Factor

1. Factor • Factor: a spanning subgraph of graph G • k-Factor: a spanning k-regular subgraph • Odd component: a component of odd order • o(H): the number of odd components of H

2. 1-Factor of K6

3. 2-Factor of K7

4. Theorem 3.3.3 • A graph G has 1-factor if and only if o(G-S)<=|S| for every SV(G). () Suppose G has 1-factor. Consider a set SV(G). every odd component of G-S has a vertex matched to one vertex of S.  o(G-S)<=|S| since these vertices of S must be distinct.

5. Theorem 3.3.3 (2/8) () 1. Let G’=G+e. Then o(G’-S)<= o(G-S)<=|S| for every SV(G).  G’ satisfies Tutte’s condition. 2. If G’ has no 1-factor, then G has no 1-factor. 3. Let S=. Then o(G)<= ||.  n(G) is even. 4. Suppose that G satisfies Tutte’s condition, G has no 1-factor and adding any missing edge to G yields a graph with 1-factor. 5. Let U be the set of vertices in G that have degree n(G)-1.

6. Theorem 3.3.3 (3/8) 6. Case 1: G-U consists of disjoint complete graphs. 7. The vertices in each component of G-U can be paired in any way, with one extra in the odd components. 8. Let S=U.  o(G-U)<=|U|.  We can match the leftover vertices to vertices of U since each vertex of U is adjacent to all of G-U.

7. Theorem 3.3.3 (4/8) 9. The remaining vertices are in U, which is a clique.  G has 1-factor since n(G) is even.  A contradiction.

8. Theorem 3.3.3 (5/8) 10. Case 2: G-U is not a disjoint union of cliques. 11. G-U has two vertices x,z at distance 2 with a common neighbor y (Exercise 1.2.23a) 12. G+xz has 1-factor and G+yw has 1-factor. 13. Let M1 be 1-factor in G+xz, and let M2 be 1-factor in G+yw.

9. Theorem 3.3.3 (6/8) 14. xzM1 and ywM2 since G has no 1-factor.  xzM1-M2 and ywM2-M1.  xzF and ywF, where F= M1M2. 15. Each vertex of G has degree 1 in each of M1 and M2, every vertex of G has degree 0 or 2 in F.  The components of F are even cycles and isolated vertices. 16. Let C be the cycle of F containing xz. 17. If C does not also contain yw, then the desired 1-factor consists of the edges of M2 from C and all of M1 not in C.  F has 1-factor avoiding xz and yw.  G has 1-factor.  Another contradiction.

10. Theorem 3.3.3 (7/8) 18. Suppose C contains both yw and xz. 19. Starting from y along yw, we use edge of M1 to avoid using yw. When we reach {x,z}, we use zy if we arrive at z; otherwise, we use xy. In the remainder of C we use the edge of M2. C+{xy,yz} has 1-factor avoiding xz and yw. We have 1-factor of G by combing with M1 or M2 outside V(C).  Also a contradiction.

11. Theorem 3.3.3 (8/8) 20. There exists no simple graph G that satisfies Tutte’s condition and has no 1-factor such that adding any missing edge to G yields a graph with 1-factor.  For any graph G that satisfies Tutte’s condition and has no 1-factor, there exists an edge e such that adding e to G yields a graph that has no 1-factor. 21 Suppose that G satisfies Tutte’s condition and has no 1-factor.  There exists an edge e such that G+e has no 1-factor. By 1, G+e satisfies Tutte’s condition.  Kn(G) has no 1-factor by repeating the same argument.  It is a contradiction since n(G) is even.

12. Join • Join: The join of simple graphs G and H, written GH, is the graph obtained from the disjoint union G+H by adding the edges {xy: xV(G), yV(H)}

13. Corollary 3.3.7. • The largest number of vertices saturated by a matching in G is minSV(G){n(G)-d(S)}, where d(S)=o(G-S)-|S|. 1.Given SV(G), at most |S| edges can match vertices of S to vertices in odd components of G-S, so every matching has at least o(G-S)-|S|=d(S) unsaturated vertices.  The least number of vertices unsaturated by a matching in G is greater than or equal to maxSV(G){d(S)}. 2. Let d= maxSV(G){d(S)}. 3. Let S=. We have d>=0. 4. Let G’=GKd.

14. Corollary 3.3.7. 5. d(S) has the same parity as n(G) for each S  n(G’) is even.  o(G’-S’)<=|S’| for S’=. 6. If S’ is nonempty but does not contain V(Kd), then G’-S’ has only one component.  o(G’-S’)<=1<=|S’|. 7. When S’ contains V(Kd), let S=S’-V(Kd).  G’-S’=G-S.  o(G’-S’)= o(G-S)<=|S|+d=|S’|. 8. G’ satisfies Tutte’s Condition by 5, 6, and 7.  G’ has a perfect matching.  G has a matching with at most d unsaturated vertices.  The least number of vertices unsaturated by a matching in G is d.  The largest number of vertices saturated by a matching in G is n(G)-d= minSV(G){n(G)-d(S)}.

15. Corollary 3.3.8 • Every 3-regular graph with no cut-edge has a 1-factor. 1. Given SV(G), let H be a odd component, and let m be the number of edges from S to H. 2. The sum of the vertex degrees in H is 3n(H)-m.  m is odd since both 3n(H)-m and n(H) are odd.  m>=3 since G has no cut-edge. 3. Let p be the number between S and the odd components of G-S.  p<=3|S| since G is 3-regular and p>=3o(G-S) since m>=3.  o(G-S)<=|S|.  G has 1-factor by Thorem 3.3.3.

16. Example 3.3.10 • Consider the Eulerian circuit in G=K5 that successively visits 1231425435. The corresponding bipartite graph H is on the right. For the 1-factor in H whose u,w-pairs are 12,43,25,31,54, the resulting 2-factor in G is the cycle (1,2,5,4,3). The remaining edges forms another 1-factor in H, which corresponds to the 2-factor (1,4,2,3,5) in G.

17. Theorem 3.3.9 • Every regular graph with positive even degree has a 2-factor. 1. Let G be a 2k-regular graph with vertices v1,…,vn.  Every component in G is Eulerian, with some Eulerian circuit C. 2. For each component, define a bipartite graph H with vertices u1,…,un and w1,…,wn by adding edge (ui, wj) if vj immediately vi follows somewhere on C.  H is k-regular because C enters and exists each vertex k times. 3. H has a 1-factor M by Corollary 3.1.13.  The 1-factor in H can be transformed into a 2-regular spanning subgraph of G.