Balanced Network Flows

Transcription

1 revied, June, 1992 Thi paper appeared in Bulletin of the Intitute of Combinatoric and it Application 7 (1993), Balanced Network Flow William Kocay* and Dougla tone Department of Computer cience Univerity of Manitoba Winnipeg, Manitoba CANADA R3T 2N2 BKOCAY@ccu.umanitoba.ca Abtract Let G be a imple, undirected graph. A pecial network N, called a balanced network, i contructed from G uch that maximum matching and f-factor in G correpond to maximum flow in N. A max-balancedflow-min-balanced-cut theorem i proved for balanced network. It i hown that Tutte Factor Theorem i equivalent to thi network flow theorem, and that f-barrier are equivalent to minimum balanced edgecut. A max-balanced-flow algorithm will olve the factor problem. 1. Balanced Network. Let G be a imple graph, directed or undirected, with vertex et V(G) and edge et E(G). A network N i a directed graph which contain two pecial vertice and t, the ource and target, repectively, and in which every edge e i aigned a poitive integervalued capacity cap(e). Terminology for graph, network, and flow i taken from Bondy and Murty [1]. Edge of a directed graph are ordered pair of vertice. If (u,v) i an edge, we indicate that u i adjacent to v by u v. Edge of an undirected graph are unordered pair, and we write the pair {u,v} a uv, where the order i unimportant. In a directed graph we alo often ue uv to indicate one of the edge (u,v) or (v,u), epecially if the direction i not explicitly given. The oppoite direction will then be given by vu. Let f be an integervalued function on E(N). Given any u V(N), the out-flow at u i f + (u) = f(uv) and the v, u v in-flow at u i f (u) = f(vu). The function f i called a flow if it atifie the two v, v u condition: i) f + (u) = f (u), for all vertice u,t, (the conervation condition) ii) 0 f(uv) cap(uv), for all edge uv (the capacity contraint). The value of the flow i val(f) = f + () f (), that i, the net out-flow at the ource. We are intereted in a pecial kind of network, called a balanced network. * Thi work wa upported by an operating grant from the Natural cience and Engineering Reearch Council of Canada. 1

2 1.1 Definition. A balanced network i a network N with the following propertie: The vertice of N conit of a ource, a target t, and two et of vertice X={x 1,x 2,...,x n } and Y={y 1,y 2,...,y n }; N contain the pair of edge (,x i ) and (y i,t) for 1 i n; The remaining edge of N are between X and Y, and occur in pair, either (x i,y j ) and (x j,y i ), or (y i,x j ) and (y j,x i ), where i j; cap(x i )=cap(y i t), cap(x i y j )=cap(x j y i ), and cap(y i x j )=cap(y j x i ), for all i,j. The vertice of N occur in complementary pair. Given any vertex u V(N), the complementary vertex will be denoted by u. Thu =t, t =, x i = y i, and y i = x i. The edge alo occur in complementary pair: (x i,y j ) =(x j,y i ) and (,x i ) =(y i,t). Notice that (u,v) =(v,u ). Complementary edge alway have the ame capacity, cap(uv)=cap(v u ). A balanced flow in N i a flow f in which every complementary pair of edge carrie the ame flow, that i, f(uv)=f(v u ). We want to find a maximum balanced flow in N. A can be een from the following example, a balanced network i bipartite. The network in Fig. 1 ha cap(x i y j )=1 and cap(x i )=cap(y i t)=d i. x 1 y 1 d 1 d 1 x 2 y 2 d 2 d 2 d 3 d 3 t d 4 x 3 y 3 d 4 N x 4 y 4 Fig 1. A balanced network. Edge capacitie are d 1, d 2, d 3, d 4 or 1. Before proceeding further, we explain the motivation for introducing balanced network and balanced flow. Given a graph G, conider the following problem. Max Matching: Find a maximum matching in G; Factor Problem: Given an integer-valued function b on V(G), doe G have a ubgraph H uch that deg H (u) = b(u), for all u V? If G were a bipartite graph, we could olve thee problem with the tandard technique of flow theory. Conider the factor problem. Let the bipartition of G be (X,Y). Create a network N from G a follow. Direct the edge of G from X to Y and aign them a capacity of 1. Add a ource and target t, plu all edge (,x) and (y,t), for all x X and y Y. Aign 2

3 capacitie b(x) and b(y) to x and yt, repectively. Find a maximum flow f in N. It i eay to ee that the ubgraph H exit if and only if the value of f i val(f) = b(x) = b(y). x X y Y To olve the maximum matching problem, jut et b(u)=1 for all vertice u, and find a maximum flow. When G i not bipartite, tandard flow theory doe not apply. However, Tutte Theorem [1,6,10] give neceary and ufficient condition for a perfect matching to exit and can be ued to obtain a formula for the ize of a maximum matching. Tutte Factor Theorem [6,8,9,10] olve the econd problem. In [10], Tutte how that finding an f-factor in G i equivalent to finding a 1-factor in a coniderably larger graph formed by tranforming G. The purpoe of balanced flow i to how that thee problem can in fact be olved by the method of flow theory. It i hown that Tutte factor theorem i equivalent to a min-max flow theorem, the Max-Balanced-Flow-Min-Balanced-Cut Theorem. In term of network flow, an f-barrier i equivalent to a minimum balanced edge-cut. Thee reult are impler to derive than the tandard tructure theory of f-factor (cf. Lovaz and Plummer [6, ection 10.2]), ince the method of network flow and edge-cut apply. In [7] chrijver tate that combinatorial min-max relation often yield elegant theorem and are cloely related to the algorithmic olution to a problem. The correponding algorithm here i the max-balancedflow algorithm [4]. Thi algorithm can alo be ued to olve the capacitated b-matching problem (ee [7]). 2. Balanced Edge-Cut. Let N be a network. Let V(N) and let denote V(N). The edge-cut K=[,] conit of all edge uv uch that u and v. Let K denote [,]. Write f + () = f(uv) and f () = f(uv). uv K uv K By the conervation condition f + ()=f () unle or t i in. We hall aume that and t, unle otherwie indicated. Then val(f)=f + () f (), for all uch. The capacity of the edge-cut i cap(k) and the total flow on K i f(k), where cap(k) = cap(uv) and f(k) = f(uv) = f + () uv K uv K By tandard flow theory, val(f) cap(k) for all flow f and all edge-cut K. The Max-Flow- Min-Cut Theorem tate that in any network, the value of a maximum flow f equal the value of a minimum edge-cut K. Furthermore, the flow f atifie f(uv)=cap(uv) for all uv K, and f(uv)=0 for all uv K. Thee reult alo hold for arbitrary flow and arbitrary edge-cut in a balanced network, of coure. However ince we want a balanced flow in N, we mut retrict attention to certain 3

4 pecial kind of edge-cut. Given an arbitrary edge cut K=[,] in a balanced network N, we divide the vertice of X and Y into 4 et: A = {x i,y i x i, y i } B = {x i,y i x i, y i } C = {x i,y i x i, y i } D = {x i,y i x i, y i } A may then be further ubdivided into A x = A X, and A y = A Y, and the other 3 et may alo be ubdivided in the ame manner. Thi i illutrated in Fig. 2. Ax Ay Bx By Cx Cy t Dx Dy Fig. 2, An edge-cut K. Edge illutrated belong to K. We know that val(f) cap(k) in general, but for balanced network, thi tatement i not a trong a it could be. With arbitrary network one can alway find a flow f uch that val(f)=cap(k) for ome edge cut K, by the method of augmenting path. But with balanced network, it i not alway poible to find a balanced flow f uch that val(f)=cap(k). It would therefore be ueful to find a pecial kind of edge cut K in a balanced network N, uch that K ha propertie in N that are analogou to thoe of the uual kind of edge-cut in an arbitrary network. The ubgraph of N induced by C i denoted N[C]. If C Ø, N[C] will be a graph with one or more connected component. Let C 1, C 2,, C k be it component. Let K i denote thoe edge of K with one endpoint in C i, for i=1,2,,k, and let K r denote the remaining edge of K, that i, thoe with no endpoint in C. We alo define K i to be thoe edge of K with one 4

5 endpoint in C i, and K r to be the remaining edge of K. 2.1 Definition. A balanced edge-cut in N i an edge-cut K that atifie the following condition: i) There are no edge between C and D; ii) If u C i, then u C i ; iii) if C Ø, then cap(k i ) i odd, for 1 i k. It will be een that there i alway an edge-cut of thi type correponding to a maximum balanced flow. The following notation i now required. Any given component C i will be denoted by I. Thu I x denote C i C x and I y denote C i C y. Let E AB denote all edge of N directed from A to B. It can be divided into thoe edge directed from A x to B y and thoe directed from A y to B x. In mot application, N will contain no edge directed from Y to X at all, but it i eay to include them in the following proof. Denote the former et by E+ AB and the latter by E AB. Then EAB =E + AB E AB. (In mot cae EAB=Ø.) Write XAB =cap(e AB ) and F AB =f(e AB ). Then a above we can write X AB = X + + AB + X AB and F AB = F AB + F AB. By + + complementarity, F AB = F BA and F AB = F BA. If EA denote all edge directed from to A, then X A denote cap(e A ) and F A denote f(e A ). Thi notation i extended to the et B,C,D, and I in the obviou way. 2.2 Lemma. Let K be a balanced edge-cut. For any component C i, (F + II F II) + + (FIA F IA) + + (FIB F IB) FI = 0. Proof. ince the C i are connected component of N[C] and there are no edge from C to D, the only edge directed out of I x are thoe from I x to I y, A y, and B y. Hence f + (I x ) = F II F + IA + F + IB. The only edge directed in to I x are thoe from I y, A y, B y, and. Hence f (I x ) = F II + FIA + F IB + F I. Taking the difference give f + (I x ) f (I x ) = 0, by the conervation condition. + + A cap XIA flow F IA B t cap X flow F I I Ix I = C i Iy edge in K i = [,] edge in K i = [,] other edge Fig. 3, The edge K i and K i. 5

6 2.3 Lemma. Let f be a balanced flow in N, and let K be a balanced edge-cut. If f(k i )=0 for ome C i, then there i an edge e K i uch that f(e)<cap(e). Proof. Referring to Fig. 3, we ee that f(k i )=F BI + F + + AI and that f(k i ) = F IB + F IA + F It. By Lemma 2.2, we can add the expreion (F + II F II) + + (FIA F IA) + + (FIB F IB) FI to f(k i ) without changing it value. Thi give f(k i ) = (F + II F II) + + FIA + 2F + IB F IB, ince FIt =F I by complementarity. If f(k i )=0 then F BI = + FAI = 0, o that f(k i ) = (F + II F II) + + 2FIB. But F + II and F II are even quantitie, ince e E II if and only if e E II. o f(k i ) i alo even. But by the definition of K, each cap(k i ) i odd. Therefore f(k i )<cap(k i ). It follow that each K i contain an edge e for which f(e)<cap(e). Thi how that for each et K i, either f(k i )>0, in which cae K i contain an edge e with f(e)>0, or ele f(k i )=0 and K i contain an edge e for which f(e)<cap(e). 2.4 Theorem. Let C 1, C 2,, C k be the connected component of C in the balanced edgecut K=[,]. Then for any balanced flow f, val(f) cap(k) k. Proof. By Lemma 2.3, one of cap(k i ) f(k i ) and f(k i ) i at leat 1, for each C i. Therefore (cap(k) f(k)) + f(k) k, which implie that cap(k) (f(k) f(k)) k. But f(k) f(k) = f + () f () = val(f). Therefore val(f) cap(k) k. The capacity of a balanced edge cut, denoted balcap(k), i therefore defined to be cap(k) k, where k i the number of connected component of N[C]. A minimum balanced edge-cut i a balanced edge-cut uch that no balanced edge-cut ha maller capacity. We have therefore proved: 2.5 Corollary. The value of a balanced flow in a balanced network i le than or equal to the capacity of every balanced edge-cut, that i, val(f) balcap(k), for all balanced f and K. Thi hold alo when f i maximum and K i minimum. In order to prove that the value of a maximum flow actually equal the capacity of a minimum balanced edge-cut, we need to conider augmenting path. 3. The Max-Flow-Min-Cut Theorem in Balanced Network. In general, a flow f in a network N i maximum if and only if there i no augmenting path in N. Becaue of thi theorem, algorithm can ue augmenting path to find a maximum flow. We need a imilar theorem when N and f are balanced. uppoe that P=(v 1,v 2,,v m ) i a path in a balanced network N, that i, P ha no repeated vertice, and conecutive vertice v i and v i+1 are adjacent. Edge can appear in both direction on P, o that either v i v i+1, in which cae v i v i+1 i a forward edge of P; or ele v i+1 v i, and v i v i+1 i a backward edge of P. Let e be an edge on P. The reidual capacity of e i defined in term of the direction in which it i travered: 6

7 recap(e) = cap(e) f(e), if e i a forward edge, f(e), if e i a backward edge. An edge e i aturated if recap(e)=0. Otherwie it i unaturated. Thi depend, of coure, on the direction of traveral. A path P i unaturated if all it edge are unaturated. The complementary path of P i defined to be P =(v m,, v 2, v 1 ). The reidual capacity of an augmenting path P i δ(p)= min recap(e). Clearly e ha the ame reidual capacity on P a e P e ha on P, o δ(p )=δ(p). If P i a uv-path, that i, if it begin at u and end at v, then P i a v u -path. If P i a vv - path, then P i alo a vv -path. Thi i of particular ue when P i an augmenting tpath, ince then P i alo an augmenting t-path. If we begin with a balanced flow f, and augment on P and on P a well, then every edge in N will carry the ame flow a it complementary edge, o that f will till be balanced. Thi i illutrated in Fig. 4. x 1 y 1 1(0) 1(1) x 2 y 2 1(0) 1(1) 1(0) 1(0) t 1(1) x 3 y 3 1(1) x 4 N y 4 aturated edge unaturated edge Fig. 4, Complementary augmenting path P=x 1 y 2 x 4 y 3 t and P =x 3 y 4 x 2 y 1 t. uppoe that P i an augmenting path. Then P i alo an augmenting path. However we may not alway be able to augment on both P and P. It i ometime the cae that augmenting on P detroy the augmenting path P. 3.1 Lemma. Given an augmenting path with reidual capacity δ, we can augment the flow on P and P by at leat one unit (but not necearily by δ unit), if and only if P doe not contain a pair of complementary edge with a reidual capacity of one. Proof. uppoe firt that P doe contain uch a pair of complementary edge. Then P ha the form P=( uv v u t), where recap(uv)=1. When we augment on P, the reidual 7

8 capacity of uv and v u on P become zero, o that P i no longer an augmenting path in N. Now uppoe that P doe not have thi form, and let e be an edge of P. If e i not on P, then augmenting on P doe not change it flow, o it reidual capacity on P remain δ. If e i on P, then recap(e ) 2 before augmenting, o that if we augment by one unit, recap(e ) on P change by one, o we can till augment on P. We mut therefore enure that any algorithm we ue to find a maximum flow in a balanced network N conider an unaturated t path to be an augmenting path only if it doe not contain a pair of complementary edge with a reidual capacity of one. 3.2 Definition. An unaturated uv-path which contain a pair of complementary edge with reidual capacity 1 i called an invalid path. Any other unaturated uv-path i called a valid path, and only a valid t-path i conidered to be an augmenting path in N. A vertex v i -reachable if there i a valid path from to v, and t-reachable if there i a valid path from v to t. Let N and f be balanced, and uppoe that there i no valid augmenting path in N. Let be the et of all -reachable vertice and conider the edge-cut K=[,]. K ha the form illutrated in Fig. 2. The vertice can be partitioned into the 4 et A,B,C, and D. A before, let C i denote the connected component of N[C], for i=1,2,,k, and let K i, K r, K i, and K r be a defined in ection Lemma. Every edge of K r K r i aturated from to. Proof. Let (u,v) be an unaturated edge from to, o that u and v. Then there i a valid u-path, but no valid v-path. Therefore the edge (uv), and hence the vertex u, mut be on every valid path to u. ince u, there i at leat one uch path, o both u and u are -reachable; thi mean that u and u are in C, o uv i in ome K i or K i. 3.4 Lemma. There are no edge between C and D. Proof. Firt notice that D, o no vertex in D i -reachable. uppoe that uch an edge e=(x i,y j ) exited, a in Fig. 5. Then e =(x j,y i ) i alo uch an edge. Either thee two edge carry flow or they do not. If they do, then x j y i i unaturated from to. But then x j would be in, not. If they do not carry flow, then x i y j i unaturated form to, o y j would be in, again a contradiction. x i y C i x j D y j Fig. 5, Edge between C and D. 8

9 3.5 Lemma. Each C i conit entirely of pair of complementary vertice. Proof. The proof i illutrated in Fig. 6. Let P be any valid path from to w C i, and uppoe that P firt enter C i along the edge uv. Then u, o every valid path to v goe through v, and uch a path exit becaue v C. But if Q i the ub path that i a vv -path, then Q i alo a vv -path. Thu every vertex on both Q and Q i in C, and therefore in C i, o Q Q C i. But if w C i then there i a path W connecting either v or v to w, ince C i i connected, o ay v. Then W connect w to v, o w i alo in C i. W u v Q Q v u W C i w w Fig. 6, Complementary vertice in C i. 3.6 Lemma. At leat one edge i unaturated from to in each K i K i, for 1 i k. Proof. The proof i illutrated in Fig. 6 and 7. Take an arbitrary component C i, and any valid path P from to a vertex in C i. Let uv be the firt edge on P where v C i. Then (uv) =v u i unaturated becaue uv i. If uv i a forward edge on P then v u i an unaturated edge from to, and v u K i. If uv i a backward edge, then f(uv)>0, ince it i unaturated. But then v u K i. In either cae, K i K i contain an unaturated edge. A B t C i edge in K i = [,] edge in K i = [,] edge into C i Fig. 7, Edge of K i K i 3.7 Theorem. Let v C i be a vertex uch that there i a valid v-path which firt enter C i at v. Then every w C i can be reached on a valid path Q w from v, where Q w C i. 9

10 Proof. By contradiction. Refer to Fig. 8 and 9. Let C* be the larget non empty et of pair of complementary vertice in C i, with the property that every vertex w C* i reachable from v on a valid path. The path Q and Q of Lemma 3.5 are in C*, o C* Ø. C * u v v u Q w w w q q R R z z C i Fig. 8, C* and C i, v P. Now uppoe that C* C i, o there i ome edge wz uch that w C* and z C i C*; thu w C* and z C i C*. If wz were aturated, then zw, and hence w z, would be unaturated. Thu either wz or w z i an unaturated edge from C* to C i C*, o uppoe it i wz. z C i, o there i a valid z -path P. uppoe firt that v P. v C* and z C*, o let P leave C* for the lat time from a vertex q, and let R be the portion of P from q to z. If wz i on R, then q=w, z R, and the ubequent portion of R i a zz -path. Then it complement would alo be a zz -path. But then we can then enlarge C* with the vertice of R R, a contradiction. Otherwie wz i not on R. Let Q w be a valid path from v to w, contained in C*. R i a valid path becaue R i. The concatenation Q w wzr i a valid path becaue if it were not, then R would contain an edge e uch that either e =wz or e i on Q w. But R contain no edge of Q w, ince Q w C*, and wz i not on R. But then Q w wzr i a valid path and we can again enlarge C* with the vertice of R R, a contradiction. Otherwie v P. Then Pz w i a valid path into C*. Let q be the firt vertex at which P enter C* (maybe q =w ), and let P q be the portion of P up to q. Then P q Q q i a valid v - path that doe not contain v. Thi i a contradiction, o C*=C i, a required. C * u v v u Q Q q q q q Pq C i Fig. 9, C* and C i, v P. 10

11 3.8 Lemma. At mot one edge i unaturated, with reidual capacity at mot 1, from to, in each K i K i, for 1 i k. Proof. If any edge e of K i K i were unaturated from to, with recap(e)>1, then a vertex in would be -reachable, a contradiction. By lemma 3.6 there i alway at leat one unaturated edge v u, o uppoe there i a econd, ay z w, a hown in Fig. 10. Then if R i a valid v-path and Q i valid vz -path contained in C i, which both exit, then RQ would be a valid path to z that doe not ue the edge wz, a contradiction. R u v Q v u w z z w C i Fig. 10, A component C i. Thi prove that there i exactly one edge in each K i K i, for 1 i k, that i unaturated from to. But by lemma 3.3, every edge in K r K r i aturated from to, o we have proved: 3.9 Corollary. Let K=[,], where i the et of all -reachable vertice, a above. Then val(f) = cap(k) k Lemma. cap(k i ) i odd, for 1 i k. Proof. By lemma 3.6 and 3.8, either f(k i )=cap(k i ) and f(k i )=1, or f(k i )=cap(k i ) 1 and f(k i )=0. Referring to the proof of lemma 2.3, we have f(k i )=F BI + + FAI and f(k i ) = (F + II F II) + F + IA + 2F + IB F IB. In the firt cae, f(ki )=F BI + + FAI =1, o one of F + AI and F BI i 0 and the other i 1. But then f(k i ) = cap(k i ) = (F + II F II) + + FIA + 2F + IB F IB, which i an odd number, ince + FII and F II are even. In the econd cae, f(ki )=F BI + + FAI =0, o that f(k i ) = cap(k i ) 1 = (F + II F II) + + i even. It follow that cap(k i ) i odd. 2F IB 3.11 Corollary. K i a balanced edge-cut. Proof. There are no edge between C i and C j, for any i and j, ince the C i are connected component, o by lemma 3.4, 3.5, and 3.10, K i a balanced edge-cut, and balcap(k)=cap(k) k. But val(f)=cap(k) k, by corollary 3.9, o by theorem 2.4, f i a maximum balanced flow in N, and K i a minimum balanced edge-cut. We therefore have: 3.12 Theorem. Let N be a balanced network with a balanced flow f. Then f i maximum if and only if there i no valid augmenting path in N. Proof. If there i a valid augmenting path, then by lemma 3.1, f i not maximum. If there i no valid augmenting path, then by corollary 3.11, the et of -reachable vertice define a 11

12 balanced edge-cut K=[,] for which val(f)=cap(k) k, o that f i maximum, by theorem 2.4. The above reult can be ummarized in a theorem. Max-Balanced-Flow-Min-Balanced-Cut Theorem. Let N be a balanced network. The value of a maximum balanced flow equal the capacity of a minimum balanced edge-cut. Proof. By theorem 2.4 and corollarie 3.9 and Thee reult mean that in a balanced network N, we can begin with the zero flow and ucceively augment the flow on complementary augmenting path until no valid augmenting path remain. At that point the flow will be maximum, and the et of -reachable vertice will define a balanced edge-cut. uch an algorithm i developed in Kocay and tone [4]. One reult i that maximum matching, f-factor, and capacitated b-matching can all be found by a ingle flow algorithm. When the flow i maximum, a balanced edge-cut will be given by the et of -reachable vertice. 4. The Factor Problem. Given the function b(v) of ection 1, a ubgraph H uch that deg H (v)=b(v) i called a factor of G. (Uually H i called an f-factor, ince the function b(v) i uually denoted f(v), but we are uing f to denote a flow.) Given arbitrary et A,B V(G), let A,B tand for the number of edge with one end in A and one in B. Let,T V(G). Write odd(,t) for the number of component C of G ( T) uch that T,C + b(v) i odd. Tutte Factor v C Theorem [10] can be tated a follow. 4.1 Factor Theorem. G i without a factor if and only if there i a ubet of V(G) and a ubet T of V uch that v b(v) < odd(,t) + (b(v) deg G (v) ). v T We how how thi can be derived from the Max-Balanced-Flow-Min-Balanced-Cut Theorem. Let V(G)={v 1, v 2,, v n }. Write 2ε= b(v). Create a balanced network N with vertice v V(G) X={x 1, x 2,, x n } and Y={x 1, x 2,, x n }, a well a and t. N ha edge (x i,y j ) and (x j,y i ) for every edge v i v j of G, and all edge (,x i ) and (y i,t) for all vertice v i of G. o G i a homomorphic image of the [X,Y]-edge of N. The capacitie are cap(x i )=cap(y i t)=b(v i ), and cap(x i y j )=cap(x j y i )=1. Given any balanced flow f in N, the flow-carrying edge of [X,Y] define a ubgraph H of G uch that deg H (v i )=f(x i ). 4.2 Lemma. G ha a factor H if and only if balcap(k) 2ε, for all balanced edge-cut K. Proof. The number of edge of H i twice val(f). 12

13 Let f be a maximum balanced flow, and let K be the correponding balanced edge-cut which define the et A,B,C, and D in N. Let A*, B*, C*, D* be the correponding et of V(G). C 1, C 2,, C k denote the connected component of N[C], and each K i ha odd capacity, by We ue the horthand notation A = b(v), and o forth, for the et B, C, D. v A* 4.3 Lemma. balcap(k)= A B + 2ε + deg G A *(v) k. v B* Proof. Referring to Fig. 2, balcap(k) = 2 A + B,B + B,C + B,D + C + D k. Notice that 2ε= A + B + C + D and that B,B + B,C + B,D = deg G A *(v). The v B* reult follow. 4.4 Proof of the Factor Theorem. uppoe that G i without a factor. Then there i a balanced edge-cut uch that balcap(k)<2ε. By lemma 4.3, thi give A B + deg G A * (v) k < 0. Each K i ha odd capacity. Notice that cap(k i )= B,C i + b(v). v B* Therefore, k odd(a*,b*). It follow that b(v) < odd(a*,b*) + (b(v) deg G A *(v) ). v A* v B* Converely, uppoe that b(v) < odd(a*,b*) + v A* v B* et A* and B* of V(G). Form a balanced edge-cut by taking C* to conit of all component v C i * (b(v) deg G A *(v) ) for dijoint of G (A* B*) uch that B,C + b(v) i odd. D* contain the remaining vertice of G. It v C* i eay to verify that thi determine a balanced edge-cut K of N for which balcap(k)<2ε. Thu we have proved the Factor Theorem with balanced network. In the cae that no factor exit in G, the maximum balanced flow algorithm actually find the et A* and B*; they are part of the edge-cut formed by the et of all -reachable vertice. That i, it either find the factor or find the two et that prove that no factor exit. uppoe that edge of G may be ued more than once by edge of H, o that H i a multigraph, with maximum multiplicity m allowed. Then balcap(k)= A B + 2ε + m deg G A * (v) k, o that G i without a factor H if and only if there are et A * and B* v B* b(v) < odd m (A*,B*) + uch that (b(v) m deg G A *(v) ), v A* v B* where odd m (A*,B*) i the number of component C* of G (A* B*) uch that m B,C* + b(v) i odd. v C* 13

14 When m i allowed to approach, o that edge may be ued any number of time in H, we obtain Tutte olubility Theorem [10]. In thi cae, [B,B]=[B,C]=[B,D]=Ø, for otherwie balcap(k) would be infinite. o G i not oluble if and only if there exit dijoint et A*,B* V(G) uch that A < B + k, where k i the number of component C* of G (A* B*) uch that b(v) i odd. v C* The Erdö-Gallai condition [2] for the exitence of a imple graph with precribed degree equence and precribed maximum edge-multiplicity can be derived directly from the maxbalanced-flow-min-balanced-cut theorem. imilarly it can be ued to find a formula for the number of edge in a maximum matching in an arbitrary graph. In ummary, Tutte Factor Theorem can be viewed a a min-max theorem of flow theory. The theory of f-barrier [6,9,10] can be viewed a the theory of edge-cut in balanced network. A a reult, the theoretical and algorithmic technique of flow theory can be ued for olving thee kind of problem. Balanced network provide a implification of the theory of f-factor, f-barrier, and capacitated b-matching by placing them into the context of network flow. Reference 1. J.A. Bondy and U..R. Murty, Graph Theory with Application, American Elevier Publihing Co., New York, P. Erdö and T. Gallai, Graph with precribed degree of vertice, (Hungarian), Mat. Lapok 11, 1960, pp Frank Harary, Graph Theory, Addion-Weley Publihing Company, Reading, Maachuett, William Kocay and Dougla tone, An algorithm for balanced flow, 1992, ubmitted. 5. Lazlo Lovaz, Three hort proof in graph theory, JCT(B) 19, 1975, pp L. Lovaz and M.D. Plummer, Matching Theory, Annal of Dicrete Mathematic 29, North-Holland, Amterdam, A. chrijver, Min-max reult in combinatorial optimization, in Mathematical Programming the tate of the Art, Ed. Grötchel, Korte, pringer Verlag, Berlin, W.T. Tutte, The factorization of linear graph, J. London Math. oc. 22, 1947, , reprinted in elected Paper of W.T. Tutte, Ed. D. McCarthy and R.G. tanton, Charle Babbage Reearch Centre, Canada, W.T. Tutte, The factor of graph, Can. J. Math., 4, 1952, pp , reprinted in elected Paper of W.T. Tutte, Ed. D. McCarthy and R.G. tanton, Charle Babbage Reearch Centre, Canada, W.T. Tutte, A hort proof of the factor theorem for finite graph, Can. J. Math., 6, 1954, pp , reprinted in elected Paper of W.T. Tutte, Ed. D. McCarthy and R.G. tanton, Charle Babbage Reearch Centre, Canada,