The Turing Machine Motivating idea Computability Build a theoretical a “human computer” Likened to a human with a paper and pencil that can solve problems in an algorithmic way The theoretical machine provides a means to determine: If an algorithm or procedure exists for a given problem What that algorithm or procedure looks like How long would it take to run this algorithm or procedure. The Church-Turing Thesis Theory Hall of Fame (1936) Alan Turing Any algorithmic procedure that can be 1912 – 1954 carried out by a human or group of b. London, England. humans can be carried out by some Turing Machine” PhD – Princeton (1938) Research Equating algorithm with running on a TM Cambridge and Manchester Turing Machine is still a valid U. computational model for most modern National Physical Lab, UK computers. Creator of the Turing Test Theory Hall of Fame Undecidability Alonso Church Informally, a problem is called unsolvable or 1903 -- 1995 undecidable if there no algorithm exists that b. Washington D.C. solves the problem. Algorithm PhD – Princeton (1927) Implies a TM that computes a solution for the Mathematics Prof (1927 – problem 1967) Solves Advisor to both Turing and Implies will always give an answer Kleene 1
Decision Problem Decision Problem Let’s formalize this a bit Running a decision problem on a TM. A decision problem is a problem that has a The problem must first be encoded yes/no answer Example: Is a given string x a palindrome (Is x ∈ pal?) Example: x is an instance of the problem Is a given context free language empty? Is a given string x a palindrome (Is x ∈ pal?) Instance of a problem is a CFG…must be encoded. Is a given context free language empty? Decision Problem Solvability Running a decision problem on a TM. In other words, a problem is solvable if Once encoded, the encoded instance in provided the language of all of its encoded “yes” as input to a TM. instances is recursive. The TM must then Determine if the input is a valid encoding There is a TM that recognizes the Run, halt, language. Place 1 on the tape if the answer for the input is yes Place 0 on the tape if the answer for the input is no If such a TM exists for a given decision problem, the problem is decidable or solvable. Otherwise the problem is called undecidable or unsolvable. Universal Language An unsolvable problem L u corresponds to the “yes encodings” Universal Language (L u ) of the decision problem: Set of all strings w i such that w i ∈ L(M i ) All strings w that are accepted by the TM Given a Turing Machine M, does it accept with w as it’s encoding. it’s own encoding. (Self-accepting) All encodings for TMs that do accept their encoding when input Since L u is not recursive, this problem is unsolvable. We showed that L u is not recursive. 2
Reducing one language to Reducing one language to another another One method of showing whether a Formally, given decision problem is unsolvable is Let L 1 and L 2 be languages over Σ 1 and Σ 2 to convert the encoding of the problem We say L 1 is reducible to L 2 if into another that we know to be either There exists a Turning computable function solvable or unsolvable. f: Σ 1 * → Σ 2 * such that This is called reducing one language to x ∈ L 1 iff f(x) ∈ L 2 another. Reducing one language to Reducing one language to another another Informally, We can take any encoded instance of one Instance problem Corresponding Conversion of P 1 TM Use a TM to compute a corresponding encoded TM Instance of P 2 recognizing instance of another problem. L 2 If this other problem has a TM that recognizes the set of “yes encodings”, we can run that TM to solve the first problem. YES NO Reducing one language to The halting problem another Key facts: Let’s consider a more general problem about TMs. If L 1 is reducible to L 2 then If L 2 is recursive then L 1 is also recursive If L 1 is not recursive then L 2 is not recursive. Given a TM, M, and a string w, is w ∈ T(M)? If P 1 and P 2 are decision problems with L 1 Given a TM, M and a string w and L 2 the languages of “yes encodings” Will M halt and accept on input w? respectively and if L 1 is reducible to L 2 then If P 2 is solvable then P 1 is also solvable We simply cannot just run the string on the TM If P 1 is unsolvable then P 2 is also unsolvable since if w ∉ L(M), M might go into an infinite loop. 3
To show a problem is The halting problem unsolvable The halting problem is unsolvable Find a problem known to be unsolvable Reduce this known unsolvable problem to the problem you wish to show is Proof: unsolvable. We can use an argument similar to that used to show that L u is not recursive. Instead, let’s use reduction Only need one to start the ball rolling Self-accepting fits the bill. Halting problem Halting problem Reduce self-accepting to halting Reduction Self-accepting Take an instance of SA and convert it to an instance of Halting Turing Machine M Does this Machine accept it’s own encoding Such that a “yes” instance of SA results in a “yes” instance of halt Halting Turing Machine M M Reduction M, e(M) String w machine Does M halt and accept on input w Halt SA State entry problem Halting problem Reduction Given: Assume Halting is solvable Turing Machine M A state q YES A string w M Reduction M, e(M) Halt TM Problem: machine Halt SA Will M enter state q on input w. The State Entry Problem is unsolvable. NO Then SA must be solvable…CONTRADICTION! 4
State entry problem State entry problem Reduce halting to state entry (SE). Reduction Halting Take an instance of Halting (M 1 , w 1 ) and Turing Machine M convert it to an instance of state-entry (M 2 , String w w 2 , q) Does M halt and accept on input w Such that a “yes” instance of halt results in a State Entry “yes” instance of self-entry Turing Machine M From M 1 create M 2 such that M 1 halts iff M 2 String w enters state q. State q Does this Machine enter state q on input w. State entry problem State entry problem Reduction Reduction M 1 will halt only there is no transition Take an instance of Halting and convert it defined (e.g. δ (q i , a) ) to an instance of state entry Take M 1, create a new state q. Define new transition in M 2 for each undefined transition in M 1 so M 1 , w Reduction M 2 , w, q (q i , a) = (q, a R) machine State-entry If M 1 halts, M 2 will enter state q Halt Strings of same length problem State entry problem Reduction Given: Assume State Entry is solvable Turing Machine M Problem: YES Will M enter accept two strings of the same M 1 ,w Reduction M 2 , w, q length. State machine Entry TM State The Same Length Problem is Halt Entry unsolvable. NO Then Halting must be solvable…CONTRADICTION! 5
Same length problem Same length problem Reduction Reduce halting to same length. Halting Take an instance of Halting (M 1 , w 1 ) and convert it to an instance of state-entry Turing Machine M String w (M w ) Does M halt and accept on input w Such that a “yes” instance of halt results in a State Entry “yes” instance of same length Turing Machine M From M 1 create M w such that M 1 halts on w iff M w accepts string of same length Does this Machine accept 2 strings of the same length. Same length problem Same length problem Reduction (q i , a) will force M w to accept “a” and “b” We can certainly create TMs that accept “a” and M 1 will halt only there is no transition “b”. defined (e.g. δ (q i , a) ) M w will do the following: Define new transition in M w for each Copy w onto it’s tape (after the input) Place the tape head at the start of w undefined transition in M 1 so Simulate M 1 (q i , a) will force M 2 to accept a and b For all halting configurations, transition back to the start of the input (where you can accept both “a” and “b”). If M 1 halts on w, M w accept “a” and You will only get back to the start of input of M w if M 1 “b” (strings of same length) halts on input w. Same length problem Same length problem Reduction Reduction Take an instance of Halting and convert it Assume Same length is solvable to an instance of same length YES M 1 ,w Reduction M w Same machine Length Same M 1 , w Reduction M w Halt TM length machine Same length Halt NO Then Halting must be solvable…CONTRADICTION! 6
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