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Lecture 21
7.3 How to count trees
A: Counting edges
How many edges does a tree have, if it has n vertices?
Answer: n − 1. Can you prove it?
Proof: by induction on n. The induction starts at n = 1, with the empty
graph on one vertex.
If n > 1, then there is at least one edge. Removing one edge disconnects
the tree into two smaller trees. If these have i and n − i vertices, then by
induction they have i − 1 and n − i − 1 edges, making a total of n − 2. Adding
in the edge we removed, we get a total of n − 1 edges in the original tree.
B: Examples
Let tn be the number of trees on n vertices, 1, 2, 3 . . . , n. It is easy to see
that t1 = t2 = 1. If n = 3, then the tree is a path, and there are three choices
for the label of the middle vertex, so t3 = 3. (Alternatively, say there are 3!
labellings, but there is no distinction between the two ends of the path, so
we have to divide by 2.)
Example 1 (n = 4) There are two possible shapes for a tree on 4 vertices.
One is a path, which has 4!/2 = 12 distinct labellings. The other is a ‘star’
(one vertex joined to all the others), which has just four labellings, as only
the labelling of the centre vertex is significant. Hence t4 = 12 + 4 = 16.
At this point you might start to see the pattern, but one more example will
make it more obvious:
Example 2 (n = 5) Again we have a path, with 5!/2 = 60 labellings, and a
star, with 5 labellings. There is one more shape of tree on 5 vertices:
u
u
u
u
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@@u
In this case the number is again 5!/2, because the two right-hand leaves (a leaf
is a vertex of degree 1) are indistinguishable, but apart from this ambiguity
all labellings are distinct. Hence t5 = 60 + 5 + 60 = 125.
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Now we see the pattern clearly:
Conjecture 1 The number of labelled trees on n vertices is tn = nn−2 .
C: A bijection
To prove this conjecture, we need to construct a bijection f between the set
T of trees on the vertex set V = {1, 2, . . . , n}, and the set S of sequences
a1 , a2 , . . . , an−2 over {1, 2, . . . , n}. Essentially, the sequence tells us where to
join the branches onto the tree. We assume that n ≥ 3, so that T has at
least two edges.
Algorithm 1 Remove the leaf with the smallest label, and write down the
label of the vertex it is joined to. Repeat until there are exactly two vertices
left.
Example 3
1u
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@
@
2u
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@u4
3u
@
@
@
6u
5u
7u
Step 1: Remove 4, write down 1
Step 2: Remove 6, write down 2
Step 3: Remove 2, write down 1
Step 4: Remove 1, write down 2
Step 5: Remove 7, write down 5
Step 6: Remove 5, write down 3
The output sequence is 1, 2, 1, 3, 5, 3.
2
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@u8
To complete the proof, you need to convince yourself that, given the
sequence, you can reconstruct the labelled tree. Then you will have proved
that the function f is a bijection, and hence that the number of labelled trees
on n vertices is nn−2 . The following algorithm defines the inverse function to
f , hence showing it is a bijection. All that is missing now is the proof that
this really is the inverse function!
Algorithm 2 Given a sequence of n−2 numbers chosen from {1, 2, 3, . . . , n},
the numbers not chosen are the labels on the leaves of the tree. More generally, the degree of each vertex is 1 more than the number of times it appears
in the sequence. Keep track of the degrees of the vertices as we go along.
Then for each number a in the sequence, in turn, join it to the smallest numbered vertex that needs only one more edge incident to it. This creates n − 2
edges altogether, and the last edge joins the only two vertices whose degrees
are not yet what they should be.
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