# Graph factorization

(Redirected from 1-factor)
Not to be confused with Factor graph.
1-factorization of Desargues graph: each color class is a 1-factor.
Petersen graph can be partitioned into a 1-factor (red) and a 2-factor (blue). However, the graph is not 1-factorable.

In graph theory, a factor of a graph G is a spanning subgraph, i.e., a subgraph that has the same vertex set as G. A k-factor of a graph is a spanning k-regular subgraph, and a k-factorization partitions the edges of the graph into disjoint k-factors. A graph G is said to be k-factorable if it admits a k-factorization. In particular, a 1-factor is a perfect matching, and a 1-factorization of a k-regular graph is an edge coloring with k colors. A 2-factor is a collection of cycles that spans all vertices of the graph.

## 1-factorization

If a graph is 1-factorable, then it has to be a regular graph. However, not all regular graphs are 1-factorable. A k-regular graph is 1-factorable if it has chromatic index k; examples of such graphs include:

• Any regular bipartite graph.[1] Hall's marriage theorem can be used to show that a k-regular bipartite graph contains a perfect matching. One can then remove the perfect matching to obtain a (k − 1)-regular bipartite graph, and apply the same reasoning repeatedly.
• Any complete graph with an even number of nodes (see below).[2]

However, there are also k-regular graphs that have chromatic index k + 1, and these graphs are not 1-factorable; examples of such graphs include:

### Complete graphs

1-factorization of K8 in which each 1-factor consists of an edge from the center to a vertex of a heptagon together with all possible perpendicular edges

A 1-factorization of a complete graph corresponds to pairings in a round-robin tournament. The 1-factorization of complete graphs is a special case of Baranyai's theorem concerning the 1-factorization of complete hypergraphs.

One method for constructing a 1-factorization of a complete graph involves placing all but one of the vertices on a circle, forming a regular polygon, with the remaining vertex at the center of the circle. With this arrangement of vertices, one way of constructing a 1-factor of the graph is to choose an edge e from the center to a single polygon vertex together with all possible edges that lie on lines perpendicular to e. The 1-factors that can be constructed in this way form a 1-factorization of the graph.

### 1-factorization conjecture

Let G be a k-regular graph with 2n nodes. If k is sufficiently large, it is known that G has to be 1-factorable:

• If k = 2n − 1, then G is the complete graph K2n, and hence 1-factorable (see above).
• If k = 2n − 2, then G can be constructed by removing a perfect matching from K2n. Again, G is 1-factorable.
• Chetwynd & Hilton (1985) show that if k ≥ 12n/7, then G is 1-factorable.

The 1-factorization conjecture[3] is a long-standing conjecture that states that k ≈ n is sufficient. In precise terms, the conjecture is:

• If n is odd and k ≥ n, then G is 1-factorable. If n is even and k ≥ n − 1 then G is 1-factorable.

The overfull conjecture implies the 1-factorization conjecture.

## 2-factorization

If a graph is 2-factorable, then it has to be 2k-regular for some integer k. Julius Petersen showed in 1891 that this necessary condition is also sufficient: any 2k-regular graph is 2-factorable.[4]

If a connected graph is 2k-regular and has an even number of edges it may also be k-factored, by choosing each of the two factors to be an alternating subset of the edges of an Euler tour.[5] This applies only to connected graphs; disconnected counterexamples include disjoint unions of odd cycles, or of copies of K2k+1.

## Notes

1. ^ Harary (1969), Theorem 9.2, p. 85. Diestel (2005), Corollary 2.1.3, p. 37.
2. ^ Harary (1969), Theorem 9.1, p. 85.
3. ^
4. ^ Petersen (1891), §9, p. 200. Harary (1969), Theorem 9.9, p. 90. See Diestel (2005), Corollary 2.1.5, p. 39 for a proof.
5. ^ Petersen (1891), §6, p. 198.