1. Computability Chapter 5

2. Overview  Turing Machine (TM) considered to be the most general computational model that can be devised (Church-Turing thesis)  TMs are a litmus test for computational feasibility – If a task can be done on a Turing machine, it is feasible If it cannot, it is considered infeasible There are well-defined tasks, notwithstanding, that cannot be done on a TM  Different flavors of TM have been devised Multi-tape, multi-head TMs Nondeterministic TMs All can be shown to be equally powerful vis-à-vis feasibility  Universal TM can be defined

3. The Standard Turing Machine Model

4. The Standard Turing Machine Model (cont.)  Two modes of TM use vis-à-vis languages:

5. Programming the Turing Machine  Programs can be expressed as a functional table:  Or as a transition diagram: Start q1q1 h ,, 0,R 1,R

6. Programming the Turing Machine (cont.)  Another program expressed as a functional table: q1q1 Start q2q2 q3q3 q4q4 q5q5 h 0,  1,  ,, 0,R 1,R ,R,R 0,R 1,R ,R,R 0,0 1,0 ,0 0,1 1,1 ,1  And as a transition diagram:

7. Extensions to the Standard Turing Machine Model  Double-ended tape Turing machine: tape is of indefinite length on both sides   

8. Extensions to the Standard Turing Machine Model (cont.)  Can be simulated by a standard TM whose one, single-ended tape has two tracks: upper track recording right half of the double-ended tape and lower track recording left half Control Unit Tape Unit   

9. Extensions to the Standard Turing Machine Model (cont.) THEOREM 5.2.0 For each double-ended Turing machine M d there is a standard Turing machine M such that a terminating T-step computation by M d can be simulated in O(T 2 ) steps by M.  Proof (sketch): Simulate each move of the double-ended tape TM M d on a standard TM M by  scanning the tape to determine the position of and symbol scanned by the head of the double-ended TM M d  O(T); and then  simulating the operation to mimic what M d would do  O(1) Control Unit Tape Unit   

10. Extensions to the Standard Turing Machine Model (cont.)  k-tape Turing machine: one control unit and k single-ended tapes of indefinite length

11. Extensions to the Standard Turing Machine Model (cont.)  k-tape Turing machine: can be simulated by a standard Turing machine whose one tape has k tracks and a large tape alphabet Control Unit

12. Extensions to the Standard Turing Machine Model (cont.)  Proof (sketch): Simulate each move of the k-tape TM on the standard TM by  scanning its single tape and collecting information that the control unit of the k- tape TM would “know” from each of its k tape heads  O(T); and then  simulating the move of the k-tape TM by altering its single tape at the appropriate cell  O(T) Control Unit

13. Extensions to the Standard Turing Machine Model (cont.)  Proof (sketch): Simulate the moves of the nondeterministic TM M ND on input w with a standard TM M D by using three tapes  one tape is read-only and contains the input w which is never altered;  another tape is a “work tape” which is used to simulate M ND  yet another tape contains the current “guess” or sequence of choices used in simulating M ND Start 1,1 1,R

14. Configuration Graphs  configuration allows for “reconstruction” of TM and its current state from a string representation x1x1 x2x2 xjxj xnxn x j+1 xj1xj1       p state

15. Configuration Graphs (cont.)  configuration allows for “reconstruction” of the k-tape TM and its current state from a string representation x1x1    x2x2 xkxk  p state h1h1 h2h2 hkhk

16. Configuration Graphs (cont.) initial state configuration final state configuration

17. Phrase-Structure Languages and Turing Machines

18. Phrase-Structure Languages and Turing Machines (cont.)  Language generated by this grammar is L(G) = {a n b n c n |n  1}  Sample derivation S  aSBC  aaBCBC  aaBBCC  aabBCC  aabbCC  aabbcC  aabbcc (a) (b) (c) (d) (e) (f) (g)

19. Phrase-Structure Languages and Turing Machines (cont.)  Proof: Design a phrase-structure grammar that would capture the (d)evolution of the acceptance of a string w by a TM.  starting with the acceptance configuration of [h  1  ], grammar rules will allow us to trace backwards towards the start configuration of [s  w 1 w 2 …w n  ]  yet other grammar rules will then strip extraneous symbols such that only the string w 1 w 2 …w n remains  establish that any string w accepted by some TM can be generated by a phrase- structure grammar establishing the theorem [s w1w2…wn][s w1w2…wn] [h1][h1] Configuration graph of TM acceptance of w S  [h  1  ]  …  [s  w 1 w 2 …w n  ] …  w1w2…wn …  w1w2…wn

20. Phrase-Structure Languages and Turing Machines (cont.)  Proof: Design a phrase-structure grammar that would capture the (d)evolution of the acceptance of a string w by a TM.  Let deterministic TM be M = ( ,Q, ,s,h)  The desired grammar G = ( N, T, R,s) is such that N = Q  {s, ,[,]} T =  and the grammar rules are:

21. Phrase-Structure Languages and Turing Machines (cont.)  Proof: Design a TM that would accept a string w if it is possible to generate it with the production rules of a phrase-structure grammar G and “hang” when it isn’t possible to do so  TM will be a 2-tape nondeterministic TM  One tape will hold the candidate string w; the other will be a “work” tape w    w’w’ Control Unit input tape work tape  TM nondeterministically simulates the production of string w’ by the phrase- structure grammar G on the work tape  If no production produces w, TM enters into an infinite loop

22. Universal Turing Machines  allows for the complete and accurate definition and description of a standard TM with a string

23. Universal Turing Machines (cont.)  Previous example would be encoded canonically as: q1q1 Start q2q2 q3q3 q4q4 q5q5 h 0,  1,  ,, 0,R 1,R ,R,R 0,R 1,R ,R,R 0,0 1,0 ,0 0,1 1,1 ,1

24. Universal Turing Machines (cont.)  A Universal Turing Machine (UTM) is capable of simulating an arbitrary Turing Machine M on an arbitrary input word w.  One possible UTM is a two-tape TM that may be described as: One tape is a work tape, starts out as containing w; Second tape is the program tape, starts out as containing  (M), the canonical encoding of M Tape symbol alphabet is: w    (M)(M) Control Unit work tape program tape

25. Universal Turing Machines (cont.)  A Universal Turing Machine (UTM) is capable of simulating an arbitrary Turing Machine M on an arbitrary input word w.  One possible UTM is a two-tape TM that may be described as: The UTM simulates M by locating on the program tape the appropriate transition given the current state of M and the symbol on the work tape being scanned by its head When the transition is located, the UTM carries out the operation on the work tape There is a standard UTM equivalent to this two-tape UTM w    (M)(M) Control Unit work tape program tape

26. Encoding of Strings and Turing Machines  Any set of strings can be arranged in lexicographic order based on the “order” of the symbols as listed in its alphabet set  = { , , }S: {, , , ,  } S’: { , , , , }  Turing machines can be listed in lexicographic order based on the “order” of their canonical encodings (strings over some alphabet)  Generating the jth Turing machine can be mechanized TM Generator j (Mj)(Mj)

27. Limits on Language Acceptance  A language L is decidable (or recursive) if there is a TM that halts on all inputs and accepts just the strings of L  A language L is recursively enumerable if there is a TM that accepts just the strings of L, possibly not halting on strings not in L.  A language L is unsolvable if it is recursively enumerable but not decidable

28. Limits on Language Acceptance (cont.)  The following results identify three languages that are decidable:  The following result identifies a language that is not recursively enumerable: Proof: By contradiction and use of diagonalization.

29. Limits on Language Acceptance (cont.)  The following results identify a language that is unsolvable (i.e., recursively enumerable but not decidable): Proof: Reclassify states, switching accepting and non-accepting. Proof: Using a Universal Turing machine, run M i with input w i. This shows the language is recursively enumerable. Since its complement is not recursively enumerable, therefore not decidable, it cannot be itself decidable.

30. Reducibility and Unsolvability  Use the “traditional” technique of reduction to establish that other languages are unsolvable: By contradiction: assume that an algorithm A exists for candidate language L then show we can use A to obtain an algorithm for a language previously known as unsolvable A (solves L) R (translates L to A) input to L input to A output Algorithm for L

31. Reducibility and Unsolvability (cont.)  Existence of such a “reducing algorithm” means one can mechanize the acceptance of one language by using the mechanism of accepting the other language:

32. Reducibility and Unsolvability (cont.)  This lemma is useful in arriving at unsolvability results: A (solves L 2 ) R (translates L 1 to L 2 ) input to L 1 input to L 2 output “Algorithm” for L 1

33. Reducibility and Unsolvability (cont.)  Use this reducibility lemma to establish that the following language is unsolvable (i.e., recursively enumerable but not decidable): Proof: Use encoding of M and w as input to universal Turing machine and simulate M with w as input. This establishes recursive enumerability. Use Lemma 5.8.1 to establish non-decidability by reducing L 1 to L H. The “reducing algorithm” generates the ith input string and the ith Turing machine and uses those as input to “imaginary” machine deciding L H and then “reverses” the polarity of the output. This decides L 1.