Concrete mathematics : a foundation for computer science

A ANSWERS TO EXERCISES 561
X, + Y,, where Y,, = -1 if the (n + 1)st particle hits a diphage receptor
(conditional probability 2X,/(n + 2)) and Y,, = +2 otherwise. Hence
EX,,+I = EX,, + EY, = EX, ~ 2EX,/(n+2) + 2(1 ~ 2EX,/(n+2)) .
The recurrence (n+2)EX,+l = (n-4)EX,+2n+4 can be solved if we multiply
both sides by the summation factor (n + 1)“; or we can guess the answer and
prove it by induction: EX, = (2n + 4)/7 for all n > 4. (Incidentally, there
are always two diphages and one triphage after five steps, regardless of the
configuration after four.)
8.48 (a) The distance between frisbees (measured so as to make it an even
number) is either 0, 2, or 4 units, initially 4. The corresponding generating
functions A, B, C (where, say, [z”] C is the probability of distance 4 after n
throws) satisfy
A = $zB, B = ;zB++zC, C = 1 + ;zB + +zC
It follows that A = z2/(16 - 202 + 5z2) = t’/F(z), and we have Mean(A) =
2 - Mean(F) = 12, Var(A) = - Var( F) = 100. (A more difficult but more
amusing solution factors A as follows:
A =
PlZ P2Z P2 PlZ + Pl P2Z
-.-=
1-q1z l-q2z Pz-Pl l-412 Pl - P2 1 - q2= ’
where p1 = a2/4 = (3 + &i/8, p2 = $j2/4 = (3 - &)/8, and p1 + q1 =
p2 + qr = 1. Thus, the game is equivalent to having two biased coins whose
heads probabilities are p1 and ~2; flip the coins one at a time until they
have both come up heads, and the total number of flips will have the same
distribution as the number of frisbee throws. The mean and variance of the
waiting times for these two coins are respectively 6 F 2fi and 50 F 22fi,
hence the total mean and variance are 12 and 100 as before.)
(b) Expanding the generating function in partial fractions makes it
possible to sum the probabilities. (Note that d/(4@) + a2/4 = 1, so the
answer can be stated in terms of powers of 4.) The game will last more than
n steps with probability 5inp11/24pn ( $nf2 - @m”p2); when n is even this is
5"124 nF,+Z. So the answer is 5504 1ooF,02 G .00006.
8.49 (a) If n > 0, PN(O,n) = i[N=Ol + ~PN j(O,n) + iPN~-i(l,n-1);
PN (m, 0) is similar; PN (0,O) = [N = 01. Hence
gm,n = $=gm l,n+l + ;=%n,n + &ll+1,n 1 ;
90 .n = i + izgg,, + igl,,-1 ; etc.
64 gk,., = l-t&, l,n+l+tg~,,+~g~sl,n~~l; sh,, = ~+$&,+-&& +?tc.
By induction on m, we have gi,, = (2m + l)gh,,,+,, - 2m2 for all m,n 3 0.