Concrete mathematics : a foundation for computer science

7.3 SOLVING RECURRENCES 335
Let’s look at some small cases. It’s pretty easy to enumerate the spanning
trees for n = 1, 2, and 3:
- 21 A &I! /I +I
f, = 1 f2 = 3 f3 = 8
(We need not show the labels on the vertices, if we always draw vertex 0 at
the left.) What about the case n = O? At first it seems reasonable to set
fc = 1; but we’ll take fo = 0, because the existence of a fan of order 0 (which
should have 2n - 1 = -1 edges) is dubious.
Our four-step procedure tells us to find a recurrence for f, that holds for
all n. We can get a recurrence by observing how the topmost vertex (vertex n)
is connected to the rest of the spanning tree. If it’s not connected to vertex 0,
it must be connected to vertex n - 1, since it must be connected to the rest of
the graph. In this case, any of the f,- 1 spanning trees for the remaining fan
(on the vertices 0 through n - 1) will complete a spanning tree for the whole
graph. Otherwise vertex n is connected to 0, and there’s some number k < n
such that vertices n, n- 1, . . , k are connected directly but the edge between
k and k - 1 is not in the subtree. Then there can’t be any edges between
0 and {n - 1,. . . , k}, or there would be a cycle. If k = 1, the spanning tree
is therefore determined completely. And if k > 1, any of the fk-r ways to
produce a spanning tree on {0, 1, . . . , k - l} will yield a spanning tree on the
whole graph. For example, here’s what this analysis produces when n = 4:
f4 f3 f3 f2
The general equation, valid for n 2 1, is
fn = f,-1 + f,-1 + f,-1 + fn-3 +. . . + f, + 1 .
k=4 k=3 k=2 k=l
+
/I
1
(It almost seems as though the ‘1’ on the end is fo and we should have chosen
fo = 1; but we will doggedly stick with our choice.) A few changes suffice to
make the equation valid for all integers n:
f, = f,-j + 2 fk + [n>O] .
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