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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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544 ANSWERS TO EXERCISES 3.3.4 Without loss of generality, assume that f = g. Tf now S is any orthogonal matrix, the matrix US defines the same form as U, since (XUS)(XUS)T = (XU)(XU)T. Choosing S so that its first column is a multiple of VI and its other columns are any suitable vectors, we have for some ~1, ~2, . . . , crt and some (t -1) X (t -1) matrix U . Hence f(zl, . . . , zt) = (~1×1 + . . + a,~,) + h(zz,. ,z,). It follows that cy1 = fi [in fact, QI~ = (VI . Uj)/Je for 1 5 j 5 t], and that h is a positive definite quadratic form defined by U , where det U = (det U)/&. By induction on t, there are integers (~2,. . . , Q) with h(xz,. . . , xt) 5 (4) - / I detU12/(t- )/e /(t-1), and for these integer values we can choose x1 so that 1×1 + (~2×2 + … + a~tzt)/a~11 5 4, i.e., ((~1~1 + … + a,~,) 5 46 . Hence 0 2 f(xl, . . . ,x,) 2 $0 + (j)( - )/ 1 det U12/(t- )/81/(t- ) and the desired inequality follows immediately. [Note: For t = 2 the result is best possible. For general t, Hermite s theorem implies that pUt < ,t 2(4/3)t(t-1) 4/(t/2)!. A fundamental theorem due to Minkowski - ( Every t-dimensional convex set symmetric about the origin with volume 2 2t contains a nonzero integer point ) gives pt 2 2t; this is stronger than Hermite s theorem for t 2 9. Even stronger results are known, cf. (41).] 10. Since yl and yz are relatively prime, we can solve zllyz - 1~2~1 = m; furthermore (2~1 + qyl)y~ -(~2 + 4yz)yl = m for all 4, so we can ensure that 212~1~1 + 2~2~21 2 yf + yi by choosing an appropriate integer 4. Now y~(u1 f ~212) = ~22~1 - yluz = 0 (modulo m), and yz must be relatively prime to m, hence ZL~ + au2 = 0. Finally let lulyl + UZYZ~ = cum, u: + ZL~ = pm, y: + yz = ym; we have 0 5 QI 2 $7, and it remains to be shown that Q 2 $p and ,@ 2 1. The identity (2~1~2 -212~1) + (%Yl + u2Y2)2 = (4 +,4(Y: + YE2 im Pl ies that 1 + ~1 = /3r. If QI > a/3, we have 2ay > 1+ cy2, i.e., y -dm < 0: 5 %r. But $7 < dm implies that y2 > 4, a contradiction. 11. Since a is odd, y1 +yz must be even. To avoid solutions with yl and y2 both even, let y1 = ~1 +x2, y2 = x1-22, and solve x:+x; = m/&–E, with (~1, x2) relatively prime and x1 even; the corresponding multiplier a will be the solution to (x2 -x,)a = 52 + x1 (modulo 2e). It is not difficult to prove that a = 1 (modulo 2k-t1) iff x1 = 0 (modulo 2k), so we get the best potency when x1 mod4 = 2. The problem reduces to finding relatively prime solutions to XT + xi = N where N is a large integer of the form 41c $1. By factoring N over the Gaussian integers, we can see that solutions exist if and only if each prime factor of N (over the usual integers) has the form 41c + 1. According to a famous theorem of Fermat, every prime p of the form 4k + 1 can be written p = u2 + 21 = (u + iv)(u -iv), v even, in a unique way except for the signs of u and V. The numbers u and v can be calculated efficiently by solving x2 = -1 (modulo p), then calculating u + iv = gcd(z + i, p) by Euclid s algorithm over the Gaussian integers. [We can take x = n(p-1)/4 modp for almost half of all integers n. This application of a Euclidean algorithm is essentially the same as finding the least nonzero u2 + 21 such that u * XV = 0 (modulo p).] If the prime factorization of N is p; . . . p: = (ul + it~~)~l(u~ -i~)~ . . . (u? + ~w~)~ (u~ -iw,)+, we get 2 - distinct solutions to xf + xi = N, gcd(xl, x2) = 1, x1 even, by letting ~XZ[ f ilxl = (u, + iwp(UZ &-i2rp.. (u, f iw7)ev; and all such solutions are obtained in this way.
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3.3.4 ANSWERS TO EXERCISES 543 3. a2 E 2a -1 and a3 = 3a -2 (modulo m). By considering all short solutions of (15), we find that V: = 6 and V: = 4, for the respective vectors (1, -2,1) and (1, -1, -1, l), except in the following cases: m = 2 q, q odd, e 2 3, a E 2e-1 (modulo 2 ), a E 1 (modulo q), vi = vi = 2; m = 3 q, gcd(3, q) = 1, e > 2, a = 1 & 3e- (modulo 3 ) a = 1 (modulo q), vi = 2; m = 9, a = 4 or 7, V; =Yi = 5. 4. (a) The unique choice for (~1, ~2) is &((Y~UZZ -~2~21, -ylu12 + ysuil), and this is = &(y1u22 + y2au22, –ylul~ -yzauiz) = (0,O) (modulo 1); i.e., ~1 and 52 are integers. (b) When (x1,52) # (O,O), we have (ziuri + ~2usi)~ f (51~2 $ ~2~22)~ = z~(u:, + ut2) + z~(u& + u&) + ~Z~ZZ(U~~UZ~ + ur2u22), and by hypothesis this is L (d +x2 -IW21)(~11 + G2) 2 $1 + 42. [Note that this is a stronger result than Lemma A, which tells us only that Z; 2 (UT, + uf2)(u& + ui2)/m2 and that d I (& +~U:2)~/m , where the latter can be 2 1. The idea is essentially Gauss s notion of a reduced binary quadratic form, Disq. Arith. (Leipzig: 1801) li l.] 5. Conditions (30) remain invariant; hence h cannot be zero in step S2, when a is relatively prime to m. Since h always decreases in that step, S2 eventually terminates with u2 + w2 2 s. Note that pp 5 0 throughout the calculation. The hinted inequality surely holds the first time step S2 is performed. The integer g that minimizes (h -qh)2 + (p -q p)2 is q = round((h h + p p)/(h2 + p )), by (24). If (h -q h)2 + (p -q p)2 < h2 + p2 we must have q # 0, q # -1, hence (p -q p)2 2 p2, hence (h -q h)2 < h2, i.e., Ih -q hJ < h, i.e., q is q or q + 1. We have hu + pv 2 h(h -q h) + p(p -q p) 2 -$(h2 f p ), so if u2 + v2 < s the next iteration of step S2 will preserve the assumption in the hint. If u2 + v2 2 s > (u -h)2 + (v -p) , we have 2/h(u -h) + p(v -p)l = 2(h(h -u) + p(p -v)) = (u -h)2 + (v -p) + h2 + p2 -(u + v ) 2 (u -h)2 + (v -P)~ 5 h2 + p2, hence (u-h)2+(v-p)2 is minimal by exercise 4. Finally if both u2+v2 and (u-h)2+(v-p)2 are 2 s, let u = h -q h, II = p -q p; then 2lhu + pv l 5 h2 + p2 5 d2 + d2, and h2 + p2 is minimal by exercise 4. 6. If u2 +v2 2 s > (u-h)2 +(v-p) in the previous answer, we have (w–P)~ > v2, hence (u -h)2 < u2; and if q = u3, so that h = u,h + u, we must have a,+1 = 1. It follows that ~2 = minc~3 2mjp,-1, hence 7. We shall prove, using condition (19), that u, . uk = 0 for all k # j iff Vj . Vk = 0 for all k # j. Assume that U, uk = 0 for all k # j, and let U, = CYIVI $- . . . + at&. Then U, . vk = ok for all k, hence U, = cyJV3, and V; . fi = cxT1(U3 . fi) = 0 for all k # j. A symmetric argument proves the converse. 8. Clearly vt+l < vt (a fact used implicitly in Algorithm S, since s is not changed when t increases). For t = 2 this is equivalent to (mp2/.rr) /2 > ($mps/.rr)1/3, i.e., P3 L 4&%2 3/2 . This reduces to $10e4 /fi with the given parameters, but for large m and fixed ~2 the bound (39) is better. 9. Let f(yl, . . . , yt) = 8; then gcd(yl, , yt) = 1, so there is an integer matrix W of determinant 1 having (2~1, . . , Wt) as its first row. (Prove the latter fact by induction on the magnitude of the smallest nonzero entry in the row.) Now if X = (21,. . . , xt) is a row vector, we have XW = X iff X = X W- , and W-l is an integer matrix of determinant 1, hence the form g defined by WU satisfies g(zl, . . . , xt) = f(zi, . . . , 2:); furthermore g(l, 0,. . . ,O) = 0.
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542 ANSWERS TO EXERCISES 3.3.3 ,l, 1) y=x-; Y++;-; / y=r-a! / l,l-a) yEz-I-a 2 2 M= g -cy 2 / / I,;-; > 1, c9 Fig. A-3. Permutation regions for a generator with potency 2; (Y = (u -l)c/m. 28. Fig. A-3 shows the various regions in the general case. The 213 region means UZ < UI < Us, if WI and UZ are chosen at random; the 321 region means that Us < U2 < Ul, etc. The probabilities for 123 and 321 are $ -cy/2 + ~ 12; the probabilities for all other cases are Q + (r/4 -cu2/4. To have all equal to &, we must have 1 -6cu + 6cu2 = 0. [This exercise establishes a theorem due to J. N. Franklin, Math. Comp. 17 (1963), 28-59, Theorem 13; other results of Franklin s paper are related to exercises 22 and 23.1 SECTION 3.3.4 1. VI is always m and ~1 = 2, for generators of maximum period. 2. Let V be the matrix whose rows are VI, . . . , vt. To minimize Y . Y, subject to the condition that Y # (0,. . . , 0) and VY is an integer column vector X, is equivalent to minimizing (V-lx). (V-lx), subject to the condition that X is a nonzero integer column vector. The columns of V- are UI, . . . , Ut.
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3.3.3 ANSWERS TO EXERCISES 541 Fig. A-l. Permutation regions for Fibonacci generator. Fig. A-2. Run-length regions for Fibonacci generator. 24. Proceed as in the previous exercise; the sum of the interval lengths is 5 c at- (a -1) = at&)(a+:-2) . ol k>l The value for a truly random sequence would be e -1; and our value is e -1 + (e/2 -1)/a + 0(1/a ). [Note: The same result holds for an ascending run, since we have U,, > Un+I if and only if 1 -U,, < 1 -Un+I. This would lead us to suspect that runs in linear congruential sequences might be slightly longer than normal, so the run test should be applied to such generators.] 25. z must be in the interval [(k + a -Q/a, (k + p -0)/a) for some k, and also in the interval [a, p). Let ko = [aa: + 0 -p l, ICI = lab + 0 -/3 l. With due regard to boundary conditions, we get the probability (kl -ko)(P -a/)/a -I- max(0, P -(kl + a -0)/a) -max(O, cx-(k. + a -.9)/a). This is (/3 -cu)(p -CY ) + E, where 1~1 < 2(p -&)/a. 26. See Fig. A-l; the orderings U1 < Us < Uz and Uz < U3 < UI are impossible; the other four each have probability f . 27. U, = {F,-~UO $ F,Ul}. We need to have both F~--~UO + Fkul < 1 and FkUo + Fk+l VI > 1. The half-unit-square in which UO > UI is broken up as shown in Fig. A-2, with various values of k indicated. The probability for a run of length k is 3, if k = 1; l/Fk-l Fk+l -l/Fk Fk+2, if k > 1. The corresponding probabilities for a random sequence are 2k/(k + l)! -2(k + l)/(k + 2)!; the following table compares the first few values. k: 1234 5 Probability in Fibonacci case: f 4 & & & Probability in random case: B s2 ti3%i&
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540 ANSWERS TO EXERCISES 3.3.3 Ss = –a(a (a -l), m, (a ) ~), if a a E 1 (modulo m); and finally ss=12 c O -1) > which is i + (1 -30 + 3e2)/6a + 0(1/a ) for large a. Note that 1 -38 + 319 2 a, so 8 can t be chosen to make this probability come out right.
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3.3.3 ANSWERS TO EXERCISES 539 At the conclusion of this algorithm, p will be equal to the original value /CO of Ic, so the desired answer will be A+ B/p. The final value of p will be h if s < 0, otherwise p will be ko -h . It would be possible to maintain B in the range 0 < B < ko , by making appropriate adjustments to A, thereby requiring only single-precision operations (with double-precision products and dividends) if ko is a single-precision number. 18. A moment s thought shows that the formula S(h, kc, 2) = _ (b/k1 lb - z)lkJ) (((W + d/k)) Co<,<, - is in fact valid for all z, not only when k 2 z. Writing [j/kJ -[(j -z)/kJ = $ + ((y)) -((f)) + fS,o -$6(q) and carrying out the sums yields S(h, k, c, z) = zd((c/d))/k + &a(!~, k, hz + c) -&o(h, k, c) + i((c/k)) -d(((hz + c)/k)), where d = gcd(h, k). [This formula allows us to express the probability that Xn+l < X, < (Y in terms of generalized Dedekind sums, given a.1 19. The desired probability is c OIzFrom our experience, we are can tell you that you can find a reliable and cheap webhost service at Java Web Hosting services.

538 ANSWERS TO EXERCISES 3.3.3 8. See Duke Math. J. 21 (1954), 391-397. 9. Begin with the handy identity ~O 0, return to D2. 1
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3.3.3 ANSWERS TO EXERCISES 537 18. (a) The numerator is -(Ue -UI)*, the denominator is (Uc -UI)*. (b) The nu- merator in this case is -(iYE + UT + U , -UOUI -VI& -U2i3); the denominator is 2(Ui +…–U~UO). (c) The denominator always equals ~o13,c-sm 2rnx)/n, which converges for all z. (The representation in Eq. (24) may be considered a finite Fourier series, for the case when x is rational.) 4. d = 21 5. Note that we have Xn+l < X, with probability a + 6, where 161 < d/(2. lOlo) = l/(2. 5 ); hence every potency-l0 generator is respectable from the standpoint of Theorem P. 5. An intermediate result: x 4x1 ;~(a, m, c) + T -& -&. c mm = OCheck Tomcat Web Hosting services for best quality webspace to host your web application.

536 ANSWERS TO EXERCISES 3.3.2 by the previous exercise and Eq. 1.2.9-28. The mean and variance are readily computed using Theorem 1.2.10A and exercise 3.4.1-17. We find that mean(G)=w+(&-l)+…+(d-:+I–I)=d(Hd–Hdw)=ii: var(G) = d (Hy) -H&2_1,) -d(Hd -H–W) = oz. The number of U s examined, as the search for a coupon set is repeated n times, therefore has the characteristics (min wn, ave pn, max 00, dev a&z). 11. 111219 8 5 31617 0141. 12. Algorithm R (Data for run test). Rl. [Initialize.] Set j + -1, and set COUNT[l] + COUNT[2] + . . . + COUNT[S] + 0. Also set U, + U,-i, for convenience in terminating the algorithm. R2. [Set r zero.] Set r + 0. R3. [Is U, < U,+I?] Increase r and j by 1. If U, < U3+1, repeat this step. R4. [Record the length.] If r 2 6, increase COUNT[G] by one, otherwise increase COUNT[r] by one. R5. [Done?] If j < n -1, return to step R2. 1 13. There are (p+ 4 + l)( t ) ways to have Ui-1 2 Ui < . . . < Uz+p-~ 3 U,++, < < Uz+p+q-1; subtract ( ~~- ; ) of th ese in which lJ-1 < Ui, and subtract ( pf;l+l) for those in which Uz+p--l < Ui+,; then add in 1 for the case that both U,-I < U, and Uz+p--l < U,+,, since this case has been subtracted out twice. (This is a special case of the inclusion-exclusion principle, which is explained further in Section 1.3.3.) 14. A run of length r occurs with probability l/r! -l/(r + l)!, assuming distinct U s. 15. This is always true of F(X) when F is continuous and S has distribution F; see Section 3.3.1C. 16. (a) Z i, = max(Z,(t-l),Z(j+l)(t-l)). If the Qt-i) are stored in memory, it is therefore a simple matter to transform this array into the set of Z,, with no auxiliary storage required. (b) With his improvement, each of the V s should indeed have the stated distribution, but the observations are no longer independent. In fact, when U, is a relatively large value, all of Z,,, Z(j+l)t, . , Z(j-t+l)t will be equal to Uj; so we almost have the effect of repeating the same data t times (and that would multiply V by t, cf. exercise 3.3.1-10). 17. (b) By Lagrange s identity, the difference is ~O has rank < 2, so its rows are linearly dependent. (A more elementary proof can be given, using the fact that UhVj -l.JiVi = 0 for 1 2 j < n implies the existence of constants cy, p such that cdJi + /3Vi = 0 for all j, provided that Vi and VA are not both zero; the latter case can be avoided by a suitable renumbering.)
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