Computation Theory

1. Computation Theory Introduction to Turing Machine

2. infinite tape control … a b B B a b B B B: blank symbol Turing Machines • Proposed by Alan Turing, 1936. England • “Turing Award”

3. Differences between finite automata and Turing machine • A Turing machine can both write on the tape and read from it. • The read-write head can move both to the left and to the right. • The tape is infinite. • The special states for rejecting and accepting take immediate effect.

4. eg. Consider a Turing machine M1 for testing membership in the language B={w#w : w{0,1}*}  0 1 1 0 0 0 # 0 1 1 0 0 0 B …  X 1 1 0 0 0 # 0 1 1 0 0 0 B … :  X 1 1 0 0 0 # X 1 1 0 0 0 B … :  X 1 1 0 0 0 # X 1 1 0 0 0 B …  X X 1 0 0 0 # X 1 1 0 0 0 B … :  X X X X X X # X X X X X X B …

5. M1 = On input string w: • Scan the input to be sure that it contains a single # symbol. If not, reject. • Zig-zap across # on either side to check on whether these positions contain the same symbol. If not, reject. Cross off symbols as they checked. • If any symbols remain, reject; otherwise accept.

6. q a Formal definition of a Turing machine: • δ: Q×Γ→ Q×Γ×{L,R} transition function. δ(q,a) = (r,b,L) r b

7. Def: A Turing Machine is a 7-tuple (Q, Σ, Γ, S, q0, qaccept, qreject), where Q, Σ, Γ are all finite sets and • Q: set of states, • Σ: the input alphabet NOT containing the special blank symbol B, • Γ: the tape alphabet, where {B} Γ and Σ Γ, • δ:Q×Γ→Q×Γ×{L,R} ~ transition func. • q0Q is the start state, • qacceptQ is the accept state, and • qrejectQ is the reject state, where qreject≠ qaccept

8. As a Turing machine computes, it may halt with ‘accept’ or ‘reject’, or it may never halt! • During computation, changes occur in • the current state, • the current tape contents, and • the current head location. • The above three items form a “configuration” of the Turing machine.

9. Write uqv for the configuration where q is the current state, uv is the current contents, the current head location is the first symbol of v. 1011qT0111 • Let C1 and C2 be two configurations. Say that C1yields C2, if the Turing machine can legally go from C1 to C2 in a single step. qT B… 1 0 1 1 0 1 1 1 1 B uaqibv yeilds uqjacv if δ(qi,b)=(qj,c,L) a,bΓ, u,vΓ*

10. uaqibv yields uacqjv if δ(qi,b)=(qj,c,R) • qibv yields qjcv if the transition is right moving. • We don’t move the head out of bound! • We don’t have configuration uaqiB. • Starting configuration: q0w • Accepting configuration: the state is qaccept • Rejecting configuration: the state is qreject

11. halting configurations • A Turing machine M accepts input w if a seq. of configurations C1, C2, …, Ck exists where • C1 is the start conf. of M on input w, • each Ci yields Ci+1, and • Ck is an accepting configuration. • The collection of strings that M accepts is the language of M, denoted L(M).

12. Def: A language is Turing-recognizable if some Turing machine recognizes it. (or recursively enumerable) r.e. Def: A language is Turing-decidable or decidable if some Turing machine decides it. (or recursive) r.

13. eg. Show a TM M that recognizes the language A={02n:n0} • M=“On input string w: • Sweep left to right across the tape, crossing off every other 0. • If in stage 1 the tape contained a single 0, accept. • If in stage 1 the tape contained more than a single 0 and the number of 0s was odd, reject. • Return the head to the left-hand and of the tape. • Goto 1.”

14. eg. (Element Distinctness problem) • M=“On input w:” 1. Place a mark on top of the leftmost tape symbol. If that symbol was not a “#”, reject. 2. Scan right to the next # and place a second mark on top of it. If no # is encountered before a blank symbol, only x1 was present, so accept.

15. 3. By zig-zagging,compare the two strings to the right of the marked #’s .If they are equal reject. 4. Move the rightmost of the two marks to the next “#” symbol to the right. If no # symbol is encountered before a blank symbol, move the leftmost mark to next # to its right. 5. Goto step 3.

16. eg. For ={a,b}, design a Turing Machine that accepts L={anbn : n1} • Sol: Q={q0,q1,q2,q3,q4}, ={a,b} ={a,b,x,y,B}, q4:accept state

17. (q0,a)=(q1,x,R) ; (q1,a)=(q1,a,R) (q1,y)=(q1,y,R) ; (q1,b)=(q2,y,L) (q2,y)=(q2,y,L) ; (q2,a)=(q2,a,L) (q2,x)=(q0,x,R) (q0,y)=(q3,y,R) ; (q3,y)=(q3,y,R) (q3,B)=(q4,B,R) q0aabb├ xq1abb├ xaq1bb├ xq2ayb├ q2xayb ├ xq0ayb├ xxq1yb├ xxyq1b├ xxq2yy ├ xq2xyy├ xxq0yy├ xxyq3y├ xxyyq3B ├ xxyyBq4

18. Variants of Turing machine • 1. Multi-tape Turing Machines a Turing Machine with several tapes.

19. Thm: Every multi-tape Turing machine has an equivalent single tape Turing machine. • Pf: Idea: 0 1 0 1 0 B M a a a B b a B S # 0 1 0 1 0 # a a a # b a # B

20. Non-deterministic Turing Machines: • Thm: Every non-deterministic TM has an equivalent deterministic TM. • Pf: 0 0 1 D x x # 0 1 x B 0 1 q 0 0 0 q 0 q q 0 0

21. Initially tape 1 contains the input w, and tapes 2 and 3 are empty. • Copy tape 1 to tape 2. • Use tape 2 to simulate N with input w on one branch of its non-det. computation. Before each step of N consult the next symbol on tape 3 to decide which branch to move. If no symbol remains or this choice is invalid goto step 4. If reject also goto 4. • Increase the count on tape 3. go to step 2.

22. Corollary:A language is Turing-recognizable if and only if some non-deterministic TM recognizes it. • Corollary: A language is decidable if and only if some non-deterministic TM decides it

23. Enumerators • A TM with an attached printer. • The language enumerated by an enumerator is the collection of all the strings that it eventually prints out. Control 0 0 1 0 … Work tape Enumerator

24. Theorem: A language is Turing-recognizable if and only if some enumerator enumerates it. • Pf: • If an enumerator E that enumerates a language A, a TM M recognizes A. M=“on input w • Run E. Every time that E outputs an string, Compare it with w. • If w ever appears in the output of E, accept.”

25. If TM M recognizes a language A, we can construct an enumerator E for A. Let S1, S2, …… be a list of all possible strings in *. E=“Ignore the input. • Repeat the following for i = 1, 2, 3, … • Run M for i steps on each input, S1, S2, …, Si, • If any computations accept, print out the corresponding Si.”

26. Definition of algorithm • Hilbert’s problems: In 1900, mathematician David Hilbert delivered a famous address at the “International congress of Mathematicians in Paris.” He proposed 23 mathematical problems. • Hilbert’s tenth problem: Design an algorithm that tests whether a multi-variable polynomial has an integral root.

27. The Church-Turing Thesis • In 1970, Yuri Matijasevič, building on work of Martin Davis, Hilary Putnam, and Julia Robinson, Showed that no algorithm exists for testing whether a poly. Has integral roots. Intuitive notion of algorithms = Turing machine algorithms Necessary to resolve Hilbert’s tenth problem

28. D = { P| P is a polynomial with an integral root} • Hilbert’s tenth problem: • Is D decidable? • A special case: • D1={P| P is a polynomial over x with an integral root} • D1 is decidable！

29. M1 = “The input is a polynomial P over x Evaluate P with x to the values 0, 1, -1, 2, -2, 3, -3,… . If at any point the polynomial evaluates to 0, accepts! ” • Problem 3.18: All the roots of P is between Thus, Di is decidable!