(i) By Corollary 1.3 (a) on P. 88, we know that the predicate. Q(x, y, z) S 1 (x, (z) 1, y, (z) 2 )

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1 COMPUTABILITY: ANSWERS TO SOME EXERCISES IN CHAPTER 5 AND , P This exercise is important for several reasons. For example, it tells us that taking inverses is an effect operation; see Example on P. 94. It also supplies a piece of crucial information in the solution to Exercise and 7 on P (i) By Corollary 1.3 (a) on P. 88, we know that the predicate S 1 (x, z, y, t) P x (z) y in t or fewer steps is decidable. Thus the predicate y E x is equivalent to z t S 1 (x, z, y, t), which is equivalent to z S 1 (x, (z) 1, y, (z) 2 ). We can easily check that Q(x, y, z) S 1 (x, (z) 1, y, (z) 2 ) satisfies both conditions (a) and (b). (ii) Since the predicate Q(x, y, z) given in part (i) is decidable, the function g(x, y) = U(µz Q(x, y, z)) is computable, where U is defined by U(z) = (z) 1. In view of (i) (a) above, when y E x, the predicate z Q(x, y, z) fails and hence g(x, y) is undefined. On the other hand, suppose we have y E x. Then it makes sense to talk about the smallest z for which Q(x, y, z) holds, say z 0 = µz Q(x, y, z). Thus g(x, y) is defined and its value is given by g(x, y) = U(z 0 ). Certainly Q(x, y, z 0 ) holds. Thus, by (i) (b) we have φ x (g(x, y)) = φ x (U(z 0 )) = φ x ((z 0 ) 1 ) = y. (iii) Since f is computable, we have f = φ n for some n. The function g(n, y) (where g is the function described in part (ii) above) is computable. Also, φ n (g(n, y)) = y for all y E n and g(n, y) is undefined for y E n, where E n is the range of φ n f. Thus f 1 (y) = g(n, y), which is computable. 2. Since f and g are computable, φ m = f and φ n = g for some m and n. We have T 1 (m, x, z) S 1 (m, x, (z) 1, (z) 2 ) P m (x) (z) 1 in (z) 2 or fewer steps. 1

2 (Notice that, if the predicate T 1 (m, x.z) holds, then (z) 1 must be f(x). But this does not concern us here.) The predicate T 1 (n, x, z) is similarly defined. Since both T 1 (m, x, z) and T 1 (n, x, z) are decidable, so is T 1 (m, x, z) or T 1 (n, x, z). Hence Φ(x) = U(µz T 1 (m, x, z) or T 1 (n, x, z) ) is computable. Notice that the domain of Φ is W m W n. (When x W m W n, the value of Φ(x) is either f(x) φ m (x) or g(x) = φ n (x), depending on which computation, P m (x) or P n (x), finishes first. From the identity h(x) = 1(Φ(x)) it follows that h is computable , P Notice that c M = 1 c M. So it is natural to consider the compuatable function f(x, y) = 1 ψ U (x, y). By the s-m-n theorem we know that there is a total computable function k such that f(x, y) = ψ k(x) (y). In case φ e = c M, we have c M (y) = 1 c M (y) = 1 φ e (y) = 1 ψ U (e, y) = f(e, y) = φ k(e) (y). Thus φ k(e) = c M. Done. 2. Since ψ U is computable, the function f(x, y) = 1(ψ U (x, y))y (= 1(φ x (y))y) is also computable. Explicitly, { y if y Wx ; f x (y) f(x, y) = undefined otherwise with Dom(f x ) = Range(f x ) = W x. The s-m-n theorem tells us that there exists some total computable function k such that φ k(x) = f x. Thus E k(x) = Range(φ k(x) ) = Range(f x ) = W x. 3. From Example on P. 93 we know that there is a total cumputable function r(x, y) such that W x W y = W r(x,y). The previous exercise tells us that there is a total computable k(x) such that E k(x) = W x. If we can find a total computable function h(x) such that W h(x) = E x, then we will get E x E y = W h(x) W h(y) = W r(h(x),h(y)) = E k(r(h(x),h(y))) = W s(x,y), where s(x, y) = k r(h(x), h(y)) is computable. It remains to find such h. To do so, we need the help of Exercise (ii) on P. 90: there is a computable function 2

3 g(x, y) such that g(x, y) is defined iff y E x. By the s-m-n theorem, we know that there is a total computable function h such that g(x, y) = φ h(x) (y). Now y W h(x) φ h(x) (y) g(x, y) is defined y E x and hence W h(x) = E x. Done. 4. This is easy. Notice that y f 1 (W x ) means that f(y) W x, that is, φ x (f(y)) is defined. So it is only natural to consider the cumputable function h(x, y) = φ x (f(y)) ψ U (x, f(y)). Applying the s-m-n theorem, we can get a total computable function k such that φ k(x) (y) = h(x, y) = φ x (f(y)). Thus W k(x) = Dom(φ k(x) ) = Dom(φ x f) = f 1 (W x ). 5. (a) The main theorem in the present chapter, namely Theorem 1.2 on P. 86, tells us that the functions ψ (m) U (e, y 1,..., y m ) and ψ (n) (e k, z 1, z 2,... z n ) (1 k m) are computable. Hence the (partial) function F on N m+1 N n N m+1+n defined by F ((e,e 1,..., e m ), (z 1,..., z n )) = ψ (m) U (e, ψ(n) U Sub(φ (m) e (e 1, z 1,..., z n ),..., ψ (n) U (e m, z 1,..., z n )) ; φ (n) e 1,..., φ (n) e m )(z 1,..., z n ) is computable. Applying the s-m-n theorem to this function F, we get a total computable function s(e, e 1,..., e m ) such that Done. F ((e, e 1,..., e m ), (z 1,..., z n )) = φ s(e,e1,...,e m )(z 1,..., z n ). (b) We know that the function f(e, x) = µy (φ n+1 e (x, y) = 0) is computable and hence there is a total computable function k such that φ (n) k(e) (x) = f(e, x) , P First we pick out those parts for which we can apply Rice s Theorem and finish them quick. (d) Let B = {0}. (f) Let B = {f C 1 : f is total and constant}. (g) Let B = {f }. (h) Let B = {f C 1 : Range(f) is infinite}. (i) Let B = {g}. (In part (g) and part (i), B is a singleton set.) 3

4 (e) From Theorem 1.6 (b) on P. 104 we know that N(y) 0 E y is undecidable. Suppose the contrary that M(x, y) x E y is decidable. Then its characteristic function c M (x, y) is computable. Consequently c M (0, y) is computable. But c M (0, y) is just c N (y), the characteristic function of N(y). So this contradicts the fact that c N is not computable. (c) (This one is a bit tricky.) Suppose the contrary that M(x) φ x (x) = 0 is decidable. [ Notice that the opposite of M(x) is M(x) either x W x, or x W x but φ x (x) 0. ] Then its characteristic function c M is computable. Hence c M = φ n for some n. Since c M is a total function, so is φ n and hence W n = N. We check that both φ n (n) = 0 and φ n (n) 0 will lead to a contradiction. If φ n (n) = 0, then M holds for n and hence c M (n) = 1, contradicting φ n (n) = c M (n). If φ n (n) 0, then M fails for n and hence c M (n) = 0, again contradicting φ n (n) = c M (n). (b) Applying Rice s theorem to B = {f C 1 : f is total} we see that the predicate M(x) φ x is total is undecidable. Hence its characteristic function c M (x) is not computable. Take any computable function g which is total (e.g. g = 0) and let n be the integer such that φ n = g. Let h(x, y) be the characteristic function of the predicate W x = W y. Then f(x, n) is the characteristic function of W x = W n, or W x = N. Thus f(x, n) is just the characteristic function of the predicate M(x). So f(x, n) = c M (x) is not computable. Hence f(x, y) is not computable; (otherwise substituting y by n in f(x, y) would give us a computable function f(x, n)). Therefore W x = W y is undecidable. (a) First proof: use a diagonal argument. Suppose the contrary that x E x is decidable. Then the function { x if x Ex g(x) = undefined otherwise; is computable. Hence g = φ n for some n. Notice that Dom(g) = Range(g) = {x: x E x }. Suppose n Dom(g) = Dom(φ n ) = W n. Then n Range(g) = Range(φ n ) = E n. On the other hand, n Dom(g) = {x: x E x } and hence n E n, a contradiction. Next suppose n Dom(g). Then n Range(g) = Range(φ n ) = E n. On the other hand, n Dom(g) = {x: x E x } and hence n E n, a contradiction again. 4

5 Second proof: use reduction. Let f and k be those functions in the proof of Theorem 1.6 on P Then W k(x) = E k(x) = N iff x W x and W k(x) = E k(x) = iff x W x, from which we deduce that k(x) E k(x) iff x W x. Since k is a total computable function and x W x is undecidable, we see that y E y is also undecidable. 2. Assume the contrary that such f exists. [Intuitively, using f we can solve the Halting problem does the computation P x (y) eventually stop or go on for ever as follows. If the computation P x (y) stops in f(x, y) or fewer steps then the answer to the Holting problem is yes and otherwise no. The point here is that there is no need to go beyond f(x, y) steps to find out the answer to P x (y)?. But converting this idea to a formal proof needs some technical knowledge to handle the situation.] Recall from Corollary on P. 88 that H(e, x, t) P e (x) in t or fewer steps is decidable. By our assumption on f, we know that H n (e, x, f(e, x)) P e (x). Notice that the characteristic function c M (e, x) of the predicate M(e, x) H n (e, x, f(e, x)) is just c Hn (e, x, f(e, x)). Since H n is decidable, c Hn (e, x, t) is computable. Hence c M (e, x) = c Hn (e, x, f(e, x)) is also computable and as a consequence the predicate M(e, x) is decidable. But M(e, x) P e (x), which is just the Halting problem. Thus we have arrived at a contradiction , P (a) E (n) x W (n) x y t H n (x, y, t), where H n (x, y, t) P (n) x (y) in t or fewer steps is knwon to be decidable. So by Corollary 6.6 on P. 115, it follows that the predicate E (n) x is partially decidable. (b) We know that M(z) z is a perfect square is decidable. So its partial characteristic function χ M (z) is computable. Hence f(x, y) = χ M (φ x (y)) is computable. 5

6 Observe that f(x, y) is just the partial characteristc function of φ x (y) is a perfect square. (c) This question is obsolete because Fermat s Last Theorem was proved in recent years by Andrew Wiles. (d) Let π = y 0. y 1 y 2 y 3 (= k=0 y k/10 k ) be the decimal expansion of π. Then the function f defined by f(k) = y k is computable (this was known at least two thousand years ago). Hence the following predicate, denoted by M(x, z), f(z) 7, f(z + 1) = f(z + 2) = = f(z + x) = 7 and f(z + x + 1) 7, is decidable. By Theorem 6.5 on P. 115, the predicate in question, which is equivalent to z M(x, z) is partially decidable. 2. Let G be a finitely presented group. Let A = {a 1,..., a m } be a finite set of generators and R = {r 1, r 2,..., r n } be a finite set of defining relations. More precisely, R is a finite subset of the free group F n generated by A and G = F n /N, where N is the normal subgroup generated by R. Let π : F n G be the quotient map. The word problem is, given a word w, decide whether π(w) = e (e is the unit element of G). Now π(w) = e if and only if we can apply finitely many operations on elements z in F n of the following types to render w to the empty word : 1. multiply z by some r in R, or r 1, on either side of z; 2. replace z by aza 1 or a 1 za for some a A. By introducing appropriate symbols for these operations (this is similar to symbols for instructions in URM) we can effectively enumerate all these operations. Using Gödel s numbers, following the same method of Chapter 5 we can effectively enumerate all finite sequences of such operations (this is similar to enumeration of all URM programs), say S x (x N). Each S x represents a sequence of operations converting any word w in F n into a new word denoted by S x (w); (unlike P x for computable functions φ x, S x (w) is always defined). Also, we can effectively enumerate all words with aphabets in A, say w x. The predicate M(x, y) S y (w x ) = is decidable. Hence π(w) = e y M(x, y) is partially decidable, in view of Theorem 6.5 on P Omitted. The idea is similar to the last exercise. 4. The partial characteristic functions χ M (x) and χ N (x) of M(x) and N(x) are computable. Putting P M and N we have χ P (x) = χ M (x)χ N (x), which is computable. 6

7 Hence M(x) and N(x) is partially decidable. The other part, showing that M(x) or N(x) is partially decidable, is more difficult. This part is essentially the same as is Exercise on P. 90. Just replace unary functions f and g by n-ary functions χ M (x) and χ N (x), and replace T 1 by T n. Then proceed in the same manner. Notice that, if M(x) x W x, then M(x) is partially decidable, but ǹot M(x) x W x is not partially decidable. 5. (a) The predicate N(x, y, z) y < z and M(x, y) is partially decidable. So, by Theorem 6.5 on P. 115, the predicate y < z N(x, y, z) y (y < z and M(x, y) is also partially decidable. (b) Let f(x, y) be the partial characteristic function of M(x, y). Since M(x, y) is partially decidable, f(x, y) is computable. By Theorem on P. 38, we know that the function F (x, z) = Π y<z f(x, y) is computable. Since F (x, z) is nothing but the partial characteristic function of the predicate y < z M(x, y), hence the last predicate is computable. (c) We know that the predicate x W x is not partially decidable. But we have x W x y (x y and y W x ) where the predicate x y and y W x is partially decidable, because x y is decidable, while y W x (x, y) Dom(ψ U ) is partially decidable. 6. (a) Indeed, x is even y (x 2y = 0). (b) Indeed, x divies y z (zx y = 0). 7. (a) Another way to say x W x iff M(k(x)) does not hold is x W x iff M(k(x)) holds. Hence the predicate x K x, which is known to be not partially decidable, is reducible to M(x). Hence M(x) is not partially decidable. (b) The computable function k constructed in the proof of Theorem 1.6 on P. 104 has the property that x W x iff φ k(x) is total. It follows from part (a) above that the opposite of φ x is total is not partially decidable. 7

8 (c) Since the predicate P x (x) does not converge in y or fewer steps is decidable, the function f(x, y) is computable. Put f x (y) = f(x, y). Then it is clear that f x is total if and only if x W x. By the s-m-n theorem we know that there is a total computable function s such that f(x, y) = φ s(x) (y). Thus we have: x W x iff φ s(x) is total. As we know, the predicate x / W x is not partially decidable. Therefore φ x is total is also not partially decidable. 8. Assume that the predicate f(x) = y is partially decidable. By Theorem 6.4 on P. 114, we get a decidable predicate R(x, y, z) such that f(x) = y z R(x, y, z). We can check that f(x) = U(µz R(x, (z) 1, (z 2 )), where U(w) = (w) 1. Hence f is computable. 9. As usual, denote by χ M (x 1,..., x n ) and χ N (y) the partial characteristic functions of M(x 1,..., x n ) and N(y) respectively. Then we can check that χ N (y) = χ M (g 1 (y),..., g n (y)), which is computable. From C.K. Fong; August 2,