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Chapter 2

Chapter 2. 2.1 Basic properties 2.2 Spanning tree and enumeration 2.3 Optimization and trees. Trees and Distance. Acyclic, tree 2.1.1. A graph with no cycle is acyclic A forest is an acyclic graph A tree is a connected acyclic graph

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Chapter 2

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  1. Chapter 2 2.1 Basic properties 2.2 Spanning tree and enumeration 2.3 Optimization and trees Trees and Distance Ch. 2. Trees and Distance

  2. Acyclic, tree2.1.1 • A graph with no cycle is acyclic • A forestis an acyclic graph • A tree is a connected acyclic graph • A leaf (or pendant vertex ) is a vertex of degree 1 leaf tree tree forest Ch. 2. Trees and Distance

  3. Spanning Subgraph 2.1.1 • A spanning subgraphof G is a subgraph with vertex setV(G) • A spanning tree is a spanning subgraph that is a tree Spanning tree Spanning subgraph Ch. 2. Trees and Distance

  4. Maximal path Lemma.Every tree with at least two vertices has at least two leaves.2.1.3 Proof: (1/2) • A connected graph with at least two vertices has an edge. • In an acyclic graph, an endpoint of a maximal nontrivial path has no neighbor other than its neighbor on the path. • Hence the endpoints of such a path are leaves. Impossible! Cycle occurs Impossible! It is Maximal. Ch. 2. Trees and Distance

  5. Lemma.Deleting a leaf from a n-vertex tree produces a tree with n-1 vertices.2.1.3 Proof: (2/2) • Letvbe a leaf of a tree G, and that G’=G-v. • A vertex of degree 1 belongs to no path connecting two other vertices. • Therefore, for u, w V(G’), every u, w-path inGis also in G’. • HenceG’is connected. • Since deleting a vertex cannot create a cycle, G’also is acyclic. • ThusG’is a tree with n-1 vertices. Ch. 2. Trees and Distance

  6. G G’ v w u • Foru, wV(G’ ), every u, w-path inGis also in G’. Ch. 2. Trees and Distance

  7. Theorem 2.1.4.For an n-vertex graphG(with n1), the following are equivalent (and characterize the trees with nvertices)2.1.4A)Gis connected and has no cyclesB) Gis connected and hasn-1 edgesC) Ghasn-1 edges and no cyclesD) Foru, vV(G), Ghas exactly one u, v-path Proof:We first demonstrate the equivalence of A, B, and C by proving that any two of {connected, acyclic,n-1 edges} together imply the third Ch. 2. Trees and Distance

  8. A{B,C}. connected, acyclic n-1 edgesTheorem 2.1.4Continue • We use induction on n. • Forn=1, an acyclic 1-vertex graph has no edge • Forn >1, we suppose that implication holds for graphs with fewer thannvertices • Given an acyclic connected graph G, Lemma 2.1.3 provides a leaf vand states that G’=G-valso is acyclic and connected(see figure in previous proof) • Applying the induction hypothesis to G’yields e(G’)=n-2 • Since only one edge is incident to v, we have e(G)=n-1 Ch. 2. Trees and Distance

  9. B{A, C}Connected andn-1 edges acyclicTheorem 2.1.4 continue • If Gis not acyclic, delete edges from cycles of Gone by one until the resulting graph G’ is acyclic • Since no edge of a cycle is a cut-edge(Theorem 1.2.14), G’is connected • Now the preceding paragraph implies that e(G’) = n-1 • Since we are given e(G)=n-1, no edges were deleted • ThusG’=G, and G is acyclic Ch. 2. Trees and Distance

  10. C{A, B} n-1 edges and no cycles  connectedTheorem2.1.4 continue • LetG1,…,Gk be the components of G • Since every vertex appears in one component, in(Gi)=n • SinceG has no cycles, each component satisfies property A. Thus e(Gi) = n(Gi) - 1 • Summing over i yieldse(G)=I[n(Gi)-1]= n-k • We are given e(G)=n-1,sok =1, and Gis connected Ch. 2. Trees and Distance

  11. ADConnected and no cycles Foru, vV(G), one and only one u, v-path exists Theorem 2.1.4 • SinceGis connected, each pair of vertices is connected by a path • If some pair is connected by more than one , we choose a shortest (total length) pair P, Qof distinct paths with the same endpoints • By this extremal choice, no internal vertex of PorQcan belong to the other path • This implies that PQis a cycle, which contradicts the hypothesis A p v u q Ch. 2. Trees and Distance

  12. DAForu, vV(G),one and only one u, v-path exists  connected and no cycles Theorem 2.1.4 • If there is a u,v-path for every u,vV(G), then Gis connected • IfGhas a cycle C, then G has two u,v-paths foru,v V(G);which contradicts the hypothesis D • HenceGis acyclic (this also forbids loops). Ch. 2. Trees and Distance

  13. Corollary: Every edge of a tree is a cut-edge 2.1.5 Proof: • A tree has no cycles • Theorem 1.2.14 implies that every edge is a cut-edge. Ch. 2. Trees and Distance

  14. Corollary: Adding one edge to a tree forms exactly one cycle 2.1.5 Proof: • A tree has a unique path linking each pair of vertices (Theorem2.1.4D) • Joining two vertices by an edge creates exactly one cycle Ch. 2. Trees and Distance

  15. Corollary: Every connected graph contains a spanning tree 2.1.5 Proof: • Similar to the proof of BA, C in Theorem 2.1.4 • Iteratively deleting edges from cycles in a connected graph yields a connected acyclic subgraph Ch. 2. Trees and Distance

  16. Proposition: IfT, T’are spanning trees of a connected graph Gand 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.2.1.6 e’ e G T T’ T-e+e’ Ch. 2. Trees and Distance

  17. Proposition:IfT, T’are spanning trees of a connected graph Gand 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. 2.1.6 Proof: • Every edge of Tis a cut-edge ofT. Let UandU’be the two components of T-e. • Since T’is connected, T’has an edge e’with endpoints in UandU’. • Now T-e+e’is connected, has n(G)-1 edges, and is a spanning tree of G. e’ e T T’ T-e+e’ Ch. 2. Trees and Distance

  18. Distance in trees and Graphs • Assume Ghas au, v-path • Then the distance from utov, writtendG(u,v) or d(u,v), is the least length of a u,v-path. • If Ghas no such path, then d(u,v)=  Ch. 2. Trees and Distance

  19. Distance in trees and Graphs • The diameter (diamG) is maxu,vV(G)d(u,v) • Upper bound of distance between every pair • The eccentricityof a vertex u is (u) = maxvV(G) d(u,v) • Upper bound of the distance from u to the others • The radius of a graph G is radG = minuV(G) (u) • Lower bound of the eccentricity Ch. 2. Trees and Distance

  20. Distance, Diameter, Eccentricity, and Radius a b f e c g d Distance(f,c) : 2 Distance(g,c): 2 Distance(a,c): 3 Radius: 2 Diameter: 3 Eccentricity(f):2 Eccentricity(a): 3 Ch. 2. Trees and Distance

  21. If G is a simple graph, then diamG 3  diam  3 2.1.11 Proof:1/3 • Since diamG> 2, there exist nonadjacent vertices u, vV(G) with no common neighbor • If any pair of nonadjacent vertices has a common neighbor, the distance of every pair is less than or equal to 2 and diamG= 2 Not every pair is of this kind u v A pair (u, v)of this kind exists Ch. 2. Trees and Distance

  22. If G is a simple graph, then diamG  3  diam 3 2.1.11 Proof:2/3 • Hence every xV( ) - {u,v} has at least one of {u,v} as a nonneighbor • Eitheruorvis not connected to x • Equivalently, this makes x adjacent to at least one of {u,v} in Everyone of these vertices has at least one of {u,v} as a nonneighbor u v Ch. 2. Trees and Distance

  23. If G is a simple graph, then diamG  3  diam  3 2.1.11 Proof:3/3 • Sincealso u v  E( ),for every pair x, y there is an x, y-path of length at most 3 in through {u,v}. Hence diam  3 u v Ch. 2. Trees and Distance

  24. Center 2.1.12 Definition:The center of a graph Gis the subgraph induced by the vertices of minimum eccentricity. • The center of a graph is the full graph if and only if the radius and diameter are equal. Center Recall: The eccentricityof a vertex u is the upper bound of the distance from u to the others Ch. 2. Trees and Distance

  25. The center of a tree is a vertex or an edge2.1.13 Proof:We use induction on the number of vertices in a tree T. • Basis step: n(T)≦ 2. With at most two vertices, the center is the entire tree. Ch. 2. Trees and Distance

  26. Theorem. 2.1.13Continue • Induction step: n(T)>2. • LetT’ = T- {leaves}. By Lemma 2.1.3, T’is a tree. • Since the internal vertices on the paths between leaves of T remain, T’has at least one vertex. • Every vertex at maximum distance in Tfrom a vertex uV(T) is a leaf (otherwise, the path reaching it from ucan be extended farther). The max path from u The end must be a leaf u T = Green Pink T’= Green Ch. 2. Trees and Distance

  27. Theorem. 2.1.13Continue • Since all the leaves have been removed and no path between two other vertices uses a leave, T’(u)= T(u)-1 for every uV(T’). • Also, the eccentricity of a leaf in Tis greater than the eccentricity of its neighbor in T. • Hence the vertices minimizing T(u) are the same as the vertices minimizing T’(u). {v| v has min (v) in T} = {v’| v’ has min (v’) in T’} T’(u)= T(u)-1 u Ch. 2. Trees and Distance

  28. Theorem. 2.1.13Continue • It is shownTand T’have the same center. • By the induction hypothesis, the center ofT’ is a vertex or an edge. T” T T’ Ch. 2. Trees and Distance

  29. Spanning Trees and Enumeration2.2 • There are 2c(n,2) simple graphs with vertex set [n]={1,…,n}. • since each pair may or may not form an edge. • How many of these are trees? Can either appear or disappear Max number of edges = c(n,2) Ch. 2. Trees and Distance

  30. Enumeration of Trees 2.2 • One or two vertices, then there is only one tree. • Three vertices, three trees. • Four vertices, then four stars and 12 paths, total 16. • Five vertices, then there are 125 trees. Ch. 2. Trees and Distance

  31. Algorithm for tree generation by using Prűfer code2.2.1 Algorithm. (Prűfer code) Production of f(T)=(a1,…,an-2) Input: A tree Twith vertex set S  N Iteration: At the i th step, delete the least remaining leaf, and say thataiis the neighborof the deleted leaf 1. Delete 2 a1= 7 2. Delete 3 a2= 4 3. Delete 5 a3= 4 4. Delete 4 a4= 1 5. Delete 6 a5= 7 6. Delete 7 a6= 1 4 1 7 3 2 5 6 8 Ch. 2. Trees and Distance

  32. Theorem: For a set SN of size n, there are nn-2trees with vertex set S 2.2.3 Proof: (sketch) • Trees with vertex set S Sn-2of lists of length n-2 • One prufer code  one tree. • There arenn-2codes. • Proved by induction Ch. 2. Trees and Distance

  33. Spanning Trees in Graphs 2.2.6 Example. A kite. To count the spanning trees: • Four are path around the outside cycle in the drawing • The remaining spanning trees use the diagonal edge • Since we must include an edge to each vertex of degree 2, we obtain four more spanning trees. • The total is eight. Ch. 2. Trees and Distance

  34. Contraction 2.2.7 • In a graph G, contraction of edge ewith endpointsu, vis the replacement of uandvwith a single vertex whose incident edges are the edges other than ethat were incident touorv • The resulting graph G·ehas one less edge thanG u e G G·e v Ch. 2. Trees and Distance

  35. Proposition. Let (G) denote the number of spanning trees of a graph G. If eE(G) is not a loop, then(G)= (G-e)+ (G·e) 2.2.8 • (G-e): The number of trees without e • (G·e): The number of trees with e • A spanning tree in G.e  A spanning tree having einG G G-e e = G·e Ch. 2. Trees and Distance

  36. Proposition. Let (G) denote the number of spanning trees of a graph G. If eE(G) is not a loop, then (G)= (G-e)+ (G·e) 2.2.8 Proof: 1/2 • The spanning trees of Gthat omit eare precisely the spanning trees ofG-e • Need to show thatGhas(G·e)spanning trees containing e • It must be shown that contraction of edefines a bijection from the set of spanning trees ofGcontaininge to the set of spanning trees of G·e Ch. 2. Trees and Distance

  37. Proposition. Let (G) denote the number of spanning trees of a graph G. If eE(G) is not a loop, then (G)= (G-e)+ (G·e) 2.2.8 Proof:2/2 • When contract e in a spanning tree having e, obtain a spanning tree of G·e because • The resulting subgraph of G·e is spanning and connected • It has the right number of edges • The other edges maintain their identity under contraction • So no two trees are mapped to the same spanning tree of G·e by this operation. Ch. 2. Trees and Distance

  38. A Matrix Tree computation. 2.2.12 • Given a loopless graph Gwith vertex set v1, …., vn • Let aijbe the number of edges with endpoints viand vj • LetQbe the matrix in which • entry (i, j) is –ai,jwhenij and is d(vi) wheni=j • Let Q* is a matrix obtained by deleting row s and column tofQ, then (G) = (-1)s+t detQ* Ch. 2. Trees and Distance

  39. A Matrix Tree computation. 2.2.11 • Theorem 2.2.12 instructs us • Form a matrix by putting the vertex degrees on the diagonal and subtracting the adjacency matrix • Delete a row and a column and take the eterminant Ch. 2. Trees and Distance

  40. A Matrix Tree computation. 2.2.11 • Consider the kite of Example 2.2.9, • The vertex degrees are 3,3,2,2 • We form the matrix on the left below and then • We take the determinant of the matrix in the middle • The result is the number of spanning trees Ch. 2. Trees and Distance

  41. Example 2.2.11 v3 v1 v4 v2 v1 v2 v3 v4 v1 v2 v3 v4 8 Ch. 2. Trees and Distance

  42. Minimum Spanning Tree 2.3 • In a connected weighted graph of possible communication links, all spanning trees have n-1edges; we seek one that minimizes or maximizes the sum of the edge weights. • Kruskal’s Algorithm • Prim’s Algorithm Ch. 2. Trees and Distance

  43. Kruskal’s Algorithmfor Minimum Spanning Tree 2.3.1 • Input: A weighted connected graph • Idea: • Maintain an acyclic spanning subgraph H. • Enlarging it by edges with low weight to form a spanning tree. • Consider edges in nondecreasing order of weight. Ch. 2. Trees and Distance

  44. Kruskal’s Algorithmfor Minimum Spanning Tree 2.3.1 • Initialization: SetE(H)=. • Iteration: If the next cheapest edge joins two components ofH, then include it; otherwise, discard it. Terminate whenHis connected. H Join Two vertices in one component. Cycle occurs. Not Allowed! Join two components. It works Ch. 2. Trees and Distance

  45. Example of using Kruskal’s Algorithm 9 9 9 2 2 8 8 4 1 2 4 1 8 4 1 6 7 6 7 6 7 5 5 5 10 10 10 11 3 11 3 11 3 12 12 12 9 9 9 2 8 4 1 2 2 1 8 4 8 4 1 7 6 6 7 7 6 5 10 10 5 5 3 11 11 3 11 3 10 12 12 12 Ch. 2. Trees and Distance

  46. Theorem: In a connected weighted graph G, Kruskal’s Algorithm constructs a minimum-weight spanning tree. 2.3.3 Proof: 1/3 • We show first that the algorithm produces a tree • We never choose an edge that completes a cycle • If the final graph has more than one component, then there is no edge joining two of them and G is not connected • Since G is connected, some such edge exists and we considered it. • Thus the final graph is connected and acyclic, which makes it a tree. Ch. 2. Trees and Distance

  47. Theorem2.3.3 Continue Proof: continue • LetTbe the resulting tree, and let T*be a spannig tree of minimum weight. • IfT=T*, we are done. • IfTT*, let e be the first edge chosen for Tthat is not in T*. Adding etoT*creates one cycle C. Since Thas no cycle, Chas an edge e’E(T). Consider the spanning tree T*+e-e’ The first edge chosen for T that is not in T* T: …………. e, ………… T*: …………. e*,………… Identical Ch. 2. Trees and Distance

  48. Theorem2.3.3 Continue Proof: continue • SinceT*containse’and all the edges of Tchosen before e, bothe’and eare available when the algorithm chooses e, and hence w(e)w(e’) • ThusT*+e-e’is a spanning tree with weight at mostT*that agrees with Tfor a longer initial list of edges thanT*does. T: ………….e, ………… T*: ………….e*,………… Identical T: …………… e, ………… T*+e-e’: …………… e, ………… Identical Ch. 2. Trees and Distance

  49. Theorem2.3.3 Continue Proof: continue • Repeating this argument eventually yields a minimum-weight spanning tree that agrees completely withT. • Phrased extremely, we have proved that the minimum spanning tree agreeing withTthe longest is Titself. Ch. 2. Trees and Distance

  50. Shortest Paths • How can we find the shortest route from one location to another? Ch. 2. Trees and Distance

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