be a path in L G ; we can associated to P the following alternating sequence of vertices and edges in G:

Transcription

1 1. The line graph of a graph. Let G = (V, E) be a graph with E. The line graph of G is the graph L G such that V (L G ) = E and E(L G ) = {ef : e, f E : e and f are adjacent}. Examples 1.1. (1) If G is a path of length n, then L G is a path of length n 1. (2) If G is a cycle of length n, then L G = G. (3) If G is a star with n vertices i.e. G 3 and G has a vertex u such that E = {vu, v V {u}}, then L G = K n 1. The converses also hold, see Exercise 1.6. For an edge uv V we write (uv) V (L G ) for the corresponding vertex of the line graph. But we abuse notation using the same symbol for edges, or subsets of edges, in G and vertices, or subsets of vertices, in L G. Lemma 1.2. Let G be a graph with E(G). Then (1) E(L G ) = ( ) d(v) ; 2 v V (2) for any F E(G) we have L G F = L G F ; (3) for any v V (G) we have L G v = L G E(v). Proof. For any v V (G), every pair of edges in E(v) corresponds to an edge in L G, and all the edges of L G are obtained in this way. Hence the first formula follows. The remaining two identitities follow trivially from the definition. Let P = e 0... e m L G be a path in L G ; we can associated to P the following alternating sequence of vertices and edges in G: w 0, e 0, w 1, e 1,... w m, e m, w m+1, where w i V (G) are such that e 0 = w 0 w 1, e m = w m w m+1 and w i is the vertex adjacent to e i 1 and e i for i = 1,..., m. By definition w 0 w 1 and w m w m+1, while we may have w i = w j for the remaining cases; see Example 1.4. From the above sequence we extract the following sequence of (not necessarily distinct) vertices in G (1.1) ω(p ) := (w 0, w 1,..., w m, w m+1 ). We shall use this notation in the next basic Lemma. Lemma - Definition 1.3. Let G be a graph and L G its line graph. (1) Let Q = v 0... v n G be a path. L Q is a path in L G. (2) Let P L G be a path such that ω(p ) consists of distinct vertices. Then there exists a path Q P G such that P = L QP. We have and V (Q P ) = ω(p ) length(q P ) = length(p ) + 1 1

2 2 (3) Fix e, f E(G) and let P L G be an ef-path of minimal length. Then P satisfies the assumption of (2). (4) Let P L G be an ef-path. Then there exists an ef-path P L G such that ω(p ) ω(p ) and such that P satisfies the assumption of (2). Proof. (1). Let Q = v 0... v n be a path in G, we already noticed that L Q is a path. Now v i v i+1 E and hence (v i v i+1 ) is a vertex of L G ; as the v i are all distinct, the (v i v i+1 ) are all distinct. Now, the edges v i v i+1 and v i 1 v i are adjacent for all i 1, hence (v i 1 v i )(v i v i+1 ) is an edge of L G. Therefore L Q L G. (2). Keeping the above notation, assume w 0,..., w m+1 are all distinct. By definition, G has the edge w i w i+1 = e i for all i m, hence Q = w 0 w 1... w m+1 is a path in G. Finally, L Q = P since, as we just noticed, the vertex (w i w i+1 ) equals the vertex e i in L G. We set Q = Q P and we are done. (3). We need to prove that if P = e 0... e m has minimal length between e = e 0 and f = e m, then the elements in ω(p ) are all distinct. By contradiction, let h be minimum such that there exists k > h with w h = w k. If h > 0 then e h 1 is adjacent to e k, hence (1.2) P 1 = e 0... e h 1 e k... e m is a path from e to f shorter than P, which is not possible. If h = 0 then k 2, hence the following (1.3) P 2 = e 0 e k... e m is too short a path from e to f, a contradiction. (4). Let P = e 0... e m be a path from e = e 0 to f = e m and ω(p ) = (w 0, w 1,..., w m+1 ) the associated sequence of vertices. If ω(p ) has no multiple vertices (i.e. if w i w j for all i j) then, by (2), we can take P = P and we are done. We shall prove (4) by induction on the number of multiple vertices in ω(p ), i.e. the number M P = {i = 0,..., m + 1 : j i : w i = w j }. The induction base, M P = 0, has already been treated. Let h be minimum such that there exists k > h with w h = w k. Then, arguing exactly as in the proof of (3), we construct a shorter path from e to f. More precisely, if h > 0, then the path is P 1, given in (1.2), and ω(p 1 ) = (w 0, w 1,..., w h, w k+1,... w m+1 ) ω(p ) with h < k; thus the multiple vertex w k is not in ω(p 1 ), hence M P1 M P 1. The iduction hypotesis gives an ef-path P 1 satisfying (2) and such that ω(p 1 ) ω(p 1 ) ω(p ). So we are done. In the case where the path is P 2, given in (1.3), we have ω(p 2 ) = (w 1, w 0, w k+1,... w m+1 ) ω(p ) with k 2, so the same argument as before applies. The next example illustrates parts of the Lemma and of its proof.

3 3 Example 1.4. In the picture below we have a star graph, G, and its line graph, where we highlighted the path P described in the sequel. In G we set e i = v i v 0 for all i = 1,..., 5. Next, consider the following path in L G joining e 1 and e 5 P = e 1 e 3 e 4 e 2 e 5 L G hence ω(p ) = (v 1, v 0, v 0, v 0, v 0, v 5 ). The proof yields ω(p ) = (v 1, v 0, v 5 ) and P = e 1 e 5. Finally, Q P = v 1 v 0 v 5 G and, of course, L v1 v 0 v 5 = (v 1 v 0 )(v 0 v 5 ) = e 1 e 5. v 1 v 2 G = v 5 v 0 e 1 e 2 v 3 L G = e 5 e 3 v 4 e 4 Figure 1 Lemma 1.5. Let G = (V, E) be a graph and L G its line graph. G is connected if and only if L G is connected and δ(g) 1. Proof. Suppose G connected, let e = uw and f = xy be two edges of G; we need to find a path in L G between e and f. Let Q = uv 1... v n x be a ux-path in G of minimal length. Consider the path L Q L G. If L Q contains e and f we are done. If L Q does not contain e, then, by the minimality of Q, the vertex w is not in Q. Hence el Q is a path in L G containing e. If this path does not contain f, arguing as above we have the path el Q f contains e and f. So we are done. Conversely, assume L G connected and G free from isolated vertices. Fix x, y V (G); we need to find a path joining x and y. Pick two edges, e x and e y, adjacent to each of them (if e x = e y then we are done). Let P = e 0... e n be a path of minimal length in L G such that e x = e 0 and e y = e n. By Lemma 1.3 there exists a path Q P G such that L QP = P ; by part (2) of the Lemma the vertices of the path Q P are those in ω(p ), which contains x and y; so Q P contains x and y, and we are done. Exercise 1.6. Let G be such that δ(g) > 0. Prove the following. (1) If L G is a path of length n, then G is a path of length n + 1. (2) If L G is a cycle, then G = L G. (3) If L G = K n, then G is a star with n edges.

4 4 Exercise 1.7. Let G = (V, E) and e, f E such that e = xy and f = uv. Set M = min{d(x, u), d(x, v), d(y, u), d(y, v)}. Show that d LG (e, f) = M Connectivity via paths The definition of a connected graph is based on the notion of path. Indeed let G = (V, E) be a graph. Then, by definition, G is 1-connected if and only if for every x, y V there exists an xy-path P G. Next, already know that G is 2-connected if and only if for every x, y V there exist two independent xy-paths P 1, P 2 G (recall that to say that P 1 and P 2 are independent is to say that i.e. P 1 P 2 = {x, y}). We shall prove in Theorem 2.7 that the analog holds for every k. For now we prove the following result, which is the easy implication of Theorem 2.7. Proposition 2.1. Let G be a graph containing k independent xy-paths for every pair of vertices x, y. Then G is k-connected. Proof. To prove that G is k-connected pick a pair of vertices x, y any a set S V {x, y} separating x from y; we need to show that S k. Now, let P 1,..., P k be k independent xy-paths in G; then S each of them must intersect S, so let s i S V (P i ); since the P i are independent, we have s i s j for all i j; hence S k. The proof of the converse is non-trivial, and it is based on an important auxiliary result, interesting in its own right. To begin with, we introduce the following notation: for any graph G = (V, E) and any A, B V (not necessarily disjoint) we set (2.1) κ G (A, B) = min{ S : S V (G) : S separates A from B} and (2.2) π G (A, B) = maximum number of disjoint AB-paths in G. Example 2.2. Assume G connected. If A = {x} and B = {y} with x y we can separate x from y with {x}; moreover it is clear that two xy-paths will never be disjoint, hence κ G (x, y) = π G (x, y) = 1. Example 2.3. Suppose B = V. Then an AB-path P must be of type P = v with v A (for otherwise P would intersect B in two points, which is not possible). Hence π G (A, V ) = A ; now a set S intersecting every AB-path must contain A, therefore κ G (A, V ) = A Lemma 2.4. Let G = (V, E) be k-connected and let x, y V be distinct and not adjacent. Then k G (N(x), N(y)) k.

5 Proof. Set A = N(x) and B = N(y); to every xy-path we can associate an AB-path as follows {xy paths} {AB paths} xv 1... v n y v 1... v n which is well defined, as xy E. We denote by P be the image of the above map; observe that P is the set of AB-paths not containing x or y, and it is not empty, as G is connected. Let now S V be a set separating A and B. Then S intersects every path in P, hence S {x, y} intersects every path in P, hence S {x, y} intersects every xy-path, hence S {x, y} separates x and y. As G is k-connected, S {x, y} k and hence, a fortiori, S k. Therefore k G (A, B) k as wanted. Now, since every S V separating A and B must contain a vertex from every AB-path is clear that π G (A, B) κ G (A, B). It is quite remarkable that, just as in the previous example, equality holds in general, as the following theorem states. Theorem 2.5. Fix a graph G = (V, E) and two subsets A, B of V. Then the minimum number of vertices separating A and B equals the maximum number of disjoint AB-paths. In symbols π G (A, B) = κ G (A, B). Proof. By what we observed above, it suffices to show that G has at least κ G (A, B) disjoint AB-paths. We shall do that by proving the following claim. Let P be a set of disjoint AB-paths with P < κ G (A, B); then there exists a set P of disjoint AB-paths such that P = P + 1 and such that the set of endpoints in B of the paths in P contains the set of endpoints of the paths in P. We shall prove the claim by descending induction on B ; observe that the claim is stronger than the theorem, but the additional requirement on the endpoints is needed for the proof to work. The basis of the induction is the case B = V, which has been treated in Example 2.3 above. We set k = κ G (A, B). Let P be a set of AB-paths as above; to simplify the terminology, by an endpoint of P we shall mean an endpoint of a path in P. Let S be the set of endpoints of P lying in B; since S < k, by hypothesis S cannot separate A and B. Hence there exists an AB path, R, not contained in P. If R is disjoint from the paths in P we are done. If not, let v be the last vertex of R lying in a path, P, in P. Now consider the following set B V B = B V (vr) V (vp ) and the following set of disjoint AB -paths P = P {P } {P v} (P is gotten from P by replacing P with its truncation at v). By construction B B, hence k G (A, B ) k G (A, B), and hence P < k G (A, B ); therefore we can apply the induction hypothesis to B and P. 5

6 6 This gives a set, P, of P + 1 disjoint AB -paths whose endpoints in B contain the endpoints of the paths in P. In particular, P contains a path, P, from A to v and a path, Q, from A to a point y B other than the endpoints of P. Hence P {P, Q } is made of disjoint AB-paths; we will conclude the proof by showing how to replace P and Q with AB-paths so as to get a set P satisfying the claim. There are two possibilities, y vp or y vp. If y vp we replace P with P = P + vp and if y B, we replace Q with Q = Q + yr; if y B we leave Q = Q. If y vp we replace P with P = P + yr and we replace Q with Q = Q + yp. Finally, the set P := P {P, Q } {P, Q } is a set of P + 1 disjoint AB-paths, as we wanted. Corollary 2.6. Let G = (V, E) be k-connected; fix A V with A k and v V A. Then G contains k independent xa-paths with different endpoints in A. Proof. Set B = N(x); we claim k G (A, B) k. Indeed, let U V be such that U k 1; pick a A U and b B U; as G is k-connected G U is connected, hence it contains an ab-path. This shows that U does not separate A and B. By Theorem 2.5, G contains k disjoint BA-paths, P 1,..., P k. Now x P i, since x A; therefore xp 1,..., xp k are independent xa-paths with different endpoints in A. The previous Theorem 2.5 is a key ingredient in the proof of the following. Theorem 2.7 (Menger). A graph G is k-connected if and only if for every pair of vertices x, y it contains k independent xy-paths. Proof. We proved one implication in Proposition 2.1. We shall now show that for any x, y V distinct, G contains k independent xy-paths. We first assume x and y not adjacent. By Lemma 2.4 and Theorem 2.5 there exist k disjoint N(x)N(y)-paths in G, written P 1,..., P k. Set P i = v 0 v 1... v n ; we claim that P i does not contain x or y. Indeed if, say, x = v h then we have h 0 (for x N(x)) and h n (for x N(y)); hence v h 1 and v h+1 are in P N(x), which is not possible as P meets N(x) only in v 0. The claim is proved. As P i does not contain x or y the following P i := xv 0 v 1... v n y is an xypath for i = 1,..., k. It is clear that P 1,..., P k are independent xy-paths, so we are done. Now suppose x and y adjacent. By contradiction, assume G has at most k 1 independent xy-paths; hence G e has at most k 2 independent xypaths. Recall now that G e is (k 1)-connected, since G is k-connected. But then, by the previous part of the proof, G e has k 1 independent xy-paths, as x and y are not adjacent in G e. This is a contradiction. Exercise 2.8. Set k 3; show that in a k-connected graphs any k vertices lie in a common cycle. (The case k = 2 is already known.)

7 Exercise 2.9. Let G be a k-connected graph with at least 2k-vertices. Then G contains a cycle of length at least 2k. 3. Edge-connectivity Definition 3.1. Let k 0. The graph G = (V, E) is k-edge-connected if V > 1 and for any F E with F k 1 the graph G F is connected. Remark 3.2. (1) If G is k-connected, G is k-edge-connnected. (2) If G is k-edge-connnected, then δ(g) k. (3) G is 1-edge-connected if and only if it is 1-connected if and only if it is connected and not K 1. (4) G is 2-connected if and only if it is connected and has no bridges. Recall that for a graph G its line graph, introduced in Section 1, is denoted by L G. Proposition 3.3. A graph G is k-edge-connected if and only if L G is k- connected and δ(g) k. Proof. The first statement follows directly from the definition of line graph. For the second, we can assume k 2 by Lemma 1.5. Assume G is k-edge-connected, then δ(g) k and hence V (L G ) = E(G) > k (here we use k 2, exercise). Let F V (L G ) be such that the graph L G F is disconnected, hence, by (2), L G F is disconnected. Now Lemma 1.5 yields that G F is disconnected, hence F k; this proves that L G is k-connected. Conversely, assume L G is k-connected and δ(g) k; hence V (G) > k. For any F E(G) such that F k 1 we have δ(g F ) 1; therefore, by Lemma 1.5 and (2), if G F is disconnected so is L G F ; now this is not possible because L G is k-connected. This proves that G is k-edgeconnected. The assumption on δ(g) in the second part of the Proposition is really needed: in Example 1.4 we have a graph G with δ(g) = 1, hence not 2- edge-connected, whereas L G is 2-connected (in fact 4-connected!). The following is a consequence of the previous result. Theorem 3.4 (Menger). A graph G is k-edge-connected if and only if every pair of vertices is joined by k edge disjoint paths. Proof. Of course, we can assume k 2. One implication is easy. If G contains k edge-disjoint paths between any two vertices, any set of edges F E such that G F is disconnected must contain at least an edge for each of those paths, hence F k, and hence G is k-edge-connected. Conversely, suppose G is k-edge connected; then, by the previous proposition, its line graph, L G, is k-connected. Consider the sets of edges adjacent to them, E(x) and E(y), and the denote the corresponding sets of vertices in the line graph by A = E(x) V (L G ) and B = E(y) V (L G ). As L G is k-connected, the minimum number of vertices separating A and B is at least k; hence, by Theorem 2.5, there exist k disjoint AB-paths, P 1,..., P k, in L G. Now we apply Lemma 1.3; we replace each P i by the path P i defined 7

8 8 in (4) of that Lemma; then P i is again an AB-path, and since P i and P j are disjoint, one easily checks that P i and P j are disjoint. Moreover, again by the Lemma, for every i = 1,..., k there is a path Q i G such that L Qi = P i. By construction the paths Q 1,..., Q k are edge-disjoint; moreover, as P i is an AB-path, Q i is an xy-path. The proof is complete. Exercise 3.5. Let G be a graph. Show that G is k-edge connected if and only if G > 1 and for every x, y V, any set F E separating x and y satisfies F k. Exercise 3.6. Let G be k-edge-connected with k 1; show that G e is (k 1)-edge-connected for any e E. Give an example of a k-edge-connected graph such that G x is not (k 1)-edge-connected for some x V.