Graph theory · Conjecture

Petersen coloring conjecture

Can every bridgeless cubic graph be colored by the Petersen graph?

DisprovedOpenMath assessment · Oct 14, 2026

Earlier version 5 of this page. View current →

Overview

Known

A source-reported 112-vertex counterexample refutes the universal coloring conjecture. A second nonisomorphic witness is separately certified; neither establishes the smallest possible order.

Related open questions

Determine the smallest possible counterexample order; continue reviewing the certificate correspondence and weaker covering assertions separately.

Latest

An edge-star mapping question, answered negatively by a finite source-reported obstruction.

Precise statement · Exact conventions and hypotheses

A Petersen coloring maps edges of a finite loopless cubic graph G to edges of the Petersen graph P. At each vertex of G, its three incident edges must map bijectively onto the three edges incident with some vertex of P. This is a local edge-star condition, not a proper three-edge-coloring. The reported witness is simple and connected; it therefore refutes the broader bridgeless cubic assertion.

Definitions

Incident edge-star
The three edges meeting a cubic vertex. A Petersen coloring sends this whole set bijectively to the star at a vertex of the Petersen graph.
Bridgeless cubic graph
Every vertex has degree three, and removing any one edge does not disconnect its component. The reported witnesses are simple and connected.

Question paraphrased by OpenMath. Original wording is available in the sources.

Landscape

Figures and explorable examples. Selecting a statement opens its exact wording.

The 112-vertex counterexample

Every vertex has degree three and no single edge disconnects the graph, yet no map to the Petersen graph preserves its edge-stars.

Main witness G = 3L + C

Positions come from a computed layout and carry no meaning; only the edges do. The picture cannot show that no Petersen coloring exists: that is the cited, formally checked theorem.

Check the witness

Recomputed from the published adjacency list: the graph is a simple bridgeless cubic graph with the size, girth and connectivity the statement gives.

PropertyRecomputedStatement
Vertices112matches
Edges168matches
Every vertex has degree 3yesmatches
Simple: no loops or parallel edgesyesmatches
Connectedyesmatches
No bridge (removing any one edge)yesmatches
Edge-connectivity at least 3 (any two edges)yesmatches
Vertex-connectivity at least 3 (any two vertices)yesmatches
Girth (shortest cycle)5matches

This checks the graph's structure only. That it has no Petersen coloring is the cited theorem and its formal proof; it is not recomputed here.

Visual setup

From the Petersen graph to the obstruction

Construction hierarchy for the main graph: F has 8 vertices, L has 36 and G has 112.

Reading boundary. This diagram shows copy counts, not every semiedge join. Use the exact adjacency list and source construction for the witness.

From the Petersen graph to the obstructionDeleting two adjacent vertices leaves F. Four F copies and C give L; three L copies and another C give G. The source tables specify the semiedge joins.Petersen · 10F · 8L = 4F + C · 36G = 3L + C · 112C · claw · 4

Select a labeled part to inspect its exact statement, scope and source evidence.

Text version

Deleting two adjacent vertices leaves F. Four F copies and C give L; three L copies and another C give G. The source tables specify the semiedge joins.

Layout is schematic; coordinates and line lengths carry no mathematical scale.

  1. The main construction

    Proved

    The main 112-vertex counterexample

    Scope & assumptions. G is the exact labeled graph in Appendix A; its normalized sorted-edge-list digest identifies the SAT input. No minimum-order claim is made.

    The main 112-vertex counterexample →

Sources & review

Authored by OpenMath source curation on 2026-10-14 (AI-assisted curation). Independent review of this visual explanation is pending; linked assertions and relationships have their own assessments.

Inspect the main graph

Inspect the exact adjacency

112 numbered vertices and 168 undirected edges.

Vertex 0 has neighbors 1, 5, 31.

Complete adjacency table
Every numbered vertex and its neighbors
VertexNeighbors
01, 5, 31
10, 2, 6
21, 7, 100
35, 6, 11
46, 7, 9
50, 3, 7
61, 3, 4
72, 4, 5
89, 10, 11
94, 8, 14
108, 12, 20
113, 8, 28
1210, 13, 17
1312, 14, 18
149, 13, 19
1517, 18, 22
1618, 19, 56
1712, 15, 19
1813, 15, 16
1914, 16, 17
2010, 21, 25
2120, 22, 26
2215, 21, 27
2325, 26, 30
2426, 27, 39
2520, 23, 27
2621, 23, 24
2722, 24, 25
2811, 29, 33
2928, 30, 34
3023, 29, 35
310, 33, 34
3234, 35, 37
3328, 31, 35
3429, 31, 32
3530, 32, 33
3637, 38, 39
3732, 36, 42
3836, 64, 92
3924, 36, 78
4041, 45, 71
4140, 42, 46
4237, 41, 47
4345, 46, 51
4446, 47, 49
4540, 43, 47
4641, 43, 44
4742, 44, 45
4849, 50, 51
4944, 48, 54
5048, 52, 60
5143, 48, 68
5250, 53, 57
5352, 54, 58
5449, 53, 59
5557, 58, 62
5616, 58, 59
5752, 55, 59
5853, 55, 56
5954, 56, 57
6050, 61, 65
6160, 62, 66
6255, 61, 67
6365, 66, 70
6438, 66, 67
6560, 63, 67
6661, 63, 64
6762, 64, 65
6851, 69, 73
6968, 70, 74
7063, 69, 75
7140, 73, 74
7274, 75, 108
7368, 71, 75
7469, 71, 72
7570, 72, 73
7677, 81, 107
7776, 78, 82
7839, 77, 83
7981, 82, 87
8082, 83, 85
8176, 79, 83
8277, 79, 80
8378, 80, 81
8485, 86, 87
8580, 84, 90
8684, 88, 96
8779, 84, 104
8886, 89, 93
8988, 90, 94
9085, 89, 95
9193, 94, 98
9238, 94, 95
9388, 91, 95
9489, 91, 92
9590, 92, 93
9686, 97, 101
9796, 98, 102
9891, 97, 103
99101, 102, 106
1002, 102, 103
10196, 99, 103
10297, 99, 100
10398, 100, 101
10487, 105, 109
105104, 106, 110
10699, 105, 111
10776, 109, 110
10872, 110, 111
109104, 107, 111
110105, 107, 108
111106, 108, 109

Construction details and sources ↓

Inspect the symmetric graph

Inspect the exact adjacency

112 numbered vertices and 168 undirected edges.

Vertex 0 has neighbors 1, 5, 31.

Complete adjacency table
Every numbered vertex and its neighbors
VertexNeighbors
01, 5, 31
10, 2, 6
21, 7, 52
35, 6, 11
46, 7, 9
50, 3, 7
61, 3, 4
72, 4, 5
89, 10, 11
94, 8, 14
108, 12, 20
113, 8, 28
1210, 13, 17
1312, 14, 18
149, 13, 19
1517, 18, 22
1618, 19, 74
1712, 15, 19
1813, 15, 16
1914, 16, 17
2010, 21, 25
2120, 22, 26
2215, 21, 27
2325, 26, 30
2426, 27, 109
2520, 23, 27
2621, 23, 24
2722, 24, 25
2811, 29, 33
2928, 30, 34
3023, 29, 35
310, 33, 34
3234, 35, 111
3328, 31, 35
3429, 31, 32
3530, 32, 33
3637, 41, 67
3736, 38, 42
3837, 43, 88
3941, 42, 47
4042, 43, 45
4136, 39, 43
4237, 39, 40
4338, 40, 41
4445, 46, 47
4540, 44, 50
4644, 48, 56
4739, 44, 64
4846, 49, 53
4948, 50, 54
5045, 49, 55
5153, 54, 58
522, 54, 55
5348, 51, 55
5449, 51, 52
5550, 52, 53
5646, 57, 61
5756, 58, 62
5851, 57, 63
5961, 62, 66
6062, 63, 110
6156, 59, 63
6257, 59, 60
6358, 60, 61
6447, 65, 69
6564, 66, 70
6659, 65, 71
6736, 69, 70
6870, 71, 109
6964, 67, 71
7065, 67, 68
7166, 68, 69
7273, 77, 103
7372, 74, 78
7416, 73, 79
7577, 78, 83
7678, 79, 81
7772, 75, 79
7873, 75, 76
7974, 76, 77
8081, 82, 83
8176, 80, 86
8280, 84, 92
8375, 80, 100
8482, 85, 89
8584, 86, 90
8681, 85, 91
8789, 90, 94
8838, 90, 91
8984, 87, 91
9085, 87, 88
9186, 88, 89
9282, 93, 97
9392, 94, 98
9487, 93, 99
9597, 98, 102
9698, 99, 111
9792, 95, 99
9893, 95, 96
9994, 96, 97
10083, 101, 105
101100, 102, 106
10295, 101, 107
10372, 105, 106
104106, 107, 110
105100, 103, 107
106101, 103, 104
107102, 104, 105
108109, 110, 111
10924, 68, 108
11060, 104, 108
11132, 96, 108

Construction details and sources ↓

Results

The question 1

Equivalent formulations 1

Counterexamples 2

Related advances 1

  • Proved
    Infinitely many Petersen-coloring counterexamples

    Corollary 5.1 combines the finite counterexample with the universal-target theorem of Ma, Mattiolo, Steffen and Wolf.

    Putman 2026

Supporting results 1

  • Proved
    Petersen and normal-five colorings are equivalent

    The paper gives an explicit bijection using P = KG(5,2).

    Putman 2026
Constructions · 2

Connections

Equivalent formulations 1

Consequences 1

Assuming The main 112-vertex counterexampleProved

Approaches

How the problem is attacked, and what each line of attack cannot yet reach.

Build the obstruction from small pieces

Delete the endpoints of one Petersen edge to obtain the eight-vertex four-pole F. Join four copies of F and a four-vertex claw six-pole C into a 36-vertex four-pole L, then join three L copies and another C into G. The source specifies every semiedge join.

Cannot yet reach Copy counts do not specify the graph without the precise joins and labeling map.

Sources & review

Supporting statements

Source-grounded explanation · independent explanation review: pending.

Timeline

  1. Two explicit counterexamples and certificatesRelated result

    Putman publishes the graph constructions, encodings and external checks; the frozen release is version 1.1.0.

    This is a selected source-supported milestone, not an exhaustive history.

  2. Jaeger’s coloring questionFoundation

    The selected paper attributes the bridgeless cubic question to Jaeger’s 1988 chapter.

    This is a selected source-supported milestone, not an exhaustive history.

  3. The normal-five formulationFoundation

    Jaeger’s earlier work establishes the coloring equivalence recalled in Proposition 2.1.

    This is a selected source-supported milestone, not an exhaustive history.

Full history with sources →

Evidence

Evidence by kind
Literature

1 preprint · 1 project

Status OpenMath assessment · Oct 14, 2026.

Formal

None linked

Reported in sources: SAT/DRAT certificates are available in the immutable archive. They are computational proof evidence, not an identified Lean or Isabelle theorem; OpenMath has not executed the certificates.

Computation

2 explicit witnesses

Review

Not yet reviewed

OpenMath assessment: A source-reported 112-vertex counterexample refutes the universal coloring conjecture. A second nonisomorphic witness is separately certified; neither establishes the smallest possible order.

Source comparison and presentation; independent mathematical review remains pending.

OpenMath pilot collection · OpenMath source curation · assessed Oct 14, 2026 · review due Jan 14, 2027

Collection last checked Oct 14, 2026.

Sources

Choose a starting point below. Each result links to its exact statement, cited passages and checks.

Start here

Bibliography 2

Selected literature; source observations and coverage limits are retained in the publication details.

Contributors

François Jaeger posed the coloring question and established its normal-five formulation. Bryce Putman authored the 2026 construction and verification release, reporting extensive OpenAI-model assistance and accepting responsibility for the final work. CaDiCaL and drat-trim supply distinct solver and certificate-checking roles. Ma, Mattiolo, Steffen and Wolf supplied the theorem used for the infinitude consequence.

Authors & roles

These are source credits; they do not imply OpenMath accounts, endorsements or independent mathematical review.

Record details
Record version
5
Classification
MSC2020 05C15 / 05C21
Prerequisite level
undergraduate

Independent mathematical review is pending. External certificates and manuscript proofs are attributed to their sources.

Source observations & version history →