1 Reducability. CSCC63 Worksheet Reducability. For your reference, A T M is defined to be the language { M, w M accepts w}. Theorem 5.

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1 CSCC63 Worksheet Reducability For your reference, A T M is defined to be the language { M, w M accepts w}. 1 Reducability Theorem 5.1 HALT TM = { M, w M is a T M that halts on input w} is undecidable. Proof. Assume that T M R decides HALT T M and we will construct T M S to decide the acceptance problem. S = On input M, w, 1. Run TM R on input M, w 2. If R rejects, reject. 3. If R accepts, simulate M on w until it halts. 4. If M has accepted, accept; if M has rejected, reject. Theorem 5.2 E T M = { M M is a TM such that L(M) = {}} is undecidable. Proof Idea. Let R be a TM that decides E T M. Construct S that decides A T M using R. Construct another T M M 1 which behaves like M on input w, but rejects all other inputs. Pass M 1 to R. If L(M 1 ) is empty, then R accepts (i.e., M 1 does not accept w.) If M 1 accepts w, then L(M 1 ) {}, so R rejects and S accepts. 1

2 Proof. Assume R decides E T M, and construct S as follows: S = On input M, w : 1. Construct M 1, from M and w: M 1 = On input x: If x w, reject; Else accept if M accepts w. 2. Run R on input M 1 and do the opposite (if R accepts, reject; if R rejects, accept). Hence, S decides A T M, a contradiction! Theorem 5.4 EQ T M = { M 1, M 2 M 1 and M 2 are T Ms such that L(M 1 ) = L(M 2 )} is undecidable. Proof Idea. Q. Which undecidable language should we use to show that EQ T M is undecidable? E T M Q. How does the reduction work? Reduce E T M to EQ T M Construct M E, a TM that doesn t accept anything. To determine if L(M) is empty, pass M and M E to EQ T M Proof. Assume R decides EQ T M and construct S to decide E T M. S = On input M : 1. Compute M E, the description of the following T M: 2. M E = On input x: reject. 3. Run R on input M, M. 4. If R accepts, accept, if R rejects, reject. Then S decides E T M, contradiction. 2

3 Theorem 5.3 REGULAR T M = { M M is a T M such that L(M) is regular} is undecidable. Exercise: Try this one (use A T M as the other language). we ll discuss in tutorial. The solution is in the textbook and 2 Mapping Reducibility and Formalizing Reduction Proofs General structure of proof of undecidability of some language A: Assume R decides A. Construct S to decide B (some undecidable language): S = On input x: Compute y such that y A x B. Run R on y and accept if R accepts and reject if R rejects. Then, S decides B because y A x B. Since the structure is always the same, concentrate on core part: construction of y from x such that y A x B. Definition 5.20 Mapping Reducibility (a.k.a reduction). Language A is mapping reducible to language B, written A m B if there is a computable function f : Σ Σ, where for every w, w A f(w) B. The function f is called the reduction of A to B. Example E T M m EQ T M. Given M, construct M 1, M 2 as follows: M 1 = M, M 2 = a fixed T M that rejects all inputs This is computable (copy string, append constant string). 3

4 Since L(M 1 ) = L(M) and L(M 2 ) = {} M E T M L(M) = {} L(M 1 ) = L(M 2 ) M 1, M 2 EQ T M. Theorem 5.22 If A m B and B is decidable, then A is decidable. Theorem 5.28 If A m B and B is recognizable, then A is recognizable. Corollary 5.23 If A m B and A is undecidable, then B is undecidable. Corollary 5.29: If A m B and A is unrecognizable, then B is unrecognizable. Properties. Q. Why? A m B A C m B C (Straight from definition, since statement w A f(w) B is equivalent to w A f(w) B and w A = w A C.) Q. Why? If A m B and B m C, then A m C. (Easy exercise, based on function composition: if f,g computable, then g(f()) computable.) 3 Let s Practice. We showed that E T M is undecidable by showing a reduction from A T M to E T M. The proof was: Proof. Assume R decides E T M, and construct S as follows: S = On input M, w : 1. Construct M 1, from M and w: M 1 = On input x: 4

5 If x w, reject; Else accept if M accepts w. 2. Run R on input M 1 and do the opposite (if R accepts, reject; if R rejects, accept). Hence, S decides A T M, a contradiction! This is actually a proof that A T M m ET C M. Why??. In fact one can prove that A T M m E T M. (done in tutorial) 3.1 Prove that HALT T M is undecidable. Q. What is the reduction? A T M m HALT T M Given M, w, construct M, w such that Define M accepts w M halts on w or equivalently ( M, w in A T M ) ( M, w in HALT T M ). M = M except replace all transitions to q reject with transition to state q new, then add transitions from q new to q new for each input symbol (q new is deliberate infinite loop) w = w Explanation. If M accepts w, then M accepts w ; if M rejects w, then M loops on w ; if M loops on w, then M loops on w, i.e., M accepts w M halts on w, as desired. 5

6 4 Mapping Reducibility Examples cont d A T M m REGULAR T M. Goal. Given M, w, construct M such that M = On input x: 1. Accept if x = 0 n 1 n for some n Else, run M on w and do the same. Explanation. M accepts w L(M ) is regular of equivalently ( M, w in A T M ) ( M in REGULAR T M ) If M accepts w, then M accepts everything, i.e., L(M ) = Σ is regular; if M does not accept w, then M accepts only 0 n 1 n, i.e., L(M ) = {0 n 1 n : n 0} is not regular. 5 Examples cont... EQ T M is neither recognizable nor co-recognizable. Recall a language is co-recognizable if its complement is recognizable. Similar definition for decidability unnecessary do you see why? Q. What is the reduction to show that EQ T M is not co-recognizable? A T M m EQ T M Q. Why does this work? Because it is equiv. to A c T M m EQ c T M and Ac T M is not recognizable. On input M, w, construct M 1, M 2 as follows: 1. M 1 = On input x: run M on w (ignore x) and do the same. 2. M 2 = On input x: accept. 6

7 Explanation Then, M, w A T M M accepts w M 1 accepts every string L(M 1 ) = L(M 2 ) M 1, M 2 EQ T M. This proves that EQ T M is not co-recognizable. Q. How can we show that EQ T M is not recognizable? Show A T M m EQ c T M (equiv. to Ac T M m EQ T M ): On input M, w, construct M 1, M 2 as follows: 1. M 1 = On input x: run M on w (ignore x) and do the same. 2. M 2 = On input x: reject. Explanation Then, M, w A T M M accepts w M 1 accepts every string L(M 1 ) L(M 2 ) M 1, M 2 EQ T M M 1, M 2 EQ c T M. This proves that EQ T M is not recognizable. 6 Examples cont... INF = { M : L(M) contains infinitely many strings } is neither recognizable nor co-recognizable. Q. What should the reduction be to prove INF is not co-recognizable? HALT T M m INF Q. Why? Since HALT T M is not decidable, but is recognizable (easy to report if M halts on w), it must be that HALTT C M is not recognizable. 7

8 On input M, w, construct M as follows: M = On input x: 1. Run M on w. 2. Accept if M halts. Explanation If M halts on w, then M accepts all strings so L(M ) is infinite. If M loops on w, then M accepts no string so L(M ) is finite. This shows INF c is unrecognizable. Q. How can we show that INF is not recognizable? We can show that HALT T M m INF c. On input M, w, construct M as follows: M = On input x: 1. Run M on w for x steps. 2. Accept if M has not halted. Explanation If M halts on w, then M accepts only the strings up to a certain length, so L(M ) is finite. If M loops on w, then M accepts all strings so L(M ) is infinite. This shows IN F is unrecognizable. 8