Concrete mathematics : a foundation for computer science

26 SUMS
where A(n), B(n), and C(n) are the coefficients of dependence on the general
parameters 01, B, and y.
The repertoire method tells us to try plugging in simple functions of n
for R,, hoping to find constant parameters 01, (3, and y where the solution is
especially simple. Setting R, = 1 implies LX = 1, (3 = 0, y = 0; hence
A(n) = 1.
Setting R, = n implies a = 0, (3 = 1, y = 0; hence
B(n) = n.
Setting R, = n2 implies a = 0, (3 = -1, y = 2; hence
2C(n) -B(n) = n2
and we have C(n) = (n2 +n)/2. Easy as pie.
Therefore if we wish to evaluate
n
E( a + bk) ,
k=O
the sum-recurrence (2.6) boils down to (2.7) with a = (3 = a, y = b, and the
answer is aA + aB(n) + bC(n) = a(n + 1) + b(n + l)n/2.
Conversely, many recurrences can be reduced to sums; therefore the special
methods for evaluating sums that we’ll be learning later in this chapter
will help us solve recurrences that might otherwise be difficult. The Tower of
Hanoi recurrence is a case in point:
To = 0;
T,, = 2T,_, +l , for n > 0.
It can be put into the special form (2.6) if we divide both sides by 2”:
To/2' = 0;
TJ2" = T,-,/2-' +l/2n, for n > 0.
Now we can set S, = T,/2n, and we have
so = 0;
s, = s,~-’ +2-n) for n > 0.
It follows that
s, = t2-k
k=l
Actually easier; n =
x 8
nx 14n+1)14n+3) .