Prof. Busch - LSU1 Undecidable Problems (unsolvable problems)

Prof. Busch - LSU2 Recall that: A language is decidable, if there is a Turing machine (decider) that accepts the language and halts on every input string Turing Machine Input string Accept Reject Decider for Decision On Halt: Decidable Languages YES NO

Prof. Busch - LSU3 A computational problem is decidable if the corresponding language is decidable We also say that the problem is solvable

Prof. Busch - LSU4 Problem: Does DFA accept the empty language ? Corresponding Language: Description of DFA as a string (For example, we can represent as a binary string, as we did for Turing machines) (Decidable)

Prof. Busch - LSU5 Determine whether there is a path from the initial state to any accepting state Decider for : On input : DFA Reject Decision: Accept

Prof. Busch - LSU6 Problem: Does DFA accept a finite language? Corresponding Language:(Decidable)

Prof. Busch - LSU7 Decider for : On input : DFA Reject Decision: Accept Check if there is a walk with a cycle from the initial state to an accepting state infinitefinite (NO) (YES)

Prof. Busch - LSU8 Problem: Does DFA accept string ? Corresponding Language:(Decidable)

Prof. Busch - LSU9 Decider for : On input string : Run DFA on input string If accepts Then accept (and halt) Else reject (and halt)

Prof. Busch - LSU10 Problem: Do DFAs and accept the same language? Corresponding Language:(Decidable)

Prof. Busch - LSU11 Let be the language of DFA Decider for : On input : Construct DFA such that: (combination of DFAs)

Prof. Busch - LSU12 and

Prof. Busch - LSU13 or

Prof. Busch - LSU14 Therefore, we only need to determine whether which is a solvable problem for DFAs:

Prof. Busch - LSU15 Undecidable Languages there is no Turing Machine which accepts the language and makes a decision (halts) for every input string undecidable language = not decidable language There is no decider: (machine may make decision for some input strings)

Prof. Busch - LSU16 For an undecidable language, the corresponding problem is undecidable (unsolvable): there is no Turing Machine (Algorithm) that gives an answer (yes or no) for every input instance (answer may be given for some input instances)

Prof. Busch - LSU17 We have shown before that there are undecidable languages: Decidable Turing-Acceptable is Turing-Acceptable and undecidable

Prof. Busch - LSU18 We will prove that two particular problems are unsolvable: Membership problem Halting problem

Prof. Busch - LSU19 Membership Problem Input:Turing Machine String Question: Does accept ? Corresponding language:

Prof. Busch - LSU20 Theorem: (The membership problem is unsolvable) Proof: We will assume that is decidable; We will then prove that every Turing-acceptable language is also decidable is undecidable Basic idea: A contradiction!

Prof. Busch - LSU21 Suppose that is decidable YES accepts NO rejects Decider for Input string

Prof. Busch - LSU22 Let be a Turing recognizable language Let be the Turing Machine that accepts We will prove that is also decidable: we will build a decider for

Prof. Busch - LSU23 NO YES accept reject Decider for (and halt) Decider for Input string accepts ? String description of This is hardwired and copied on the tape next to input string s, and then the pair is input to H

Prof. Busch - LSU24 Therefore,is decidable But there is a Turing-Acceptable language which is undecidable Contradiction!!!! Since is chosen arbitrarily, every Turing-Acceptable language is decidable END OF PROOF

Prof. Busch - LSU25 We have shown: Decidable Undecidable

Prof. Busch - LSU26 We can actually show: Decidable Turing-Acceptable

Prof. Busch - LSU27 Turing machine that accepts : 1. Run on input 2. If accepts then accept is Turing-Acceptable

Prof. Busch - LSU28 Halting Problem Input:Turing Machine String Question: Does halt while processing input string ? Corresponding language:

Prof. Busch - LSU29 Theorem: (The halting problem is unsolvable) Proof: Suppose that is decidable; we will prove that every Turing-acceptable language is also decidable is undecidable Basic idea: A contradiction!

Prof. Busch - LSU30 YES halts on input doesn’t halt on input NO Suppose that is decidable Decider for Input string

Prof. Busch - LSU31 Let be a Turing-Acceptable language Let be the Turing Machine that accepts We will prove that is also decidable: we will build a decider for

Prof. Busch - LSU32 halts and accepts halts on ? YES NO Run with input reject accept reject Decider for Input string Decider for and halt halts and rejects and halt

Prof. Busch - LSU33 Contradiction!!!! END OF PROOF Therefore,is decidable But there is a Turing-Acceptable language which is undecidable Since is chosen arbitrarily, every Turing-Acceptable language is decidable

Prof. Busch - LSU34 Theorem: Proof: Assume for contradiction that the halting problem is decidable; (The halting problem is unsolvable) is undecidable we will obtain a contradiction using a diagonilization technique An alternative proof Basic idea:

Prof. Busch - LSU35 YEShalts on doesn’t halt on NO Suppose that is decidable Decider for Input string

Prof. Busch - LSU36 Input string: YES NO Looking inside Decider for halts on ?

Prof. Busch - LSU37 NO Loop forever YES Construct machine : If halts on input Then Loop Forever Else Halt halts on ?

Prof. Busch - LSU38 Construct machine : Copy on tape If halts on input Then loop forever Else halt

Prof. Busch - LSU39 Run with input itself Copy on tape If halts on input Then loops forever on input Else halts on input END OF PROOF CONTRADICTION!!!

Prof. Busch - LSU40 We have shown: Decidable Undecidable

Prof. Busch - LSU41 We can actually show: Decidable Turing-Acceptable

Prof. Busch - LSU42 Turing machine that accepts : 1. Run on input 2. If halts on then accept is Turing-Acceptable