Concrete mathematics : a foundation for computer science

486 ANSWERS TO EXERCISES
move the bottom k, then finishing with the top n - k.) The stated relation
turns out to be based on the unique value of k that minimizes the righthand
side of this general inequality, when n = n(n + 1)/2. (However, we
cannot conclude that equality holds; many other strategies for transferring
the tower are conceivable.) If we set Y,, = (W,,(n+ll,Z - 1)/2n, we find that
Y,, 6 Y,-1 + 1; hence W,(,+1),2 < 2n(n- 1) + 1.
1.18 It suffices to show that both of the lines from (n2j,0) intersect both of
the lines from (n 2k , 0) , and that all these intersection points are distinct.
A line from (xi, 0) through (xi - oj, 1) intersects a line from (xk, 0)
through (Xk - ok, 1) at the point (Xj - toj,t) where t = (xk - Xj)/(CIk - oj).
Let Xj = n2j and oj = nj + (0 or nP”). Then the ratio t = (nZk - n2j)/
(nk - nj + ( -nPn or 0 or nPn )) lies strictly between nj+nk-1 and nj+nk+l;
hence the y coordinate of the intersection point uniquely identifies j and k.
Also the four intersections that have the same j and k are distinct.
1.19 Not when n > 11. A bent line whose half-lines run at angles 8 and
8 + 30” from its apex can intersect four times with another whose half-lines
run at angles 4 and @ + 30” only if 10 - +I > 30”. We can’t choose more
than 11 angles this far apart from each other. (Is it possible to choose ll?)
1.20 Let h(n) = a(n)o1+ b(n)l& + c(n)01 + d(n)yc + e(n)y,. We know
from (1.18) that a(nb+b(n)Bo+c(n)Bl = (aPb,,-, fib,,-2 . . . obl pbo)4 when
n = (1 b,,-l . . . bl bo)z; this defines a(n), b(n), and c(n). Setting h(n) = n in
the recurrence implies that a(n)+c(n)-2d(n)-2e(n) = n; setting h(n) = n2
implies that a(n) + c(n) + 4e(n) = n2. Hence d(n) = (3a(n) + 3c(n) - n2 -
2n)/4; e(n) = (n2 - a(n) - c(n))/4.
1.21 We can let m be the least (or any) common multiple of 2n, 2n - 1,
. * * ) n + 1. [A non-rigorous argument suggests that a “random” value of m
will succeed with probability
n n-l 1 2n &ii
z---&-i..“‘- = 1
n+l A n )
-7)
so we might expect to find such an m less than 4n.]
1.22 Take a regular polygon with 2n sides and label the sides with the Ioncerode a
elements of a “de Bruijn cycle” of length 2”. (This is a cyclic sequence of de Brudl] ,cyc’e
O’s and l’s in which all n-tuples of adjacent elements are different; see [173, ~Sh~~~~~&~~e,,
exercise 2.3.4.2-231 and [174, exercise 3.2.2-171.) Attach a very thin convex
extension to each side that’s labeled 1. The n sets are copies of the resulting
polygon, rotated by the length of k sides for k = 0, 1, . . . , n - 1.
The Netherlands).
1.23 Yes. (We need principles of elementary number theory from Chapter
4.) Let L(n) = lcm(l,2,. , . , n). We can assume that n > 2; hence by
’