5.4 Shortest-path problem

1. Theorem 5.9: Let G be a simple graph with n vertices, where n>2. G has a Hamilton circuit if for any two vertices u and v of G that are not adjacent, d(u)+d(v)≥n. • Corollary 1: Let G be a simple graph with n vertices, n>2. G has a Hamilton circuit if each vertex has degree greater than or equal to n/2. • Proof: If any two vertices of G are adjacent ,then G has a Hamilton circuit v1,v2,v3,…vn,v1。 • If G has two vertices u and v that are not adjacent, then d(u)+d(v)≥n. • By the theorem 5.9, G has a Hamilton circuit. • Kn has a Hamilton circuit where n≥3

2. Theorem 5.10: Let the number of edges of G be m. Then G has a Hamilton circuit if m≥(n2-3n+6)/2,where n is the number of vertices of G. • Proof: If any two vertices of G are adjacent ,then G has a Hamilton circuit v1,v2,v3,…vn,v1. • Suppose that u and v are any two vertices of G that are not adjacent. • Let H be the graph produced by eliminating u and v from G. • Thus H has n-2 vertices and m-d(u)-d(v) edges.

3. Theorem 5. 11：Let G be a simple graph with n vertices, n>2. G has a Hamilton path if for any two vertices u and v of G that are not adjacent, d(u)+d(v)n-1. • G has a Hamilton path If Ghas a Hamiltoncircuit • If G has a Hamilton path,then Ghas a Hamiltoncircuit or has not a Hamiltoncircuit

4. 5.4 Shortest-path problem • Let G=(V,E,w) be a weighted connected simple graph, w is a function from edges set E to position real numbers set. We denoted the weighted of edge {i,j} by w(i,j), and w(i,j)=+ when {i,j}E • Definition 21: Let the length of a path p in a weighted graph G =(V,E,w) be the sum of the weights of the edges of this path. We denoted by w(p). The distance between two vertices u and v is the length of a shortest path between u and v, we denoted by d(u,v).

5. Dijkstra’s algorithm(E.W.Dijkstra) • In 1959

6. Let G=(V,E,w) and |V|=n where w>0. Suppose that S is a nonempty subset of V and v1S. Let T=V-S. Example: Suppose that (u,v',v'',v''',v) is a shortest path between u and v. Then (u,v',v'',v''') is a shortest path between u and v'''.

7. Let vT. l(v) be the length of a shortest path from v1 to v containing only vertices in S.

8. Example：S={a,b}, T={c,d,e,z} l(c)： ac：a,c4 a,b,c3, l(c)=3。 • l(e)=? • a,b,e6; • a,b,c,e4, l(e)=4?。 • Note:l(v) is the length of a shortest path from v1 to v containing only vertices in S. • l(e)=6. • l(z)=。

9. l(c)=3，l(e)=6, l(d)=8, l(z)=。 • minvT{l(v)}=l(c)=3 is the length of a shortest path from a to c. • Theorem 5.12：Suppose that minvT {l(v)}= l(v'). Then the length of the shortest path from v1 to v' is l (v').

10. Proof: Suppose that there is a path p from v1 to v', and the length of this path less than l(v'). • Then the path p must conclude some vertices of T-{v'}.

11. l(v)? • (1)S={v1},T=V-{v1},for vT (2)For SV, T=V-S, suppose that these vertices of the shortest paths from v1 to any vertices of S are in S. By Theorem 5.12(minvT{l(v)}=l(vk), vkT), we gained the result which l(vk) is the length of the shortest path from v1 to vk(distance)。The vertex vk is added to S. (3)Let S'=S∪{vk}, T'=T-{vk}, vT'. Suppose that l'(v) is the length of a shortest path from v1 to v containing only vertices in S'. Then l'(v)=min{l(v),l(vk)+w(vk,v))} (4)Let S=S',T=T', l(v)=l'(v), goto (2)

12. S={a,b},T={c,d,e,z} • l(c)=3，l(e)=6, l(d)=8, l(z)=+。 • minvT{l(v)}=l(c)=3 • S’={a,b,c}, T’={d,e,z} • l'(e)=min{ l(e), l(c)+w(c,e)}=4, • l'(d)=min{l(d), l(c)+w(c,d)}=8, • w(c,d)= + • l'(z)=min{l(z), l(c)+w(c,z)}= +

13. Theorem 5.13: For vT‘, l’(v)= min{l(v), l(vk)+w(vk, v)} • Proof: Let S'=S∪{vk}. • Suppose that l‘(v) is the length of a shortest path from v1 to v containing only vertices in S’. • (1)There are some paths from v1 to v, but these paths don’t contain the vertex vk and other vertices of T'. Then l(v) is the length of the shortest path of these paths, i.e. l'(v)=l(v)。 • (2)There are some paths from v1 to v, these paths from v1 to vk don’t contain other vertices of T', and the vertex vk adjacent edge {vk,v}. Then l(vk)+w(vk,v) is the length of the shortest path of these paths, viz l'(v)= l(vk)+w(vk,v).

14. (1)Let S=,T=V,l(v1)=0,for vv1 and l(v)=+ • Let k=1 • (2)S'=S∪{vk},T'=T-{vk}， • For each vertex v of T', • l'(v)=min{l(v),l(vk)+w(vk,v)}; • l'(v)l(v),S'S,T'T。 • (3)minvT{l(v)}=l(vk+1)。 • (4)if k=n-1,then stop • if k<n-1,then k+1k goto (2)。

15. Example:Find the length of a shortest path between a and z in the given weighted graph

17. The length of a shortest path between a and z is 13 and the path is (a,c,g,e,f,z)

18. 5.5 Trees • 5.5.1 Trees and their properties • Definition 22: A tree is a connected undirected graph with no simple circuit, and is denoted by T. A vertex of T is a leaf if only if it has degree one. Vertices are called internal vertices if the degrees of the vertices are more than 1. A graph is called a forest if the graph is not connected and each of the graph’s connected components is a tree.

20. Theorem 5.14: Let T be a graph with n vertices. The following assertions are equivalent. • (1)T is a connected graph with no simple circuit. • (2)T is a graph with no simple circuit and e=n-1 where e is number of edges of T. • (3)T is a connected graph with e=n-1 where e is number of edges of T. • (4)T is a graph with no simple circuit, and if x and y are nonadjacent vertices of T then T+{x,y} contains exactly a simple circuit. T+{x,y} is a new graph which is obtained from T by joining x to y. • (5)T is connected and if {x,y}E(T) then T-{x,y} is disconnected. Where T-{x,y} is a new graph which is obtained from T by removing edge {x,y}. • (6)There is a unique simple path between any two of vertices of T.

21. Proof：(1)→(2): If T is a connected graph with no simple circuit, then T is a graph with no simple circuit and e=n-1. i.e prove e=n-1 • Let us apply induction on the number of vertices of T. • When n=2, T is a connected graph with no simple circuit, the result holds

22. Suppose that result holds for n=k • n=k+1? • (Theorem 5.4:Let  (G)≥2, then there is a simple circuit in the graph G.) • By the theorem 5.4, and T is connected and no simple circuit. • There is a vertex that has degree one. Let the vertex be u, and suppose that u is incident with edge {u,v}. • We remove the vertex u and edge {u,v} from T, and get a connected graph T’ with no simple circuit, and T’ has k vertices. • By the inductive hypothesis, T’ has k-1 edges. • e(T)=e(T’)+1=k

23. (2)→(3): T is an acyclic graph with e=n-1.Now we prove T is connected and e=n-1. i.e. prove T is connected • Suppose T is disconnected. Then T have (>1) connected components T1,T2,…,T. The number of vertices of Ti is ni for i=1,2,…, and n1+n2+…+n=n.

24. (3)→(4): T is a connected graph with e=n-1, we prove “T is a graph with no simple circuit, and if x and y are nonadjacent vertices of T then T+{x,y} contains exactly a simple circuit”. • 1)We first prove that T doesn't contain simple circuit. • Let us apply induction on the number of vertices of T. • T is connected with n=2 and e=1, Thus T doesn't contain any simple circuit. The result holds when n=2 and e=1

25. Suppose that result holds for n=k-1 • For n=k, (T)1 • There is a vertex that has degree one. The vertex is denoted by u. i.e d(u)=1. • Why? • 2e2k,i.e. ek( e=n-1=k-1), contradiction • We has a new graph T’ which is obtained from T by removing the vertex u and incident with edge {u,v} • By the inductive hypothesis, T' doesn't contain any simple circuit. Thus T doesn't contain any simple circuit

26. 2) If x and y are nonadjacent vertices of T, then T+{x,y} contains a simple circuit • There is a simple path from x to y in T. • (x=vi,vi1,…, vis,vj=y)。 • (x=vi,vi1,…, vis,vj=y,vi=x)。 • 3)Next, we prove T+{x,y} contains exactly a simple circuit. • Suppose that there are two (or more than) simple circuit in T+{x,y}.

27. Exercise • 1.Find the length of a shortest path between a and z in the given weighted graph.