Graph factorization

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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.


If a graph is 1-factorable (ie, has a 1-factorization), 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[edit]

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 on an even number of vertices 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.

The number of distinct 1-factorizations of K2, K4, K6, K8, ... is 1, 1, 6, 6240, 1225566720, 252282619805368320, 98758655816833727741338583040, ... OEISA000438.

1-factorization conjecture[edit]

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) 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.

Perfect 1-factorization[edit]

A perfect pair from a 1-factorization is a pair of 1-factors whose union induces a Hamiltonian cycle.

A perfect 1-factorization (P1F) of a graph is a 1-factorization having the property that every pair of 1-factors is a perfect pair. A perfect 1-factorization should not be confused with a perfect matching (also called a 1-factor).

In 1964, Anton Kotzig conjectured that every complete graph K2n where n ≥ 2 has a perfect 1-factorization. So far, it is known that the following graphs have a perfect 1-factorization:[4]

  • the infinite family of complete graphs K2p where p is an odd prime (by Anderson and also Nakamura, independently),
  • the infinite family of complete graphs Kp + 1 where p is an odd prime,
  • and sporadic additional results, including K2n where 2n ∈ {16, 28, 36, 40, 50, 126, 170, 244, 344, 730, 1332, 1370, 1850, 2198, 3126, 6860, 12168, 16808, 29792}. Some newer results are collected here.

If the complete graph Kn + 1 has a perfect 1-factorization, then the complete bipartite graph Kn,n also has a perfect 1-factorization.[5]


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.[6]

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.[7] This applies only to connected graphs; disconnected counterexamples include disjoint unions of odd cycles, or of copies of K2k+1.

The Oberwolfach problem concerns the existence of 2-factorizations of complete graphs into isomorphic subgraphs. It asks for which subgraphs this is possible. This is known when the subgraph is connected (in which case it is a Hamiltonian cycle and this special case is the problem of Hamiltonian decomposition) but the general case remains unsolved.


  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. (Chetwynd Hilton). (Niessen 1994). (Perkovic Reed). West.
  4. Wallis, W. D. (1997), "16. Perfect Factorizations", One-factorizations, Mathematics and Its Applications, 390 (1 ed.), Springer US, p. 125, doi:10.1007/978-1-4757-2564-3_16, ISBN 978-0-7923-4323-3 
  5. Bryant, Darryn; Maenhaut, Barbara M.; Wanless, Ian M. (May 2002), "A Family of Perfect Factorisations of Complete Bipartite Graphs", Journal of Combinatorial Theory, A 98 (2): 328–342, doi:10.1006/jcta.2001.3240, ISSN 0097-3165 
  6. (Petersen 1891), §9, p. 200. (Harary 1969), Theorem 9.9, p. 90. See (Diestel 2005), Corollary 2.1.5, p. 39 for a proof.
  7. (Petersen 1891), §6, p. 198.


Further reading[edit] factorization was the original source. Read more.