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

2. Graph Theory Ch. 2. Trees and Distance 2 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  A leaf (or pendant vertex ) is a vertex of degree 1 tree forest leaf

3. Graph Theory Ch. 2. Trees and Distance 3 Spanning Subgraph 2.1.1  A spanning subgraph of G is a subgraph with vertex set V(G)  A spanning tree is a spanning subgraph that is a tree Spanning subgraph Spanning tree

4. Graph Theory Ch. 2. Trees and Distance 4 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. Maximal path Impossible! Cycle occurs Impossible! It is Maximal.

5. Graph Theory Ch. 2. Trees and Distance 5 Proof: (2/2) –Let v be 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 in G is also in G’. –Hence G’ is connected. –Since deleting a vertex cannot create a cycle, G’ also is acyclic. –Thus G’ is a tree with n -1 vertices. Lemma. Deleting a leaf from a n-vertex tree produces a tree with n-1 vertices. 2.1.3

6. Graph Theory Ch. 2. Trees and Distance 6 For u, w  V(G ’ ), every u, w-path in G is also in G’. u w G G’G’ v

7. Graph Theory Ch. 2. Trees and Distance 7 Theorem 2.1.4. For an n-vertex graph G (with n  1 ), the following are equivalent (and characterize the trees with n vertices) 2.1.4 A) G is connected and has no cycles B) G is connected and has n-1 edges C) G has n-1 edges and no cycles D) For u, v  V(G), G has 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} to gether imply the third

8. Graph Theory Ch. 2. Trees and Distance 8 A  {B,C}. connected, acyclic  n-1 edges Theorem 2.1.4 Continue  We use induction on n.  For n =1, an acyclic 1-vertex graph has no edge  For n >1, we suppose that implication holds for graphs with fewer than n vertices –Given an acyclic connected graph G, Lemma 2.1.3 provides a leaf v and states that G’=G-v also 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

9. Graph Theory Ch. 2. Trees and Distance 9 B  {A, C} Connected and n-1 edges  acyclic Theorem 2.1.4 continue  If G is not acyclic, delete edges from cycles of G one 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  Thus G’=G, and G is acyclic

10. Graph Theory Ch. 2. Trees and Distance 10 C  {A, B} n-1 edges and no cycles  connected Theorem 2.1.4 continue  Let G 1,…,G k be the components of G  Since every vertex appears in one component,  i n(G i )=n –Since G has no cycles, each component satisfies property A. Thus e(G i ) = n(G i ) - 1 –Summing over i yields e(G)=  I [n(G i )-1]= n-k  We are given e(G)=n-1,so k =1, and G is connected

11. Graph Theory Ch. 2. Trees and Distance 11 A  D Connected and no cycles  For u, v  V(G), one and only one u, v-path exists Theorem 2.1.4  Since G is 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, Q of distinct paths with the same endpoints  By this extremal choice, no internal vertex of P or Q can belong to the other path  This implies that P  Q is a cycle, which contradicts the hypothesis A p v u q

12. D  A For u, 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 G is connected  If G has a cycle C, then G has two u,v-paths for u,v  V(G) ; which contradicts the hypothesis D  Hence G is acyclic (this also forbids loops). Graph Theory Ch. 2. Trees and Distance 12

13. Graph Theory 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.

14. Graph Theory 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

15. Graph Theory 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

16. Graph Theory Ch. 2. Trees and Distance 16 Proposition: 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. 2.1.6 T’ T T-e+e’ e e’ G

17. Graph Theory Ch. 2. Trees and Distance 17 Proposition: 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. 2.1.6 Proof:  Every edge of T is a cut-edge of T. Let U and U’ be the two components of T-e.  Since T’ is connected, T’ has an edge e’ with endpoints in U and U’.  Now T-e+e’ is connected, has n(G)-1 edges, and is a spanning tree of G. T’ T T-e+e’ e e’

18. Graph Theory Ch. 2. Trees and Distance 18 Distance in trees and Graphs  Assume G has a u, v-path  Then the distance from u to v, written d G (u,v) or d(u,v), is the least length of a u,v-path.  If G has no such path, then d(u,v)= 

19. Graph Theory Ch. 2. Trees and Distance 19 Distance in trees and Graphs  The diameter ( diam G ) is max u,v  V(G) d(u,v) –Upper bound of distance between every pair  The eccentricity of a vertex u is  (u ) = max v  V(G) d(u,v) –Upper bound of the distance from u to the others  The radius of a graph G is rad G = min u  V(G)  (u) –Lower bound of the eccentricity

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

21. Graph Theory Ch. 2. Trees and Distance 21 If G is a simple graph, then diam G  3  diam  3 2.1.11 Proof: 1/3  Since diam G > 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 diam G = 2 u v Not every pair is of this kind A pair (u, v) of this kind exists

22. Graph Theory Ch. 2. Trees and Distance 22 If G is a simple graph, then diam G  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 –Either u or v is not connected to x  Equivalently, this makes x adjacent to at least one of {u,v} in u v Everyone of these vertices has at least one of {u,v} as a nonneighbor

23. Graph Theory Ch. 2. Trees and Distance 23 If G is a simple graph, then diam G  3  diam  3 2.1.11 Proof: 3/3  Since also 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

24. Graph Theory Ch. 2. Trees and Distance 24 Center 2.1.12 Definition: The center of a graph G is 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 eccentricity of a vertex u is the upper bound of the distance from u to the others

25. Graph Theory Ch. 2. Trees and Distance 25 The center of a tree is a vertex or an edge 2.1.13 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.

26. Graph Theory Ch. 2. Trees and Distance 26 Theorem. 2.1.13 Continue  Induction step: n(T)>2. –Let T’ = 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 T from a vertex u  V(T) is a leaf (otherwise, the path reaching it from u can be extended farther). T = Green  Pink T’ = Green u The max path from u The end must be a leaf

27. Graph Theory Ch. 2. Trees and Distance 27 Theorem. 2.1.13 Continue –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 T is greater than the eccentricity of its neighbor in T. –Hence the vertices minimizing  T (u) are the same as the vertices minimizing  T’ (u). u  T ’ (u)=  T (u)-1 {v| v has min  (v) in T} = {v’| v’ has min  (v’) in T’}

28. Graph Theory Ch. 2. Trees and Distance 28 Theorem. 2.1.13 Continue  It is shown T and T’ have the same center.  By the induction hypothesis, the center of T’ is a vertex or an edge. TT’ T”

29. Graph Theory Ch. 2. Trees and Distance 29 Spanning Trees and Enumeration 2.2  There are 2 c (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? Max number of edges = c(n,2) Can either appear or disappear

30. Graph Theory 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.

31. Graph Theory Ch. 2. Trees and Distance 31 Algorithm for tree generation by using Prűfer code 2.2.1 Algorithm. (Prűfer code) Production of f(T)=(a 1,…,a n-2 ) Input: A tree T with vertex set S  N Iteration: At the i th step, delete the least remaining leaf, and say that a i is the neighbor of the deleted leaf 2 7 1 4 3 68 5 1. Delete 2 a 1 = 7 2. Delete 3 a 2 = 4 3. Delete 5 a 3 = 4 4. Delete 4 a 4 = 1 5. Delete 6 a 5 = 7 6. Delete 7 a 6 = 1

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

33. Graph Theory 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.

34. Graph Theory Ch. 2. Trees and Distance 34 Contraction 2.2.7  In a graph G, contraction of edge e with endpoints u, v is the replacement of u and v with a single vertex whose incident edges are the edges other than e that were incident to u or v  The resulting graph G·e has one less edge than G u e v G G·e

35. Graph Theory 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 e in G e G G·e G-e =

36. Graph Theory 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 G that omit e are precisely the spanning trees of G-e  Need to show that G has  (G·e) spanning trees containing e –It must be shown that contraction of e defines a bijection from the set of spanning trees of G containing e to the set of spanning trees of G·e

37. Graph Theory 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.

38. Graph Theory Ch. 2. Trees and Distance 38 A Matrix Tree computation. 2.2.12  Given a loopless graph G with vertex set v 1, …., 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 i,j when i  j and is d(v i ) when i=j  Let Q* is a matrix obtained by deleting row s and column t of Q, then  (G) = (-1) s+t detQ*

39. Graph Theory 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

40. Graph Theory 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

41. Graph Theory Ch. 2. Trees and Distance 41 Example 2.2.11 8 v1v1 v2v2 v3v3 v4v4 v 1 v 2 v 3 v 4 v1v2v3v4v1v2v3v4

42. Graph Theory 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 - 1 edges; we seek one that minimizes or maximizes the sum of the edge weights. –Kruskal’s Algorithm –Prim’s Algorithm

43. Graph Theory Ch. 2. Trees and Distance 43 Kruskal’s Algorithm for 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.

44. Graph Theory Ch. 2. Trees and Distance 44 Kruskal’s Algorithm for Minimum Spanning Tree 2.3.1  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. H Join Two vertices in one component. Cycle occurs. Not Allowed! Join two components. It works

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

46. Graph Theory 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.

47. Graph Theory Ch. 2. Trees and Distance 47 Theorem 2.3.3 Continue Proof: continue  Let T be the resulting tree, and let T * be a spannig tree of minimum weight.  If T=T *, we are done.  If T  T *, let e be the first edge chosen for T that is not in T *. Adding e to T * creates one cycle C. Since T has no cycle, C has an edge e’  E(T). Consider the spanning tree T * +e-e’ T: …………. e, ………… T*: …………. e*,………… Identical The first edge chosen for T that is not in T*

48. Graph Theory Ch. 2. Trees and Distance 48 Theorem 2.3.3 Continue Proof: continue  Since T * contains e’ and all the edges of T chosen before e, both e’ and e are available when the algorithm chooses e, and hence w(e)  w(e’)  Thus T * +e-e’ is a spanning tree with weight at most T * that agrees with T for a longer initial list of edges than T * does. T: …………. e, ………… T*: …………. e*,………… Identical T: …………… e, ………… T*+e-e’: …………… e, ………… Identical

49. Graph Theory Ch. 2. Trees and Distance 49 Theorem 2.3.3 Continue Proof: continue  Repeating this argument eventually yields a minimum- weight spanning tree that agrees completely with T.  Phrased extremely, we have proved that the minimum spanning tree agreeing with T the longest is T itself.

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

51. Graph Theory Ch. 2. Trees and Distance 51 Dijkstra’s Algorithm 2.3.5 Input: A graph (or digraph) with nonnegative edge weights and a starting vertex u. –The weight of edge xy is w(xy) –Let w(xy)=  if xy is not an edge u a d e c b 1 5 2 6 4 4 5 3

52. Graph Theory Ch. 2. Trees and Distance 52 Dijkstra’s Algorithm 2.3.5 Idea: –Maintain the set S of vertices to which a shortest path from u is known –Enlarge S to include all vertices. To do this, maintain a tentative distance t(z) from u to each z  S, being the length of the shortest u, v-path yet found. u S u S The one nearest to u v

53. Graph Theory Ch. 2. Trees and Distance 53 Dijkstra’s Algorithm 2.3.5 Initialization: Set S={u}; t(u)=0; t(z)=w(uz) for z  u Iteration: 1.Select a vertex v outside S such that t(v)=min z  S t(z). 2.Add v to S. 3.Explore edges from v to update tentative distance : for each edge vz with z  S, update t(z) to min{ t(z), t(v)+w(vz) } The iteration continues until S=V(G) or until t(z)=  for every z  S. At the end, set d(u, v)=t(v) for all v.

54. Graph Theory Ch. 2. Trees and Distance 54 Tentative distance v.s. Shortest distance 2.3.5  The tentative distance of a vertex may not be the shortest distance, for example: –t (d ) = 6 and not the minimum –Actually d (d ) = 5 u a d e b 1 5 2 6 2 3 d=0 d=1 t=3 t=6 u a d e b 1 5 2 6 2 3 d=0 d=1 t=5 d=3 t=9

55. Graph Theory Ch. 2. Trees and Distance 55 Example: Find Shortest paths by using Dijkstra’s Algorithm 2.3.6 u a d e c b 1 5 2 6 4 4 5 3 u a d e c b 1 5 2 6 4 4 5 3 d=0 t=1 t=3

56. Graph Theory Ch. 2. Trees and Distance 56 Example: Find Shortest paths by using Dijkstra’s Algorithm continue 2.3.6 u a d e c b 1 5 2 6 4 4 5 3 d=0 d=1 t=3 t=6 t=5 u a d e c b 1 5 2 6 4 4 5 3 d=0 d=1 d=3 t=6 t=5

57. Graph Theory Ch. 2. Trees and Distance 57 Example: Find Shortest paths by using Dijkstra’s Algorithm continue 2.3.6 u a d e c b 1 5 2 6 4 4 5 3 d=0 d=1 d=3 t=11 d=5 t=6 u a d e c b 1 5 2 6 4 4 5 3 d=0 d=1 d=3 t=8 d=5 d=6

58. Graph Theory Ch. 2. Trees and Distance 58 Theorem: Given a (di)graph G and a vertex u  V(G), Dijkstra’s algorithm compute d(u, z) for every z  V(G) 2.3.7 Proof:  We prove the stronger statement that at each iteration, 1)For z  S, t(z)=d(u, z), and 2)For z  S, t(z) is the least length of a u, z-path reaching z directly from S.

59. Graph Theory Ch. 2. Trees and Distance 59 Theorem 2.3.7 Continue We use induction on k=|S|.  Basis step : k=1  From the initialization, – S={u}, d(u, u)=t(u)=0, –the least length of a u, z-path reaching z directly from S is t(z)=w(u,z), which is infinite when uz is not an edge

60. Graph Theory Ch. 2. Trees and Distance 60 Theorem 2.3.7 Continue  Induction step : Suppose that when |S|=k, (1) and (2) are true. –Let v be a vertex among z  S such that t(z) is smallest. –The algorithm now choose v ; let S’=S  {v}. We first argue that d(u, v)=t(v). A shortest u, v-path must exit S before reaching v. The induction hypothesis states that the length of the shortest path going directly to v from S is t(v). The induction hypothesis and choice of v also guarantee that a path visiting any vertex outside S and later reaching v has length at least t(v). –Hence d(u, v)=t(v), and (1) holds for S’.

61. Graph Theory Ch. 2. Trees and Distance 61 Theorem 2.3.7 Continue  To prove (2) for S’, let z be a vertex outside S other than v. –By the hypothesis, the shortest u, v-path reaching z directly from S has length t(z) (  if there is no such path). –When we add v to S, we must also consider paths reaching z from v. Since we have now computed d(u, v)=t(v), the shortest such path has length t(v)+w(vz), and we compare this with the previous value of t(z) to find the shortest path reaching z directly from S’.  We have verified that (1) and (2) hold for the new set S’ of size k+1 ; this completes the induction step.

62. Graph Theory Ch. 2. Trees and Distance 62 Prim’s Algorithm for Minimum Spanning Tree Input: A graph (or digraph) with nonnegtive edge weights and a starting vertex u. The weight of edge xy is w(xy) ; let w(xy)=  if xy is not an edge. Idea : Maintain a tree T including u at the beginning, enlarging T to include all vertices. To do this, select the cheapest edge from the edges E={(x,v) |x  T, v  T }.

63. Graph Theory Ch. 2. Trees and Distance 63 Prim’s Algorithm Initialization: Set T={u}. Iteration: – Select a vertex v outside T such that w(x,v)=min x  T, v  T w(x,v) – Add v to T – The iteration continues until V(T )=V(G).

64. Graph Theory Ch. 2. Trees and Distance 64 Dijkstra’s Algorithm v.s. Prim’s Algorithm u a d e c b d=0 d=1 t=3 t=6 t=5 The vertex which is the nearest to u 8 The end vertex of the cheapest edge 3 6

65. Graph Theory Ch. 2. Trees and Distance 65 Algorithm - Breadth First Search 2.3.8 Input: An unweighted graph (or digraph) and a start vertex u. Idea: Maintain a set R of vertices that have been reached but not searched and a set S of vertices that have been searched The set R is maintained as a First-In First-Out list (queue), so the first vertices found are the first vertices explored

66. Graph Theory Ch. 2. Trees and Distance 66 Algorithm - Breadth First Search 2.3.8 Initialization: R={u}, S= , d(u, u)=0 Iteration: 1.As long as R , we search from the first vertex v of R 2.The neighbors of v not in S  R are added to the back of R and assigned distance d (u, v)+1 3.Then v is removed from the front of R and placed in S

67. Graph Theory Ch. 2. Trees and Distance 67 Example of BFS a b c d e f g h i j Order of BFS : a, b, c, d, e, f, g, h, i, j, … a, c, b, d, f, h, e, g, i, j, … b, a, g, f, e, c, d, … h, i, j, …

68. Graph Theory Ch. 2. Trees and Distance 68 Breadth-F-S vs. Depth-F-S a b c d e f g h i j a b c d e f g h i j a,b,c,d,e,f,g,h,i,j, … a,b,e,f, …c, h,…

69. Graph Theory Ch. 2. Trees and Distance 69 Application: Chinese Postman Problem 2.3.9  A mail carrier must traverse all edges in a road network, starting and ending at the Post Office. –The edges has nonnegative weights representing distance or time. –We seek a closed walk of minimum total length that uses all the edges –This is the Chinese Postman Problem, named in honor of the Chinese mathematician Guan Meigu [1962], who proposed it.

70. Graph Theory Ch. 2. Trees and Distance 70 Application: Chinese Postman Problem continue 2.3.9  If every vertex is even, then the graph is Eulerian and the answer is the sum of the edge weights.  Otherwise, we must repeat edges. Every traversal is an Eulerian circuit of a graph obtained by duplicating edges.  Finding the shortest traversal is equivalent to finding the minimum total weight of edges whose duplication will make all vertex degrees even.

71. Graph Theory Ch. 2. Trees and Distance 71 Application: Chinese Postman Problem continue 2.3.9  We say “duplication” because we need not use an edge more than twice  If we use an edge three or more times in making all vertices even, then deleting two of those copies will leave all vertices even  There may be many ways to choose the duplicated edges

72. Graph Theory Ch. 2. Trees and Distance 72 Application: Chinese Postman Problem continue 2.3.9  If there are only two odd vertices, then –We can use Dijkstra’s Algorithm to find the shortest path between them and solve the problem  If there are 2 k odd vertices, then –Use Dijkstra’s algorithm to find the shortest paths connecting each pair of odd vertices –Use these lengths as weights on the edges of K 2k –Find the minimum total weight of k edges that pair up these 2 k vertices. This is a maximum matching problem.

73. Graph Theory Ch. 2. Trees and Distance 73 Application: Chinese Postman Problem continue 2.3.9 A vertex: An odd vertex An edge between u and v: The shortest path between u and v in the given graph 16 20 40 15 20 10 A matching in a general graph with minimum total weight

74. Graph Theory Ch. 2. Trees and Distance 74 Rooted Tree 2.3.11  A rooted tree is a tree with one vertex r chosen as root.  For each vertex v, let P(v) be the unique v, r-path. –The parent of v is its neighbor on P(v); –Its children are its other neighbors. –Its ancestors are the vertices of P(v)-v. –Its descendants are the vertices u such that P(u) contains v. –The leaves are the vertices with no children. –A rooted plane tree or planted tree is a rooted tree with a left-to–right ordering specified for the children of each vertex.

75. Graph Theory Ch. 2. Trees and Distance 75 Rooted Tree 2.3.11 root v Parent of v, Ancester of v A leaf, A child of v

76. Binary Tree 2.3.12  A binary tree is a rooted plane tree where each vertex has at most two children, and each child of a vertex is designated as its left child or right child  The subtrees rooted at the children of the root are the left subtree and the right subtree of the tree  A k-ary tree allows each vertex up to k children Graph Theory Ch. 2. Trees and Distance 76

77. Graph Theory Ch. 2. Trees and Distance 77 Huffman’s Algorithm 2.3.13 Input: Weights (frequencies or probabilities) p 1,…,p n Output: Prefix-free code (equivalently, a binary tree) Idea: Infrequent items should have longer codes; put infrequent items deeper by combining them into parent nodes

78. Graph Theory Ch. 2. Trees and Distance 78 Huffman’s Algorithm 2.3.13 Initial case: When n=2, the optimal length is one, with 0 and 1 being the codes assigned to the two items (the tree has a root and two leaves; n=1 can also be used as the initial case)

79. Graph Theory Ch. 2. Trees and Distance 79 Huffman’s Algorithm 2.3.13 Recursion :  Replace the two least likely items p, p’ with a single item q of weight p+p’ when n>2  Treat the smaller set as a problem with n-1 items.  After solving it, give children with weights p, p’ to the resulting leaf with weight q. –Equivalently, replace the code computed for the combined item with its extensions by 1 and 0, assigned to the items that were replaced.

80. Graph Theory Ch. 2. Trees and Distance 80 Example of Huffman coding 2.3.14  Consider eight items with frequencies 5, 1, 1, 7, 8, 2, 3, 6  Algorithm 2.3.13 combines items according to the tree on the left below, working from the bottom up –First the two items of weight 1 combine to from one of weight 2 2 511 7 82 36

81. Graph Theory Ch. 2. Trees and Distance 81 Huffman coding 2.3.14 4 2 511 7 82 36 –Now this and the original item of weight 2 are the least likely and combine to form an item of weight 4.

82. Graph Theory Ch. 2. Trees and Distance 82 Huffman coding 2.3.14 2 511 7 82 36 4 2 511 7 82 36 7 4 2 511 7 82 36 7 11 4 2 511 7 82 36

83. Graph Theory Ch. 2. Trees and Distance 83 Huffman coding 2.3.14 33 19 14 7 11 4 2 511 7 82 36 6:101 8:11 5:100 7:01 3:001 2:0001 1:00001 1:00000 0 1 0 0 0 1 1 1 0 1 0 1 1 0

84. Graph Theory Ch. 2. Trees and Distance 84 Theorem 2.3.15  Given a probability distribution {p i } on n items, Huffman’s Algorithm produces the prefix-free code with minimum excepted length