1 / 60

Graph Theory Chapter 3 Tree

Graph Theory Chapter 3 Tree. 大葉大學 (Da-Yeh Univ.) 資訊工程系 (Dept. CSIE) 黃鈴玲 (Lingling Huang). Outline. 3.1 Properties of trees 3.2 Rooted trees 3.3 Depth-first search 3.5 Breadth-first search 3.6 The minimum spanning tree problem. 3.1 Properties of trees. Definition:

harmon
Download Presentation

Graph Theory Chapter 3 Tree

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Graph TheoryChapter 3 Tree 大葉大學(Da-Yeh Univ.)資訊工程系(Dept. CSIE)黃鈴玲(Lingling Huang)

  2. Outline 3.1 Properties of trees 3.2 Rooted trees 3.3 Depth-first search 3.5 Breadth-first search 3.6 The minimum spanning tree problem

  3. 3.1 Properties of trees Definition: tree: a connected graph without cycles. ( every edge is a bridge) forest: a graph (connected or disconnected) without cycles . A spanning treeof a graph G is a spanning subgraph of G that is a tree . See Figure 3.1.

  4. Pf. By strong from of induction onp. p = 1 : k1 is the only tree of order 1, |E(k1)| = 0. ok! Assume the result is true for all trees of order p < k (k  2 , k  ). consider p = k : Let T be a tree of order p = k, and e be an edge of T. Let T1, T2 be the two components of T-{e}, and |V(T1)| = p1, |V(T2)| = p2 (p1+p2 =p) Thm 3.1: Let T be a tree of order p and size q, then q=p-1 .

  5. ∵p1,p2<k ∴|E(T1)| = p1-1, |E(T2)| = p2-1 by assumption.  |E(T)| = p1-1 + p2-1 +1 = p1+p2-1 = p-1. By induction, the result is true for all trees. #

  6. Cor 3.1: Let G be a graph of order p, the following are equivalent : (i) G is a tree. (ii) G is connected and has size p-1. (iii) G has size p-1 and no cycles. Pf. (i)  (ii) (by Thm 3.1) (ii)  (iii) (證明 connected + size p-1 no cycle)反證,若有cycle可扣掉邊後製造tree, 邊數不對 (iii)  (i) (證明 size p-1 + no cycle  connected)反證,每一個component本身都是connected + no cycle  tree邊數總和不對

  7. Thm 3.2: Every nontrivial tree contains at least two end-vertices. Pf. Suppose T is a tree of order p and size q, ∵ T is connected and nontrivial. ∴ deg(v)  1, v  V(T). ∴ There are at least two vertices of degree 1.#

  8. Thm 3.3: If T is a tree, u≠vV(T), then there is exactly one u-v path. Pf. (Bondy)(跟書上不同) Suppose T has two u-v paths, say P and Q. Since P≠Q, there is an edge e=xy E(P) but eE(Q). Clearly the graph (P∪Q) – e is connected. It therefore contains an x-y path P’. But then P’ + e is a cycle in T, a contradiction.

  9. Note: T: a tree. If uv E(T), then T+uv has exactly one cycle C. (which must contains the edge uv) If some edge e (未必是uv) is deleted from C, a tree results again. Thm 3.4 : Let T be a tree of order m, and let G be a graph with δ(G)m-1. Then T is isomorphic to some subgraph of G. (證明跳過)

  10. Homework Exercise : 2, 5, 8, 11, 12, 14, 15, 16, 19 Exercise 8.Let T be a tree of order 21 having degree set {1, 3, 5, 6}. Suppose that T has 15 end-vertices and one vertex of degree 6. How many verticesof degree 5 does T have?

  11. Exercise 15.Prove or disprove: (c) Every vertex of a tree is a cut-vertex.(d) A tree of order p3 has more cut-vertices then bridges.(e) There exist exactly two regular trees.

  12. Outline 3.1 Properties of trees 3.2 Rooted trees 3.3 Depth-first search 3.5 Breadth-first search 3.6 The minimum spanning tree problem

  13. 3.2 Rooted Trees Definition: A directed tree is an asymmetric digraph whose underlying graph is a tree. A directed tree T is called a rooted tree if there exists a vertex r of T, called the root, s. t.  vertex v of T, there is an r-v path in T.

  14. ※為方便起見,將tree的點分層畫出: level r root r在 level 0 0 1 2 3 height = 3 tree 的 height 即最大的level 值

  15. Thm 3.5 : A directed tree T is a rooted tree iff T contains a vertex r with id r = 0, and id v = 1 for all other vertices v of T. pf : () root 即 r, 每點往上一層只連接到一點,否則不是tree, 故 id v = 1  vr. () (證明 u1V(T),  r-u1 path) 因 id v = 1  vr,從u1開始往回走, 最後一定停在r,否則表示有 cycle, 與 T 是directed tree 矛盾.

  16. Definition: ※因rooted tree 中所有arc都是向下,所以可省略方向性. ※ parent child 更上層: ancestor 更下層: descendant r T ※ maximal subtree of T rooted at v (由v及其子孫構成的induced subtree) v See Figure 3.8.

  17. left child right child ※ leaf : a vertex with no children internal vertex: leaf 之vertex. ※ A rooted tree ism-ary if every vertex has at most m children. m = 2  binary ※ complete m-ary tree: every vertex has exactly m children or no children.

  18. Thm 3.6 : A complete m-ary tree with i internal vertices has order m i +1. Pf. 每個 internal vertex 有 m個children, 故共有 m i 個 children (每個child只有一個parent,所以只會被計算一次),root 沒有parent,還沒被計算到 ∴共有 m i +1 個點

  19. Cor 3.6 : Every complete binary tree with i internal vertices has i+1 leaves. pf. By Thm 3.6, m = 2, i internal vertex ∴ 點數 = 2i + 1 ,  i +1 leaves. Thm 3.7 : If T is a binary tree of height h and order p, then h+1 p 2h+1- 1. Pf. (h+1 p) 當每個 level 只有一點時 (p 2h+1– 1)當 tree 是 complete 時

  20. Definition: A rooted tree T of height h is balanced if every leaf is at level h or h-1. Cor 3.7 : If T is a binary tree of height h and order p, then h log2 ((p+1)/2) .Equality holds if T is a balanced complete binary tree. Pf. p 2h+1- 1  h log2 ((p+1)/2)

  21. Pf of Cor 3.7: (continued) If T is a balanced complete binary tree, then p > 1+2+22+…+2h-1 = 2h-1 2h-1 < p 2h+1- 1 2h-1 < (p+1)/2 2h  h= log2 ((p+1)/2) #

  22. Homework: 4, 5 Exercise 5:A complete m-ary tree T of height h is called a full m-ary tree if every leaf is at the level h. If T has l leaves, show that h = logml.

  23. Outline 3.1 Properties of trees 3.2 Rooted trees 3.3 Depth-first search 3.5 Breadth-first search 3.6 The minimum spanning tree problem

  24. G T1 T3 T2 3.3 Depth-first Search A spanning tree of a graph G is a tree that is a subgraph of G with vertex set V(G). Some spanning trees of G :

  25. ※ Many graph algorithms require a systematic method of visiting the vertices of a graph  depth-first search (DFS, 深先搜尋) bread-first search (BFS, 廣先搜尋)

  26. v2 1 v1 v6 v5 v4 v3 v7 Depth-first search G 假設G用其adjacency list來表示,而list中,每個點的neighbor依下標由小排到大 v1 v2 v3 v4 v6 v5 v7 Step 1: visit v1

  27. v2 1 v1 v6 2 v5 v4 v3 v7 v2 1 v1 v6 2 3 v5 v4 v3 v7 G v1 v2 v3 v4 v6 v5 v7 Step 2: visit v3 Step 3: visit v5

  28. v2 1 v1 v6 2 3 v5 v4 v3 4 v7 5 v2 1 v1 v6 2 3 v5 v4 v3 4 v7 G v1 v2 v3 v4 v6 v5 v7 Step 4: visit v7 Step 5: visit v2

  29. 5 v2 1 v1 v6 2 3 v5 6 v4 v3 4 v7 depth-first search index 5 v2 1 v1 7 v6 2 3 v5 6 v4 v3 dfi(v5)=3 4 v7 G v1 v2 v3 v4 v6 v5 v7 Step 6: visit v4 Step 7: visit v6 DFS forest

  30. DFS forest G F v5 v5 v13 v13 v8 v8 v2 v2 v12 v7 v12 v7 v6 v6 v1 v1 v10 v10 v15 v15 v14 v14 v4 v4 v9 v9 v3 v3 v11 v11 3 12 9 2 5 4 11 10 1 7 8 15 14 13 6

  31. rooted forest DFS forest 1 v1 6 v3 G F F’ 3 v5 12 v13 v5 v2 2 v13 7 8 v8 v4 v8 v2 v9 v2 9 2 5 v12 v7 4 v12 v7 v6 v5 v6 3 11 10 v1 v1 4 5 v10 v10 1 v7 9 v6 v12 v15 v15 7 v14 v14 8 15 v4 v10 v4 v9 v9 10 14 v3 v8 v11 v3 v11 13 v11 13 11 6 14 15 v14 v15 12 v13 back edge

  32. Algorithm 3.1 (DFS algorithm) 1. dfi(v) 0 vV(G) 2. i 1 /* 用來assign 點的 dfi值 */ 3. S  /* 用來存放 DFS forest 的所有 arc */ 4. If dfi(r)  0  r V(G), then output S and stop; otherwise, let r be the first vertex s. t. dfi(r) = 0 and let w  r. /* r是這個 component 的tree 的 root */ 5. dfi(w)  i 6. i  i + 1 7. (search) 7.1. If dfi(v) = 0 for some v in the adjacency list of w, then continue; otherwise, go to Step 7.4. 7.2 S S∪ {(w, v)} and assign parent(v) w. 7.3 w  v and return to Step 5. 7.4 If w  r, thenw parent(w) and return to Step7.1; otherwise, return to Step 4.

  33. ※ Time complexity of Alg. 3.1: O(max{p, q}) Thm 3.8. Every back edge e of a graph G joins an ancestor and a descendant. In particular, if e = uv and u is visited before v in the depth-first search of G, then v is a descendant of u.

  34. Exercise1. For the graph G shown below, find thedepth-first search forest. (不必為rooted形式) G v5 v12 v3 v9 v4 v2 v1 v11 v7 v6 v8 v10

  35. Outline 3.1 Properties of trees 3.2 Rooted trees 3.3 Depth-first search 3.5 Breadth-first search 3.6 The minimum spanning tree problem

  36. 3.5 Breadth-first Search BFS: visit過點v後,先visit完點v的每個neighbor, 將之存入queue中,再從queue中選取第一個 點visit,依此類推

  37. v5 v13 G v8 v2 v12 v7 v6 v1 v10 v15 v14 v4 v9 v3 v11 Step 1: visit v1 Queue 1 v1 v1

  38. v5 v13 G v8 v2 v12 v7 v6 v1 v10 v15 v14 v4 v9 v3 v11 7 v12 5 v6 2 v2 1 v1 6 v9 4 v4 3 v3 Step 2: visit neighbors of v1 Queue v1 v2 v3 v4 v6 v9 v12

  39. v5 v13 G v8 v2 v12 v7 v6 v1 v10 v15 v14 v4 v9 v3 v11 8 v5 7 v12 2 v2 v1 v2 v3 v4 v6 v9 v12 5 v6 1 v1 6 v9 4 v4 3 v3 Step 3: visit neighbors of v2 Queue v5

  40. v5 v13 G v8 v2 v12 v7 v6 v1 v10 v15 v14 v4 v9 v3 v11 8 v5 7 v12 2 v2 5 v6 9 v7 1 v1 6 v9 4 v4 3 v3 Step 4: visit v7 Queue v7

  41. v5 v13 G v8 v2 v12 v7 v6 v1 v10 v15 v14 v4 v9 v3 v11 8 v5 11 v13 7 v12 2 v2 5 v6 9 v7 1 v1 10 v10 6 v9 4 v4 3 v3 Step 5: visit neighbors ofv7 Queue v7 v10 v13

  42. v5 v13 G v8 v2 v12 v7 v6 v1 v10 v15 v14 v4 v9 v3 v11 8 v5 11 v13 v8 12 9 v7 7 v12 2 v2 5 v6 10 v10 1 v1 6 v9 4 v4 15 13 14 3 v3 v15 v11 v14 Step 6: visit neighbors ofv10 Queue v10 v13 v8 v11 v14 v15 所有點皆已visit BFS forest

  43. Exercise2. For the graph G shown below, find thebreadth-first search forest. G v5 v12 v3 v9 v4 v2 v1 v11 v7 v6 v8 v10

  44. Outline 3.1 Properties of trees 3.2 Rooted trees 3.3 Depth-first search 3.5 Breadth-first search 3.6 The minimum spanning tree problem

  45. 3.6 The minimum spanning tree problem Definition: (1) weighted graph:graph 中每個 edge 上有 weight (2) weightw(H) of a subgraphH of a weighted graph: the sum of the weights of the edges of H The Minimum Spanning Tree Problem : Find a minimum spanning tree in all connected weighted graph.

  46. Algorithm 3.3 (Kruskal’s Algorithm) 1. S ← . 2. Let e be an edge of minimum weight such that eS and <S∪{e}> is acyclic, and let S ← S∪{e}. 3. If |S| = p – 1, then output S; otherwise, return to Step 2. 建議:在 step2 執行前先將 G 的 edge 依 weight 由小至大排列 O(p2)

  47. y FIGURE 3-18 2 5 G 5 x 3 4 6 w 2 2 1 v u y y 2 x x w w 1 v 1 v u u Sort the edges of G: uv, xy, vw, uw, ux, vx, vy, uy, wx Constructing the tree by Kruskal algorithm Step 2 Step 1 S={uv} S={uv, xy}

  48. y FIGURE 3-18 2 5 G 5 x 3 4 6 w 2 2 y y 1 v u 2 2 x x 3 w w 2 2 1 v 1 v u u Sort the edges of G: uv, xy, vw, uw, ux, vx, vy, uy, wx Constructing the tree Step 3 Step 4 |S|=p-1 stop! S={uv, xy, vw} S={uv, xy, vw, ux}

  49. Theorem 3.12. Kruskal’s algorithm produces a minimum spanning tree in a nontrivial connected weighted graph. Pf : Let G be a nontrivial connected weighted graph of order p, and let T be a subgraph produced by Kruskal’s algorithm. Certainly, T is a spanning tree of G (and so has size p – 1), say E(T) = { e1, e2, …, ep-1 } where w(e1) w(e2)  …  w(ey-1). Therefore, the weight of T is

  50. Proof of Thm 3.12 (continued): Suppose, to the contrary, that T is not a minimum spanning tree. Then, among the minimum spanning trees of G, choose one, call it H, which has a maximum number of edges in common with T. Now HT; so T has at least one edge that does not belong to H. Let ei(1 ip-1) be the first edge with ei E(T),ei E(H). {e1, e2, …, ei-1}E(H). Define G0 = H+ei. Then G0has exactly one cycle C. Since T has no cycles, there is an edge e0 of C that is not in T.  e0 E(T),e0 E(H). The graph T0 = G0-e0 is also a spanning tree of G and w(T0) = w(H) + w(ei) -w(e0).

More Related