C241 Homework Assignment 9

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1 C41 Homework Assignment 9 1. The language L and functions R, A, and T defined below are the same as in Section 7.6. L {a, b, } + 1. L a. u L au L b. u L bu L 3. n. e. A: L L R: L L T : L L 1. A(, v) = v R( ) = T (, v) = v a. A(au, v) = aa(u, v) R(au) = A(R(u), a ) T (au, v) = T (u, av) b. A(bu, v) = ba(u, v) R(bu) = A(R(u), b ) T (bu, v) = T (u, bv) It is proved in the book that Proposition 7.6. A is associative; that is, for all u, v, w L, A(u, A(v, w)) = A(A(u, v), w). Proposition 7.9. Assuming Proposition 7.8, below, R is self-cancelling; that is, for all u L, R(R(u)) = u. Prove the following: (a) Proposition 7.7. For all u L, A(u, ) = u. (b) Proposition 7.8. For all u, v L, R(A(u, v) = A(R(v), R(u)). (c) Proposition For all u, v L, T (u, v) = A((R(u), v). (d) Proposition For all u L, T (u, ) = R(u). (e) Proposition 7.1. For all u L, T (T (u, ), ) = u. (a) Proposition 7.7. For all u L, A(u, ) = u. Proof. By induction on u L. Base Case. A(, ) A.1 =. Induction: Assume IH A(u, ) = u. A(au, ) A.a = a A(u, ) IH = au

2 Similarly, for bu, A(bu, ) A.b = b A(u, ) IH = bu (b) Proposition 7.8. For all u, v L, R(A(u, v) = A(R(v), R(u)). Proof. By induction on u L. Base Case. Induction. R(A(, v)) = R(v) (A.1) = A(R(v), ) (Prop 7.7) = A(R(v), R( )) (R.1) (c) Proposition For all u, v L, T (u, v) = A((R(u), v). Proof. By induction on u L. Base Case. T (, v) = v (T.1) = A(, v) (A.1) = A(R( ), v) (R.1) Induction: Assume T (u, v) = A(R(u), v). Similarly for T (bu, v). T (au, v) = T (u, av) (T.a) = A(R(u), av) (I.H.) = A(R(u), A(a, v)) (A.; A.1) = A(A(R(u), a ), v) (Prop. 7.6) = A(R(au), v) (R.a) (d) Proposition For all u L, T (u, ) = R(u). Proof. Induction is not needed. By Proposition 7.11, T (u, ) = A(R(u), ), and by Proposition 7.7, A(R(u), ) = R(u). (e) Proposition 7.1. For all u L, T (T (u, ), ) = u. Proof. Induction is not needed. By Proposition 7.10, used twice, T (T (u, ), ) = T (R(u), ) = R(R(u)). And by Proposition 7.9, R(R(u)) = u.

3 . The program below is called Wensley s algorithm for computing the quotient of real numbers x and y to within tolerance t. Use the Theorem on Loop Invariants to prove the this program satisfies the post-condition {z x/y < z + t}. {0 x < y 1} begin z := 0; d := 1; u := 0; v := 1 y ; while d > t do { Inv z x/y < z + d u = zy v = 1 dy } begin d := 1 d; if u + v > x then skip else begin z := z + d; u := u + v ; end ; v := 1 v end end {z x/y < z + t} There are three things to prove: Initialization. The program starts with 0 x < y 1, and first reaches the while-loop with z = 0, d = 1, u = 0, and v = 1 y. The invariant Inv is the conjunction of three propositions. All three are true when the variables current values are substituted. 1. z x/y z+d 0 x/y 0+1, which holds because 0 x < y 1.. u = zy 0 = 0 y, which is true. 3. v = 1 dy 1 1 y = 1 y, which is the initial value assigned to v. Invariance. If Inv z x/y < z + d u = zy v = 1 dy holds and d > t then the loop body executes, always computing new values for d = 1 d and v = 1 v. The values of x and y are unchanged througout. A. If conditional test u + v > x is true, all other variables are unchanged. 1. z x/y z + d z x/y z + 1 d. The first part, z x/y, is given by Inv. Also by Inv, x < u + v = zy + 1 dy = (z + 1 d)y = (z + d )y. Dividing through by y preserves the inequality because y is positive and gives x/y z + d as desired.. u = z y u = zy, which is given by Inv. 3. v = 1 d y 1 v = 1 ( 1 d)y. Dividing this equation by 1, we get v = 1 dy, which is given by Inv.

4 B. If the conditional test, u + v > x fails z and u are also changed and in addition to d and v we have. z = z + d because new assignment to d has already occurred = z + 1 d u = u + v x because the conditional test fails Looking again at Inv one conjuct at a time, 1. z x/y z + d z + 1 d x/y (z + 1 d) + 1 d. z + 1 d + 1 d = z + d, and Inv gives us x/y < z + d, so the right-hand inequality is valid. To show z + 1 d x/y, Inv gives us that u = zy and v = 1 dy, so u + v = (z + 1 d)y. At this point of the program, u + v x and y > 0. Thus, (z + 1 d)y x, and dividing both sides by y gives us what we want.. u = z y u+v = (z + 1 d)y. By Inv we have u+v = zy + 1 dy = (z + 1 d)y as needed. 3. v = 1 d y 1 v = 1 ( 1 d)y. Dividing this equation by 1, we get v = 1 dy, which is given by Inv. termination. On termination we have Inv d t. So it is immediate that z x/y < z + d z + t, thus satisfying the postcondition.

5 3. Define F : N N and G: N N as follows: F (0) = 1 F (k + 1) = (k + 1) F (k) G(0, m) = m G(k + 1, m) = G(k, m (k + 1)) (a) Prove by induction on n N: For all n, m N, G(n, m) = m G(n, 1). (b) Prove: For all n N, F (n) = G(n, 1). comment. If you try to prove the result directly by induction, you ll get stuck: Proposition 0. For all n N, F (n) = G(n, 1). base case. F (0) = 1 = G(0, 1) induction. Assume that F (k) = G(k, 1). F (k + 1) = (k + 1) F (k) (Defn. F ) = (k + 1) G(k, 1) (I.H.).?! (Nowhere to go from here...? = G(k + 1, 1)... to reach the goal.) You need to find and prove a more general result about G. Proposition 1. For all n, m N, G(n, m) = m G(n, 1). proof. The proof is by induction on k N. base case. G(0, m) G = m = m 1 F = m F (0) induction. Assume that G(k, m) = m G(k, 1). Then G(k + 1, m) = G(k, m(k + 1)) (Defn. G) = [ m(k + 1) ] G(k, 1) (I.H. = m [ (k + 1) G(k, 1) ] (algebra) = m G(k, k + 1) (I.H.) = m G(k + 1, 1) (Defn. G) comment. A more elegant version of Proposition 1 would be For all n, m, k N, G(n, m k) = m G(n, k). However, we don t need that much generality for our purposes.

6 Corollary. For all n N, F (n) = G(n, 1). proof. The proof is by induction on k N. base case. F (0) = 1 = G(0, 1) induction. Assume that F (k) = G(k, 1). F (k + 1) = (k + 1) F (k) (Defn. F ) = (k + 1) G(k, 1) (I.H.) = G(k, k + 1) (Prop. 1) = G(k + 1, 1) (Defn. G)

7 4. Performance estimation for recursive programs often involves recurrence relations like the one below. Let a N. The function T : N N is defined recursively by T (0) = a T (k + 1) = T (k) + k + 1 We would like to find a closed form for T, that is, and an algebraic expression that does not involve recursion Prove that for all n N, T (n) = a + n + n. proof. The proof is by induction on n N. base case. T (0) = a = a induction case. Assume that T (k) = a + k + k. T (k + 1) = T (k) + k + 1 (Defn. T ) = a + k + k = a + k + k = a + k + 3k + + k + 1 (I.H.) + k + = a + (k + k + 1) + (k + 1) = a + (k + 1) + (k + 1) as needed. This completes the induction. (multiply k + 1 by 1 = ) (adding fractions) (algebra) (factoring k + k + 1)

8 Supplemental Problem. H41 students should attempt this programming problem, but don t spend more than two or three hours on it. In a programming language of your choice, write a program that takes no input and outputs its own source code. No provided.