Theory of Computer Science - Post Correspondence Problem

Math- I, II, III, IV, RSA & TCS by Parmar Sir [9764546384]
MU | TCS/FA | Un-decidability | Feb 2014 | Parmar Sir Page 1
Un-decidability
 Recursive Languages
 A language L over the alphabet is called recursive if it is
accepted by some Turing Machine, M that accepts every word
in L and reject every word in (i.e. not belonging to L)
i.e. Accept (M) =L
Reject (M) =
Loop (M) = (No output)
 Example
TM accepting L over that starts with “a” and reject for in all cases. So L is recursive.
 For recursive languages we can define algorithm.
 The union of two recursive languages is recursive language.
 The complements of recursive language is recursive
 Recursively Enumerable Languages
 A language L over the alphabet is called recursive
enumerable languages if it is accepted by some Turing
Machine, M that accepts every word in L and either rejects
or loops for every word in
i.e. Accept (M) =L
Reject (M)+ Loop (M) = (No output)
 Example
A TM accepting language and reject or loops all words not in L. So L is recursive
enumerable.
 The universal language is recursively enumerable. [Universal Language: this language
consists of strings that are interpreted as TM followed by an Input for that TM. The string is in
if the TM accepts that input.]
 But diagonalisation language is not recursively enumerable. [Diagonalisation Language: this
language is the set of strings of 0’s and 1’s that when interpreted as a TM, are in the language of
that TM]
 For recursive enumerable languages we cannot define algorithm.
 The union of two recursive enumerable languages is recursive enumerable language.

Accept
Reject
L
L’

AcceptL
L’

 Difference between Recursive and Recursive Enumerable languages
 There is a profound difference between recursive and recursively enumerable language.
 If there is a regular language, we always have a FA to accept it.
 Now if the string w is given, and we want to know whether w belongs to this languages, w will
run it on the FA, and within n number of steps, where , as each state will
consume one character, we will get the answer YES if w belongs to L, otherwise NO. This is
called effective decision procedure.
 With TM accepting the language which is recursively enumerable and if w is executed on TM, we
are having a harder time. If then the machine will halt, otherwise we will have to wait,
perhaps for a very long time.
 In the worst case if w is in loop set for this machine, we shall never get a answer.
 A recursive language has the advantage that we shall at least some day get the answer, even
though we may not know how long it will take.
 An interesting observation is that, not all TM’s that accept a recursive language have no loop set.
 A language is recursive if at least one TM accepts it and rejects its complement. Some TM’s
which accept the same language but loops on some inputs.
 Undesirability and Un-solvability
 A problem whose language is recursive is said to be decidable otherwise it is un-decidable (RE
languages).
 If the problem is un-decidable then there is no algorithm which take the input and find the answer
either YES or NO.
 The problem “Whether a TM will halt for the given input word” is un-decidable problem
 If there is a TM which, when applied to any problem in the class, always eventually terminates
with the correct (YES/NO) answer, we call the problem solvable.
 If there is a TM which, when applied to any problem in the class, always eventually terminates
with the correct answer when the answer is YES and may or may not terminate when the correct
answer is NO, we call the problem semi-solvable or partially solvable.
 If there is no TM which, when applied to a problem in the class, eventually with the correct
answer YES, we call the problem un-solvable.
 Rice’s Theorem
 Every non-trivial property of the RE (Recursive Enumerable) languages is un-decidable.
Proof
 Let P be any nontrivial property of the RE languages. That means at least one language has a
property and the at least one language does not.

 Initially we will assume an empty language is not having the property P
 Since P is non-trivial, there should be some language L having property P i.e. And
therefore there exists a coded TM accepting language L let this TM is
 Now to prove construct a Universal Turing Machine that simulate TM, M and
 The design of be given in figure, if M does not accept w, and if M
accept w.
M
M L
M’Accept
AcceptAccept
Start
x
w
 Let reduce a universal language to and is un-decidable, so is also un-decidable (by
reduction theorem).
 The Turing machine can be produced on reduction. For the reduction algorithm the pair (M,
w) can be given input.
 is a two tape TM.
 One tape is used to simulate M on input w. then input is used for designing the transitions
of
 Second tape is used to simulate on input x. again the pair is used for designing the
transitions of
 The TM, is constructed to do the following things-
1. Simulate M on input w. note that w is not the input to ; rather writes M and w onto one
of its tapes and simulate the Universal Turing Machine on this pair.
2. If M does not accept w, then does nothing else. never accepts its own input, x so
. Since we assume is not in property P, that means the code for is not in
3. If M accepts w, then begins simulating on its own input x. Thus, will accept
exactly the language L. since L is in P, the code for is in P.
 Since constructing from M and w can be carried out by an algorithm. Since this algorithm
turns (M, w) into an that is in if and only if (M, w) is in , this reduction of to ,
and hence proves that the property P is un-decidable.
 Now another possibility when is in P.
 If so, consider the complements property of P i.e. , the set of RE languages that do not have
property P.

 By the foregoing, is un-decidable. However, since every TM accepts an RE language, , the
set of (codes for) Turing machines that do not accept a languages in P is the same as , the set
of TM’s that accepts a language in .
 Suppose were decidable, then so would be is decidable. since the complement of a
recursive language is recursive.
 Post Correspondence Problem
 Post’s correspondence Problem (PCP) was first introduced by Emil Post in 1946. Later, the
problem was found to have many applications in the theory of formal languages.
 PCP is uses to check un-decidability of strings, (another way to check by Turing machine)
 PCP define as
 “the post’s correspondence problem consists of two lists of strings that are of equal length over
the input .
The two lists are and then there exists a non empty set
of integers such that ”
 Another way we can say “The PCP is to determine whether or not there exist ,
where , such that
 To solve the PCP, we try to find all combinations of to find the , then we say
that PCP has a solution.
 PCP is un-decidable.
Example:
Does the PCP with two lists and have a solution?
Answer:
We have to find out such a sequence that strings formed by x and y are identical.
Such sequences are 2, 1, 1, and 3.
Hence from x and y list equal
2 1 1 3 2 1 1 3
= (combined and see)
Which forms thus PCP has a solution.
........................................................................................................................................................
Theorem: It is Un-decidable a CFG is ambiguous.
Proof:

Theory of Computer Science - Post Correspondence Problem

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