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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. - PowerPoint PPT Presentation

5.8 Graph MatchingExample: Set of worker assign to a set of taskFour 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 womenPeter knows Mary and AnnKevin knows Mary, Ann, Rose and TinaBrian knows Mary and AnnFred knows AnnIs 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 : vM1-unsaturated: uM3-saturated:u,v

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={e1,e3,e7,e8},M1={e1,e3,e5,e6} M1 and M are perfect matching.

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 5.25M 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|
(2)If M is a maximal matching of G then there is no augmenting path relative to MAssume that there exists an M-augmenting path p. To see that Mp is a matching of G and |Mp|>|M|1)Mp is a matching of G e1,e2Mp, e1 and e2 are not adjacent. 2) |Mp|>|M||Mp|=|(Mp)-(Mp)|= |(Mp)|-|(Mp)|=|M|+|p|-2|(Mp)|=|M|+1

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) ={vV|uS,{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|. 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 SV1Example: Let G be a k-regular bipartite graph. Then there exists a perfect matching of G, where k>0.k-regular For AV1,E1={e|e incident a vertex of A}, E2={e|e incident a vertex of N(A)}For eE1, eE2. Thus E1E2. Therefore |E1||E2|.Because k|A|=|E1||E2|=k|N(A)|, |N(A)||A|.By Halls theorem, G has a complete matching M from V1 to V2.Because |V1|=|V2|, Thus M is a perfect matching.

5.9 Planar Graphs 5.9.1 Eulers FormulaDefinitions 41: Intuitively, a graph G is planar if it can be embedded in the plane, that is, if it can be drawn in the plane without any two edges crossing each other. If a graph is embedded in the plane, it is called a planar graph.

Definition 42:A planar embedded of a graph splits the plane into connected regions, including an unbounded region. The unbounded region is called outside region, the other regions are called inside regions.

Theorem 5.28(Eulers formula) If G is a connected plane graph with n vertices, e edges and f regions, then n -e+f= 2.Proof. Induction on e, the case e = 0 being as in this case n = 1, e = 0 and f =1 n-e+f=1-0+1=2

Assume the result is true for all connected plane graphs with fewer than e edges,e 1, and suppose G has e edges. If G is a tree, then n =e+1 and f= 1, so the result holds. If G is not a tree, let a be an edge of a cycle of G and consider G-a.Clearly, G-a is a connected plane graph with n vertices, e-1 edges and f-1 regions, so by the induction hypothesis, n-(e-1) + (f- 1) = 2, from which it follows that n -e +f = 2.

Corollary 5.1 If G is a plane graph with n vertices, e edges, k components and f regions, then n-e +f= 1+k.Corollary 5.2: If G is a connected planar simple graph with e edges and n vertices where n 3, then e3n-6.Proof: A connected planar simple graph drawn in the plane divides the plane into regions, say f of them. The degree of each region is at least three(Since the graphs discussed here are simple graphs, no multiple edges that could produce regions of degree two, or loops that could produce regions of degree one, are permitted).The degree of a region is defined to be number of edges on the boundary of this region.We denoted the sum of the degree of the regions by s.

Suppose that K5 is a planar graph, by the Corollary 5.2, n=5,e=10103*5-6=9, contradiction K3,3n=6,e=9, 3n-6=3*6-6=12>9=eBut K3,3 is a nonplanar graph

Corollary 5.3: If a connected planar simple graph G has e edges and n vertices with n 3 and no circuits of length three, then e2n-4.Proof: Now, if the length of every cycle of G is at least 4, then every region of (the plane embodied of) G is bounded by at least 4 edges. K3,3 is a nonplanar graphProof: Because K3,3 is a bipartite graph, it is no odd simple circule.

Corollary 5.4:Every connected planar simple graph contains a vertex of degree at most five.

ProofIf n2 the result is trivial

For n3, if the degree of every vertex were at least six, then we would have 2e=

.

By the Corollary 5.2, we would have 2e6n-12. contradiction.

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Corollary 5.5: Every connected planar simple graph contains at least three vertices of degree at most five, where n3. Next: Characterizations of Planar Graphs Graph Colourings P320

- Exercise: 1. Let G be a bipartite graph. Then G has a perfect matching iff |N(A)||A| for AV.2.Suppose that G is a planar simple graph. If the number of edges of G less than 30, then there exists a vertex so that its degree less than 5.3.Let G be a connected planar graph with (G)3 and f