Concrete mathematics : a foundation for computer science

3.3 FLOOR/CEILING RECURRENCES 79
Exercise 25 asks for a proof or disproof that K, > n, for all n 3 0. The
first few K’s just listed do satisfy the inequality, so there’s a good chance that
it’s true in general. Let’s try an induction proof: The basis n = 0 comes
directly from the defining recurrence. For the induction step, we assume
that the inequality holds for all values up through some fixed nonnegative n,
and we try to show that K,+l > n + 1. From the recurrence we know that
K n+l = 1 + minWl,pJ ,3Kln/31 1. The induction hypothesis tells us that
2KL,,/~J 3 2Ln/2J and 3Kln/3~ 3 3 [n/31. However, 2[n/2J can be as small
as n - 1, and 3 Ln/3J can be as small as n - 2. The most we can conclude
from our induction hypothesis is that Kn+l > 1 + (n - 2); this falls far short
of K,+l 3 n + 1.
We now have reason to worry about the truth of K, 3 n, so let’s try to
disprove it. If we can find an n such that either 2Kl,,zl < n or 3Kl,,31 < n,
or in other words such that
we will have K,+j < n + 1. Can this be possible? We’d better not give the
answer away here, because that will spoil exercise 25.
Recurrence relations involving floors and/or ceilings arise often in computer
science, because algorithms based on the important technique of “divide
and conquer” often reduce a problem of size n to the solution of similar problems
of integer sizes that are fractions of n. For example, one way to sort
n records, if n > 1, is to divide them into two approximately equal parts, one
of size [n/21 and the other of size Ln/2]. (Notice, incidentally, that
n = [n/21 + Ln/2J ; (3.17)
this formula comes in handy rather often.) After each part has been sorted
separately (by the same method, applied recursively), we can merge the
records into their final order by doing at most n - 1 further comparisons.
Therefore the total number of comparisons performed is at most f(n), where
f(1) = 0;
f(n)=f([n/21)+f([n/2J)+n-1, for n > 1
A solution to this recurrence appears in exercise 34.
(3.18)
The Josephus problem of Chapter 1 has a similar recurrence, which can
be cast in the form
J(1) = 1;
J(n) = 2J( LnI2J) - (-1)” ,
for n > 1.