3.3.4 THE SPECTRAL TEST 95 vl,…, (Web server setup) Vt such
3.3.4 THE SPECTRAL TEST 95 vl,…, Vt such that For example, in the special form (16) that arises in the spectral test, we have Ul = ( m, 070,. . . ,017 Vl =+$,a$2 ,…) at- ), u2 = ( –a, 1, 0, . . . , O), v2 = (0, 1, 0,. . . , O), us = ( -u2, 0, 1, . . . ) O), v, = (O,O, 1,. . . , O), (20) . . . . . . . . . . . . ut = (-d-l, 0, 0, . . . , l), vt = (O,O, 0,. . . , 1). These V, are precisely the vectors (8), (9) that we used to define our original lattice Lo. As the reader may well suspect, this is not a coincidence-indeed, if we had begun with an arbitrary lattice LO, defined by any set of linearly independent vectors VI, . . ,V,, the argument we have used above can be generalized to show that the maximum separation between hyperplanes in a covering family is equivalent to minimizing (17), where the coefficients uij are defined by (19). (See exercise 2.) Our first step in minimizing (18) is to reduce it to a finite problem, i.e., to show that we won t need to test infinitely many vectors (~1, . . . , Q) to find the minimum. This is where the vectors VI, . . . , Vt come in handy; we have xk = (xl& + +xtut) vk, and Cauchy s inequality tells us that ((xlul + + xtut) vk)2 2 j-(x1,. . . > xt)(Vk vk). Hence we have derived a useful upper bound on each coordinate xk: Lemma A. Let (xl,…, q) be a nonaero vector that minimizes (18) and let (Yl,.. . , yt) be any nonzero integer vector. Then 2: 2 (vk vk/k)f(Yl , . . , Yt ), for 1 < k 5 t. (21) In particular, letting yz = Sij for all i, 2; 5 (vk vk)(uy u,), for 15 j,k 5 t. 1 (22) Lemma A reduces the problem to a finite search, but the right-hand side of (21) is usually much too large to make an exhaustive search feasible; we need at least one more idea. On such occasions, an old maxim provides sound advice: If you can t solve a problem as it is stated, change it into a simpler problem that
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