4. Connectivity Connectivity Connectivity. Whitney's s connectivity theorem: (G) (G) (G) for special

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1 4. Connectivity 4.. Connectivity Vertex-cut and vertex-connectivity Edge-cut and edge-connectivty Whitney' connectivity theorem: Further theorem for the relation of and graph 4.. The Menger Theorem and it conequence The Whitney' characterization Application (Reliable network) 4.3. Flow in a graph The value of a flow and the capacity of a cut The Max-Flow Flow-Min-Cut Theorem A new proof for the Menger Theorem 4.4. The thoughne of a grpah for pecial Graph Theory Connectivity We recall the definition connected, diconnected and component. If G i connected and G W i diconnected,, where W i a et of vertice, then we ay that W eparate G,, or o that W i a vertex-cut. If in G W two vertice and t belong to different component then W eparate from t. A graph G i k-vertex-connected (k ) ) if either K k+ or it ha at leat k+ vertice and no et of k vertice eparate G. Graph Theory 3 A connected graph i alo aid to be -connected. The maximal value of k for which a connected graph G i k-vertex- connected i the vertex-connectivity of G, denoted by. If G i diconnected,, we put =0. Every graph that i not complete ha a vertex-cut cut: : the et of all vertice ditinct from two nonadjacent vertice i a vertex-cut cut. The vertex-connectivity of a graph G i the minimum cardinality of a vertex-cut if G i not complete,, and = n if G = K n for ome poitive integer n. κ( G ) = min The two definition i equivalent. { W : W V ( G ) and W i a vertex - cut} If a graph G i k-vertex-connected then k. Graph Theory 3 3 Graph Theory 3 4

2 Example-: A nontrivial graph ha vertex-connectivity 0 iff it i diconnected. A graph G ha vertex-connectivity iff G = K or G i connected with cut-vertice vertice. iff G i noneparable of order 3 or more. The edge-connectivity i defined analogouly: An edge-cut in a graph G i a et X of edge of G uch that G X i diconnected. An edge-cut X i minimal if no proper ubet of X i alo an edge- cut. If X i a minimal edge-cut of a connected graph G, then G X contain exactly two component. A graph G i k-edge-connected (k ) ) if it ha at leat two vertice and no et of at mot k edge eparate G. The edge-connectivity connectivity,, of a graph G i the minimum cardinality of an edge-cut of G if G i nontrivial, and (K ) = 0. Graph Theory 3 Graph Theory 3 6 Example : A graph G i -edge-connected iff it i connected, ha at leat two vertice and contain no bridge. = 0 iff G i diconnected or trivial. = iff G i connected and contain a bridge. Example 3: It i often eay to determine the connectivity of a given graph. If j n then (P j ) = (K n ) = (P j ) = (C n ) = (K n ) = n (K j,n ) = (C n ) = (K j,n ) = j = for any graph G? K j x K j = = j Graph Theory 3 7 Graph Theory 3 8

3 Theorem 4.. (R. A. Brualdi,, J. Cima,, 99): Let G be a connected graph of order n 3 that i not complete,, For each edge-cut X there exit a vertex-cut W of G uch that W X. W.l.o.g.. we can uppoe that X i a minimal edge-cut of G.. Then G X i diconnected, which contain exactly two component, ay G and G with order n and n, repectively.. Then n + n = n. We have two cae: Cae : Every vertex of G i adjacent to every vertex of G in G. Then X = n n. Since (n( )(n ) 0,, it follow that n n n +n = n and o X n. Since G i not complete,, it contain two non-adjacent vertice u and v.. Then W = V {u,v}} i a vertex-cut of cardinality n, and o W < X. Graph Theory 3 9 Cae : There are vertice u in G and v in G that are not adjacent in G. For each edge e way: X,, we elect a vertex for W in the following If u i incident with e,, then chooe the other vertex (in G ) incident with e for W. Otherwie, elect for W the vertex that i incident with e and belong to G. Now W X. Since u,v V(G W), and G W contain no u v path, o W i a vertex-cut cut. Graph Theory 3 0 G: u G : G : v Theorem 4.. (H. Whitney, 93) : If G i a non-trivial graph then If we delete all the edge incident with a vertex, the graph become diconnected, o the econd inequality hold. Let u ee the other inequality: If G i a complete graph then = = G. If then =. Suppoe G i not complete and = k and {x y, x y,, x k y k } i a et of edge diconnecting G. If G {x, x,, x k } i diconnected then k. If G {x, x,, x k } i connected then each vertex x i ha degree at mot k.. Deleting the neighbour of x, we diconnect G.. Hence k. Graph Theory 3 Graph Theory 3 3

4 Theorem 4.. (H. Whitney, 93) : If G i a non-trivial graph then If we delete all the edge incident with a vertex, the graph become diconnected, o the econd inequality hold. Let u ee the other inequality: If G i a complete graph then = = G. If then =. Suppoe G i not complete and = k edge-cut cut.. Let X = k be an Then by the Theorem 4.. there exit a vertex-cut U uch that U X. Thu U X = Graph Theory 3 3 A graph G with =, = 3 and = 4. Graph Theory 3 4 Theorem 4.3.(G. Chartrand, F. Harary, 968 ): Let G be a graph of order n,, and let k be an integer that k n.. If deg v n + k for every v V, then G i k-vertex-connected. Suppoe that the theorem i fale. Then there i a graph G atifying the hypothei of the theorem uch that G i not k-vertex-connected.. Since G i not a complete graph. Then there exit a vertex-cut W uch that W = l So, the graph G W i diconnected of order n l. k. Let G be a component of G W of mallet order, ay n. Thu n Let v be a vertex of G. Necearily, v i adjacent in G only to vertice of W or to other vertice of G. Hence contrary to the hypothei. n l n l deg v l + ( n ) l + n + l n + k 3 = Graph Theory 3 Graph Theory 3 6 4

5 Theorem 4.4.: A nontrivial graph G i k-edge-connected iff there exit no nonempty proper ubet W of V uch that the number of edge joining W and V W i le than k. A: Aume that there exit no nonempty proper ubet W of V for which the number of edge joining W and V W le than k but that G i not k-edge connected. Since G i nontrivial, thi implie that there exit l edge, 0 < l < k, uch that their deletion from G reult in a diconnected graph H. Let H be a component of H. The number of edge joining V(H ) and V W i at mot l, where l < k, and thi i a contradiction. Graph Theory 3 7 B: Converely, uppoe that G i a k-edge-connected graph. If there hould exit a ubet W of V uch that j edge, j < k, join W and V W, then the deletion of thee j edge produce a diconnected graph, which i again a contradiction. The following theorem of Plenik give a ufficient condition for equality to hold between and. Graph Theory 3 8 Theorem 4.. (Plenik( Plenik, 97): If G i a graph of diameter,, then =. Let S be an edge-cut with S =,, and let H and H be the two component of G S,, where n i i the order of H i (i=,)) with n n. Since diam G =, every vertex of H i adjacent to ome vertex of H. So either each vertex of H i adjacent to ome vertex of H, or each vertex of H i adjacent to ome vertex of H. Therefore = S min{n,n } = n. For u V(H ), let d i (u) denote the number of vertice of H i (i=,) adjacent to u in G. Graph Theory 3 9 Graph Theory 3 0

6 Thu deg G u = d (u) + d (u), and o δ( G ) degg u = d( u ) + d(u ) n + d( u ) λ( G ) + d (u ) δ(g ) + d(u ). Therefore, d (u) for each u V; that i, each vertex of H i adjacent to ome vertex of H. Let V(H ) = {u, u,,u }. Then n n λ( G ) S = d ( ui ) = d ( ui ) + d ( un = i= i= n + d ( un ). ),u n Corollary 4.6.: If G i a graph of order n deg u + deg v n, uch that for each pair of u,v of nonadjacent vertice in G,, then = Proof If deg u + deg v n,, then u and v have at leat one common neighbour. So, the diameter of the graph i and we can apply the theorem 4.. and o n + d (u n ) Thu =. n + d (u ). n Graph Theory 3 Graph Theory 3 Exercie.. (G. Chartrand and L. Leniak page 74.). Determine the connectivity complete k-partite graph. and edge-connectivity of each. Show that every k-connected graph contain every tree of order k+ a a ubgraph. 3. Let H = G+K, where G i k-connected. Prove that H i (k+)- connected. 4. Prove the Corollary Let a,b and c be poitive integer with a b c. Prove that there exit a graph G with = a, = b, and = c. Graph Theory 3 3 A nontrivial graph i connected if between every two ditinct vertice of G there exit at leat one path. Thi fact can be generalized in many way,, mot of which involve, either directly or indirectly,, a theorem due to Menger. In thi ection we dicu the major one of thee, beginning with Dirac` proof (966)) of Menger` theorem itelf. We recall to the definition of eparating ubet of vertice: A et S of vertice (or( edge) of a graph G i aid to eparate two vertice u and v if the removal of the element of S from G produce a diconnected graph in which u and v lie in different component. Certainly then, S i a vertex-cut (edge-cut)) of G. In the next graph there i a et S = {w{,w,w 3 } of vertice that eparate the vertice u and v.. No et with fewer than three vertice eparate u and v.. A i gauranteed by Menger` theorem, tated next,, there are three internally dijoint u v path in G. Graph Theory 3 4 6

7 4.. Menger' Theorem u w w v A et S of vertice (or( edge) of a graph G i aid to eparate two vertice u and v if the removal of the element of S from G produce a diconnected graph in which u and v lie in different component. Certainly then, S i a vertex-cut (edge-cut)) of G. Two u v path are independent if they have only the vertice u and v in common. Theorem 4.7. (Menger' theorem,, 97): w 3 Let u and v be ditinct non-adjacent vertice of a graph G.. Then the minimum number of vertice eparating u from v i equal to the maximum number of independent u v path. A graph illutrating the Menger' theorem. Graph Theory 3 We denote by S k (u,v) the tatement that the minimum number of vertice that eparate u and v i k. Graph Theory 3 6 The reult i true if u and v lie in different component of G or if u and v lie in different block of G; So, we may aume that the graph under conideration are connected and that u and v lie in the ame block. If S k (u,v) i true, then the maximum number of internally dijoint u v path in G i at mot k. Thu, if k = then the reult i true. Suppoe that the theorem i fale. Then there exit a mallet poitive integer t ( ) uch that S t (u,v) i true in ome graph G but the maximum number of internally dijoint u v path i le than t. Among all uch graph G of mallet order, let H be one of minimum ize. We now etablih three propertie of the graph H. Graph Theory 3 7 Lemma 4.8.: For every two adjacent vertice v and v of H,, where neither v nor v i u or v,, there exit a et of U of t vertice of H uch that U {v i }, i =,,, eparate u and v. Firt, we will how that if e = v v then S t (u,v) i fale for H e. To prove thi we claim that S t- (u,v) i true for H e. If not then there exit a et W of vertice that eparate u and v in H e,, where W t. Then W eparate u and v in both H v and H v. So W {v i }, i=, eparate u and v in H, which i impoible. Thu, a claimed, S t- (u,v) i true in H e.. So, there exit a et U which eparate u and v in H e,, where U = t. However, then U {v i }, i =,,, eparate u and v in H. Graph Theory 3 8 7

8 H: v t vertice Lemma 4.9. For each vertex w ( u, v ) in H,, not both uw and vw are edge of H. u v v Suppoe that it i not true. Then S t- (u,v) i true for H w. Then, however, H w contain t internally dijoint u v path. So, H contain t internally dijoint u v path, which i a contradict- ion. S t (u,v) i true in H, but the maximum number of internally dijoint u v path i le than t. NINCS KÉSZ Graph Theory 3 30 Graph Theory 3 9 Lemma 4.0.: If W={w,w,,w t } i a et of vertice which eparate u and v in H,, then either uw i E(H) for all i ( i t ) or vw i E(H) for all i ( i t ). Suppoe that the tatement i fale. Define H u a the ubgraph induced by the edge on all u w i path in H that contain only one vertex of W. Define H v imilarly. We will how that V(H u )V(H v )=W. Suppoe that the above tatement i not true. Then both H u and H v have order at leat t+. * Define H to conit of H u, a new vertex v * u together with all edge v * w i. * Similarly, we define H to conit of H v, a new vertex u * v together with all edge v * w i. Graph Theory 3 3 H u * H v * Oberve that and have maller order than H. * H u So, S t (u,v) i true in and S t (u,v) i true in. Therefore, there exit t internally dijoint u v * * path in and t internally dijoint u * * v path in. H v * H v Thee t path produce t internally dijoint u v path in H. Thi contradict the condition that H i among thoe of graph for which the number of internally dijoint path i le then t. Graph Theory 3 3 H u 8

9 Now, we can return to prove the tatement of the Menger' theorem: Let P be a u v path in H of length d(u,v). By Lemma 4.9. d(u,v) 3. Thu we may write P: u,u,u,,v,v where u,u v. By Lemma 4.8.,., there exit a et U of t vertice uch that both U {u } and U {u } eparate u and v. Since v i not adjacent to u, therefore every vertex of U {u } i adjacent to u. Since u i not adjacent to u, therefore every vertex of U {u } i adjacent to v. Thu d(u,v) ) =, which i a contradiction. Theorem 4.. (Whitney' characterization): A graph G of order n i k-(vertex)connected ( k n ) ) iff for each pair u,v of ditinct vertice there are at leat k internally dijoint u v path in G. A: Let G be a k-connected graph. Aume to the contrary, that there are two vertice u and v uch that the maximum number of internally dijoint u v path in G i l,, where l < k. k If uv E then, by Theorem 4.7 contrary to hypothei. l < k, k which i If uv E, then the maximum number of internally dijoint u v path in G uv i l < k. k Therefore there exit a et U of fewer than k vertice uch that G uv U i a diconnected graph. Graph Theory 3 33 Graph Theory 3 34 > Therefore,, at leat one of G (U { {u}) and G (U { {v}) i diconnected, implying that < k. k Thi alo produce a contradiction. B: Converely, uppoe that G i a nontrivial graph that i not k- connected but in which every pair of ditinct vertice are connected by at leat k internally dijoint path. Certainly, G i not complete. Since G i not k-connected, k < k. Let W be a et of vertice of G uch that G W i diconnected,, and let u and v be in different component of G W. The vertice u and v are necearily nonadjacent; however,, by the hypothei,, there are at leat k internally dijoint u v path. By the Menger' theorem, u and v cannot be eparated by fewer than k vertice, o a contradiction arie. Graph Theory 3 3 Theorem 4..: If G i a k-connected graph and v,v,v,,v k are k+ ditinct vertice of G,, then there exit internally dijoint v v i path ( i k). Contruct a new graph H from G by adding a new vertex u together with the edge uv i, i =,,,k. Since G i k-connected, H i k-connected. By Theorem 4.,, there exit k internally dijoint u v path in H. The retriction of thee path to G yield the deired internally dijoint v v i path. One of the intereting propertie of -connected graph i that every two vertice of uch graph lie on a common cycle. There i a genaralization to k-connected graph by Dirac: Graph Theory

10 Theorem 4.3.: Let G be a k-connected graph, k. Then every k vertice of G lie on a common cycle of G. For k = the reult follow from the Theorem.; hence we aume that k 3. Let W be a et of vertice of G.. Among all cycle of G,, let C be a cycle containing a maximum number, ay l,, vertice of W. It i clear that l. We will to how that l = k. Aume to the contrary, that l < k, and let w be a vertex of W which doe not lie on C. Graph Theory 3 37 Lemma 4.4.: C contain at leat l+ vertice. Suppoe the contrary. Then C can be labeled o that C:w,w,,w l,w, where w i W for i l. l By Theorem 4..,., there exit internally dijoint w w i path, denoted by Q i, i l. Replacing the edge w w on C by the w w path determined by Q and Q. So, we get a cycle containing at leat l+ vertice of W,, which i a contradiction. Therefore, C contain at leat l+ vertice. Graph Theory 3 38 Thu, we may aume that C contain vertice w,w,,w l,w l+, uch that w i W for i l. and w l+ W Since k l+,, we may apply Theorem 4. again to conclude that there exit l+ internally dijoint w w i path, ay P i, i l+. Let v i be the firt vertex on P i that belong to C (poible v i =w i ) and let P denote the w v i ubpath of P i. i Since C contain exactly l vertice of W,, there are ditinct integer and t,,t l+, uch that one of the two v v t path, ay P, determined by C contain no interior vertex belonging to W. Replacing P by the v v t path determined by P i, we obtain a cycle of G containing at leat l+ vertice of W. Thi contradiction erve the deired reult that l = k. k Both the Menger theorem and the Whitney characterization have edge -analogue: Theorem 4..(Ford.(Ford and Fulkeron,, 96): Let u and v be ditinct vertice of G.. Then the minimal number of edge eparating u from v i equal to the maximal number of edge-dijoint dijoint u v path. Corollary 4.6. : A graph i k-edge-connected (k ) ) iff it ha at leat two vertice and any two vertice can be joined by k edge- dijoint path. Graph Theory 3 39 Graph Theory

11 With the aid of the earlier theorem it i eay to prove an edge analogue of the Menger theorem. Graph Theory 3 4 Application of the connectivity. If we think of a graph a repreenting a communication network, the connectivity become the mallet number of communication tation whoe breakdown would jeopardie communication in the ytem. The higher the connectivity and edge connectivity, the more reliable the ytem. Corollary 4.7. : A graph i k-edge-connected (k ) ) iff it ha at leat two vertice and any two vertice can be joined by k edge- dijoint path. The higher the connectivity and edge connectivity, the more expen- ive i the ytem. So, the tree network which one can obtained by Krukal' algorithm i cheap, but it i not very reliable. Thi lead to conider the following generaliation of the panning ng tree problem. Graph Theory 3 4 The Reliable Communication Network Problem: Let k be given poitive integer and let G be a weighted graph. Determined a minimum-weight k-connected panning ubgraph of G. If k = then thi problem reduce to the panning tree problem. If k then the problem i unolved and i known to be difficult. If G i a complete graph in which edge i aigned unit weight, then the problem ha a imple olution. For a weighted complete graph on n vertice in which each edge i aigned unit weight,, a minimum-weight m-connected panning ubgraph i an m-connected graph on n vertice with a few edge a poible. We hall denote by by f(m,n) the leat number of edge that an m- connected graph on n vertice can have. (It i clear that m < n.) n From the Handhaking Lemma and the Whitney' Connectivity Theorem it follow immediately that mn f (m,n). We hall prove that equality hold by contructing an m-connected graph H m,n on n vertice that ha exatly mn/ edge. The tructure of the contruction depend on the paritie of n and m. Graph Theory 3 43 Graph Theory 3 44

12 Cae. m i even.. Let m = r.. Then H r,n i contructed a follow. It ha vertice 0,,,n- and two vertice i and j are joined if i r j i + r (where addition i taken modulo n.) Cae. m odd, n even.. Let m = r +. Then H r+,n contructed by firt drawing H r,n and then adding edge joining vertex i to vertex i+(n/) for i n/. Cae 3. m odd, n odd.. Let m = r +. Then H r+,n i contructed by firt drawing H r,n then adding edge joining vertex 0 to vertice (n )/ and (n + )/ and vertex i + (n+)/ for i (n )/ Theorem 4.7. (Harary( Harary,, 96): The graph H m,n i m-connected. H 4,8 If i = 0 then j = 6,7,,. If i = then j = 7,0,,3. H,8 H,9 Graph Theory 3 4 Graph Theory Flow and Connectivity Let G be a (finite) directed graph with vertex et V and edge et E. We will tudy (tatic) flow in G from a vertex (the ource) ) to a vertex t (the ink). A flow f i a non-negative negative function defined on the edge; the value f(xy) i the flow or current in the edge xy. For notational implicity we hall write f(x,y) and a imilar convention will be ued for other function. The only condition a flow from to t ha to atify the Kirchoff' current law: the total current flowing into each intermediate vertex (that i different from and t) i equal to the total current leaving the vertex. More formally: : if for x V then we denote by + Γ ( x ) = ( y V : xy E ), Γ ( x ) = ( y V : yx E ), are the outgoing neighbour, are the ingoing neighbour then a flow from to t atifie the following condition: for each x V {,tv,t}. f ( x,y ) = + y Γ ( x ) z Γ ( x ) f ( z,x) Graph Theory 3 47 Graph Theory 3 48

13 Since 0 = f ( x, y ) f ( z, x ) x V {,t } y Γ ( x ) z Γ ( x ) + What i the maximal flow value from to t while we have capacity contraint on edge which retrict the current through the edge? If xy E, then c(x,y) 0 i the capacity of the edge. = f ( z,u ) f ( u, z ) u {,t } z Γ ( u ) y Γ ( u ) + We hall aume that the current flowing through the edge cannot be more than the capacity c(x,y). xy o we find that f (, y ) f ( y, ) = f ( y,t ) + + y Γ ( ) y Γ ( ) y Γ ( t ) y Γ ( t ) f ( t, y ) The common value, denoted by v(f), i called the value of f or the amount of flow from to t. Graph Theory 3 49 Given two ubet X,Y of V we write E( X,Y ) for the et of directed X Y edge: E( X,Y ) = { xy : x X, y Y }. Whenever g:e i a function,, we put g( X,Y ) = g( x, y ) Graph Theory 3 0 e E ( X,Y ) If S i a ubet of V containing but not t then E( S,S ) i called a cut eparating from t. The capacity of a cut E ( S,S ) i denoted by c(s,s ). Some eay tatement: If we delete the edge of a cut then no poitive-valued valued flow from to t can be defined on the remainder. 6 t Converely, if F i a et of edge after whoe deletion there i no flow (v(f)) = 0) 0 from to t then F contain a cut. A cut with capacity Graph Theory 3 Graph Theory 3 3

14 The capacity of a cut i at leat a large a the value of any flow: min S i cut Do thee value exit? c( S,S ) max v( f ) v i flow Since there are only finitely many cut, there i a cut whoe capacity i minimal. The exitence of a flow with maximal value i lighly le trivial. For every flow f we know that and o v = up v(f) ) <. v( f ) xy E c( x, y ) Let f,f, be a equence of flow with lim n v(f n ) = v. Then, by paing to a ubequence, we may aume that for each xy E the equence (f n (x,y)) i convergent, ay to f(x,y). The function f i a flow with value c,, that i a flow with maximal capacity. In thi way one can how that even if ome of the edge have infinite inite capacity, there i a flow with maximal value which can be either finite or infinite. Graph Theory 3 3 Graph Theory 3 4 Theorem 4.8. (Max( Max-flow min-cut theorem.) The maximal flow value from to t i equal to the minimum of the capacitie of cut epatating from t. We have already een that there i a flow f with maximal value, ay v,, and the capacity of every cut i at leat v. So, in order to prove the theorem we have to how that there i a cut with capacity v. We will contruct the cut in hand. Define a ubet S V recurively a follow: Let S. If x S and c(x,y) ) > f(x,y) or f(y,x) ) > 0 then y S. We claim that E(S,S ) i a cut eparating from t with capacity v = v(f). Firt, we will prove that S i a cut, i.e. t Suppoe the contrary: t belong to S. Becaue of the recurive definition there i a equence of vertice = x 0, x, x,, x k = t uch that ε max c(x,x ) f(x,x ),f(x,x ) for every i, 0 i k-. Let = min i. Then f can be augmented to a flow f * in the following way: if i > f(x i+,x i ) then increae the flow in x ix i+ by. if f(x,x i ) then decreae the flow in x by. i f(x i+,x S. { } 0 i = i i+ i i+ i+ i > x i + i Graph Theory 3 Graph Theory 3 6 4

15 Clearly, f * i a flow,, and it value i v(f * ) = v(f) ) + the maximality of f. So, t S and therefore E(S,S ) eparating from t. i a cut, eparating, contradicting We know that v(f) i equal to the value of the flow from S to S and it i defined in the obviou way: f ( x, y ) f ( x, y ) x S,y S x S,y S By the definition of S the firt um i exactly x S,y S c( x, y ) = c( S,S ) and each ummand in the econd um i zero.. Hence c ( S,S ) = v( f ), a required. Graph Theory 3 7 The max-flow min-cut theorem remain valid if ome of the edge have infinite capacity but the maximal flow value i finite. The proof of the previou theorem alo provide a urpriingly efficient algorithm for finding a flow with maximal value if the capacity function i integral: f 0 ( x,y ) = 0 xy E Suppoe we have contructed f i, and we find the et S belonging to f i : If t S i a maximal flow,, the algorithm terminate. f i If t S f i can be augmented to a flow f i+ by increaing the flow along a path from to t (like in the theorem). Since v(f i ) i integer, o v(f i- ) < v(f i ), and the equence mut end in at mot x,yc(x,y c(x,y) tep. Graph Theory 3 8 Theorem 4.9. (Integrality( Theorem) : If the capacity function i integral then there i a maximal flow which i alo integral. It i important that the algorithm doe not tate uniquene: : it find one of the maximal flow (uually there are many) ) and the theorem claim that one of maximal flow i integral. If intead of one ource and one ink we take everal of each, the max-flow min-cut theorem remain valid. H G t Let u cetch the proof: If,,, k are the ource and t,t,,t l are the ink then E(S,S ) i a cut if i S and t j S for every i,j,, i k, j l. ource ink Graph Theory 3 9 Graph Theory 3 60

16 Theorem 4.0. : The maximum of the flow value from a et of ource to a et of ink i equal to the minimum of the capacity of cut eparating the ource from the ink. What happen if we have capacity retriction on the vertice, except the ource and the ink? We are given a function c: V {,tv,t} + and every flow f from to t ha to atify the following inequality: f ( x,y ) = + y Γ ( x ) z Γ ( x ) f ( z,x ) c( x ) for every x V {,t} () 6 (4) 6 v () 6 () () (0) 3 v 3 (0) v (0) 4 () (3) (0) () 3 v 6 () (0) 6 v 8 () (4) 4 t v v 4 () 3 (4) 4 v 7 We can extend the original theorem to thi ituation a well. Graph Theory v ( f ) = f ( ) = f ( t ) = 6 Graph Theory 3 6 Let a = (x,y) be an arc in a directed graph G and, c(a) be the capacity of thi edge and let f(a) be the flow in the arc. Then we call a f-zero f-poitive f-unaturated f-aturated The Labelling Algorithm if if if if f(a) ) = 0, f(a) ) > 0, f(a) ) < c(a), f(a) ) = c(a). How can we decide that a given flow i a maximum flow or not? Let G be a directed graph and let f be a flow in G. With each h path P in the graph we aociate a non-negative negative integer (P) defined by ε( P ) = min ε( a ) a A( P ) where ε( a ) c( a ) = f ( a ) f ( a ) if a if a i a forward arc of i revere arc of (P) i the larget amount by which the flow along P can be increa- ed without violating the condition f(a) c(a). The path P i aid to be f-aturated if (P) > 0. P P (P) = 0 an f-unaturated if An f-incrementing path i an f-unaturated path from the ource to the ink t. Graph Theory 3 63 Graph Theory

17 The exitence of an f-incrementing path P in a network i ignificant ince it implie that f i not a maximum flow. By ending an additional flow (P) along P one obtain a new flow f' defined by f ( a ) + ε( P ) if a i forwarded arc of P f = f ( a ) ε( P ) if a i revere arc of P f ( a ) otherwie for which v(f ) ) = v(f) ) + (P). () 6 v 3 (4) 4 (0) 4 () () (0) () 3 (3) v v () 3 v () 4 v () t (3) 6 v 3 (4) 4 () 4 (3) () (0) (0) 3 (3) v v () 3 v () 4 v () t Theorem 4.. A flow in a directed graph i a maximum flow iff the t graph contain no f-incrementing path. f-incrementing path P. (,v ) = (v 3,v ) = (v,v ) = (v 3,t) = 4 revied flow baed on P. (P) = Graph Theory 3 6 Graph Theory 3 66 The algorithm i ready: Staring with a known flow, for intance the zero flow, it recurively contruct a equence of flow of increaing value, and terminate with a maximum flow. How can we find an f-incrementing path? L. R. Ford and D. L. Fulkeron (97): The Labelling Method. J. Edmond and R. M. Karp (97): Firt-labelled Firt-canned canned Labelling Method A tree T in a directed graph i an f-unaturated tree if x V(T) for every vertex v of T,, the unique (,v) path in T i an f-unatur- ated path. () 6 v 3 (4) 4 (0) 4 () () (0) () 3 (3) v v () 3 v () 4 v An f-unatrated tree. t () Graph Theory 3 67 Graph Theory

18 Initially, T conit of jut the ource. At any tage, there are two way in which the tree may grow: If there exit an f-atureted arc a in (S,S ), where S = V(T), then both a and it head are adjoined to T. If there exit an f-poitive arc a in (S,S ), then both a and it tail are adjoined to T. The reulting T either reache the ink t, or top growing before reaching t (breakthrough). The earch for an f-incrementing path involve growing an f-unat- urated tree in G. In the event of breakthrough, the (,t(,t)-path in T i our deired f-in- crementing path. If T top growing before reaching t,, then f i a maximum flow. The labelling procedure: Aign to the ource the label l() ) =. If a i an f-unaturated arc whoe tail u i already labelled but whoe head v i not, then v i labelled l(v) ) = min{l(u l(u), c(a) f(a)}. If a i an f-poitive arc whoe head u i already labelled but whoe tail v i not, then v i labelled l(v) ) = min{l(u l(u), f(a)}. In the above cae v i aid to be labelled baed on u. Graph Theory 3 69 Graph Theory 3 70 To can a labelled vertex u i to label all unlabelled vertice that can be labelled baed on u. The labelling procedure i continued until either the ink t i labelled (breakthrough) or all labelled vertice have been canned and no more vertice can be labelled (implying that f i a maximum flow). The original Labelling Method i not an efficient procedure, ince the number of earching an incrementing path may be equal with the t maximum value of the capacity function on E. The Edmond Karp improvement ue during the canning tep a firt-labelled firt-canned canned bai: before canning a labelled vertex u can the vertice that were labelled before u. f := f ' find a revied flow f ' baed on P while Y f := L := {} S := l() ) := vertex u L S can u L := L L(u) t L S := S { {u} L: et of labelled vertice. S: et of canned vertice. L(u): et of vertice labelled during canning of u. Graph Theory 3 7 Stop Y L S = Graph Theory 3 7 N 8

19 () 6 v 3 (4) 4 (0) 4 () () (0) () 3 v(3) () 3 v () 4 v v () t (3) 6 v 3 (4) 4 () 4 (3) () (0) (0) 3 (3) v v () 3 v () 4 v () t A flow can be interpreted to a flow in a vertex a well, namely from the part where all the current enter it to the part where all the current leave it. We can turn each vertex of G into an edge in uch a way that that any current entering (and leaving) ) the vertex will be forced through the edge. Replace each vertex x V {,t} by two vertice, ay x - and x +, end each incoming edge to x - and end each outgoing edge to x +. Finally, add edge from x - to x + with capacity c(x -,x + ) = c(x). Theorem 4..: Let G be a directed graph with capacity bound on the vertice other then the ource and the ink t.. Then the minimum of the capacity of a vertex-cut i equal to the maximum of the flow value from to t. Graph Theory 3 73 Graph Theory 3 74 Now,, we are ready to give a new proof for the Menger theorem. G x x - x + u - H Two t path are independent if they have only the vertice and t in common. Theorem 4.3. (Menger' theorem): y u z t y - y + u + z - z + t Let and t be ditinct non-adjacent vertice of a graph G.. Then the minimum number of vertice eparating from t i equal to the maximum number of independent t path. Let and t be ditinct vertice of G.. Then the minimal number of edge eparating from t i equal to the maximal number of edge- dijoint t path. () Let () Let Replacing the vertice by edge Graph Theory 3 7 Graph Theory

20 () Replace each edge xy of G by two directed edge, xy and yx, and give each vertex other than and t capacity. By the Theorem the maximal flow value from to t i equal to the minimum of the capacity of a cut eparating from t. By the Integrality Theorem there i a maximal flow with current or 0 in each edge. Therefore the maximum flow value from to t i equal equal to the maximal number of independent t path. The minimum of the cut capacity i clearly the minimal number of vertice eparating from t. () Proving the econd tatement of the theorem one can proceed a in (), except intead of retricting the capacity of the vertice, give each directed edge capacitx The toughne of a graph If G i a noncomplete graph and t i a nonnegative real number uch that t S / k(g S) for every vertex-cut S of G,, then G i defined t- tough. (Here k mean the number of component of a graph G.) If G i t-toughtough and i a nonnegative real number uch that < t, t then G i alo -tough. The maximum real number t for which a graph G i t-toughtough i called the toughne of G and denoted by t. Since complete graph do not contain vertex-cut cut, thi definition doe not apply to uch graph. We define t(k n ) = + for every poitive integer n. Graph Theory 3 77 Graph Theory 3 78 Simple tatement: The toughne of a noncomplete graph i a rational number. t = 0 iff G i diconnected. If G i noncomplete graph then t ) = min S /k(g S), where the minimum i taken over all vertex-cut S of G. The toughne of a graph G how the meaure that how tightly the ubgraph of G are held together. The maller the toughne the more vulnerable the graph i: a -tough graph ha the property that breaking the graph into k component require the removal of at leat k vertice. Breaking a -tough graph into k component require the removal of at leat k vertice. Graph Theory 3 79 a b c d u v S ={u,v,w}, S ={w}, S 3 ={u,v} } are three (of many) vertex-cut S /k(g S ) = 3/8, S /k(g S ) = /3, S 3 /k(g S 3 ) = /7 A graph of toughne /7. Graph Theory 3 80 w x z y 0

21 A parameter that play an important role in the tudy of toughne i the independence number: Two nonadjacent vertice in a graph are aid to be independent. A et S of vertice i independent if every two vertice of S are independent. The vertex independence number ) of a graph G i the maximum cardinality among the independent et of vertice of G. Example: (K r, r, ) = max {r,} (C n ) = n/ (K n ) =. The independence number i related to the toughne in the ene that among all the vertex-cut S of a noncomplete graph G,, the maximum value of k(g S) ) i, o for every vertex-cut S of G,, we have that S and k(g S). Theorem 4.4. (Chvatal( Chvatal,, 973): For every noncomplete graph G, t. According to the imple tatement earlier we get S t = min k(g S). Graph Theory 3 8 Graph Theory 3 8 Let S' be a vertex-cut with S' S' =. Thu k(g S'),, o S S t = min. k(g S) k(g S) Theorem 4.. (Matthew and Summer, 984): If G i a noncomplete claw-free graph then t =. Let F be a graph. A graph G i F-free if G contain no induced ubgraph iomorphic to F. Example: : a K -free graph i empty. A K,3 -free graph i called a a claw-free graph. Graph Theory 3 83 Graph Theory 3 84