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# TM Design Universal TM

MA/CSSE 474 Theory of Computation. TM Design Universal TM. Your Questions?. I have included some slides online that we will not have time to do in class, but may be helpful to you anyway. Previous class days' material Reading Assignments. HW 12 problems Anything else. The CFL Hierarchy. Download Presentation ## TM Design Universal TM

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1. MA/CSSE 474 Theory of Computation TM Design Universal TM

2. Your Questions? I have included some slides online that we will not have time to do in class, but may be helpful to you anyway. • Previous class days' material • Reading Assignments • HW 12 problems • Anything else

3. The CFL Hierarchy

4. Context-Free Languages Over a Single-Letter Alphabet Theorem: Any context-free language over a single-letter alphabet is regular. Proof: Requires Parikh’s Theorem, which we are skipping

5. Algorithms and Decision Procedures for Context-Free Languages Chapter 14

6. Decision Procedures for CFLs Membership: Given a language L and a string w, is w in L? Two approaches: ● If L is context-free, then there exists some context-free grammar G that generates it. Try derivations in G and see whether any of them generates w. Problem (later slide): ● If L is context-free, then there exists some PDA M that accepts it. Run M on w. Problem (later slide):

7. Decision Procedures for CFLs Membership: Given a language L and a string w, is w in L? Two approaches: ● If L is context-free, then there exists some context-free grammar G that generates it. Try derivations in G and see whether any of them generates w. S S T | a Try to derive aaa S ST S T

8. Decision Procedures for CFLs Membership: Given a language L and a string w, is w in L? ● If L is context-free, then there exists some PDA M that accepts it. Run M on w. Problem:

9. Using a Grammar decideCFLusingGrammar(L: CFL, w: string) = 1. If given a PDA, build G so that L(G) = L(M). 2. If w =  then if SG is nullable then accept, else reject. 3. If w then: 3.1 Construct G in Chomsky normal form such that L(G) = L(G) – {}. 3.2 If G derives w, it does so in 2|w| - 1 steps. Try all derivations in G of 2|w| - 1 steps. If one of them derives w, accept. Otherwise reject. Alternative O(n3) algorithm: CKY

10. Membership Using a PDA Recall CFGtoPDAtopdown, which built: M = ({p, q}, , V, , p, {q}), where  contains: ● The start-up transition ((p, , ), (q, S)). ● For each rule Xs1s2…sn. in R, the transition ((q, , X), (q, s1s2…sn)). ● For each character c, the transition ((q, c, c), (q, )). Can we make this work so there are no -transitions? If every transition consumes an input character then M would have to halt after |w| steps. • Put the grammar into Greibach Normal form:All rules are of the following form: • ● X  a A, where a  and A  (V - )*. • We can combime pushing the RHS of the production withmatching the first character a. Details on p 316-317.

11. Greibach Normal Form All rules are of the following form: ● XaA, where a and A (V - )*. No need to push the a and then immediately pop it. So M = ({p, q}, , V, , p, {q}), where  contains: 1. The start-up transitions: For each rule Scs2…sn, the transition: ((p, c, ), (q, s2…sn)). 2. For each rule Xcs2…sn (where c and s2 through sn are elements of V - ), the transition: ((q, c, X), (q, s2…sn))

12. A PDA Without -Transitions Must Halt Consider the execution of M on input w: ● Each individual path of M must halt within |w| steps. ● The total number of paths pursued by M must be less than or equal to P = B|w|, where B is the maximum number of competing transitions from any state in M. ● The total number of steps that will be executed by all paths of M is bounded by P |w|. So all paths must eventually halt.

13. Emptiness Given a context-free language L, is L = ? decideCFLempty(G: context-free grammar) = 1. Let G = removeunproductive(G). 2. If S is not present in G then return True else return False.

14. Finiteness Given a context-free language L, is L infinite? decideCFLinfinite(G: context-free grammar) = 1. Lexicographically enumerate all strings in * of length greater than bn and less than or equal to bn+1 + bn. 2. If, for any such string w, decideCFL(L, w) returns True then return True. L is infinite. 3. If, for all such strings w, decideCFL(L, w) returns False then return False. L is not infinite. Why these bounds?

15. Some Undecidable Questions about CFLs ● Is L = *? ● Is the complement of L context-free? ● Is L regular? ● Is L1 = L2? ● Is L1L2? ● Is L1L2 = ? ● Is L inherently ambiguous? ● Is G ambiguous?

16. Regular and CF Languages Regular Languages Context-Free Languages ● regular exprs. ● context-free grammars ● or ● regular grammars ● = DFSMs ● = NDPDAs ● recognize ● parse ● minimize FSMs ● try to find unambiguous grammars ● try to reduce nondeterminism in PDAs ● find efficient parsers ● closed under: ● closed under: ♦ concatenation ♦ concatenation ♦ union ♦ union ♦ Kleene star ♦ Kleene star ♦ complement ♦ intersection ♦ intersection w/ reg. langs ● pumping theorem ● pumping theorem ● D = ND ● D  ND

17. Languages and Machines SD D Context-Free Languages Regular Languages reg exps FSMs cfgs PDAs unrestricted grammars Turing Machines

18. Grammars, SD Languages, and Turing Machines SD Language L Unrestricted Grammar Accepts Turing Machine

19. Turing Machines We want a new kind of automaton: ● powerful enough to describe all computable things unlike FSMs and PDAs. ● simple enough that we can reason formally about it like FSMs and PDAs, unlike real computers. Goal: Be able to prove things about what can and cannot be computed.

20. Turing Machines At each step, the machine must: ● choose its next state, ● write on the current square, and ● move left or right.

21. A Formal Definition A (deterministic) Turing machine M is (K, , , , s, H): ● K is a finite set of states; ● is the input alphabet, which does not contain q; ● is the tape alphabet, which must contain q and have  as a subset. ● sK is the initial state; ● HK is the set of halting states; ● is the transition function: (K - H)  to K  {, } non-halting  tape  state  tape  direction to move state char char (R or L)

22. Notes on the Definition 1. The input tape is infinite in both directions. 2.  is a function, not a relation. So this is a definition for deterministic Turing machines. 3.  must be defined for all (state, tape symbol) pairs unless the state is a halting state. 4. Turing machines do not necessarily halt (unlike FSM's and most PDAs). Why? To halt, they must enter a halting state. Otherwise they loop. 5. Turing machines generate output, so they can compute functions.

23. An Example M takes as input a string in the language: {aibj, 0 ji}, and adds b’s as required to make the number of b’s equal the number of a’s. The input to M will look like this: The output should be:

24. The Details K = {1, 2, 3, 4, 5, 6},  = {a, b},  = {a, b, q, \$, #}, s = 1, H = {6},  =

25. Notes on Programming The machine has a strong procedural feel, with one phase coming after another. There are common idioms, like scan left until you find a blank There are two common ways to scan back and forth marking things off. Often there is a final phase to fix up the output. Even a very simple machine is a nuisance to write.

26. Halting ● A DFSM M, on input w, is guaranteed to halt in |w| steps. ● A PDA M, on input w, is not guaranteed to halt. To see why, consider again M = But there exists an algorithm to construct an equivalent PDA M that is guaranteed to halt. A TM M, on input w, is not guaranteed to halt. And there is no algorithm to construct an equivalent TM that is guaranteed to halt.

27. Formalizing the Operation A configuration of a Turing machine M = (K, , , s, H) is an element of: K  ((- {q}) *)  {}    (* (- {q}))  {} state up current after to current square current square square

28. Example Configurations b (1) (q, ab, b, b) = (q, abbb) (2) (q, , q, aabb) = (q, qaabb) Initial configuration is (s, qw).

29. Yields (q1, w1) |-M (q2, w2) iff (q2, w2) is derivable, via, in one step. For any TM M, let |-M* be the reflexive, transitive closure of |-M. Configuration C1yields configuration C2 if: C1|-M*C2. A path through M is a sequence of configurations C0, C1, …, Cn for some n 0 such that C0 is the initial configuration and: C0 |-MC1 |-MC2 |-M … |-MCn. A computation by M is a path that halts. If a computation is of lengthn (has n steps), we write: C0 |-MnCn

30. Exercise A TM to recognize { wwR : w  {a, b}* }. If the input string is in the language, the machine should halt with y as its current tape symbol If not, it should halt with n as its current tape symbol. The final symbols on the rest of the tape may be anything.

31. TMs are complicated • … and low-level! • We need higher-level "abbreviations". • Macros • Subroutines

32. A Macro language for Turing Machines (1) Define some basic machines ● Symbol writing machines For each x, define Mx, written just x, to be a machine that writes x. ● Head moving machines R: for each x, (s, x) = (h, x, ) L: for each x, (s, x) = (h, x, ) ● Machines that simply halt: h, which simply halts (don't care whether it accepts). n, which halts and rejects. y, which halts and accepts.

33. Checking Inputs and Combining Machines Next we need to describe how to: ● Check the tape and branch based on what character we see, and ● Combine the basic machines to form larger ones. To do this, we need two forms: ● M1M2 ● M1 <condition> M2

34. Turing Machines Macros Cont'd Example: >M1aM2 b M3 ● Start in the start state of M1. ● Compute until M1 reaches a halt state. ● Examine the tape and take the appropriate transition. ● Start in the start state of the next machine, etc. ● Halt if any component reaches a halt state and has no place to go. ● If any component fails to halt, then the entire machine may fail to halt.

35. More macros a M1M2 becomes M1a, bM2 b M1all elems of  M2 becomes M1M2 or M1M2 Variables M1all elems of  M2 becomes M1xaM2 except a and x takes on the value of the current square M1a, bM2 becomes M1xa, bM2 and x takes on the value of the current square M1x = yM2 if x = y then take the transition e.g., > xqRx if the current square is not blank, go right and copy it.

36. Blank/Non-blank Search Machines Find the first blank square to the right of the current square. Find the first blank square to the left of the current square. Find the first nonblank square to the right of the current square. Find the first nonblank square to the left of the current square Rq Lq Rq Lq

37. More Search Machines La Find the first occurrence of a to the left of the current square. Ra,b Find the first occurrence of a or b to the right of the current square. La,baM1 Find the first occurrence of a or b to the left of the current square, b then go to M1 if the detected character is a; go to M2 if the M2 detected character is b. Lxa,b Find the first occurrence of a or b to the left of the current square and set x to the value found. Lxa,bRx Find the first occurrence of a or b to the left of the current square, set x to the value found, move one square to the right, and write x (a or b).

38. An Example Input: qww {1}* Output: qw3 Example: q111qqqqqqqqqqqqqq

39. What does this machine do?

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