Internally Disjoint Paths

• Internally Disjoint Paths : Two paths u to v are internally disjoint if they have no common internal vertex.

Theorem 4.2.2

• A graph G having at least three vertices is 2-connected if and only if for each pair u,vV(G) there exists internally disjoint u,v-paths in G.

Proof. () Consider any two vertices u,vV(G). G has internally disjoint u,v-paths. Deletion of any vertex in V(G) cannot separate u from v. G is 2-connected.

() Suppose G is 2-connected. That G has internally disjoint u,v-paths is proved by induction on d(u,v).

Theorem 4.2.2 (2/3)

Basis Step: d(u,v)=1. The graph G-uv is connected since ’(G)>=(G)>=2. A u,v-path in G-uv is internally disjoint in G from the u,v-path formed by the edge uv itself.

Induction Step: d(u,v)>1. 1. Let k=d(u,v). Let w be the vertex before v on a shorte

st u,v-path.2. d(u,w)=k-1. G has internally disjoint u,w-paths P a

nd Q.3. If vV(P)V(Q), then we find the desired paths in th

e cycle PQ.

Theorem 4.2.2 (3/3)4. Otherwise, G-w is connected and contains a u,v-path R s

ince G is 2-connected.

5. If R avoids P or Q, we are done.

6. Otherwise, let z be the last vertex of R (before v) belonging to PQ. We assume that zP by symmetry.

We combine the u,z-subpath of P with the z,v-subpath of R to obtain a u,v-path internally disjoint from Qwv.

Lemma 4.2.3 (Expansion Lemma)If G is a k-connected graph, and G’ is obtained from G by

adding a new vertex y with at least k neighbors in G, then G’ is k-connected.

Proof. 1. Let S be a separating set of G’.

2. If yS, then S-{y} separates G. |S|>=k+1.

3. If yS and N(y)S, then |S|>=k since y has at least k neighbors.

4. Otherwise, y and N(y)-S lie in a component of G’-S. |S|>=k since S must separate G.

Theorem 4.2.4

For a graph G with at least three vertices, the following conditions are equivalent (and characterize 2-connected graphs).

A) G is connected and has no cut-vertex.

B) For all x,yV(G), there are internally disjoint x,y-paths.

C) For all x,yV(G), there is a cycle through x and y.

D) (G)>=1, and every pair of edges in G lies on a common cycle.

Theorem 4.2.4 (2/3)Proof. 1. Theorem 4.2.2 proves AB.

2. For BC, the cycles containing x and y corresponds to pairs of internally disjoint x,y-paths.

3. For DC, (G)>=1 implies that vertices x and y are not isolated. Consider edges incident to x and y.

4 If there are at least two such edges e and f, then e and f lies on a common cycle. There is a cycle through x and y.

5 Otherwise, only one such edge e. Let f be an edge incident to the third vertex. e and f lies on a common cycle. There is a cycle through x and y.

Theorem 4.2.4 (3/3)6. Suppose G satisfies condition C. G satisfies condition A.

G is connected. (G)>=1.

7. Consider two edges uv and xy. Add to G the vertices w with neighborhood {u,v} and z with neighborhood {x,y} to form G’. Since G is 2-connected, Lemma 4.2.3 implies G’ is 2-connected. w and z lie on a cycle C in G’.

8. Since w,z each have degree 2, C must contain the paths u,w,v and x,z,y but not the edges uv or xy.

9. Replacing the path u,w,v and x,z,y in C with the edges uv and xy yields the desired cycle through uv and xy in G.

x,y-cut

• x,y-cut: Given x,yV(G), a set SV(G)-{x,y} is an x,y-separator or x,y-cut if G-S has no x,y-path.

(x,y): the minimum size of x,y-cut. (x,y): the maximum size of a set of pairwise internal

ly disjoint x,y-paths.

Example 4.2.16

• {b,c,z,d} is an x,y-cut of size 4. (x,y)<=4.• G has four internally disjoint x,y-paths. (x,y)>=4.• {b,c,x} is an w,z-cut of size 3. (w,z)<=3.• G has three internally disjoint w,z-paths. (w,z)>=3.

Theorem 4.2.17 (Menger Theorem)• If x,y are vertices of a graph G and xyE(G), then (x,

y)= (x,y).Proof. An x,y-cut must contain an internal vertex of ever

y internally disjoint x,y-paths, and no vertex can cut two internally disjoint x,y-paths. (x,y)>= (x,y). We prove equality by induction on n(G).

Basis Step: n(G)=2. xyE(G) yields (x,y)= (x,y)=0.Induction Step: n(G)>2.

1. Let k= G(x,y).2. N(x) and N(y) are x,y-cuts. no minimum cut proper

ly contains N(x) or N(y).

Theorem 4.2.17 (2/5)3. Case 1: G has a minimum x,y-cut S other than N(x) or

N(y). Let V1 be the set of vertices on x,S-path, and let V2 be the set of vertices on S,y-path.

4. SV1 and SV2 SV1V2.

5. If there exists v such that vV1V2 –S, then combing x,v-portion of some x,S-path and v,y-portion of some S,y-path yields an x,y-path that avoids the x,y-cut S. It contradicts that S is a minimum x,y-cut. S=V1V2.

6. V1 omits N(y)-S and V2 omits N(x)-S by the same argument in 5.

7. Form H1 by adding to G[V1] a vertex y’ with edges from S. Form H2 by adding to G[V2] a vertex x’ with edges from S.

Theorem 4.2.17 (3/5)8. Every x,y-path in G starts with an x,S-path (contained in H

1). Every x,y’cut in H1 is an x,y-cut in G. H1(x,y’)= k.

9. H2(x’,y)= k by the same argument in 8.

10. V1 omits N(y)-S and V2 omits N(x)-S. H1 and H2 are smaller than G. H1

(x,y’)=k= H2(x’,y).

11. S=V1V2. Deleting y’ from the k paths in H1 and x’ from the k paths in H2 yields the desired x,S-paths and S,y-paths in G that combine to form k internally disjoint x,y-paths in G.

Theorem 4.2.17 (4/5)

12. Case 2: N(x) is minimum x,y-cut.13. If there exists node uN(x)N(y), then S-u is x,y-cu

t in G-u. G-u(x,y)=k-1. G-u has k-1 internally disjoint x,y-paths by induction hypothesis. Combining these k-1 x,y-path and the path x,u,y yields k internally disjoint x,y-paths in G.

14. If there exists node v{x}N(x)N(y){y}, then S is minimum x,y-cut in G-v. G-v(x,y)=k. G-v has k internally disjoint x,y-paths by induction hypothesis. These are k internally disjoint x,y-paths in G.

Theorem 4.2.17 (5/5)15. In the remaining case, N(x) and N(y) partition V(G)-

{x,y}. Let G’ be the bipartite graph with bipartition N(x), N(y) and edge set [N(x),N(y)].

16. Every x,y-path in G uses some edge from N(x) to N(y). x,y-cuts in G are the vertex covers of G’. (G’)=k. G’ has a matching of size k by Theorem 3.1.16. These k edges yield k internally disjoint x,y-paths of length 3.

Line Graph (Digraph)

Line Graph (Digraph): The line graph (digraph) of a graph (digraph) G, written L(G), is the graph (digraph) whose vertices are the edges of G, with efE(L(G)) when e=uv and f=vw in G.

Theorem 4.2.19

If x and y are distinct vertices of a graph or digraph G, then the minimum size of an x,y-disconnecting set of edges equals the maximum number of pairwise edge-disjoint x,y-paths.

Proof. 1. Modify G to obtain G’ by adding two new vertices s, t and two new edges sx and yt. Cleary, ’G(x,y)= ’G’(x,y) and ’G(x,y)= ’G’(x,y).

Theorem 4.2.19 (2/2)

2. Edge-disjoint x,y-paths in G become internally disjoint sx,yt-paths in L(G’), and vice versa. ’G(x,y)= L(G’)(sx,yt).

3. A set of edges disconnects y from x if and only if the corresponding vertices of L(G’) form an sx,yt-cut. ’G(x,y)=L(G’)(sx,yt).

4. L(G’)(sx,yt)= L(G’)(sx,yt). ’G(x,y)= ’G(x,y).

Lemma 4.2.20

Deletion of an edge reduces connectivity by at most 1.

Proof. 1. Every separating set of G is a separating set of G-xy. (G-xy)<=(G).

2. If G-xy has no separating set S of size smaller than (G), then (G-xy)=(G).

3. Suppose that G-xy has a separating set S of size smaller than (G).

4. G-S is connected.

Lemma 4.2.20 (2/2)5. G-xy-S has two components G[X] and G[Y], with x

X and yY. In G-S, the only edge joining X and Y is xy.

6. If |X|>=2, then S{x} is a separating set of G, and (G)<= (G-xy)+1.

7. If |Y|>=2, then S{y} is a separating set of G, and (G)<= (G-xy)+1.

8. In the remaining case, |S|=n(G)-2. (G)>n(G)-1 since (G)>|S|. G is a complete graph. (G-xy)=n(G)-2= (G)-1.

Theorem 4.2.21The connectivity of G equals the maximum k such that

(x,y)>=k for all x,yV(G). The edge-connectivity of G equals the maximum k such that ’(x,y)>=k for all x,yV(G).

Proof. 1. ’(G)=minx,yV(G) ’(x,y). ’(G)= minx,yV(G) ’(x,y) by Theorem 4.2.19.

2. (G)=minx,yV(G), xyE(G) (x,y).

3. By Theorem 4.2.17, (x,y)= (x,y) for xyE(G). (G)=minx,yV(G), xyE(G) (x,y).

4. For xyE(G), G(x,y)=1+G-xy(x,y)= 1+ G-xy(x,y)>=1+(G-xy)>=(G).