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

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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,v  V(G) there exists internally disjoint u,v-paths in G. Proof. (  ) Consider any two vertices u,v  V(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 shortest u,v-path. 2. d(u,w)=k-1.  G has internally disjoint u,w-paths P and Q. 3. If v  V(P)  V(Q), then we find the desired paths in the cycle P  Q.

Theorem 4.2.2 (3/3) 4. Otherwise, G-w is connected and contains a u,v-path R since 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 P  Q. We assume that z  P 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 Q  wv.

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 y  S, then S-{y} separates G.  |S|>=k+1. 3. If y  S 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,y  V(G), there are internally disjoint x,y- paths. C)For all x,y  V(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 A  B. 2. For B  C, the cycles containing x and y corresponds to pairs of internally disjoint x,y-paths. 3. For D  C,  (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,y  V(G), a set S  V(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 internally 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 xy  E(G), then  (x,y)= (x,y). Proof. An x,y-cut must contain an internal vertex of every 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. xy  E(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 properly 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 V 1 be the set of vertices on x,S-path, and let V 2 be the set of vertices on S,y-path. 4. S  V 1 and S  V 2  S  V 1  V 2. 5. If there exists v such that v  V 1  V 2 –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=V 1  V 2. 6. V 1 omits N(y)-S and V 2 omits N(x)-S by the same argument in 5. 7. Form H 1 by adding to G[V 1 ] a vertex y’ with edges from S. Form H 2 by adding to G[V 2 ] 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 H 1 is an x,y-cut in G.   H 1 (x,y’)= k. 9.  H 2 (x’,y)= k by the same argument in 8. 10. V 1 omits N(y)-S and V 2 omits N(x)-S.  H 1 and H 2 are smaller than G.  H 1 (x,y’)=k= H 2 (x’,y). 11. S=V 1  V 2.  Deleting y’ from the k paths in H 1 and x’ from the k paths in H 2 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 u  N(x)  N(y), then S-u is x,y- cut 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 ef  E(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 y  Y. 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 4.2.21 The connectivity of G equals the maximum k such that (x,y)>=k for all x,y  V(G). The edge-connectivity of G equals the maximum k such that ’(x,y)>=k for all x,y  V(G). Proof. 1.  ’(G)=min x,y  V(G)  ’(x,y).   ’(G)= min x,y  V(G) ’(x,y) by Theorem 4.2.19. 2.  (G)=min x,y  V(G), xy  E(G)  (x,y). 3. By Theorem 4.2.17,  (x,y)= (x,y) for xy  E(G).   (G)=min x,y  V(G), xy  E(G) (x,y). 4. For xy  E(G), G (x,y)=1+ G-xy (x,y)= 1+  G- xy (x,y)>=1+  (G-xy)>=  (G).

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