5.7.2 A Maximum flow algorithm Lemma 5.3: Let f be a conservation flow, E(P,V-P) be a cut. If Vf=C(P,V-P)， then Vfmax=Vf ，Cmin(P,V-P)=C(P,V-P). Proof: By the theorem 5.23, Theorem 5.23: For every conservation flow f and any cut E(P,V-P), the result holds: Vf C(P,V-P).

Frod,Falkerson 1956 1）We construct a initial conservation flow in N(V,E,C) Generally, we set fij0=0 for every edge (i,j) of N. The conservation flow is called zero flow. 2）We shall construct an increasing sequence of flows f 1, f 2,…, f n, that has to terminate in a maximal flow. How do we construct the increasing sequence?

Let u be an undirected path from s to t, (1)When u is a directed path from s to t, if fij<cij for every edge of the path, then we change fij for every edge of the path, which equals min{cij-fij} 1)Label s with (-,Δs), where Δs=+∞ 2)Suppose that vertex i is labeled, Let j be an adjacent vertex of i, and no labeled. If fij<cij，then j is labeled (i+, Δj), where Δj = min{Δi，cij- fij} 3)If t is labeled, then an increasing flow is constructed. We change fij to fij +Δt for every edge of the path u.

In the path (s,b,c,t) from s to t, edge (c,b) is reverse order .

Suppose that vertex b is labeled, If fcb>0，then c is labeled (b-,Δc), where Δc = min{Δb,fcb} If t is labeled, then an increasing flow is constructed. We change fij to fij +Δt when (i,j)E, if (i,j)E then we change fji to fji -Δt.

0) Construct a initial conservation flow in N(V,E,C). 1) Label s with (-,+∞). U={x|x is an adjacent vertex of s} 2)Suppose that vertex i is labeled, and j is no labeled, where jU. U=U-{j} i) If (i,j)E and fij<cij， then { j is labeled (i+, Δj), where Δj = min{Δi，cij- fij}, U=U∪{x|x is an adjacent vertex of j}. goto 3) } ii)If (j,i)E and fji>0，then {j is labeled (i-, Δj), where Δj = min{Δi，fji}. U=U∪{x|x is an adjacent vertex of j} } If j is not labeled, then goto 4) 3)If t is labeled then { We change fij to fij +Δt . if j is labeled with i+. If j is labeled with i-, then fji is changed to fji –Δt goto 1) else goto 2) 4)If U then goto 2) else stop.

Theorem 5.24: The labeling algorithm produces a maximum flow. Proof: P={x|x is labeled when algorithm end},thus V-P={ x|x is not labeled when algorithm end}. By the labeling algorithm, sP and tV-P. Thus E(P,V-P)is a cut. (1) (i,j) E(P,V-P)(i.e. iP. jV-P) fij=cij， (2) (j,i) E (i.e. iP. jV-P) fji=0. By lemma 5.3, the labeling algorithm produces a maximum flow.

5.8 Graph Matching Example: Set of worker assign to a set of task Four tasks are to be assigned to four workers. – Worker 1 is qualified to do tasks B and C – Worker 2 is qualified to do tasks A,C and D – Worker 3 is qualified to do tasks B and D – Worker 4 is qualified to do task A and C. Can all 4 workers be assigned to different tasks for which they are qualified?

Example 2: The Marriage Problem:Given a set of men, each of whom knows some women from a given set of women, under what conditions is it possible for all men to marry women they know? Four men each know some of four women Peter knows Mary and Ann Kevin knows Mary, Ann, Rose and Tina Brian knows Mary and Ann Fred knows Ann Is it possible for all the men to marry women they know? Graph Matching

Definition 36: A matching M in a graph G(V;E) is a subset of the edge set E such that no two edges in M are incident on the same vertex. The size of a matching M is the number of edges in M. For a graph G(V;E), a matching of maximum size is called a maximum matching. M1={e1,e7},M2={e1,e2,e5},M3={e1,e2,e5,e6}and M4= {e1,e2,e7,e8} are matching. M3 and M4 are maximum matching.

Definition 37: If M is a matching in a graph G, a vertex v is said to be M-saturated if there is an edge in M incident on v. Vertex v is said to be M-unsaturated if there is no edge in M incident on v. M1={e1,e7}, M3={e1,e2,e5,e6} M1-saturated : v M1-unsaturated: u M3-saturated:u,v

M={e1,e3,e7,e8}, M1={e1,e3,e5,e6} M1 and M are perfect matching. Definition 38: A matching M of G is perfect if all vertices of G are M-saturated. If G(V1;V2) is a bipartite graph then a matching M of G that saturates all the vertices in V1 is called a complete matching from V1 to V2.

M={{v1,u1},{v2,u2},{v3,u3}} M is a complete matching from V1 to V2, but it is not a complete matching from V2 to V1.

Definition 39: Given a matching M in a graph G, a M-alternating path (cycle) is a path (cycle) in G whose edges are alternately in M and outside of M (i.e. if an edge of the path is in M, the next edge is outside M and vice versa). A M-alternating path whose end vertices are M-unsaturated is called an M-augmenting path.

Theorem 5.25：M is a maximal matching of G iff there is no augmenting path relative to M. Proof: (1) There is no augmenting path relative to M, we prove M is a maximal matching of G . Suppose M and N are matching with |M|<|N|. To see that MN contains an augmenting path relative to M we consider G’ = (V’, MN ) 1≤dG’(v)≤2 Since |M|<|N|, MN has more edges from N than M and hence has at least one augmenting path relative to M contradiction

(2)If M is a maximal matching of G then there is no augmenting path relative to M Assume that there exists an M-augmenting path p. To see that Mp is a matching of G and |Mp|>|M| 1)Mp is a matching of G e1,e2Mp, e1 and e2 are not adjacent. 2) |Mp|>|M| |Mp|=|(M∪p)-(M∩p)|= |(M∪p)|-|(M∩p)| =|M|+|p|-2|(M∩p)| =|M|+1

N(S) ={vV|uS,{u,v}E(G)} Definition 40: Given a bipartite graph G(V1;V2), and a subset of vertices S V, the neighborhood N(S) is the subset of vertices of V that are adjacent to some vertex in S, i.e. N(S) ={vV|uS,{u,v}E(G)} A={v1,v3},N(A)={v2,v6,v4} A1={v1,v4},N(A1)={v2,v6,v4,v3,v5,v1}

Theorem 5. 26: Let G(V1,V2) be a bipartite graph with |V1|=|V2| Theorem 5.26: Let G(V1,V2) be a bipartite graph with |V1|=|V2|. Then a complete matching of G from V1 to V2 is a perfect matching

Theorem 5.27 (Hall's Theorem) Let G(V1; V2) be a bipartite graph with |V1|≤|V2|. Then G has a complete matching saturating every vertex of V1 iff |S|≤|N(S)| for every subset SV1 Example: Let G be a k-regular bipartite graph. Then there exists a perfect matching of G, where k>0. k-regular For AV1,E1={e|e incident a vertex of A}, E2={e|e incident a vertex of N(A)} For eE1, eE2. Thus E1E2. Therefore |E1|≤|E2|. Because k|A|=|E1|≤|E2|=k|N(A)|, |N(A)|≥|A|. By Hall’s theorem, G has a complete matching M from V1 to V2. Because |V1|=|V2|, Thus M is a perfect matching.

Exercise: P314 (Sixth) OR P (Fifth) 7,9,10,11,17,19, 20,21 NEXT Planar Graphs Exercise: P314 (Sixth) OR P (Fifth) 7,9,10,11,17,19, 20,21 1. Let G be a bipartite graph. Then G has a perfect matching iff |N(A)|≥|A| for AV.