The document discusses the post correspondence problem and uses it to show that the halting problem is undecidable. It introduces the concept of mapping reducibility to show how one problem can be reduced to another to prove properties like decidability. Several examples are provided to illustrate mapping reducibility and how it can be used to prove that problems like the equivalence of Turing machines are neither recognizable nor co-recognizable.

1. EECS 2001: Introduction to the
Theory of Computation
Muhammad Umair Khan

2. Reduction via computation histories
EECS 2001 Introduction to Theory of Computation 2
 Post correspondence problem
 Suppose we have two sets of strings over the alphabet {a,b}.
 Suppose we make a longer string by concatenating members of the first set (repetition is
allowed)
 Suppose we make another longer string by concatenating members of the second set (repetition
is allowed)
 Is it possible to make the above two longer strings identical using the members of the sets
provided to us – is it decidable?

3. Reduction via computation histories
EECS 2001 Introduction to Theory of Computation 3
 Post correspondence problem
 Suppose Set 1 = {bb, ab, c}
 Suppose Set 2 = {b, ba, bc}
 The longer string from Set 1 = bbababababababc (we repeated the second string from set 1)
 The longer string from Set 2 = bbababababababc (we repeated the second string from set 2)
 Is this possible for all sets?

4. Reduction via computation histories
EECS 2001 Introduction to Theory of Computation 4
 Post correspondence problem
 Suppose Set 3 = {a, ab, bba}
 Suppose Set 4 = {baa, aa, bb}
 The longer string from Set 3 = bbaabbbaa
 The longer string from Set 4 = bbaabbbaa
 Is this possible for all sets?

5. Reduction via computation histories
EECS 2001 Introduction to Theory of Computation 5
 Post correspondence problem
PCP = {⟨P⟩| P is an instance of the Post Correspondence Problem with a match}
 Overview of the proof
 The strings will be accepted by the TM if there is a computation history. If there is no computation
history, the TM can say reject. This is the functionality we need for deciding ATM. Contradiction.
Hence PCP is undecidable.
Detailed proof in the book. Read – not in the exam.

6. Reducibility
EECS 2001 Introduction to Theory of Computation 6
 Until now, we have been using reducibility to prove whether a
language is decidable or not
 Can we use reducibility to check a language for Turing-
recognizability?

7. Mapping Reducibility
EECS 2001 Introduction to Theory of Computation 7
 Mapping reducibility (not a new concept – just formalizing
what we have already discussed)
 If we can reduce problem A to B by using mapping reducibility, then
this means that there is a computable function which converts
instances of A into instances of B
 There are other ways to formally define reducing one problem
to another
 Mapping reducibility is just one way to define
 Depends on the application of reducibility

8. Mapping Reducibility
EECS 2001 Introduction to Theory of Computation 8
 Mapping reducibility (not a new concept – just formalizing
what we have already discussed)
 If we can reduce problem A to B by using mapping reducibility, then
this means that there is a computable function which converts
instances of A to instances of B
 Such a function is called a reduction
 We can solve A with a solver for B

9. Mapping Reducibility
EECS 2001 Introduction to Theory of Computation 9
 Computable function
 A TM computes a function by starting with the input to the function and
stopping (halts) when the output of the function is on the tape
 Formally, f: Σ*Σ*, where f is a computable function if there is a TM M
such that for every input w, M halts with only f(w) on the tape

10. Mapping Reducibility
EECS 2001 Introduction to Theory of Computation 10
 Computable function example
 Arithmetic operation on integers are computable functions
 Think of a TM which takes as input one or more integers (m1,
m2, m3, m4, …) and returns
 m1+m2+…,
 m1-m2,
 m1/m2,
 m1*m2,
 - m1

11. Mapping Reducibility
EECS 2001 Introduction to Theory of Computation 11
 More complex computable function example
 Takes a description of a TM as a string, and converts it into another
TM (output is the description of the new TM)
 e.g.,
 one which does not attempt to move left of the leftmost cell,
 one that does not use any symbol to mark the start of the tape,
 …

12. Mapping Reducibility
EECS 2001 Introduction to Theory of Computation 12
 Formal definition
 Language A is mapping reducible to Language B (A ≤m B), if there is a
computable function f: Σ*Σ*, where
w  A  f(w)  B

13. Mapping Reducibility
EECS 2001 Introduction to Theory of Computation 13
 Theorem 5.22
 If A ≤m B and B is decidable, then A is decidable
 Proof
 Let M be the decider for B
 Let f be the reduction from A to B
 N (a decider for A) is described as
 When w is received as input
 Compute f(w)
 Run M on f(w) and let N output whatever M outputs
 In essence, If f(w) belongs to B, then w belongs to A

14. Mapping Reducibility
EECS 2001 Introduction to Theory of Computation 14
 Previously, we have been using the following
 Corollary 5.23
 If A ≤m B and A is undecidable, then B is undecidable
 Using the proven undecidability of something (ATM) to decide the
undecidability of some other problems
 As compared to
 Theorem 5.22 (previous slide)
 If A ≤m B and B is decidable, then A is decidable
 Using the proven decidability of something to decide the decidability of
some other problems
 Note: the results will not change

15. Mapping Reducibility
EECS 2001 Introduction to Theory of Computation 15
 Revisiting HALTTM
 Previously: a decider for HALTTM was used to decide ATM, and we
ended up with a contradiction
 Mapping reducibility from ATM to HALTTM
 Computable function:
 Input ⟨M, w⟩, Output ⟨M’, w’⟩
 ⟨M, w⟩ ∈ ATM if and only if ⟨M′, w′⟩ ∈ HALTTM.
 Why is that?
 Answer: if the HALT machine stops on the complement of the input then
only our input will be accepted by ATM
 Remember: complement of something decidable should be recognizable
otherwise the original will not be decidable (HALT is a recognizer)

16. Mapping Reducibility
EECS 2001 Introduction to Theory of Computation 16
 Revisiting HALTTM
 Previously: a decider for HALTTM was used to decide ATM, and we
ended up with a contradiction
 Mapping reducibility from ATM to HALTTM
 Let F be a TM which computes the reduction f as follows
 On input ⟨M, w⟩
 F Constructs M’ which works as follows (on input x)
 Run M on x (any string)
 If M accepts, M’ accepts. If M rejects, M’ enters a loop.
 Let D be the decider for HALT
 Run D on ⟨M’, w⟩ as follows
 If D accepts ⟨M’, w⟩ then F accepts ⟨M, w⟩
 If D rejects ⟨M’, w⟩ then F rejects ⟨M, w⟩

17. Mapping Reducibility
EECS 2001 Introduction to Theory of Computation 17
 Theorem 5.28
 If A ≤m B and B is Turing-recognizable, then A is Turing-recognizable
 Proof (similar to 5.22)
 Let M be the Turing-recognizer for B
 Let f be the reduction from A to B
 N (a Turing-recognizer for A) is described as
 When w is received as input
 Compute f(w)
 Run M on f(w) and output whenever M outputs

18. Mapping Reducibility
EECS 2001 Introduction to Theory of Computation 18
 Corollary 5.29
If A ≤m B and A is not Turing-recognizable, then B is not
Turing-recognizable.

19. Mapping Reducibility
EECS 2001 Introduction to Theory of Computation 19
 Corollary 4.23
 ATM’ is not Turing recognizable
 ATM is Turing recognizable
 If ATM’ was also Turing recognizable, then ATM would be decidable
 Since we have already proved that ATM is not decidable (Theorem 4.11), ATM’
must not be Turing-recognizable

20. EQTM is neither T-recognizable nor co-T-recognizable
EECS 2001 Introduction to Theory of Computation 20
 Step 1: EQTM is not T-recognizable
 Reducing ATM to EQTM’
 Reducing function f
 TM F takes (M, w) as input
 1. Construct M1 and M2
 M1 rejects on all inputs
 M2 runs M on all inputs (w)
 If M accepts w, then M2 accepts
 2. Output (M1, M2)
 The above steps mean that M1 accepts nothing and M2 accepts everything
(provided M accepts w). Hence the machines are not equivalent
 Conversely, if M does not accept w, then M2 accepts nothing and they are
equivalent
 Hence f reduces ATM to EQTM

21. EQTM is neither T-recognizable nor co-T-recognizable
EECS 2001 Introduction to Theory of Computation 21
 Step 2: EQTM’ is not T-recognizable
 Reducing ATM to EQTM (complement of EQTM’)
 Showing ATM ≤m EQTM
 Let the reducing function be g
 TM G takes (M, w) as input
 1. Construct M1 and M2
 M1 Accept on all inputs
 M2 run M on all inputs
 If M accepts, then accept
 2. Output (M1, M2)
 The above steps mean that M1 accepts everything and M2 accepts
everything (provided M accepts w)

22. EQTM is neither T-recognizable nor co-T-recognizable
EECS 2001 Introduction to Theory of Computation 22
 Difference between f and g
 f
 M1 always rejects
 M accepts w iff M1 and M2 are equivalent
 g
 M1 always accepts
 f and g
 M accepts iff M2 always accepts

23. References
EECS 2001 Introduction to Theory of Computation 23
Ideas, problems and their solutions in this lecture/tutorial have been taken from
• Prof. Jeffery Edmonds’ Lecture notes for EECS 2001 at York University
• Prof. Suprakash Datta’ Lecture notes for EECS 2001 at York University
• Introduction to the Theory of Computation (3rd edition) by Michael Sipser
• Introduction to Theory of Computation by Anil Maheshwari and Michiel Smid
• Wikipedia and other webpages of different professors/universities