Characterizations of Trees

Transcription

1 Characterizations of Trees Lemma Every tree with at least two vertices has at least two leaves. Proof. 1. A connected graph with at least two vertices has an edge. 2. In an acyclic graph, an end point of a maximal nontrivial path has no neighbor other than its neighbor on the path, and so the two endpoints of a such a path are leaves.

2 Characterizations of Trees Lemma Every tree with at least two vertices has at least two leaves. Proof. 1. A connected graph with at least two vertices has an edge. 2. In an acyclic graph, an end point of a maximal nontrivial path has no neighbor other than its neighbor on the path, and so the two endpoints of a such a path are leaves.

3 Characterizations of Trees Lemma Deleting a leaf from an n-vertex tree produces a tree with n 1 vertices. Proof. 1. Let v be a leaf of a tree. 2. Let G = G v. 3. Claim: G is a tree, that is, connected and acyclic: 3.1 Observation: A vertex of degree 1 cannot belong to any path connecting two other vertices. 3.2 Let u, w V (G ). 3.3 Since G is connected, there is a u, w-path in G. 3.4 This path must be in G because of the observation above, and so G is connected. 3.5 Deleting a vertex cannot create a cycle and so G is acyclic.

4 Characterizations of Trees Lemma Deleting a leaf from an n-vertex tree produces a tree with n 1 vertices. Proof. 1. Let v be a leaf of a tree. 2. Let G = G v. 3. Claim: G is a tree, that is, connected and acyclic: 3.1 Observation: A vertex of degree 1 cannot belong to any path connecting two other vertices. 3.2 Let u, w V (G ). 3.3 Since G is connected, there is a u, w-path in G. 3.4 This path must be in G because of the observation above, and so G is connected. 3.5 Deleting a vertex cannot create a cycle and so G is acyclic.

5 Characterizations of Trees Lemma Deleting a leaf from an n-vertex tree produces a tree with n 1 vertices. Proof. 1. Let v be a leaf of a tree. 2. Let G = G v. 3. Claim: G is a tree, that is, connected and acyclic: 3.1 Observation: A vertex of degree 1 cannot belong to any path connecting two other vertices. 3.2 Let u, w V (G ). 3.3 Since G is connected, there is a u, w-path in G. 3.4 This path must be in G because of the observation above, and so G is connected. 3.5 Deleting a vertex cannot create a cycle and so G is acyclic.

6 Characterizations of Trees with n-vertices Theorem TFAE: 1. G is connected and has no cycles (that is, it is a tree) 2. G is connected and has n 1 edges. 3. G has n 1 edges and no cycles 4. For u, v V (G), G has exactly one u, v-path. Plan of proof: We will show that any two of (connected, acyclic, n 1 edges) together imply the third, and so the first three statements are equivalent to one another. Then we will show that the first and the last are equivalent to each other.

7 Characterizations of Trees with n-vertices Theorem TFAE: 1. G is connected and has no cycles (that is, it is a tree) 2. G is connected and has n 1 edges. 3. G has n 1 edges and no cycles 4. For u, v V (G), G has exactly one u, v-path. Plan of proof: We will show that any two of (connected, acyclic, n 1 edges) together imply the third, and so the first three statements are equivalent to one another. Then we will show that the first and the last are equivalent to each other.

8 Connected and no cycles n 1 edges 1. Use induction on n. 2. Base step (n = 1): an acyclic 1-vertex graph has no edge. 3. Inductive step: Suppose the statement is true for graphs with fewer than n vertices. 4. Since G = G v for a leaf v of G is also connected and acyclic, e(g ) = n 2 by the induction hypothesis. 5. Since only one edge is incident to v, e(g) = n 1.

9 Connected and n 1 edges acyclic 1. Suppose G has at least one cycle. 2. Delete edges from cycles until the resulting graph G is acyclic. 3. Since no edge of a cycle is a cut-edge, G is connected. 4. Thus G is connected and acyclic with n-vertices and so e(g ) = n But e(g) = n 1 = e(g ), a contradiction. 6. Thus G is acyclic.

10 n 1 edges and acylic connected 1. Let G 1,..., G k be the components of G. 2. Each component G i is connected and acyclic and so e(g i ) = n(g i ) n 1 = e(g) = i e(g i) = i n(g i) k = n k 4. Thus k = 1, that is, G is connected.

11 connected and acyclic exactly one u, v-path for u, v V (G) 1. ( ): By contradiction. 1.1 Since G is connected, there is at least one u, v-path for any pair u, v V (G) 1.2 Suppose that the statement is wrong. 1.3 Choose a shortest (total length) pair P, Q of distinct paths with the same endpoints. 1.4 By the extremal choice, no internal vertex of P or Q can belong to the other path. 1.5 Thus P Q must be a cycle, which contradicts to the hypothesis that G is acyclic. 2. ( ): By contradiction 2.1 G is obviously connected. 2.2 Suppose that G has a cycle C, then G has two u, v-paths for u, v V (C), a contradiction. 2.3 Thus G must be acyclic.

12 connected and acyclic exactly one u, v-path for u, v V (G) 1. ( ): By contradiction. 1.1 Since G is connected, there is at least one u, v-path for any pair u, v V (G) 1.2 Suppose that the statement is wrong. 1.3 Choose a shortest (total length) pair P, Q of distinct paths with the same endpoints. 1.4 By the extremal choice, no internal vertex of P or Q can belong to the other path. 1.5 Thus P Q must be a cycle, which contradicts to the hypothesis that G is acyclic. 2. ( ): By contradiction 2.1 G is obviously connected. 2.2 Suppose that G has a cycle C, then G has two u, v-paths for u, v V (C), a contradiction. 2.3 Thus G must be acyclic.

13 More Properties of Trees 1. Every edge of a tree is a cut-edge 2. Adding one edge to a tree forms exactly one cycle. 3. Every connected graph contains a spanning tree. 4. If T, T are spanning trees of a connected graph G and e E(T ) E(T ), then there is an edge e E(T ) E(T ) such that T e + e is a spanning tree of G. 5. If T, T are spanning trees of a connected graph G and e E(T ) E(T ), then there is an edge e E(T ) E(T ) such that T + e e is a spanning tree of G. 6. If T is a tree with k edges and G is a simple graph with δ(g) k, then T is a subgraph of G. (This is sharp: the graph K k has minimum degree k 1 but it contains no tree with k edges.)

14 More Properties of Trees 1. Every edge of a tree is a cut-edge: Recall (Theorem ): An edge is a cut-edge iff it belongs to no cycle. 2. Adding one edge to a tree forms exactly one cycle.: unique u, v-path 3. Every connected graph contains a spanning tree. :Delete edges from cycles until we get a tree

15 More Properties of Trees If T, T are spanning trees of a connected graph G and e E(T ) E(T ), then there is an edge e E(T ) E(T ) such that T e + e is a spanning tree of G: (Deleting e from T ): 1. Note that e is a cut-edge. 2. Let U, U be the two components of T e. 3. Since T is connected, T has an edge e with endpoints in U and U. 4. Now T e + e is connected with n(g) 1 edges and thus it is a spanning tree of G.

16 More Properties of Trees If T, T are spanning trees of a connected graph G and e E(T ) E(T ), then there is an edge e E(T ) E(T ) such that T e + e is a spanning tree of G: (Deleting e from T ): 1. Note that e is a cut-edge. 2. Let U, U be the two components of T e. 3. Since T is connected, T has an edge e with endpoints in U and U. 4. Now T e + e is connected with n(g) 1 edges and thus it is a spanning tree of G.

17 If T, T are spanning trees of a connected graph G and e E(T ) E(T ), then there is an edge e E(T ) E(T ) such that T + e e is a spanning tree of G: (Adding e to T ): 1. Note that T + e contains a unique cycle C. 2. Since T is acyclic, there is an edge e E(C) E(T ). 3. Now T + e e is connected and acyclic because deleting e breaks the only cycle in T + e and so T + e e is a spanning tree.

18 If T, T are spanning trees of a connected graph G and e E(T ) E(T ), then there is an edge e E(T ) E(T ) such that T + e e is a spanning tree of G: (Adding e to T ): 1. Note that T + e contains a unique cycle C. 2. Since T is acyclic, there is an edge e E(C) E(T ). 3. Now T + e e is connected and acyclic because deleting e breaks the only cycle in T + e and so T + e e is a spanning tree.

19 If T is a tree with r edges and G is a simple graph with δ(g) r, then T is a subgraph of G: 1. strong induction on r. 2. Basis step: r = 0, a tree with no edges which is just K 1. clearly it holds. 3. Induction step: r > 0: 3.1 T has at least two vertices and so we can choose a leaf v in T. 3.2 Let u be the neighbor of v. Consider the small tree T = T v. By the strong induction hypothesis, G contains T as a subgraph. 3.3 Let x be the vertex in this copy of T in G that corresponds to u. Then x has a neighbor y in G that is not in this copy of T because Adding the edge xy expands this copy of T into a copy of T in G. Remark: The inequality δ r is sharp: consider K r.

20 If T is a tree with r edges and G is a simple graph with δ(g) r, then T is a subgraph of G: 1. strong induction on r. 2. Basis step: r = 0, a tree with no edges which is just K 1. clearly it holds. 3. Induction step: r > 0: 3.1 T has at least two vertices and so we can choose a leaf v in T. 3.2 Let u be the neighbor of v. Consider the small tree T = T v. By the strong induction hypothesis, G contains T as a subgraph. 3.3 Let x be the vertex in this copy of T in G that corresponds to u. Then x has a neighbor y in G that is not in this copy of T because Adding the edge xy expands this copy of T into a copy of T in G. Remark: The inequality δ r is sharp: consider K r.

21 Conjecture 1. Every n-vertex simple graph G with more than n(r 1) edges has T (a tree with r edges) as a subgraph. 2. (Er dos, Sós) Conjecture: If e(g) > n(r 1)/2, then G has T as a subgraph. 3. It has been proved for graphs without 4-cycles (Saclé-Woźniak,1997). 4. Ajtai, Komlós, and Szemerédi proved an asymptotic version. (2000)

22 If H is a subgraph of G, then d G (u, v) d H (u, v). Proof. Every u, v-path in H appears in G, so the shortest u, v path in G is no longer than the shortest u, v-path in H.

23 If H is a subgraph of G, then d G (u, v) d H (u, v). Proof. Every u, v-path in H appears in G, so the shortest u, v path in G is no longer than the shortest u, v-path in H.

24 Examples (Pop-quiz) 1. The diameter of the Petersen graph is ( ). 2. The diameter of the hypercube Q k is ( ). 3. The diameter of the cycle C n is ( ). 4. Let G be any of the above, then rad(g) = ( ). 5. The n-vertex tree (n 3) of least diameter is ( ) with diameter ( ) and radius ( ). 6. The n-vertex tree (n 3) of largest diameter is ( ) with diameter ( ) and radius ( ). 7. The diameter of a tree is the length of its ( ). (Uniquess of paths...)

25 If G is a simple graph, then diamg 3 diamḡ Since diamg 3, we can find u, v V (G) which are nonadjacent and have no commnon neighbor. 2. Hence every x V (G) \ {u, v} has at least one of {u, v} as a nonneighbor. 3. Thus x is adjacent in Ḡ to at least on of {u, v}. 4. Since uv E(Ḡ), for every pair x, y there is an x, y-path of length at most 3 in Ḡ (through {u, v}). 5. Hence diamḡ 3.

26 If G is a simple graph, then diamg 3 diamḡ Since diamg 3, we can find u, v V (G) which are nonadjacent and have no commnon neighbor. 2. Hence every x V (G) \ {u, v} has at least one of {u, v} as a nonneighbor. 3. Thus x is adjacent in Ḡ to at least on of {u, v}. 4. Since uv E(Ḡ), for every pair x, y there is an x, y-path of length at most 3 in Ḡ (through {u, v}). 5. Hence diamḡ 3.

27 (Jordan, 1869) The center of a tree is a vertex or an edge. Proof. 1. Strong induction on the number of vertices in a tree T. 2. Basis step: n(t ) 2: the center is the entire tree. 3. Induction step: n(t ) > 2:Let T be a tree formed from T by deleting all leaves of T. claim: T and T have the same center: 3.1 The internal vertices on paths between leaves of T remain in T and thus T has at least one vertex. 3.2 Also note that given a vertex u V (T ), every vertex at maximum distance in T from u is a leaf. Thus the eccentricity of a leaf in T is greater than the eccentricity of its neighbor in T. 3.3 Also ɛ T (u) = ɛ T (u) 1 for every u V (T ) because Thus the claim follows. Thus, by the induction hypothesis, the center of T which is

28 Among trees with n vertices, the Wiener index D(T ) = u,v V (T ) d(u, v) is minimized by stars and maximized by paths. D(K 1,n 1 ) = D(P n ) = Remark: Pascal s formula, ( ) ( n 1 k + n 1 ( k 1) = n ) k

29 Among trees with n vertices, the Wiener index D(T ) = u,v V (T ) d(u, v) is minimized by stars and maximized by paths. ( ) n 1 D(K 1,n 1 ) = (n 1) + 2 = (n 1) 2. 2 ( ) ( ) n n + 1 D(P n ) = D(P n 1 ) + =. 2 3

30 If G is a connected n-vertex graph, then D(G) D(P n ). Proof. Let T be a spanning tree of G. Since T is a subgraph of G, D(G) D(T ). But n(t ) = n(g) = n since T is a spanning tree of G. Thus D(G) D(T ) D(P n ).

31 Enumeration of Trees How many are trees among 2 (n 2) simple graphs with vertex set [n]? n = 1? n = 2? n = 3? n = 4?

32 (Cayley s formula [1889]) There are n n 2 trees with vertex set [n] How to prove it? Idea: Get a bijection to a set of known size. Let S n 2 be the set of lists of length n 2 with entries in a set S (allowing repetitions). Note that S n 2 = n n 2 if S = [n]

33 Prüfer code Let T be a tree with vertex set S N of size n. f (T ) = (a 1,..., a n 2 ) is called the Prüfer code of T which is defined as follows: At the ith step, delete the least remaining leaf and let a i be the neighbor of this leaf. f is a function from the set of trees with vertex set S of size n to S n 2. We need to show that it is a bijection. f 1?

34 Prüfer code Let T be a tree with vertex set S N of size n. f (T ) = (a 1,..., a n 2 ) is called the Prüfer code of T which is defined as follows: At the ith step, delete the least remaining leaf and let a i be the neighbor of this leaf. f is a function from the set of trees with vertex set S of size n to S n 2. We need to show that it is a bijection. f 1?

35 Count trees by vertex degrees. Given positive integers d 1,..., d n summing to 2n 2, there are exactly (n 2)! (di 1)! trees with vertex set [n] such that vertex i has degree d i for each i. 1. Each vertex v of tree T appears d T (v) 1 times in the Prüfer code. Why? 2. Thus it is enough to count lists of length n 2 which have d i 1 copies of i for each i. 3. the number of permutations.

36 Count trees by vertex degrees. Given positive integers d 1,..., d n summing to 2n 2, there are exactly (n 2)! (di 1)! trees with vertex set [n] such that vertex i has degree d i for each i. 1. Each vertex v of tree T appears d T (v) 1 times in the Prüfer code. Why? 2. Thus it is enough to count lists of length n 2 which have d i 1 copies of i for each i. 3. the number of permutations.

37 Counting the number of spanning trees in a graph The number of spanning trees in a complete graph on n vertices is n n 2. what about for general graphs?

38 If e E(G) is not a loop, then τ(g) = τ(g e) + τ(g e). Proof. 1. It is clear that τ(g e) counts the number of spanning trees of G which do not contain e. 2. It is enough to construct a bijection between the set of spanning tress of G e and the set of spanning trees of G containing e...

39 If e E(G) is not a loop, then τ(g) = τ(g e) + τ(g e). Proof. 1. It is clear that τ(g e) counts the number of spanning trees of G which do not contain e. 2. It is enough to construct a bijection between the set of spanning tress of G e and the set of spanning trees of G containing e...

40 Remarks 1. The previous statement does not hold if e is a loop. Why? 2. A disconnected graph has no spanning tree. 3. If G is connected and loopless, and any cycle in G is of length 2 (multiple edge), then τ(g) is equal to the product of the edge multiplicities. Why? 4. Loops do not affect τ(g), and so we can delete loops as they arise. 5. However, the amount of computation may grow exponentially with the size of the graph: impractical. 6. We have a faster computation: Matrix Tree computation.

41 Matrix Tree Theorem Let G be a loopless graph with V (G) = {v 1,..., v n }. Let D be the diagonal matrix with D i,i = d(v i ). Let Q = D A, where A is the adjacency matrix of G. Let Q be a matrix obtained by deleting row s and column t of Q, then τ(g) = ( 1) s+t detq

42 class work: 2.2.1, (left by recurrence), (left by Matrix tree theorem),

43 Directed Matrix Tree Theorem-Tutte[1948] Definition. A branching (out-tree) is an orientation of a tree with a root of indegree 0 and all other vertices of indegree 1. ( An in-tree is an out-tree with edges reversed.) Theorem. Let G be a loopless digraph. Let Q = D A t, where D is the diagonal matrices of indegrees. (Similarly, Q + = D + A t. Then the number of spanning out-trees (in-trees) of G rooted at v i is the value of each cofactor in the ith row of Q. (ith column of Q + )

44 Decomposition of G into copies of a tree T A necessary condition: e(t ) divides e(g). What if G is e(t )-regular? (e.g. Petersen graph) No. (ex. 20) Conjecture 1 (Häggkvist): if G is a 2m-regular graph and e(t ) = m, then G decomposes into n(g) copies of T. Conjecture 2 (Ringel): The previous conjecture only for K 2m+1. Theorem (Rosa [1967]) Conjecture 2 holds true if a tree T with e(t ) = m has a graceful labeling. Conjecture 3 (Graceful Tree Conjecture-Kotzig, Ringel) Every tree has a graceful labeling.

45 Decomposition of G into copies of a tree T A necessary condition: e(t ) divides e(g). What if G is e(t )-regular? (e.g. Petersen graph) No. (ex. 20) Conjecture 1 (Häggkvist): if G is a 2m-regular graph and e(t ) = m, then G decomposes into n(g) copies of T. Conjecture 2 (Ringel): The previous conjecture only for K 2m+1. Theorem (Rosa [1967]) Conjecture 2 holds true if a tree T with e(t ) = m has a graceful labeling. Conjecture 3 (Graceful Tree Conjecture-Kotzig, Ringel) Every tree has a graceful labeling.

46 Decomposition of G into copies of a tree T A necessary condition: e(t ) divides e(g). What if G is e(t )-regular? (e.g. Petersen graph) No. (ex. 20) Conjecture 1 (Häggkvist): if G is a 2m-regular graph and e(t ) = m, then G decomposes into n(g) copies of T. Conjecture 2 (Ringel): The previous conjecture only for K 2m+1. Theorem (Rosa [1967]) Conjecture 2 holds true if a tree T with e(t ) = m has a graceful labeling. Conjecture 3 (Graceful Tree Conjecture-Kotzig, Ringel) Every tree has a graceful labeling.

47 Decomposition of G into copies of a tree T A necessary condition: e(t ) divides e(g). What if G is e(t )-regular? (e.g. Petersen graph) No. (ex. 20) Conjecture 1 (Häggkvist): if G is a 2m-regular graph and e(t ) = m, then G decomposes into n(g) copies of T. Conjecture 2 (Ringel): The previous conjecture only for K 2m+1. Theorem (Rosa [1967]) Conjecture 2 holds true if a tree T with e(t ) = m has a graceful labeling. Conjecture 3 (Graceful Tree Conjecture-Kotzig, Ringel) Every tree has a graceful labeling.

48 Decomposition of G into copies of a tree T A necessary condition: e(t ) divides e(g). What if G is e(t )-regular? (e.g. Petersen graph) No. (ex. 20) Conjecture 1 (Häggkvist): if G is a 2m-regular graph and e(t ) = m, then G decomposes into n(g) copies of T. Conjecture 2 (Ringel): The previous conjecture only for K 2m+1. Theorem (Rosa [1967]) Conjecture 2 holds true if a tree T with e(t ) = m has a graceful labeling. Conjecture 3 (Graceful Tree Conjecture-Kotzig, Ringel) Every tree has a graceful labeling.

49 Decomposition of G into copies of a tree T A necessary condition: e(t ) divides e(g). What if G is e(t )-regular? (e.g. Petersen graph) No. (ex. 20) Conjecture 1 (Häggkvist): if G is a 2m-regular graph and e(t ) = m, then G decomposes into n(g) copies of T. Conjecture 2 (Ringel): The previous conjecture only for K 2m+1. Theorem (Rosa [1967]) Conjecture 2 holds true if a tree T with e(t ) = m has a graceful labeling. Conjecture 3 (Graceful Tree Conjecture-Kotzig, Ringel) Every tree has a graceful labeling.

50 Graceful labeling A graceful labeling of a graph G with m edges is an injective function f : V (G) {0,..., m} such that { f (u) f (v) : uv E(G)} = {1,..., m}.

51 Stars, paths, caterpillars... Examples

52 (Rosa, 1967) If a tree T with m edges has a graceful labeling, then K 2m+1 has a decomposition into 2m + 1 copies of T. Proof. 1. View the vertices of K 2m+1 as elements of Z 2m+1, arranged circularly. 2. Divide the edges of K 2m+1 into m groups by the difference between the endpoints. 3. Let T 0 be the subgraph of K 2m+1 defined as follows: 3.1 V (T 0 ) = {0, 1,..., m}(mod2m + 1). 3.2 (i, j) E(T 0 ) iff i, j are adjacent in the graceful labeling of T 4. Similarly, define T 1,..., T 2m. More precisely, 4.1 V (T k ) = {k, k + 1,..., k + m}(mod2m + 1). 4.2 (k + i, k + j) E(T k ) iff i, j are adjacent in the graceful labeling of T 5. Note that each difference class of edges has one edge in each T k, and thus T 0,..., T 2 m decompose K 2m+1.

53 (Rosa, 1967) If a tree T with m edges has a graceful labeling, then K 2m+1 has a decomposition into 2m + 1 copies of T. Proof. 1. View the vertices of K 2m+1 as elements of Z 2m+1, arranged circularly. 2. Divide the edges of K 2m+1 into m groups by the difference between the endpoints. 3. Let T 0 be the subgraph of K 2m+1 defined as follows: 3.1 V (T 0 ) = {0, 1,..., m}(mod2m + 1). 3.2 (i, j) E(T 0 ) iff i, j are adjacent in the graceful labeling of T 4. Similarly, define T 1,..., T 2m. More precisely, 4.1 V (T k ) = {k, k + 1,..., k + m}(mod2m + 1). 4.2 (k + i, k + j) E(T k ) iff i, j are adjacent in the graceful labeling of T 5. Note that each difference class of edges has one edge in each T k, and thus T 0,..., T 2 m decompose K 2m+1.

54 (Rosa, 1967) If a tree T with m edges has a graceful labeling, then K 2m+1 has a decomposition into 2m + 1 copies of T. Proof. 1. View the vertices of K 2m+1 as elements of Z 2m+1, arranged circularly. 2. Divide the edges of K 2m+1 into m groups by the difference between the endpoints. 3. Let T 0 be the subgraph of K 2m+1 defined as follows: 3.1 V (T 0 ) = {0, 1,..., m}(mod2m + 1). 3.2 (i, j) E(T 0 ) iff i, j are adjacent in the graceful labeling of T 4. Similarly, define T 1,..., T 2m. More precisely, 4.1 V (T k ) = {k, k + 1,..., k + m}(mod2m + 1). 4.2 (k + i, k + j) E(T k ) iff i, j are adjacent in the graceful labeling of T 5. Note that each difference class of edges has one edge in each T k, and thus T 0,..., T 2 m decompose K 2m+1.

55 Optimization and Trees The best spanning tree Weighted graphs (e.g. costs of links, distances) (we consider only nonnegative weights.)

56 Kruskal s Algorithm-for minimum spanning trees Input: A weighted connected graph. Idea: Maintain an acyclic spanning subgraph H, enlarging it by edges with low weight to form a spanning tree. Initialization: Set E(H) =. Iteration: If the next cheapest edge joins two components of H, then include it; otherwise, discard it. Terminate when H is connected. Why does it produce a tree?

57 Example (ex 2.3.3) There are 5 cities in a network. The cost of building a road directly between i and j is the entry a i,j in the matrix below. An infinite entry indicates that there is a mountain in the way and the road cannot be built. Determine the least cost of making all the cities reachable from each other