k-Factor

• Factor: a spanning subgraph of graph G

• k-Factor: a spanning k-regular subgraph

1-Factor of K6

k-Factor

2-Factor of K7

Odd Component

• Odd component: a component of odd order

• o(H): the number of odd components of HOdd component

Even component

o(H)=4

Tutte’s Condition

• A graph G has 1-factor if and only if o(G-S)<=|S| for every SV(G)

Proof of Tutte’s Condition () Suppose G has 1-factor. Consider a set SV(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

Proof of Tutte’s Condition (⇐ )

1. n(G) is even.

2. Let G’=G+e.

G’ satisfies Tutte’s condition.

SOdd component

Even component

e

e

Let S=. Then o(G)<= ||.

Then o(G’-S) <= |S| for every SV(G).

<= o(G-S)

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).

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.

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.

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. xzM1 and ywM2 since G has no 1-factor. xzM1-M2 and ywM2 -M1. xzF and ywF, where F= M1M2.

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.

xz

yw xz

yw : in M1 – M2

: in M2 – M1

: in both (hence not in F)

C

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.

Join

• Join: The join of simple graphs G and H, written GH, is the graph obtained from the disjoint union G+H by adding the edges {xy: xV(G), yV(H)}

Berge-Tutte Formula• The largest number of vertices saturated by a matching in

G is minSV(G){n(G)-d(S)}, where d(S)=o(G-S)-|S|.

1. Given SV(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.

or less than

Every matching has at least maxSV(G){d(S)} unsaturated vertices. Every matching has at most minSV(G){n(G)-d(S)} saturated vertices. The largest number of vertices saturated by a matching in G is equal to minSV(G){n(G)-d(S)}.

Berge-Tutte Formula2. Let d=maxSV(G){d(S)}. We need to show there exists a matching in G that has exactly d

unsaturated vertices.

It suffices to show G’ satisfies the Tutte’s Condition.

3. Let S=. We have d>=0.

4. Let G’=GKd.

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.

Berge-Tutte Formula6. 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).

o(G-S) |S| d(S) n(G) odd odd even even odd even odd odd even odd odd odd even even even even

n(G’) is even.

d(S) = o(G-S)-|S| has the same parity as n(G) for each S.

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’|.

d has the same parity as n(G).

Corollary 3.3.8• Every 3-regular graph with no cut-edge has a 1-

factor.1. It suffices to show2. Let H be a odd component, and let m be the number

of edges from S to H.3. The sum of the vertex degrees in H is 3n(H)-m. m is odd m>=34. Let p be the number of edges between S and the odd

components of G-S. p<=3|S| and p>=3o(G-S) o(G-S)<=|S|.

since 3n(H)-m is even and n(H) is odd.since G has no cut-edge.

since G is 3-regular

since m>=3.

o(G-S)<=|S| for any SV(G),

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. (see 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.

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.

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.