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[[Image:Desargues graph 3color edge.svg|thumb|200px|1-factorization of [[Desargues graph|Desargues graph]]: each color class is a 1-factor.]]
{{Short description|Partition of a graph into spanning subgraphs}}
[[Image:Petersen-graph-factors.svg|right|thumb|200px|[[Petersen graph|Petersen graph]] can be partitioned into a 1-factor (red) and a 2-factor (blue). However, the graph is not 1-factorable.]]
{{Distinguish|Factor graph}}
In [[Graph theory|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 graph|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|perfect matching]], and a 1-factorization of a ''k''-[[Regular graph|regular graph]] is a [[Edge coloring|proper edge coloring]] with ''k'' colors. A '''2-factor''' is a collection of [[Cycle (graph theory)|cycles]] that spans all vertices of the graph.
 
[[Image:Desargues graph 3color edge.svg|thumb|200px|1-factorization of the [[Desargues graph]]: each color class is a {{nowrap|1-factor}}.]]
[[Image:Petersen-graph-factors.svg|right|thumb|200px|The [[Petersen graph]] can be partitioned into a {{nowrap|1-factor}} (red) and a {{nowrap|2-factor}} (blue). However, the graph is not {{nowrap|1-factorable}}.]]
{{unsolved|mathematics|'''Conjecture:''' 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.}}
In [[Graph theory|graph theory]], a '''factor''' of a [[Graph (discrete mathematics)|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 graph|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|perfect matching]], and a 1-factorization of a ''k''-regular graph is a [[Edge coloring|proper edge coloring]] with ''k'' colors. A '''2-factor''' is a collection of disjoint [[Cycle (graph theory)|cycle]]s that spans all vertices of the graph.


==1-factorization==
==1-factorization==


If a graph is 1-factorable (i.e., has a 1-factorization), then it has to be a [[Regular graph|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:
If a graph is 1-factorable then it has to be a [[Regular graph|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|bipartite graph]].<ref>{{harvtxt|Harary|1969}}, Theorem 9.2, p. 85. {{harvtxt|Diestel|2005}}, Corollary 2.1.3, p. 37.</ref> [[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''&nbsp;&minus;&nbsp;1)-regular bipartite graph, and apply the same reasoning repeatedly.
* Any regular [[Bipartite graph|bipartite graph]].<ref>{{harvtxt|Harary|1969}}, Theorem 9.2, p. 85. {{harvtxt|Diestel|2005}}, Corollary 2.1.3, p. 37.</ref> [[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''&nbsp;&minus;&nbsp;1)-regular bipartite graph, and apply the same reasoning repeatedly.
* Any [[Complete graph|complete graph]] with an even number of nodes (see below).<ref>{{harvtxt|Harary|1969}}, Theorem 9.1, p. 85.</ref>
* Any [[Complete graph|complete graph]] with an [[Parity (mathematics)|even]] number of nodes (see below).<ref>{{harvtxt|Harary|1969}}, Theorem 9.1, p. 85.</ref>
However, there are also ''k''-regular graphs that have chromatic index ''k''&nbsp;+&nbsp;1, and these graphs are not 1-factorable; examples of such graphs include:
However, there are also ''k''-regular graphs that have chromatic index ''k''&nbsp;+&nbsp;1, and these graphs are not 1-factorable; examples of such graphs include:
* Any regular graph with an odd number of nodes.
* Any regular graph with an [[Parity (mathematics)|odd]] number of nodes.
* The [[Petersen graph]].
* The [[Petersen graph]].


===Complete graphs===
===Complete graphs===
[[File:Complete-edge-coloring.svg|thumb|200px|1-factorization of ''K''<sub>8</sub> in which each 1-factor consists of an edge from the center to a vertex of a [[Heptagon|heptagon]] together with all possible perpendicular edges]]
[[File:Complete-edge-coloring.svg|thumb|200px|1-factorization of ''K''<sub>8</sub> in which each {{nowrap|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|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 [[Hypergraph|hypergraph]]s.
A 1-factorization of a [[Complete graph|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 [[Hypergraph|hypergraph]]s.


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|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.
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|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 ''K''<sub>2</sub>, ''K''<sub>4</sub>, ''K''<sub>6</sub>, ''K''<sub>8</sub>, ... is 1, 1, 6, 6240, 1225566720, 252282619805368320, 98758655816833727741338583040, ... {{oeis|A000438}}.
The number of distinct 1-factorizations of ''K''<sub>2</sub>, ''K''<sub>4</sub>, ''K''<sub>6</sub>, ''K''<sub>8</sub>, ... is 1, 1, 6, 6240, 1225566720, 252282619805368320, 98758655816833727741338583040, ... ({{oeis|A000438}}).


===1-factorization conjecture===
===1-factorization conjecture===
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* If ''k''&nbsp;=&nbsp;2''n''&nbsp;&minus;&nbsp;1, then ''G'' is the complete graph ''K''<sub>2''n''</sub>, and hence 1-factorable (see above).
* If ''k''&nbsp;=&nbsp;2''n''&nbsp;&minus;&nbsp;1, then ''G'' is the complete graph ''K''<sub>2''n''</sub>, and hence 1-factorable (see above).
* If ''k''&nbsp;=&nbsp;2''n''&nbsp;&minus;&nbsp;2, then ''G'' can be constructed by removing a perfect matching from ''K''<sub>2''n''</sub>. Again, ''G'' is 1-factorable.
* If ''k''&nbsp;=&nbsp;2''n''&nbsp;&minus;&nbsp;2, then ''G'' can be constructed by removing a perfect matching from ''K''<sub>2''n''</sub>. Again, ''G'' is 1-factorable.
* {{harvtxt|Chetwynd|Hilton|1985}} show that if ''k''&nbsp;≥&nbsp;12n/7, then ''G'' is 1-factorable.
* {{harvtxt|Chetwynd|Hilton|1985}} show that if ''k''&nbsp;≥&nbsp;12''n''/7, then ''G'' is 1-factorable.
The '''1-factorization conjecture'''<ref>{{harvtxt|Chetwynd|Hilton|1985}}. {{harvtxt|Niessen|1994}}. {{harvtxt|Perkovic|Reed|1997}}. West.</ref> is a long-standing [[Conjecture|conjecture]] that states that ''k''&nbsp;≈&nbsp;''n'' is sufficient. In precise terms, the conjecture is:
The '''1-factorization conjecture'''<ref>{{harvtxt|Chetwynd|Hilton|1985}}. {{harvtxt|Niessen|1994}}. {{harvtxt|Perkovic|Reed|1997}}. West.</ref> is a long-standing [[Conjecture|conjecture]] that states that ''k''&nbsp;≈&nbsp;''n'' is sufficient. In precise terms, the conjecture is:
* If ''n'' is odd and ''k''&nbsp;≥&nbsp;''n'', then ''G'' is 1-factorable. If ''n'' is even and ''k''&nbsp;≥&nbsp;''n''&nbsp;&minus;&nbsp;1 then ''G'' is 1-factorable.
* If ''n'' is odd and ''k''&nbsp;≥&nbsp;''n'', then ''G'' is 1-factorable. If ''n'' is even and ''k''&nbsp;≥&nbsp;''n''&nbsp;&minus;&nbsp;1 then ''G'' is 1-factorable.
The [[Overfull conjecture|overfull conjecture]] implies the 1-factorization conjecture.
The [[Overfull conjecture|overfull conjecture]] implies the 1-factorization conjecture.
The conjecture was confirmed by Csaba, Kühn, Lo, Osthus and Treglown for sufficiently large ''n''.<ref name="csaba">
{{Citation
| last1  = Csaba
| first1 = Béla
| last2  = Kühn
| first2 = Daniela
| last3  = Lo
| first3 = Allan
| last4  = Osthus
| first4 = Deryk
| last5  = Treglown
| first5 = Andrew
| title  = Proof of the 1-factorization and Hamilton decomposition conjectures
| journal = Memoirs of the American Mathematical Society
| date    = June 2016
| doi    = 10.1090/memo/1154
}}
</ref>


===Perfect 1-factorization===
===Perfect 1-factorization===
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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).
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, [[Biography:Anton Kotzig|Anton Kotzig]] conjectured that every [[Complete graph|complete graph]] ''K''<sub>2''n''</sub> where ''n'' ≥ 2 has a perfect 1-factorization. So far, it is known that the following graphs have a perfect 1-factorization:<ref name="wallis">
In 1964, [[Biography:Anton Kotzig|Anton Kotzig]] conjectured that every complete graph ''K''<sub>2''n''</sub> where ''n'' ≥ 2 has a perfect 1-factorization. So far, it is known that the following graphs have a perfect 1-factorization:<ref name="wallis">
{{Citation
{{Citation
  | first = W. D. | last = Wallis
  | first = W. D. | last = Wallis
Line 50: Line 73:
</ref>
</ref>


* the infinite family of complete graphs ''K''<sub>2''p''</sub> where ''p'' is an odd prime (by Anderson and also Nakamura, independently),
* the infinite family of complete graphs ''K''<sub>2''p''</sub> where ''p'' is an odd [[Prime number|prime]] (by Anderson and also Nakamura, independently),
* the infinite family of complete graphs ''K''<sub>''p'' + 1</sub> where ''p'' is an odd prime,
* the infinite family of complete graphs ''K''<sub>''p''+1</sub> where ''p'' is an odd prime,
* and sporadic additional results, including ''K''<sub>2''n''</sub> where 2''n'' ∈ {16, 28, 36, 40, 50, 126, 170, 244, 344, 730, 1332, 1370, 1850, 2198, 3126, 6860, 12168, 16808, 29792}. Some newer results are collected [http://users.monash.edu.au/~iwanless/data/P1F/newP1F.html here].
* and sporadic additional results, including ''K''<sub>2''n''</sub> where 2''n'' ∈ {16, 28, 36, 40, 50, 126, 170, 244, 344, 730, 1332, 1370, 1850, 2198, 3126, 6860, 12168, 16808, 29792}. Some newer results are collected [http://users.monash.edu.au/~iwanless/data/P1F/newP1F.html here].
<!-- Related OEIS sequences: A005702 A120488 A120489 -->
<!-- Related OEIS sequences: A005702 A120488 A120489 -->


If the complete graph ''K''<sub>''n'' + 1</sub> has a perfect 1-factorization, then the [[Complete bipartite graph|complete bipartite graph]] ''K''<sub>''n'',''n''</sub> also has a perfect 1-factorization.<ref name="wanless">
If the complete graph ''K''<sub>''n''+1</sub> has a perfect 1-factorization, then the [[Complete bipartite graph|complete bipartite graph]] ''K''<sub>''n'',''n''</sub> also has a perfect 1-factorization.<ref name="wanless">
{{Citation
{{Citation
| last1  = Bryant
| last1  = Bryant
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==2-factorization==
==2-factorization==


If a graph is 2-factorable, then it has to be 2''k''-regular for some integer ''k''. [[Biography:Julius Petersen|Julius Petersen]] showed in 1891 that this necessary condition is also sufficient: any 2''k''-regular graph is 2-factorable.<ref>{{harvtxt|Petersen|1891}}, §9, p. 200. {{harvtxt|Harary|1969}}, Theorem 9.9, p. 90. See {{harvtxt|Diestel|2005}}, Corollary 2.1.5, p. 39 for a proof.</ref>
If a graph is 2-factorable, then it has to be 2''k''-regular for some integer ''k''. [[Biography:Julius Petersen|Julius Petersen]] showed in 1891 that this necessary condition is also sufficient: [[2-factor theorem|any 2''k''-regular graph is 2-factorable]].<ref>{{harvtxt|Petersen|1891}}, §9, p. 200. {{harvtxt|Harary|1969}}, Theorem 9.9, p. 90. See {{harvtxt|Diestel|2005}}, Corollary 2.1.5, p. 39 for a proof.</ref>


If a connected graph is 2''k''-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.<ref>{{harvtxt|Petersen|1891}}, §6, p. 198.</ref>  This applies only to connected graphs; disconnected counterexamples include disjoint unions of odd cycles, or of copies of ''K''<sub>2''k''+1</sub>.
If a [[Connectivity (graph theory)|connected graph]] is 2''k''-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.<ref>{{harvtxt|Petersen|1891}}, §6, p. 198.</ref>  This applies only to connected graphs; disconnected [[Counterexample|counterexample]]s include disjoint unions of odd cycles, or of copies of ''K''<sub>2''k''+1</sub>.


The [[Oberwolfach problem]] concerns the existence of 2-factorizations of [[Complete graph|complete graph]]s 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.
The [[Oberwolfach problem]] concerns the existence of 2-factorizations of complete graphs into [[Graph isomorphism|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:Open problem|open]].


==References==
==References==
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}}.
}}.
*{{citation
*{{citation
| last=Diestel | first=Reinhard
  | title=Graph Theory
  | title=Graph Theory
  | publisher=[[Company:Springer Science+Business Media|Springer]]
  | publisher=[[Company:Springer Science+Business Media|Springer]]
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  | pages = 193–220
  | pages = 193–220
  | doi = 10.1007/BF02392606| doi-access=free
  | doi = 10.1007/BF02392606| doi-access=free
| url=https://zenodo.org/record/2304433/files/article.pdf
  }}.
  }}.
*{{cite web
*{{cite web

Latest revision as of 15:46, 13 February 2026

Short description: Partition of a graph into spanning subgraphs
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.
Unsolved problem in mathematics:
Conjecture: 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.
(more unsolved problems in mathematics)

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 disjoint 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 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, ... (OEISA000438).

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) 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. The conjecture was confirmed by Csaba, Kühn, Lo, Osthus and Treglown for sufficiently large n.[4]

Perfect 1-factorization

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:[5]

  • 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.[6]

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.[7]

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.[8] 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

  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. Csaba, Béla; Kühn, Daniela; Lo, Allan; Osthus, Deryk; Treglown, Andrew (June 2016), "Proof of the 1-factorization and Hamilton decomposition conjectures", Memoirs of the American Mathematical Society, doi:10.1090/memo/1154 
  5. 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 
  6. 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 
  7. (Petersen 1891), §9, p. 200. (Harary 1969), Theorem 9.9, p. 90. See (Diestel 2005), Corollary 2.1.5, p. 39 for a proof.
  8. (Petersen 1891), §6, p. 198.

Bibliography

Further reading