3.2.2 ANSWERS TO EXERCISES 525 8. Since X, (Web hosting faq)
3.2.2 ANSWERS TO EXERCISES 525 8. Since X, is always odd, Xn+2 = (234 + 3 . 218 f 9)X, mod 235 = (234 + 6X,+1 -9X,) mod 235. Given Y, and Yn+i, the possibilities for Yn+2 M (5 i- 6(Y,+l + ~1) - 9(Y, + a)) mod 10, with 0 2 ~1 < 1, 0 2 ~2 < 1, are limited and nonrandom. Note: If the multiplier suggested in exercise 3 were, say, 233 +218 +22 + 1, instead of 223 + 214 + 22 + 1, we would similarly find X n+:! -10X,+1 $25X, z constant (modulo 235). In general, we do not want a f 6 to be divisible by high powers of 2 when 6 is small, else we get second order impotency. See Section 3.3.4 for a more detailed discussion. The generator that appears in this exercise is discussed in an article by MacLaren and Marsaglia, JACM 12 (1965), 83-89. The deficiencies of such generators were first demonstrated by M. Greenberger, CACM 8 (1965), 177-179. Yet generators like this were still in widespread use more than ten years later (cf. the discussion of RANDU in Section 3.3.4). SECTION 3.2.2 1. The method is useful only with great caution. In the first place, aU, is likely to be so large that the addition of c/m that follows will lose almost all significance, and the mod 1 operation will nearly destroy any vestiges of significance that might remain. We conclude that double-precision floating point arithmetic is necessary. Even with double precision, one must be sure that no rounding, etc., occurs to affect the numbers of the sequence in any way, since that would destroy the theoretical grounds for the good behavior of the sequence. (But see exercise 23.) 2. Xn+i equals either X,-i +X, or X,-l + X, -m. If Xn+i < X, we must have X,+1 = X,.-l +X, -m; hence Xn+l < X,-l. 3. (a) The underlined numbers are V[f after step M3. Output: initial / 0 4 5 6 2 0 3(2 7 4 1 6 3 0 5) and repeats. V[O]: 0 jp77777741777777/7… V[l]: 3 333333255555552555… V[2]: 2 2222QCl333333033333… V[3]: 5 556~1111116~111111… x: 476103254761032547… Y: 016745230167452301… So the potency has been reduced to l! (See further comments in the answer to exercise 15.)