## Graph Theory Lecture Notes 7

### Trees

Def: Tree, Forest

Theorem 3.14: In a tree T, there is one and only one simple chain joining any two vertices x and y.

Pf: Two vertices, x and y are joined in T since T is connected. The shortest chains between x and y must be simple, if not the section between repeated vertices could be removed to obtain a shorter chain joining x and y. Suppose then that there are two simple chains joining x to y. Let xp be the first vertex at which the two chains differ, and let xq be the next vertex at which they agree. Then following the first chain from xp to xq and then the second chain from xq back to xp forms a circuit. > <.

Theorem 3.10: If T is a tree with n vertices and e edges then n = e + 1.

Pf: By induction on n.
If n = 1, there is only 1 vertex and no edges, so 1 = 0 + 1.
Assume true for all trees with fewer than k vertices (strong form of induction). Remove an edge from a tree G with k vertices. This will give us two disconnected graphs, both of which are trees (this follows from Thm 3.14). Since they each have fewer than k vertices, we can apply the induction hypothesis to each, and in the first tree we have n1 = e1 + 1 and in the second we have n2 = e2 + 1. Thus, in G we have k = n1 + n2 = e1 + e2 + 2 = e + 1, since the number of edges in G is e = e1 + e2 + 1.

Theorem 3.12: In any graph G with e edges, the sum of the degrees of all the vertices = 2e.

Def: Leaf (pendant vertex)

Theorem 3.13: If T is a tree with more than 1 vertex, there are at least 2 pendant vertices.

Pf: Since T is connected, every vertex has degree at least 1. The sum of the degrees of all the vertices = 2e = 2(n-1) = 2n - 2. If n-1 of the vertices of T had degree at least 2 we would get a contradiction.

Theorem: If T is a tree with n vertices, then the chromatic polynomial, P(T,x) = x(x-1)n-1.

Pf: By induction on n.
If n = 1 then the tree is a single isolated vertex with chromatic polynomial x = x(x-1)0.
Assume the statement is true for all trees with k or fewer vertices (strong form of induction). Let G be a tree with k + 1 vertices. Since G has at least two pendant vertices, let ß be an edge to one of these leaves. By the fundamental reduction theorem,

P(G, x) = P(G'ß, x) - P(G"ß, x),
where G'ß is an isolated vertex together with a tree on k vertices, and G"ß is a tree on k vertices. By the induction hypothesis we can then say that P(G'ß, x) = x(x(x-1)k-1) and P(G"ß, x) = x(x-1)k-1. So,
P(G, x) = x2(x-1)k-1 - x(x-1)k-1 = x(x-1)k.
Def: Spanning subgraph, spanning tree

Theorem 3.15: A graph G is connected if and only if it has a spanning tree.

Pf: <= Since any two vertices are joined in the spanning tree, they are joined in G, so G is connected.

=> If G has no circuits then it is a tree (and so, is its own spanning tree). If it has a circuit, remove one edge from the circuit. The result is still connected. Repeat this until there are no circuits left. This gives a graph having all the vertices of G and no circuits, i.e., it is a spanning tree for G.

Theorem 3.11: Suppose that G is a graph with n vertices and e edges. Then G is a tree if and only if G is connected and n = e + 1.

Pf: => Follows from the definition of a tree and Theorem 3.10.

<= Since G is connected it has a spanning tree (Thm. 3.15), say T. T has n vertices and n-1 = e edges since it is a tree. Since G also has e edges, we must have G = T.

Theorem 3.16: Suppose that G is a graph with n vertices and e edges. Then G is a tree if and only if G has no circuits and n = e + 1.

Pf: => Follows from the definition of a tree and Theorem 3.10.

<= If G is not connected, consider its connected components K1, ..., Kp with p > 1. Since each of these is connected and has no circuits, each is a tree. By counting the edges in all of these components, we get n1 - 1 + n2 - 1 + ... + np - 1 = n - p which can not equal e = n - 1 if p > 1.