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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. u u v v Common internal vertex Internally disjoint paths

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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 of if part) 1.It suffices to show for each pair u,v V(G), deletion of any vertex in V(G) cannot separate u from v. 2. This is clearly true because G has internally disjoint u,v-paths.

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Theorem 4.2.2 (Proof of only if part) 1.G is 2-connected. (Premise) 2.That G has internally disjoint u,v-paths is proved by induction on d(u,v). 3. Basis Step: d(u,v)=1. 3.1. The graph G-uv is connected since ’(G)>= (G)>=2. 3.2. There exists a u,v-path in G-uv, which is internally disjoint in G from the u,v-path formed by the edge uv itself.

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Theorem 4.2.2 4. Induction Step: d(u,v)>1. 4.1. Let k=d(u,v). 4.2. Let w be the vertex before v on a shortest u,v-path. d(u,w)=k-1. 4.3 G has internally disjoint u,w-paths P and Q. (Induction Hypothesis)

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Theorem 4.2.2 vu w P Q 4.4. If v V(P) V(Q), then we find the desired paths in the cycle P Q.

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Theorem 4.2.2 4.5. Otherwise, G-w is connected 4.6. G-w contains a u,v-path R. 4.7. If R avoids P or Q, we are done. v u w Q P R since G is 2-connected.

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Theorem 4.2.2 4.8. Otherwise, let z be the last vertex of R (before v) belonging to P Q. We assume that z P by symmetry. 4.9. 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. vu w z P Q R

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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.

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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.

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Theorem 4.2.4 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.

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Theorem 4.2.4 4. Consider edges incident to x and y. 5. Case 1: 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. 6. Case 2: 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. xy e f u x y z f y x e

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Theorem 4.2.4 7. For C D. G satisfies condition C. G satisfies condition A. G is connected. (G)>=1. 8. We need to show any two edges, uv and xy, lie on a common cycle. 9. Add to G the vertices w with neighborhood {u,v} and z with neighborhood {x,y} to form G’.

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Theorem 4.2.4 10. Since G is 2-connected, Lemma 4.2.3 implies G’ is 2-connected. 11. w and z lie on a cycle C in G’. 12. 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. 13. 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. u v x y w z

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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.

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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.

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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. 1. 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). 2. We prove equality by induction on n(G).

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Theorem 4.2.17 (Menger Theorem) 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.No minimum cut properly contains N(x) or N(y) since N(x) and N(y) are x,y-cuts. 3. Case 1: G has a minimum x,y-cut S other than N(x) or N(y). 4. Case 2: Every minimum x,y-cut is N(x) or N(y).

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Theorem 4.2.17 5. For Case 1, let V1 be the set of vertices on x,S-path, and let V2 be the set of vertices on S,y-path. 6. S V1 and S V2 S V1 V2. 7. If there exists v such that v V1 V2 –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. 8. This implies S=V1 V2. G V1V1 V2V2 xy S v

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Theorem 4.2.17 H1 y’ x H2 x’ y 9. Form H1 by adding to G[V1] a vertex y’ with edges from S, and form H2 by adding to G[V2] a vertex x’ with edges from S.

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Theorem 4.2.17 10. Every x,y-path in G starts with an x,S-path (contained in H1). Every x,y’cut in H1 is an x,y-cut in G. H1 (x,y’)= k. 11. H2 (x’,y)= k by the same argument in 10. 12. H1 and H2 are smaller than G since N(y) S and N(x) S. H1 (x,y’)=k= H2 (x’,y).

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Theorem 4.2.17 13. S=V1 V2. 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.

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Theorem 4.2.17 14. For Case 2, 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. 15. Combining these k-1 x,y-path and the path x,u,y yields k internally disjoint x,y-paths in G.

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Theorem 4.2.17 16. If there exists node v {x} N(x) N(y) {y}, then S is minimum x,y-cut in G-v. (If there exists a x,y-cut, S’, in G-v whose size is smaller than |S|, then S’ {v} is a x,y-cut in G. It is a contradiction.) 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.

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Theorem 4.2.17 17. We may assume that N(x) and N(y) partition V(G)- {x,y}. 18. Let G’ be the bipartite graph with bipartition N(x), N(y) and edge set [N(x),N(y)]. 19. 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.

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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.

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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. x y a b c d e f g ts Proof. 1. Modify G to obtain G’ by adding two new vertices s, t and two new edges sx and yt. 2. Cleary, ’ G (x,y)= ’ G’ (x,y) and ’ G (x,y)= ’ G’ (x,y).

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Theorem 4.2.19 3. A x,y-path exists in G’ that traverses edges p, q, r if and only if a sx,yt-path exists in L(G’) that traverses vertices p, q, r.

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Theorem 4.2.19 4. 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). 5. A set of edges disconnects y from x in G’ if and only if the corresponding vertices of L(G’) form an sx,yt-cut. ’ G’ (x,y)= L(G’) (sx,yt). 6. L(G’) (sx,yt)= L(G’) (sx,yt). ’ G’ (x,y)= ’ G’ (x,y). ’ G (x,y)= ’ G (x,y).

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Lemma 4.2.20 Deletion of an edge reduces connectivity at most 1. Proof. 1. (G-xy) ≤ (G). 2. (G-xy)< (G) only if G-xy has a separating set S that has size less than (G) and is not a separating set of G. 3. G-S-xy has two components G[X] and G[Y], with x X and y Y since G-S is connected. 4. If |X| ≥ 2 (or |Y| ≥ 2), S {x} (or S {y}) is a separating set of G, and (G) ≤ (G-xy)+1. 5. Otherwise, |S|=n(G)-2. 6. Since |S|< (G), (G) = n(G)-1. 7. G is complete graph and (G-xy) = n(G)-2 = (G)-1.

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Theorem 4.2.21 The edge-connectivity of G equals the maximum k such that ’(x,y)>=k for all x,y V(G). That is, ’(G)= min x,y V(G) ’(x,y). Proof. 1. ’(G)= min x,y V(G) ’(x,y). 2. ’(x,y)= ’(x,y) for all x,y V(G) by Theorem 4.2.19.

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Theorem 4.2.21 The connectivity of G equals the maximum k such that (x,y)>=k for all x,y V(G). That is, (G)=min x,y V(G) (x,y). Proof. 1. (G)=min xy E(G) (x,y). 2. (x,y)= (x,y) for xy E(G). 3. It suffices to show (x,y)≥ (G) if xy E(G). 4. G-xy (x,y)= G-xy (x,y) ≥ (G-xy) by Menger’s Theorem. 5. (G-xy) ≥ (G) -1 by Lemma 4.2.20. 6. (x,y)= G-xy (x,y)+1 ≥ (G).

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