Computational Complexity: Introduction-Turing Machines-Undecidability

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Computational Complexity
Graduate Course
Antonis Antonopoulos
Computation and Reasoning Laboratory
National Technical University of Athens
This work is licensed under a Creative Commons Attribution-
NonCommercial- NoDerivatives 4.0 International License.
£
¢

¡3f311e480301e69f7d7081fe3b91ec87f3480230

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Bibliography
Textbooks
1 C. Papadimitriou, Computational Complexity, Addison Wesley,
1994
2 S. Arora, B. Barak, Computational Complexity: A Modern
Approach, Cambridge University Press, 2009
3 O. Goldreich, Computational Complexity: A Conceptual
Perspective, Cambridge University Press, 2008
Lecture Notes
1 L. Trevisan, Lecture Notes in Computational Complexity,
2002, UC Berkeley
2 J. Katz, Notes on Complexity Theory, 2011, University of
Maryland

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Algorithms & Complexity Turing Machines Undecidability
Contents
Introduction
Turing Machines
Undecidability
Complexity Classes
Oracles & The Polynomial Hierarchy
The Structure of NP
Randomized Computation
Non-Uniform Complexity
Interactive Proofs
Inapproximability
Derandomization of Complexity Classes
Counting Complexity
Epilogue

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Why Complexity?
Computational Complexity: Quantifying the amount of
computational resources required to solve a given task.
Classify computational problems according to their inherent
diﬃculty in complexity classes, and prove relations among them.
Structural Complexity: “The study of the relations between
various complexity classes and the global properties of individual
classes. [...] The goal of structural complexity is a thorough
understanding of the relations between the various complexity
classes and the internal structure of these complexity classes.” [J.
Hartmanis]

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Problems....
Decision Problems
Have answers of the form “yes” or “no”.
Encoding: each instance x of the problem is represented as a string
of an alphabet Σ (|Σ| ≥ 2).
Decision problems have the form “Is x in L?”, where L is a
language, L ⊆ Σ∗.
So, for an encoding of the input, using the alphabet Σ, we
associate the following language with the decision problem Π:
L(Π) = {x ∈ Σ∗
| x is a representation of a “yes” instance of the problem Π}
Example
Given a number x, is this number prime? (x
?
∈ PRIMES)
Given graph G and a number k, is there a clique with k (or more) nodes
in G?

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Problems....
Search Problems
Have answers of the form of an object.
Relation R(x, y) connecting instances x with answers (objects) y
we wish to ﬁnd for x.
Given instance x, ﬁnd a y such that (x, y) ∈ R.

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Problems....
Search Problems
Have answers of the form of an object.
Relation R(x, y) connecting instances x with answers (objects) y
we wish to find for x.
Given instance x, find a y such that (x, y) ∈ R.
Example
FACTORING: Given integer N, find its prime decomposition:
N = pk1
1 pk2
2 · · · pkm
m

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Problems....
Optimization Problems
For each instance x there is a set of Feasible Solutions F(x).
To each s ∈ F(x) we map a positive integer c(x), using the
objective function c(s).
We search for the solution s ∈ F(x) which minimizes (or
maximizes) the objective function c(s).

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Problems....
Optimization Problems
For each instance x there is a set of Feasible Solutions F(x).
To each s ∈ F(x) we map a positive integer c(x), using the
objective function c(s).
We search for the solution s ∈ F(x) which minimizes (or
maximizes) the objective function c(s).
Example
The Traveling Salesperson Problem (TSP):
Given a ﬁnite set C = {c1, . . . , cn} of cities and a distance
d(ci, cj) ∈ Z+
, ∀(ci, cj) ∈ C2
, we ask for a permutation π of C, that
minimizes this quantity:
n−1∑
i=1
d(cπ(i), cπ(i+1)) + d(cπ(n), cπ(1))

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Problems....
A Model Discussion
There are many computational models (RAM, Turing Machines
etc).
The Church-Turing Thesis states that all computation models
are equivalent. That is, every computation model can be simulated
by a Turing Machine.

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Problems....
A Model Discussion
etc).
In Complexity Theory, we consider eﬃciently computable the
problems which are solved (aka the languages that are decided) in
polynomial number of steps (Edmonds-Cobham Thesis).

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Problems....
A Model Discussion
etc).
In Complexity Theory, we consider eﬃciently computable the
problems which are solved (aka the languages that are decided) in
polynomial number of steps (Edmonds-Cobham Thesis).
§
¦
¤
¥
Eﬃciently Computable ≡ Polynomial-Time Computable

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Problems....
Computational Complexity classiﬁes problems into classes,
and studies the relations and the structure of these classes.
We have decision problems with boolean answer, or
function/optimization problems which output an object as an
answer.
Given some nice properties of polynomials, we identify
polynomial-time algorithms as eﬃcient algorithms.
Summary

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Definitions
Definition
A Turing Machine M is a quintuple M = (Q, Σ, δ, q0, F):
Q = {q0, q1, q2, q3, . . . , qn, qyes, qno} is a finite set of states.
Σ is the alphabet. The tape alphabet is Γ = Σ ∪ {⊔}.
q0 ∈ Q is the initial state.
F ⊆ Q is the set of final states.
δ : (Q F) × Γ → Q × Γ × {S, L, R} is the transition function.
A TM is a “programming language” with a single data structure (a
tape), and a cursor, which moves left and right on the tape.
Function δ is the program of the machine.

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Deﬁnitions
Turing Machines and Languages
Deﬁnition
Let L ⊆ Σ∗ be a language and M a TM such that, for every string
x ∈ Σ∗:
If x ∈ L, then M(x) = “yes”
If x /∈ L, then M(x) = “no”
Then we say that M decides L.
Alternatively, we say that M(x) = L(x), where L(x) = χL(x) is the
characteristic function of L (if we consider 1 as “yes” and 0 as
“no”).
If L is decided by some TM M, then L is called a recursive
language.

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Deﬁnitions
Deﬁnition
If for a language L there is a TM M, which if x ∈ L then M(x) = “yes”,
and if x /∈ L then M(x) ↑, we call L recursively enumerable.
*By M(x) ↑ we mean that M does not halt on input x (it runs forever).
Theorem
If L is recursive, then it is recursively enumerable.
Proof: Exercise

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Definitions
Definition
If for a language L there is a TM M, which if x ∈ L then M(x) = “yes”,
and if x /∈ L then M(x) ↑, we call L recursively enumerable.
*By M(x) ↑ we mean that M does not halt on input x (it runs forever).
Theorem
If L is recursive, then it is recursively enumerable.
Proof: Exercise
Definition
If f is a function, f : Σ∗ → Σ∗, we say that a TM M computes f if, for
any string x ∈ Σ∗, M(x) = f(x). If such M exists, f is called a
recursive function.
Turing Machines can be thought as algorithms for solving string
related problems.

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Definitions
Multitape Turing Machines
We can extend the previous Turing Machine definition to obtain a
Turing Machine with multiple tapes:
Definition
A k-tape Turing Machine M is a quintuple M = (Q, Σ, δ, q0, F):
Q = {q0, q1, q2, q3, . . . , qn, qhalt, qyes, qno} is a finite set of states.
δ : (Q F) × Γk → Q × (Γ × {S, L, R})k is the transition function.

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Properties of Turing Machines
Bounds on Turing Machines
We will characterize the “performance” of a Turing Machine by
the amount of time and space required on instances of size n, when
these amounts are expressed as a function of n.
Definition
Let T : N → N. We say that machine M operates within time T(n) if,
for any input string x, the time required by M to reach a final state is at
most T(|x|). Function T is a time bound for M.
Definition
Let S : N → N. We say that machine M operates within space S(n) if,
for any input string x, M visits at most S(|x|) locations on its work tapes
(excluding the input tape) during its computation. Function S is a space
bound for M.

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Theorem
Given any k-tape Turing Machine M operating within time T(n), we can
construct a TM M′ operating within time O
(
T2(n)
)
such that, for any
input x ∈ Σ∗, M(x) = M′(x).
Proof: See Th.2.1 (p.30) in [1].

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Theorem
Given any k-tape Turing Machine M operating within time T(n), we can
construct a TM M′ operating within time O
(
T2(n)
)
such that, for any
input x ∈ Σ∗, M(x) = M′(x).

This is a strong evidence of the robustness of our model:
Adding a bounded number of strings does not increase their
computational capabilities, and aﬀects their eﬃciency only polynomially.

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Algorithms Complexity Turing Machines Undecidability
Linear Speedup
Theorem
Let M be a TM that decides L ⊆ Σ∗, that operates within time T(n).
Then, for every ε 0, there is a TM M′ which decides the same
language and operates within time T′(n) = εT(n) + n + 2.

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Linear Speedup
Theorem
If, for example, T is linear, i.e. something like cn, then this theorem
states that the constant c can be made arbitrarily close to 1. So, it is fair
to start using the O (·) notation in our time bounds.
A similar theorem holds for space:

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Linear Speedup
Theorem
If, for example, T is linear, i.e. something like cn, then this theorem
states that the constant c can be made arbitrarily close to 1. So, it is fair
to start using the O (·) notation in our time bounds.
A similar theorem holds for space:
Theorem
Let M be a TM that decides L ⊆ Σ∗, that operates within space S(n).
language and operates within space S′(n) = εS(n) + 2.

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NTMs
Nondeterministic Turing Machines
We will now introduce an unrealistic model of computation:
Deﬁnition
A Turing Machine M is a quintuple M = (Q, Σ, δ, q0, F):
Q = {q0, q1, q2, q3, . . . , qn, qhalt, qyes, qno} is a ﬁnite set of states.
δ : (Q F) × Γ → Pow(Q × Γ × {S, L, R}) is the transition
relation.

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NTMs
In this model, an input is accepted if there is some sequence of
nondeterministic choices that results in “yes”.
An input is rejected if there is no sequence of choices that lead to
acceptance.
Observe the similarity with recursively enumerable languages.

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NTMs
acceptance.
Deﬁnition
We say that M operates within bound T(n), if for every input x ∈ Σ∗
and every sequence of nondeterministic choices, M reaches a ﬁnal state
within T(|x|) steps.

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NTMs
acceptance.
Definition
We say that M operates within bound T(n), if for every input x ∈ Σ∗
and every sequence of nondeterministic choices, M reaches a final state
within T(|x|) steps.
The above definition requires that M does not have computation
paths longer than T(n), where n = |x| the length of the input.
The amount of time charged is the depth of the computation tree.

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NTMs
Examples of Nondeterministic Computations
Example
✓ ✓
Accepting computation Rejecting Computation
Without loss of generality, the computation trees are binary, full
and complete. (why?)

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NTMs
A recursive language is decided by a TM.
A recursive enumerable language is accepted by a TM that
halts only if x ∈ L.
Multiple tape TMs can be simulated by a one-tape TM with
quadratic overhead.
Linear speedup justiﬁes the O (·) notation.
Nondeterministic TMs move in “parallel universes”, making
diﬀerent choices simultaneously.
A Deterministic TM computation is a path.
A Nondeterministic TM computation is a tree, i.e.
exponentially many paths ran simultaneously.
Summary

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Contents
Introduction
Turing Machines
Undecidability
Complexity Classes
Oracles The Polynomial Hierarchy
The Structure of NP
Randomized Computation
Non-Uniform Complexity
Interactive Proofs
Inapproximability
Derandomization of Complexity Classes
Counting Complexity
Epilogue

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Diagonalization
Diagonalization
Suppose there is a town with just one
barber, who is male. In this town, the
barber shaves all those, and only
those, men in town who do not shave
themselves. Who shaves the barber?
Diagonalization is a technique that was used in many diﬀerent cases:

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Diagonalization
Diagonalization
Theorem
The functions from N to N are uncountable.
Proof: Let, for the sake of contradiction that are countable: ϕ1, ϕ2, . . . .
Consider the following function: f(x) = ϕx(x) + 1. This function must
appear somewhere in this enumeration, so let ϕy = f(x). Then
ϕy(x) = ϕx(x) + 1, and if we choose y as an argument, then
ϕy(y) = ϕy(y) + 1. □

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Diagonalization
Diagonalization
Theorem
The functions from N to N are uncountable.
Proof: Let, for the sake of contradiction that are countable: ϕ1, ϕ2, . . . .
Consider the following function: f(x) = ϕx(x) + 1. This function must
appear somewhere in this enumeration, so let ϕy = f(x). Then
ϕy(x) = ϕx(x) + 1, and if we choose y as an argument, then
ϕy(y) = ϕy(y) + 1. □
Using the same argument:
Theorem
The functions from {0, 1}∗ to {0, 1} are uncountable.

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Simulation
Machines as strings
It is obvious that we can represent a Turing Machine as a string:
just write down the description and encode it using an alphabet, e.g.
{0, 1}.
We denote by ⌞M⌟ the TM M’s representation as a string.
Also, if x ∈ Σ∗, we denote by Mx the TM that x represents.
Keep in mind that:
Every string represents some TM.
Every TM is represented by inﬁnitely many strings.
There exists (at least) an uncomputable function from {0, 1}∗ to
{0, 1}, since the set of all TMs is countable.

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Simulation
The Universal Turing Machine
So far, our computational models are speciﬁed to solve a single
problem.
Turing observed that there is a TM that can simulate any other TM
M, given M’s description as input.
Theorem
There exists a TM U such that for every x, w ∈ Σ∗, U(x, w) = Mw(x).
Also, if Mw halts within T steps on input x, then U(x, w) halts within
CT log T steps, where C is a constant independent of x, and depending
only on Mw’s alphabet size number of tapes and number of states.
Proof: See section 3.1 in [1], and Th. 1.9 and section 1.7 in [2].

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Undecidability
The Halting Problem
Consider the following problem: “Given the description of a TM M,
and a string x, will M halt on input x? ” This is called the
HALTING PROBLEM.
We want to compute this problem ! ! ! (Given a computer
program and an input, will this program enter an inﬁnite loop?)
In language form: H = {⌞M⌟; x | M(x) ↓}, where “ ↓ ” means that
the machine halts, and “ ↑ ” that it runs forever.
Theorem
H is recursively enumerable.
Proof: See Th.3.1 (p.59) in [1]
In fact, H is not just a recursively enumerable language:
If we had an algorithm for deciding H, then we would be able to derive
an algorithm for deciding any r.e. language (RE-complete).

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Undecidability
The Halting Problem
But....
Theorem
H is not recursive.
Suppose, for the sake of contradiction, that there is a TM MH that
decides H.
Consider the TM D:
D(⌞M⌟) : if MH(⌞M⌟; ⌞M⌟) = “yes” then ↑ else “yes”
What is D(⌞D⌟)?

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Undecidability
The Halting Problem
But....
Theorem
H is not recursive.
Suppose, for the sake of contradiction, that there is a TM MH that
decides H.
Consider the TM D:
D(⌞M⌟) : if MH(⌞M⌟; ⌞M⌟) = “yes” then ↑ else “yes”
What is D(⌞D⌟)?
If D(⌞D⌟) ↑, then MH accepts the input, so ⌞D⌟; ⌞D⌟ ∈ H, so
D(D) ↓.
If D(⌞D⌟) ↓, then MH rejects ⌞D⌟; ⌞D⌟, so ⌞D⌟; ⌞D⌟ /∈ H, so
D(D) ↑. □

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Undecidability
Recursive languages are a proper subset of recursive enumerable
ones.
Recall that the complement of a language L is deﬁned as:
L = {x ∈ Σ∗
| x /∈ L} = Σ∗
L
Theorem
1 If L is recursive, so is L.
2 L is recursive if and only if L and L are recursively enumerable.
Proof: Exercise

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Undecidability
Recursive languages are a proper subset of recursive enumerable
ones.
Recall that the complement of a language L is deﬁned as:
L = {x ∈ Σ∗
| x /∈ L} = Σ∗
L
Theorem
1 If L is recursive, so is L.
2 L is recursive if and only if L and L are recursively enumerable.
Proof: Exercise
Let E(M) = {x | (q0, ▷, ε)
M∗
→ (q, y ⊔ x⊔, ε}
E(M) is the language enumerated by M.
Theorem
L is recursively enumerable iﬀ there is a TM M such that L = E(M).

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Undecidability
More Undecidability
The HALTING PROBLEM, our ﬁrst undecidable problem, was
the ﬁrst, but not the only undecidable problem. Its spawns a wide
range of such problems, via reductions.
To show that a problem A is undecidable we establish that, if there
is an algorithm for A, then there would be an algorithm for H,
which is absurd.

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Undecidability
More Undecidability
The HALTING PROBLEM, our ﬁrst undecidable problem, was
the ﬁrst, but not the only undecidable problem. Its spawns a wide
range of such problems, via reductions.
To show that a problem A is undecidable we establish that, if there
is an algorithm for A, then there would be an algorithm for H,
which is absurd.
Theorem
The following languages are not recursive:
1 {⌞M⌟ | M halts on all inputs}
2 {⌞M⌟; x | There is a y such that M(x) = y}
3 {⌞M⌟; x | The computation of M uses all states of M}
4 {⌞M⌟; x; y | M(x) = y}

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Undecidability
Rice’s Theorem
The previous problems lead us to a more general conclusion:

Any non-trivial property of
Turing Machines is undecidable
If a TM M accepts a language L, we write L = L(M).
Theorem (Rice’s Theorem)
Suppose that C is a proper, non-empty subset of the set of all recursively
enumerable languages. Then, the following problem is undecidable:
Given a Turing Machine M, is L(M) ∈ C?

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Undecidability
Rice’s Theorem
We can assume that ∅ /∈ C (why?).
Since C is nonempty, ∃ L ∈ C, accepted by the TM ML.
Let MH the TM accepting the HALTING PROBLEM for an
arbitrary input x. For each x ∈ Σ∗, we construct a TM M as
follows:
M(y) : if MH(x) = “yes” then ML(y) else ↑
We claim that: L(M) ∈ C if and only if x ∈ H.

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Undecidability
Rice’s Theorem
We can assume that ∅ /∈ C (why?).
Since C is nonempty, ∃ L ∈ C, accepted by the TM ML.
Let MH the TM accepting the HALTING PROBLEM for an
arbitrary input x. For each x ∈ Σ∗, we construct a TM M as
follows:
M(y) : if MH(x) = “yes” then ML(y) else ↑
We claim that: L(M) ∈ C if and only if x ∈ H.
Proof of the claim:
If x ∈ H, then MH(x) = “yes”, and so M will accept y or never halt,
depending on whether y ∈ L. Then the language accepted by M is
exactly L, which is in C.
If MH(x) ↑, M never halts, and thus M accepts the language ∅,
which is not in C. □

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Undecidability
TMs are encoded by strings.
The Universal TM U(x, ⌞M⌟) can simulate any other TM M
along with an input x.
The Halting Problem is recursively enumerable, but not
recursive.
Many other problems can be proved undecidable, by a
reduction from the Halting Problem.
Rice’s theorem states that any non-trivial property of TMs is an
undecidable problem.
Summary

Computational Complexity: Introduction-Turing Machines-Undecidability

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