Theory of Computation

Transcription

1 Theory of Computation Lecture #10 Sarmad Abbasi Virtual University Sarmad Abbasi (Virtual University) Theory of Computation 1 / 35

2 Lecture 10: Overview Undecidability of the Halting Problem Russell s Paradox Reducibility Undecidable Problems Linear bounded automata Sarmad Abbasi (Virtual University) Theory of Computation 2 / 35

3 Acceptance Problem for TMs Let us define A TM = { M, w : M is a TM that accepts w}. This is called the acceptance problem for TMs or the halting problem. We also call this the Halting Problem. We showed that this problem is undecidable. Sarmad Abbasi (Virtual University) Theory of Computation 3 / 35

4 Undecidability of the Halting Problem Theorem A TM is not decidable. The language A TM is undecidable. There is no TH H such that on input M, w 1 H accepts if M accepts w. 2 H rejects if M does not accept w. Sarmad Abbasi (Virtual University) Theory of Computation 4 / 35

5 Undecidability of the Halting Problem If H exists we can use H to construct D as follows: 1 On input M. 2 Convert M into M, M 3 Run H on M, M. 4 Output the opposite of what H does; that is If H accepts reject, if H rejects accept. Once we have constructed D we will ask how it behaves on its own description. Sarmad Abbasi (Virtual University) Theory of Computation 5 / 35

6 Undecidability of the Halting Problem Assume that H exists. We use H to construct D. This is easy D simply converts its input M into M, M and calls H. It returns the answer that is opposite of the one given by H. 1 H accepts M, w exactly when M accepts w. 2 D rejects M exactly when M accepts M. 3 D rejects D exactly when D accepts D. This leads to a contradiction. Sarmad Abbasi (Virtual University) Theory of Computation 6 / 35

7 Russell s Paradox In a town called Seville there lives a barber. A man Seville is shaved by the barber if and only if the man does not shave himself?" Question Does the barber of Seville shaves himself. Sarmad Abbasi (Virtual University) Theory of Computation 7 / 35

8 Russell s Paradox Students are asked to make a list in which they can include the names of anyone (including themselves) in a class room. Is there a particular student wrote the names of those people in the class who did not include their own names in their own list? Question Does that particular student write his/her own name on his/her list? Sarmad Abbasi (Virtual University) Theory of Computation 8 / 35

9 Russell s Paradox Is it possible to have an enumerator that enumerates the description of all the enumerators that do not enumerate their own description. Suppose such an enumerator F exists. Question Does F enumerate its own description? Sarmad Abbasi (Virtual University) Theory of Computation 9 / 35

10 Russell s Paradox 1 D that accepts (the description of) those TMs that do not accept their own description. 2 Barber who shaves all those people who do not shave themselves. 3 Student who lists all those students who do not list themselves. 4 Enumerator that enumerates (the description of) all those enumerators that do not enumerate themselves. Such objects cannot exist. Sarmad Abbasi (Virtual University) Theory of Computation 10 / 35

11 Reducibility Suppose u have a language B in your mind and you can prove that If I can make a decider for B the I can make a decider for the halting problem. Sarmad Abbasi (Virtual University) Theory of Computation 11 / 35

12 Reducibility Now suppose some one claims that they have a decider for the language B. Then I can claim that I have a decider for A TM. Which is absurd. From this we can conclude that B is also undecidable as we know that A TM is undecidable. Sarmad Abbasi (Virtual University) Theory of Computation 12 / 35

13 Reducibility Theorem HALT TM is undecidable. HALT TM = { M, w : M is a TM that halts on w}. Sarmad Abbasi (Virtual University) Theory of Computation 13 / 35

14 Reducibility Suppose that HALT TM is decidable and R decides HALT TM. If we have R then we have no problem on TM that loop endlessly. We can always check if a TM will loop endlessly on a given input using R. We can use R to make a decider for A TM. We know that A TM is undecidable. We have a contradiction and that is why R cannot exist. Sarmad Abbasi (Virtual University) Theory of Computation 14 / 35

15 Reducibility Given R we can construct S which decides A TM that operates as follows: 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 w, accept; if M has rejected w, reject. Note that S will reject M, w even if M loops on w. Sarmad Abbasi (Virtual University) Theory of Computation 15 / 35

16 Emptiness problem for TMs Theorem E TM is undecidable. E TM = { M : L(M) = }. Sarmad Abbasi (Virtual University) Theory of Computation 16 / 35

17 Emptiness problem Suppose that E TM is decidable and R decides E TM. R accepts M if the language of M is empty. We can use R to make a decider S for A TM. We know that A TM is undecidable. We have a contradiction and that is why R cannot exist. How can we construct S? Sarmad Abbasi (Virtual University) Theory of Computation 17 / 35

18 Emptiness problem S gets M, w as an input. The first idea is to run R on input M and check if it accepts. If it does, we know that L(M) is empty and therefore that M does not accept w. But, if R rejects we only know that the language of R is not empty. We do not know anything about what it does on w. So the idea does not work. We have to come up with another idea. Sarmad Abbasi (Virtual University) Theory of Computation 18 / 35

19 Emptiness problem Given M and w let us consider M 1 : 1 On input x 2 If x w, reject. 3 If x = w, run M on input w and accept if M accepts w. What is L(M 1 )? Sarmad Abbasi (Virtual University) Theory of Computation 19 / 35

20 Emptiness problem L(M 1 ) = {, if M does not accept w; {w}, if M accepts w. Sarmad Abbasi (Virtual University) Theory of Computation 20 / 35

21 Emptiness problem Given R we can now we can construct S which decides A TM that operates as follows: On input M, w 1 Run TM R on input M, w. 2 Construct M 1 3 Run R on M 1 4 If R accepts M 1, reject. If it rejects M 1 accept. Sarmad Abbasi (Virtual University) Theory of Computation 21 / 35

22 Emptiness problem S is a decider for A TM. Since S does not exist we conclude that R does not exist. So E TM is undecidable. Sarmad Abbasi (Virtual University) Theory of Computation 22 / 35

23 Regular languages and TMs REG TM = { M : L(M) is regular}. Theorem REG TM is undecidable. Sarmad Abbasi (Virtual University) Theory of Computation 23 / 35

24 Regular languages and TMs Suppose that REG TM is decidable and R decides REG TM. R accepts all TMs that accept a regular language and reject all TMs that do not accept a regular language. How can we use R to construct a decider for A TM? Note that S has to get as input M, w and has to decide if M accepts w. The idea is as follows: Sarmad Abbasi (Virtual University) Theory of Computation 24 / 35

25 Regular languages and TMs S gets M, w as an input. Given M and w we will construct a machine M 2 such that { is regular, if M accpets w; L(M 2 ) = not regular, if M does not w. Sarmad Abbasi (Virtual University) Theory of Computation 25 / 35

26 Regular languages and TMs Given M and w let us consider M 2 : 1 On input x 2 If x = 0 n 1 n then accept. 3 else run M on input w and accept x if M accepts w. What is L(M 2 )? Sarmad Abbasi (Virtual University) Theory of Computation 26 / 35

27 Regular languages and TMs L(M 2 ) = { {0 n 1 n : n 0}, if M does not accept w; {0, 1}, if M accepts w. Sarmad Abbasi (Virtual University) Theory of Computation 27 / 35

28 Regular languages and TMs Given R we can now we can construct S which decides A TM that operates as follows: On input M, w 1 Construct M 2 2 Run R on M 2 3 If R accepts M 2, accept. If it rejects M 2 reject. Sarmad Abbasi (Virtual University) Theory of Computation 28 / 35

29 Regular languages and TMs S is a decider for A TM. Since S does not exist we conclude that R does not exist. So REG TM is undecidable. Sarmad Abbasi (Virtual University) Theory of Computation 29 / 35

30 Equivalence problem for TMs So far we have only used A TM to show that other problems are undecidable. Let us do something different. Let us define: EQ TM = { M 1, M 2 : L(M 1 ) = L(M 2 )}. Theorem EQ TM is undecidable. Sarmad Abbasi (Virtual University) Theory of Computation 30 / 35

31 Equivalence problem for TMs Suppose that EQ TM is decidable and R decides EQ TM. E TM is the problem of determining whether the language of a TM is empty. EQ TM is the problem of determining whether the languages of two TMs are the same. If one of these languages is forced to be the empty set, we end up with the problem of determining whether the language of the other machine is empty. E TM is a special case of EQ TM. Sarmad Abbasi (Virtual University) Theory of Computation 31 / 35

32 Equivalence problem for TMs Suppose R decides EQ TM. We design S to decide E TM. S gets M as an input. Let M 0 be a machine that rejects all inputs. Sarmad Abbasi (Virtual University) Theory of Computation 32 / 35

33 Equivalence problem for TMs On input M 1 Run R on M, M 0 2 If R accepts, accept. If it rejects, reject. It is easy to see that the language of S is E TM. As, E TM is undecidable so is EQ TM. Sarmad Abbasi (Virtual University) Theory of Computation 33 / 35

34 Linear Bounded Automata An LBA L is a restricted type of TM. Its tape head is not permitted to move off the portion of the tape containing the input. Here is a diagram of an LBA: Sarmad Abbasi (Virtual University) Theory of Computation 34 / 35

35 Linear Bounded Automata As you can see the input portion of the tape is the only tape that the LBA is allowed to move its head on. Sarmad Abbasi (Virtual University) Theory of Computation 35 / 35