Relationship between Maximum Flows and Minimum Cuts

Transcription

1 128 Flows and Connectivity Recall Flows and Maximum Flows A connected weighted loopless graph (G,w) with two specified vertices x and y is called anetwork. If w is a nonnegative capacity function c, then the network (G xy,c) is called a capacity network. An (x, y)-flow in (G xy,c) is a function f E (G) satisfying The value of an (x, y)-flow f 0 f(a) c(a), a E(G), f + (u) = f (u), u V (G) \ {x, y}. valf = f + (x) f (x) = f (y) f + (y) is call The value of f. An (x, y)-flow f is maximum if val f val f for any (x, y)-flow f. Cuts and Minimum Cuts An (x, y)-cut in N is a set of edges of the form (S, S), where x S and y S. The capacity cap B of an (x, y)-cut B is the sum of the capacities of edges in B. An (x, y)-cut B in N is called to be minimum if cap B cap B for any (x, y)-cut B. Relationship between Maximum Flows and Minimum Cuts Theorem 4.1 (max-flow min-cut theorem) In any capacity network, the value of a maximum flow is equal to the capacity of a minimum cut. Corollary 4.1 In any integral capacity network, there must be an integral maximum flow, and its value is equal to the capacity of a minimum cut.

2 4.2. MENGER S THEOREM 129 Edge Version of Menger s Theorem η G (x, y): the maximum numbers of edge-disjoint (x, y)-paths in G λ G (x, y): the minimum number of edges in an (x, y)-cut in G, which is call the local edge-connectivity of G. Theorem 4.2 Let x and y be two distinct vertices in a graph G. Then η G (x, y) = λ G (x, y). Vertex Version of Menger s Theorem A nonempty set S V (G) \ {x, y} is said to be an (x, y)-separating set in G if there exists no (x, y)-path in G S. κ G (x, y): the minimum cardinality of an (x, y)-separating set in G, which is call the local (vertex-)connectivity of G. ζ G (x, y): the maximum numbers of internally disjoint (x, y)-paths in G. Theorem 4.3 (Menger s theorem) Let x and y be two distinct vertices of G without edges from x to y. Then ζ G (x, y) = κ G (x, y). Equivalence of Three Theorems Corollary 4.1 Theorem 4.2 Theorem 4.3. Direct proofs of Theorem 4.3 and Theorem 4.2 Theorem 4.3 Theorem 4.2 Corollary 4.1.

3 130 Flows and Connectivity 4.3 Connectivity Global Connectivity In preceding section, we have defined two parameters of a graph G, κ G (x, y) and λ G (x, y), called local connectivity and local edge-connectivity, respectively. We, in this section, consider the global connectivity of G. Vertex-Connectivity Let G be a strongly connected digraph. A nonempty proper subset S of V (G) is said to be a separating set if G S is not strongly connected. It is clear that every strongly connected digraph contains a separating set provided it contains no complete graph as a spanning subgraph. The parameter 0, if G is not strongly connected; κ(g) = v 1, if G contains a complete spanning subgraph; min{ S : S is a separating set of G}, otherwise is defined as the (vertex-)connectivity of G. Clearly, two parameters κ(g) and κ G (x, y) satisfy the following relation. κ(g) = min{κ G (x, y) : x, y V (G), E G (x, y) = }. For example, κ(k n ) = n 1, κ(c n ) = 1 if C n is directed and κ(c n ) = 2 if C n is undirected for n 3. A separating set S of G is a κ-separating set if S = κ(g). A graph G is said to be k-connected if κ(g) k. All nontrivial connected undirected graphs and strongly connected digraphs are 1-connected.

4 4.3. CONNECTIVITY 131 Edge-Connectivity A nonempty proper subset B of E(G) is said to be a directed cut if G B is not strongly connected. Clearly, every nontrivial strongly connected digraph must contain a directed cut. The parameter λ(g) = { 0, if G is trivial or not strongly connected; min{ B : B is a directed cut of G}, otherwise is defined as the edge-connectivity of G. It is also clear that λ(g) = min{λ G (x, y) : x, y V (G)}. For example, λ(k n ) = n 1, λ(c n ) = 1 if C n is directed and λ(c n ) = 2 if C n is undirected for n 3. A directed cut B of G is a λ-cut if B = λ(g). G is said to be k-edge-connected if λ(g) k. All nontrivial connected undirected graphs and strongly connected digraphs are 1-edge-connected. Remarks By definition, if B is a directed cut of G, then there exists a nonempty proper subset S V (G) such that (S, S) = B. Recall that, in Section 2.2, a cut of G is a subset of E(G) of the form [S, S]. These two concepts are identical if G is undirected. However, if G is directed, then its cut contains certainly a directed cut, but the converse may be not always true.

5 132 Flows and Connectivity Whitney s Inequality The following fundamental inequality relating to κ(g), λ(g), and δ(g), is often referred to as Whitney s inequality, found first by Whitney (1932) for undirected graphs, and generalized to digraphs by Geller and Harary (1971). Theorem 4.4 κ(g) λ(g) δ(g) for any graph G. Proof: Without loos of generality, we need to only prove this theorem for a nontrivial, nonempty and loopless digraph G with δ(g) = δ + (G). Let x V (G) such that d + G (x) = δ(g). Since the set of out-going edges E+ G (x) of x is a directed cut of G, it follows that λ(g) E + G (x) = δ(g). We now prove κ(g) λ(g) by induction on λ(g) 0. Let λ = λ(g). If λ = 0, then G is not strongly connected, i.e., κ(g) = 0 and, thus, the equality holds. Suppose κ(h) λ(h) for any digraph H with λ(h) < λ, and suppose λ 1. Let G be a directed graph with λ(g) = λ. Then, there exists a directed cut B of G such that B = λ. Let a B and H = G a. Then λ(h) λ 1. By the induction hypothesis, we have κ(h) λ(h) λ 1. If H contains a complete spanning subgraph, then so does G. Thus κ(g) = v 1 = κ(h) λ(h) λ 1 < λ(g). If H does not contain a complete graph as a spanning subgraph, then there exists a κ-separating set S in H. If G S is not strongly connected, then κ(g) S = κ(h) λ(h) < λ(g). Suppose that G S is strongly connected below. If v(g S) = 2, then κ(g) v 1 = v(g S) + S 1 = S + 1 = κ(h) + 1 λ(h) + 1 λ(g). If v(g S) > 2, let ψ G (a) = (x, y), then S {x} or S {y} is a separating set of G. Thus, we have κ(g) S + 1 = κ(h) + 1 λ(h) + 1 λ(g). Thus in each case we have κ(g) λ(g). The theorem follows by the principle of induction.

6 4.3. CONNECTIVITY 133 Remarks The inequality in Theorem 4.4 are often strict. For example, the graph G in Figure 4.7 has κ(g) = 2, λ(g) = 3 and δ(g) = 4. In fact, it has been proved that for all integers κ, λ, δ with 0 < κ λ δ, there exists a graph G with κ(g) = κ, λ(g) = λ and δ(g) = δ by Chartrand and Harary (1968) for the undirected case and by Geller and Harary (1971) for the directed case (see the exercise ). Figure 4.7: An illustration of Whitney s inequality Examples It is clearly of importance to know which conditions can ensure the equalities in Theorem 4.4. We present a sufficient condition for the latter equality to hold, due to Jolivet (1972) for the undirected case, and Plesnik (1975) for the directed case. Example Let G be a simple graph with diameter d(g) 2. Then λ(g) = δ(g). Proof: We prove the equality for a digraph G. If d(g) = 0 or 1, then G is trivial or a complete graph since G is simple, and so the equality holds clearly. We suppose d(g) = 2 below. By Whitney s inequality, we need to only prove λ(g) δ(g) for any simple digraph G. Let B = (S, S) be a λ-cut of G, and let X S and Y S be sets of the tails and the heads of the edges in B, respectively. We first note S \ X S \ Y = 0. Otherwise, for any x S \ X and y S \ Y, any (x, y)-path in G must contains an edge in B, that is, d G (x, y) 3, which contradicts the hypothesis of d(g) = 2. Without loss of generality, we suppose that S \ Y is an empty set. Let Y = {y 1, y 2,, y b } and let s i = N G (y i) X, i = 1, 2,, b. Since G is simple, it follows that b s i = λ(g) b 1. i=1

7 134 Flows and Connectivity Let δ = δ(g) and λ = λ(g). Noting δ d G (y i) s i + (b 1) for all i = 1, 2,,b, we have that is, b bδ s i + b(b 1) = λ + b(b 1), i=1 λ bδ b(b 1) = δ + (δ b)(b 1) δ as desired. The following example gives a sufficient condition for a undirected graph to be k-connected, due to Bondy (1969) and, independently, Boesch (1974). Example Let G be a simple undirected graph of order v 2, degrees d i of vertices satisfy d 1 d 2 d v, and k be an integer with 1 k v 1. If 1 d i k + i 2 d v k+1 v i for i = 1, 2,, (v k + 1), 2 then κ(g) k. Proof: To the contrary suppose κ(g) < k. Then G is not complete, and there exists a separating set S in G such that κ(g) S = k 1. Let H be a component of G S with the minimum number of vertices i. Then i 1 2 (v k + 1), that is, k v 2i + 1. Thus, for each x V (H), we have d G (x) v(h) 1 + S = i + k 2 v i 1. On the other hand, d G (y) v i 1 for any y V (G) \ (V (H) S) clearly. This implies that all of vertices of degree at least v i are in S. Since d v d v 1 d v k+1 v i, it follows that S v (v k + 1) + 1 = k, a contradiction.

8 4.3. CONNECTIVITY 135 Criterion With the aid of Menger s theorem, we now present a sufficient and necessary condition for a graph to be k-connected, its version of the undirected case was first found by Whitney (1932) in Theorem 4.5 Let G be a graph of order at least k + 1. Then (a) κ(g) k ζ G (x, y) k, x, y V (G); (b) λ(g) k η G (x, y) k, x, y V (G). Proof: We prove the conclusion (a) for a digraph G. For k = 1, the conclusion holds clearly. Suppose k 2 below. ( ) Let x and y be two distinct vertices of G, and let µ = E G (x, y). If µ = 0, then, by Menger s theorem, we have ζ G (x, y) = κ G (x, y) κ(g) k. Suppose now that µ 1. If µ k, then ζ G (x, y) µ k. Thus, we suppose µ < k below, and let H = G E G (x, y). We need to only prove ζ H (x, y) k µ. To the contrary suppose ζ H (x, y) < k µ. By Menger s theorem, there exists an (x, y)-separating set S in H such that S = ζ H (x, y) k µ 1. Since v(h S) = v(h) S k + 1 (k µ 1) = µ + 2, it follows that there exists z V (H S) other than x and y. We now prove ζ H S (x, z) 1. If E H (x, z), then the assertion holds clearly. Suppose E H (x, z) = below. Then E G (x, z) =. By the hypothesis and Menger s theorem, we have ζ G (x, z) k. Thus, ζ H (x, z) k µ. Since S k µ 1, we have ζ H S (x, z) 1. Similarly, we can prove ζ H S (z, y) 1. Thus, we have ζ H S (x, y) 1, which contradicts the choice of S. The necessity follows. ( ) If G contains a complete spanning subgraph, then κ(g) = v 1 k. Suppose now that G contains no complete spanning subgraph. Let S be a κ-separating set of G. Then G S is not strongly connected. Thus, there are two distinct vertices x and y in G S such that there is no (x, y)-path in G S. This implies that S is an (x, y)-separating set in G. It follows that S κ G (x, y). By the hypothesis

9 136 Flows and Connectivity ζ G (x, y) k, and by Menger s theorem, we have κ(g) = S κ G (x, y) = ζ G (x, y) k as desired. Similarly, we can prove the conclusion (b), left to the reader as an exercise for the details (the exercise 4.3.1). Application of Theorem 4.5 Theorem 4.6 If the connectivity κ(g i ) > 0 for each i = 1, 2, then connectivity κ(g 1 G 2 ) κ(g 1 ) + κ(g 2 ). Particularly, κ(q n ) = n. Proof: Let κ(g i ) = k i for i = 1, 2. Assume that x and y are two distinct vertices in G 1 G 2. By Theorem 4.5, we need only show there are k 1 +k 2 internally disjoint (x, y)-paths in G 1 G 2. To this purpose, Let x = x 1 x 2 and y = y 1 y 2, where x 1, y 1 V (G 1 ) and x 2, y 2 V (G 2 ). If x 1 y 1, then by Theorem 4.5 there must exist k 1 internally disjoint (x 1, y 1 )- paths P 1, P 2,, P k1 in G 1 since G 1 is k 1 -connected. Let P be a shortest (x 1, y 1 )- path in G 1. We can, without loss of generality, suppose that P 1 has the first edge in common with P. Thus ε(p i ) 2 for each i = 2, 3,,k 1. Let v i be the first internal vertex in P i (v 1 = y 1 if (x 1, y 1 ) E(G 1 )). Then v i cuts the path P i into two subpaths a i and P i, where a i is the first edge (x 1, v i ) in P i and P i is the subpath of P i from v i to y 1. And so the (x 1, y 1 )-path P i can be expressed as a P i = x i P i 1 vi y 1, i = 2, 3,,k 1. Similarly, if x 2 y 2 then there are k 2 internally disjoint (x 2, y 2 )-paths W 1, W 2,, W k2 in G 2. Let W be a shortest (x 2, y 2 )-path in G 2. Assume that W has the first edge in common with W 1. Let u j be the first internal vertex in W j for each j = 1, 2,, k 2 (u 1 = y 2 if (x 2, y 2 ) E(G 2 )). Then W j can be represented as b j W j W j = x 2 uj y 2, j = 2, 3,,k 2, where b j is the first edge (x 2, u j ) in W j and W j is the subpath of W j from u j to y 2. Since G i is k i -connected for i = 1, 2, there must be k 1 internally disjoint (v i, x 1 )- paths T 1, T 2,, T k1 in G 1 and k 2 internally disjoint (u i, x 2 )-paths U 1, U 2,,U k2 in G 2 (the exercise ).

10 4.3. CONNECTIVITY 137 Using the above notation, we can construct k 1 +k 2 internally disjoint (x 1 x 2, y 1 y 2 )- paths R 1, R 2,, R k1+k 2 in G 1 G 2 as follows. If y 1 x 1 and y 2 x 2, then let Px R 1 = x 1 x 2 y 1W 1 2 y1 x 2 y1 y 2, a R i = x 1 x ix 2 v 2 vi x iw P 2 vi y i 2 y2 y 1 y 2, i = 2,,k 1, x R k1+1 = x 1 x 1W P 1y 2 2 x1 y 2 y1 y 2, x 1b j Pu j y 1W j R k1+j = x 1 x 2 x1 u j y1 u j y 1 y 2, j = 2,,k 2. If y 1 = x 1 and y 2 x 2, then let a R i = x 1 x ix 2 v 2 vi x iw T iy 2 2 vi y 2 x1 y 2 = y 1 y 2, i = 1, 2,, k 1 x 1W j R k1+j = x 1 x 2 x1 y 2 = y 1 y 2, j = 1, 2,,k 2. If y 1 x 1 and y 2 = x 2, then let P R i = x 1 x 2 ix 2 y1 x 2 = y 1 y 2, i = 1, 2,,k 1, x 1b j Pu j x 1U j R k1+j = x 1 x 2 x1 u j y1 u j y1 x 2 = y 1 y 2, j = 1, 2,,k 2. It is easy to check that these (x, y)-paths R 1, R 2,, R k1+k 2 constructed above are internally disjoint in G 1 G 2 whichever case occurs. Immediately, κ(q n ) = n since κ(k 2 ) = 1 = δ(k 2 ). Exercises: 4.3.3; ; ; Thank You!