Decidable and undecidable languages

Transcription

1 The Chinese University of Hong Kong Fall 2011 CSCI 3130: Formal languages and automata theory Decidable and undecidable languages Andrej Bogdanov

2 Problems about automata Does a b a q 0 q 1 b accept input abb? We can formulate this question as a language: A DFA = { D, w : D is a DFA that accepts input w} Is A DFA decidable? ((q0,q1)(a,b)((q0,a,q0)(q0,b,q1)(q1,a,q0)(q1,b,q1)(q0)(q1))(abb) D = (Q, Σ, δ, q 0, F) w

3 Problems about automata A DFA = { D, w : D is a DFA that accepts input w} pseudocode: On input D, w, where D = (Q, Σ, δ, q 0, F): Set q := q 0 For i := 1 to length(w): q := δ(q, w i ) If q F accept, else reject TM description: On input D, w, where D is a DFA, w is a string Simulate D on input w If simulation ends in acc state, accept. Otherwise, reject.

4 Problems about automata A DFA = { D, w : D is a DFA that accepts input w} Turing Machine details: Check input is in correct format. (Transition function is complete, no duplicate transitions) Perform simulation: state input symbol ((q0,q1)(a,b)((q0,a,q0)(q0,b,q1)(q1,a,q0)(q1,b,q1)(q0)(q1))(abb) q acc

5 Problems about automata A DFA = { D, w : D is a DFA that accepts input w} Turing Machine details: Check input is in correct format. (Transition function is complete, no duplicate transitions) Perform simulation: Put markers on start state of D and first symbol of w Until marker for w reaches last symbol: Update both markers If state marker is on accepting state, accept. Else reject.

6 Acceptance problems about automata A DFA = { D, w : D is a DFA that accepts input w} A NFA = { N, w : N is an NFA that accepts w} A REX = { R, w : R is a regular expression that generates w} Which of these is decidable?

7 Acceptance problems about automata A DFA = { D, w : D is a DFA that accepts input w} The following TM decides A DFA : M := On input D, w, where D is a DFA and w is a string Simulate D on input w If the simulation ends in acc state of D, accept. If it doesn t, reject.

8 Acceptance problems about automata A NFA = { N, w : N is an NFA that accepts input w} The following TM decides A NFA : N := On input N, w, where N is an NFA and w is a string Convert N to a DFA D using the conversion procedure from Lecture 2 Run the TM M for A DFA on input D, w If M accepts, accept. Otherwise, reject.

9 Acceptance problems about automata A REX = { R, w : R is a regular expression that generates w} The following TM decides A REX : P := On input R, w, where R is a reg exp and w is a string Convert R to an NFA N using the conversion procedure from Lecture 4 Run the TM N for A NFA on input N, w If N accepts, accept. Otherwise, reject.

10 Other problems about automata MIN DFA = { D : D is a minimal DFA} The following TM decides MIN DFA : R := On input D, where D is a DFA Run the distinguishable states algorithm If every pair of states is distinguishable, accept. Otherwise, reject.

11 Other problems about automata EQ DFA = { D 1, D 2 : D 1, D 2 are DFAs and L(D 1 ) = L(D 2 )} The following TM decides EQ DFA : S := On input D 1, D 2, where D 1 and D 2 are DFAs Run the DFA minimization algorithm on D 1 to obtain a DFA D 1 Run the DFA minimization algorithm on D 2 to obtain a DFA D 2 If D 1 = D 2 accept, otherwise reject.

12 Other problems about automata E DFA = { D : D is a DFAs and L(D) is empty} The following TM decides E DFA : T := On input D, where D is a DFA Run the TM S for EQ DFA on input D, If S accepts accept, otherwise reject.

13 Problems about context-free grammars A CFG = { G, w : G is a CFG that generates w} V := On input G, w, where G is a CFG and w is a string Eliminate the nullable and unit productions from G Convert G to Chomsky Normal Form Run the Cocke-Younger-Kasami algorithm on G, w If the CYK algorithm produces a parse tree, accept. Otherwise, reject.

14 Decidability of context-free languages Every context-free language is decidable. Let L be a context-free language. There is a CFG G for L. The following TM decides L: M G := On input w, Run TM V on input G, w. If V accepts accept, otherwise reject.

15 Are all languages about CFGs decidable? EQ CFG = { G 1, G 2 : G 1 and G 2 are context-free grammars that generate the same strings} What s the difference between EQ DFA and EQ CFG? To decide EQ DFA, we minimized both DFAs But there is no method that, given a CFG or PDA, produces a unique equivalent minimal CFG or PDA

16 The universal Turing Machine and undecidability

17 Turing Machines versus computers program data computer output A computer is a machine that manipulates data according to a list of instructions. How does a Turing Machine take a program as part of its input?

18 The Universal Turing Machine program M input x for M U whatever M does on x The universal TM U takes as inputs a program M and a string x and simulates M on x The program M itself is specified as a TM!

19 Turing Machines as strings / R q q a q r M A Turing Machine is (Q, Σ, Γ, δ, q 0, q acc, q rej ) This Turing Machine can be described by the string M = (q,qa,qr)(0,1)(0,1, ) ((q,q, / R) (q,qa,0/0r) (q,qr,1/1r)) (q)(qa)(qr)

20 The universal Turing Machine U (q,qa,qr)(0,1)(0,1, 001 program M input w for M U := On input M, w, Simulate M on input w If M enters accept state, accept. If M enters reject state, reject.

21 Acceptance of Turing Machines A TM = { M, w : M is a TM that accepts w} U := On input M, w, Simulate M on input w M accepts w M rejects w M loops on w U accepts M, w U rejects M, w U loops on M, w TM U recognizes but does not decide A TM

22 Recognizing versus deciding q acc q rej accept reject loop halt A TM decides language L if it recognizes L and halts (does not loop) on every input

23 Undecidability Turing s Theorem: The language A TM is undecidable. Before we show this, let s observe one thing: A Turing Machine M can be given its own description M as an input!

24 Proof of Turing s Theorem Proof by contradiction: Suppose A TM is decidable. Then there is a TM H that decides A TM : M, M, M w accept if M accepts wm H What happens when w = M? reject if M rejects w M or M loops on w M

25 Proof of undecidability M, M accept if M accepts M H reject if M rejects M or M loops on M Let H be a TM that does the opposite of H H acc rej H acc To go from H to H, rej just switch its accept and reject states

26 Proof of undecidability M, M accept if M accepts M H reject if M rejects M or M loops on M M, M H accept if M rejects M or M loops on M reject if M accepts M

27 Proof of undecidability M, M H accept if M rejects M or M loops on M reject if M accepts M Let D be the following TM: M M, M copy H

28 Proof of undecidability M D D if DM rejects M D accept or DM loops on M D reject if MD accepts M D What happens when M = D? If D accepts D then D rejects D If D rejects D then D accepts D H never loops, so D never loops either so D does not exist!

29 Proof of undecidability: conclusion Proof by contradiction We assumed A TM was decidable Then we built Turing Machines H, H, D But D does not exist! Conclusion The language A TM is undecidable.

30 What happened? all possible inputs w ε all possible Turing Machines M 1 M 2 M 3 acc rej rej acc rej acc loop rej rej loop rej rej We can write an infinite table for every pair (M, w)

31 What happened? M 1 M 2 M 3 M 4 M 1 M 2 M 3 acc loop rej acc rej acc rej acc loop rej loop rej Now let s look only at those w that describe a TM M

32 What happened? M 1 M 2 M 3 M 4 M 1 M 2 acc loop rej acc rej acc rej acc D rej rej acc rej If A TM is decidable, then TM D is in the table

33 What happened? M 1 M 2 M 3 M 4 M 1 M 2 acc loop rej acc rej acc rej acc D rej rej acc rej D does the opposite of the diagonal entries M D accept if M rejects or loops on M reject if M accepts M

34 What happened? M 1 M 2 M 1 M 2 M 3 M 4 acc loop rej acc rej acc rej acc D loop acc D rej rej acc rej? We run into trouble when we look at (D, D )!

35 Unrecognizable languages The language A TM is recognizable but not decidable. How about languages that are not recognizable? A TM = { M, w : M is a TM that does not accept w} = { M, w : M rejects or loops on input w} The language A TM is not recognizable.

36 Unrecognizable languages Theorem If L and L are both recognizable, then L is decidable. We know A TM is recognizable, so if A TM were also, then A TM would be decidable But Turing s Theorem says A TM is not decidable

37 Unrecognizable languages If L and L are both recognizable, then L is decidable. Proof idea w M accept if w L rej/loop if w L accept w w accept if w L M rej/loop if w L reject

38 Unrecognizable languages If L and L are both recognizable, then L is decidable. Let M = TM for L, M = TM for L On input w, 1 2 Problem: If M loops on w, we will never get to step 2! Simulate M on input w. If it accepts, accept. Simulate M on input w. If it accepts, reject.

39 Bounded simulation If L and L are both recognizable, then L is decidable. Turing Machine that decides L: Let M = TM for L, M = TM for L On input w, For t := 0 to infinity Do t transitions of M on w. If it accepts, accept. Do t transitions of M on w. If it accepts, reject.

40 The Chinese University of Hong Kong Fall 2011 CSCI 3130: Formal languages and automata theory Reducibility Andrej Bogdanov

41 Summary of last lecture program M input w for M U accept, if M accepts w reject, if M rejects w loops, if M loops on w The universal TM U takes as inputs a program M and a string x and simulates M on x The program M itself is specified as a TM!

42 Summary of last lecture q acc q rej accept reject loop halt A TM = { M, w : M is a TM that accepts input w} recognizable: undecidable: By universal TM U Turing s Theorem

43 Another undecidable language HALT TM = { M, w : M is a TM that halts on input w} HALT TM is an undecidable language We will argue that If HALT TM is decidable, so is A TM. but by Turing s Theorem it is not.

44 Undecidability of halting If HALT TM can be decided, so can A TM. Suppose R decides HALT TM M, w R accept if M halts on w reject if M loops on w We want a TM S that decides A TM M, w accept if M accepts w? reject if M rejects or loops on w

45 Undecidability of halting M, w R accept if M halts on w reject if M loops on w M, w R acc simulate U M on w acc rej S accept if M accepts w rej reject if M rejects or loops on w

46 Undecidability of halting HALT TM = { M, w : M is a TM that halts on input w} A TM = { M, w : M is a TM that accepts input w} Suppose that HALT TM is decidable. Let H be a TM that decides HALT TM. Then this TM decides A TM : S := On input M, w, Run R on input M, w. If R rejects, reject. If R accepts, run U on input M, w If U accepts, accept. If U rejects, reject. Impossible, because A TM is undecidable.

47 Reducibility Suppose you think L is undecidable. How do you know for sure? Show: If some TM R decides L then using R, I can build another TM S so that S decides A TM...but this is impossible, because A TM is undecidable.

48 Example 1 AEPS TM = { M : M is a TM that accepts input ε} decidable undecidable Step 1: You gotta believe it To know if M accepts ε, it looks like we have to simulate it But then we might end up in a loop Step 2: Prove it!

49 Example 1: Figuring out the reduction M R accept if M accepts ε reject if not M, w M R S accept if M accepts w reject if not M should be a Turing Machine such that: If M accepts w, then M accepts ε M on input ε = M on input w If M does not accept w, then M does not accept ε

50 Example 1: Implementing the reduction M, w construct M M M should be a Turing Machine such that: M on input ε = M on input w M := On input z, Simulate M on input w If M accepts w, accept Otherwise, reject

51 Example 1: Impementing the reduction M, w construct M M R accept if M accepts w reject if not S := On input M, w : Construct the following TM M : M := On input z, Simulate M on input w and output answer Run R on input M and output its answer.

52 Example 1: Writing it up AEPS TM = { M : M is a TM that accepts input ε} A TM = { M, w : M is a TM that accepts input w} Suppose AEPS TM is decidable and R decides it. Let S := On input M, w where M is a TM and w is a string Construct the following TM M : M := On input z, Simulate M on input w and output answer Run R on input M and output its answer. Then S accepts M, w if and only if M accepts w So S decides A TM, which is impossible.

53 Example 2 SOME TM = { M : M is a TM that accepts some input} decidable undecidable Step 1: You gotta believe it To know if M is in SOME TM, looks like we have to simulate But then we might end up in a loop Step 2: Prove it!

54 Example 2: Setting up the reduction SOME TM = { M : M is a TM that accepts some input} Show: If SOME TM can be decided by some TM R, M R accept if M accepts some input reject if M accepts no inputs... then A TM can be decided by some other TM S M, w S accept if M accepts w reject if not

55 Example 2: Setting up the reduction M, w construct M M R accept if M accepts w reject if not Task Given M, w, construct M so that: If M accepts w, then M accepts some input If M does not accept w, then M accepts no inputs

56 Example 2: Implementing the reduction Task Given M, w, construct M so that: If M accepts w, then M accepts some input If M does not accept w, then M accepts no inputs M := On input z, Simulate M on input w If M accepts, accept Otherwise, reject

57 Example 2: Writing it up SOME TM = { M : M is a TM that accepts some input} A TM = { M, w : M is a TM that accepts input w} Suppose SOME TM is decidable and R decides it. Let S := On input M, w where M is a TM and w is a string Construct the following TM M : M := On input z, Simulate M on input w and output answer Run R on input M and output its answer. Then S accepts M, w if and only if M accepts w So S decides A TM, which is impossible.

58 Example 3 E TM = { M : M is a TM that accepts no input} decidable undecidable Show: If E TM can be decided by some TM R,... then SOME TM can be decided by another TM S SOME TM = { M : M is a TM that accepts some input}

59 Example 3 E TM = { M : M is a TM that accepts no input} SOME TM = { M : M is a TM that accepts some input} Suppose E TM can be decided by some TM R. Consider the following TM S: S := On input M where M is a TM Run R on input M. If R accepts, reject. If R rejects, accept. Then S decides SOME TM, a contradiction.

60 Example 4 EQ TM = { M 1, M 2 : M 1 and M 2 accept the same inputs} decidable undecidable Show: If EQ TM can be decided by some TM R,... then E TM can be decided by another TM S

61 Example 4: Setting up the reduction EQ TM = { M 1, M 2 : M 1 and M 2 accept the same inputs} E TM = { M : M is a TM that accepts no input} Task Idea Given M, construct M 1, M 2 so that: If M accepts no input, then M 1, M 2 accept same inputs If M accepts some input, then M 1, M 2 do not accept same inputs Make M 1 = M Make M 2 accept nothing

62 Example 2: Writing it up EQ TM = { M 1, M 2 : M 1 and M 2 accept the same inputs} E TM = { M : M is a TM that accepts no input} Suppose EQ TM is decidable and R decides it. Let S := On input M where M is a TM Construct the following TM M 2 : M 2 := On input z, reject. Run R on input M, M 2 and output its answer. Then S accepts M if and only if M accepts no input So S decides E TM, which is impossible.