Data Representation

Computer Organization and Assembly Language
Data Representation in
Computer Systems

Chapter 2 Objectives
 Understand the fundamentals of numerical data
representation and manipulation in digital
computers.
 Master the skill of converting between various
important radix systems.
 Understand how errors can occur in computations
because of overflow and truncation.
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Chapter 2 Objectives
 Understand the fundamental concepts of floating-
point representation.
 Gain familiarity with the most popular character
codes.
 Understand the concepts of error detecting and
correcting codes.
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 A bit is the most basic unit of information in a
computer.
 It is a state of “on” or “off” in a digital circuit.
 Sometimes these states are “high” or “low” voltage
instead of “on” or “off..”
 A byte is a group of eight bits.
 A byte is the smallest possible addressable unit of
computer storage.
 The term, “addressable,” means that a particular byte
can be retrieved according to its location in memory.
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 A word is a contiguous group of bytes.
 Words can be any number of bits or bytes.
 Word sizes of 16, 32, or 64 bits are most common.
 In a word-addressable system, a word is the smallest
addressable unit of storage.
 A group of four bits is called a nibble.
 Bytes, therefore, consist of two nibbles: a “high-order
nibble,” and a “low-order” nibble.
 Nibble is sometimes spelled nybble.
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 Bytes store numbers using the position of each
bit to represent a power of 2.
 The binary system is also called the base-2 system.
 Our decimal system is the base-10 system. It uses
powers of 10 for each position in a number.
 Any integer quantity can be represented exactly using
any base (or radix).
1 0 1 1 0 0 0 1 = 177
27 26 25 24 23 22 21 20
128 + 0 + 32 + 16 + 0 + 0 + 0 + 1 = 177
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 The decimal number 947 in powers of 10 is:
 The decimal number 5836.47 in powers of 10 is:
 To avoid confusion, the base is often denoted by
a subscript:
110012 = 2510
5  103 + 8  102 + 3  101 + 6  100
+ 4  10-1 + 7  10-2
9  102 + 4  101 + 7  100
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 Suppose we want to
convert the decimal
number 8763 to base 8.
 We know that 8 5 = 32768
so our result will be less
than six digits wide. The
largest power of 8 that we
need is therefore 8 4 =
4096 and 4096  2 = 8192.
 Write down the 2 and
subtract 8192 from 8763,
giving 571.
Converting Between Bases
8763
-8192 = 8 4  2
571

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 Converting 8763 to base 8…
 The next power of 8 is 8 3 =
512. We’ll need one of these,
so we subtract 512 and write
down 59 in our result.
 The next power of 8, 8 2 =
64, is too large, but we have
to assign a placeholder of
zero and carry down the 59.
8763
-8192 = 8 4  2
571
-512 = 8 3  1
59
-0 = 8 2  0
59

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 8 1 = 8 goes 7 times, leaving
us with 3.
 The last power of 8, 8 0 = 1,
is our last choice, and times
3 gives us a difference of
zero.
 Our result, reading from
top to bottom is:
876310 = 210738
8763
-8192 = 8 4  2
571
-512 = 8 3  1
59
-0 = 8 2  0
59
-56 = 8 1  7
3
-3 = 8 0  3
0

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 Another method uses division
instead of subtraction.
 First we take the number
that we wish to convert and
divide it by the radix in
which we want to express our
result.
 In this case, 8 divides 8763
1095 times, with a remainder
of 3.
 Record the quotient and the
remainder
8 |8763 3
1095

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 Divide 1095 by 8.
 Our remainder is 7, and the
quotient is 136.
8 |8763 3
8 |1095 7
136

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 Continue in this way until
the quotient is zero.
 In the final calculation, we
note that 8 divides 2 zero
times with a remainder of 2.
 Our result, reading from
bottom to top is:
876310 = 210738
8 |8763 3
8 |1095 7
8 |136 0
8 }17 1
8 |2 2
0

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 Fractional values can be approximated in all base
systems.
 Unlike integer values, fractions do not necessarily
have exact representations under all radices.
 The quantity ½ is exactly representable in the
binary and decimal systems, but is not in the
ternary (base 3) numbering system.
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 Fractional decimal values have nonzero digits to
the right of the decimal point.
 Fractional values of other radix systems have
nonzero digits to the right of the radix point.
 Numerals to the right of a radix point represent
negative powers of the radix:
0.4710 = 4  10 -1 + 7  10 -2
0.112 = 1  2 -1 + 1  2 -2
= ½ + ¼
= 0.5 + 0.25 = 0.75
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 As with whole-number conversions, you can use
either of two methods: a subtraction method or an
easy multiplication method.
 The subtraction method for fractions is identical to
the subtraction method for whole numbers. Instead
of subtracting positive powers of the target radix,
we subtract negative powers of the radix.
 We always start with the largest value first, n -1,
where n is our radix, and work our way along using
larger negative exponents.
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 The calculation to the
right is an example of
using the subtraction
method to convert the
decimal 0.8125 to binary.
 Our result, reading
from top to bottom is:
0.812510 = 0.11012
 Of course, this method
works with any base,
not just binary.
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 Using the multiplication
method to convert the
decimal 0.8125 to binary,
we multiply by the radix 2.
 The first product carries
into the units place.
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 Converting 0.8125 to binary . . .
– Ignoring the value in the units
place at each step, continue
multiplying each fractional part
by the radix.
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 Converting 0.8125 to binary . . .
 You are finished when the
product is zero, or until you
have reached the desired
number of binary places.
 Our result, reading from top to
bottom is:
0.812510 = 0.11012
 This method also works with
any base. Just use the target
radix as the multiplier.
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 The binary numbering system is the most
important radix system for digital computers.
 However, it is difficult to read long strings of binary
numbers -- and even a modestly-sized decimal
number becomes a very long binary number.
 For example: 110101000110112 = 1359510
 For compactness and ease of reading, binary
values are usually expressed using the
hexadecimal, or base-16, numbering system.
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 The hexadecimal numbering system uses the
numerals 0 through 9 and the letters A through F.
 The decimal number 12 is C16.
 The decimal number 26 is 1A16.
 It is easy to convert between base 16 and base 2,
because 16 = 24.
 Thus, to convert from binary to hexadecimal, all we
need to do is group the binary digits into groups of
four.
A group of four binary digits is called a hextet
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 Using groups of hextets, the binary number
110101000110112 (= 1359510) in hexadecimal is:
 Octal (base 8) values are derived from binary by
using groups of three bits (8 = 23):
Octal was very useful when computers used six-bit words.
If the number of bits is
not a multiple of 4, pad
on the left with zeros.
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 To convert back to binary from hexidecimal or
octal, simply expand each digit to its equivalent
binary and then string the binary together
 The leading 0s are often dropped unless you wish
to represent all the bits in a particular memory or
register location.
 The easiest way to convert from octal to hex or
vice-versa is to first convert to binary.
3B7A16: 3 = 0011, B = 1011, 7 = 0111, and A = 1010.
That gives us 00111011011110102

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Binary Addition
 Binary addition is as easy as it gets. You need to
know only four rules:
0 + 0 = 0 0 + 1 = 1
1 + 0 = 1 1 + 1 = 10
 The simplicity of this system makes it possible for
digital circuits to carry out arithmetic operations.
 Everything is fine as long as our sums remain
less than the space we have available
 For unsigned 8-bit math, that would be 29 or 256
 For signed 8-bit math, that would be +/-28 or 128
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Binary Addition
 Actually, digital computers ‘know’ memory as high
or low voltage, not 0s or 1s, so they don’t really do
math like adding or subtracting.
 Instead, they use boolean algebra to simulate math.
 For simple 1-bit adder you have 2 inputs and two
outputs, the result and a carry bit.
 The outputs are derived using truth tables:
RESULT
0 0 0
0 1 1
1 0 1
1 1 0
CARRY
0 0 0
0 1 0
1 0 0
1 1 1

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Binary Addition
 Actually, digital computers ‘know’ memory as high
or low voltage, not 0s or 1s, so they don’t really do
math like adding or subtracting.
 Instead, they use boolean algebra to simulate math.
 For simple 1-bit adder you have 2 inputs and two
outputs, the result and a carry bit.
 The outputs are derived using truth tables:
XOR RESULT
0 0 0
0 1 1
1 0 1
1 1 0
AND CARRY
0 0 0
0 1 0
1 0 0
1 1 1

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Binary Addition
 So our simple adder (also known as a half-adder)
consists of an XOR circuit and an AND circuit.
 If we want more bits, the ones to the left will have an
extra input: the carry from the bit to the right. But it’s
still the same idea: do the truth tables to get the
circuits for carry and result. We call these full-adders.
 An adder for X bits simply strings together X of these
full-adders by sending the carry from a bit to the next
leftmost adder.
 The rightmost adder always gets a 0 carry
 The leftmost adder outputs the carry for the entire set
 The diagrams for these circuits are in Chapter 3.

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Binary Multiplication
 The rules for binary multiplication are even easier:
 Anything times 0 = 0
 Anything times 1 = itself
 For example: 19 X 21 = 399
 Of course, each time you have
to add and check for overflow
 You can create a circuit that
strings together multi-bit adders that get the contents
of the multiplications but it’s a little complicated
10011
× 10101
10011
00000
10011
00000
10011
110001111

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Binary Mathematics
 Subtraction and division can be constructed using
logic circuits similarly.
 Boundary conditions add the most complications.
 Much research has gone into finding better
algorithms for binary math and is still ongoing.
 Booth’s algorithm is one such method that works for
binary multiplication.
 Take away: computers do math differently than we
do. The way we represent numbers might have a
beneficial effect on the way we perform binary math.

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Signed Integers
 To represent a signed integer, we use the high-
order bit to indicate the sign.
 The high-order bit is the leftmost bit
 0 means positive, 1 means negative
 The simplest scheme has the remaining bits
signifying the value of the number.
 This is called signed magnitude
 However, this means that there are 2 zeros: +0
(000000002) and -0 (100000002)

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Signed Integers
 With unsigned addition, large sums leave us with
extra digits which trigger an overflow error. Signed
addition can also work that way.
 Example:
 Using signed magnitude
binary arithmetic, find the
sum of 107 and 46.
 We see that the carry from the
seventh bit overflows and is
discarded, giving us the
erroneous result: 107 + 46 = 25.
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Signed Integers
 The signs in signed
magnitude representation
work just like the signs in
pencil and paper arithmetic.
 Example: Using signed
magnitude binary arithmetic,
find the sum of - 46 and - 25.
• Because the signs are the same, all we do is
add the numbers and supply the negative sign
when we are done.
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Signed Integers
 Mixed sign addition (or
subtraction) is done the same
way.
 Example: Using signed
magnitude binary arithmetic,
find the sum of 46 and - 25.
• The sign of the result gets the sign of the number
that is larger.
– Note the “borrows” from the second and sixth bits.
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Signed Integers
 Because of the special way we have to treat the
sign bits for signed magnitude integers, it is
awkward and costly to implement in digital logic.
 Let’s not forget the two values for 0, too, which can
confuse things.
 And recall the difficulties creating logical circuits to
do binary math.
 Perhaps there’s a better way?
 Computers often employ a complement system to
represent signed integers.

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Signed Integers
 In complement systems, negative values are
represented by some difference between a
number and its base.
 The diminished radix complement of a non-zero
number N in base r with d digits is (rd – 1) – N
 In the binary system, this gives us one’s
complement. It amounts to little more than flipping
the bits of a binary number.
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Signed Integers
 For example, using 8-bit one’s complement
representation:
+ 3 is: 00000011
- 3 is: 11111100
 In one’s complement representation, as with
signed magnitude, negative values are
indicated by a 1 in the high order bit.
 Complement systems are useful because they
eliminate the need for subtraction. The
difference of two values is found by adding the
minuend to the complement of the subtrahend.
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Signed Integers
 With one’s complement
addition, the carry bit is
“carried around” and added
to the sum.
 Example: Using one’s
complement binary
arithmetic, find the sum of 48
and – 19
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Signed Integers
 Although the “end carry around” adds some
complexity, one’s complement is simpler to
implement than signed magnitude.
 But it still has the disadvantage of having two
different representations for zero: positive zero
and negative zero.
 Two’s complement solves this problem.
 Two’s complement is the radix complement of the
binary numbering system; the radix complement
of a non-zero number N in base r with d digits is rd
– N.
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Signed Integers
 To express a value in two’s complement
representation:
 If the number is positive, just convert it to binary and
you’re done.
 If the number is negative, find the one’s complement
of the number and then add 1.
 Example:
 In 8-bit binary, 3 is:
00000011
 -3 using one’s complement representation is:
11111100
 Adding 1 gives us -3 in two’s complement form:
11111101.
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Signed Integers
 With two’s complement arithmetic, all we do is add
our two binary numbers. Just discard any carries
emitting from the high order bit.
– Example: Using one’s
complement binary
arithmetic, find the sum
of 48 and - 19.

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Signed Integer Representation
 Overflow and carry are tricky ideas.
 Signed number overflow means nothing in the
context of unsigned numbers, which set a carry
flag instead of an overflow flag.
 If a carry out of the leftmost bit occurs with an
unsigned number, overflow has occurred.
 Carry and overflow occur independently of each
other.
The table on the next slide summarizes these ideas.

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 We can do binary multiplication and division by 2
very easily using an arithmetic shift operation
 A left arithmetic shift inserts a 0 in for the
rightmost bit and shifts everything else left one bit;
in effect, it multiplies by 2
 A right arithmetic shift shifts everything one bit to
the right, but copies the sign bit; it divides by 2
 Let’s look at some examples.
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Example:
Multiply the value 11 (expressed using 8-bit signed two’s
complement representation) by 2.
We start with the binary value for 11:
00001011 (+11)
We shift left one place, resulting in:
00010110 (+22)
The sign bit has not changed, so the value is valid.
To multiply 11 by 4, we simply perform a left shift twice.
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Example:
Divide the value 12 (expressed using 8-bit signed two’s
complement representation) by 2.
We start with the binary value for 12:
00001100 (+12)
We shift left one place, resulting in:
00000110 (+6)
(Remember, we carry the sign bit to the left as we shift.)
To divide 12 by 4, we right shift twice.
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 Shifting is easier and faster to implement in logic
circuits than implementing the algorithm we discussed
back on slide 29.
 However, shifting only works in powers of 2.
 Booth’s algorithm combines shifting with addition and
subtraction where necessary to bring the speed of
shifting to general multiplication.

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In Booth’s algorithm:
 If the current multiplier bit is 1
and the preceding bit was 0,
subtract the multiplicand from
the product
 If the current multiplier bit is 0
and the preceding bit was 1,
we add the multiplicand to
the product
 If we have a 00 or 11 pair, we
simply shift.
 Assume a mythical “0”
starting bit
 Shift after each step
0011
x 01100<- mythical “0”
+ 0000 (shift)
- 0011 (subtract)
+ 0000 (shift)
+ 0011 (add) .
00010010
We see that 3  6 = 18!2012 Jones and Bartlett Learning, LLC
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 Here is a larger
example.
00110101
x 01111110
+ 0000000000000000
+ 111111111001011
+ 00000000000000
+ 0000000000000
+ 000000000000
+ 00000000000
+ 0000000000
+ 000110101_______
10001101000010110
Ignore all bits over 2n.
53  126 = 6678!
Subtract (add negative).

FLOATING POINT REPRESENTATION
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Floating Point Representation
 The operations we’ve discussed work for integer
representations, but imagine having to keep track of a
radix point at the same time.
 Instead, we will use a representation of floating point
numbers and leverage some useful mathematical
properties.
 This is known as floating point emulation because the
values aren’t stored as such (no radix point in binary
storage).
 Most computers today have hardware that manages
this floating point emulation in hardware, speeding it
up immensely.

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 Floating point numbers may be expressed in Scientific
Notation: 0.125 = 1.25  10-1
 This notation consists of three parts: sign, mantissa,
and exponent.
 Computer representation of a floating point number
consists of three fixed-size fields:
Note: Although “significand” and “mantissa” do not technically mean the
same thing, many people use these terms interchangeably. We use the term
“significand” to refer to the fractional part of a floating point number.
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 The one-bit sign field is the sign of the stored value.
 The size of the exponent field determines the range
of values that can be represented.
 The size of the significand determines the precision
of the representation.

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 The significand is always preceded by an implied
binary point.
 Thus, the significand always contains a fractional
binary value.
 The exponent indicates the power of 2 by which the
significand is multiplied.

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 We introduce a hypothetical “Simple Model” to
explain the concepts
 In this model:
 A floating-point number is 14 bits in length
 The exponent field is 5 bits
 The significand field is 8 bits

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 The significand is always preceded by an implied
binary point.
 Thus, the significand always contains a fractional
binary value.
 The exponent indicates the power of 2 by which the
significand is multiplied.

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 Example:
 Express 3210 in the simplified 14-bit floating-point
model.
 We know that 32 is 25. So in (binary) scientific
notation 32 = 1.0 x 25 = 0.1 x 26.
 We’re using the second version arbitrarily to
illustrate a point on the next slide.
 Using this information, we put 110 (= 610) in the
exponent field and 1 in the significand as shown.

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 The illustrations shown at
the right are all equivalent
representations for 32
using our simplified model.
 Not only do these
synonymous
representations waste
space, but they can also
cause confusion.
 Obviously, we need a
standard.

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 To resolve the problem of synonymous forms, we
establish a rule that the first digit of the
significand must be 1, with no ones to the left of
the radix point.
 This process, called normalization, results in a
unique pattern for each floating-point number.
 In our simple model, all significands must have the
form 0.1xxxxxxxx
 For example, 4.5 = 100.1 x 20 = 1.001 x 22 = 0.1001 x 23.
The last expression is correctly normalized.

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 Another problem with our system is that we have
made no allowances for negative exponents. We
have no way to express 0.5 (=2 -1)! (Notice that
there is no sign in the exponent field.)
 Don’t be confused by the sign bit for the entire
number.

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 To provide for negative exponents, we will use a
biased exponent.
 A bias is a number that is approximately midway
in the range of values expressible by the
exponent. We subtract the bias from the value in
the exponent to determine its true value.
 In our case, we have a 5-bit exponent. We will use 16
for our bias. This is called excess-16 representation.
 In our model, exponent values less than 16 are
negative, representing fractional numbers.

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 Example:
 Express 0.062510 in the revised 14-bit floating-point
model.
 We know that 0.0625 is 2-4. So in (binary) scientific
notation 0.0625 = 1.0 x 2-4 = 0.1 x 2 -3.
 To use our excess 16 biased exponent, we add 16 to -
3, giving 1310 (=011012).

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 Example:
 Express -26.62510 in the revised 14-bit floating-point
model.
 We find 26.62510 = 11010.1012. Normalizing, we have:
26.62510 = 0.11010101 x 2 5.
 To use our excess 16 biased exponent, we add 16 to 5,
giving 2110 (=101012). We also need a 1 in the sign bit.

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 The IEEE has established a standard for
floating-point numbers
 The IEEE-754 single precision floating point
standard uses an 8-bit exponent (with a bias of
127) and a 23-bit significand.
 The IEEE-754 double precision standard uses
an 11-bit exponent (with a bias of 1023) and a
52-bit significand.

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 In both the IEEE single-precision and double-
precision floating-point standard, the significant has
an implied 1 to the LEFT of the radix point.
 The format for a significand using the IEEE format is:
1.xxx…
 For example, 4.5 = .1001 x 23 in IEEE format is 4.5 = 1.001
x 22. The 1 is implied, which means is does not need to
be listed in the significand (the significand would
include only 001).

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 Example: Express -3.75 as a floating point number
using IEEE single precision.
 First, let’s normalize according to IEEE rules:
 3.75 = -11.112 = -1.111 x 21
 The bias is 127, so we add 127 + 1 = 128 (this is our
exponent)
 The first 1 in the significand is implied, so we have:
 Since we have an implied 1 in the significand, this equates
to
-(1).1112 x 2 (128 – 127) = -1.1112 x 21 = -11.112 = -3.75.
(implied)

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 Using the IEEE-754 single precision floating point
standard:
 An exponent of 255 indicates a special value.
 If the significand is zero, the value is  infinity.
 If the significand is nonzero, the value is NaN, “not a number,”
often used to flag an error condition.
 Using the double precision standard:
 The “special” exponent value for a double precision
number is 2047, instead of the 255 used by the single
precision standard.

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 Both the 14-bit model that we have presented
and the IEEE-754 floating point standard allow
two representations for zero.
 Zero is indicated by all zeros in the exponent and the
significand, but the sign bit can be either 0 or 1.
 This is why programmers should avoid testing a
floating-point value for equality to zero.
 Negative zero does not equal positive zero.

FLOATING POINT REPRESENTATION
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Floating Point Calculation
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 Floating-point addition and subtraction are done
using methods analogous to how we perform
calculations using pencil and paper.
 The first thing that we do is express both
operands in the same exponential power, then
add the numbers, preserving the exponent in the
sum.
 If the exponent requires adjustment, we do so at
the end of the calculation.

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 Example:
 Find the sum of 1210 and 1.2510 using the 14-bit “simple”
floating-point model.
 We find 1210 = 0.1100 x 2 4. And 1.2510 = 0.101 x 2 1 =
0.000101 x 2 4.
• Thus, our sum is
0.110101 x 2 4.

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 Floating-point multiplication is also carried out in
a manner akin to how we perform multiplication
using pencil and paper.
 We multiply the two operands and add their
exponents.
 If the exponent requires adjustment, we do so at
the end of the calculation.

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 Example:
 Find the product of 1210 and 1.2510 using the 14-bit
floating-point model.
 We find 1210 = 0.1100 x 2 4. And 1.2510 = 0.101 x 2 1.
• Thus, our product is
0.0111100 x 2 5 =
0.1111 x 2 4.
• The normalized
product requires an
exponent of 2210 =
101102.

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 No matter how many bits we use in a floating-point
representation, our model must be finite.
 The real number system is, of course, infinite, so our
models can give nothing more than an approximation
of a real value.
 At some point, every model breaks down, introducing
errors into our calculations.
 By using a greater number of bits in our model, we
can reduce these errors, but we can never totally
eliminate them.

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 Our job becomes one of reducing error, or at least
being aware of the possible magnitude of error in
our calculations.
 We must also be aware that errors can compound
through repetitive arithmetic operations.
 For example, our 14-bit model cannot exactly
represent the decimal value 128.5. In binary, it is 9
bits wide:
10000000.12 = 128.510

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 When we try to express 128.510 in our 14-bit model,
we lose the low-order bit, giving a relative error of:
 If we had a procedure that repetitively added 0.5 to
128.5, we would have an error of nearly 2% after only
four iterations.
128.5 -
128
128.5
 0.39%

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 Floating-point errors can be reduced when we use
operands that are similar in magnitude.
 If we were repetitively adding 0.5 to 128.5, it would
have been better to iteratively add 0.5 to itself and
then add 128.5 to this sum.
 In this example, the error was caused by loss of
the low-order bit.
 Loss of the high-order bit is more problematic.

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 Floating-point overflow and underflow can cause
programs to crash.
 Overflow occurs when there is no room to store
the high-order bits resulting from a calculation.
 Underflow occurs when a value is too small to
store, possibly resulting in division by zero.
Experienced programmers know that it’s better for a
program to crash than to have it produce incorrect,
but plausible, results.

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 Floating-point overflow and underflow can cause
programs to crash.
 Overflow occurs when there is no room to store
the high-order bits resulting from a calculation.
 Underflow occurs when a value is too small to
store, possibly resulting in division by zero.
Experienced programmers know that it’s better for a
program to crash than to have it produce incorrect,
but plausible, results.
Pat’s note: really experienced programmers know it’s
better not to crash, either.

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 When discussing floating-point numbers, it is
important to understand the terms range,
precision, and accuracy.
 The range of a numeric integer format is the
difference between the largest and smallest values
that can be expressed.
 Accuracy refers to how closely a numeric
representation approximates a true value.
 The precision of a number indicates how much
information we have about a value

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 Most of the time, greater precision leads to better
accuracy, but this is not always true.
 For example, 3.1333 is a value of pi that is accurate to
two digits, but has 5 digits of precision.
 There are other problems with floating point
numbers.
 Because of truncated bits, you cannot always
assume that a particular floating point operation is
commutative or distributive.

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Character Codes
 If computer’s don’t really store numbers, how
can we tell devices what we want printed or
displayed?
 Beyond numbers, how about the alphabet or
other characters?
 Basically, we need a code.
 Computer ‘knows’ bit patterns
 We must match patterns to characters in I/O context.

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Character Codes
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 As computers have evolved, character codes
have evolved.
 Larger computer memories and storage
devices permit richer character codes.
 The earliest computer coding systems used six
bits.
 Binary-coded decimal (BCD) was one of these
early codes. It was used by IBM mainframes in
the 1950s and 1960s.

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Character Codes
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 In 1964, BCD was extended to an 8-bit code,
Extended Binary-Coded Decimal Interchange
Code (EBCDIC).
 EBCDIC was one of the first widely-used
computer codes that supported upper and
lowercase alphabetic characters, in addition to
special characters, such as punctuation and
control characters.
 EBCDIC and BCD are still in use by IBM
mainframes today.

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Character Codes
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 Other computer manufacturers chose the 7-bit
ASCII (American Standard Code for Information
Interchange) as a replacement for 6-bit codes.
 While BCD and EBCDIC were based upon
punched card codes, ASCII was based upon
telecommunications (Telex) codes.
 Until recently, ASCII was the dominant character
code outside the IBM mainframe world.

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 Many of today’s systems embrace Unicode, a 16-
bit system that can encode the characters of
every language in the world.
 The Java programming language, and some operating
systems now use Unicode as their default character
code.
 The Unicode codespace is divided into six parts.
The first part is for Western alphabet codes,
including English, Greek, and Russian.

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Character Codes
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 The Unicode codes-
pace allocation is
shown at the right.
 The lowest-numbered
Unicode characters
comprise the ASCII
code.
 The highest provide for
user-defined codes.

EROOR DETECT
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ERROR DETECTION AND CORRECTION

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Error Detection and Correction
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 It is physically impossible for any data recording or
transmission medium to be 100% perfect 100% of
the time over its entire expected useful life.
 As more bits are packed onto a square centimeter
of disk storage, as communications transmission
speeds increase, the likelihood of error increases--
sometimes geometrically.
 Thus, error detection and correction is critical to
accurate data transmission, storage and retrieval.

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 Checksums and CRCs are examples of systematic
error detection.
 In systematic error detection a group of error control
bits is appended to the end of the block of transmitted
data.
 This group of bits is called a syndrome.
 CRCs are polynomials over the modulo 2 arithmetic
field.
The mathematical theory behind modulo 2 polynomials is
beyond our scope. However, we can easily work with it
without knowing its theoretical underpinnings.

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 Modulo 2 subtraction is easy.
 Ignore carry or borrow.
 0 - 0 = 0
 1 – 1 = 0
 1 – 0 = 1
 0 – 1 = 1
 So 1010
–1100
0110

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 Find the quotient and
remainder when 1111101 is
divided by 1101 in modulo 2
arithmetic.
 As with traditional division,
we note that the dividend is
divisible once by the divisor.
 We place the divisor under
the dividend and perform
modulo 2 subtraction.

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arithmetic…
 Now we bring down the next
bit of the dividend.
 We see that 00101 is not
divisible by 1101. So we place a
zero in the quotient.

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arithmetic…
 1010 is divisible by 1101 in
modulo 2.
 We perform the modulo 2
subtraction.

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arithmetic…
 We find the quotient is 1011,
and the remainder is 0010.
 This procedure is very useful
to us in calculating CRC
syndromes.

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 Suppose we want to transmit the
information string: 1111101.
 The receiver and sender decide to
use the (arbitrary) polynomial
pattern, 1101.
 The information string is shifted
left by one position less than the
number of positions in the divisor.
 The remainder is found through
modulo 2 division (at right) and
added to the information string:
1111101000 + 111 = 1111101111.

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 If no bits are lost or
corrupted, dividing the
received information string by
the agreed upon pattern will
give a remainder of zero.
 We see this is so in the
calculation at the right.
 Real applications use longer
polynomials to cover larger
information strings.
 If remainder = 0, receiver
truncates to get original msg.

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 Data transmission errors are easy to fix once an error
is detected.
 Just ask the sender to transmit the data again.
 In computer memory and data storage, however, this
cannot be done.
 Too often the only copy of something important is in
memory or on disk.
 Thus, to provide data integrity over the long term,
error correcting codes are required.

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 Hamming codes and Reed-Solomon codes are two
important error correcting codes.
 Reed-Solomon codes are particularly useful in
correcting burst errors that occur when a series of
adjacent bits are damaged.
 Because CD-ROMs are easily scratched, they employ a
type of Reed-Solomon error correction.
 Because the mathematics of Hamming codes is much
simpler than Reed-Solomon, we discuss Hamming
codes in detail.

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 Let’s suppose we have 2-bit data and we add a parity
bit to make up 3-bit code words: 00 + 1 = 001, 01 + 0
= 010, 10 + 0 = 100, and 11 + 1 = 111.
 We only have 4 code words but there are 8 possible
combinations of 3 bits: 000, 001, 010, 011, 100, 101,
110, and 111 (our code words are in red).
 The Hamming Distance between words is how many
bits must be changed to get from one word to
another: 001 to 010 takes two bit changes so the
distance is two.

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000, 001, 010, 011, 100, 101, 110, and 111
 The minimum Hamming Distance D(min) is the
smallest distance between any two code words (the
non-used codes don’t count).
 In this example, D(min) = 2
 Two single-bit errors in a code word in our example
could give us another code word but one error will
always give us a non-used word, so we can always
detect a single error.
 More generally, to detect k errors, D(min) = k + 1

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000, 001, 010, 011, 100, 101, 110, and 111
 Hamming Codes have n-bit code words which have m
data bits and r (redundant) parity bits. Our example is
3-bit words with 2 data bits and 1 parity bit.
 To detect and correct errors, each code word must
have n invalid words at a Hamming Distance of 1.
001 is associated with 000, 011, and 101.
 However, correction is only possible if the invalid
words aren’t shared: 011 –> 001 or 010?

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 Hamming codes can detect D(min) - 1 errors
and correct errors
 Thus, a Hamming distance of 2k + 1 is
required to be able to correct k errors in any
data word.
 Hamming distance is provided by adding a
suitable number of parity bits to a data word.

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 Using n bits, we have 2 n possible bit patterns. We
have 2 m valid code words with r check bits (where
n = m + r).
 For each valid codeword, we have (n+1) bit
patterns (1 legal and N illegal).
 This gives us the inequality:
(n + 1)  2 m  2 n
 Because n = m + r, we can rewrite the inequality
as:
(m + r + 1)  2 m  2 m + r or (m + r + 1)  2 r
 This inequality gives us a lower limit on the number of
check bits that we need in our code words.

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 Suppose we have data words of length m = 8.
Then:
(8 + r + 1)  2 r
implies that r must be greater than or equal to 4.
 We should always use the smallest whole value of r that
makes the inequality true.
 This means to build a code with 8-bit data words
that will correct single-bit errors, we must add 4
check bits, creating code words of length 12.
 Finding the number of check bits is the hard part.
The rest is easy.

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 Using our code words of length 12, number each
bit position starting with 1 in the low-order bit.
 Each bit position corresponding to a power of 2
will be occupied by a check bit.
 The remaining bits (our data) can all be
expressed as sums of powers of 2.
 Bit 10, for example, is 28 + 22 .

111
 Each data bit contributes to the parity of its
corresponding powers of 2. Bit 10 contributes to
bit 8 and 2.
 Put the other way around, each parity bit takes
parity from it’s corresponding data bits. Bit 4
takes parity from 12, 7, 6, and 5.

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 Lets insert the data word 11010110 and use even
parity for our check bits.
 Since 1 (= 20) contributes to the values 3, 5, 7, 9,
and 11, bit 1 will check parity over these bits.
 Since 2 (= 21) contributes to the values 3, 6, 7, 10,
and 11, bit 2 will check parity over these bits.

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 The completed code word is shown above.
 Bit 4 checks the bits 5, 6, 7, and 12, so its value is 1.
 Bit 8 checks the bits 9, 10, 11, and 12, so its value is also
1.
 Using the Hamming algorithm, we can not only
detect single bit errors in this code word, but also
correct them!

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 Suppose an error occurs in bit 5, as shown above.
Our parity bit values are:
 Bit 1 checks 3, 5, 7, 9, and 11. This is incorrect as we have a
total of 2 ones (which is not odd parity).
 Bit 2 checks bits 3, 6, 7, 10, and 11. The parity is correct.
 Bit 4 checks bits 5, 6, 7, and 12. This parity is incorrect, as
we have 2 ones.
 Bit 8 checks bit 9, 10, 11, and 12. This parity is correct.

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 We have erroneous parity for check bits 1 and 4.
 With two parity bits that don’t check, we know that
the error is in the data, and not in a parity bit.
 Each data bit position is the sum of two or more
powers of 2, so at least 2 parity bits would be
affected if it was wrong.
 If only one parity bit doesn’t check, it must be the
parity bit that is wrong.

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 Which data bits are in error? Bit 1 points to 3, 5, 7,
9, and 11. Bit 4 points to 5, 6, 7, and 12. Together
they point to 5 and 7. But bit 2, which also points
to 7, shows no problem so the error must be bit 5.
 More simply, we find out by adding the bit positions
of the erroneous bits.
 1 + 4 = 5. This tells us that the error is in bit 5. Flip
bit 5 to a 1 and our data is restored.

Data Representation

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