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

1-Factor of K 6

2-Factor of K 7

Theorem 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.

Theorem (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.

Theorem (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.

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

Theorem (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 a) 12. G+xz has 1-factor and G+yw has 1-factor. 13. Let M 1 be 1-factor in G+xz, and let M 2 be 1-factor in G+yw.

Theorem (6/8) 14. xz  M 1 and yw  M 2 since G has no 1-factor.  xz  M 1 -M 2 and yw  M 2 -M 1.  xz  F and yw  F, where F= M 1  M Each vertex of G has degree 1 in each of M 1 and M 2, 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 M 2 from C and all of M 1 not in C.  F has 1-factor avoiding xz and yw.  G has 1-factor.  Another contradiction.

Theorem (7/8) 18. Suppose C contains both yw and xz. 19. Starting from y along yw, we use edge of M 1 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 M 2. C+{xy,yz} has 1-factor avoiding xz and yw. We have 1-factor of G by combing with M 1 or M 2 outside V(C).  Also a contradiction.

Theorem (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.  K n(G) has no 1-factor by repeating the same argument.  It is a contradiction since n(G) is even.

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)}

Corollary The largest number of vertices saturated by a matching in G is min S  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 max S  V(G) {d(S)}. 2. Let d= max S  V(G) {d(S)}. 3. Let S= . We have d>=0. 4. Let G’=G  K d.

Corollary 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(K d ), then G’-S’ has only one component.  o(G’- S’)<=1<=|S’|. 7. When S’ contains V(K d ), let S=S’-V(K d ).  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= min S  V(G) {n(G)-d(S)}.

Corollary 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 =3o(G-S) since m>=3.  o(G-S)<=|S|.  G has 1-factor by Thorem

Example Consider the Eulerian circuit in G=K 5 that successively visits 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.

Theorem Every regular graph with positive even degree has a 2-factor. 1. Let G be a 2k-regular graph with vertices v 1,…,v n.  Every component in G is Eulerian, with some Eulerian circuit C. 2. For each component, define a bipartite graph H with vertices u 1,…,u n and w 1,…,w n by adding edge (u i, w j ) if v j immediately v i 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  The 1- factor in H can be transformed into a 2-regular spanning subgraph of G.