550 ANSWERS (Free web hosts) TO EXERCISES 3.4.1 7. If Ic
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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