Concrete mathematics : a foundation for computer science

484 ANSWERS TO EXERCISES
Venn [294] claimed that there is no way to do the five-set case with ellipses,
but a five-set construction with ellipses was found by Griinbaum [137].
1.6 If the nth line intersects the previous lines in k > 0 distinct points, we This answer asget
k- 1 new bounded regions (assuming that none of the previous lines were SumeS that n > 0.
mutually parallel) and two new infinite regions. Hence the maximum number
of bounded regions is (n-2)+(n-3)+.. . = S-1 = (n-l)(n-2)/2 = L,-2n.
1.7 The basis is unproved; and in fact, H(1) # 2.
1.8 Qz = (1 + B)/a; 43 = (1 + LX+ B)/ct/3; 44 = (1 + x)/B; Qs = cx;
46 = b. So the sequence is periodic!
1.9 (a) We get P(n - 1) from the inequality
x1 + . . . + X,-l < x1 +. . . $X,-l n
x1 . . . xn-1
( n-l >-( n-l > .
(b) XI . . .x,x,+1 . ..xz,, 6 (((XI + ... + xnl/n)((xn+l + ... + x2nl/n))n by
P(n); the product inside is 6 ((x1 +...+xzn)/2n)’ by P(2). (c) For example,
P(5) follows from P(6) from P(3) from P(4) from P(2).
1.10 First show that R, = R,-l + 1 + Q+l + 1 + R+l, when n > 0.
Incidentally, the methods of Chapter 7 will tell us that Q,, = ((1 + v’?)~+’ -
(1 - fi)““)/(2fi) - 1.
1.11 (a) We cannot do better than to move a double (n - 1)-tower, then
move (and invert the order of) the two largest disks, then move the double
(n - 1)-tower again; hence A, = 2Anpl + 2 and A,, = 2T,, = 2n+1 - 2. This
solution interchanges the two largest disks but returns the other 2n - 2 to
their original order.
(b) Let B, be the minimum number of moves. Then B1 = 3, and it can
be shown that no strategy does better than B, = A,-1 + 2 + A,_1 + 2 + B,-1
when n > 1. Hence B, = 2n+2 -5, for all n > 0. Curiously this is just 2A,-1,
and we also have B, = A,-1 + 1 + A,-1 + 1 + A,-1 + 1 + A,-l.
1.12 Ifallmk >O, thenA(mi ,..., m,) =2A(ml,.,,, m,-1)$-m,. Thisis
an equation of the “generalized Josephus” type, with solution (ml , , . m,,)2 =
2n-1 ml + . . .+2m,-1 +m,.
Incidentally, the corresponding generalization of exercise llb appears
to satisfy the recurrence
B(ml,...,m,) =
. . . , A(ml, md,
if m, = 1;
2m,-1,
2A(ml,. . . , m,-1) + 2mn
ifn=l;
+B(m,...,m,-l), ifn>l andm,>l.