Chapter 4. Trees. 4.1 Basics


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1 Chapter 4 Trees 4.1 Basics A tree is a connected graph with no cycles. A forest is a collection of trees. A vertex of degree one, particularly in a tree, is called a leaf. Trees arise in a variety of applications. Perhaps most important is the use of trees in computer science, where they are useful as data structures which support information searches. We had an indication of such applications in our discussion of the phone book problem. Theorem If T is a tree of order N, then T has N 1 edges. Proof. Use induction. The cases N = 1, 2 are trivial. Suppose the result is true for trees with fewer than N vertices, and T has N vertices. Discard one edge to get two smaller trees. Use the hypothesis. Theorem A graph G of order N is a tree if and only if it is connected and contains N 1 edges. Proof. The previous result gives half the result. Suppose G is connected and contains N 1 edges. If it has a cycle we can remove an edge without destroying connectivity. Continue until there are no cycles left. The result is a tree with N vertices, but fewer than N 1 edges, which is impossible. So G had no cycles. Theorem If F is a forest of order N containing K connected components, then F contains N K edges. 37
2 38 CHAPTER 4. TREES Proof. Suppose the components, which are trees, have orders N 1,..., N K and sizes N 1 1,...,N K 1. Then F has size (N 1 1) + + (N K 1) = N K. Theorem A graph G of order N is a tree if and only if it is acyclic and contains N 1 edges. Proof. We ve already established that a tree is acyclic, by definition, and has N 1 edges. Suppose G is acyclic with N 1 edges. Then G is a forest with K = 1 components. Theorem Let T be a tree of order N 2. Then T has at least 2 leaves. Proof. Pick a maximal length path v 1,...,v K. Suppose v 1 (or v K ) is not a leaf. Let v 0 v 2 be adjacent to v 1. If v 0 {v 1,...,v K } then T contains a cycle, which it doesn t, so we can extend the path to v 0, contradicting the distance maximality. Theorem Every pair of distinct vertices in a tree T is connected by a unique path. Proof. Pick two distinct vertices u 1 and u 2. Since T is connected there is path from u 1 to u 2. Suppose there are two different paths, u 1 = v 1, v 2,..., v M = u 2, u 1 = w 1, w 2,...,w N = u 2. Pick j as large as possible so that v i = w i for i j. If j < M, then v j, v j+1, w j+1 are distinct vertices. The walk v j,...,v M, w N 1,...,w j+1 contains a path from v j to w j+1, and so we have a cycle by adding the edge w j+1 v j. But T has no cycles, so the path is unique. Recall that the center of a graph is the set of vertices for which ecc(v) = rad(g). Theorem In any tree T, the center is either a single vertex or a pair of adjacent vertices.
3 4.2. SPANNING TREES 39 Proof. Recursively define a sequence of trees, starting with T 0 = T. Given a nonempty T n, let T n+1 be the tree obtained from T n be removing all leaves and their incident edges. Let T N be the last nonempty tree in the sequence. Using the uniqueness of paths, if T n+1 is nonempty, then the center of T n+1 is the center of T n, since the eccentricity of each vertex in T n+1 is reduced by 1 from its eccentricity in T n. For any tree, a leaf v can be in the center if and only if the distance d(v, w) to the most distant vertex w is 0 or 1, so T N is K 1 or K 2. Theorem Let T be a tree with K edges. If G is a graph whose minimum degree is δ(g) K, then G contains T as a subgraph. Proof. Again use induction on K. Check the cases K = 0, 1. Suppose the result is true for trees with fewer than K edges. Now let T be a tree with K edges. Let v be a leaf of T and let w be adjacent to v. Consider the tree T 1 which has vertex set V (T) \ {v}, and the edges of T except vw. Then T 1 is a subgraph of G. The vertex w G has degree K, so some vertex u G adjacent to w is not in T 1. Adding the edge uv to T 1 gives T. 4.2 Spanning trees A spanning tree of a graph G is a subgraph T which is a tree containing all vertices of G. A weighted graph is a graph with a positive number assigned to each edge. A minimum weight spanning tree is a spanning tree with the property that the sum of the edge weights is a minimum. Weighted spanning trees arise in problems where one wants to connect a set of vertices at minimal cost. The text mentions the problem of building new rail lines that connect a group of cities. In this case the weights could be the construction cost for a citytocity rail link. Similar problems arise for establishing communications lines, like fiberoptic cable, etc. Theorem Every connected graph contains a spanning tree. Every weighted graph contains a minimum weight spanning tree. Proof. Remove edges from cycles as long a possible. In practice it may be important to have an efficient procedure for finding minimum weight spanning trees. Before discussing Kruskal s algorithm, here a few points glossed over in the text.
4 40 CHAPTER 4. TREES As part of the algorithm we have to be able to recognize when adding an edge to a tree produces a new tree. Lemma Suppose T is a subtree of G, and e is an edge of G with exactly one vertex in V (T). Then adding e to T gives a tree. Proof. Use the previous result that a graph G of order N is a tree if and only if it is connected and contains N 1 edges. Adding e gives a connected graph, with exactly one new vertex and exactly one new edge. Here is a method to find the connected components of a graph. Connected component algorithm: Initially give the vertices of G distinct numbers. Check all vertices to see if a vertex v has a higher number n(v) than an adjacent vertex w. If it does, reduce the number of v to that of w, that is set n(v) := n(w). Count the number of vertices whose numbers have been changed. Repeat until no vertices change numbers. The vertices with the same numbers lie in the same component. Each round of vertex checking reduces the sum of the vertex numbers by one or more, since at least one vertex number drops until the algorithm terminates. The sum of the vertex numbers starts at N(N 1)/2, so the algorithm could take roughly N 2 (N 1)/2 steps, and each step can involve checking (N 1) adjacent vertices, giving a possible cost of N 2 (N 1) 2 /2. Kruskal s algorithm: (i) Find an edge of minimum weight, and mark it. (ii) While possible, find a minimal weight unmarked edge that can be added to the marked edges without producing a (marked) cycle. (For instance, we can add the edge, compute connected components, and check if the number of edges for each component is one less than the number of vertices.) Theorem Starting with a connected weighted graph, Kruskal s algorithm produces a minimal weight spanning tree. Proof. At each stage we can consider adding an edge by tentatively adding it, finding components, and checking the components to make sure they are trees. Thus at each step we have a forest. If at some stage there is a vertex not in our forest, then by Lemma there will be an addable edge. Thus when the algorithm stops all vertices are included. If two forest components
5 4.3. COUNTING TREES 41 have adjacent vertices, then adding the edge between them won t create a cycle, so when the algorithm stops there will be only one component. Consider the edges e 1,...,e n added by the algorithm in order. If the resulting tree T is not a minimal weight tree, let T be a minimal weight spanning tree which contains the set of edges {e 1,...,e k }, and such that k is as big as possible. Since T is a spanning tree, T + e k+1 contains a cycle C, and some edge e of the cycle is not in T. Removing e from T + e k+1 leaves a connected graph with N 1 edges, which is thus a spanning tree, which contains edges e 1,...,e k+1. Since the algorithm chose e k+1 rather than e, we must have w(e k+1 ) w(e ). This means T + e k+1 e is minimal weight, and agrees with T for more edges, a contradiction. 4.3 Counting trees We ll start by counting all the possible trees with a fixed set of N vertices. In this count we pay attention to edge indexing, not isomorphism classes. Looking ahead, we can think of this as counting the distinct spanning trees of the complete graph on N vertices. Theorem (Cayley s Tree Formula) There are N N 2 distinct labeled trees of order N. Proof. Notice that this is the number of all sequences of length N 2 whose terms come from the set {1,...,N}. In fact, following Prüfer, we ll try to establish a onetoone correspondence between the two sets. Assigning a Sequence to a Labeled Tree For N 3 we re given a tree T with vertices 1,...,N. We want to form a sequence a 1,...,a N Let i = 0 and take T 0 = T. 2. Find the leaf v on T i with the smallest label. 3. Find the label n(w) of the vertex w adjacent to v Let a i+1 = n(w). 4. Remove v from T i to create the tree T i If T i+1 = K 2, stop. Otherwise increment i and go to step 2. This finishes step 1 of the proof, which constructs a function from trees to sequences of length n 2. Let s check that distinct trees result in different sequences. Use induction on the number of vertices in the tree. The first case is N = 3. In this case
6 42 CHAPTER 4. TREES exactly one vertex has degree 2, and the sequences are length 1, giving the number of the vertex of degree 2. Suppose the result is true for distinct trees on the same vertex set of size less than N, and let S and T be two distinct trees on a set of N vertices. Suppose the minimum label of a leaf is i for S and j for T. If i = j there are two cases. If the adjacent vertex labels are different, then the sequences are different in the first position, and we re done. If the adjacent vertex labels are the same, remove the vertices v j and incident edge. The remaining trees are distinct trees on the same N 1 vertices, so have different sequences by the induction hypothesis. Now consider the case i j, and suppose i < j. As the tree shrinks in the algorithm, each vertex is or becomes a leaf. Since v i is not a leaf of T, there is a last stage before it becomes a leaf, and at that stage v i will be adjacent to a leaf. Thus i will appear in the sequence. But as a leaf of S, i will never appear in the S sequence, so the sequences are different. Assigning a Labeled Tree to a Sequence We re given a a sequence σ = a 1,...,a K with terms from 1,..., K Let S = {1,..., K + 2}. 2. Initialize i = 0, σ 0 = σ, S 0 = S. 3. Let j be the smallest number in S i that does not appear in σ i. 4. Add the edge e j joining j and the first entry of σ i. 5. Remove the first entry of σ i to get σ i+1. Remove j from S i to get S i If σ i+1 is empty, add an edge between the two elements of S i+1 and stop. Otherwise, increment i by 1 and return to step 3. We have to check that the graph we produced is a tree. This is done by induction on the number N = K + 2. If N = 2 there is only one edge, so the result is true. Suppose the result is true for orders less than N. The first vertex j 1 selected from S is not in σ. By the induction hypothesis the graph produced using S \ j 1 and σ 1 is a tree, and adding the first edge does not create a cycle since j 1 is excluded from the tree vertices. Now we need to show that distinct sequences produce different trees. Again argue by induction on the number of vertices N = K + 2. In the first case the sequence has length 1, N = 3, and S = {1, 2, 3}. In each case the one term sequence k produces a 3vertex tree with degree two vertex k, and these are distinct. Suppose that distinct sequences produce different trees for fewer than N vertices. Let σ = a 1,...a K and τ = b 1,...,b K be distinct sequences. Let j σ
7 4.3. COUNTING TREES 43 be the first jvalue selected by the algorithm if the sequence σ is used, and j τ be the first jvalue in the τ case. If j σ = j τ and a 1 = b 1, the induction hypothesis insures that we get different trees from the reduced sequences σ 1 and τ 1, and these remain different when the leaves j σ = j τ and edges j σ a 1 = j τ b 1 are added. If j σ = j τ and a 1 b 1, then one tree has a leaf j σ adjacent to a 1, while the other tree has j σ = j τ adjacent to b 1, so these trees are different. If j σ j τ, it suffices to consider j σ < j τ. Since j σ is not available in the τ sequence case, it must be that j σ is in the τ sequence. Every time j σ appears at the front of τ i, an incident edge is added. After the last occurence of j σ in the τ sequence, another edge is added to j σ as an element of S. Thus the degree of j σ in the tree described by the τ sequence is one more than the number of times it appears in the τ sequence. In particular it is not a leaf, as it was in the tree determined by the σ sequence. Here is a more general, if less explicit, way to count spanning trees. The result (due to Kirchhoff 1847) is called the Matrix Tree Theorem. Recall the adjacency matrix A and the degree matrix D. Also recall from Linear Algebra that the i, j cofactor of an n n matrix B is ( 1) i+j det(b(i j), where B(i j) is the (n 1) (n 1) matrix obtained from B by deleting row i and column j. Theorem (Matrix Tree Theorem) If G is a connected labeled graph with adjacency matrix A and degree matrix D, then the number of spanning trees of G is equal to the value of any cofactor of D A.
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