trees discrete mathematics

path graph

A tree that contains only vertices of degree one or two

spanning tree

If H is a subgraph of G such that V (H) = V (G), then H is called a spanning subgraph of G. If H is a spanning subgraph which is also a tree, then H is said to be a spanning tree of G.

Recursive construction of all trees

Initial Step Let T1 be the tree with one vertex and put S1 = {T1}. Recursive Step Suppose we have constructed a set Si containing all trees with i vertices, for some integer i 1. For each tree Ti 2 Si and each vertex u 2 V (Ti) we construct a tree T on i + 1 vertices by adding a new vertex v to Ti and joining v to u by a single edge. Let Si+1 be the set of all trees constructed in this way. We can be sure that this algorithm will give us all trees, since Lemmas 3.1 and 3.2 tell us that if T is a tree on i + 1 vertices, then T has a vertex of degree one and so can be constructed from a tree on i vertices by the recursive step in the algorithm. Note however that we have glossed over one difficulty. When we present the set Si containing all trees on i vertices we would like it to contain only one copy of each distinct (i.e. non-isomorphic) tree. To do this we require a 'sub-routine' for checking when two trees are isomorphic. This can be done easily by inspection when the trees are small. It is not so easy for large trees.

Let T be a tree with at least two vertices. Then T has at least two vertices of degree one.

Proof. Let P = v1v2 . . . vm be a path of maximum length in T. Since T is connected and has at least two vertices, P has length at least one. Thus v1 not equal to vm Since T has no cycles, the only vertex of the path P which is adjacent to v1 in T is v2 Since P is a path of maximum length in T, no vertex of V (T)−V (P) is adjacent to v1 the only vertex of T which is adjacent to v1 is v2. Hence degT (v1) = 1. Applying a similar argument to vm we deduce that degT (vm) = 1. Thus T has at least two vertices of degree one

Let T be a tree and v be a vertex of T of degree one. Then T − v is a tree

Proof. To show that T − v is a tree we need to show that (a) T − v has no cycles and is (b) connected (a) If T −v had a cycle C, then C would also be a cycle in T. This is impossible since T has no cycles. Hence T − v has no cycles. (b) To see that T − v is connected, we choose two vertices x and y of T − v. Since T is connected, there is a path P from x to y in T. Since degT (v) = 1 and P does not repeat vertices, P cannot pass through v. Thus P is also a path from x to y in T − v. Since x and y can be any vertices of T − v, it follows that T − v is connected. Thus T − v is a tree.

Let T be a tree with n vertices. Then T has n − 1 edges

Proof. We shall use induction on the number of vertices n. Base Case. If T is a tree with one vertex then T has no edges3. Thus 3Because T has no cycles and the theorem is true when n = 1. Induction Hypothesis. Suppose that k >= 1 is an integer, and that all trees with n vertices have n − 1 edges when n = 1, 2, . . . , k. Induction Step. Let T be any tree on k + 1 vertices4. .Since k + 1 >= 2, it follows from Lemma 3.1 that T has a vertex of degree one. Using Lemma 3.2, we can delete this vertex and obtain a tree T1 with k vertices

tree properties

tree contains no cycles, it has no loops (because a loop is a cycle of length 1) and no multiple edges (because two edges with the same endpoints form a cycle of length 2).

what is tree

tree is a connected graph that contains no cycles