Web site templates - 94 RANDOM NUMBERS 3.3.4 hence the minimum in
94 RANDOM NUMBERS 3.3.4 hence the minimum in (12) occurs when each xj = uj/(uf + .. . + uz); the distance between neighboring hyperplanes is 1 dm = l/length(U). (14I In other words, the quantity ut we seek is precisely the length of the shortest ve&or U that defines a family of hyperplanes { X. U = q 1 integer q } containing all the elements of Lo. Such a vector U = (~1,. . . , ut) must be nonzero, and it must satisfy V.U = integer for all V in Lo. In particular, since the points (1, 0, . . . , 0), (0, 1, . . . , 0), . . . . (0,O )…) 1) are all in Lo, all of the uj must be integers. Furthermore since VI is in Lo, we must have &(ul + au2 + . . . + atP1ut) = integer, i.e., Ul + au2 + . . . + &lut s 0 (modulo m). (15) Conversely, any nonzero integer vector U = (ul, . . . , ut) satisfying (15) defines a family of hyperplanes with the required properties, since all of LO will be covered: (y1Vl + . . + + yt&) . U will be an integer for all integers yl, . . . , yt. We have proved that u:= min {uf+-..+uF Iu1+uu2+..-+cP1ut =O(modulom)} (~1,..-,~t)#(O,…,O) =(%I,…,$; (O,…,0)(( mx~-ux~-u2×3-*. . -ut-1xt)2+x;+x;+~~~ +xt ). (16) C. Deriving a computational method. We have now reduced the spectral test to the problem of finding the minimum value (16); but how on earth can we determine that minimum value in a reasonable amount of time? A brute-force search is out of the question, since m is very large in cases of practical interest. It will be interesting and probably more useful if we develop a computational method for solving an even more general problem: Find the minimum value of the quantity f(Xl,… , xt) = (%1X1 + . . . + Utlxt)2 + . . . + (%X1 + * *. + %xt)2 (17) over all nonzero integer vectors (xl,. . . , xt), given any nonsingular matrix of oefficients U = (uij). The expression (17) is called a positive definite quadratic . form m t variables. Since U is nonsingular, (17) cannot be zero unless the xj are all zero. Let us write VI, .,., Ut for the rows of U. Then (17) may be written f(Xl,.-. , xt) = (Xl& + . . . + XtUt) . (Xl& + … $-x&s), (18) the square of the length of the vector x1 VI + . . . + xtUt . The nonsingular matrix U has an inverse, which means that we can find uniquely determined vectors
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