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. 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. S
Odd component
Even component

6. Proof of Tutte’s Condition (⇐)
 n(G) is even.
2. Let G’=G+e.
 G’ satisfies Tutte’s condition.
Let S=. Then o(G)<= ||.
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 Eulerian with some Eulerian circuit C by Theorem
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
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’s Condition.

18. Berge-Tutte Formula 6. o(G’-S’)<=|S’| for S’=.  n(G’) is even.
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).
 n(G’) is even.
 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’|.