GRAPH THEORY and APPLICATIONS. Trees

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1 GRAPH THEORY and APPLICATIONS Trees

2 Properties Tree: a connected graph with no cycle (acyclic) Forest: a graph with no cycle Paths are trees. Star: A tree consisting of one vertex adjacent to all the others. Graph Theory and Applications 2007 A. Yayimli 2

3 Trees as Models Trees are used in many applications: analysis of algorithms, compilation of algebraic expressions, theoretical models of computation, etc. Search trees Abstract models: sort techniques like heapsort. Graph Theory and Applications 2007 A. Yayimli 3

4 Number of Edges Theorem: In every tree T = (V, E), V = E + 1 Proof: by induction on number of edges. If E = 0 then the tree consists of a single isolated vertex. V = 1 = E + 1 Assume that the theorem is true for trees of at most k edges. Graph Theory and Applications 2007 A. Yayimli 4

5 Number of Edges Consider a tree T, where E = k + 1 y z The edge (y,z) is removed: Two subtrees T 1 and T 2. V = V 1 + V 2 and E = E 1 + E Since 0 E 1 k and 0 E 2 k, V 1 = E and V 2 = E Consequently, V = V 1 + V 2 = E E = E 1 + E = E + 1 Graph Theory and Applications 2007 A. Yayimli 5

6 Definition of Tree Theorem: The following statements are equivalent for a loop-free undirected graph G = (V, E): A. G is a tree. B. G is connected, but the removal of any edge from G disconnects G into two subgraphs that are trees. C. G contains no cycle, and V = E + 1. D. G is connected, and V = E + 1. E. G contains no cycle, and if a, b V with (a, b) E, then the graph obtained by adding (a, b) to G has precisely one cycle. Graph Theory and Applications 2007 A. Yayimli 6

7 Proof A B If G is a tree, then G is connected. Let e = (a,b) be any edge of G. Then, if G-e is connected, there are at least two paths in G from a to b. From two such paths we can form a cycle. But G has no cycle. Hence, G-e is disconnected and the vertices in G-e can be partitioned into two subsets: Vertex a and those vertices that can be reached from a. Vertex b and those vertices that can be reached from b. These two connected components are trees, because a loop or cycle in either component would also be in G. Graph Theory and Applications 2007 A. Yayimli 7

8 Proof (cont.) B C If G contains a cycle then let e = (a,b) be an edge of the cycle. But then, G-e is connected, contradicting the hypothesis in part B. So, G contains no cycle. Since G is a loop-free connected graph, we know that G is a tree. Then, V = E + 1. Graph Theory and Applications 2007 A. Yayimli 8

9 Proof (cont.) C D Assume G is not connected. Let G 1, G 2,, G r be components of G. For 1 i r, select a vertex v i G i and add the r-1 edges (v 1,v 2 ), (v 2,v 3 ),, (v r-1,v r ) to G to form G. G is a tree. V = E + 1 From C, V = E + 1, so E = E and r-1 = 0 With r = 1, it follows that G is connected. To complete the proof: D E E A Graph Theory and Applications 2007 A. Yayimli 9

10 More Definitions on Graphs Distance: If G has a (u,v) path, then the distance from u to v, d(u,v) is the least length of a (u,v) path. Eccentricity ɛ(u) of a vertex u is max vεv d(u,v). The radius of a graph G is min uεv ɛ(u) Each vertex is labeled with its eccentricity. Radius: 2 Diameter: 4 (maximum eccentricity) Graph Theory and Applications 2007 A. Yayimli 10

11 Center of a Tree Center: The subgraph induced by the vertices of minimum eccentricity. Theorem: The center of a tree is a vertex or an edge (two adjacent vertices) Graph Theory and Applications 2007 A. Yayimli 11

12 Branch A branch at a node u of a tree is a maximal subtree containing u as an endnode. The number of branches at u is the degree of u. The weight at a node u is the maximum number of edges in any branch at u. u Graph Theory and Applications 2007 A. Yayimli 12

13 Centroid of a Tree A node v is a centroid, if v has minimum weight. The centroid of a tree consist of all centroid nodes. Theorem: The centroid of a tree is a vertex or an edge (two adjacent vertices) Graph Theory and Applications 2007 A. Yayimli 13

14 Center Centroid Smallest trees with one and two central and centroid nodes: 1 CENTER 2 1 CENT- ROID 2 Graph Theory and Applications 2007 A. Yayimli 14

15 Wiener Index In a communication network, large diameter is acceptable if most pairs can communicate via shortest paths. We study the average distance. Average: Sum divided by n(n-1)/2. (all pairs) It is equivalent to study: DG ( ) = dg ( uv, ) uv, V This sum is called the Wiener Index of G. Graph Theory and Applications 2007 A. Yayimli 15

16 Directed Tree Edges of a tree may be directed. If <u,v> is a directed edge, then: u is the parent of v, v is the child of u. A vertex v is the root of a directed tree, if there are paths from v to every other vertex in the tree. Graph Theory and Applications 2007 A. Yayimli 16

17 Rooted Tree Definition: A rooted tree is a tree in which we identify a vertex v as root (indegree: 0). Level of a vertex: Vertices at distance i from the root lie at level i+1. The height of the rooted tree is its maximum level. root Graph Theory and Applications 2007 A. Yayimli 17

18 Ordered Trees Ordered tree: A directed tree in which the set of children of each vertex is ordered. Binary Tree: An ordered tree in which no vertex has more than two children: Left child and right child. Complete binary tree: Every vertex has either two children or none. Balanced complete binary tree: Every endpoint (leaf) has the same level. Graph Theory and Applications 2007 A. Yayimli 18

19 Complete Trees Theorem: A complete balanced binary tree of height h has 2 h 1 vertices. A complete balanced N-ary tree of height h has N N h 1 1 vertices. Graph Theory and Applications 2007 A. Yayimli 19

20 Cut Edge A cut edge of G is an edge e such that G e is disconnected. This graph has 3 cut edges. Theorem: A connected graph is a tree if and only if every edge is a cut edge. Graph Theory and Applications 2007 A. Yayimli 20

21 Edge Cut For subsets S and S of V, [S,S ] is the set of edges with one end in S, the other in S. Edge cut: A subset of E of the form [S,S ], where S is a nonempty proper subset of V, S = V S. Bond: A minimal nonempty edge cut of G. If G is connected, then a bond B is a minimal subset of E such that G B is disconnected. an edge cut a bond Graph Theory and Applications 2007 A. Yayimli 21

22 Cut Vertex A vertex v is a cut vertex if: E can be partitioned into two nonempty subsets E 1 and E 2, G[E 1 ] and G[E 2 ] have just the vertex v in common. cut vertices Graph Theory and Applications 2007 A. Yayimli 22

23 Spanning Tree A spanning tree of a connected undirected graph G is a subgraph which is a tree and which contains all the vertices of G. The construction of a communication network A road map or railway system Graph Theory and Applications 2007 A. Yayimli 23

24 Minimum-weighted Spanning Tree Problem: Given the cost of directly connecting any two nodes, problem is to find a network: at minimum cost and providing route between every two nodes Solution: The solution is the minimum-weighted spanning tree of the associated weighted graph. Minimum-weighted spanning tree can be found by an efficient algorithm. Graph Theory and Applications 2007 A. Yayimli 24

25 Steiner Tree A generalization of the minimum-weighted spanning tree problem: Given a proper subset V of the vertices of a graph find a minimum-weighted tree which spans the vertices of V. Such a tree is called Steiner Tree. No efficient algorithm is known for Steiner tree problem. Graph Theory and Applications 2007 A. Yayimli 25

26 Enumeration of Trees With one or two vertices, only one tree can be formed. With three vertices there is one isomorphism class. The adjacency matrix is determined by which vertex is the center. So, there are 3 trees with 3 vertices. Graph Theory and Applications 2007 A. Yayimli 26

27 Enumeration of Trees With 4 vertices: There are 4 stars and 12 paths This yields to 16 trees. a b d b c d a c With 5 vertices, a careful study yields 125 trees. With n vertices, there are n n-2 trees: this is Cayley s Formula. Graph Theory and Applications 2007 A. Yayimli 27

28 Spanning Trees in a Graph The complete graph with n vertices has all the edges that can be used in forming trees with n vertices. The number of spanning trees in a complete graph with n vertices is n n-2. Can we find a method to compute the number of spanning trees in any graph? Not containing the diagonal Containing the diagonal Graph Theory and Applications 2007 A. Yayimli 28

29 Contraction Definition: In a graph G, contraction of edge e with end points u and v is the replacement of u and v with a single vertex the incident edges of this vertex are the edges other than e that were incident to u and v. The resulting graph G e has one less edge than G. u e G v G e Graph Theory and Applications 2007 A. Yayimli 29

30 Recursive Solution Proposition: Let τ(g) denote the number of spanning trees of a graph G. If e E is not a loop, then: τ(g) = τ(g-e) + τ(g e) Example: a b e + c d G G e 4 spanning trees G e 4 spanning trees Graph Theory and Applications 2007 A. Yayimli 30

31 Recursive Solution This may lead to a recursive algorithm. We cannot apply the recurrence when e is a loop. The loops do not affect the number of spanning trees. Hence, we can delete loops as they arise. If we try to compute by deleting and contracting every edge, the amount of computation grows exponentially with the size of the graph. Graph Theory and Applications 2007 A. Yayimli 31

32 Matrix Tree Computation Form a matrix: Put the vertex degrees on the diagonal The remaining elements are 0. Substract the adjacency matrix from it. Example: a b c d a b c d a b c d a b = c d a b c d a b c d Matrix for the Kite Graph Theory and Applications 2007 A. Yayimli 32

33 Matrix Tree Computation Delete a row and a column of the resulting matrix. Take the determinant. a b c d a b c d a b d a c d det: = -8 Graph Theory and Applications 2007 A. Yayimli 33

34 Matrix Tree Theorem Given a loopless graph G: Vertex set: v 1, v 2,, v n Let a ij be the number of edges with endpoints v i and v j. Let Q be the matrix in which entry (i,j) is: -a ij when i j d(v i ) when i=j. If Q* is obtained by deleting rows s and column t of Q, then: τ(g) = (-1) s+t det(q*) Graph Theory and Applications 2007 A. Yayimli 34

35 The Connector Problem A railway network connecting a number of towns is to be set up. Given: the cost c ij of constructing a direct line between towns i and j Design: a network minimizing the total cost of construction. Graph Theory and Applications 2007 A. Yayimli 35

36 Representing the connector problem Town = vertex direct line = edge Represent the map of possible lines as a graph. The problem becomes: In a weighted (c ij ) graph, find a spanning subgraph of minimum weight. As costs are positive numbers, this is equivalent to finding a minimum spanning tree. Graph Theory and Applications 2007 A. Yayimli 36

37 Minimum Spanning Tree: Kruskal s Algorithm Edges: e 1, e 2,, e n Weights: w(e i ) Choose a link e 1 such that w(e 1 ) is minimum. count = 0 repeat if edges e 1, e 2,, e i have been chosen then choose an edge e i+1 from E {e 1,e 2,,e i } so that: G[{e 1, e 2,, e i+1 }] contains no cycle w(e i+1 ) is minimum. endif count = count + 1 until the tree has n-1 edges (count == n-1). Graph Theory and Applications 2007 A. Yayimli 37

38 Example Lon Tok 70 Mex Bei 68 NY Par Graph Theory and Applications 2007 A. Yayimli 38

39 Kruskal s Algorithm Theorem: Any spanning tree constructed by Kruskal s algorithm is an optimal (minimum) tree. What about the complexity? The edges can be sorted in increasing order of weights. This takes O(e loge) time. At each step one edge is added to the tree, the algorithm ends when no more edges can be added. Although the tree will contain n 1 edges for an n node graph, we may need to examine e edges. Hence, the number of steps necessary to construct the tree is e. Graph Theory and Applications 2007 A. Yayimli 39

40 Complexity of Kruskal s Algorithm At each step we check that the new edge doesn t create a cycle. The vertices are labeled so that at any stage, two vertices belong to the same component if they have the same label. Initially, v 1 belongs to component 1, and so on. Once e i is added to the tree, the vertices at the ends are relabeled with the smaller of their two labels. So, we can check whether a new edge creates a cycle, by checking the labels of its endpoints. Relabeling may take O(n) comparisons. Therefore, the algorithm is O(e.n + eloge) = O(e.n) Graph Theory and Applications 2007 A. Yayimli 40

41 The Directed Minimum Spanning Tree Problem Problem Statement Consider a directed graph, G(V,A). Associated with each arc (i,j) is a cost c(i,j). Let V =n and A =m. The problem is to find: A rooted directed spanning tree, G(V,S) where: S is a subset of A such that the sum of c(i,j) for all (i,j) in S is minimized. The rooted directed spanning tree: A graph which connects, without any cycle, all nodes with n-1 arcs. Each node, except the root, has one and only one incoming arc. Graph Theory and Applications 2007 A. Yayimli 41

42 Chu-Liu/Edmonds Algorithm Discard the arcs entering the root if any. For each node other than the root select the entering arc with the smallest cost If no cycle formed, G(V,S) is a MST. Otherwise, continue. For each cycle formed: contract the nodes in the cycle into a pseudo-node k modify the cost of each arc which enters a node j in the cycle from some node i outside the cycle according to the following equation: c(i,k) = c(i,j)-( c(x(j),j) - min(c(in-cycle edges) ) where c(x(j),j) is the cost of the arc in the cycle which enters j. For each pseudo-node: select the entering arc with the smallest modified cost Replace the arc in S (to same real node) by the new selected arc. Go to step 3. Graph Theory and Applications 2007 A. Yayimli 42

43 Directed MST Example Root: Graph Theory and Applications 2007 A. Yayimli 43 6

44 Tree Application: Branch-and-Bound Method Knapsack problem: A container Several items, each associated with: a size, and a value. Which items should we choose to pack in the container, so that: The total value is maximized The total size do not exceed container s size. Graph Theory and Applications 2007 A. Yayimli 44

45 Tree Application: Branch-and-Bound Method To find the optimal solution, we need to examine all possible combinations. Difficult problem Suppose we have 5 items: Item A B C D E Weight Value Container size = 9 Graph Theory and Applications 2007 A. Yayimli 45

46 Tree Application Find a packing of: Largest possible total value. Total weight should not exceed 9. List all possible packings: 32 possibility Choose the one with maximum value. Not practical for large problem size. A more efficient procedure: Branch-and-bound method: Search through a tree of possible solutions. Graph Theory and Applications 2007 A. Yayimli 46

47 Solution: First Step List the items in decreasing order of value per unit weight: Order Item E B C D A Weight Value Value per unit Graph Theory and Applications 2007 A. Yayimli 47

48 Solution Denote each possible packing by a solution vector (x 1, x 2, x 3, x 4, x 5 ) x i = 1, if item i is packed x i = 0, otherwise (0,0,1,1,0) includes items 3 and 4 (C and D). Feasible solution: A solution which satisfies the weight constraint. (0,0,1,1,0) is infeasible. Weight = 10 Graph Theory and Applications 2007 A. Yayimli 48

49 Branching out From vector (0,1,0,0,0) we can branch out: (0,1,1,0,0) (0,1,0,0,0) (0,1,0,1,0) (0,1,0,0,1) New solutions have 1 more item. We may change only the positions to the right of the last 1. The branch-and-bound starts with a null solution Graph Theory and Applications 2007 A. Yayimli 49

50 Example (1,0,0,0,0) (0,1,0,0,0) (0,0,0,0,0) (0,0,1,0,0) (0,0,0,1,0) (0,0,0,0,1) w = 2, v = 5 w = 8, v = 12 w = 6, v = 9 w = 4, v = 3 w = 3, v = 2 Store: v = 12, solution = (0,1,0,0,0) (0,0,0,0,1) is marked with a square: We cannot continue the branching from this vertex. Graph Theory and Applications 2007 A. Yayimli 50

51 Example Delete the marked vertex. Continue the branching from the solution with the highest value. (1,0,0,0,0) v = 5 (0,1,1,0,0) w = 14 (0,1,0,0,0) (0,1,0,1,0) w = 12 (0,0,0,0,0) (0,0,1,0,0) v = 9 (0,1,0,0,1) w = 11 (0,0,0,1,0) v = 3 All three new solutions are infeasible. Continue from (0,0,1,0,0) Graph Theory and Applications 2007 A. Yayimli 51

52 Example (1,0,0,0,0) v = 5 (0,0,0,0,0) (0,0,1,0,0) not feasible (0,0,1,1,0) w = 10 (0,0,0,1,0) v = 3 (0,0,1,0,1) w = 9, v = 11 we cannot branch further Cut the marked branches. Continue from vertex: (1,0,0,0,0) Graph Theory and Applications 2007 A. Yayimli 52

53 Home study: Finish the branch-and-bound example. Research Prim s Algorithm Graph Theory and Applications 2007 A. Yayimli 53

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