Web domain - 546 ANSWERS TO EXERCISES 3.3.4 matrices; thus the
546 ANSWERS TO EXERCISES 3.3.4 matrices; thus the two-dimensional case will be handled by the general method. A special procedure for t = 2 could, of course, be devised.) In steps S4 and S8, set s + min(s, ]Uk]). In step S8, set zk + lmaxl~j~t IwkjIs/m]. In step slo, set s + min(s, ]Y( -6); and in step Sll, output s = Nt. Otherwise leave the algorithm as it stands, since it already produces suitably short vectors. [Math. Comp. 29 (1975) 827-833.1 17. When k > t in SlO, and if Y .Y 2 s, output Y and -Y; furthermore if Y .Y < s, take back the previous output of vectors for this t. [In the author s experience preparing Table 1, there was exactly one vector (and its negative) output for each vt, except when yl = 0 or yt = 0.1 18. (a) Let 5 = m, y = (1 -m)/3, vZj = y + x&j, uij = -y + 6ii. Then V, . Vj = *(m -1) for j # Ic, vk. vk = $(m2 + JJ), u, . u, = $(m + 2), zk M Am. (This example satisfies (28) with a = 1 and works for all m G 1 (modulo 3).) (b) Interchange the roles of U and V in step S5. Also set s + min(s, Ui. Vi) for all U, that change. For example, when m = 64 this transformation with j = 1, applied to the matrices of (a), reduces v= -21 y: 2; Z), u=(f :; Z) ( to v= ( -21 l 4: -2q, u=g ; p). -21 -21 43 [Since the transformation can increase the length of V,, an algorithm that incorporates both transformations must be careful to avoid infinite looping. See also exercise 23.1 19. No, since a product of non-identity matrices with all off-diagonal elements non- negative and all diagonal elements 1 cannot be the identity. [However, looping would be possible if a subsequent transformation with CJ = -1 were performed when -2V,. V, = V, . V.; the rounding rule must be asymmetric with respect to sign if non-shortening transformations are allowed.] 20. Use the ordinary spectral test for a and m = 2e-2; cf. exercise 3.2.1.2-9. [On intuitive grounds, the same answer should apply also when a mod 8 = 3.1 21. Xdn+4 E X4, (modulo 4) so it is now appropriate to let VI = (4, 4a2, 4a3)/m, V2 = (0, 1, 0), V3 = (0, 0,l) define the corresponding lattice LO. 24. Let m = p; an analysis paralleling the text can be given. For example, when t = 4 we have Xn+s = ((u + b)Xn+l + abX,)modn, and we want to minimize uf + uz + uz + ui # 0 such that ui + bug + abud E u2 + aus + (a + b)uq = 0 (modulo m). Replace steps Sl through S3 by the operations of setting and outputting ~2 = m. Replace step S4 by
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