By Richard Brent, Paul Zimmermann
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The input is two integers a ≥ b > 0. The output is a 2 × 2 matrix R and integers a′ , b′ such that [a′ b′ ]t = R[a b]t . If the inputs have size n bits, then the elements of R should have at most n/2+O(1) bits, and the outputs a′ , b′ should have at most 3n/4 + O(1) bits. (b) Construct a plain GCD algorithm which calls HalfGcd until the arguments are small enough to call a naive algorithm. 3. 32 (Galbraith, Sch¨onhage, Stehl´e) The Jacobi symbol (a|b) of an integer a and a positive odd integer b satisfies (a|b) = (a mod b|b), the law of quadratic reciprocity (a|b)(b|a) = (−1)(a−1)(b−1)/4 for a odd and posi2 tive, together with (−1|b) = (−1)(b−1)/2 , and (2|b) = (−1)(b −1)/8 .
Similarly for x = au−cw, y = av−bw, z = bu−cv. 1), prove that qj∗ ≤ β+1. Can this bound be reached? In the case qj∗ ≥ β, prove that the while-loop at 42 Integer arithmetic steps 6–8 is executed at most once. e. 2). 20 (Granlund, M¨oller) In Algorithm BasecaseDivRem, estimate the probability that A < 0 is true at step 6, assuming the remainder rj from the division of an+j β + an+j−1 by bn−1 is uniformly distributed in [0, bn−1 − 1], A mod β n+j−1 is uniformly distributed in [0, β n+j−1 − 1], and B mod β n−1 is uniformly distributed in [0, β n−1 −1].
We have n = 9, ℓ = 2, a3 = 123, a2 = 45, a1 = 67, and a0 = 89. The recursive call for a3 β ℓ + a2 = 12 345 yields s′ = 111 and r′ = 24. The DivRem call yields q = 11 and u = 25, which gives s = 11 111 and r = 2 468. e. 2). Still with input 123 456 789, we successively get s = 61 728 395, 30 864 198, 15 432 100, 7 716 053, 3 858 034, 1 929 032, 964 547, 482 337, 241 296, 120 903, 60 962, 31 493, 17 706, 12 339, 11 172, 11 111, 11 111. Convergence is slow because the initial value of u assigned at line 1 is much too large.