Concrete mathematics : a foundation for computer science

7.6 EXPONENTIAL GENERATING FUNCTIONS 355
for all n > 1. Here”s why: There are (k, ,:,,‘,,k_) ways to assign n- 1 elements
to a sequence of TTL components of respective sizes kl, k2, . . . , k,; there are
tk, tk1 . . . tk, ways to connect up those individual components with spanning
trees; there are kr k.2 . . . k, ways to connect vertex n to those components; and
we divide by m! because we want to disregard the order of the components.
For example, when n = 4 the recurrence says that
t4 = 3t3 + ;((,32)2W2 + (23,)2tzt,) + ;((, ; ,)tf) = 3t3 + 6tzt, + t;.
3 I
The recurrence for t, looks formidable at first, possibly even frightening;
but it really isn’t bad, only convoluted. We can define
u n = nt,
and then everything simplifies considerably:
%I
n! = m>O m!
IL 1 - -
-
uk, ukj
-
uk
m
t k,! k2! “’ k,! ’
kl+kJ+...+k,=n-1
ifn>l. (7.82)
The inner sum is the coefficient of z+’ .m the egf 0 (z) , raised to the mth
power; and we obtain the correct formula also when n = 1, if we add in the
term fi(z)O that corresponds to the case m = 0. So
WI
- = [P’] t ; ti(p
n!
In>0 .
for all n > 0, and we have the equation
Progress! Equation (7,83) is almost like
E(z) = erEcri,
= [z”-‘] ,w = [zn] ,,w
(7.83)
which defines the generalized exponential series E(z) = El (z) in (5.59) and
(7.70); indeed, we have
cl(z) = z&(z)
So we can read off the answer to our problem:
t, = X = z [zn] Cl(z) = (n-l)! [z”~‘] E(z) = nnp2 (7.84)
The complete graph on {l ,2, . . . , n} has exactly nn ’ spanning trees, for all
n > 0.