k-Factor

1. k-Factor • Factor: a spanning subgraph of graph G • k-Factor: a spanning k-regular subgraph 1-Factor of K6

2. k-Factor 2-Factor of K7

3. Odd Component • Odd component: a component of odd order • o(H): the number of odd components of H Odd component o(H)=4 Even component

4. Tutte’s Condition • A graph G has 1-factor if and only if o(G-S)<=|S| for every SV(G)

5. S Proof of Tutte’s Condition () 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. Odd component Even component

6. Proof of Tutte’s Condition (⇐) Let S=. Then o(G)<= ||. 1.  n(G) is even. 2. Let G’=G+e.  G’ satisfies Tutte’s condition. Then o(G’-S) <= |S| for every SV(G). <= o(G-S) e e Odd component S Even component

7. Proof of Tutte’s Condition (⇐) 3. Let U be the set of vertices in G that have degree n(G)-1. 4. Two cases are discussed: (Case 1) G-U consists of disjoint complete graphs and (Case 2) G-U is not a disjoint union of cliques (complete graph).

8. Proof of Tutte’s Condition (⇐) 5. Case 1: G-U consists of disjoint complete graphs. 6. The vertices in each component of G-U can be paired in any way, with one extra in the odd components.  We can match the leftover vertices to vertices of U since each vertex of U is adjacent to all of G-U. 8. The remaining vertices in U can be matched because U is a clique and n(G) is even. 7. Let S=U.  o(G-U)<=|U|.  G has 1-factor.

9. Proof of Tutte’s Condition (⇐) 9. Case 2: G-U is not a disjoint union of cliques. 10. It suffices to show if G satisfies Tutte’s condition and G has no 1-factor, there exists an edge e such that G+e has no 1-factor. 11. Suppose 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. Now, let’s show 10.

10. Proof of Tutte’s Condition (⇐) 12. 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. 13. G-U has two vertices x,z at distance 2 with a common neighbor y not in U. 14. G-U has another vertex w not adjacent to y since y is not in U. 15. G+xz has 1-factor and G+yw has 1-factor. 16. Let M1 be 1-factor in G+xz, and let M2 be 1-factor in G+yw. 17. 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.

11. Proof of Tutte’s Condition (⇐) 18. 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. 19. Let C be the cycle of F containing xz. 20. 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.  A contradiction. : in M1 – M2 : in M2 – M1 : in both (hence not in F) C

12. Proof of Tutte’s Condition (⇐) 21. Suppose C contains both yw and xz. 22. 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).  Another contradiction.

13. Theorem 3.3.9 • Every regular graph with positive even degree has a 2-factor. • Let G be a 2k-regular graph with vertices v1,…,vn.  Every component in G is Eulerianwith some Eulerian circuit C by Theorem 1.2.26. 2. For each component, define a bipartite graph H with vertices u1,…,un and w1,…,wn by adding edge (ui, wj) if vj immediately follows vi somewhere on C. 8 1 5 10 6 2 9 7 4 3

14. Theorem 3.3.9 3. H is k-regular because C enters and exists each vertex k times. 4. H has a 1-factor M by Corollary 3.1.13. 5. The 1-factor in H can be transformed into a 2-regular spanning subgraph of G.

15. 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)}

16. Berge-Tutte Formula • 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|. • 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. • Every matching has at least maxSV(G){d(S)} unsaturated vertices.  Every matching has at most minSV(G){n(G)-d(S)} saturated vertices.  The largest number of vertices saturated by a matching in G is equal to minSV(G){n(G)-d(S)}. or less than

17. Berge-Tutte Formula 2. Let d=maxSV(G){d(S)}. We need to show there exists a matching in G that has exactly d unsaturated vertices. 3. Let S=. We have d>=0. 4. Let G’=GKd. 5. If G’ has a perfect matching, then we obtain a matching in G having at most d unsaturated vertices from a perfect matching in G’, because deleting the d added vertices eliminates edges that saturate at most d vertices of G. It suffices to show G’ satisfies Tutte’sCondition.

18. Berge-Tutte Formula  n(G’) is even. 6. o(G’-S’)<=|S’| for S’=. 7. o(G’-S’)<=|S’| if S’ is nonempty but does not contain V(Kd). 8. o(G’-S’)<=|S’| if S’ contains V(Kd).  d has the same parity as n(G).  d(S) = o(G-S)-|S| has the same parity as n(G) for each S. o(G-S) |S| d(S) n(G) odd odd odd even even odd even even even even odd odd odd odd even even  G’-S’ has only one component. Let S=S’-V(Kd).  G’-S’=G-S.  o(G’-S’)= o(G-S)= d(S)+|S|<=|S|+d=|S’|.