Chapter 4: A Universal Program 1

Coding programs Example : For our programs P we have variables that are arranged in a certain order: Y 1 X 1 Z 1 X 2 Z 2 X 3 Z 3 … Similarly labels are ordered: A 1 B 1 C 1 D 1 E 1 A 1 B 1 C 1 D 1 … Definition : #(V) is the position of a variable or label in the given ordering, where V is the label or variable. Example: #(X 3 ) = 6 #(C) = 3 #(X 1 ) = #(X) = 2 2

Coding programs Definition : Let I be (labeled or unlabeled) instruction of the language L. Thus:, where the following points are satisfied: If I is unlabeled, then a = 0; if I is labeled with L, then a = #(L). If the variable V is mentioned in I, then c = #(V) – 1. If the statement in I is either:  V ← V, then b = 0  V ← V + 1, b = 1  V ← V – 1, b = 2 If the statement in I is IF V ≠ 0 GOTO L', then b = #(L') + 2 3

Coding programs Reminder from Chapter 3: Given the following variable ordering: Y X 1, Z 1, X 2, Z 2 … Provided the following label ordering: A 1 B 1 C 1 D 1 E We have a program as follows: [A] X ← X + 1 IF X ≠ 0 GOTO A The instruction numbers for this program are thus: #(I 1 ) = > = > = = 21 #(I 2 ) = > = > = = 46 Definition: The program P number is then calculated as: #(P) = [ #(I 1 ), #(I 2 ), #(I 3 ),... ] - 1 In the program above: #(P) = 2^21 * 3^46 – 1; 4

Coding programs In order to avoid ambiguity of unlabeled Y ← Y instruction, which is > an additional rule is imposed: The final instruction in a program is not permitted to be the unlabeled statement Y ← Y. N.B: The program number thus determines a unique program, which can reconstructed. Example: Given #(P) = 576 = = 5^2 + 23^1 = [ 0, 0, 2, 0, 0, 0, 0, 0, 1 ] – 9 instructions For 2 = = > so unlabeled: Y ← Y + 1 For 1 = = > so labeled: [A] Y ← Y For 0 = >, this will be unlabeled: Y ← Y Empty program has the number 0. 5

The Halting Problem Definition: For a given y, let P be the program such that #(P) = y. Then HALT(x, y) is true if is defined and false if is undefined, so: HALT( x, y ) program number y eventually halts on input x Theorem: HALT( x, x ) is not a computable predicate. Proof: Construct a program P: [A] IF HALT( X, X ) GOTO A Let #(P) = y 0, then HALT( x, y 0 ) ~HALT( x, x ) So we can set x=y 0, thus HALT (y 0,y 0 ) ~HALT(y 0,y 0 ) Contradiction! 6

Insolvability of Halting Problem “There is no algorithm that, given a program of P and an input to that program, can determine whether or not the given program will eventually halt on the given input” (Chapter 4, page 68) Church’s Thesis: Any algorithm for computing on numbers can be carried out by a program of P. See Golbach’s conjecture (every even number greater or equal to 4 is the sum of two prime numbers) for an example of computable program for which it is hard to determine whether it will ever halt. 7

Universality Universality Theorem: For each n > 0, where For each n >0, the function is partially computable. Step-Counter Theorem: For each n > 0, the predicate is primitive recursive, where:, where #(P)=y Program number y halts after t or fewer Steps on inputs x 1, …, x n. 8

Normal Form Theorem Normal Form Theorem: Let f( x 1,…, x n ) be a partially computable function. Then there is a primitive recursive predicate R( x 1, …, x n, y ) such that: Theorem: A function is computable iff can be obtained from the initial functions by a finite number of applications of composition, recursion and (proper ) minimalization 9

Normal Form Theorem Normal Form Theorem Proof: Let y 0 be the program number for f(x 1,…,x n ). when the right-hand side of the equation is defined, then there exists a number z: For any z, the program with number y 0 has reached a terminal snapshot in r(z) or fewer steps and l(z) is in output variable Y. If the right side is undefined, then is false for all t i.e. 10

Recursively Enumerable Sets Theorem: Let the sets B, C belong to some PRC class C. Then so do the sets (prove using predicate union and intersection) Theorem: Let C be a PRC class, and let B be a subset of. Then B belongs to C iff: Definition: The set is called recursively enumerable if there is a partially computable function g(x) such that: belongs to C 11

Recursively Enumerable Sets If B is a recursive set, then B is recursively enumerable. The set B is recursive iff B and are both recursively enumerable. If B and C are recursively enumerable sets so are Enumeration Theorem: A set B is recursively enumerable iff there is an n for which B = W n 12

Recursively Enumerable Sets Definition: K is a set of all numbers n such that program number n eventually halts on input n. K is recursively enumerable but not recursive. Theorem: Let B be a recursively enumerable set. Then there is a primitive recursive predicate R(x,t) such that 13

Recursively Enumerable Sets Theorem: Let S be a nonempty recursively enumerable set. Then there is a primitive recursive function f(u) such that S = { f(n) | n N} = { f(0), f(1), … }, where S is the range of f. Theorem: Let f(x) be a partially computable function and let S = { f(x) | f(x) ↓}, where S is the range. Then S is recursively enumerable. 14

Recursively Enumerable Sets Theorem: Suppose that, then the following statements are all equivalent: –S is recursively enumerable –S is the range of a primitive recursive function –S is the range of a recursive function –S is the range of a partial recursive function 15

The Parameter Theorem s-m-n Theorem: For each n, m > 0 there is a primitive recursive function such that: 16

The Parameter Theorem s-m-n proof: using induction on n. Base case: For program P take n = 1, then the following must be shown to be true: should be the number of the program with m inputs, which must be the same as program number y having m + 1 inputs. #(P) = y So will be the number of the program, that will provide the variable before proceeding with P. 17

The Parameter Theorem s-m-n proof: continued… The instruction will have a number associated with it as follows: > = 16m + 10 Thus may be defined as a primitive recursive function: Induction step: let n = k, then: 18

Diagonalization and Reducibility Definition: Let TOT be the set of all numbers p such that p is the number of a program that computes a total function f(x) of one variable. That is,, where Theorem: TOT is not recursively enumerable.,s 19

Diagonalization and Reducibility Theorem: TOT is not recursively enumerable. Proof: Assume that TOT is recursively enumerable. Then given that, then there is a computable function g(x): and let: Since g(x) is the number of a program that computes a total function, thus h is a computable function. Let P be the program that computes h and p=#(P). Then Contradiction! 20

Diagonalization and Reducibility Theorem: Suppose A ≤ m B, then: –If B is recursive, then A is recursive. –If B is recursively enumerable, then A is as well. Definition: A set A is m-complete if: –A is recursively enumerable and –For every recursively enumerable set B, B ≤ m A 21

Diagonalization and Reducibility If A ≤ m B and B ≤ m C, then A ≤ m C (transitivity) If A is m-complete, B is recursively enumerable and A ≤ m B then B is m-complete. Definition: A B means that A ≤ m B and B ≤ m A Theorem: K and K 0 are m-complete K K 0 Theorem: EMPTY is not recursively enumerable, where 22

Rice's Theorem Examples: Г is the set of computable functions Г is the set of primitive recursive functions Г is the set of partially computable functions that are defined for all but a finite number of values of x Rice’s Theorem: Let Г be a collection of partially computable functions of one variable, such that there are functions f(x), g(x) with f(x) in Г, g(x) not in Г. Then R Г is not recursive. Corollary: There are no algorithms for testing a given program P of the language L to determine whether belongs to any of the classes described in the above examples. What does it mean? If Г is a non-trivial property of functions, then Г is undecidable. 23

Rice's Theorem Proof (using recursion theorem): Let f(x) and g(x) be partially computable functions such that: f(x) Г g(x) Г Suppose is computable. Let and Then, since - is partially computable. Iis not recursive

Rice's Theorem Proof (cont’d): Since is partially computable, by recursion theorem there is a number e s.t.: As and : Contradiction!

26 The Recursion Theorem Recursion Theorem: Let g(z, x 1, …, x m ) be a partially computable function of m + 1 variables. Then, there is a number e such that: Discussion: Given a partially computable function g, we must produce a “program” e that is supplied m arguments and one of the implicit arguments is its own encoding e. We may attempt to prove the theorem by first incrementing a variable e times. But then the problem is that the encoding of such a program will be larger than e! The program P with #(P) = e must contain a “partial description” of itself built-in so that that it can computer its own encoding e from that description.

27 Let Q be the program: Z = Y = g(Z, X 1, X 2, …, X m ) Now, the program P will consist of #(Q) copies of the instruction X m+1 = X m+1 + 1, followed by the program Q. After executing the first #(Q) increments as well as the instruction Z = …, Z holds the number of the program consisting of the #(Q) copies of the instruction followed by the program Q (Remember that as defined in the proof of the parameter theorem computes the # of the program consisting of #(Q) copies of the increment instruction X m+1 = X m followed by program Q). But this is exactly the program P.

The Recursion Theorem Proof: Let g be a partially computable function such as: is the function in the parameter theorem. Then for some z 0 and using s-m-n theorem: Set v = z 0 and, then: Contradiction! 28

The Recursion Theorem Corollary: There is a number e such that for all x: Proof: Let g be a computable function: Using the recursion theorem: Fixed Point Theorem: Let f(z) be a computable function. Then there is a number e such that: Take g(z, x) = 29

Computable functions that are not primitive recursive The primitive recursive functions are precisely the functions from the initial functions (see Chapter 3) Theorem: The unary primitive recursive functions are precisely those obtained from the initial functions s(x) = x + 1, n(x) = 0, l(x), r(x) by applying the following three operations on unary functions: –To go from f(x) and g(x) to f(g(x)) –To go from f(x) and g(x) to –To go from f(x) and g(x) to the function defined by the recursion Theorem: The function f( x, x ) + 1 is a computable function that is not primitive recursive. 30