ELEMENTS IN MATHEMATICS FOR INFORMATION SCIENCE NO.6 TURING MACHINE AND COMPUTABILITY. Tatsuya Hagino

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1 1 ELEMENTS IN MATHEMATICS FOR INFORMATION SCIENCE NO.6 TURING MACHINE AND COMPUTABILITY Tatsuya Hagino

2 2 So far Computation flow chart program while program recursive function primitive recursive function minimization operator Turing machine flow chart while program recursive function Turing machine lambda calculus

3 Decidable vs Undecidable Problems Decidable Problem A problem for which a program can say yes or no. The program needs to terminate. The corresponding recursive function needs to be total. Undecidable Problem A problem which is not decidable. There might be a problem which may say yes, but it does not termination if the answer is no. The corresponding function is not recursive, or it is recursive but not total. Halting Problem: Is there a program which tells whether a given program P for a given input a 1,, a n will eventually terminate and return a value or will run forever? 3 Undecidable Problems Decidable Problems

4 4 Encoding Programs In order to make a program as an input to another program, we need to represent a program as a number (i.e. encoding) Encoding flow chart programs: Boxes are connected by arrows Put a number to each box Each box is one of the following: input(x 1, x 2,, x n ) x i = m x i = x j + x k x i = x j x k x i = x j x k x i = x j x k if x i = x j output(x i )

5 5 Encoding Let x 1,, x n be input variables and x n+1, x n+2,, x t be other variables. Let A 1, A 2,, A l be boxes of program P where A 1 is the input box and A l is the output box. Using Gödel number, encode each box as #A: A a input(x 1, x 2,, x n ) x i = m x i = x j + x k x i = x j x k x i = x j x k x i = x j x k if x i = x j output(x i ) #A a 1, n, a 2, i, m, a 3, i, j, k, a 4, i, j, k, a 5, i, j, k, a 6, i, j, k, a 7, i, j, k, a 8, i The program can be encoded as: #P = #A 1, #A 2,, #A l

6 6 Interpreter for P Theorem: The following partial function comp n : N n+1 N is computable. comp n z, x 1,, x n = y when z = #P and y = f P (x 1, x n ) undefined otherwise where f P is the recursive function for program P. Proof: Write a program which computes comp n by simulating the flow chart program represented by #P.

7 7 comp n (z, x 1,, x n ) input(z, x 1, x 2,, x n ) a 1 v 0 b G a z c G 1 b c = 1 no c = 2 no c = 3 no yes n G 2 (b) v G(x 1,, x n, 0,, 0) a G 3 b yes i G 2 (b) m G 3 (b) v G G 1 v,, m,, G n v a G 4 b v yes i G 2 (b) j G 3 (b) k G 4 b G G 1 v,, G i v + G j (v),, G n v a G 5 b

8 8 Is comp Total? Theorem: If comp n : N n+1 N is extended to a total function g: N n+1 N, g is not recursive. Proof: Show the case for n = 1: Proof by contradiction and use Cantor s diagonal argument. Assume comp 1 z, x = g(z, x) and g: N 2 N is a recursive total function. Let h x = g x, x + 1. Then, h is recursive total function. There is a program which calculates h. Let c be the code. Then, from the definition of comp 1, h x = comp 1 (c, x). Give h an input c. h c = comp 1 c, c = g(c, c) This contradicts with h c = g c, c + 1. Therefore, a recursive total function g does not exist. (QED)

9 9 Recursive Predicate Definition: Predicate p: N n T, F is a recursive predicate if its characteristic function C p : N n N is recursive. C p is total. p is decidable. If p x 1,, x n, q(x 1,, x n ) and r x 1,, x n, y are recursive, the following predicates are also recursive: p(x 1,, x n ) q(x 1,, x n ) p(x 1,, x n ) q(x 1,, x n ) p x 1,, x n z < y r x 1,, x n, z z < y r x 1,, x n, z

10 10 Halting Problem is Undecidable Define predicate halt n z, x 1,, x n N n+1 {T, F} as follows: halt n z, x 1,, x n = T when comp n z, x 1,, x n is defined F when comp n z, x 1,, x n is undefined Theorem: halt n z, x 1,, x n is not recursive (i.e. undecidable). Proof: If halt n z, x 1,, x n is a recursive predicate, its characteristic function C haltn is recursive and total. Then, g z, x 1,, x n = C haltn z, x 1,, x n comp n z, x 1,, x n is a total recursive function and this contradicts with the previous theorem. (QED)

11 Totality Problem is Undecidable Theorem: For n > 0, there is no total recursive function g: N n+1 N which satisfies the following: g c, x 1,, x n : N n+1 N c N = f: N n N f is total and recursive } comp n z, x 1,, x n : N n+1 N is the universal function for recursive functions (both partial and total), but there is no universal function for total recursive functions. Proof: In the case for n = 1, if g: N 2 N exists, f x = g x, x + 1 is a total recursive function. Let c be the code of f, g c, x = f x = g x, x + 1 and this contradicts when x = c. In the case for n > 1, the proof can be similar. (QED) 11 Corollary: total n (z) x 1 x n halt n z, x 1,, x n is undecidable. is not a recursive predicate, i.e. total n (z) Proof: If C totaln is the characteristic function of total n, g z, x 1,, x n = C totaln (z) comp n z, x 1,, x n g is a total recursive function and this contradicts with previous theorem. (QED)

12 12 Undecidable Predicates halt n z, x 1,, x n whether a give program z terminates for the input x 1,, x n or not. total n (z) whether a given program z always terminates or not. x 1 x n comp n z, x 1,, x n = 0 whether a given program z always outputs 0 or not. x 1 x n comp n z, x 1,, x n = 0 whether a given program z outputs 0 for some input or not. For z, the domain of comp n z, x 1,, x n is finite. whether a program z terminates for finite sets of input or not. For z, comp n z, x 1,, x n is a constant function. whether a program z outputs always the same number or not. For z and z, comp n z, x 1,, x n whether two programs z and z are same or not. = comp n z, x 1,, x n

13 13 s-m-n Theorem Theorem: For natural numbers m and n, there is a primitive recursive function S m,n : N m+1 N which satisfies: comp m+n z, x 1,, x n, y 1,, y m = comp n S m,n (z, y 1,, y m ), x 1,, x n Proof: S m,n (z, u 1,, u m ) is the function which converts z = #A 1, #A 2,, #A l into z = # input x 1,, x n, #(y 1 u 1 ),, #(y m u m ), #A 2,, #A l which represents: input x 1,, x n y 1 u 1 y m u m A 2 A l The conversion function can be written as a primitive recursive function. (QED)

14 14 Recursion Theorem Theorem: For n and a total recursive function f: N N, there is a natural number c which makes the following equation true: comp n f c, x 1,, x n = comp n c, x 1,, x n Proof: Let a be the code for comp n+1 y, x 1,, x n, y. comp n+1 y, x 1,, x n, y = comp n+1 a, x 1,, x n, y = comp n S 1,n a, y, x 1,, x n Let b be the code for comp n f S 1,n a, y, x 1,, x n comp n f S 1,n a, y, x 1,, x n = comp n+1 b, x 1,, x n, y comp n f S 1,n a, b, x 1,, x n = comp n+1 b, x 1,, x n, b = comp n S 1,n a, b, x 1,, x n c = S 1,n a, b (QED)

15 15 Rice Theorem Theorem: Let n be a natural number. If a predicate p(z) satisfies the following two conditions, p(z) is not recursive (i.e. p(z) is undecidable). (1) c c x 1 x n comp n (c, x 1,, x n ) = comp n c, x 1,, x n p c p c (2) c c p c p c (1) means that p(z) truth value is the same for the same program. (2) means that p(z) is true for certain number and is false for a different number. Proof: If p is a recursive predicate, let C p be its characteristic function. Let define f: N N using c and c which satisfy (2) as follows: f z = C p z c + 1 C p z From the definition, p f z p z Since f is a total recursive function, using recursion theory there exists c which makes comp n f c, x 1,, x n = comp n (c, x 1,, x n ). From (1), p(f(c )) = p(c ), but this contradicts. (QED) Using this theorem, we can prove many predicates are undecidable. p(z) "comp n z, x 1,, x n is a constant function." p(z) is same for the same program, and there are a constant program and a not-constant one. c

16 16 Post Correspondence Problem Problem: Given a finite set of string pairs, s 1, t 1, s 2, t 2,, s n, t n using string concatenation, determine whether there is a number sequence i 1,, i m which makes the following equality hold: s i1 s i2 s im = t i1 t i2 t im Example: {(e, abcde), (ababc, ab), (d, cab)} This problem (post correspondence problem) is undecidable. There is no program which gives a solution to the problem or none if there is no solution.

17 17 Summary Decidable Problem A problem for which a program can say yes or no. Undecidable Problem A problem which is not decidable. Undecidable predicates: Halting problem Totality problem Post correspondence problem