Nonseparable Graphs. March 18, 2010

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1 Nonseparable Graphs March 18, Cut Vertices A cut ertex of graph G is a ertex such that when the ertex and the edges incident with are remoed, the number of connected components are increased, i.e., c(g ) > c(g). A connected graph is said to be 2-connected if it is a single ertex, or a single loop, or it has at least two ertices and any two ertices lie on a common cycle. Theorem 1.1. Let G be a connected graph G with at least three ertices. Then G has no cut ertex if and only if any two distinct ertices are connected by two internally disjoint paths. Proof. We first proe the sufficiency. Since any two ertices of G are connected by two internally disjoint paths, then for each ertex of G, any two ertices of G are connected by at least one path, i.e., c(g ) = c(g). ence G has no cut ertex. Next we proe the necessity. Let u, be two distinct ertices of G. To show that there are internally disjoint paths between u and, we apply induction on the distance d(u, ) between u and. When d(u, ) = 1, i.e., u, are end-ertices of an edge e in G. Since both u and are not cut ertices, the edge e is not a cut edge. So e is contained in a cycle C. Thus ue and C\e are two internally disjoint paths between u and. Now assume that any two ertices haing distance less than d are connected by two internally disjoint paths, where d 2. Let d(u, ) = d. Let P := 0 e 1 1 d 1 e d d be a path from u = 0 to = d. Since d( 0, d 1 ) = d 1, there are two internally disjoint paths P 1 and P 2 from 0 to d 1 in G. Since G has no cut ertex, the subgraph G d 1 is connected. Then there is a u-path P 3 in G d 1. Let w be the last ertex of P 3 that meets P 1 P 2. Without loss of generality, we may assume that w lies in P 1. Write P 1 = P 1 Q 1, P 3 = P 3 Q 3, where P 1 is the sub-path of P from 0 to w, and Q 3 is the sub-path of P 3 from w to d ; see Figure below. w 1 Q Q 1 d 1 e d d P 2 3 Then P := P 1 Q 3 and Q := P 2 e d d are two internally disjoint paths from u to. 2 Separation and Blocks A separation of a connected graph G is a decomposition of G into two connected subgraphs G 1, G 2 that hae exactly one ertex in common, and no one is contained in another. This common ertex is called a separating ertex of G. A cut ertex is a separating ertex, and a separating ertex is not necessarily a cut ertex. 1

2 A connected graph G is said to be separable if it has at least one separating ertex; otherwise it is said to be nonseparable. A loop is a nonseparable graph. If a graph G is nonseparable and is not a single loop, then G contains no loops. A nonseparable graph is either a single ertex, or a loop, or a link edge, or a 2-connected graph. A block of a graph is a maximal nonseparable subgraph. Theorem 2.1. Let G be a connected graph. Then G is nonseparable if and only if any two edges lie on a common cycle. Proof. : Suppose G is separable, i.e., G can be decomposed into two connected subgraphs G 1, G 2, no one is contained in another and hae exactly one ertex in common. Let e i be edges of G i (i = 1, 2) incident with. If one of e 1, e 2 is a loop, it is clear that there is no cycle containing both e 1, e 2 ; this is a contradiction. Thus e i are non-loops. Let i be another end-ertex of e i (i = 1, 2) other than respectiely. It is clear that there is no 1 2 -path in G. ence there is no cycle in G that contains both e 1, e 2. : If G is a loop, then nothing is to be proed. If G is not a loop, then G has no loops. We may assume that G has at least two edges. Let e be an edge with end-ertices 1, 2. Subdiide e into two edges by introducing a new ertex w on e to obtain a new graph G. We claim that G is also nonseparable. In fact, suppose G is separable. Then G must be separated at the ertex w into two connected subgraphs G 1, G 2. We may assume that i belongs to G i (i = 1, 2). Then w is a cut ertex of G ; subsequently, the edge e is a cut edge of G. Thus G\e has two connected components G 1, G 2 with i V (G i ). Since G has at least two edges, then either G 1 has an edge at 1 or G 2 has an edge at 2, say, G 1 has an edge at 1. Therefore G can be separated at 1 into G 1 and G 2 e 1. This is a contradiction. Now, let e 1, e 2 be two edges of G. Subdiide e i by introducing a new ertex i on e i to obtain a new graph G (i = 1, 2). Then G is nonseparable. The resulted graph G is also nonseparable and has at least three ertices. Since Nonseparable graphs hae no cut ertices, then by Theorem 1.1, there are two internally disjoint 1 2 -paths. This means that there is a cycle containing both edges e 1, e 2. Theorem 2.2 (Block-Tree Decomposition). Let G be a connected graph. Then G can be decomposed into blocks such that (a) Any two blocks of G hae at most one ertex in common. (b) Eery cycle is contained in a block of G. (c) There is no block cycle, i.e., there is no blocks B 0, B 1,..., B l such that V (B i ) V (B i+1 ) = (0 i l), where G l+1 = G 0. Proof. (a) Suppose there are two distinct blocks B 1, B 2 haing k ertices 1, 2,..., k in common and k 2. Since B i are not single loops, then B i hae no loops. Consider the subgraph B := B 1 B 2. We shall see that B cannot be separated at a ertex other than i. In fact, suppose B is separated at into G 1, G 2 and V (B 1 ). Then both G 1 and G 2 contain edges of B 1. Thus B 1 is separated at into B 1 G 1 and B 2 G 2 ; this is a contradiction. Note that B 1 = (B 1 1 ) (B 2 1 ). Since B 1 1, B 2 1 are connected and hae the ertex 2 in common, then B cannot be separated at 1. Likewise, B cannot be separated at i. So B is a block containing both B 1 and B 2. This is contradict to the maximality of B 1, B 2. (b) and (c) are equialent. We proe (c). Suppose there is a sequence B 0, B 1,..., B l of blocks such that V (B i ) V (B i+1 ) = { i }, 0 i l, where B l+1 = B 0. Consider the subgraph B := l i=0 B i. It is clear that B is connected and cannot be separated at any ertex other than i, 0 i l. Likewise, B cannot be separated at the ertices i. So B is nonseparable. This is contradict to the maximality of B i. Let B denote the set of blocks of a connected graph G, and S the set of separating ertices. Let B(G) denote the bipartite graph whose ertex set has the bipartition {S, B}, and whose edges are the pairs {, B}, where belongs to B. Then B(G) is a tree, called the block tree of G. The blocks of G corresponding to the leaes of the block tree B(G) are called end blocks. Any ertex of a block of G other than the separating ertex is called an internal ertex of the block. 2

3 3 Ear Decomposition Eery nonseparable graph other than a single ertex or a link edge contains a cycle. An ear of a subgraph of a graph G is a path P in G such that P is not closed, the initial and terminal ertices of P lie in, and internal ertices of P lie outside. Proposition 3.1. Let be a nontriial subgraph of a nonseparable graph G. Then is neither a single ertex nor a loop, but has an ear in G. Proof. Since is a nontriial subgraph, then is not a single ertex and cannot be G. If is a loop e at a ertex, then G must hae an edge not in ; thus G is separated at into and G\e; this is a contradiction. If is a spanning subgraph of G, then V () = V (G) and E(G) E(). Since is neither a single ertex nor a loop, then eery edge e E(G) E() is a link edge, and subsequently, is an ear of in G. If is not a spanning subgraph, since G is connected, there is an edge e with an end-ertex u V () and an end-ertex V (); see Figure below. P u u e Since G is nonseparable, then G\e is connected, subsequently, there is a (, u)-path P in G\e. Thus Q := uep is an ear of in G. Proposition 3.2. Let be a nonseparable subgraph of a graph G, and let P be an ear of in G. Then P is nonseparable. Proof. It is clear that P cannot be separated at any ertex of. It is also clear that P cannot be separated at any internal ertex of P. A sequence G 0, G 1,..., G k of graphs is said to be nested if G i G i+1, where 0 i k 1. An ear decomposition of a nonseparable graph G is a nested sequence G 0, G 1,..., G k of nonseparable subgraphs of G such that (i) G 0 is a cycle; (ii) G i+1 = G i P i, where P i is an ear of G i in G, 0 i k 1; and (iii) G k = G. Theorem 3.3. Let G be a nonseparable graph. If G is neither a single ertex nor a link edge, then G has an ear decomposition. Proof. Since G is neither a single ertex nor a link edge, then G contains at least two edges. Applying Theorem 2.1, any two edges of G lie on a common cycle. Thus G contains a cycle G 0 with at least two ertices. If G 0 G, then G 0 is a proper subgraph of G. Since G 0 is neither a single ertex nor a link edge, then G 0 has an ear P 0 in G by Theorem 3.1. Applying Proposition 3.2, G 1 := G 0 P 0 is nonseparable. Similarly, if G 1 is a proper subgraph of G, then G 1 has an ear in G. Continue this procedure; we obtain nonseparable subgraphs G i and its ears P i in G such that G i+1 := G i P i. Since G i G i+1 and G is finite, the procedure must end up with G k = G at some step k. Recall that a digraph D is said to be strongly connected (or just strong) if for any proper subset X V (D), the set (X, X c ), consisitng of edges whose orientations hae tails in X and heads in X c, is nonempty. Proposition 3.4. A digraph D is strong if and only if for any two ertices u, of D, there is a directed path from u to and a directed path from to u, i.e., the ertices u, are strongly connected. 3

4 Proof. The sufficiency is triial. For necessity, let V u be the set of all ertices w such that there exists a directed path from u to w in D. Then for any ertex w V u c there is no directed path from u to w. Clearly, V u, since we allow directed path of length zero. If V u V (D), then (V u, V u c ) is nonempty, for D is strong. Let e (V u, V u c ) be an edge with an end-ertex w 1 V u and an end-ertex w 2 V c u, and let P be a directed path from u to w 1. Then Q := P ew 2 is a directed path from u to w 2 ; this is contradict to that there is no directed path from u to w 2. ence V u = V (D); analogously, V = V (D). This means that u, are strongly connected. Proposition 3.5. A connected digraph is strong if and only if each of its block is strong. Proof. Triial. Proposition 3.6. Let P be an ear of a subgraph in a digraph D. If is strongly connected and P is a directed path, then P is also strongly connected. Proof. Let P = 0 e 1 1 e l l be directed from 0 to l. For two ertices u, P, if u,, nothing is to be proed, since is strongly connected. If u, P, say, u = u i and = j with i < j, let P 0 be a directed path from l to 0 in. Then P 1 := i e i+1 i+1 e j j is a directed path from u to, and P 2 := j e j+1 j+1 e l l P 0 e 1 1 e i 1 i is a directed from to u in P. So P is strongly connected. Theorem 3.7. Eery connected graph G without cut edges has a strong orientation. Proof. It suffices to show that each block of G has a strong orientation. Without loss of generality, we may assume that G is nonseparable and has no cut edges. If G is a single ertex or a loop, it is triial that G can be oriented to be strongly connected. Note that G cannot be a link edge; applying Theorem 3.3, G has an ear decomposition (G 0, G 1,..., G k ), where P i is an ear of G i in G and G i+1 = G i P i, 0 i k 1. Now orient edges of G 0 and P i so that G 0 becomes a directed cycle and P i becomes a directed path. Initially, G 0 is strongly connected. Applying Lemma 3.6, we see that all G i+1 = G i P i are strongly connected. ence G = G k is strongly connected. 4 Ear Decomposition of Digraphs Let D = (G, ε) be a digraph and a subdigraph of D. A directed ear of in D is a directed path P in D whose distinct initial and terminal ertices lie in and internal ertices lie outside. A direction of a path P = 0 e 1 1 e l l is an orientation ε P on P such that ε P (e i, i )ε P (e i+1, i ) = 1, i.e., i is neither a source nor a sink, 1 i l 1. A direction ε P of P with ε P (e 1, 0 ) = 1 is usually called a positie direction of P. A direction of a walk W = 0 e 1 1 e l l is a function ε W on the pairs (e i, i 1 ), (e i, i ) (1 i l) such that ε W (e i, i )ε W (e i+1, i ) = 1, 1 i l 1. A direction ε W of W with ε W (e 1, 0 ) = 1 is usually called a positie direction of W. Proposition 4.1. Let be a nontriial strong subdigraph of a nonseparable strong digraph D. Then has a directed ear in D. Proof. Since D is nonseparable and is a nontriial subgraph of D, then has ears in D by Proposition 3.1. Among these ears we choose an ear P haing minimal number of reersing edges. We claim that such an ear P is actually a directed ear of in D. Let P = 0 e 1 1 e k k. If k = 1, then either P or P 1 is a directed ear for. Suppose P is not a directed path. We must hae k 2. Let e i be an edge whose orientation is from i to i 1. Then one of the ertices i 1 and i is outside. Since D is strong, there is a directed path P from i 1 to i in D. Then Q := 0 e 1 1 e i 1 P e i+1 i+1 e k k is a walk from 0 to k, haing less number of reersing edges comparing with the walk P. If P =, then Q is disjoint from, except 0, k. Let Q be a path followed the walk Q from 0 to k, haing no ertices repeating. Then Q is an ear of in D, haing less number of reersing edges comparing with P ; this is contradict to the choice of P. If P, let u, be the first and last ertices in Q such that u, respectiely, let P 1 be the subpath of P from i 1 to u, and let P 2 be the subpath of P from to i. If u, then 4

5 Q := P 2 e ip 1 is a directed ear of in D; see the left Figure below. If u =, we hae two cases: (i) u 0, then the walk Q 1 := 0 e 1 1 i 2 e i 1 P 1 contains an ear of in D, haing less number of reersing edges comparing with P ; (ii) u k, then the walk Q 2 := P 2 e i+1 i+1 e k k contains a directed ear of in D, haing less number of reersing edges comapring with P ; all these are contradict to the choice of P ; see the right Figure below. u 0 l P e i i i 1 u 0 l P e i i i 1 A directed ear decomposition of a nonseparable strong digraph D is nested sequence (D 0, D 1,..., D k ) of nonseparable strong subdigraphs of D such that (i) D 0 is a directed cycle, (ii) D i+1 = D i P i, where P i is a directed ear of D i in D, 0 i k 1, and (iii) D k = D. Theorem 4.2. Eery nontriial, nonseparable, strong digraph D has a directed ear decomposition. Proof. It is obiously true if D is a directed loop. If D is not a directed loop, then D contains at least two ertices u,. There is a directed path P from u to and a directed path Q from to u. Then W := P Q 1 is a closed directed walk containing both u,. Of course, W contains a directed cycle D 0 that contains both u,. Clearly, D 0 is strongly connected and non separable. By Proposition 4.1, D 0 has a directed ear P 0 in D. Then by Proposition 3.2 and Proposition 3.6, D 1 := D 0 P 0 is strongly connected and nonseparable. Continue this procedure; we obtain a nested sequence (D 0, D 1,..., D k ) of strongly connected nonseparable subdigraphs of D such that D i+1 = D i i, where P i is a directed ear of D i in D, 0 i k 1, and D k = D. This is a directed ear decomposition of D. A feedback set of a digraph D is an edge subset S of D such that D\S contains no directed cycles. A feedback set S of a digraph D is said to be minimal if for each edge e S the subdigraph D\S + e contains at least one directed cycle. Each such directed cycle intersects S at the only edge e, and is called a fundamental directed cycle of D with respect to S. A minimal feedback set S of a digraph D is said to be coherent if eery edge of D is contained in some fundamental directed cycle of D with respect to S. If a digraph D admits a coherent feedback set, then eery component of D must be strongly connected, for each edge of D is contained in a fundamental directed cycle. Theorem 4.3. Eery strongly connected digraph D admits a coherent feedback set. Proof. If D is separable, then each of its block is strongly connected, and we may consider each of its blocks. So without loss of generality, we may assume that D is nonseparable. By Theorem 4.2, D has a directed ear decomposition (D 0, D 1,..., D k ), where D 0 is a directed cycle, D i+1 = D i P i, P i is a directed ear of D i in D, 0 i k 1, and D k = D. Choose an edge e 0 from D 0 and set S 0 := {e 0 }. If D 1 \e 0 contains no directed cycles, set S 1 := S 0. If D 1 \e 0 contains a directed cycle, then the directed cycle must contain the path P 0 ; choose an edge e 1 from P 0 and set S 1 := S 0 e 1. Thus D\S 1 contains no directed cycles. In general, if D i \S i 1 contains no directed cycles, set S i := S i 1. If D i \S i 1 contains a directed cycle, then the directed cycle must contain the whole path P i 1 ; choose an edge e i from P i 1 and set S i := S i 1 e i. Then D i \S i contains no directed cycles. Finally, we hae a coherent feedback set S = S k for D. Proposition 4.4. Eery strong digraph D has a strong spanning subgraph of at most 2 V (D) 2 edges. 5

6 Proof. Delete all loops of D if necessary; so we may assume that D contains no loops. If D is a single ertex, it is clearly true. If D is not a single ertex, then each block B of D is strong. Consider a directed ear decomposition of B. Delete from B the edges in the directed ears of length one; we obtain a strong spanning subdigraph of B, and a directed ear decomposition (D 0, D 1,..., D k ) of, where D 0 is a directed cycle, D i+1 = D i P i, P i is a directed ear of length at least two, and D k =. Since each ear contains at least one internal ertex and V (D) 2, we see that k V () V (D 0 ) V () 2. Since D 0 is a cycle and P i are paths, then E(D 0 ) = V (D 0 ), E(P i ) = V (P i ) 1, 0 i k 1. Thus k 1 k 1 ( E() = E(D 0 ) + E(P i ) = V (D 0 ) + V (Pi ) 1 ) i=0 = V () + k 2 V () 2. Now the union of the strong subdigraphs (one for each block B of D) is a strong spanning subdigraph of D. Since each block B has a strong spanning subdigraph and E() 2 V () 2, it follows that the union has the number of edges: E ( ) = E() ( ) ( ) 2 V () 2 = 2 V () 1 = 2 V (D) 2. i=0 6