5.4 shortest-path problem

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

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.

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 G has a Hamilton circuit

If G has a Hamilton path, then G has a Hamilton circuit or has not a Hamilton circuit

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).

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

Let G=(V,E,w) and |V|=n where w>0. Suppose that S is a nonempty subset of V and v1S. 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'''.

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

l(e)=? a,b,e 6; a,b,c,e 4, 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)=。

Example ：S={a,b}, T={c,d,e,z}l(c)：ac ： a,c 4a,b,c 3,l(c)=3。

l(c)=3 ， l(e)=6, l(d)=8, l(z)=。 minvT{l(v)}=l(c)=3 is the length of a

shortest path from a to c. Theorem 5.12 ： Suppose that minvT

{l(v)}= l(v'). Then the length of the shortest path from v1 to v' is l (v').

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'}.

l(v)? (1) S={v1},T=V-{v1},for vT

(2)For SV, 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(minvT{l(v)}=l(vk), vkT), 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 {v∪ k}, T'=T-{vk}, vT'. 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)

otherwise

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S={a,b},T={c,d,e,z} l(c)=3 ， l(e)=6, l(d)=8, l(z)=+。 minvT{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)}= +

Theorem 5.13: For vT‘, 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).

(1)Let S=,T=V,l(v1)=0,for vv1 and l(v)=+

Let k=1 (2)S'=S {v∪ k},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)minvT{l(v)}=l(vk+1)。 (4)if k=n-1,then stop if k<n-1,then k+1k goto (2)。

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

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

5.5 Trees5.5.1 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.

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.

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

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

(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.

(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

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? 2e2k, i.e. ek( 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

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}.

Exercise1.Find the length of a shortest path

between a and z in the given weighted graph.