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  1. 15-453 FORMAL LANGUAGES, AUTOMATA, AND COMPUTABILITY * Read chapter 4 of the book for next time * Lecture9x.ppt

  2. REVIEW

  3. A Turing Machine is represented by a 7-tuple T = (Q, Σ, Γ, , q0, qaccept, qreject): Q is a finite set of states Σ is the input alphabet, where   Σ Γ is the tape alphabet, a superset of Σ;   Γ  : Q Γ→ Q Γ  {L, R}is the transition func q0 Q is the start state qaccept Q is the accept state qreject Q is the reject state, and qreject qaccept

  4. CONFIGURATION:(1) tape contents, (2) current state, (3) location of read/write head. 11010q700110

  5. A TM recognizesa language iff it accepts all and only those strings in the language. A language L is called Turing-recognizable or recursively enumerable iff some TM recognizes L. A TM decides a language L iff it accepts all strings in L andrejects all strings not in L. A language L is called decidable or recursiveiff some TM decides L.

  6. A language is called Turing-recognizable or recursively enumerable (r.e.) if some TM recognizes it. A language is called decidable or recursiveif some TM decides it.

  7. Theorem: If A and A are r.e. then A is decidable. Given: a Turing Machine TMA that recognizes A and a Turing Machine TMR that recognizes A, we can build a new machine that decides A. How can we prove this? • Run TMA and TMR in parallel. (Or more precisely, interleave them.) • One of them will eventually recognize the input string. • If TMA recognizes it, then accept. • If TMR recognizes it, then reject.

  8. 2n A TM that decides { 0 | n ≥ 0 } • We want to accept iff: • the input string consists entirely of zeros, and • the number of zeros is a power of 2. • High-Level Idea. • Repeatedly divide the number of zeros in half until it becomes an odd number. • If we are left with a single zero, then accept. • Otherwise, reject.

  9. 2n A TM that decides { 0 | n ≥ 0 } PSEUDOCODE: • Sweep from left to right, cross out every other 0.(Divides number in half.) • If in step 1, the tape had only one 0, accept. • Else if the tape had an odd number of 0’s, reject. • Move the head back to the first input symbol. • Go to step 1.

  10. C = {aibjck | k = i×j, and i, j, k ≥ 1} Example aaabbbbcccccccccccc 3 4 3×4 = 12

  11. C = {aibjck | k = i×j, and i, j, k ≥ 1} • High-Level Idea. • For each occurrence of a: { • For each occurrence ofb: { • Delete an occurrence of c. • } • }

  12. C = {aibjck | k = i×j, and i, j, k ≥ 1} PSEUDOCODE: • If the input doesn’t match a*b*c*, reject. • Move the head back to the leftmost symbol. • Cross off an a, scan to the right until b. Sweep between b’s and c’s, crossing out one of each until all b’s are out. If too few c’s, reject. • Uncross all the b’s. If there’s another aleft, then repeat stage 3.If all a’s are crossed out, Check if all c’s are crossed off. If yes, then accept, else reject.

  13. C = {aibjck | k = i×j, and i, j, k ≥ 1} aabbbcccccc xabbbcccccc xayyyzzzccc xabbbzzzccc xxyyyzzzzzz

  14. TURING-MACHINE VARIANTS Turing machines can be extended in various ways, but so long as a new TM only reads and writes a finite number of symbols in each step,an old TM can still simulate it! Example: Turing machines with multiple tapes. Input comes in on one tape, and other tapes are used for scratch work.

  15. FINITE STATE CONTROL MULTITAPE TURING MACHINES  : Q Γk→ Q Γk  {L,R}k

  16. FINITE STATE CONTROL FINITE STATE CONTROL Theorem: Every Multitape Turing Machine can be transformed into a single tape Turing Machine 1 0 0 . . . # # 1 0 0 #

  17. FINITE STATE CONTROL FINITE STATE CONTROL Theorem: Every Multitape Turing Machine can be transformed into a single tape Turing Machine 1 0 0 . . . # # 1 0 0 #

  18. THE CHURCH-TURING THESIS Anything that can be computed by algorithm (in our intuitive sense of the term “algorithm”) can be computed by a Turing Machine.

  19. We can encode a TM as a string of 0s and 1s start state reject state n states 0n10m10k10s10t10r10u1… m tape symbols (first k are input symbols) blank symbol accept state ( (p, a), (q, b, L) ) = 0p10a10q10b10 ( (p, a), (q, b, R) ) = 0p10a10q10b11

  20. Similarly, we can encode DFAs, NFAs, CFGs, etc. into strings of 0s and 1s So we can define the following languages: ADFA = { (B, w) | B is a DFA that accepts string w } ANFA = { (B, w) | B is an NFA that accepts string w } ACFG = { (G, w) | G is a CFG that generates string w }

  21. ADFA = { (B, w) | B is a DFA that accepts string w } Theorem: ADFA is decidable Proof Idea: Simulate B on w ANFA = { (B, w) | B is an NFA that accepts string w } Theorem: ANFA is decidable ACFG = { (G, w) | G is a CFG that generates string w } Theorem: ACFG is decidable Proof Idea: Transform G into Chomsky Normal Form. Try all derivations of length 2|w|-1

  22. UNDECIDABLE PROBLEMS

  23. w  Σ* w  Σ* TM TM w  L ? w  L ? no yes no yes accept reject or no output accept reject L is semi-decidable (recursively enumerable, Turing-recognizable) L is decidable (recursive) Theorem: L is decidable if both L and L are recursively enumerable

  24. There are languages over {0,1} that are not decidable. If we believe the Church-Turing Thesis, this is major: it means there are things that formal computational models inherently cannot do. We can prove this using a counting argument. We will show there is no functionfrom the set of all Turing Machines ontothe set of all languages over {0,1}. (Works for any Σ.) Then we will prove something stronger: There are semi-decidable (r.e.) languages that are NOT decidable.

  25. Languages over {0,1} Turing Machines

  26. Cantor’s Theorem Let L be any set and 2L be the power set of L Theorem: There is no map from L onto 2L Assume, for a contradiction, that there is an onto map f : L  2L Proof: Let S = { x  L | x  f(x) } We constructed S so that, for every elem x in L, the set S differs from f(x): S ≠ f(x) because x  S iff x  f(x)

  27. Theorem: There is no onto function from the positive integers to the real numbers in (0, 1) Proof: Suppose f is such a function: 0.28347279… 2 1 2 3 4 5 : 0.88388384… 8 0.77635284… 6 0.11111111… 1 5 0.12345678… : 1 if [ nth digit of f(n) ]  1 [ nth digit of r ] = 0 otherwise f(n)  r for all n ( Here, r = 0.11101... )

  28. Sidenote Let Z+ = {1,2,3,4…}. There exists a bijection between Z+ and Z+ Z+ (or Q+) (1,1) (1,2) (1,3) (1,4) (1,5) … (2,1) (2,2) (2,3) (2,4) (2,5) … (3,1) (3,2) (3,3) (3,4) (3,5) … (4,1) (4,2) (4,3) (4,4) (4,5) … (5,1) (5,2) (5,3) (5,4) (5,5) … : : : : : ∙.

  29. THE MORAL: For any set L, 2Lalways has ‘more’ elements than L

  30. Not all languages over {0,1} are decidable, in fact: not all languages over {0,1} are semi-decidable {decidable languages over {0,1}} {semi-decidable langs over{0,1}} {Languages over {0,1}} {Turing Machines} {Strings of 0s and 1s} {Sets of strings of 0s and 1s} Powerset of L: 2L Set L

  31. THE ACCEPTANCE PROBLEM ATM = { (M, w) | M is a TM that accepts string w } Theorem: ATM is semi-decidable (r.e.) but NOTdecidable ATM is r.e. : Define a TM U as follows: On input (M, w), U runs M on w. If M ever accepts, accept. If M ever rejects, reject. U is a universal TM Therefore, U accepts (M,w) M accepts w (M,w)  ATM Therefore, U recognizes ATM

  32. Accept if M accepts w Reject if M does not accept w Reject if M accepts M Accept if M does not accept M ATM = { (M,w) | M is a TM that accepts string w } ATM is undecidable: (proof by contradiction) Assume machine H decides ATM H( (M,w) ) = Construct a new TM D as follows: on input M, run H on (M,M) and output the opposite of H D D D( M ) = Contradiction! D D D

  33. Theorem: ATM is r.e. but NOT decidable Theorem: ATM is not even r.e.! The Halting Problem is Not Decidable

  34. We have shown: Given any presumed machine H for ATM,we can effectively construct a TM D such that (D,D) ATM but H fails to tell us that. In other words, For any machine H that recognizes ATMwe can effectively give an instance where H fails to decide ATM In other words, Given any good candidate for deciding the Halting Problem, we can effectively construct an instance where the machine fails.

  35. THE HALTING PROBLEM HALTTM = { (M,w) | M is a TM that halts on string w } Theorem: HALTTM is undecidable Assume, for a contradiction, that TM H decides HALTTM Proof: We use H to construct a TM D that decides ATM On input (M,w), D runs H on (M,w) If H rejects then reject If H accepts, run M on w until it halts: Accept if M accepts and Reject if M rejects

  36. In many cases, one can show that a language L is undecidable by showing that if it is decidable, then so is ATM We reduce deciding ATM to deciding the language in question ATM ≤ L We just showed:ATM ≤ HaltTM Is HaltTM ≤ ATM ?

  37. Read chapter 4 of the book for next time