Concrete mathematics : a foundation for computer science

556 ANSWERS TO EXERCISES
8.32 By symmetry, we can reduce each month’s situation to one of four
possibilities: “Toto, I have a
D, the states are diagonally opposite;
A, the states are adjacent and not Kansas;
K, the states are Kansas and one other;
S, the states are the same.
Considering the Markovian transitions, we get four equations
D = 1 +z($D++)
A = z(;A+ AK)
4 4 4
K = z(~D+~A+~~K)
S = z(fD + ;A + $K)
whosesumisD+K+A+S=l+z(D+A+K). Thesolutionis
s=
812-452= -- 423
243-243z+24z2 +8z3 '
but the simplest way to find the mean and variance may be to write z = 1 + w
and expand in powers of w, ignoring multiples of w2:
D = g+‘593w+.,..
16 512
A = ?+?115w+..,.’
8 256
K = fi+&i!iiw+.,..
8 256
Now S’(1) = g + g + $ = $, and is”(l) = s + z + $$! = w. The
mean is $ and the variance is y. (Is there a simpler way?)
8.33 First answer: Clearly yes, because the hash values hl, . . . , h, are
independent. Second answer: Certainly no, even though the hash values hi,
h, areindependent. WehavePr(Xj=O) =xt=, sk([j#k](m-1)/m) =
(;‘lsj)(m-1)/m, but Pr(XI=Xz=O) =xL=, sk[k>2](m-l)2/m2 = (1 -
s1 - sl)(m- 1)=/m= # Pr(XI =0) Pr(X2 =O).
8.34 Let [z”] S,(z) be the probability that Gina has advanced < m steps
after taking n turns. Then S,( 1) is her average score on a par-m hole;
[z”‘] S,(z) is the probability that she loses such a hole against a steady player;
and 1 - [z’“~~‘] S,(z) is the probability that she wins it. We have the recurrence
So(z) = 0;
S,(z) = (1 + P&l~~Z(~) + q&lpl (z))/(l - =I, for m > 0.
feeling we’re not in
Kansas anymore. ”
-Dorothy