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3.4.2 ANSWERS TO EXERCISES 555 11. Arguing as in Section 1.2.10, which considers the special case n = 1, we see that the generating function is >( The mean is n + znctcN (n/t) = n(1 + H,v -Hn); and the variance turns out to be n(HN -If,) -n2(H~)-Ha)). 12. (Note that ~-l = (btt). . (bs3)(&2), so we seek an algorithm that goes from the representation of VT to that for r-l.) Set b, + j for 1 5 j 2 t. Then for j = 2, 3, t (in this order) interchange b, c) ba3. Finally for j = t, , 3, 2 (in this order) sei baJ + b,. (The algorithm is based on the fact that (att)nl = nl(btt).) 13. Renumbering the deck 0, 1, . , 2n -2, we find that s takes card number z into card number (22) mod (2n -I), while c takes card z into (zz f 1) mod (2n -1). We have (c followed by s) = cs = sc2. Therefore any product of c s and s s can be transformed into the form sZck. Also 2 p(2n-1) = 1 (modulo (2n -1)); since .s~( ~- ) and c2n-1 are the identity permutation, at most (2n -l)p(2n -1) arrangements are possible. (The exact number of different arrangements is (2n -1)/c, where k is the order of 2 modulo (2n -1). For if sk = c3, then c3 fixes the card 0, so sk = c3 = identity.) For further details, see SIAM Review 3 (1961), 293-297. 14. Set Y, +-j for t -n < j 2 t. Then for j = t, t -1, . . , t -n f 1 do the following operations: Set k + Lju] + 1. If k > t -n then set X, + Yk and Yk + Y,; otherwise if Ic = X, for some i > j (a symbol table algorithm could be used), then set X, + Y, and Yi + Y3; otherwise set X, + k. (The idea is to let Ytpn+l, , Yj represent Xt–n+l, . . . , X,, and if i > j and X, 2 t -n also to let Y, represent Xx,, in the execution of Algorithm P.) 15. We may assume that n 5 iN, otherwise it suffices to find the N-n elements not in the sample. Using a hash table of size 2n, the idea is to generate random numbers between 1 and N, storing them in the table and discarding duplicates, until n distinct numbers have been generated. The average number of random numbers generated is N/N+N/(N-l)+…+N/(N-n+l) < 2n, by exercise 3.3.2-10, and the average time to process each number is O(1). We want to output the results in increasing order, and this can be done as follows: Using an ordered hash table (exercise 6.4-66) with linear probing, the hash table will appear as if the values had been inserted in increasing order. Thus if we use a monotonic hash address such as [nlc/iV] for the key k, it will be a simple matter to output the keys in sorted order by making at most two passes over the table. 16. In most cases it will be best simply to select n records one at a time by Walker s alias method, with probabilities proportional to their weights, rejecting an element that occurs more than once. But if some weights are significantly larger than others, the following algorithm will sometimes be better [cf. Wong and Easton, SIAM J. Cornput. 9 (1980), ill-1131. Begin with a complete binary tree of weights x2, where x2 = ~2~ + zzi+l for 1 5 i < N and z,Afz-l = w, for 1 5 i 2 N; then do the following operation n times: Set j + 1, and generate X = Uxl. If X < XZ~, set j +--2j, otherwise set X + X -2~~ and j + 2j f 1; repeat this until j > N. Then select elementj-N+l, seti + [j/2], and while i 2 1 set x2 +- zz -x3 and i +-[i/ZJ. Finally set q + 0.
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554 ANSWERS TO EXERCISES 3.4.1 29. Let Xn+l = 1, then set Xk + Xk+lUi k or Xk + Xk+tlePYklk for k = n, n-l, . . . . 1, where uk is uniform or Yk is exponential. [ACM Trans. Math. Software, to appear.] SECTION 3.4.2 1. There are (rr$) ways to pick n -m records from the last N -t; (:!,I:) ways to pick n -m -1 from N -t -1 after selecting the (t + 1)st item. 2. Step S3 will never go to step S5 when the number of records left to be examined is equal to n -m. 3. We should not confuse conditional and unconditional probabilities. The quantity m depends randomly on the selections that took place among the first t elements; if we take the average over all possible choices that could have occurred among these elements, we will find that (n -m)/(N -t) is exactly n/N on the average. For example, consider the second element; if the first element was selected in the sample (this happens with probability n/N), the second element is selected with probability (n -l)/(N -1); if the first element was not selected, the second is selected with probability n/(N -1). The overall probability of selecting the second element is (nlN)((n -l)l(N -1)) i- (1 -nlN)(nlW -1)) = n/N. 4. From the algorithm, p(m,t + 1) = (1 - E) p(m, t) + n -i , ) dm - 1, t), The desired formula can be proved by induction on t. In particular, p(n, N) = 1. 5. In the notation of exercise 4, the probability that t = k at termination is qk = p(n, k) -p(n, k -1) = (:I:)/(:). The average is ~,,,,, kqk = (N $- l)n/(n f 1). 6. Similarly, co,,,, k(k + l)qk = (N + 2)(N $-l)n/(n + 2); the variance is therefore (N + l)(N -n)n/(n + 2)(n + 1) . 7. Suppose the choice is 1 5 ~1 < 22 < ... < zn 2 N. Let 20 = 0, zn+l = N+l. The choice is obtained with probability p = n,,,,N--pt, where p = N-((t-1)-n+m t for xm < t < xm+l; N -(t -1) n-m pt = N-@-l) for t = zm+l. The denominator of the product p is N!; the numerator contains the terms N -n, N-n -1, , 1 for those t s that are not z s, and the terms n, n -1, . . , 1 for those t s that are z s. Hence p = (N -n)!n!/N!. Example: n = 3, N = 8, (~1~22,~s) = (2,3,7); p = i$$$$$++. 9. The reservoir gets seven records: 1, 2, 3, 5, 9, 13, 16. The final sample consists of records 2, 5, 16. 10. Delete step R6 and the variable m. Replace the I table by a table of records, initialized to the first n records in step Rl, and with the new record replacing the Mth table entry in step R4.
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3.4.1 ANSWERS TO EXERCISES 553 This approach applies also to the binomial distribution, with 1 G(x) = U=- (l -U)n–2 dur(t + l)/r(z)r(t + 1 -x), s P since [G- (U)] is binomial with parameters (t, p) and G is approximately normal. [See also the alternative method proposed by Ahrens and Dieter in Computing (1980), to appear.] 23. Yes. The second method calculates Icos201, where 0 is uniformly distributed between 0 and 7rf 2. (Let U = r cos 0, V = r sin 0.) 25. #& = (.10101)2. In general, the binary representation is formed by using 1 for V and 0 for A, from left to right, then suffixing 1. This technique [cf. K. D. Tocher, J. Roy. Stat. Sot El6 (1954), 491 can lead to efficient generation of independent bits having a given probability p, and it can also be applied to the geometric and binomial distributions. 26. (a) True, c, Pr(N1 = k)Pr(Nz = 12 - k) = e-p1-@2(p1 + pz) /n!. (b) False, unless ~2 = 0; otherwise iV1 -N2 might be negative. 27. Let the binary representation of p be (&lb&.. .)2, and proceed according to the following rules: Bl. [Initialize.] Set m t t, N + 0, j + 1. (During this algorithm, m represents the number of simulated uniform deviates whose relation to p is still unknown, since they match p in their leading j-l bits; and N is the number of simulated deviates known to be less than p.) B2. [Look at next column of bits.] Generate a random integer M with the binomial distribution (m, a). (Now M represents the number of unknown deviates that fail to match b3.) Set m t m -M, and if b3 = 1 set N + N+M. B3. [Done?] If m = 0, or if the remaining bits (.b3+lb3+2.. .)z of p are all zero, the algorithm terminates. Otherwise, set j + j f 1 and return to step B2. 1 [When bJ = 1 for infinitely many j, the average number of iterations At satisfies A,, = 0; An=l+&~ ; Ak, for n 2 1. 0 k LettingA = zAnzn/n!, we haveA = e -l+A(~z)e /2. ThereforeA(z)e- = 1 - e- + A( $z)e-Z/2 = Ck,0(l-e-Z/2 ) = 1-e-Z-C,,1(-z)n/(7z!(2n -l)), and A,=l+C ; E =1+t V 0 n+l k>l in the notation of exercise 5.2.2.-48.1 28. Generate a random point (~1,. . . , yn) on the unit sphere, and let p = dm. Generate an independent uniform deviate U, and if pn+ U < K dm, output the point (yl/p, . . . , y,/p); otherwise start over. Here K2 = min{ (c aky~) + /(~ a:~:) 1 Cy&l}=a;--l f I nan >_ al, ((n + l)/(al f a7L))n+1(ulan/n)n otherwise.
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552 ANSWERS TO EXERCISES 3.4.1 Bl. [Initialize.] Set N +-0, q + pt, r +-q, and generate a random uniform deviate U. (We will have q = pN and r = po + . . . + pN during this algorithm, which stops as soon as U < r.) B2. [Search.] If ZJ 2 r, set N +-- N + 1, q t q(l -p)(t -1 $ N)/iV, r +-r $ q, and repeat this step. m [An interesting technique for the negative binomial distribution, for arbitrarily large real values of t, has been suggested by R. L6ger: First generate a random gamma deviate X of order t, then let iV be a random Poisson deviate of mean X(1 -p)/p.] 20. Rl = 1 + (1 -A/R) . Rl. When R2 is performed, the algorithm terminates with probability I/R; when R3 is performed, it goes to Rl with probability E/R. We have Rl RIA RIA RIA WA R2 0 WA 0 RIA R3 0 0 WA R/A -I/A R4 WA R/A -I/A R/A-E/A R/A -I/A -E/A 21. R = fi z 1.71153; A = fl F=Z1.25331. Since Ju&=&i du = (a -b~)~ ( $(a -bu) -3)/b2, we have I = soaib ud= du = =a 5f2/b2 where a = 4(1 + In c) and b = 4 4c; when c = e114, I has its maximum value g fi Z=Z 1.13020. Finally the following integration formulas are needed for E: JJbu--au2du = ib2a-3/2arcsin(2ua/b -1) + $ba- dn(2ua/b -l), Jdwdu= -&b2a-3f21n(~~+u&+b/2JSi) + +ba- dv(2ua/b + l), where a, b > 0. Let the test in step R3 be X2 2 4e - /U -4x ; then the exterior region hits the top of the rectangle when u = T(X) = (eZ -de2= -2es)/2ez. (Incidentally, T(Z) reaches its maximum value at z = l/2, a point where it is not, differentiable!) We have E = ~~ (~ -m) du where b = 4eZe1 and a = 4x. The maximum value of E occurs near x = –.35, where we have E FZ .29410. 22. (Solution by G. Marsaglia.) Consider the continuous Poisson distribution de- fined by G(s) = s, emttz- dt/r(z), for z > 0; if X has this distribution then 1×1 is Poisson distributed, since G(z + 1) -G(z) = e- @Z/s!. If p is large, G is approximately normal, hence G-l(F,(x)) is approximately linear, where F,(Z) is the distribution function for a normal deviate with mean and variance p; i.e., Fp(z) = F((x -~)ldF), w here F(z) is the normal distribution function (10). Let g(x) be an efficiently computable function such that IG- (F,(x)) -g(x)/ < 6 for -co < 3: < 00; we can now generate Poisson deviates efficiently as follows: Generate a normal deviate X,andset~tg(~++X),Nt~YJ,M~~Y+~~.If(Y-~~>~,outputN; otherwise output M -1 or M, according as G- (F(X)) < M or not.
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3.4.1 ANSWERS TO EXERCISES 551 13. Take b, = p3; consider now the problem with pL3 = 0 for each j. In matrix notation, if Y = AX, where A = (a,,), we need AAT = C = (ct3). (In other notation, if Y3 = c alkXk, then the average value of YiY3 is c a,kajk.) If this matrix equation can be solved for A, it can be solved when A is triangular, since A = BU for some orthogonal matrix U and some triangular B, and BBT = C. The desired triangular solution can be obtained by solving the equations ufl = ~11, alla21 = c12, a21 2 + a22 2 = c22, alla31 = c13, a21a31 $-a22a32 = c23, . . , successively for all, a21, ~22, ~31, ~32, etc. [Note: The covariance matrix must be positive semidefinite, since the average value of (c y,Y,) is c c%,y,yj, which must be nonnegative. And there is always a solution when C is positive semidefinite, since C = U- diag(Al, . . . , X,)U, where the eigenvalues X, are nonnegative, and U- diag(&, . . . , &)U is a solution.] 14. F(s/c) if c > 0, a step function if c = 0, and 1 -F(z/c) if c < 0. 15. Distribution s- , Fl(z -t) &72(t). Density s- , fl(l: -t)fi(t) dt. This is called the convolution of the given distributions. 16. It is clear that f(t) 2 cg(t) f or all t as required. Since sooo g(t) dt = 1 we have g(t) = CtaP1 for 0 5 t < 1, Cc-t for t 2 1, where C = ue/(u + e). A random variable with density g is easy to obtain as a mixture of two distributions, Gl(z) = za for 0 5 z < 1, and Gz(s) = 1 -e -% for 5 2 1: Gl. [Initialize.] Set p +-e/(u + e). (This is the probability that G1 should be used.) G2. [Generate G deviate.] Generate independent uniform deviates U, V, where V # 0. If U < p, set X +-V lfa and q + ePX; otherwise set X + 1 -In V and q + X - . (Now X has density g, and q = f(X)/cg(X).) G3. [Reject?] Generate a new uniform deviate U. If U 2 q, return to G2. m The average number of iterations is c = (a + e)/(er(a + 1)) < 1.4. It is possible to streamline this procedure in several ways. First, we can replace V by an exponential deviate Y of mean 1, generated by Algorithm S, say, and then we set X + ePy or X + 1 + Y in the two cases. Moreover, if we set q t pe-* in the first case and q t p + (1 -p)X - in the second, we can use the original U instead of a newly generated one in step G3. Finally if U < p/e we can accept V la immediately, avoiding the calculation of q about 30 percent of the time. 17. (a) F(z) = 1 -(1 -p) , for z 2 0. (b) G(z) = pz/(l -(1 -p)z). (c) Mean l/p, standard deviation m/p. To do the latter calculation, observe that if H(z) = q + (1 -q)z, then H (1) = 1 -q and H (1) + H (1) -(H (1))2 = q(l -q), so the mean and variance of l/H(z) are q -1 and q(q -l), respectively. (See Section 1.2.10.) In this case, q = l/p; the extra factor z in the numerator of G(z) increases the mean by one. 18. Set N c Nl + NZ -1, where N1 and NZ independently have the geometric distribution for probability p. (Consider the generating function.) 19. Set N + N1 +. . . + Nt -t, where the Nj have the geometric distribution for p. (This is the number of failures before the tth success, when a sequence of independent trials are made each of which succeeds with probability p.) Fort=p= f, and in general when the mean value (namely t(1 -p)/p) of the distribution is small, we can simply evaluate the probabilities pn = ( -iWn)p (l -p) consecutively for n = 0, 1, 2, . . . as in the following algorithm:
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550 ANSWERS TO EXERCISES 3.4.1 7. If Ic = 1, then nl = n and the problem is trivial. Otherwise it is always possible to find i # j such that n, 5 n 5 nj. Fill Bi with n, cubes of color Ci and n -n2 of color Cj, then decrease n3 by n -ni and eliminate color C,. We are left with the same sort of problem but with k reduced by 1; by induction, it s possible. The following algorithm can be used to compute the P and Y tables: Form a list of pairs (pi, 1). . . (pk, Ic) and sort it by first components, obtaining a list (91, al). (qk, ak) where q1 2 … 2 qk and n = k. Repeat the following operation until n = 0: Set P[al -l] t kql and Y[ai -l] + za,. Delete (41, ai) and (qn, a,), then insert the new entry (q,, -(l/k -ql), a,) into its proper place in the list and decrease n by 1. (If p, 5 l/k the algorithm will never put xj in the Y table; this fact is used implicitly in Algorithm M. The algorithm attempts to maximize the probability that V < PK in (3), by always robbing from the richest remaining element and giving it to the poorest. However, it is very difficult to determine the absolute minimum of this probability, since such a task is at least as difficult as the bin-packing problem ; cf. Chapter 7.) 8. Replace P3 by (j + Pj)/k for 0 5 j < k. 9. Consider the sign of f (s) = m (x2 -l)e-Z2/2. 10. Let S j = (j -1)/5 for 1 2 j 2 16 and pj+ls = F(S,+i) -F(Sj) -p, for 1 < j 5 15; also let ps1 = 1 -F(3) and psz = 0. (Eq. (15) defines pl, . . . , ~~5.) The algorithm of exercise 7 can now be used with k = 32 to compute P3 and Yj, after which we will have 1 5 Y3 2 15 for 1 < j 2 32. Set PO + P32 (which is 0) and Yc + Y32. Then set 2, + l/(5 -5Pj) and Yj + SYj -2, for 0 < j < 32; Q3 + 1/(5Pj) for 1 2 j 5 15. Let h = 3 and f3+15(x) = ~(~--za 2-~-jz 50)/pj+~~ for S j 5 x 5 S,+h. Then let a3 = f3+is(SJ) for 1 5 j 5 5, bj = fJ+ls(Sj) for 6 2 j 2 15; also b, = 44+,5 (Sj +h) for 1 5 j 5 5, and o3 = fj+15(xJ)+(xj-Sj)bj/h for 6 5 j 2 15, where x3 is the root of the equation f:+ls(Zj) = -b,lh. Finally set D3+15 t ajlbj for 1 5 j 5 15 and &+ls + 25/j for 1 5 j 5 5, E3+i5 t l/(e(2j-1)/50 -1) for 6 5 j 5 15. Table 1 was computed while making use of the following intermediate values: (PI,. . . ,psi) = (.156, .147, .133, .116, .097, .078, .060, .044, .032, .022, .014, ,009, .005, .003, .002, .002, .005, .007, .009, .OlO, .009, .009, .008, .006, .005, ,004, .002, .002, .OOl, .OOl, .003); (X6,. . . ,x15) = (1.115, 1.304, 1.502, 1.700, 1.899, 2.099, 2.298, 2.497, 2.697, 2.896); (al,. . . ,u15) = (7.5, 9.1, 9.5, 9.8, 9.9, 10.0, 10.0, 10.1, 10.1, 10.1, 10.1, 10.2, 10.2, 10.2, 10.2); bl,..., bls) = (14.9, 11.7, 10.9, 10.4, 10.1, 10.1, 10.2, 10.3, 10.4, 10.5, 10.6, 10.7, 10.7, 10.8, 10.9). 11. Let g(t) = eg12te-t2/2 for t > 3. Since G(x) = 1: g(t) dt = 1 -e-(22-g)/2, a random variable X with density g-can be computed by setting X + G- (1 -V) = 49 -21nV. Now e- / 2 (t/3)e-t2/2 for t > 3, so we obtain a valid rejection method if we accept X with probability f(X)/cg(k) = 3/X. 12. We have f (x) = xf(x)-1 < 0 for x 2 0, since f(x) = x-1-e22/2 ~zme-t2/2 dt/t2 for z > 0. Let x = u3-l and y2 = x2 + 2ln2; then fiSyo0 e-t2/2 dt = 3 fie-z 12f(y) < 3fie-Z2/2f(x) = 2-j, hence y > oj.
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3.4.1 ANSWERS TO EXERCISES 549 Circle U2+ V2= 1 Fig. A-4. Region of acceptance for the algorithm of exercise 6. SECTION 3.4.1 1. Q: + (p -cx)U. 2.LetU=Xlm;~kU]=riffr~kX/mIf you are looking for cheap and quality webhost to host and run your website check Jboss Web Hosting services.

548 ANSWERS TO EXERCISES 3.3.4 28. Let 5 = ezrz (m-l) and let Ski = CO In fact, DEL1 5 H c T(ZL~, , Ut) [summed over nonzero solutions of (15)] + G&C?-(Ul,…, ut) [summed over all nonzero (pi, . . . , ut)]. The latter sum is O(log m)t by exercise 25 with d = 1, and the former sum is treated as in exercise 27. Let us now consider the quantity R(a) = cr(~l,. . . , ut) summed over nonzero solutions of (15). Since m is prime, each (~1,. , ut) can be a solution to (15) for at most t -1 values of a, hence xOFrom our experience, we are can tell you that you can find a reliable and cheap webhost service at Java Web Hosting services.

3.3.4 ANSWERS TO EXERCISES 547 S4 . [Advance t.] If t = T, the algorithm terminates. Otherwise set t c t + 1 and R t R( T i)modm. Set Ut to the new row (-Qz, -r22,0,. . . ,0, 1) of t elements, and set uZt e 0 for 1 2 i < t. Set Vt to the new row (0,. . . ,O, m). For 1 5 i < t, set q + round((v,lrls + v,m2)/m), vat + 0~1~2 + ~~2~22 -qm, and Ut + Ut + qJX. Finally set s + min(s, Ut Ut), Jc + t, j + 1. [A similar generalization applies to all sequences of length pk -1 defined by Eq. 3.2.2-8. Additional numerical examples have been given by A. Grube, Zeitschrift fiir angewandte Math. und Mechanik 53 (1973), T223-T225.1 25. The given sum is at most two times the quantity ~O 1, and replace uj by trunc(u,); but do nothing if p(u3) = 1 for all j. (This operation essentially throws away one bit of information about (~1, . . . , it).) If (u:, . . . , u:) and (u:, . . , uy) are two vectors of the same class having the same p-fold truncation, we say they are similar; in this case it follows that p(u: -Us). . . p(u: -u:) < 2 < pmax. For example, any two vectors of the form ((lxzxr)z, 0, m -(1x3)2, (lOlxsx4)2, (1101)~) are similar when m is large and p = 5; the p-fold truncation operator successively removes xi, x2, xs, x4, 5s. Since the difference of two bad vectors satisfies (15), it is impossible for two unequal bad vectors to be similar. Therefore class (Li, . , Lt) can contain at most max(1, J(L1). J(Lt)/2P) bad vectors. If class (Li, . . . , Lt) contains exactly one bad vector (ul,. . . , Q), we have r(ul,. . ,~t) 5 rmax 5 l/pmax; if it contains 5 J(L1). . . J(Lt)/2* bad vectors, each of them has T(u~, . . . ,ut) 5 l/p(~i). . .p(ut) 5 l/J&l). J&t).
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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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