3.3.4 ANSWERS TO EXERCISES 545 Note: (Submit web site) When m
3.3.4 ANSWERS TO EXERCISES 545 Note: When m = 10e, a similar procedure can be used, but it is five times as much work since we must keep trying until finding a solution with 21 G 0 (module 10). For example, when m = lOlo we have [m/&l = 5773502691, and 5773502689 = 53.108934013 = (7 + 2i)(7 -2i)(2203 + 102022)(2203 -10202i). Of the two solutions 1×21 + ilxll = (7 + 2i)(2203 + 10202i) or (7 + 23 )(2203 -10202i), the former gives Izi 1 = 67008 (no good) and the latter gives 1~1 I = 75820, 1~1 = 4983 (which is usable). Line 9 of Table 1 was obtained by taking 51 = 75820, 2s = -4983. Line 20 of the table was obtained as follows: [2 /&] = 19837604196; we drop down to 19837604193, which is divisible by 3 so it is ineligible. Similarly, 19837604189 is divisible by 19, and 19837604185 by 7, and 19837604181 by 3; but 19837604177 is prime and equals 1318842 + 4943g2. The corresponding multiplier is 1175245817; a better one could be found if we continued searching. The multiplier on line 24 is the best of the first sixteen multipliers found by this procedure when m = 232. 12. U, U, = u, . u, + 2 -&, s4Ut . U,) + czzj xkfj qm@, . Uk). The partial derivative with respect to ok is twice the lef&hand side of (26). If the minimum can be achieved, these partial derivatives must all vanish. 13. ~~11 = 1, 2~21 = irrational, ~12 = 2~22 = 0. 14. After three Euclidean steps we find V; = 52 + 52, then S4 produces Transformations (j, 41, q2, q3) = (1, *, 0,2), (2, -4, *, l), (3,0,0, *), (1, *, 0,O) result in Thus vs = &, as we already knew from exercise 3. 15. The largest achievable q in (ll), minus the smallest achievable, plus 1, is lull + . . + lutl–6, where 6 = 1 if uluJ < 0 for some i and j, otherwise 6 = 0. For example if t = 5, u1 > 0, ILZ > 0, 11s > 0, 2~4 = 0, and 2~5 < 0, the largest achievable value is q = ~1 + u2 + us -1 and the smallest is q = 2~5 + 1 = -/usI+ 1. [Note that the number of hyperplanes is unchanged when c varies, hence the same answer applies to the problem of covering L instead of LO. However, the stated formula is not always exact for covering LO, since the hyperplanes that intersect the unit hypercube may not all contain points of LO. In the example above, we can never achieve the value q = u1 + ua + us -1 in LO if ZL~ + 2~s + us > m; it is achievable iff there is a solution to m -ul-2~2 -us = xiui + 22~s + 5sus + z4 Ius I in nonnegative integers (21, ~2, xs, x4). It may be true that the stated limits are always achievable when IZL~ I + . + 1~~ I is minimal, but this does not appear to be obvious.] 16. It suffices to determine all solutions to (15) having minimum Iui I + . . . + 1~~1, subtracting 1 if any one of these solutions has components of opposite sign. Instead of positive definite quadratic forms, we work with the somewhat similar function f(xi, . . . , xt) = IxlUl +. . .+xtUtl, defining IYI = I~il+~~~+l~tl. Inequality (21) can be replaced by 1% I (maxlsjst lvkjl) f(y~, . . . ,yt). Thus a workable algorithm can be obtained as follows. Replace steps Sl through S3 by: Set JJ c (m), V t (l), r +-1, s +- m, t +- 1. (Here U and V are 1 x 1
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