj for some K0(λ) > 0. Since Pn j=1 D−xj = 1, our original problem will have a stationarising solution when D−xj = pj, that is to say xj = − log pj log D and n X j=1 pjxj = − n X j=1 pj log pj log D . It is not hard to convince oneself that the stationarising solution just found is, in fact, maximising, but it is an unfortunate fact that IB variational calculus is suggestive rather than conclusive. The next two exercises (which will be done in lectures and form part of the course) provide a rigorous proof. Exercise 4.4. (i) Show that log t ≤ t − 1 for t > 0 with equality if and only if t = 1. (ii) [Gibbs’ inequality] Suppose that pj, qj > 0 and n X j=1 pj = n X j=1 qj = 1. By applying (i) with t = qj/pj, show that n X j=1 pj log pj ≥ n X j=1 pj log qj with equality if and only if pj = qj. 12
- 13. Exercise 4.5. We use the notation of Problem 4.3. (i) Show that, if x∗ j = − log pj/ log D, then x∗ j > 0 and n X j=1 D−x∗ j = 1. (ii) Suppose that yj > 0 and n X j=1 D−yj = 1. Set qj = D−yj . By using Gibbs’ inequality from Exercise 4.4 (ii), show that n X j=1 pjx∗ j ≤ n X j=1 pjyj with equality if and only if yj = x∗ j for all j. Analysts use logarithms to the base e, but the importance of two-symbol alphabets means that communication theorists often use logarithms to the base 2. Exercise 4.6. (Memory jogger.) Let a, b > 0. Show that loga b = log b log a . The result of Problem 4.3 is so important that it gives rise to a definition. Definition 4.7. Let A be a non-empty finite set and A a random variable taking values in A. If A takes the value a with probability pa we say that the system has Shannon entropy4 (or information entropy) H(A) = − X a∈A pa log2 pa. Theorem 4.8. Let A and B be finite alphabets and let B have D symbols. If A is an A-valued random variable, then any decodable code c : A → B must satisfy E|c(A)| ≥ H(A) log2 D . 4 It is unwise for the beginner and may or may not be fruitless for the expert to seek a link with entropy in physics. 13
- 14. Here |c(A)| denotes the length of c(A). Notice that the result takes a particularly simple form when D = 2. In Problem 4.3 the xj are just positive real numbers but in Problem 4.2 the dj are integers. Choosing dj as close as possible to the best xj may not give the best dj, but it is certainly worth a try. Theorem 4.9. [Shannon–Fano encoding] Let A and B be finite alphabets and let B have D symbols. If A is an A-valued random variable, then there exists a prefix-free (so decodable) code c : A → B which satisfies E|c(A)| ≤ 1 + H(A) log2 D . Proof. By Lemma 3.2 (which states that given lengths satisfying Kraft’s inequality we can construct an associated prefix-free code), it suffices to find strictly positive integers la such that X a∈A D−la ≤ 1, but X a∈A pala ≤ 1 + H(A) log2 D . If we take la = ⌈− logD pa⌉, that is to say, we take la to be the smallest integer no smaller than − logD pa, then these conditions are satisfied and we are done. It is very easy to use the method just indicated to find an appropriate code. (Such codes are called Shannon–Fano codes5 . Fano was the professor who set the homework for Huffman. The point of view adopted here means that for some problems there may be more than one Shannon–Fano code.) Exercise 4.10. (i) Let A = {1, 2, 3, 4}. Suppose that the probability that letter k is chosen is k/10. Observing that log 2−n = −n write down an appropriate Shannon–Fano code c. (ii) We found a Huffman code ch for the system in Example 2.4. Show6 that the entropy is approximately 1.85, that E|c(A)| = 2.4 and that E|ch(A)| = 1.9. Check that these results are consistent with our previous theorems. Putting Theorems 4.8 and Theorem 4.9 together, we get the following remarkable result. 5 Wikipedia and several other sources give a definition of Shannon–Fano codes which is definitely inconsistent with that given here. Within a Cambridge examination context you may assume that Shannon–Fano codes are those considered here. 6 Unless you are on a desert island in which case the calculations are rather tedious. 14
- 15. Theorem 4.11. [Shannon’s noiseless coding theorem] Let A and B be finite alphabets and let B have D symbols. If A is a A-valued random variable, then any decodable code c which minimises E|c(A)| satisfies H(A) log2 D ≤ E|c(A)| ≤ 1 + H(A) log2 D . In particular, Huffman’s code ch for two symbols satisfies H(A) ≤ E|ch(A)| ≤ 1 + H(A). Exercise 4.12. (i) Sketch h(t) = −t log t for 0 ≤ t ≤ 1. (We define h(0) = 0.) (ii) Let Γ = ( p ∈ Rn : pj ≥ 0, n X j=1 pj = 1 ) and let H : Γ → R be defined by H(p) = n X j=1 h(pj). Find the maximum and minimum of H and describe the points where these values are attained. (iii) If n = 2r +s with 0 ≤ s < 2r and pj = 1/n, describe the Huffman code ch for two symbols and verify directly that (with notation of Theorem 4.11) H(A) ≤ E|ch(A)| ≤ 1 + H(A). Waving our hands about wildly, we may say that ‘A system with low Shannon entropy is highly organised and, knowing the system, it is usually quite easy to identify an individual from the system’. Exercise 4.13. The notorious Trinity gang has just been rounded up and Trubshaw of the Yard wishes to identify the leader (or Master, as he is called). Sam the Snitch makes the following offer. Presented with any collection of members of the gang he will (by a slight twitch of his left ear) indicate if the Master is among them. However, in view of the danger involved, he demands ten pounds for each such encounter. Trubshaw believes that the probability of the jth member of the gang being the Master is pj [1 ≤ j ≤ n] and wishes to minimise the expected drain on the public purse. Advise him. 15
- 16. 5 Non-independence (This section is non-examinable.) In the previous sections we discussed codes c : A → B∗ such that, if a letter A ∈ A was chosen according to some random law, E|c(A)| was about as small as possible. If we choose A1, A2, . . . independently according to the same law, then it is not hard to convince oneself that E|c∗ (A1A2A3 . . . An)| = nE|c(A)| will be as small as possible. However, in re*l lif* th* let*ers a*e often no* i*d*p***ent. It is sometimes possible to send messages more efficiently using this fact. Exercise 5.1. Suppose that we have a sequence Xj of random variables taking the values 0 and 1. Suppose that X1 = 1 with probability 1/2 and Xj+1 = Xj with probability .99 independent of what has gone before. (i) Suppose we wish to send ten successive bits XjXj+1 . . . Xj+9. Show that if we associate the sequence of ten zeros with 0, the sequence of ten ones with 10 and any other sequence a0a1 . . . a9 with 11a0a1 . . . a9 we have a decodable code which on average requires about 5/2 bits to transmit the sequence. (ii) Suppose we wish to send the bits XjXj+106 Xj+2×106 . . . Xj+9×106 . Ex- plain why any decodable code will require on average at least 10 bits to trans- mit the sequence. (You need not do detailed computations.) If we transmit sequences of letters by forming them into longer words and coding the words, we say we have a block code. It is plausible that the longer the blocks, the less important the effects of non-independence. In more advanced courses it is shown how to define entropy for systems like the one discussed in Exercise 5.1 (that is to say Markov chains) and that, provided we take long enough blocks, we can recover an analogue of Theorem 4.11 (the noiseless coding theorem). In the real world, the problem lies deeper. Presented with a photograph, we can instantly see that it represents Lena wearing a hat. If a machine reads the image pixel by pixel, it will have great difficulty recognising much, apart from the fact that the distribution of pixels is ‘non-random’ or has ‘low entropy’ (to use the appropriate hand-waving expressions). Clearly, it ought to be possible to describe the photograph with many fewer bits than are required to describe each pixel separately, but, equally clearly, a method that works well on black and white photographs may fail on colour photographs and a method that works well on photographs of faces may work badly when applied to photographs of trees. 16
- 17. Engineers have a clever way of dealing with this problem. Suppose we have a sequence xj of zeros and ones produced by some random process. Someone who believes that they partially understand the nature of the pro- cess builds us a prediction machine which, given the sequence x1, x2, . . . xj so far, predicts the next term will be x′ j+1. Now set yj+1 ≡ xj+1 − x′ j+1 mod 2. If we are given the sequence y1, y2, . . . we can recover the xj inductively using the prediction machine and the formula xj+1 ≡ yj+1 + x′ j+1 mod 2. If the prediction machine is good, then the sequence of yj will consist mainly of zeros and there will be many ways of encoding the sequence as (on average) a much shorter code word. (For example, if we arrange the sequence in blocks of fixed length, many of the possible blocks will have very low probability, so Huffman’s algorithm will be very effective.) Build a better mousetrap, and the world will beat a path to your door. Build a better prediction machine and the world will beat your door down. There is a further real world complication. Engineers distinguish be- tween irreversible ‘lossy compression’ and reversible ‘lossless compression’. For compact discs, where bits are cheap, the sound recorded can be recon- structed exactly. For digital sound broadcasting, where bits are expensive, the engineers make use of knowledge of the human auditory system (for ex- ample, the fact that we can not make out very soft noise in the presence of loud noises) to produce a result that might sound perfect (or nearly so) to us, but which is, in fact, not. For mobile phones, there can be greater loss of data because users do not demand anywhere close to perfection. For digital TV, the situation is still more striking with reduction in data content from film to TV of anything up to a factor of 60. However, medical and satellite pictures must be transmitted with no loss of data. Notice that lossless coding can be judged by absolute criteria, but the merits of lossy coding can only be judged subjectively. Ideally, lossless compression should lead to a signal indistinguishable (from a statistical point of view) from a random signal in which the value of each bit is independent of the value of all the others. In practice, this is only possible in certain applications. As an indication of the kind of problem involved, consider TV pictures. If we know that what is going to be transmit- ted is ‘head and shoulders’ or ‘tennis matches’ or ‘cartoons’ it is possible to obtain extraordinary compression ratios by ‘tuning’ the compression method 17
- 18. to the expected pictures, but then changes from what is expected can be dis- astrous. At present, digital TV encoders merely expect the picture to consist of blocks which move at nearly constant velocity remaining more or less un- changed from frame to frame7 . In this, as in other applications, we know that after compression the signal still has non-trivial statistical properties, but we do not know enough about them to exploit them. 6 What is an error correcting code? In the introductory Section 1, we discussed ‘telegraph codes’ in which one five letter combination ‘QWADR’, say, meant ‘please book quiet room for two’ and another ‘QWNDR’ meant ‘please book cheapest room for one’. Obviously, also, an error of one letter in this code could have unpleasant consequences8 . Today, we transmit and store long strings of binary sequences, but face the same problem that some digits may not be transmitted or stored correctly. We suppose that the string is the result of data compression and so, as we said at the end of the last section, although the string may have non-trivial statistical properties, we do not know enough to exploit this fact. (If we knew how to exploit any statistical regularity, we could build a prediction device and compress the data still further.) Because of this, we shall assume that we are asked to consider a collection of m messages each of which is equally likely. Our model is the following. When the ‘source’ produces one of the m possible messages µi say, it is fed into a ‘coder’ which outputs a string ci of n binary digits. The string is then transmitted one digit at a time along a ‘com- munication channel’. Each digit has probability p of being mistransmitted (so that 0 becomes 1 or 1 becomes 0) independently of what happens to the other digits [0 ≤ p < 1/2]. The transmitted message is then passed through a ‘decoder’ which either produces a message µj (where we hope that j = i) or an error message and passes it on to the ‘receiver’. The technical term for our model is the binary symmetric channel (binary because we use two symbols, symmetric because the probability of error is the same whichever symbol we use). Exercise 6.1. Why do we not consider the case 1 ≥ p > 1/2? What if p = 1/2? 7 Watch what happens when things go wrong. 8 This is a made up example, since compilers of such codes understood the problem. 18
- 19. For most of the time we shall concentrate our attention on a code C ⊆ {0, 1}n consisting of the codewords ci. (Thus we use a fixed length code.) We say that C has size m = |C|. If m is large then we can send a large number of possible messages (that is to say, we can send more information) but, as m increases, it becomes harder to distinguish between different messages when errors occur. At one extreme, if m = 1, errors cause us no problems (since there is only one message) but no information is transmitted (since there is only one message). At the other extreme, if m = 2n , we can transmit lots of messages but any error moves us from one codeword to another. We are led to the following rather natural definition. Definition 6.2. The information rate of C is log2 m n . Note that, since m ≤ 2n the information rate is never greater than 1. Notice also that the values of the information rate when m = 1 and m = 2n agree with what we might expect. How should our decoder work? We have assumed that all messages are equally likely and that errors are independent (this would not be true if, for example, errors occurred in bursts9 ). Under these assumptions, a reasonable strategy for our decoder is to guess that the codeword sent is one which differs in the fewest places from the string of n binary digits received. Here and elsewhere the discussion can be illuminated by the simple notion of a Hamming distance. Definition 6.3. If x, y ∈ {0, 1}n , we write d(x, y) = n X j=1 |xj − yj| and call d(x, y) the Hamming distance between x and y. Lemma 6.4. The Hamming distance is a metric. 9 For the purposes of this course we note that this problem could be tackled by permut- ing the ‘bits’ of the message so that ‘bursts are spread out’. In theory, we could do better than this by using the statistical properties of such bursts to build a prediction machine. In practice, this is rarely possible. In the paradigm case of mobile phones, the properties of the transmission channel are constantly changing and are not well understood. (Here the main restriction on the use of permutation is that it introduces time delays. One way round this is ‘frequency hopping’ in which several users constantly swap transmission channels ‘dividing bursts among users’.) One desirable property of codes for mobile phone users is that they should ‘fail gracefully’, so that as the error rate for the channel rises the error rate for the receiver should not suddenly explode. 19
- 20. We now do some very simple IA probability. Lemma 6.5. We work with the coding and transmission scheme described above. Let c ∈ C and x ∈ {0, 1}n . (i) If d(c, x) = r, then Pr(x received given c sent) = pr (1 − p)n−r . (ii) If d(c, x) = r, then Pr(c sent given x received) = A(x)pr (1 − p)n−r , where A(x) does not depend on r or c. (iii) If c′ ∈ C and d(c′ , x) ≥ d(c, x), then Pr(c sent given x received) ≥ Pr(c′ sent given x received), with equality if and only if d(c′ , x) = d(c, x). This lemma justifies our use, both explicit and implicit, throughout what follows of the so-called maximum likelihood decoding rule. Definition 6.6. The maximum likelihood decoding rule states that a string x ∈ {0, 1}n received by a decoder should be decoded as (one of) the code- word(s) at the smallest Hamming distance from x. Notice that, although this decoding rule is mathematically attractive, it may be impractical if C is large and there is often no known way of finding the codeword at the smallest distance from a particular x in an acceptable number of steps. (We can always make a complete search through all the members of C but unless there are very special circumstances this is likely to involve an unacceptable amount of work.) 7 Hamming’s breakthrough Although we have used simple probabilistic arguments to justify it, the max- imum likelihood decoding rule will often enable us to avoid probabilistic considerations (though not in the very important part of this concerned with Shannon’s noisy coding theorem) and concentrate on algebra and combi- natorics. The spirit of most of the course is exemplified in the next two definitions. Definition 7.1. We say that C is d error detecting if changing up to d digits in a codeword never produces another codeword. 20
- 21. Definition 7.2. We say that C is e error correcting if knowing that a string of n binary digits differs from some codeword of C in at most e places we can deduce the codeword. Here are some simple schemes. Some of them use alphabets with more than two symbols but the principles remain the same. Repetition coding of length n. We take codewords of the form c = (c, c, c, . . . , c) with c = 0 or c = 1. The code C is n − 1 error detecting, and ⌊(n − 1)/2⌋ error correcting. The maximum likelihood decoder chooses the symbol that occurs most often. (Here and elsewhere ⌊α⌋ is the largest integer N ≤ α and ⌈α⌉ is the smallest integer M ≥ α.) Unfortunately, the information rate is 1/n which is rather low10 . The Cambridge examination paper code Each candidate is asked to write down a Candidate Identifier of the form 1234A, 1235B, 1236C, . . . (the eleven11 possible letters are repeated cyclically) and a desk number. The first four numbers in the Candidate Identifier identify the candidate uniquely. If the letter written by the candidate does not correspond to to the first four numbers the candidate is identified by using the desk number. Exercise 7.3. Show that if the candidate makes one error in the Candidate Identifier, then this will be detected. Would this be true if there were 9 possible letters repeated cyclically? Would this be true if there were 12 possible letters repeated cyclically? Give reasons. Show that, if we also use the Desk Number then the combined code Can- didate Number/Desk Number is one error correcting The paper tape code. Here and elsewhere, it is convenient to give {0, 1} the structure of the field F2 = Z2 by using arithmetic modulo 2. The codewords have the form c = (c1, c2, c3, . . . , cn) with c1, c2, . . . , cn−1 freely chosen elements of F2 and cn (the check digit) the element of F2 which gives c1 + c2 + · · · + cn−1 + cn = 0. The resulting code C is 1 error detecting since, if x ∈ Fn 2 is obtained from c ∈ C by making a single error, we have x1 + x2 + · · · + xn−1 + xn = 1. 10 Compare the chorus ‘Oh no John, no John, no John, no’. 11 My guess. 21
- 22. However it is not error correcting since, if x1 + x2 + · · · + xn−1 + xn = 1, there are n codewords y with Hamming distance d(x, y) = 1. The informa- tion rate is (n − 1)/n. Traditional paper tape had 8 places per line each of which could have a punched hole or not, so n = 8. Exercise 7.4. If you look at the inner title page of almost any book published between 1970 and 2006 you will find its International Standard Book Number (ISBN). The ISBN uses single digits selected from 0, 1, . . . , 8, 9 and X rep- resenting 10. Each ISBN consists of nine such digits a1, a2, . . . , a9 followed by a single check digit a10 chosen so that 10a1 + 9a2 + · · · + 2a9 + a10 ≡ 0 mod 11. (*) (In more sophisticated language, our code C consists of those elements a ∈ F10 11 such that P10 j=1(11 − j)aj = 0.) (i) Find a couple of books12 and check that (∗) holds for their ISBNs13 . (ii) Show that (∗) will not work if you make a mistake in writing down one digit of an ISBN. (iii) Show that (∗) may fail to detect two errors. (iv) Show that (∗) will not work if you interchange two distinct adjacent digits (a transposition error). (v) Does (iv) remain true if we replace ‘adjacent’ by ‘different’? Errors of type (ii) and (iv) are the most common in typing14 . In communi- cation between publishers and booksellers, both sides are anxious that errors should be detected but would prefer the other side to query errors rather than to guess what the error might have been. (vi) After January 2007, the appropriate ISBN is a 13 digit number x1x2 . . . x13 with each digit selected from 0, 1, . . . , 8, 9 and the check digit x13 computed by using the formula x13 ≡ −(x1 + 3x2 + x3 + 3x4 + · · · + x11 + 3x12) mod 10. Show that we can detect single errors. Give an example to show that we cannot detect all transpositions. 12 In case of difficulty, your college library may be of assistance. 13 In fact, X is only used in the check digit place. 14 Thus a syllabus for an earlier version of this course contained the rather charming misprint of ‘snydrome’ for ‘syndrome’. 22
- 23. Hamming had access to an early electronic computer but was low down in the priority list of users. He would submit his programs encoded on paper tape to run over the weekend but often he would have his tape returned on Monday because the machine had detected an error in the tape. ‘If the machine can detect an error’ he asked himself ‘why can the machine not correct it?’ and he came up with the following scheme. Hamming’s original code. We work in F7 2. The codewords c are chosen to satisfy the three conditions c1 + c3 + c5 + c7 = 0 c2 + c3 + c6 + c7 = 0 c4 + c5 + c6 + c7 = 0. By inspection, we may choose c3, c5, c6 and c7 freely and then c1, c2 and c4 are completely determined. The information rate is thus 4/7. Suppose that we receive the string x ∈ F7 2. We form the syndrome (z1, z2, z4) ∈ F3 2 given by z1 = x1 + x3 + x5 + x7 z2 = x2 + x3 + x6 + x7 z4 = x4 + x5 + x6 + x7. If x is a codeword, then (z1, z2, z4) = (0, 0, 0). If c is a codeword and the Hamming distance d(x, c) = 1, then the place in which x differs from c is given by z1 + 2z2 + 4z4 (using ordinary addition, not addition modulo 2) as may be easily checked using linearity and a case by case study of the seven binary sequences x containing one 1 and six 0s. The Hamming code is thus 1 error correcting. Exercise 7.5. Suppose we use eight hole tape with the standard paper tape code and the probability that an error occurs at a particular place on the tape (i.e. a hole occurs where it should not or fails to occur where it should) is 10−4 . A program requires about 10 000 lines of tape (each line containing eight places) using the paper tape code. Using the Poisson approximation, direct calculation (possible with a hand calculator but really no advance on the Poisson method), or otherwise, show that the probability that the tape will be accepted as error free by the decoder is less than .04%. Suppose now that we use the Hamming scheme (making no use of the last place in each line). Explain why the program requires about 17 500 lines of tape but that any particular line will be correctly decoded with probability about 1 − (21 × 10−8 ) and the probability that the entire program will be correctly decoded is better than 99.6%. 23
- 24. Hamming’s scheme is easy to implement. It took a little time for his com- pany to realise what he had done15 but they were soon trying to patent it. In retrospect, the idea of an error correcting code seems obvious (Hamming’s scheme had actually been used as the basis of a Victorian party trick) and indeed two or three other people discovered it independently, but Hamming and his co-discoverers had done more than find a clever answer to a ques- tion. They had asked an entirely new question and opened a new field for mathematics and engineering. The times were propitious for the development of the new field. Before 1940, error correcting codes would have been luxuries, solutions looking for problems, after 1950, with the rise of the computer and new communica- tion technologies, they became necessities. Mathematicians and engineers returning from wartime duties in code breaking, code making and general communications problems were primed to grasp and extend the ideas. The mathematical engineer Claude Shannon may be considered the presiding ge- nius of the new field. The reader will observe that data compression shortens the length of our messages by removing redundancy and Hamming’s scheme (like all error correcting codes) lengthens them by introducing redundancy. This is true, but data compression removes redundancy which we do not control and which is not useful to us and error correction coding then replaces it with carefully controlled redundancy which we can use. The reader will also note an analogy with ordinary language. The idea of data compression is illustrated by the fact that many common words are short16 . On the other hand the redund of ordin lang makes it poss to understa it even if we do no catch everyth that is said. 8 General considerations How good can error correcting and error detecting17 codes be? The following discussion is a natural development of the ideas we have already discussed. Later, in our discussion of Shannon’s noisy coding theorem we shall see an- other and deeper way of looking at the question. Definition 8.1. The minimum distance d of a code is the smallest Hamming 15 Experienced engineers came away from working demonstrations muttering ‘I still don’t believe it’. 16 Note how ‘horseless carriage’ becomes ‘car’ and ‘telephone’ becomes ‘phone’. 17 If the error rate is low and it is easy to ask for the message to be retransmitted, it may be cheaper to concentrate on error detection. If there is no possibility of retransmission (as in long term data storage), we have to concentrate on error correction. 24
- 25. distance between distinct code words. We call a code of length n, size m and distance d an [n, m, d] code. Less briefly, a set C ⊆ Fn 2 , with |C| = m and min{d(x, y) : x, y ∈ C, x 6= y} = d is called an [n, m, d] code. By an [n, m] code we shall simply mean a code of length n and size m. Lemma 8.2. A code of minimum distance d can detect d − 1 errors18 and correct ⌊d−1 2 ⌋ errors. It cannot detect all sets of d errors and cannot correct all sets of ⌊d−1 2 ⌋ + 1 errors. It is natural, here and elsewhere, to make use of the geometrical insight provided by the (closed) Hamming ball B(x, r) = {y : d(x, y) ≤ r}. Observe that |B(x, r)| = |B(0, r)| for all x and so, writing V (n, r) = |B(0, r)|, we know that V (n, r) is the number of points in any Hamming ball of radius r. A simple counting argument shows that V (n, r) = r X j=0 n j . Theorem 8.3. [Hamming’s bound] If a code C is e error correcting, then |C| ≤ 2n V (n, e) . There is an obvious fascination (if not utility) in the search for codes which attain the exact Hamming bound. Definition 8.4. A code C of length n and size m which can correct e errors is called perfect if m = 2n V (n, e) . 18 This is not as useful as it looks when d is large. If we know that our message is likely to contain many errors, all that an error detecting code can do is confirm our expectations. Error detection is only useful when errors are unlikely. 25
- 26. Lemma 8.5. Hamming’s original code is a [7, 16, 3] code. It is perfect. It may be worth remarking in this context that, if a code which can correct e errors is perfect (i.e. has a perfect packing of Hamming balls of radius e), then the decoder must invariably give the wrong answer when presented with e + 1 errors. We note also that, if (as will usually be the case) 2n /V (n, e) is not an integer, no perfect e error correcting code can exist. Exercise 8.6. Even if 2n /V (n, e) is an integer, no perfect code may exist. (i) Verify that 290 V (90, 2) = 278 . (ii) Suppose that C is a perfect 2 error correcting code of length 90 and size 278 . Explain why we may suppose without loss of generality that 0 ∈ C. (iii) Let C be as in (ii) with 0 ∈ C. Consider the set X = {x ∈ F90 2 : x1 = 1, x2 = 1, d(0, x) = 3}. Show that, corresponding to each x ∈ X, we can find a unique c(x) ∈ C such that d(c(x), x) = 2. (iv) Continuing with the argument of (iii), show that d(c(x), 0) = 5 and that ci(x) = 1 whenever xi = 1. If y ∈ X, find the number of solutions to the equation c(x) = c(y) with x ∈ X and, by considering the number of elements of X, obtain a contradiction. (v) Conclude that there is no perfect [90, 278 ] code. The result of Exercise 8.6 was obtained by Golay. Far more importantly, he found another case when 2n /V (n, e) is an integer and there does exist an associated perfect code (the Golay code). Exercise 8.7. Show that V (23, 3) is a power of 2. Unfortunately the proof that the Golay code is perfect is too long to be given in the course, We obtained the Hamming bound, which places an upper bound on how good a code can be, by a packing argument. A covering argument gives us the GSV (Gilbert, Shannon, Varshamov) bound in the opposite direction. Let us write A(n, d) for the size of the largest code with minimum distance d. 26
- 27. Theorem 8.8. [Gilbert, Shannon, Varshamov] We have A(n, d) ≥ 2n V (n, d − 1) . Until recently there were no general explicit constructions for codes which achieved the GSV bound (i.e. codes whose minimum distance d satisfied the inequality A(n, d)V (n, d − 1) ≥ 2n ). Such a construction was finally found by Garcia and Stichtenoth by using ‘Goppa’ codes. 9 Some elementary probability Engineers are, of course, interested in ‘best codes’ of length n for reasonably small values of n, but mathematicians are particularly interested in what happens as n → ∞. We recall some elementary probability. Lemma 9.1. [Tchebychev’s inequality] If X is a bounded real valued random variable and a 0, then Pr(|X − EX| ≥ a) ≤ var X a2 . Theorem 9.2. [Weak law of large numbers] If X1, X2, . . . is a sequence of independent identically distributed real valued bounded random variables and a 0, then Pr n−1 n X j=1 Xj − EX ≥ a ! → 0 as N → ∞. Applying the weak law of large numbers, we obtain the following impor- tant result. Lemma 9.3. Consider the model of a noisy transmission channel used in this course in which each digit has probability p of being wrongly transmitted independently of what happens to the other digits. If ǫ 0, then Pr number of errors in transmission for message of n digits ≥ (1 + ǫ)pn → 0 as n → ∞. 27
- 28. By Lemma 8.2, a code of minimum distance d can correct ⌊d−1 2 ⌋ errors. Thus, if we have an error rate p and ǫ 0, we know that the probability that a code of length n with error correcting capacity ⌈(1 + ǫ)pn⌉ will fail to correct a transmitted message falls to zero as n → ∞. By definition, the biggest code with minimum distance ⌈2(1 + ǫ)pn⌉ has size A(n, ⌈2(1 + ǫ)pn⌉) and so has information rate log2 A(n, ⌈2(1+ǫ)pn⌉)/n. Study of the behaviour of log2 A(n, nδ)/n will thus tell us how large an information rate is possible in the presence of a given error rate. Definition 9.4. If 0 δ 1/2 we write α(δ) = lim sup n→∞ log2 A(n, nδ) n . Definition 9.5. We define the entropy function H : [0, 1] → R by H(0) = H(1) = 0 and H(t) = −t log2(t) − (1 − t) log2(1 − t). Exercise 9.6. (i) We have already met Shannon entropy in Definition 4.7. Give a simple system such that, using the notation of that definition, H(A) = H(t). (ii) Sketch H. What is the value of H(1/2)? Theorem 9.7. With the definitions just given, 1 − H(δ) ≤ α(δ) ≤ 1 − H(δ/2) for all 0 ≤ δ 1/2. Using the Hamming bound (Theorem 8.3) and the GSV bound (Theo- rem 8.8), we see that Theorem 9.7 follows at once from the following result. Theorem 9.8. We have log2 V (n, nδ) n → H(δ) as n → ∞. Our proof of Theorem 9.8 depends, as one might expect, on a version of Stirling’s formula. We only need the very simplest version proved in IA. Lemma 9.9 (Stirling). We have loge n! = n loge n − n + O(log2 n). 28
- 29. We combine this with the remarks that V (n, nδ) = X 0≤j≤nδ n j and that very simple estimates give n m ≤ X 0≤j≤nδ n j ≤ (m + 1) n m where m = ⌊nδ⌋. Although the GSV bound is very important, Shannon showed that a stronger result can be obtained for the error correcting power of the best long codes. 10 Shannon’s noisy coding theorem In the backstreets of Cambridge (Massachusetts) there is a science museum devoted to the glory of MIT. Since MIT has a great deal of glory and since much thought has gone into the presentation of the exhibits, it is well worth a visit. However, for any mathematician, the highlight is a glass case con- taining such things as a juggling machine, an electronic calculator19 that uses Roman numerals both externally and internally, the remnants of a machine built to guess which of heads and tails its opponent would choose next20 and a mechanical maze running mouse. These objects were built by Claude Shannon. In his 1937 master’s thesis, Shannon showed how to analyse circuits us- ing Boolean algebra and binary arithmetic. During the war he worked on gunnery control and cryptography at Bell labs and in 1948 he published A Mathematical Theory of Communication21 . Shannon had several predeces- sors and many successors, but it is his vision which underlies this course. Hamming’s bound together with Theorem 9.7 gives a very strong hint that it is not possible to have an information rate greater than 1 − H(δ) for an error rate δ 1/2. (We shall prove this explicitly in Theorem 10.3.) On the other hand the GSV bound together with Theorem 9.7 shows that it is 19 THROBAC the THrifty ROman numeral BAckwards-looking Computer. Google ‘MIT Museum’, go to ‘objects’ and then search ‘Shannon’. 20 That is to say a prediction machine. Google ‘Shannon Mind-Reading Machine’ for sites giving demonstrations and descriptions of the underlying program. 21 This beautiful paper is available on the web and in his Collected Works. 29
- 30. always possible to have an information rate greater than 1 − H(2δ) for an error rate δ 1/4. Although we can use repetition codes to get a positive information rate when 1/4 ≤ δ 1/2 it looks very hard at first (and indeed second) glance to improve these results. However, Shannon realised that we do not care whether errors arise be- cause of noise in transmission or imperfections in our coding scheme. By allowing our coding scheme to be less than perfect (in this connection, see Question 25.13) we can actually improve the information rate whilst still keeping the error rate low. Theorem 10.1. [Shannon’s noisy coding theorem] Suppose 0 p 1/2 and η 0. Then there exists an n0(p, η) such that, for any n n0, we can find codes of length n which have the property that (under our standard model of a symmetric binary channel with probability of error p) the probability that any codeword is mistaken is less than η and still have information rate 1 − H(p) − η. Shannon’s theorem is a masterly display of the power of elementary prob- abilistic arguments to overcome problems which appear insuperable by other means22 . However, it merely asserts that good codes exist and gives no means of finding them apart from exhaustive search. More seriously, random codes will have no useful structure and the only way to use them is to ‘search through a large dictionary’ at the coding end and ‘search through an enor- mous dictionary’ at the decoding end. It should also be noted that n0(p, η) will be very large when p is close to 1/2. Exercise 10.2. Why, in the absence of suitable structure, is the dictionary at the decoding end much larger than the dictionary at the coding end? It is relatively simple to obtain a converse to Shannon’s theorem. Theorem 10.3. Suppose 0 p 1/2 and η 0. Then there exists an n0(p, η) such that, for any n n0, it is impossible to find codes of length n which have the property that (under our standard model of a symmetric binary channel with probability of error p) the probability that any codeword is mistaken is less than 1/2 and the code has information rate 1 − H(p) + η. 22 Conway says that in order to achieve success in a mathematical field you must either be first or be clever. However, as in the case of Shannon, most of those who are first to recognise a new mathematical field are also clever. 30
- 31. As might be expected, Shannon’s theorem and its converse extend to more general noisy channels (in particular, those where the noise is governed by a Markov chain M). It is possible to define the entropy H(M) associated with M and to show that the information rate cannot exceed 1 − H(M) but that any information rate lower than 1−H(M) can be attained with arbitrarily low error rates. However, we must leave something for more advanced courses, and as we said earlier, it is rare in practice to have very clear information about the nature of the noise we encounter. There is one very important theorem of Shannon which is not covered in this course. In it, he reinterprets a result of Whittaker to show that any continuous signal whose Fourier transform vanishes outside a range of length R can be reconstructed from its value at equally spaced sampling points provided those points are less than A/R apart. (The constant A depends on the conventions used in defining the Fourier transform.) This enables us to apply the ‘digital’ theory of information transmission developed here to continuous signals. 11 A holiday at the race track Although this section is examinable23 , the material is peripheral to the course. Suppose a very rich friend makes you the following offer. Every day, at noon, you may make a bet with her for any amount k you chose. You give her k pounds which she keeps whatever happens. She then tosses a coin and, if it shows heads, she pays you ku and, if it shows tails, she pays you nothing. You know that the probability of heads is p. What should you do? If pu 1, you should not bet, because your expected winnings are neg- ative. If pu 1, most mathematicians would be inclined to bet, but how much? If you bet your entire fortune and win, you will be better off than if you bet a smaller sum, but, if you lose, then you are bankrupt and cannot continue playing. Thus your problem is to discover the proportion w of your present fortune that you should bet. Observe that your choice of w will always be the same (since you expect to go on playing for ever). Only the size of your fortune will vary. If your fortune after n goes is Zn, then Zn+1 = ZnYn+1 23 When the author of the present notes gives the course. This is his interpretation of the sentence in the schedules ‘Applications to gambling and the stock market.’ Other lecturers may view matters differently. 31
- 32. where Yn+1 = uw + (1 − w) if the n + 1st throw is heads and Yn+1 = 1 − w if it is tails. Using the weak law of large numbers, we have the following result. Lemma 11.1. Suppose Y , Y1, Y2, . . . are identically distributed independent random variables taking values in [a, b] with 0 a b. If we write Zn = Y1 . . . Yn, then Pr(|n−1 log Zn − E log Y | ǫ) → 0 as n → 0. Thus you should choose w to maximise E log Yn = p log uw + (1 − w) + (1 − p) log(1 − w). Exercise 11.2. (i) Show that, for the situation described, you should not bet if up ≤ 1 and should take w = up − 1 u − 1 if up 1. (ii) We write q = 1−p. Show that, if up 1 and we choose the optimum w, E log Yn = p log p + q log q + log u − q log(u − 1). We have seen the expression −(p log p + q log q) before as (a multiple of) the Shannon information entropy of a simple probabilistic system. In a paper entitled A New Interpretation of Information Rate24 Kelly showed how to interpret this and similar situations using communication theory. In his model a gambler receives information over a noisy channel about which horse is going to win. Just as Shannon’s theorem shows that information can be transmitted over such a channel at a rate close to channel capacity with negligible risk of error (provided the messages are long enough), so that the gambler can (with arbitrarily high probability) increase her fortune at a certain optimum rate provided that she can continue to bet long enough. Although the analogy between betting and communication channels is very pretty, it was the suggestion that those making a long sequence of bets should aim to maximise the expectation of the logarithm (now called Kelly’s criterion) which made the paper famous. Although Kelly seems never to have used his idea in practice, mathematicians like Thorp, Berlekamp and 24 Available on the web. The exposition is slightly opaque because the Bell company which employed Kelly was anxious not draw attention to the use of telephones for betting fraud. 32
- 33. Shannon himself have made substantial fortunes in the stock market and claim to have used Kelly’s ideas25 . Kelly is also famous for an early demonstration of speech synthesis in which a computer sang ‘Daisy Bell’. This inspired the corresponding scene in the film 2001. Before rushing out to the race track or stock exchange26 , the reader is invited to run computer simulations of the result of Kelly gambling for various values of u and p. She will observe that although, in the very long run, the system works, the short run can be be very unpleasant indeed. Exercise 11.3. Returning to our original problem, show that, if you bet less than the optimal proportion, your fortune will still tend to increase but more slowly, but, if you bet more than some proportion w1, your fortune will decrease. Write down the equation for w1. [Moral: If you use the Kelly criterion veer on the side under-betting.] 12 Linear codes The next few sections involve no probability at all. We shall only be inter- ested in constructing codes which are easy to handle and have all their code words at least a certain Hamming distance apart. Just as Rn is a vector space over R and Cn is a vector space over C, so Fn 2 is a vector space over F2. (If you know about vector spaces over fields, so much the better, if not, just follow the obvious paths.) A linear code is a subspace of Fn 2 . More formally, we have the following definition. Definition 12.1. A linear code is a subset of Fn 2 such that (i) 0 ∈ C, (ii) if x, y ∈ C, then x + y ∈ C. Note that, if λ ∈ F, then λ = 0 or λ = 1, so that condition (i) of the definition just given guarantees that λx ∈ C whenever x ∈ C. We shall see that linear codes have many useful properties. Example 12.2. (i) The repetition code with C = {x : x = (x, x, . . . x)} is a linear code. 25 However, we hear more about mathematicians who win on the stock market than those who lose. 26 A sprat which thinks it’s a shark will have a very short life. 33
- 34. (ii) The paper tape code C = ( x : n X j=0 xj = 0 ) is a linear code. (iii) Hamming’s original code is a linear code. The verification is easy. In fact, examples (ii) and (iii) are ‘parity check codes’ and so automatically linear as we see from the next lemma. Definition 12.3. Consider a set P in Fn 2 . We say that C is the code defined by the set of parity checks P if the elements of C are precisely those x ∈ Fn 2 with n X j=1 pjxj = 0 for all p ∈ P. Lemma 12.4. If C is a code defined by parity checks, then C is linear. We now prove the converse result. Definition 12.5. If C is a linear code, we write C⊥ for the set of p ∈ Fn such that n X j=1 pjxj = 0 for all x ∈ C. Thus C⊥ is the set of parity checks satisfied by C. Lemma 12.6. If C is a linear code, then (i) C⊥ is a linear code, (ii) (C⊥ )⊥ ⊇ C. We call C⊥ the dual code to C. In the language of the course on linear mathematics, C⊥ is the annihilator of C. The following is a standard theorem of that course. Lemma 12.7. If C is a linear code in Fn 2 then dim C + dim C⊥ = n. 34
- 35. Since the treatment of dual spaces is not the most popular piece of math- ematics in IB, we shall give an independent proof later (see the note after Lemma 12.13). Combining Lemma 12.6 (ii) with Lemma 12.7, we get the following corollaries. Lemma 12.8. If C is a linear code, then (C⊥ )⊥ = C. Lemma 12.9. Every linear code is defined by parity checks. Our treatment of linear codes has been rather abstract. In order to put computational flesh on the dry theoretical bones, we introduce the notion of a generator matrix. Definition 12.10. If C is a linear code of length n, any r × n matrix whose rows form a basis for C is called a generator matrix for C. We say that C has dimension or rank r. Example 12.11. As examples, we can find generator matrices for the repe- tition code, the paper tape code and the original Hamming code. Remember that the Hamming code is the code of length 7 given by the parity conditions x1 + x3 + x5 + x7 = 0 x2 + x3 + x6 + x7 = 0 x4 + x5 + x6 + x7 = 0. By using row operations and column permutations to perform Gaussian elimination, we can give a constructive proof of the following lemma. Lemma 12.12. Any linear code of length n has (possibly after permuting the order of coordinates) a generator matrix of the form (Ir|B). Notice that this means that any codeword x can be written as (y|z) = (y|yB) where y = (y1, y2, . . . , yr) may be considered as the message and the vector z = yB of length n − r may be considered the check digits. Any code whose codewords can be split up in this manner is called systematic. We now give a more computational treatment of parity checks. 35
- 36. Lemma 12.13. If C is a linear code of length n with generator matrix G, then a ∈ C⊥ if and only if GaT = 0T . Thus C⊥ = (ker G)T . Using the rank, nullity theorem, we get a second proof of Lemma 12.7. Lemma 12.13 enables us to characterise C⊥ . Lemma 12.14. If C is a linear code of length n and dimension r with gen- erator the n×r matrix G, then, if H is any n×(n−r)– matrix with columns forming a basis of ker G, we know that H is a parity check matrix for C and its transpose HT is a generator for C⊥ . Example 12.15. (i) The dual of the paper tape code is the repetition code. (ii) Hamming’s original code has dual with generator matrix 1 0 1 0 1 0 1 0 1 1 0 0 1 1 0 0 0 1 1 1 1 We saw above that the codewords of a linear code can be written (y|z) = (y|yB) where y may be considered as the vector of message digits and z = yB as the vector of check digits. Thus encoders for linear codes are easy to construct. What about decoders? Recall that every linear code of length n has a (non-unique) associated parity check matrix H with the property that x ∈ C if and only if xH = 0. If z ∈ Fn 2 , we define the syndrome of z to be zH. The following lemma is mathematically trivial but forms the basis of the method of syndrome decoding. Lemma 12.16. Let C be a linear code with parity check matrix H. If we are given z = x + e where x is a code word and the ‘error vector’ e ∈ Fn 2 , then zH = eH. Suppose we have tabulated the syndrome uH for all u with ‘few’ non- zero entries (say, all u with d(u, 0) ≤ K). When our decoder receives z, it computes the syndrome zH. If the syndrome is zero, then z ∈ C and the decoder assumes the transmitted message was z. If the syndrome of the received message is a non-zero vector w, the decoder searches its list until it 36
- 37. finds an e with eH = w. The decoder then assumes that the transmitted message was x = z − e (note that z − e will always be a codeword, even if not the right one). This procedure will fail if w does not appear in the list, but, for this to be case, at least K + 1 errors must have occurred. If we take K = 1, that is we only want a 1 error correcting code, then, writing e(i) for the vector in Fn 2 with 1 in the ith place and 0 elsewhere, we see that the syndrome e(i) H is the ith row of H. If the transmitted message z has syndrome zH equal to the ith row of H, then the decoder assumes that there has been an error in the ith place and nowhere else. (Recall the special case of Hamming’s original code.) If K is large the task of searching the list of possible syndromes becomes onerous and, unless (as sometimes happens) we can find another trick, we find that ‘decoding becomes dear’ although ‘encoding remains cheap’. We conclude this section by looking at weights and the weight enumera- tion polynomial for a linear code. The idea here is to exploit the fact that, if C is linear code and a ∈ C, then a + C = C. Thus the ‘view of C’ from any codeword a is the same as the ‘view of C’ from the particular codeword 0. Definition 12.17. The weight w(x) of a vector x ∈ Fn 2 is given by w(x) = d(0, x). Lemma 12.18. If w is the weight function on Fn 2 and x, y ∈ Fn 2 , then (i) w(x) ≥ 0, (ii) w(x) = 0 if and only if x = 0, (iii) w(x) + w(y) ≥ w(x + y). Since the minimum (non-zero) weight in a linear code is the same as the minimum (non-zero) distance, we can talk about linear codes of minimum weight d when we mean linear codes of minimum distance d. The pattern of distances in a linear code is encapsulated in the weight enumeration polynomial. Definition 12.19. Let C be a linear code of length n. We write Aj for the number of codewords of weight j and define the weight enumeration polyno- mial WC to be the polynomial in two real variables given by WC(s, t) = n X j=0 Ajsj tn−j . Here are some simple properties of WC. 37
- 38. Lemma 12.20. Under the assumptions and with the notation of Defini- tion 12.19, the following results are true. (i) WC is a homogeneous polynomial of degree n. (ii) If C has rank r, then WC(1, 1) = 2r . (iii) WC(0, 1) = 1. (iv) WC(1, 0) takes the value 0 or 1. (v) WC(s, t) = WC(t, s) for all s and t if and only if WC(1, 0) = 1. Lemma 12.21. For our standard model of communication along an error prone channel with independent errors of probability p and a linear code C of length n, WC(p, 1 − p) = Pr(receive a code word | code word transmitted) and Pr(receive incorrect code word | code word transmitted) = WC(p, 1−p)−(1−p)n . Example 12.22. (i) If C is the repetition code, WC(s, t) = sn + tn . (ii) If C is the paper tape code of length n, WC(s, t) = 1 2 ((s+t)n +(t−s)n ). Example 12.22 is a special case of the MacWilliams identity. Theorem 12.23. [MacWilliams identity] If C is a linear code WC⊥ (s, t) = 2− dim C WC(t − s, t + s). We give a proof as Exercise 26.9. (The result is thus not bookwork though it could be set as a problem with appropriate hints.) 13 Some general constructions However interesting the theoretical study of codes may be to a pure mathe- matician, the engineer would prefer to have an arsenal of practical codes so that she can select the one most suitable for the job in hand. In this section we discuss the general Hamming codes and the Reed-Muller codes as well as some simple methods of obtaining new codes from old. Definition 13.1. Let d be a strictly positive integer and let n = 2d − 1. Consider the (column) vector space D = Fd 2. Write down a d × n matrix H whose columns are the 2d − 1 distinct non-zero vectors of D. The Hamming (n, n − d) code is the linear code of length n with HT as parity check matrix. 38
- 39. Of course the Hamming (n, n−d) code is only defined up to permutation of coordinates. We note that H has rank d, so a simple use of the rank nullity theorem shows that our notation is consistent. Lemma 13.2. The Hamming (n, n−d) code is a linear code of length n and rank n − d [n = 2d − 1]. Example 13.3. The Hamming (7, 4) code is the original Hamming code. The fact that any two rows of H are linearly independent and a look at the appropriate syndromes gives us the main property of the general Hamming code. Lemma 13.4. The Hamming (n, n − d) code has minimum weight 3 and is a perfect 1 error correcting code [n = 2d − 1]. Hamming codes are ideal in situations where very long strings of binary digits must be transmitted but the chance of an error in any individual digit is very small. (Look at Exercise 7.5.) Although the search for perfect codes other than the Hamming codes produced the Golay code (not discussed here) and much interesting combinatorics, the reader is warned that, from a practical point of view, it represents a dead end27 . Here are a number of simple tricks for creating new codes from old. Definition 13.5. If C is a code of length n, the parity check extension C+ of C is the code of length n + 1 given by C+ = ( x ∈ Fn+1 2 : (x1, x2, . . . , xn) ∈ C, n+1 X j=1 xj = 0 ) . Definition 13.6. If C is a code of length n, the truncation C− of C is the code of length n − 1 given by C− = {(x1, x2, . . . , xn−1) : (x1, x2, . . . , xn) ∈ C for some xn ∈ F2}. 27 If we confine ourselves to the binary codes discussed in this course, it is known that perfect codes of length n with Hamming spheres of radius ρ exist for ρ = 0, ρ = n, ρ = (n− 1)/2, with n odd (the three codes just mentioned are easy to identify), ρ = 3 and n = 23 (the Golay code, found by direct search) and ρ = 1 and n = 2m − 1. There are known to be non-Hamming codes with ρ = 1 and n = 2m − 1, it is suspected that there are many of them and they are the subject of much research, but, of course they present no practical advantages. The only linear perfect codes with ρ = 1 and n = 2m − 1 are the Hamming codes. 39
- 40. Definition 13.7. If C is a code of length n, the shortening (or puncturing) C′ of C by the symbol α (which may be 0 or 1) is the code of length n − 1 given by C′ = {(x1, x2, . . . , xn−1) : (x1, x2, . . . , xn−1, α) ∈ C}. Lemma 13.8. If C is linear, so is its parity check extension C+ , its trunca- tion C− and its shortening C′ (provided that the symbol chosen is 0). How can we combine two linear codes C1 and C2? Our first thought might be to look at their direct sum C1 ⊕ C2 = {(x|y) : x ∈ C1, y ∈ C2}, but this is unlikely to be satisfactory. Lemma 13.9. If C1 and C2 are linear codes, then we have the following relation between minimum distances. d(C1 ⊕ C2) = min d(C1), d(C2) . On the other hand, if C1 and C2 satisfy rather particular conditions, we can obtain a more promising construction. Definition 13.10. Suppose C1 and C2 are linear codes of length n with C1 ⊇ C2 (i.e. with C2 a subspace of C1). We define the bar product C1|C2 of C1 and C2 to be the code of length 2n given by C1|C2 = {(x|x + y) : x ∈ C1, y ∈ C2}. Lemma 13.11. Let C1 and C2 be linear codes of length n with C1 ⊇ C2. Then the bar product C1|C2 is a linear code with rank C1|C2 = rank C1 + rank C2. The minimum distance of C1|C2 satisfies the equality d(C1|C2) = min(2d(C1), d(C2)). We now return to the construction of specific codes. Recall that the Hamming codes are suitable for situations when the error rate p is very small and we want a high information rate. The Reed-Muller are suitable when the error rate is very high and we are prepared to sacrifice information rate. They were used by NASA for the radio transmissions from its planetary 40
- 41. probes (a task which has been compared to signalling across the Atlantic with a child’s torch28 ). We start by considering the 2d points P0, P1, . . . , P2d−1 of the space X = Fd 2. Our code words will be of length n = 2d and will correspond to the indicator functions IA on X. More specifically, the possible code word cA is given by cA i = 1 if Pi ∈ A cA i = 0 otherwise. for some A ⊆ X. In addition to the usual vector space structure on Fn 2 , we define a new operation cA ∧ cB = cA∩B . Thus, if x, y ∈ Fn 2 , (x0, x1, . . . , xn−1) ∧ (y0, y1, . . . , yn−1) = (x0y0, x1y1, . . . , xn−1yn−1). Finally we consider the collection of d hyperplanes πj = {p ∈ X : pj = 0} [1 ≤ j ≤ d] in Fn 2 and the corresponding indicator functions hj = cπj , together with the special vector h0 = cX = (1, 1, . . . , 1). Exercise 13.12. Suppose that x, y, z ∈ Fn 2 and A, B ⊆ X. (i) Show that x ∧ y = y ∧ x. (ii) Show that (x + y) ∧ z = x ∧ z + y ∧ z. (iii) Show that h0 ∧ x = x. (iv) If cA + cB = cE , find E in terms of A and B. (v) If h0 + cA = cE , find E in terms of A. We refer to A0 = {h0 } as the set of terms of order zero. If Ak is the set of terms of order at most k, then the set Ak+1 of terms of order at most k +1 is defined by Ak+1 = {a ∧ hj : a ∈ Ak, 1 ≤ j ≤ d}. Less formally, but more clearly, the elements of order 1 are the hi , the ele- ments of order 2 are the hi ∧ hj with i j, the elements of order 3 are the hi ∧ hj ∧ hk with i j k and so on. 28 Strictly speaking, the comparison is meaningless. However, it sounds impressive and that is the main thing. 41
- 42. Definition 13.13. Using the notation established above, the Reed-Muller code RM(d, r) is the linear code (i.e. subspace of Fn 2 ) generated by the terms of order r or less. Although the formal definition of the Reed-Muller codes looks pretty impenetrable at first sight, once we have looked at sufficiently many examples it should become clear what is going on. Example 13.14. (i) The RM(3, 0) code is the repetition code of length 8. (ii) The RM(3, 1) code is the parity check extension of Hamming’s orig- inal code. (iii) The RM(3, 2) code is the paper tape code of length 8. (iii) The RM(3, 3) code is the trivial code consisting of all the elements of F3 2. We now prove the key properties of the Reed-Muller codes. We use the notation established above. Theorem 13.15. (i) The elements of order d or less (that is the collection of all possible wedge products formed from the hi ) span Fn 2 . (ii) The elements of order d or less are linearly independent. (iii) The dimension of the Reed-Muller code RM(d, r) is d 0 + d 1 + d 2 + · · · + d r . (iv) Using the bar product notation, we have RM(d, r) = RM(d − 1, r)|RM(d − 1, r − 1). (v) The minimum weight of RM(d, r) is exactly 2d−r . Exercise 13.16. The Mariner mission to Mars used the RM(5, 1) code. What was its information rate? What proportion of errors could it correct in a single code word? Exercise 13.17. Show that the RM(d, d − 2) code is the parity extension code of the Hamming (N, N − d) code with N = 2d − 1. (This is useful because we often want codes of length 2d .) 42
- 43. 14 Polynomials and fields This section is starred. Its object is to make plausible the few facts from modern29 algebra that we shall need. They were covered, along with much else, in various post-IA algebra courses, but attendance at those courses is no more required for this course than is reading Joyce’s Ulysses before going for a night out at an Irish pub. Anyone capable of criticising the imprecision and general slackness of the account that follows obviously can do better themselves and should rewrite this section in an appropriate manner. A field K is an object equipped with addition and multiplication which follow the same rules as do addition and multiplication in R. The only rule which will cause us trouble is If x ∈ K and x 6= 0, then we can find y ∈ K such that xy = 1. ⋆ Obvious examples of fields include R, C and F2. We are particularly interested in polynomials over fields, but here an interesting difficulty arises. Example 14.1. We have t2 + t = 0 for all t ∈ F2. To get round this, we distinguish between the polynomial in the ‘indeter- minate’ X P(X) = n X j=0 ajXj with coefficients aj ∈ K and its evaluation P(t) = Pn j=0 ajtj for some t ∈ K. We manipulate polynomials in X according to the standard rules for polynomials, but say that n X j=0 ajXj = 0 if and only if aj = 0 for all j. Thus X2 + X is a non-zero polynomial over F2 all of whose values are zero. The following result is familiar, in essence, from school mathematics. Lemma 14.2. [Remainder theorem] (i) If P is a polynomial over a field K and a ∈ K, then we can find a polynomial Q and an r ∈ K such that P(X) = (X − a)Q(X) + r. (ii) If P is a polynomial over a field K and a ∈ K is such that P(a) = 0, then we can find a polynomial Q such that P(X) = (X − a)Q(X). 29 Modern, that is, in 1920. 43
- 44. The key to much of the elementary theory of polynomials lies in the fact that we can apply Euclid’s algorithm to obtain results like the following. Theorem 14.3. Suppose that P is a set of polynomials which contains at least one non-zero polynomial and has the following properties. (i) If Q is any polynomial and P ∈ P, then the product PQ ∈ P. (ii) If P1, P2 ∈ P, then P1 + P2 ∈ P. Then we can find a non-zero P0 ∈ P which divides every P ∈ P. Proof. Consider a non-zero polynomial P0 of smallest degree in P. Recall that the polynomial P(X) = X2 + 1 has no roots in R (that is P(t) 6= 0 for all t ∈ R). However, by considering the collection of formal expressions a + bi [a, b ∈ R] with the obvious formal definitions of addition and multiplication and subject to the further condition i2 + 1 = 0, we obtain a field C ⊇ R in which P has a root (since P(i) = 0). We can perform a similar trick with other fields. Example 14.4. If P(X) = X2 +X +1, then P has no roots in F2. However, if we consider F2[ω] = {0, 1, ω, 1 + ω} with obvious formal definitions of addition and multiplication and subject to the further condition ω2 + ω + 1 = 0, then F2[ω] is a field containing F2 in which P has a root (since P(ω) = 0). Proof. The only thing we really need prove is that F2[ω] is a field and to do that the only thing we need to prove is that ⋆ holds. Since (1 + ω)ω = 1 this is easy. In order to state a correct generalisation of the ideas of the previous paragraph we need a preliminary definition. Definition 14.5. If P is a polynomial over a field K, we say that P is reducible if there exists a non-constant polynomial Q of degree strictly less than P which divides P. If P is a non-constant polynomial which is not reducible, then P is irreducible. Theorem 14.6. If P is an irreducible polynomial of degree n ≥ 2 over a field K, then P has no roots in K. However, if we consider K[ω] = (n−1 X j=0 ajωj : aj ∈ K ) 44
- 45. with the obvious formal definitions of addition and multiplication and subject to the further condition P(ω) = 0, then K[ω] is a field containing K in which P has a root. Proof. The only thing we really need prove is that K[ω] is a field and to do that the only thing we need to prove is that ⋆ holds. Let Q be a non-zero polynomial of degree at most n − 1. Since P is irreducible, the polynomials P and Q have no common factor of degree 1 or more. Hence, by Euclid’s algorithm, we can find polynomials R and S such that R(X)Q(X) + S(X)P(X) = 1 and so R(ω)Q(ω) + S(ω)P(ω) = 1. But P(ω) = 0, so R(ω)Q(ω) = 1 and we have proved ⋆. In a proper algebra course we would simply define K[ω] = K[X]/(P(X)) where (P(X)) is the ideal generated by P(X). This is a cleaner procedure which avoids the use of such phrases as ‘the obvious formal definitions of addition and multiplication’ but the underlying idea remains the same. Lemma 14.7. If P is a polynomial over a field K which does not factorise completely into linear factors, then we can find a field L ⊇ K in which P has more linear factors. Proof. Factor P into irreducible factors and choose a factor Q which is not linear. By Theorem 14.6, we can find a field L ⊇ K in which Q has a root α say and so, by Lemma 14.2, a linear factor X − α. Since any linear factor of P in K remains a factor in the bigger field L, we are done. Theorem 14.8. If P is a polynomial over a field K, then we can find a field L ⊇ K in which P factorises completely into linear factors. We shall be interested in finite fields (that is fields K with only a finite number of elements). A glance at our method of proving Theorem 14.8 shows that the following result holds. Lemma 14.9. If P is a polynomial over a finite field K, then we can find a finite field L ⊇ K in which P factorises completely. In this context, we note yet another useful simple consequence of Euclid’s algorithm. 45
- 46. Lemma 14.10. Suppose that P is an irreducible polynomial over a field K which has a linear factor X − α in some field L ⊇ K. If Q is a polynomial over K which has the factor X − α in L, then P divides Q. We shall need a lemma on repeated roots. Lemma 14.11. Let K be a field. If P(X) = Pn j=0 ajXj is a polynomial over K, we define P′ (X) = Pn j=1 jajXj−1 . (i) If P and Q are polynomials, (P + Q)′ = P′ + Q′ and (PQ)′ = P′ Q + PQ′ . (ii) If P and Q are polynomials with P(X) = (X − a)2 Q(X), then P′ (X) = 2(X − a)Q(X) + (X − a)2 Q′ (X). (iii) If P is divisible by (X − a)2 , then P(a) = P′ (a) = 0. If L is a field containing F2, then 2y = (1+1)y = 0y = 0 for all y ∈ L. We can thus deduce the following result which will be used in the next section. Lemma 14.12. If L is a field containing F2 and n is an odd integer, then Xn − 1 can have no repeated linear factors as a polynomial over L. We also need a result on roots of unity given as part (v) of the next lemma. Lemma 14.13. (i) If G is a finite Abelian group and x, y ∈ G have coprime orders r and s, then xy has order rs. (ii) If G is a finite Abelian group and x, y ∈ G have orders r and s, then we can find an element z of G with order the lowest common multiple of r and s. (iii) If G is a finite Abelian group, then there exists an N and an h ∈ G such that h has order N and gN = e for all g ∈ G. (iv) If G is a finite subset of a field K which is a group under multiplica- tion, then G is cyclic. (v) Suppose n is an odd integer. If L is a field containing F2 such that Xn − 1 factorises completely into linear terms, then we can find an ω ∈ L such that the roots of Xn − 1 are 1, ω, ω2 , . . . ωn−1 . (We call ω a primitive nth root of unity.) Proof. (ii) Consider z = xu yv where u is a divisor of r, v is a divisor of s, r/u and s/v are coprime and rs/(uv) = lcm(r, s). (iii) Let h be an element of highest order in G and use (ii). (iv) By (iii) we can find an integer N and a h ∈ G such that h has order N and any element g ∈ G satisfies gN = 1. Thus XN − 1 has a linear factor 46
- 47. X − g for each g ∈ G and so Q g∈G(X − g) divides XN − 1. It follows that the order |G| of G cannot exceed N. But by Lagrange’s theorem N divides G. Thus |G| = N and g generates G. (v) Observe that G = {ω : ωn = 1} is an Abelian group with exactly n elements (since Xn − 1 has no repeated roots) and use (iv). Here is another interesting consequence of Lemma 14.13 (iv). Lemma 14.14. If K is a field with m elements, then there is an element k of K such that K = {0} ∪ {kr : 0 ≤ r ≤ m − 2} and km−1 = 1. Proof. Observe that K {0} forms an Abelian group under multiplication. We call an element k with the properties given in Lemma 14.14 a primitive element of K. Exercise 14.15. Find all the primitive elements of F7. With this hint, it is not hard to show that there is indeed a field with 2n elements containing F2. Lemma 14.16. Let L be some field containing F2 in which X2n−1 − 1 = 0 factorises completely. Then K = {x ∈ L : x2n = x} is a field with 2n elements containing F2. Lemma 14.14 shows that there is (up to field isomorphism) only one field with 2n elements containing F2. We call it F2n . 15 Cyclic codes In this section, we discuss a subclass of linear codes, the so-called cyclic codes. Definition 15.1. A linear code C in Fn 2 is called cyclic if (a0, a1, . . . , an−2, an−1) ∈ C ⇒ (a1, a2, . . . , an−1, a0) ∈ C. 47
- 48. Let us establish a correspondence between Fn 2 and the polynomials on F2 modulo Xn − 1 by setting Pa = n−1 X j=0 ajXj whenever a ∈ Fn 2 . (Of course, Xn − 1 = Xn + 1 but in this context the first expression seems more natural.) Exercise 15.2. With the notation just established, show that (i) Pa + Pb = Pa+b, (ii) Pa = 0 if and only if a = 0. Lemma 15.3. A code C in Fn 2 is cyclic if and only if PC = {Pa : a ∈ C} satisfies the following two conditions (working modulo Xn − 1). (i) If f, g ∈ PC, then f + g ∈ PC. (ii) If f ∈ PC and g is any polynomial, then the product fg ∈ PC. (In the language of abstract algebra, C is cyclic if and only if PC is an ideal of the quotient ring F2[X]/(Xn − 1).) From now on we shall talk of the code word f(X) when we mean the code word a with Pa(X) = f(X). An application of Euclid’s algorithm gives the following useful result. Lemma 15.4. A code C of length n is cyclic if and only if (working modulo Xn −1, and using the conventions established above) there exists a polynomial g such that C = {f(X)g(X) : f a polynomial} (In the language of abstract algebra, F2[X] is a Euclidean domain and so a principal ideal domain. Thus the quotient F2[X]/(Xn −1) is a principal ideal domain.) We call g(X) a generator polynomial for C. Lemma 15.5. A polynomial g is a generator for a cyclic code of length n if and only if it divides Xn − 1. Thus we must seek generators among the factors of Xn − 1 = Xn + 1. If there are no conditions on n, the result can be rather disappointing. Exercise 15.6. If we work with polynomials over F2, then X2r + 1 = (X + 1)2r . In order to avoid this problem and to be able to make use of Lemma 14.12, we shall take n odd from now on. (In this case, the cyclic codes are said to be separable.) Notice that the task of finding irreducible factors (that is factors with no further factorisation) is a finite one. 48
- 49. Lemma 15.7. Consider codes of length n. Suppose that g(X)h(X) = Xn −1. Then g is a generator of a cyclic code C and h is a generator for a cyclic code which is the reverse of C⊥ . As an immediate corollary, we have the following remark. Lemma 15.8. The dual of a cyclic code is itself cyclic. Lemma 15.9. If a cyclic code C of length n has generator g of degree n − r then g(X), Xg(X), . . . , Xr−1 g(X) form a basis for C. Cyclic codes are thus easy to specify (we just need to write down the generator polynomial g) and to encode. We know that Xn + 1 factorises completely over some larger finite field and, since n is odd, we know, by Lemma 14.12, that it has no repeated factors. The same is therefore true for any polynomial dividing it. Lemma 15.10. Suppose that g is a generator of a cyclic code C of odd length n. Suppose further that g factorises completely into linear factors in some field K containing F2. If g = g1g2 . . . gk with each gj irreducible over F2 and A is a subset of the set of all the roots of all the gj and containing at least one root of each gj [1 ≤ j ≤ k], then C = {f ∈ F2[X] : f(α) = 0 for all α ∈ A}. Definition 15.11. A defining set for a cyclic code C is a set A of elements in some field K containing F2 such that f ∈ F2[X] belongs to C if and only if f(α) = 0 for all α ∈ A. (Note that, if C has length n, A must be a set of zeros of Xn − 1.) Lemma 15.12. Suppose that A = {α1, α2, . . . , αr} is a defining set for a cyclic code C in some field K containing F2. Let B be the n × r matrix over K whose jth column is (1, αj, α2 j , . . . , αn−1 j )T Then a vector a ∈ Fn 2 is a code word in C if and only if aB = 0 in K. 49
- 50. The columns in B are not parity checks in the usual sense since the code entries lie in F2 and the computations take place in the larger field K. With this background we can discuss a famous family of codes known as the BCH (Bose, Ray-Chaudhuri, Hocquenghem) codes. Recall that a primitive nth root of unity is an root α of Xn − 1 = 0 such that every root is a power of α. Definition 15.13. Suppose that n is odd and K is a field containing F2 in which Xn −1 factorises into linear factors. Suppose that α ∈ K is a primitive nth root of unity. A cyclic code C with defining set A = {α, α2 , . . . , αδ−1 } is a BCH code of design distance δ. Note that the rank of C will be n − k, where k is the degree of the product of those irreducible factors of Xn − 1 over F2 which have a zero in A. Notice also that k may be very much larger than δ. Example 15.14. (i) If K is a field containing F2, then (a + b)2 = a2 + b2 for all a, b ∈ K. (ii) If P ∈ F2[X] and K is a field containing F2, then P(a)2 = P(a2 ) for all a ∈ K. (iii) Let K be a field containing F2 in which X7 − 1 factorises into linear factors. If β is a root of X3 +X +1 in K, then β is a primitive root of unity and β2 is also a root of X3 + X + 1. (iv) We continue with the notation (iii). The BCH code with {β, β2 } as defining set is Hamming’s original (7,4) code. The next theorem contains the key fact about BCH codes. Theorem 15.15. The minimum distance for a BCH code is at least as great as the design distance. Our proof of Theorem 15.15 relies on showing that the matrix B of Lemma 15.12 is of full rank for a BCH. To do this we use a result which every undergraduate knew in 1950. Lemma 15.16. [The van der Monde determinant] We work over a field K. The determinant 1 1 1 . . . 1 x1 x2 x3 . . . xn x2 1 x2 2 x2 3 . . . x2 n . . . . . . . . . ... . . . xn−1 1 xn−1 2 xn−1 3 . . . xn−1 n = Y 1≤ji≤n (xi − xj). 50
- 51. How can we construct a decoder for a BCH code? From now on, until the end of this section, we shall suppose that we are using the BCH code C described in Definition 15.13. In particular, C will have length n and defining set A = {α, α2 , . . . , αδ−1 } where α is a primitive nth root of unity in K. Let t be the largest integer with 2t + 1 ≤ δ. We show how we can correct up to t errors. Suppose that a codeword c = (c0, c1, . . . , cn−1) is transmitted and that the string received is r. We write e = r − c and assume that E = {0 ≤ j ≤ n − 1 : ej 6= 0} has no more than t members. In other words, e is the error vector and we assume that there are no more than t errors. We write c(X) = n−1 X j=0 cjXj , r(X) = n−1 X j=0 rjXj , e(X) = n−1 X j=0 ejXj . Definition 15.17. The error locator polynomial is σ(X) = Y j∈E (1 − αj X) and the error co-locator is ω(X) = n−1 X i=0 eiαi Y j∈E, j6=i (1 − αj X). Informally, we write ω(X) = n−1 X i=0 eiαi σ(X) 1 − αiX . We take ω(X) = P j ωjXj and σ(X) = P j σjXj . Note that ω has degree at most t − 1 and σ degree at most t. Note that we know that σ0 = 1 so both the polynomials ω and σ have t unknown coefficients. 51
- 52. Lemma 15.18. If the error locator polynomial is given the value of e and so of c can be obtained directly. We wish to make use of relations of the form 1 1 − αjX = ∞ X r=0 (αj X)r . Unfortunately, it is not clear what meaning to assign to such a relation. One way round is to work modulo Z2t (more formally, to work in K[Z]/(Z2t )). We then have Zu ≡ 0 for all integers u ≥ 2t. Lemma 15.19. If we work modulo Z2t then (1 − αj Z) 2t−1 X m=0 (αj Z)m ≡ 1. Thus, if we work modulo Z2t , as we shall from now on, we may define 1 1 − αjZ = 2t−1 X m=0 (αj Z)m . Lemma 15.20. With the conventions already introduced. (i) ω(Z) σ(Z) ≡ 2t−1 X m=0 Zm e(αm+1 ). (ii) e(αm ) = r(αm ) for all 0 ≤ m ≤ 2t − 1. (iii) ω(Z) σ(Z) ≡ 2t−1 X m=0 Zm r(αm+1 ). (iv) ω(Z) ≡ P2t−1 m=0 Zm r(αm+1 )σ(Z). (v) ωj = X u+v=j r(αu+1 )σv for all 0 ≤ j ≤ t − 1. (vi) 0 = X u+v=j r(αu+1 )σv for all t ≤ j ≤ 2t − 1. (vii) The conditions in (vi) determine σ completely. Part (vi) of Lemma 15.20 completes our search for a decoding method, since σ determines E, E determines e and e determines c. It is worth noting that the system of equations in part (v) suffice to determine the pair σ and ω directly. Compact disc players use BCH codes. Of course, errors are likely to occur in bursts (corresponding to scratches etc) and this is dealt with by 52
- 53. distributing the bits (digits) in a single codeword over a much longer stretch of track. The code used can correct a burst of 4000 consecutive errors (2.5 mm of track). Unfortunately, none of the codes we have considered work anywhere near the Shannon bound (see Theorem 10.1). We might suspect that this is be- cause they are linear, but Elias has shown that this is not the case. (We just state the result without proof.) Theorem 15.21. In Theorem 10.1 we can replace ‘code’ by ‘linear code’. The advance of computational power and the ingenuity of the discover- ers30 have lead to new codes which appear to come close to the Shannon bounds. But that is another story. Just as pure algebra has contributed greatly to the study of error correct- ing codes, so the study of error correcting codes has contributed greatly to the study of pure algebra. The story of one such contribution is set out in T. M. Thompson’s From Error-correcting Codes through Sphere Packings to Simple Groups [9] — a good, not too mathematical, account of the discovery of the last sporadic simple groups by Conway and others. 16 Shift registers In this section we move towards cryptography, but the topic discussed will turn out to have connections with the decoding of BCH codes as well. Definition 16.1. A general feedback shift register is a map f : Fd 2 → Fd 2 given by f(x0, x1, . . . , xd−2, xd−1) = (x1, x2, . . . , xd−1, C(x0, x1, . . . , xd−2, xd−1)) with C a map C : Fd 2 → F2. The stream associated to an initial fill (y0, y1, . . . , yd−1) is the sequence y0, y1, . . . , yj, yj+1, . . . with yn = C(yn−d, yn−d+1, . . . , yn−1) for all n ≥ d. Example 16.2. If the general feedback shift f given in Definition 16.1 is a permutation, then C is linear in the first variable, i.e. C(x0, x1, . . . , xd−2, xd−1) = x0 + C′ (x1, x2, . . . , xd−2, xd−1). 30 People like David MacKay, now better known for his superb ‘Sustainable Energy Without the Hot Air’ — rush out and read it. 53
- 54. Definition 16.3. We say that the function f of Definition 16.1 is a linear feedback register if C(x0, x1, . . . , xd−1) = a0x0 + a1x1 + . . . + ad−1xd−1, with a0 = 1. Exercise 16.4. Discuss briefly the effect of omitting the condition a0 = 1 from Definition 16.3. The discussion of the linear recurrence xn = a0xn−d + a1xn−d+1 + · · · + ad−1xn−1 over F2 follows the IA discussion of the same problem over R but is compli- cated by the fact that n2 = n in F2. We assume that a0 6= 0 and consider the auxiliary polynomial C(X) = Xd − ad−1Xd−1 − · · · − a1X − a0. In the exercise below, n v is the appropriate polynomial in n. Exercise 16.5. Consider the linear recurrence xn = a0xn−d + a1xn−d+1 + . . . + ad−1xn−1 ⋆ with aj ∈ F2 and a0 6= 0. (i) Suppose K is a field containing F2 such that the auxiliary polynomial C has a root α in K. Show that xn = αn is a solution of ⋆ in K. (ii) Suppose K is a field containing F2 such that the auxiliary polynomial C has d distinct roots α1, α2, . . . , αd in K. Show that the general solution of ⋆ in K is xn = d X j=1 bjαn j for some bj ∈ K. If x0, x1, . . . , xd−1 ∈ F2, show that xn ∈ F2 for all n. (iii) Work out the first few lines of Pascal’s triangle modulo 2. Show that the functions fj : Z → F2 fj(n) = n j 54
- 55. are linearly independent in the sense that m X j=0 bjfj(n) = 0 for all n implies bj = 0 for 0 ≤ j ≤ m. (iv) Suppose K is a field containing F2 such that the auxiliary polynomial C factorises completely into linear factors. If the root αu has multiplicity m(u), [1 ≤ u ≤ q], show that the general solution of ⋆ in K is xn = q X u=1 m(u)−1 X v=0 bu,v n v αn u for some bu,v ∈ K. If x0, x1, . . . , xd−1 ∈ F2, show that xn ∈ F2 for all n. A strong link with the problem of BCH decoding is provided by Theo- rem 16.7 below. Definition 16.6. If we have a sequence (or stream) x0, x1, x2, . . . of elements of F2 then its generating function G is given by G(Z) = ∞ X n=0 xjZj . If the recurrence relation for a linear feedback generator is d X j=0 cjxn−j = 0 for n ≥ d with c0, cd 6= 0 we call C(z) = d X j=0 cjZj the auxiliary polynomial of the generator. Theorem 16.7. The stream (xn) comes from a linear feedback generator with auxiliary polynomial C if and only if the generating function for the stream is (formally) of the form G(Z) = B(Z) C(Z) with B a polynomial of degree strictly smaller than that of C. 55
- 56. If we can recover C from G then we have recovered the linear feedback generator from the stream. The link with BCH codes is established by looking at Lemma 15.20 (iii) and making the following remark. Lemma 16.8. If a stream (xn) comes from a linear feedback generator with auxiliary polynomial C of degree d, then C is determined by the condition G(Z)C(Z) ≡ B(Z) mod Z2d with B a polynomial of degree at most d − 1. We thus have the following problem. Problem Given a generating function G for a stream and knowing that G(Z) = B(Z) C(Z) with B a polynomial of degree less than that of C and the constant term in C is c0 = 1, recover C. The Berlekamp–Massey method In this method we do not assume that the degree d of C is known. The Berlekamp–Massey solution to this problem is based on the observation that, since d X j=0 cjxn−j = 0 (with c0 = 1) for all n ≥ d, we have xd xd−1 . . . x1 x0 xd+1 xd . . . x2 x1 . . . . . . ... . . . . . . x2d x2d−1 . . . xd+1 xd 1 c1 . . . cd = 0 0 . . . 0 . ⋆ The Berlekamp–Massey method tells us to look successively at the ma- trices A1 = (x0), A2 = x1 x0 x2 x1 , A3 = x2 x1 x0 x3 x2 x1 x4 x3 x2 , . . . starting at Ar if it is known that r ≥ d. For each Aj we evaluate det Aj. If det Aj 6= 0, then j − 1 6= d. If det Aj = 0, then j − 1 is a good candidate 56
- 57. for d so we solve ⋆ on the assumption that d = j − 1. (Note that a one dimensional subspace of Fd+1 contains only one non-zero vector.) We then check our candidate for (c0, c1, . . . , cd) over as many terms of the stream as we wish. If it fails the test, we then know that d ≥ j and we start again31 . As we have stated it, the Berlekamp–Massey method is not an algorithm in the strict sense of the term although it becomes one if we put an upper bound on the possible values of d. (A little thought shows that, if no upper bound is put on d, no algorithm is possible because, with a suitable initial stream, a linear feedback register with large d can be made to produce a stream whose initial values would be produced by a linear feedback register with much smaller d. For the same reason the Berlekamp–Massey method will produce the B of smallest degree which gives G and not necessarily the original B.) In practice, however, the Berlekamp–Massey method is very effective in cases when d is unknown. By careful arrangement of the work it is possible to cut down considerably on the labour involved. The solution of linear equations gives us a method of ‘secret sharing’. Problem 16.9. It is not generally known that CMS when reversed forms the initials of of ‘Secret Missile Command’. If the University is attacked by HEFCE32 , the Faculty Board will retreat to a bunker known as Meeting Room 23. Entry to the room involves tapping out a positive integer S (the secret) known only to the Chairman of the Faculty Board. Each of the n members of the Faculty Board knows a certain pair of numbers (their shadow) and it is required that, in the absence of the Chairman, any k members of the Faculty can reconstruct S from their shadows, but no k − 1 members can do so. How can this be done? Here is one neat solution. Suppose S must lie between 0 and N (it is sensible to choose S at random). The chairman chooses a prime p N, n. She then chooses integers a1, a2, . . . , ak−1 at random and distinct integers x1, x2, . . . , xn at random subject to 0 ≤ aj ≤ p − 1, 1 ≤ xj ≤ p − 1, sets a0 = S and computes P(r) ≡ a0 + a1xr + a2x2 r + · · · + ak−1xk−1 r mod p choosing 0 ≤ P(r) ≤ p − 1. She then gives the rth member of the Faculty Board the pair of numbers xr, P(r) (the shadow pair), to be kept secret 31 Note that, over F2, det Aj can only take two values so there will be many false alarms. Note also that the determinant may be evaluated much faster using reduction to (rear- ranged) triangular form than by Cramer’s rule and that once the system is in (rearranged) triangular form it is easy to solve the associated equations. 32 An institution like SPECTRE but without the charm. 57
- 58. from everybody else) and tells everybody the value of p. She then burns her calculations. Suppose that k members of the Faculty Board with shadow pairs yj, Q(j) = xrj , P(rj) [1 ≤ j ≤ k] are together. By the properties of the Van der Monde determinant (see Lemma 15.16) 1 y1 y2 1 . . . yk−1 1 1 y2 y2 2 . . . yk−1 2 1 y3 y2 3 . . . yk−1 3 . . . . . . . . . ... . . . 1 yk y2 k . . . yk−1 k ≡ 1 1 1 . . . 1 y1 y2 y3 . . . yk y2 1 y2 2 y2 3 . . . y2 k . . . . . . . . . ... . . . yk−1 1 yk−1 2 yk−1 3 . . . yk−1 k ≡ Y 1≤ji≤k−1 (yi − yj) 6≡ 0 mod p. Thus the system of equations z0 + y1z1 + y2 1z2 + . . . + yk−1 1 zk−1 ≡ Q1 z0 + y2z1 + y2 2z2 + . . . + yk−1 2 zk−1 ≡ Q2 z0 + y3z1 + y2 3z2 + . . . + yk−1 3 zk−1 ≡ Q3 . . . z0 + ykz1 + y2 kz2 + . . . + yk−1 k zk−1 ≡ Qk has a unique solution z. But we know that a is a solution, so z = a and the secret S = z0. On the other hand, y1 y2 1 . . . yk−1 1 y2 y2 2 . . . yk−1 2 y3 y2 3 . . . yk−1 3 . . . . . . ... . . . yk−1 y2 k−1 . . . yk−1 k−1 ≡ y1y2 . . . yk−1 Y 1≤ji≤k−1 (yi − yj) 6≡ 0 mod p, so the system of equations z0 + y1z1 + y2 1z2 + . . . + yk−1 1 zk−2 ≡ Q1 z0 + y2z1 + y2 2z2 + . . . + yk−1 2 zk−2 ≡ Q2 z0 + y3z1 + y2 3z2 + . . . + yk−1 3 zk−2 ≡ Q3 . . . z0 + yk−1z1 + y2 k−1z2 + . . . + yk−1 k−1zk−2 ≡ Qk−1 58
- 59. has a solution, whatever value of z0 we take, so k −1 members of the Faculty Board have no way of saying that any possible values of S is more likely than any other. One way of looking at this method of ‘secret sharing’ is to note that a polynomial of degree k − 1 can be recovered from its value at k points but not from its value at k −1 points. However, the proof that the method works needs to be substantially more careful. Exercise 16.10. Is the secret compromised if the values of the xj become known? 17 A short homily on cryptography Cryptography is the science of code making. Cryptanalysis is the art of code breaking. Two thousand years ago, Lucretius wrote that ‘Only recently has the true nature of things been discovered’. In the same way, mathematicians are apt to feel that ‘Only recently has the true nature of cryptography been discovered’. The new mathematical science of cryptography with its promise of codes which are ‘provably hard to break’ seems to make everything that has gone before irrelevant. It should, however, be observed that the best cryptographic systems of our ancestors (such as diplomatic ‘book codes’) served their purpose of ensuring secrecy for a relatively small number of messages between a relatively small number of people extremely well. It is the modern requirement for secrecy on an industrial scale to cover endless streams of messages between many centres which has made necessary the modern science of cryptography. More pertinently, it should be remembered that the German Naval Enigma codes not only appeared to be ‘provably hard to break’ (though not against the modern criteria of what this should mean) but, considered in isolation, probably were unbreakable in practice33 . Fortunately the Submarine codes formed part of an ‘Enigma system’ with certain exploitable weaknesses. (For an account of how these weaknesses arose and how they were exploited see Kahn’s Seizing the Enigma [4].) Even the best codes are like the lock on a safe. However good the lock is, the safe may be broken open by brute force, or stolen together with its contents, or a key holder may be persuaded by fraud or force to open the lock, or the presumed contents of the safe may have been tampered with before they go into the safe, or . . . . The coding schemes we shall consider, 33 Some versions remained unbroken until the end of the war. 59
- 60. are at best, cryptographic elements of larger possible cryptographic systems. The planning of cryptographic systems requires not only mathematics but also engineering, economics, psychology, humility and an ability to learn from past mistakes. Those who do not learn the lessons of history are condemned to repeat them. In considering a cryptographic system, it is important to consider its purpose. Consider a message M sent by A to B. Here are some possible aims. Secrecy A and B can be sure that no third party X can read the message M. Integrity A and B can be sure that no third party X can alter the message M. Authenticity B can be sure that A sent the message M. Non-repudiation B can prove to a third party that A sent the message M. When you fill out a cheque giving the sum both in numbers and words you are seeking to protect the integrity of the cheque. When you sign a traveller’s cheque ‘in the presence of the paying officer’ the process is intended, from your point of view, to protect authenticity and, from the bank’s point of view, to produce non-repudiation. Another point to consider is the level of security aimed at. It hardly matters if a few people use forged tickets to travel on the underground, it does matter if a single unauthorised individual can gain privileged access to a bank’s central computer system. If secrecy is aimed at, how long must the secret be kept? Some military and financial secrets need only remain secret for a few hours, others must remain secret for years. We must also, to conclude this non-exhaustive list, consider the level of security required. Here are three possible levels. (1) Prospective opponents should find it hard to compromise your system even if they are in possession of a plentiful supply of encoded messages Ci. (2) Prospective opponents should find it hard to compromise your system even if they are in possession of a plentiful supply of pairs (Mi, Ci) of messages Mi together with their encodings Ci. (3) Prospective opponents should find it hard to compromise your system even if they are allowed to produce messages Mi and given their encodings Ci. Clearly, safety at level (3) implies safety at level (2) and safety at level (2) implies safety at level (1). Roughly speaking, the best Enigma codes satisfied (1). The German Navy believed on good but mistaken grounds that they satisfied (2). Level (3) would have appeared evidently impossible to attain until a few years ago. Nowadays, level (3) is considered a minimal requirement for a really secure system. 60
- 61. 18 Stream ciphers One natural way of enciphering is to use a stream cipher. We work with streams (that is, sequences) of elements of F2. We use a cipher stream k0, k1, k2 . . . . The plain text stream p0, p1, p2, . . . is enciphered as the cipher text stream z0, z1, z2, . . . given by zn = pn + kn. This is an example of a private key or symmetric system. The security of the system depends on a secret (in our case the cipher stream) k shared be- tween the cipherer and the encipherer. Knowledge of an enciphering method makes it easy to work out a deciphering method and vice versa. In our case a deciphering method is given by the observation that pn = zn + kn. (Indeed, writing α(p) = p + z, we see that the enciphering function α has the property that α2 = ι the identity map. Ciphers like this are called symmetric.) In the one-time pad, first discussed by Vernam in 1926, the cipher stream is a random sequence kj = Kj, where the Kj are independent random vari- ables with Pr(Kj = 0) = Pr(Kj = 1) = 1/2. If we write Zj = pj + Kj, then we see that the Zj are independent random variables with Pr(Zj = 0) = Pr(Zj = 1) = 1/2. Thus (in the absence of any knowledge of the ciphering stream) the code- breaker is just faced by a stream of perfectly random binary digits. Deci- pherment is impossible in principle. It is sometimes said that it is hard to find random sequences, and it is, indeed, rather harder than might appear at first sight, but it is not too difficult to rig up a system for producing ‘sufficiently random’ sequences34 . The secret services of the former Soviet Union were particularly fond of one- time pads. The real difficulty lies in the necessity for sharing the secret 34 Take ten of your favourite long books, convert to binary sequences xj,n and set kn = P10 j=1 xj,1000+j+n + sn where sn is the output of your favourite ‘pseudo-random number generator’ (in this connection see Exercise 27.16). Give a memory stick with a copy of k to your friend and, provided both of you obey some elementary rules, your correspondence will be safe from MI5. The anguished debate in the US about codes and privacy refers to the privacy of large organisations and their clients, not the privacy of communication from individual to individual. 61
- 62. sequence k. If a random sequence is reused it ceases to be random (it becomes ‘the same code as last Wednesday’ or the ‘the same code as Paris uses’) so, when there is a great deal of code traffic35 , new one-time pads must be sent out. If random bits can be safely communicated, so can ordinary messages and the exercise becomes pointless. In practice, we would like to start from a short shared secret ‘seed’ and generate a ciphering string k that ‘behaves like a random sequence’. This leads us straight into deep philosophical waters36 . As might be expected, there is an illuminating discussion in Chapter III of Knuth’s marvellous The Art of Computing Programming [7]. Note, in particular, his warning: . . . random numbers should not be generated with a method chosen at random. Some theory should be used. One way that we might try to generate our ciphering string is to use a gen- eral feedback shift register f of length d with the initial fill (k0, k1, . . . , kd−1) as the secret seed. Lemma 18.1. If f is a general feedback shift register of length d, then, given any initial fill (k0, k1, . . . , kd−1), there will exist N, M ≤ 2d such that the output stream k satisfies kr+N = kr for all r ≥ M. Exercise 18.2. Show that the decimal expansion of a rational number must be a recurrent expansion. Give a bound for the period in terms of the quo- tient. Conversely, by considering geometric series, or otherwise, show that a recurrent decimal represents a rational number. Lemma 18.3. Suppose that f is a linear feedback register of length d. (i) f(x0, x1, . . . , xd−1) = (x0, x1, . . . , xd−1) if (x0, x1, . . . , xd−1) = (0, 0, . . . , 0). (ii) Given any initial fill (k0, k1, . . . , kd−1), there will exist N, M ≤ 2d − 1 such that the output stream k satisfies kr+N = kr for all r ≥ M. We can complement Lemma 18.3 by using Lemma 14.16 and the associ- ated discussion. 35 In 1941, the Soviet Union’s need for one-time pads suddenly increased and it appears that pages were reused in different pads. If the reader reflects, she will see that, though this is a mistake, it is one which it is very difficult to exploit. However, under the pressure of the cold war, US code-breakers managed to decode messages which, although several years old, still provided useful information. After 1944, the Soviet Union’s one-time pads became genuinely one-time again and the coded messages became indecipherable. 36 Where we drown at once, since the best (at least, in my opinion) modern view is that any sequence that can be generated by a program of reasonable length from a ‘seed’ of reasonable size is automatically non-random. 62
- 63. Lemma 18.4. A linear feedback register of length d attains its maximal pe- riod 2d − 1 (for a non-trivial initial fill) when the roots of the auxiliary poly- nomial37 are primitive elements of F2d . (We will note why this result is plausible, but we will not prove it. See Exercise 27.19 for a proof.) It is well known that short period streams are dangerous. During World War II the British Navy used codes whose period was adequately long for peace time use. The massive increase in traffic required by war time con- ditions meant that the period was now too short. By dint of immense toil, German naval code breakers were able to identify coincidences and crack the British codes. Unfortunately, whilst short periods are definitely unsafe, it does not follow that long periods guarantee safety. Using the Berlekamp–Massey method we see that stream codes based on linear feedback registers are unsafe at level (2). Lemma 18.5. Suppose that an unknown cipher stream k0, k1, k2 . . . is pro- duced by an unknown linear feedback register f of unknown length d ≤ D. The plain text stream p0, p1, p2, . . . is enciphered as the cipher text stream z0, z1, z2, . . . given by zn = pn + kn. If we are given p0, p1, . . . p2D−1 and z0, z1, . . . z2D−1 then we can find kr for all r. Thus if we have a message of length twice the length of the linear feedback register together with its encipherment the code is broken. It is easy to construct immensely complicated looking linear feedback registers with hundreds of registers. Lemma 18.5 shows that, from the point of view of a determined, well equipped and technically competent opponent, cryptographic systems based on such registers are the equivalent of leaving your house key hidden under the door mat. Professionals say that such systems seek ‘security by obscurity’. However, if you do not wish to baffle the CIA, but merely prevent little old ladies in tennis shoes watching subscription television without paying for it, systems based on linear feedback registers are cheap and quite effective. Whatever they may say in public, large companies are happy to tolerate a certain level of fraud. So long as 99.9% of the calls made are paid for, the 37 In this sort of context we shall sometimes refer to the ‘auxiliary polynomial’ as the ‘feedback polynomial’. 63
- 64. profits of a telephone company are essentially unaffected by the .1% which ‘break the system’. What happens if we try some simple tricks to increase the complexity of the cipher text stream? Lemma 18.6. If xn is a stream produced by a linear feedback system of length N with auxiliary polynomial P and yn is a stream produced by a linear feedback system of length M with auxiliary polynomial Q, then xn + yn is a stream produced by a linear feedback system of length N + M with auxiliary polynomial P(X)Q(X). Note that this means that adding streams from two linear feedback system is no more economical than producing the same effect with one. Indeed the situation may be worse since a stream produced by linear feedback system of given length may, possibly, also be produced by another linear feedback system of shorter length. Lemma 18.7. Suppose that xn is a stream produced by a linear feedback system of length N with auxiliary polynomial P and yn is a stream produced by a linear feedback system of length M with auxiliary polynomial Q. Let P have roots α1, α2, . . . αN and Q have roots β1, β2, . . . βM over some field K ⊇ F2. Then xnyn is a stream produced by a linear feedback system of length NM with auxiliary polynomial Y 1≤i≤N Y 1≤i≤M (X − αiβj). We shall probably only prove Lemmas 18.6 and 18.7 in the case when all roots are distinct, leaving the more general case as an easy exercise. We shall also not prove that the polynomial Q 1≤i≤N Q 1≤i≤M (X −αiβj) obtained in Lemma 18.7 actually lies in F2[X] but (for those who are familiar with the phrase in quotes) this is an easy exercise in ‘symmetric functions of roots’. Here is an even easier remark. Lemma 18.8. Suppose that xn is a stream which is periodic with period N and yn is a stream which is periodic with period M. Then the streams xn +yn and xnyn are periodic with periods dividing the lowest common multiple of N and M. Exercise 18.9. One of the most confidential German codes (called FISH by the British) involved a complex mechanism which the British found could be simulated by two loops of paper tape of length 1501 and 1497. If kn = xn +yn where xn is a stream of period 1501 and yn is a stream of period 1497, what 64
- 65. is the longest possible period of kn? How many consecutive values of kn would you need to to find the underlying linear feedback register using the Berlekamp–Massey method if you did not have the information given in the question? If you had all the information given in the question how many values of kn would you need? (Hint, look at xn+1497 − xn.) You have shown that, given kn for sufficiently many consecutive n we can find kn for all n. Can you find xn for all n? It might be thought that the lengthening of the underlying linear feed- back system obtained in Lemma 18.7 is worth having, but it is bought at a substantial price. Let me illustrate this by an informal argument. Suppose we have 10 streams xj,n (without any peculiar properties) produced by linear feedback registers of length about 100. If we form kn = Q10 j=1 xj,n, then the Berlekamp–Massey method requires of the order of 1020 consecutive values of kn and the periodicity of kn can be made still more astronomical. Our cipher key stream kn appears safe from prying eyes. However it is doubtful if the prying eyes will mind. Observe that (under reasonable conditions) about 2−1 of the xj,n will have the value 1 and about 2−10 of the kn = Q10 j=1 xj,n will have value 1. Thus, if zn = pn +kn, in more than 999 cases out of a 1000 we will have zn = pn. Even if we just combine two streams xn and yn in the way suggested we may expect xnyn = 0 for about 75% of the time. Here is another example where the apparent complexity of the cipher key stream is substantially greater than its true complexity. Example 18.10. The following is a simplified version of a standard satel- lite TV decoder. We have 3 streams xn, yn, zn produced by linear feedback registers. If the cipher key stream is defined by kn =xn if zn = 0, kn =yn if zn = 1, then kn = (yn + xn)zn + xn and the cipher key stream is that produced by linear feedback register. We must not jump to the conclusion that the best way round these dif- ficulties is to use a non-linear feedback generator f. This is not the easy way out that it appears. If chosen by an amateur, the complicated looking f so produced will have the apparent advantage that we do not know what is wrong with it and the very real disadvantage that we do not know what is wrong with it. 65
- 66. Another approach is to observe that, so far as the potential code breaker is concerned, the cipher stream method only combines the ‘unknown secret’ (here the feedback generator f together with the seed (k0, k1, . . . , kd−1)) with the unknown message p in a rather simple way. It might be better to consider a system with two functions F : Fm 2 × Fn 2 → Fq 2 and G : Fm 2 × Fq 2 → Fn 2 such that G(k, F(k, p)) = p. Here k will be the shared secret, p the message, and z = F(k, p) the encoded message which can be decoded by using the fact that G(k, z) = p. In the next section we shall see that an even better arrangement is possi- ble. However, arrangements like this have the disadvantage that the message p must be entirely known before it is transmitted and the encoded message z must have been entirely received before it can be decoded. Stream ciphers have the advantage that they can be decoded ‘on the fly’. They are also much more error tolerant. A mistake in the coding, transmission or decoding of a single element only produces an error in a single place of the sequence. There will continue to be circumstances where stream ciphers are appropriate. There is one further remark to be made. Suppose, as is often the case, that we know F, that n = q and we know the ‘encoded message’ z. Suppose also that we know that the ‘unknown secret’ or ‘key’ k ∈ K ⊆ Fm 2 and the ‘unknown message’ p ∈ P ⊆ Fn 2 . We are then faced with the problem:- Solve the system z = F(k, p) where k ∈ K, p ∈ P. ⋆ Speaking roughly, the task is hopeless unless ⋆ has a unique solution38 Speaking even more roughly, this is unlikely to happen if |K||P| 2n and is likely to happen if 2n is substantially greater than |K||P|. (Here, as usual, |B| denotes the number of elements of B.) Now recall the definition of the information rate given in Definition 6.2. If the message set M has information rate µ and the key set (that is the shared secret set) K has information rate κ, then, taking logarithms, we see that, if n − mκ − nµ 38 ‘According to some, the primordial Torah was inscribed in black flames on white fire. At the moment of its creation, it appeared as a series of letters not yet joined up in the form of words. For this reason, in the Torah rolls there appear neither vowels nor punctuation, nor accents; for the original Torah was nothing but a disordered heap of letters. Furthermore, had it not been for Adam’s sin, these letters might have been joined differently to form another story. For the kabalist, God will abolish the present ordering of the letters, or else will teach us how to read them according to a new disposition only after the coming of the Messiah.’ ([1], Chapter 2.) A reader of this footnote has directed me to the International Torah Codes Society. 66
- 67. is substantially greater than 0, then ⋆ is likely to have a unique solution, but, if it is substantially smaller, this is unlikely. Example 18.11. Suppose that, instead of using binary code, we consider an alphabet of 27 letters (the English alphabet plus a space). We must take logarithms to the base 27, but the considerations above continue to apply. The English language treated in this way has information rate about .4. (This is very much a ball park figure. The information rate is certainly less than .5 and almost certainly greater than .2.) (i) In the Caesar code, we replace the ith element of our alphabet by the i + jth (modulo 27). The shared secret is a single letter (the code for A say). We have m = 1, κ = 1 and µ ≈ .4. Thus n − mκ − nµ ≈ .6n − 1. If n = 1 (so n − mκ − nµ ≈ −.4) it is obviously impossible to decode the message. If n = 10 (so n − mκ − nµ ≈ 5) a simple search through the 27 possibilities will almost always give a single possible decode. (ii) In a simple substitution code, a permutation of the alphabet is chosen and applied to each letter of the code in turn. The shared secret is a sequence of 26 letters (given the coding of the first 26 letters, the 27th can then be deduced). We have m = 26, κ = 1 and µ ≈ .4. Thus n − mκ − nµ ≈ .6n − 26. In The Dancing Men, Sherlock Holmes solves such a code with n = 68 (so n − mκ − nµ ≈ 15) without straining the reader’s credulity too much and I would think that, unless the message is very carefully chosen, most of my audience could solve such a code with n = 200 (so n − mκ − nµ ≈ 100). (iii) In the one-time pad m = n and κ = 1, so (if µ 0) n − mκ − nµ = −nµ → −∞ as n → ∞. (iv) Note that the larger µ is, the slower n − mκ − nµ increases. This corresponds to the very general statement that the higher the information rate of the messages, the harder it is to break the code in which they are sent. The ideas just introduced can be formalised by the notion of unicity distance. Definition 18.12. The unicity distance of a code is the number of bits of message required to exceed the number of bits of information in the key plus the number of bits of information in the message. 67
- 68. (The notion of information content brings us back to Shannon whose paper Communication theory of secrecy systems39 , published in 1949, forms the first modern treatment of cryptography in the open literature.) If we only use our code once to send a message which is substantially shorter than the unicity distance, we can be confident that no code breaker, however gifted, could break it, simply because there is no unambiguous de- code. (A one-time pad has unicity distance infinity.) However, the fact that there is a unique solution to a problem does not mean that it is easy to find. We have excellent reasons, some of which are spelled out in the next section, to believe that there exist codes for which the unicity distance is essentially irrelevant to the maximum safe length of a message. For these codes, even though there may be a unique solution, the amount of work required to find the solutions makes (it is hoped) any attempt impractical. 19 Asymmetric systems Towards the end of the previous section, we discussed a general coding scheme depending on a shared secret key k known to the encoder and the decoder. The scheme can be generalised still further by splitting the secret in two. Consider a system with two functions F : Fm 2 ×Fn 2 → Fq 2 and G : Fp 2×Fq 2 → Fn 2 such that G(l, F(k, p)) = p. Here (k, l) will be be a pair of secrets, p the message and z = F(k, p) the encoded message which can be decoded by using the fact that G(l, z) = p. In this scheme, the encoder must know k, but need not know l and the decoder must know l, but need not know k. Such a system is called asymmetric. So far the idea is interesting but not exciting. Suppose, however, that we can show that (i) knowing F, G and k it is very hard to find l (ii) if we do not know l then, even if we know F, G and k, it is very hard to find p from F(k, p). Then the code is secure at what we called level (3). Lemma 19.1. Suppose that the conditions specified above hold. Then an opponent who is entitled to demand the encodings zi of any messages pi they choose to specify will still find it very hard to find p when given F(k, p). Let us write F(k, p) = pKA and G(l, z) = zK−1 A and think of pKA as participant A’s encipherment of p and zK−1 A as participant B’s decipherment 39 Available on the web and in his Collected Papers. 68
- 69. of z. We then have (pKA )K−1 A = p. Lemma 19.1 tells us that such a system is secure however many messages are sent. Moreover, if we think of A as a spy-master, he can broadcast KA to the world (that is why such systems are called public key systems) and invite anybody who wants to spy for him to send him secret messages in total confidence40 . It is all very well to describe such a code, but do they exist? There is very strong evidence that they do, but, so far, all mathematicians have been able to do is to show that provided certain mathematical problems which are believed to be hard are indeed hard, then good codes exist. The following problem is believed to be hard. Problem Given an integer N, which is known to be the product N = pq of two primes p and q, find p and q. Several schemes have been proposed based on the assumption that this fac- torisation is hard. (Note, however, that it is easy to find large ‘random’ primes p and q.) We give a very elegant scheme due to Rabin and Williams. It makes use of some simple number theoretic results from IA and IB. The reader may well have seen the following results before. In any case, they are easy to obtain by considering primitive roots. Lemma 19.2. If p is an odd prime the congruence x2 ≡ d mod p is soluble if and only if d ≡ 0 or d(p−1)/2 ≡ 1 modulo p. Lemma 19.3. Suppose p is a prime such that p = 4k − 1 for some integer k. Then, if the congruence x2 ≡ d mod p has any solution, it has dk as a solution. We now call on the Chinese remainder theorem. Lemma 19.4. Let p and q be primes of the form 4k − 1 and set N = pq. Then the following two problems are of equivalent difficulty. (A) Given N and d find all the m satisfying m2 ≡ d mod N. (B) Given N find p and q. 40 Although we make statements about certain codes along the lines of ‘It does not matter who knows this’, you should remember the German naval saying ‘All radio traffic is high treason’. If any aspect of a code can be kept secret, it should be kept secret. 69
- 70. (Note that, provided that d 6≡ 0, knowing the solution to (A) for any d gives us the four solutions for the case d = 1.) The result is also true but much harder to prove for general primes p and q. At the risk of giving aid and comfort to followers of the Lakatosian heresy, it must be admitted that the statement of Lemma 19.4 does not really tell us what the result we are proving is, although the proof makes it clear that the result (whatever it may be) is certainly true. However, with more work, everything can be made precise. We can now give the Rabin–Williams scheme. The spy-master A selects two very large primes p and q. (Since he has only done an undergraduate course in mathematics, he will take p and q of the form 4k − 1.) He keeps the pair (p, q) secret, but broadcasts the public key N = pq. If B wants to send him a message, she writes it in binary code and splits it into blocks of length m with 2m N 2m+1 . Each of these blocks is a number rj with 0 ≤ rj N. B computes sj such that r2 j ≡ sj modulo N and sends sj. The spy-master (who knows p and q) can use the method of Lemma 19.4 to find one of four possible values for rj (the four square roots of sj). Of these four possible message blocks it is almost certain that three will be garbage, so the fourth will be the desired message. If the reader reflects, she will see that the ambiguity of the root is gen- uinely unproblematic. (If the decoding is mechanical then fixing 50 bits scattered throughout each block will reduce the risk of ambiguity to negli- gible proportions.) Slightly more problematic, from the practical point of view, is the possibility that someone could be known to have sent a very short message, that is to have started with an m such that 1 ≤ m ≤ N1/2 but, provided sensible precautions are taken, this should not occur. If I Google ‘Casino’, then I am instantly put in touch with several of the world’s ‘most trusted electronic casinos’ who subscribe to ‘responsible gambling’ and who have their absolute probity established by ‘internation- ally recognised Accredited Test Facilities’. Given these assurances, it seems churlish to introduce Alice and Bob who live in different cities, can only communicate by e-mail and are so suspicious of each other that neither will accept the word of the other as to the outcome of the toss of a coin. If, in spite of this difficulty, Alice and Bob wish to play heads and tails (the technical expression is ‘bit exchange’ or ‘bit sharing’), then the ambiguity of the Rabin–Williams scheme becomes an advantage. Let us set out the steps of a ‘bit sharing scheme’ based on Rabin–Williams. STEP 1 Alice chooses at random two large primes p and q such that p ≡ q ≡ 3 mod 4. She computes n = pq and sends n to Bob. STEP 2 Bob chooses a random integer r with 1 r n/2. (He wishes to hide r from Alice, so he may take whatever other precautions he wishes in 70
- 71. choosing r.) He computes m ≡ r2 mod n and sends m to Alice. STEP 3 Since Alice knows p and q she can easily compute the 4 square roots of m modulo n. Exactly two of the roots r1 and r2 will satisfy 1 r n/2. (If s is a root, so is −s.) However, Alice has no means of telling which is r. Alice writes out r1 and r2 in binary and chooses a place (the kth digit say) where they differ. She then tells Bob ‘I choose the value u for the kth bit’. STEP 4 Bob tells Alice the value of r. If the value of the kth bit of r is u, then Alice wins. If not, Bob wins. Alice checks that r2 ≡ m mod n. Since, r1r−! 2 is a square root of unity which is neither 1 nor −1, knowing r1 and r2 is equivalent to factoring n, she knows that Bob could not lie about the value of r. Thus Alice is happy. STEP 5 Alice tells Bob the values of p and q. He checks that p and q are primes (see Exercise 27.12 for why he does this) and finds r1 and r2. After Bob has verified that r1 and r2 do indeed differ in the kth bit, he also is happy, since there is no way Alice could know from inspection of m which root he started with. 20 Commutative public key systems In the previous sections we introduced the coding and decoding functions KA and K−1 A with the property that (pKA )K−1 A = p, and satisfying the condition that knowledge of KA did not help very much in finding K−1 A . We usually require, in addition, that our system be commutative in the sense that (pK−1 A )KA = p. and that knowledge of K−1 A does not help very much in finding KA. The Rabin–Williams scheme, as described in the last section, does not have this property. Commutative public key codes are very flexible and provide us with simple means for maintaining integrity, authenticity and non-repudiation. (This is not to say that non-commutative codes can not do the same; simply that commutativity makes many things easier.) Integrity and non-repudiation Let A ‘own a code’, that is know both KA and K−1 A . Then A can broadcast K−1 A to everybody so that everybody can decode but only A can encode. (We say that K−1 A is the public key and KA the private key.) Then, for example, A could issue tickets to the castle ball 71
- 72. carrying the coded message ‘admit Joe Bloggs’ which could be read by the recipients and the guards but would be unforgeable. However, for the same reason, A could not deny that he had issued the invitation. Authenticity If B wants to be sure that A is sending a message then B can send A a harmless random message q. If B receives back a message p such that pK−1 A ends with the message q then A must have sent it to B. (Anybody can copy a coded message but only A can control the content.) Signature Suppose now that B also owns a commutative code pair (KB, K−1 B ) and has broadcast K−1 B . If A wants to send a message p to B he computes q = pKA and sends pK−1 B followed by qK−1 B . B can now use the fact that (qK−1 B )KB = q to recover p and q. B then observes that qK−1 A = p. Since only A can produce a pair (p, q) with this property, A must have written it. There is now a charming little branch of the mathematical literature based on these ideas in which Albert gets Bertha to authenticate a message from Caroline to David using information from Eveline, Fitzpatrick, Gilbert and Harriet whilst Ingrid, Jacob, Katherine and Laszlo play bridge without using a pack of cards. However, a cryptographic system is only as strong as its weakest link. Unbreakable password systems do not prevent computer systems being regularly penetrated by ‘hackers’ and however ‘secure’ a trans- action on the net may be it can still involve a rogue at one end and a fool at the other. The most famous candidate for a commutative public key system is the RSA (Rivest, Shamir, Adleman) system. It was the RSA system41 that first convinced the mathematical community that public key systems might be feasible. The reader will have met the RSA in IA, but we will push the ideas a little bit further. Lemma 20.1. Let p and q be primes. If N = pq and λ(N) = lcm(p−1, q−1), then Mλ(N) ≡ 1 (mod N) for all integers M coprime to N. 41 A truly patriotic lecturer would refer to the ECW system, since Ellis, Cocks and Williamson discovered the system earlier. However, they worked for GCHQ and their work was kept secret. 72
- 73. Since we wish to appeal to Lemma 19.4, we shall assume in what follows that we have secretly chosen large primes p and q. We choose an integer e and then use Euclid’s algorithm to check that e and λ(N) are coprime and to find an integer d such that de ≡ 1 (mod λ(N)). If Euclid’s algorithm reveals that e and λ(N) are not coprime, we try another e. Since others may be better psychologists than we are, we would be wise to use some sort of random method for choosing p, q and e. The public key includes the value of e and N, but we keep secret the value of d. Given a number M with 1 ≤ M ≤ N − 1, we encode it as the integer E with 1 ≤ E ≤ N − 1 E ≡ Md (mod N). The public decoding method is given by the observation that Ee ≡ Mde ≡ M for M coprime to N. (The probability that M is not coprime to N is so small that it can be neglected.) As was observed in IA, high powers are easy to compute. Exercise 20.2. Show how M2n can be computed using n multiplications. If 1 ≤ r ≤ 2n show how Mr can be computed using at most 2n multiplications. To show that (providing that factoring N is indeed hard) finding d from e and N is hard we use the following lemma. Lemma 20.3. Suppose that d, e and N are as above. Set de−1 = 2a b where b is odd. (i) a ≥ 1. (ii) If y ≡ xb (mod N) and y 6≡ 1 then there exists an r with 0 ≤ r ≤ a−1 such that z = y2r 6≡ 1 but z2 ≡ 1 (mod N). Combined with Lemma 19.4, the idea of Lemma 20.3 gives a fast prob- abilistic algorithm where, by making random choices of x, we very rapidly reduce the probability that we can not find p and q to as close to zero as we wish. Lemma 20.4. The problem of finding d from the public information e and N is essentially as hard as factorising N. 73
- 74. Remark 1 At first glance, we seem to have done as well for the RSA code as for the Rabin–Williams code. But this is not so. In Lemma 19.4 we showed that finding the four solutions of M2 ≡ E (mod N) was equivalent to factorising N. In the absence of further information, finding one root is as hard as finding another. Thus the ability to break the Rabin-Williams code (without some tremendous stroke of luck) is equivalent to the ability to factor N. On the other hand it is, a priori, possible that someone may find a decoding method for the RSA code which does not involve knowing d. They would have broken the RSA code without finding d. It must, however, be said that, in spite of this problem, the RSA code is much used in practice and the Rabin–Williams code is not. Remark 2 It is natural to ask what evidence there is that the factorisation problem really is hard. Properly organised, trial division requires O(N1/2 ) op- erations to factorise a number N. This order of magnitude was not bettered until 1972 when Lehman produced a O(N1/3 ) method. In 1974, Pollard42 produced a O(N1/4 ) method. In 1979, as interest in the problem grew be- cause of its connection with secret codes, Lenstra made a breakthrough to a O(ec((log N)(log log N))1/2 ) method with c ≈ 2. Since then some progress has been made (Pollard reached O(e2((log N)(log log N))1/3 ) but, in spite of intense efforts, mathematicians have not produced anything which would be a real threat to codes based on the factorisation problem. A series of challenge numbers is hosted on the Wikipedia article entitled RSA. In 1996, it was possible to factor 100 (decimal) digit numbers routinely, 150 digit numbers with immense effort but 200 digit numbers were out of reach. In May 2005, the 200 digit challenge number was factored by F. Bahr, M. Boehm, J. Franke 42 Although mathematically trained, Pollard worked outside the professional mathemat- ical community. 74
- 75. and T. Kleinjunge as follows 27997833911221327870829467638722601621 07044678695542853756000992932612840010 76093456710529553608560618223519109513 65788637105954482006576775098580557613 57909873495014417886317894629518723786 9221823983 = 35324619344027701212726049781984643 686711974001976250236493034687761212536 79423200058547956528088349 × 7925869954478333033347085841480059687 737975857364219960734330341455767872818 152135381409304740185467 but the 210 digit challenge 24524664490027821197651766357308801846 70267876783327597434144517150616008300 38587216952208399332071549103626827191 67986407977672324300560059203563124656 12184658179041001318592996199338170121 49335034875870551067 remains (as of mid-2008) unfactored. Organisations which use the RSA and related systems rely on ‘security through publicity’. Because the problem of cracking RSA codes is so notorious, any breakthrough is likely to be publicly announced43 . Moreover, even if a breakthrough occurs, it is unlikely to be one which can be easily exploited by the average criminal. So long as the secrets covered by RSA-type codes need only be kept for a few months rather than forever44 , the codes can be considered to be one of the strongest links in the security chain. 43 And if not, is most likely to be a government rather than a Mafia secret. 44 If a sufficiently robust ‘quantum computer’ could be built, then it could solve the factorisation problem and the discrete logarithm problem (mentioned later) with high probability extremely fast. It is highly unlikely that such a machine would be or could be kept secret, since it would have many more important applications than code breaking. 75
- 76. 21 Trapdoors and signatures It might be thought that secure codes are all that are needed to ensure the security of communications, but this is not so. It is not necessary to read a message to derive information from it45 . In the same way, it may not be necessary to be able to write a message in order to tamper with it. Here is a somewhat far fetched but worrying example. Suppose that, by wire tapping or by looking over people’s shoulders, I discover that a bank creates messages in the form M1, M2 where M1 is the name of the client and M2 is the sum to be transferred to the client’s account. The messages are then encoded according to the RSA scheme discussed after Lemma 20.1 as Z1 = Md 1 and Z2 = Md 2 . I then enter into a transaction with the bank which adds $ 1000 to my account. I observe the resulting Z1 and Z2 and then transmit Z1 followed by Z3 2 . Example 21.1. What will (I hope) be the result of this transaction? We say that the RSA scheme is vulnerable to ‘homomorphism attack’ that is to say an attack which makes use of the fact our code is a homomorphism. (If θ(M) = Md , then θ(M1M2) = θ(M1)θ(M2).) One way of increasing security against tampering is to first code our message by a classical coding method and then use our RSA (or similar) scheme on the result. Exercise 21.2. Discuss briefly the effect of first using an RSA scheme and then a classical code. However there is another way forward which has the advantage of wider applicability since it also can be used to protect the integrity of open (non- coded) messages and to produce password systems. These are the so-called signature systems. (Note that we shall be concerned with the ‘signature of the message’ and not the signature of the sender.) Definition 21.3. A signature or trapdoor or hashing function is a mapping H : M → S from the space M of possible messages to the space S of possible signatures. (Let me admit, at once, that Definition 21.3 is more of a statement of notation than a useful definition.) The first requirement of a good signature function is that the space M should be much larger than the space S so that H is a many-to-one function (in fact a great-many-to-one function) and we can not 45 During World War II, British bomber crews used to spend the morning before a night raid testing their equipment, this included the radios. 76
- 77. work back from H(M) to M. The second requirement is that S should be large so that a forger can not (sensibly) hope to hit on H(M) by luck. Obviously we should aim at the same kind of security as that offered by our ‘level 2’ for codes:- Prospective opponents should find it hard to find H(M) given M even if they are in possession of a plentiful supply of message– signature pairs (Mi, H(Mi)) of messages Mi together with their encodings Ci. I leave it to the reader to think about level 3 security (or to look at section 12.6 of [10]). Here is a signature scheme due to Elgamal46 . The message sender A chooses a very large prime p, some integer 1 g p and some other integer u with 1 u p (as usual, some randomisation scheme should be used). A then releases the values of p, g and y = gu (modulo p) but keeps the value of u secret. Whenever he sends a message m (some positive integer), he chooses another integer k with 1 ≤ k ≤ p − 2 at random and computes r and s with 1 ≤ r ≤ p − 1 and 0 ≤ s ≤ p − 2 by the rules47 r ≡ gk (mod p), (*) m ≡ ur + ks (mod p − 1). (**) Lemma 21.4. If conditions (*) and (**) are satisfied, then gm ≡ yr rs (mod p). If A sends the message m followed by the signature (r, s), the recipient need only verify the relation gm ≡ yr rs (mod p) to check that the message is authentic48 . Since k is random, it is believed that the only way to forge signatures is to find u from gu (or k from gk ) and it is believed that this problem, which is known as the discrete logarithm problem, is very hard. Needless to say, even if it is impossible to tamper with a message–signature pair it is always possible to copy one. Every message should thus contain a unique identifier such as a time stamp. 46 This is Dr Elgamal’s own choice of spelling according to Wikipedia. 47 There is a small point which I have glossed over here and elsewhere. Unless k and and p − 1 are coprime the equation (**) may not be soluble. However the quickest way to solve (**), if it is soluble, is Euclid’s algorithm which will also reveal if (**) is insoluble. If (**) is insoluble, we simply choose another k at random and try again. 48 Sometimes, m is replaced by some hash function H(m) of m so (∗∗) becomes H(m) ≡ ur + ks (mod p − 1). In this case the recipient checks that gH(m) ≡ yr rs (mod p). 77
- 78. The evidence that the discrete logarithm problem is very hard is of the same kind of nature and strength as the evidence that the factorisation prob- lem is very hard. We conclude our discussion with a description of the Diffie– Hellman key exchange system which is also based on the discrete logarithm problem. The modern coding schemes which we have discussed have the disadvan- tage that they require lots of computation. This is not a disadvantage when we deal slowly with a few important messages. For the Web, where we must deal speedily with a lot of less than world shattering messages sent by im- patient individuals, this is a grave disadvantage. Classical coding schemes are fast but become insecure with reuse. Key exchange schemes use modern codes to communicate a new secret key for each message. Once the secret key has been sent slowly, a fast classical method based on the secret key is used to encode and decode the message. Since a different secret key is used each time, the classical code is secure. How is this done? Suppose A and B are at opposite ends of a tapped telephone line. A sends B a (randomly chosen) large prime p and a randomly chosen g with 1 g p − 1. Since the telephone line is insecure, A and B must assume that p and g are public knowledge. A now chooses randomly a secret number α and tells B the value of gα (modulo p). B chooses randomly a secret number β and tells A the value of gβ (modulo p). Since gαβ ≡ (gα )β ≡ (gβ )α , both A and B can compute k = gαβ modulo p and k becomes the shared secret key. The eavesdropper is left with the problem of finding k ≡ gαβ from knowl- edge of g, gα and gβ (modulo p). It is conjectured that this is essentially as hard as finding α and β from the values of g, gα and gβ (modulo p) and this is the discrete logarithm problem. 22 Quantum cryptography In the days when messages were sent in the form of letters, suspicious people might examine the creases where the paper was folded for evidence that the letter had been read by others. Our final cryptographic system has the advantage that it too will reveal attempts to read it. It also has the advantage that, instead of relying on the unproven belief that a certain mathematical task is hard, it depends on the fact that a certain physical task is impossible49 . 49 If you believe our present theories of the universe. 78
- 79. We shall deal with a highly idealised system. The business of dealing with realistic systems is a topic of active research within the faculty. The system we sketch is called the BB84 system (since it was invented by Bennett and Brassard in 1984) but there is another system invented by Ekert. Quantum mechanics tells us that a polarised photon has a state φ = α| li + β| ↔i where α, β ∈ R, α2 + β2 = 1, li is the vertically polarised state and ↔i is the horizontally polarised state. Such a photon will pass through a vertical polarising filter with probability α2 and its state will then be li. It will pass through a horizontal polarising filter with probability β2 and its state will then be ↔i. We have an orthonormal basis consisting of li and ↔i by +. We now consider a second basis given by i = 1 √ 2 | li + 1 √ 2 | ↔i and i = 1 √ 2 | li − 1 √ 2 | ↔i in which the states correspond to polarisation at angles π/4 and −π/4 to the horizontal. Observe that a photon in either state will have a probability 1/2 of passing through either a vertical or a horizontal filter and will then be in the appropriate state. Suppose Eve50 intercepts a photon passing between Alice and Bob. If Eve knows that it is either horizontally or vertically polarised, then she can use a vertical filter. If the photon passes through, she knows that it was vertically polarised when Alice sent it and can pass on a vertically polarised photon to Bob. If the photon does not pass through through, she knows that the photon was horizontally polarised and can pass on a horizontally polarised photon to Bob. However, if Alice’s photon was actually diagonally polarised (at angle ±π/4), this procedure will result in Eve sending Bob a photon which is horizontally or vertically polarised. It is possible that the finder of a fast factorising method would get a Field’s medal. It is certain that anyone who can do better than Eve would get the Nobel prize for physics since they would have overturned the basis of Quantum Mechanics. Let us see how this can (in principle) be used to produce a key exchange scheme (so that Alice and Bob can agree on a random number to act as the basis for a classical code). STEP 1 Alice produces a secret random sequence a1a2 . . . of bits (zeros and ones) and Bob produces another secret random sequence b1b2 . . . of bits. STEP 2 Alice produces another secret random sequence c1c2 . . . . She trans- mits it to Bob as follows. 50 This is a traditional pun. 79
- 80. If aj = 0 and cj = 0, she uses a vertically polarised photon. If aj = 0 and cj = 1, she uses a horizontally polarised photon. If aj = 1 and cj = 0, she uses a ‘left diagonally’ polarised photon. If aj = 1 and cj = 1, she uses a ‘right diagonally’ polarised photon. STEP 3 If bj = 0, Bob uses a vertical polariser to examine the jth photon. If he records a vertical polarisation, he sets dj = 0, if a horizontal he sets dj = 1. If bj = 1, Bob uses a π/4 diagonal polariser to examine the jth photon. If he records a left diagonal polarisation, he sets dj = 0, if a right he sets dj = 1. STEP 4 Bob and Alice use another communication channel to tell each other the values of the aj and bj. Of course, they should try to keep these com- munication secret, but we shall assume that worst has happened and these values become known to Eve. STEP 5 If the sequences are long, we can be pretty sure, by the law of large numbers, that aj = bj in about half the cases. (If not, Bob and Alice can agree to start again.) In particular, we can ensure that, with probability of at least 1 − ǫ/4 (where ǫ is chosen in advance), the number of agreements is sufficiently large for the purposes set out below. Alice and Bob only look at the ‘good cases’ when aj = bj. In such cases, if Eve does not examine the associated photon, then dj = cj. If Eve does examine the associated photon, then with probability 1/4, dj 6= cj. To see this, we examine the case when cj = 0 and Eve uses a diagonal polariser. (The other cases may be treated in exactly the same way.) With probability 1/2, aj = 1 so the photon is diagonally polarised, Eve records the correct polarisation and sends Bob a correctly polarised photon. Thus dj = cj. With probability 1/2, aj = 0 so the photon is vertically or horizontally polarised. Since Eve records a diagonal polarisation she will send a diagonally polarised photon to Bob and, since Bob’s polariser is vertical, he will record a vertical polarisation with probability 1/2. STEP 6 Alice uses another communication channel to tell Bob the value of a randomly chosen sample of good cases. Standard statistical techniques tell Alice and Bob that, if the number of discrepancies is below a certain level, the probability that Eve is intercepting more than a previously chosen proportion p of photons is less than ǫ/4. If the number of discrepancies is greater than the chosen level, Alice and Bob will abandon the attempt to communicate. STEP 7 If Eve is intercepting less than a proportion p of photons and q p (with q chosen in advance) the probability that she will have intercepted more than a proportion q of the remaining ‘good’ photons is less than ǫ/4. Although we shall not do this, the reader who has ploughed through these 80
- 81. notes will readily accept that Bob and Alice can use the message conveyed through the remaining good photons to construct a common secret such that Eve has probability less than ǫ/4 of guessing it. Thus, unless they decide that their messages are being partially read, Alice and Bob can agree a shared secret with probability less than ǫ that an eavesdropper can guess it. There are various gaps in the exposition above. First we have assumed that Eve must hold her polariser at a small fixed number of angles. A little thought shows that allowing her a free choice of angle will make little dif- ference. Secondly, since physical systems always have imperfections, some ‘good’ photons will produce errors even in the absence of Eve. This means that p in STEP 5 must be chosen above the ‘natural noise level’ and the sequences must be longer but, again, this ought to make little difference. There is a further engineering problem that it is very difficult just to send single photons every time. If there are too many groups of photons, then Eve only need capture one and let the rest go, so we can not detect eavesdrop- ping. If there are only a few, then the values of p and q can be adjusted to take account of this. There are several networks in existence which employ quantum cryptography. Quantum cryptography has definite advantages when matched individu- ally against RSA, secret sharing (using a large number of independent chan- nels) or one-time pads. It is less easy to find applications where it is better than the best choice of one of these three ‘classical’ methods51 . Of course, quantum cryptography will appeal to those who need to per- suade others that they are using the latest and most expensive technology to guard their secrets. However as I said before coding schemes are at best, cryptographic elements of larger possible cryptographic systems. If smiling white coated technicians install big gleaming machines with ‘Unbreakable Quantum Code Company’ painted in large letters above the keyboard in the homes of Alice and Bob, it does not automatically follow that their commu- nications are safe. Money will buy the appearance of security. Only thought will buy the appropriate security for a given purpose at an appropriate cost. And even then we can not be sure. As we know, There are known knowns. There are things we know we know. We also know 51 One problem is indicated by the first British military action in World War I which was to cut the undersea telegraph cables linking Germany to the outside world. Complex systems are easier to disrupt than simple ones. 81
- 82. There are known unknowns. That is to say We know there are some things We do not know. But there are also unknown unknowns, The ones we don’t know We don’t know52 . 23 Further reading For many students this will be one of the last university mathematics course they will take. Although the twin subjects of error-correcting codes and cryptography occupy a small place in the grand panorama of modern math- ematics, it seems to me that they form a very suitable topic for such a final course. Outsiders often think of mathematicians as guardians of abstruse but set- tled knowledge. Even those who understand that there are still problems un- settled, ask what mathematicians will do when they run out of problems. At a more subtle level, Kline’s magnificent Mathematical Thought from Ancient to Modern Times [5] is pervaded by the melancholy thought that, though the problems will not run out, they may become more and more baroque and inbred. ‘You are not the mathematicians your parents were’ whispers Kline ‘and your problems are not the problems your parents’ were.’ However, when we look at this course, we see that the idea of error- correcting codes did not exist before 1940. The best designs of such codes depend on the kind of ‘abstract algebra’ that historians like Kline and Bell consider a dead end, and lie behind the superior performance of CD players and similar artifacts. In order to go further into the study of codes, whether secret or error correcting, we need to go into the question of how the information content of a message is to be measured. ‘Information theory’ has its roots in the code breaking of World War II (though technological needs would doubtless have led to the same ideas shortly thereafter anyway). Its development required a level of sophistication in treating probability which was simply not available in the 19th century. (Even the Markov chain is essentially 20th century53 .) The question of what makes a calculation difficult could not even have 52 Rumsfeld 53 We are now in the 21st century, but I suspect that we are still part of the mathematical ‘long 20th century’ which started in the 1880s with the work of Cantor and like minded contemporaries. 82
- 83. been thought about until Gödel’s theorem (itself a product of the great ‘foun- dations crisis’ at the beginning of the 20th century). Developments by Turing and Church of Gödel’s theorem gave us a theory of computational complex- ity which is still under development today. The question of whether there exist ‘provably hard’ public codes is intertwined with still unanswered ques- tions in complexity theory. There are links with the profound (and very 20th century) question of what constitutes a random number. Finally, the invention of the electronic computer has produced a cultural change in the attitude of mathematicians towards algorithms. Before 1950, the construction of algorithms was a minor interest of a few mathematicians. (Gauss and Jacobi were considered unusual in the amount of thought they gave to actual computation.) Today, we would consider a mathematician as much as a maker of algorithms as a prover of theorems. The notion of the probabilistic algorithm which hovered over much of our discussion of secret codes is a typical invention of the last decades of the 20th century. Although both the subjects of error correcting and secret codes are now ‘mature’ in the sense that they provide usable and well tested tools for prac- tical application, they still contain deep unanswered questions. For example How close to the Shannon bound can a ‘computationally easy’ error cor- recting code get? Do provably hard public codes exist? Even if these questions are too hard, there must surely exist error cor- recting and public codes based on new ideas54 . Such ideas would be most welcome and, although they are most likely to come from the professionals, they might come from outside the usual charmed circles. Those who wish to learn about error correction from the horse’s mouth will consult Hamming’s own book on the matter [2]. For the present course, the best book I know for further reading is Welsh [10]. After this, the book of Goldie and Pinch [8] provides a deeper idea of the meaning of information and its connection with the topic. The book by Koblitz [6] develops the number theoretic background. The economic and practical importance of transmitting, storing and processing data far outweighs the importance of hiding it. However, hiding data is more romantic. For budding cryptologists and cryptographers (as well as those who want a good read), Kahn’s The Codebreakers [3] has the same role as is taken by Bell’s Men of Mathematics for budding mathematicians. I conclude with a quotation from Galbraith (referring to his time as am- bassador to India) taken from Koblitz’s entertaining text [6]. I had asked that a cable from Washington to New Delhi . . . be 54 Just as quantum cryptography was. 83
- 84. reported to me through the Toronto consulate. It arrived in code; no facilities existed for decoding. They brought it to me at the airport — a mass of numbers. I asked if they assumed I could read it. They said no. I asked how they managed. They said that when something arrived in code, they phoned Washington and had the original read to them. References [1] U. Eco The Search for the Perfect Language (English translation), Black- well, Oxford 1995. [2] R. W. Hamming Coding and Information Theory (2nd edition) Prentice Hall, 1986. [3] D. Kahn The Codebreakers: The Story of Secret Writing MacMillan, New York, 1967. (A lightly revised edition has recently appeared.) [4] D. Kahn Seizing the Enigma Houghton Mifflin, Boston, 1991. [5] M. Kline Mathematical Thought from Ancient to Modern Times OUP, 1972. [6] N. Koblitz A Course in Number Theory and Cryptography Springer, 1987. [7] D. E. Knuth The Art of Computing Programming Addison-Wesley. The third edition of Volumes I to III is appearing during this year and the next (1998–9). [8] G. M. Goldie and R. G. E. Pinch Communication Theory CUP, 1991. [9] T. M. Thompson From Error-correcting Codes through Sphere Packings to Simple Groups Carus Mathematical Monographs 21, MAA, Wash- ington DC, 1983. [10] D. Welsh Codes and Cryptography OUP, 1988. 84
- 85. There is a widespread superstition, believed both by supervisors and su- pervisees, that exactly twelve questions are required to provide full under- standing of six hours of mathematics and that the same twelve questions should be appropriate for students of all abilities and all levels of diligence. I have tried to keep this in mind, but have provided some extra questions in the various exercise sheets for those who scorn such old wives’ tales. 24 Exercise Sheet 1 Q 24.1. (Exercises 1.1 and 1.2.) (i) Consider Morse code. A 7→ • − ∗ B 7→ − • • • ∗ C 7→ − • − • ∗ D 7→ − • •∗ E 7→ •∗ F 7→ • • − • ∗ O 7→ − − −∗ S 7→ • • •∗ 7 7→ − − • • •∗ Decode − • − • ∗ − − − ∗ − • • ∗ • ∗. (ii) Consider ASCII code. A 7→ 1000001 B 7→ 1000010 C 7→ 1000011 a 7→ 1100001 b 7→ 1100010 c 7→ 1100011 + 7→ 0101011 ! 7→ 0100001 7 7→ 0110111 Encode b7!. Decode 110001111000011100010. Q 24.2. (Exercises 1.3, 1.4 and 1.7.) Consider two alphabets A and B and a coding function c : A → B∗ (i) Explain, without using the notion of prefix-free codes, why, if c is injective and fixed length, c is decodable. Explain why, if c is injective and fixed length, c is prefix-free. (ii) Let A = B = {0, 1}. If c(0) = 0, c(1) = 00 show that c is injective but c∗ is not. (iii) Let A = {1, 2, 3, 4, 5, 6} and B = {0, 1}. Show that there is a variable length coding c such that c is injective and all code words have length 2 or less. Show that there is no decodable coding c such that all code words have length 2 or less Q 24.3. The product of two codes cj : Aj → B∗ j is the code g : A1 × A2 → (B1 ∪ B2)∗ given by g(a1, a2) = c1(a1)c2(a2). Show that the product of two prefix-free codes is prefix free, but the prod- uct of a decodable code and a prefix-free code need not even be decodable. 85
- 86. Q 24.4. (Exercises 2.5 and 2.7) (i) Apply Huffman’s algorithm to the nine messages Mj where Mj has probability j/45 for 1 ≤ j ≤ 9. (ii) Consider 4 messages with the following properties. M1 has probability .23, M2 has probability .24, M3 has probability .26 and M4 has probability .27. Show that any assignment of the code words 00, 01, 10 and 11 produces a best code in the sense of this course. Q 24.5. (Exercises 2.6 and 4.6.) (i) Consider 64 messages Mj. M1 has probability 1/2, M2 has probability 1/4 and Mj has probability 1/248 for 3 ≤ j ≤ 64. Explain why, if we use code words of equal length, then the length of a code word must be at least 6. By using the ideas of Huffman’s algorithm (you should not need to go through all the steps) obtain a set of code words such that the expected length of a code word sent is no more than 3. (ii) Let a, b 0. Show that loga b = log b log a . Q 24.6. (Exercise 4.10) (i) Let A = {1, 2, 3, 4}. Suppose that the probability that letter k is chosen is k/10. Use your calculator to find ⌈− log2 pk⌉ and write down a Shannon–Fano code c. (ii) We found a Huffman code ch for the system in Example 2.4. Show that the entropy is approximately 1.85, that E|c(A)| = 2.4 and that E|ch(A)| = 1.9. Check that these results are consistent with the appropriate theorems of the course. Q 24.7. (Exercise 5.1) Suppose that we have a sequence Xj of random vari- ables taking the values 0 and 1. Suppose that X1 = 1 with probability 1/2 and Xj+1 = Xj with probability .99 independent of what has gone before. (i) Suppose we wish to send 10 successive bits XjXj+1 . . . Xj+9. Show that if we associate the sequence of ten zeros with 0, the sequence of ten ones with 10 and any other sequence a0a1 . . . a9 with 11a0a1 . . . a9, we have a decodable code which on average requires about 5/2 bits to transmit the sequence. (ii) Suppose we wish to send the bits XjXj+106 Xj+2×106 . . . Xj+9×106 . Ex- plain why any decodable code will require on average at least 10 bits to transmit the sequence. (You need not do detailed computations.) Q 24.8. In Bridge, a 52 card pack is dealt to provide 4 hands of 13 cards each. 86
- 87. (i) Purely as a matter of interest, we consider the following question. If the contents of a hand are conveyed by one player to their partner by a series of nods and shakes of the head how many movements of the head are required? Show that at least 40 movements are required. Give a simple code requiring 52 movements. [You may assume for simplicity that the player to whom the information is being communicated does not look at her own cards. (In fact this does not make a difference since the two players do not acquire any shared information by looking at their own cards.)] (ii) If instead the player uses the initial letters of words (say using the 16 most common letters), how many words will you need to utter55 ? Q 24.9. (i) In a comma code, like Morse code, one symbol from an alphabet of m letters is reserved to end each code word. Show that this code is prefix-free and give a direct argument to show that it must satisfy Kraft’s inequality. (ii) Give an example of a code satisfying Kraft’s inequality which is not decodable. Q 24.10. Show that if an optimal binary code has word lengths s1, s2, . . . sm then m log2 m ≤ s1 + s2 + · · · + sm ≤ (m2 + m − 2)/2. Q 24.11. (i) It is known that exactly one member of the starship Emphasise has contracted the Macguffin virus. A test is available that will detect the virus at any dilution. However, the power required is such that the ship’s force shields must be switched off56 for a minute during each test. Blood samples are taken from all crew members. The ship’s computer has worked out that the probability of crew member number i harbouring the virus is pi. (Thus the probability that the captain, who is, of course, number 1, has the disease is p1.) Explain how, by testing pooled samples, the expected number of tests can be minimised. Write down the exact form of the test when there are 2n crew members and pi = 2−n . (ii) Questions like (i) are rather artificial, since they require that exactly one person carries the virus. Suppose that the probability that any member of a population of 2n has a certain disease is p (and that the probability 55 ‘Marked cards, M. l’Anglais?’ I said, with a chilling sneer. ’They are used, I am told, to trap players–not unbirched schoolboys.’ ’Yet I say that they are marked!’ he replied hotly, in his queer foreign jargon. ’In my last hand I had nothing. You doubled the stakes. Bah, sir, you knew! You have swindled me!’ ’Monsieur is easy to swindle – when he plays with a mirror behind him,’ I answered tartly. Under the Red Robe S. J. Weyman 56 ‘Captain, ye canna be serious.’ 87
- 88. is independent of the health of the others) and there exists an error free test which can be carried out on pooled blood samples which indicates the presence of the disease in at least one of the samples or its absence from all. Explain why there cannot be a testing scheme which can be guaranteed to require less than 2n tests to diagnose all members of the population. How does the scheme suggested in the last sentence of (i) need to be modified to take account of the fact that more than one person may be ill (or, indeed, no one may be ill)? Show that the expected number of tests required by the modified scheme is no greater than pn2n+1 + 1. Explain why the cost of testing a large population of size x is no more than about 2pcx log2 x with c the cost of a test. (iii) In practice, pooling schemes will be less complicated. Usually a group of x people are tested jointly and, if the joint test shows the disease, each is tested individually. Explain why this is not sensible if p is large but is sensible (with a reasonable choice of x) if p is small. If p is small, explain why there is an optimum value for x Write down (but do not attempt to solve) an equation which indicates (in a ‘mathematical methods’ sense) that optimum value in terms of p, the probability that an individual has the disease. Schemes like these are only worthwhile if the disease is rare and the test is both expensive and will work on pooled samples. However, these circumstances do occur together from time to time and the idea then produces public health benefits much more cheaply than would otherwise be possible. Q 24.12. (i) Give the appropriate generalisation of Huffman’s algorithm to an alphabet with a symbols when you have m messages and m ≡ 1 mod a − 1. (ii) Prove that your algorithm gives an optimal solution. (iii) Extend the algorithm to cover general m by introducing messages of probability zero. Q 24.13. (i) A set of m apparently identical coins consists of m − 1 coins and one heavier coin. You are given a balance in which you can weigh equal numbers of the coins and determine which side (if either) contains the heavier coin. You wish to find the heavy coin in the fewest average number of weighings. If 3r + 1 ≤ m ≤ 3r+1 show that you can label each coin with a ternary number a1a2 . . . ar+1 with aj ∈ {0, 1, 2} in such a way that the number of coins having 1 in the jth place equals the number of coins with 2 in the jth place for each j (think Huffman ternary trees). By considering the Huffman algorithm problem for prefix-free codes on an alphabet with three letters, solve the problem stated in the first part 88
- 89. and show that you do indeed have a solution. Show that your solution also minimises the maximum number of weighings that you might have to do. (ii) Suppose the problem is as before but m = 12 and the odd coin may be heavier or lighter. Show that you need at least 3 weighings. [In fact you can always do it in 3 weighings, but the problem of showing this ‘is said to have been planted during the war . . . by enemy agents since Operational Research spent so many man-hours on its solution.’57 ] Q 24.14. Extend the definition of entropy to a random variable X taking values in the non-negative integers. (You must allow for the possibility of infinite entropy.) Compute the expected value EY and entropy H(Y ) in the case when Y has the geometric distribution, that is to say Pr(Y = k) = pk (1 − p) [0 p 1]. Show that, amongst all random variables X taking values in the non-negative integers with the same expected value µ [0 µ ∞], the geometric distribution maximises the entropy. Q 24.15. A source produces a set A of messages M1, M2, . . . , Mn with non-zero probabilities p1, p2, . . . pn. Let S be the codeword length when the message is encoded by a decodable code c : A → B∗ where B is an alphabet of k letters. (i) Show that n X i=1 √ pi !2 ≤ E(kS ) [Hint: Cauchy–Schwarz, p 1/2 i = p 1/2 i ksi/2 k−si/2 .] (ii) Show that min E(kS ) ≤ k n X i=1 √ pi !2 . where the minimum is taken over all decodable codes. [Hint: Look for a code with codeword lengths si = ⌈− logk p 1/2 i /λ⌉ for an appropriate λ.] 57 The quotation comes from Pedoe’s The Gentle Art of Mathematics which also gives a very pretty solution. As might be expected, there are many accounts of this problem on the web. 89
- 90. 25 Exercise Sheet 2 Q 25.1. (Exercise 7.3.) In an exam each candidate is asked to write down a Candidate Number of the form 3234A, 3235B, 3236C,. . . (the eleven possible letters are repeated cyclically) and a desk number. (Thus candidate 0004 sitting at desk 425 writes down 0004D − −425.) The first four numbers in the Candidate Identifier identify the candidate uniquely. Show that if the candidate makes one error in the Candidate Identifier then that error can be detected without using the Desk Number. Would this be true if there were 9 possible letters repeated cyclically? Would this be true if there were 12 possible letters repeated cyclically? Give reasons. Show that if we combine the Candidate Number and the Desk Number the combined code is one error correcting. Q 25.2. (Exercise 6.1) In the model of a communication channel, we take the probability p of error to be less than 1/2. Why do we not consider the case 1 ≥ p 1/2? What if p = 1/2? Q 25.3. (Exercise 7.4.) If you look at the inner title page of almost any book published between 1974 and 2007, you will find its International Standard Book Number (ISBN). The ISBN uses single digits selected from 0, 1, . . . , 8, 9 and X representing 10. Each ISBN consists of nine such digits a1, a2, . . . , a9 followed by a single check digit a10 chosen so that 10a1 + 9a2 + · · · + 2a9 + a10 ≡ 0 mod 11. (*) (In more sophisticated language, our code C consists of those elements a ∈ F10 11 such that P10 j=1(11 − j)aj = 0.) (i) Find a couple of books and check that (∗) holds for their ISBNs. (ii) Show that (∗) will not work if you make a mistake in writing down one digit of an ISBN. (iii) Show that (∗) may fail to detect two errors. (iv) Show that (∗) will not work if you interchange two distinct adjacent digits (a transposition error). (v) Does (iv) remain true if we replace ‘adjacent’ by ‘different’? Errors of type (ii) and (iv) are the most common in typing. In communication between publishers and booksellers, both sides are anx- ious that errors should be detected but would prefer the other side to query errors rather than to guess what the error might have been. (vi) Since the ISBN contained information such as the name of the pub- lisher, only a small proportion of possible ISBNs could be used58 and the 58 The same problem occurs with telephone numbers. If we use the Continent, Country, 90
- 91. system described above started to ‘run out of numbers’. A new system was introduced which was is compatible with the system used to label most con- sumer goods. After January 2007, the appropriate ISBN became a 13 digit number x1x2 . . . x13 with each digit selected from 0, 1, . . . , 8, 9 and the check digit x13 computed by using the formula x13 ≡ −(x1 + 3x2 + x3 + 3x4 + · · · + x11 + 3x12) mod 10. Show that we can detect single errors. Give an example to show that we cannot detect all transpositions. Q 25.4. (Exercise 7.5.) Suppose we use eight hole tape with the standard paper tape code and the probability that an error occurs at a particular place on the tape (i.e. a hole occurs where it should not or fails to occur where it should) is 10−4 . A program requires about 10 000 lines of tape (each line containing eight places) using the paper tape code. Using the Poisson approximation, direct calculation (possible with a hand calculator but really no advance on the Poisson method), or otherwise, show that the probability that the tape will be accepted as error free by the decoder is less than .04%. Suppose now that we use the Hamming scheme (making no use of the last place in each line). Explain why the program requires about 17 500 lines of tape but that any particular line will be correctly decoded with probability about 1 − (21 × 10−8 ) and the probability that the entire program will be correctly decoded is better than 99.6%. Q 25.5. If 0 δ 1/2, find an A(δ) 0 such that, whenever 0 ≤ r ≤ nδ, we have r X j=0 n j ≤ A(δ) n r . (We use weaker estimates in the course but this is the most illuminating. The particular value of A(δ) is unimportant so do not waste time trying to find a ‘good’ value.) Q 25.6. Show that the n-fold repetition code is perfect if and only if n is odd. Q 25.7. (i) What is the expected Hamming distance between two randomly chosen code words in Fn 2 . (As usual we suppose implicitly that the two choices are independent and all choices are equiprobable.) Town, Subscriber system we will need longer numbers than if we just numbered each member of the human race. 91
- 92. (ii) Three code words are chosen at random from Fn 2 . If kn is the expected value of the distance between the closest two, show that n−1 kn → 1/2 as n → ∞. [There are many ways to do (ii). One way is to consider Tchebychev’s in- equality.] Q 25.8. (Exercises 11.2 and 11.3.) Consider the situation described in the first paragraph of Section 11. (i) Show that for the situation described you should not bet if up ≤ 1 and should take w = up − 1 u − 1 if up 1. (ii) Let us write q = 1 − p. Show that, if up 1 and we choose the optimum w, E log Yn = p log p + q log q + log u − q log(u − 1). (iii) Show that, if you bet less than the optimal proportion, your fortune will still tend to increase but more slowly, but, if you bet more than some proportion w1, your fortune will decrease. Write down the equation for w1. [Moral: If you use the Kelly criterion veer on the side under-betting.] Q 25.9. Your employer announces that he is abandoning the old-fashioned paternalistic scheme under which he guarantees you a fixed sum Kx (where, of course, K, x 0) when you retire. Instead, he will empower you by giving you a fixed sum x now, to invest as you wish. In order to help you and the rest of the staff, your employer arranges that you should obtain advice from a financial whizkid with a top degree from Cambridge. After a long lecture in which the whizkid manages to be simultaneously condescending, boring and incomprehensible, you come away with the following information. When you retire, the world will be in exactly one of n states. By means of a piece of financial wizardry called ditching (or something like that) the whizkid can offer you a pension plan which for the cost of xi will return Kxiq−1 i if the world is in state i, but nothing otherwise. (Here qi 0 and Pn i=1 qi = 1.) The probability that the world will be in state i is pi. You must invest the entire fixed sum. (Formally, Pn i=1 xi = x. You must also take xi ≥ 0.) On philosophical grounds you decide to maximise the expected value S of the logarithm of the sum received on retirement. Assuming that you will have to live off this sum for the rest of your life, explain, in your opinion, why this choice is reasonable or explain why it is unreasonable. Find the appropriate choices of xi. Do they depend on the qi? 92
- 93. Suppose that K is fixed, but the whizkid can choose qi. We may suppose that what is good for you is bad for him so he will seek to minimise S for your best choices. Show that he will choose qi = pi. Show that, with these choices, S = log Kx. Q 25.10. Let C be the code consisting of the word 10111000100 and its cyclic shifts (that is 01011100010, 00101110001 and so on) together with the zero code word. Is C linear? Show that C has minimum distance 5. Q 25.11. (i) The original Hamming code was a 7 bit code used in an 8 bit system (paper tape). Consider the code c : {0, 1}4 → {0, 1}8 obtained by using the Hamming code for the first 7 bits and the final bit as a check digit so that x1 + x2 + · · · + x8 ≡ 0 mod 2. Find the minimum distance for this code. How many errors can it detect? How many can it correct? (ii) Given a code of length n which corrects e errors can you always construct a code of length n + 1 which detects 2e + 1 errors? Q 25.12. In general, we work under the assumption that all messages sent through our noisy channel are equally likely. In this question we drop this assumption. Suppose that each bit sent through a channel has probability 1/3 of being mistransmitted. There are 4 codewords 1100, 0110, 0001, 1111 sent with probabilities 1/4, 1/2, 1/12, 1/6. If you receive 1001 what will you decode it as, using each of the following rules? (i) The ideal observer rule: find b ∈ C so as to maximise Pr(b sent | u received}. (ii) The maximum likelihood rule: find b ∈ C so as to maximise Pr(u received | b sent}. (iii) The minimum distance rule: find b ∈ C so as to minimise the Ham- ming distance d(b, u) from the received message u. Q 25.13. (i) Show that −t ≥ log(1 − t) for 0 ≤ t 1. (ii) Show that, if δN 0, 1 − NδN 0 and N2 δN → ∞, then N−1 Y m=1 (1 − mδN ) → 0. 93
- 94. (iii) Let V (n, r) be the number of points in a Hamming ball of radius r in Fn 2 and let p(n, N, r) be the probability that N such balls chosen at random do not intersect. By observing that if m non-intersecting balls are already placed, then an m + 1st ball which does not intersect them must certainly not have its centre in one of the balls already placed, show that, if N2 n2−n V (n, rn) → ∞, then p(n, Nn, rn) → 0. (iv) Show that, if 2β + H(α) 1, then p(n, 2βn , αn) → 0. Thus simply throwing balls down at random will not give very good sys- tems of balls with empty intersections. 94
- 95. 26 Exercise Sheet 3 Q 26.1. A message passes through a binary symmetric channel with prob- ability p of error for each bit and the resulting message is passed through a second binary symmetric channel which is identical except that there is probability q of error [0 p, q 1/2]. Show that the result behaves as if it had been passed through a binary symmetric channel with probability of error to be determined. Show that the probability of error is less than 1/2. Can we improve the rate at which messages are transmitted (with low error) by coding, sending through the first channel, decoding with error correction and then recoding, sending through the second channel and decoding with error correction again or will this produce no improvement on treating the whole thing as a single channel and coding and decoding only once? Q 26.2. Write down the weight enumerators of the trivial code (that is to say, Fn 2 ), the zero code (that is to say, {0}), the repetition code and the simple parity code. Q 26.3. List the codewords of the Hamming (7,4) code and its dual. Write down the weight enumerators and verify that they satisfy the MacWilliams identity. Q 26.4. (a) Show that if C is linear, then so are its extension C+ , truncation C− and puncturing C′ , provided the symbol chosen to puncture by is 0. Give an example to show that C′ may not be linear if we puncture by 1. (b) Show that extension followed by truncation does not change a code. Is this true if we replace ‘truncation’ by ‘puncturing’? (c) Give an example where puncturing reduces the information rate and an example where puncturing increases the information rate. (d) Show that the minimum distance of the parity extension C+ is the least even integer n with n ≥ d(C). (e) Show that the minimum distance of the truncation C− is d(C) or d(C) − 1 and that both cases can occur. (f) Show that puncturing cannot decrease the minimum distance, but give examples to show that the minimum distance can stay the same or increase. Q 26.5. If C1 and C2 are linear codes of appropriate type with generator matrices G1 and G2, write down a generator matrix for C1|C2. Q 26.6. Show that the weight enumerator of RM(d, 1) is y2d + (2d+1 − 2)x2d−1 y2d−1 + x2d . 95
- 96. Q 26.7. (i) Show that every codeword in RM(d, d − 1) has even weight. (ii) Show that RM(m, m − r − 1) ⊆ RM(m, r)⊥ . (iii) By considering dimension, or otherwise, show that RM(m, r) has dual code RM(m, m − r − 1). Q 26.8. (Exercises 8.6 and 8.7.) We show that, even if 2n /V (n, e) is an integer, no perfect code may exist. (i) Verify that 290 V (90, 2) = 278 . (ii) Suppose that C is a perfect 2 error correcting code of length 90 and size 278 . Explain why we may suppose, without loss of generality, that 0 ∈ C. (iii) Let C be as in (ii) with 0 ∈ C. Consider the set X = {x ∈ F90 2 : x1 = 1, x2 = 1, d(0, x) = 3}. Show that, corresponding to each x ∈ X, we can find a unique c(x) ∈ C such that d(c(x), x) = 2. (iv) Continuing with the argument of (iii), show that d(c(x), 0) = 5 and that ci(x) = 1 whenever xi = 1. If y ∈ X, find the number of solutions to the equation c(x) = c(y) with x ∈ X and, by considering the number of elements of X, obtain a contradiction. (v) Conclude that there is no perfect [90, 278 ] code. (vi) Show that V (3, 23) is a power of 2. (In this case a perfect code exists called the binary Golay code.) Q 26.9. [The MacWilliams identity for binary codes] Let C ⊆ Fn 2 be a linear code of dimension k. (i) Show that X x∈C (−1)x.y = ( 2k if y ∈ C⊥ 0 if y / ∈ C⊥ . (ii) If t ∈ R, show that X y∈Fn 2 tw(y) (−1)x.y = (1 − t)w(x) (1 + t)n−w(x) . (iii) By using parts (i) and (ii) to evaluate X x∈C X y∈Fn 2 (−1)x.y s t w(y) 96
- 97. in two different ways, obtain the MacWilliams identity WC⊥ (s, t) = 2− dim C WC(t − s, t + s). Q 26.10. An erasure is a digit which has been made unreadable in trans- mission. Why are they easier to deal with than errors? Find a necessary and sufficient condition on the parity check matrix for it to be always possible to correct t erasures. Find a necessary and sufficient condition on the parity check matrix for it never to be possible to correct t erasures (ie whatever message you choose and whatever t erasures are made the recipient cannot tell what you sent). Q 26.11. Consider the collection K of polynomials a0 + a1ω with aj ∈ F2 manipulated subject to the usual rules of polynomial arithmetic and to the further condition 1 + ω + ω2 = 0. Show by finding a generator and writing out its powers that K∗ = K {0} is a cyclic group under multiplication and deduce that K is a finite field. [Of course, this follows directly from general theory but direct calculation is not uninstructive.] Q 26.12. (i) Identify the cyclic codes of length n corresponding to each of the polynomials 1, X − 1 and Xn−1 + Xn−2 + · · · + X + 1. (ii) Show that there are three cyclic codes of length 7 corresponding to irreducible polynomials of which two are versions of Hamming’s original code. What are the other cyclic codes? (iii) Identify the dual codes for each of the codes in (ii). Q 26.13. (Example 15.14.) Prove the following results. (i) If K is a field containing F2, then (a + b)2 = a2 + b2 for all a, b ∈ K. (ii) If P ∈ F2[X] and K is a field containing F2, then P(a)2 = P(a2 ) for all a ∈ K. (iii) Let K be a field containing F2 in which X7 − 1 factorises into linear factors. If β is a root of X3 + X + 1 in K, then β is a primitive root of unity and β2 is also a root of X3 + X + 1. (iv) We continue with the notation of (iii). The BCH code with {β, β2 } as defining set is Hamming’s original (7,4) code. 97
- 98. Q 26.14. Let C be a binary linear code of length n, rank k and distance d. (i) Show that C contains a codeword x with exactly d non-zero digits. (ii) Show that n ≥ d + k − 1. (iii) Prove that truncating C on the non-zero digits of x produces a code C′ of length n − d, rank k − 1 and distance d′ ≥ ⌈d 2 ⌉. [Hint: To show d′ ≥ ⌈d 2 ⌉, consider, for y ∈ C, the coordinates where xj = yj and the coordinates where xj 6= yj.] (iv) Show that n ≥ d + k−1 X u=1 ⌈ d 2u ⌉. Why does (iv) imply (ii)? Give an example where n d + k − 1. Q 26.15. Implement the secret sharing method of page 57 with k = 2, n = 3, xj = j + 1 p = 7, a0 = S = 2, a1 = 3. Check directly that any two people can find S but no single individual can. If we take k = 3, n = 4, p = 6, xj = j+1 show that the first two members and the fourth member of the Faculty Board will be unable to determine S uniquely. Why does this not invalidate our method? 98
- 99. 27 Exercise Sheet 4 Q 27.1. (Exercise 18.2.) Show that the decimal expansion of a rational number must be a recurrent expansion. Give a bound for the period in terms of the quotient. Conversely, by considering geometric series, or otherwise, show that a recurrent decimal represents a rational number. Q 27.2. A binary non-linear feedback register of length 4 has defining rela- tion xn+1 = xnxn−1 + xn−3. Show that the state space contains 4 cycles of lengths 1, 2, 4 and 9 Q 27.3. A binary LFR was used to generate the following stream 110001110001 . . . Recover the feedback polynomial by the Berlekamp–Massey method. [The LFR has length 4 but you should work through the trials for length r for 1 ≤ r ≤ 4.] Q 27.4. (Exercise 16.5.) Consider the linear recurrence xn = a0xn−d + a1xn−d+1 + . . . + ad−1xn−1 ⋆ with aj ∈ F2 and a0 6= 0. (i) Suppose K is a field containing F2 such that the auxiliary polynomial C has a root α in K. Show that xn = αn is a solution of ⋆ in K. (ii) Suppose K is a field containing F2 such that the auxiliary polynomial C has d distinct roots α1, α2, . . . , αd in K. Show that the general solution of ⋆ in K is xn = d X j=1 bjαn j for some bj ∈ K. If x0, x1, . . . , xd−1 ∈ F2, show that xn ∈ F2 for all n. (iii) Work out the first few lines of Pascal’s triangle modulo 2. Show that the functions fj : Z → F2 fj(n) = n j are linearly independent in the sense that m X j=0 bjfj(n) = 0 99
- 100. for all n implies bj = 0 for 1 ≤ j ≤ m. (iv) Suppose K is a field containing F2 such that the auxiliary polynomial C factorises completely into linear factors. If the root αu has multiplicity m(u) [1 ≤ u ≤ q], show that the general solution of ⋆ in K is xn = q X u=1 m(u)−1 X v=0 bu,v n v αn u for some bu,v ∈ K. If x0, x1, . . . , xd−1 ∈ F2, show that xn ∈ F2 for all n. Q 27.5. Consider the recurrence relation un+p + n−1 X j=0 cjuj+p = 0 over a field (if you wish, you may take the field to be R but the algebra is the same for all fields.) We suppose c0 6= 0. Write down an n × n matrix M such that u1 u2 . . . un = M u0 u1 . . . un−1 . Find the characteristic and minimal polynomials for M. Would your answers be the same if c0 = 0? Q 27.6. (Exercise 18.9.) One of the most confidential German codes (called FISH by the British) involved a complex mechanism which the British found could be simulated by two loops of paper tape of length 1501 and 1497. If kn = xn +yn where xn is a stream of period 1501 and yn is a stream of period 1497, what is the longest possible period of kn? How many consecutive values of kn would you need to to find the underlying linear feedback register using the Berlekamp–Massey method if you did not have the information given in the question? If you had all the information given in the question how many values of kn would you need? (Hint, look at xn+1497 − xn.) You have shown that, given kn for sufficiently many consecutive n we can find kn for all n. Can you find xn for all n? Q 27.7. We work in F2. I have a secret sequence k1, k2, . . . and a message p1, p2, . . . , pN . I transmit p1 + k1, p2 + k2, . . . pN + kN and then, by error, transmit p1 + k2, p2 + k3, . . . pN + kN+1. Assuming that you know this and that my message makes sense, how would you go about finding my message? Can you now decipher other messages sent using the same part of my secret sequence? 100
- 101. Q 27.8. Give an example of a homomorphism attack on an RSA code. Show in reasonable detail that the Elgamal signature scheme defeats it. Q 27.9. I announce that I shall be using the Rabin–Williams scheme with modulus N. My agent in X’Dofdro sends me a message m (with 1 ≤ m ≤ N − 1) encoded in the requisite form. Unfortunately, my cat eats the piece of paper on which the prime factors of N are recorded, so I am unable to decipher it. I therefore find a new pair of primes and announce that I shall be using the Rabin–Williams scheme with modulus N′ N. My agent now recodes the message and sends it to me again. The dreaded SNDO of X’Dofdro intercept both code messages. Show that they can find m. Can they decipher any other messages sent to me using only one of the coding schemes? Q 27.10. Extend the Diffie–Hellman key exchange system to cover three participants in a way that is likely to be as secure as the two party scheme. Extend the system to n parties in such a way that they can compute their common secret key by at most n2 −n communications of ‘Diffie–Hellman type numbers’. (The numbers p and g of our original Diffie-Hellman system are known by everybody in advance.) Show that this can be done using at most 2n − 2 communications by including several ‘Diffie–Hellman type numbers’ in one message. Q 27.11. St Abacus, who established written Abacan, was led, on theological grounds, to use an alphabet containing only three letters A, B and C and to avoid the use of spaces. (Thus an Abacan book consists of single word.) In modern Abacan, the letter A has frequency .5 and the letters B and C both have frequency .25. In order to disguise this, the Abacan Navy uses codes in which the 3r + ith number is x3r+i + yi modulo 3 [0 ≤ i ≤ 2] where xj = 0 if the jth letter of the message is A, xj = 1 if the jth letter of the message is B, xj = 2 if the jth letter of the message is C and y0, y1 and y2 are the numbers 0, 1, 2 in some order. Radio interception has picked up the following message. 120022010211121001001021002021 Although nobody in Naval Intelligence reads Abacan, it is believed that the last letter of the message will be B if the Abacan fleet is at sea. The Admiralty are desperate to know the last letter and send a representative to your rooms in Baker Street to ask your advice. Give it. 101
- 102. Q 27.12. Consider the bit exchange scheme proposed at the end of Sec- tion 19. Suppose that we replace STEP 5 by:- Alice sends Bob r1 and r2 and Bob checks that r2 1 ≡ r2 2 ≡ m mod n. Suppose further that Alice cheats by choosing 3 primes p1, p2, p3, and sending Bob p = p1 and q = p2p3. Explain how Alice can shift the odds of heads to 3/4. (She has other ways of cheating, but you are only asked to consider this one.) Q 27.13. (i) Consider the Fermat code given by the following procedure. ‘Choose N a large prime. Choose e and d so that ade ≡ a mod N, encrypt using the publicly known N and e, decrypt using the secret d.’ Why is this not a good code? (ii) In textbook examples of the RSA code we frequently see e = 65537. How many multiplications are needed to compute ae modulo N? (iii) Why is it unwise to choose primes p and q with p − q small when forming N = pq for the RSA method? Factorise 1763. Q 27.14. The University of Camford is proud of the excellence of its privacy system CAMSEC. To advertise this fact to the world, the Vice-Chancellor decrees that the university telephone directory should bear on its cover a number N (a product of two very large secret primes) and each name in the University Directory should be followed by their personal encryption number ei. The Vice-Chancellor knows all the secret decryption numbers di but gives these out on a need to know basis only. (Of course each member of staff must know their personal decryption number but they are instructed to keep it secret.) Messages a from the Vice-Chancellor to members of staff are encrypted in the standard manner as aei modulo N and decrypted as bdi modulo N. (i) The Vice-Chancellor sends a message to all members of the University. An outsider intercepts the encrypted message to individuals i and j where ei and ej are coprime. How can the outsider read the message? Can she read other messages sent from the Vice-Chancellor to the ith member of staff only? (ii) By means of a phone tapping device, the Professor of Applied Numis- matics (number u in the University Directory) has intercepted messages from the Vice-Chancellor to her hated rival, the Professor of Pure Numismatics (number v in the University Directory). Explain why she can decode them. What moral should be drawn? Q 27.15. The Poldovian Embassy uses a one-time pad to communicate with the notorious international spy Ivanovich Smith. The messages are coded 102
- 103. in the obvious way. (If the pad has C the 3rd letter of the alphabet and the message has I the 9th then the encrypted message has L the 3 + 9th. Work modulo 26.) Unknown to them, the person whom they employ to carry the messages is actually the MI5 agent ‘Union’ Jack Caruthers in disguise. MI5 are on the verge of arresting Ivanovich when ‘Union’ Jack is given the message LRPFOJQLCUD. Caruthers knows that the actual message is FLY XATXONCE and suggests that ‘the boffins change things a little’ so that Ivanovich deci- phers the message as REMAINXHERE. The only boffin available is you. Advise MI5. Q 27.16. Suppose that X and Y are independent random variables taking values in Zn. Show that H(X + Y ) ≥ max{H(X), H(Y )}. Why is this remark of interest in the context of one-time pads? Does this result remain true if X and Y need not be independent? Give a proof or counterexample. Q 27.17. I use the Elgamal signature scheme described on page 77. Instead of choosing k at random, I increase the value used by 2 each time I use it. Show that it will often be possible to find my privacy key u from two successive messages. Q 27.18. Confident in the unbreakability of RSA, I write the following. What mistakes have I made? 0000001 0000000 0002048 0000001 1391142 0000000 0177147 1033288 1391142 1174371. Advise me on how to increase the security of messages. Q 27.19. Let K be the finite field with 2d elements and primitive root α. (Recall that α is a generator of the cyclic group K{0} under multiplication.) Let T : K → F2 be a non-zero linear map. (Here we treat K as a vector space over F2.) 103
- 104. (i) Show that the map S : K × K → F2 given by S(x, y) = T(xy) is a symmetric bilinear form. Show further that S is non-degenerate (that is to say S(x, y) = 0 for all x implies y = 0). (ii) Show that the sequence xn = T(αn ) is the output from a linear feedback register of length at most d. (Part (iii) shows that it must be exactly d.) (iii) Show that the period of the system (that is to say the minimum period of T) is 2d − 1. Explain briefly why this is best possible. 104