Concrete mathematics : a foundation for computer science

The solution is given by the super generating function
G(w,z) = 2 G,(z)w” = A(w)/(l - zB(w)) ,
n30
A ANSWERS TO EXERCISES 559
where B(w) = w(qf-(qf-p,,)w)/(l -qhw) and A(w) = (1 -B(w))/(l -w).
Now tnaO GA(l)w” = cxw/(l -w)~+ B/(1 -w) - B/(1 - (qf -ph)w) where
OL=Ph
Ph+Pf ’
B = Pf(qf -Ph) .
(Ph+Pf)’ ’
hence G;(l) = cxnf fi(l - (qf-ph)n). (Similarly G:(l) = K2n2 + O(n), so
the variance is O(n).)
8.43 G,(z) = 1k20 [L].zk/n! = z?/n!, by (6.11). This is a product of
binomialpgf’s, nE=, ((k-l +z)/k), wh ere the kth has mean 1 /k and variance
(k- 1)/k’; hence Mean = H, and Var(G,) = H, - Ht).
8.44 (a) The champion must be undefeated in n rounds, so the answer is pn.
(b,c) Players xl, . . . , X~L must be “seeded” (by chance) in distinct subtournaments
and they must win all 2k(n - k) of their matches. The 2” leaves of
the tournament tree can be filled in 2n! ways; to seed it we have 2k!(2”-k)2k
ways to place the top 2k players, and (2” - 2k)! ways to place the others.
Hence the probability is (2p)2k’n-k)/(iE). When k = 1 this simplifies to
(2~~)“~l/(2” - 1). (d) Each tournament outcome corresponds to a permutation
of the players: Let y1 be the champ; let y2 be the other finalist; let y3 and
y4 be the players who lost to yr and y2 in the semifinals; let (ys, . . . ,ys) be
those who lost respectively to (y, , . . . ,y4) in the quarterfinals; etc. (Another
proof shows that the first round has 2n!/2n-1! essentially different outcomes;
the second round has 2np1!/2np2!; and so on.) (e) Let Sk be the set of 2kp1
potential opponents of x2 in the kth round. The conditional probability that
x2 Wins, given that xl belongs to Sk, is
Pr(xl plays x2) .pn-’ (1 -p) + Pr(xl doesn’t play x2) .p”
= pkP’pnP’(l -p) + (1 -pypn.
The chance that x1 E Sk is 2kp1/(2n - 1); summing on k gives the answer:
*;, -$&(pk-‘p+p) + (I-pk-‘)pn) = pn - (‘,“n’“, ’ pnp’ .
t
(f) Each of the 2”! tournament outcomes has a certain probability of occurring,
and the probability that xj wins is the sum of these probabilities over
all (2” - 1 )! tournament outcomes in which xj is champion. Consider interchanging
xj with xj+l in all those outcomes; this change doesn’t affect the