82 RANDOM NUMBERS 3.3.3 Here ai = lmjlm,+~l, (Cheapest web hosting)
82 RANDOM NUMBERS 3.3.3 Here ai = lmjlm,+~l, bj = lcjlmi+d (33) m,+2 = rnj modmj+r, ci+l = cy modmj+l, and it follows that 0 I m3+l < mj, 0 2 Cj < mj. (34) We have assumed for convenience that Euclid s algorithm terminates in (32) after four iterations; this assumption will reveal the pattern that holds in the general case. Since h and k were relatively prime to start with, we must have m5 = 1 and cg = 0 in (32). Let us further assume that cs # 0 but c4 = 0, in order to get a feeling for the effect this has on the recurrence. Equations (30) and (31) yield c(h, k, c) = o(m2, ml, cl) = f(m2, ml, cl) -4m3, m2, c2) = . . = f(m2, ml, cd -f(m3, m2, c2) + f(m4, m3, c3) -f(m5, m4, c4). The first part h/k + k/h of the formula for f(h, k, c) in (19) contributes ?$+Tp?$-$+~+~-~-~ to the total, and this simplifies to !!+(al+~)-~-(a~+~)+~+(a3+~)-~-a4 = h/k + al -a2 + a3 -a4. _ The next part l/hk of (19) also leads to a simple contribution; according to Eq. 4.5.3-9 and other formulas in Section 4.5.3, we have l/mlmz -l/m2m3 + l/m3m4 -l/m4m5 = h /k -1, (35) where h is the unique integer satisfying h h c 1 (modulo k), 0 < h < k. (36) Adding up all the contributions, and remembering our assumption that c4 = 0 (so that e(m4,cs) = 0, cf. (20)), we find that o(h, k, c) = y + (al -a2 + a3 -4 -6(bl -b2 + b3 - b4) 4 4 -++-- m2m3 m3m4 in terms of the assumed tableau (32). Similar results hold in general:
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