Wikipedia

Graph factorization

(Redirected from 1-factor)
Not to be confused with Factor graph.

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 a proper edge coloring with k colors. A 2-factor is a collection of cycles that spans all vertices of the graph.

1-factorization of the Desargues graph: each color class is a 1-factor.
The Petersen graph can be partitioned into a 1-factor (red) and a 2-factor (blue). However, the graph is not 1-factorable.

1-factorization

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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:

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

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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 in a regular polygon, with the remaining vertex at the center. 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, ... (OEIS: A000438).

1-factorization conjecture

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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.

Perfect 1-factorization

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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]

2-factorization

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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 open.

References

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  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 (1985). Niessen (1994). Perkovic & Reed (1997). West.
  4. ^ Wallis, W. D. (1997), "16. Perfect Factorizations", One-factorizations, Mathematics and Its Applications, vol. 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.

Bibliography

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Further reading

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