Combinatorics Probabilistic and discrete-geometric methods Conjecture
Cycle double cover conjecture
Statement
Let be a finite graph, possibly with multiple edges, that has no bridge. The conjecture: there is a finite list of cycles of such that each edge of lies in exactly two cycles of the list. The list may contain a cycle more than once. Such a list is called a cycle double cover of .
Remarks
The conjecture is open. It was posed by Szekeres and, independently, by Seymour. The hypothesis is necessary: a bridge lies on no cycle, so a graph with a bridge has no cycle double cover. For a bridgeless planar graph, each face boundary of a plane embedding splits into edge-disjoint cycles, and each edge lies on exactly two faces. So these cycles form a cycle double cover. The general case reduces to cubic bridgeless graphs, and further to snarks, the cubic bridgeless graphs with no proper edge colouring with colours. If a cubic graph has a proper edge colouring with colours, the unions of two colour classes give three families of disjoint cycles, which together cover each edge twice.
Reading path
To understand this page from the ground up, read these 4 pages in order. The foundations come first.
Show the path as a graph
- no evidence level
- Bridge concept
- Cycle concept
- Graph concept
- Cycle double cover conjecture conjecture
Further reading
About this page
| Tag | G2RYQMGG1R |
|---|---|
| URI | m2://G2RYQMGG1R |
| Kind | conjecture |
| Writing | W0 stub |
| MSC | 05C38, 05C70 |
| Status | This copy of the site has no build data. Run m2 status G2RYQMGG1R. |
| About | Cycle |
| Uses | Graph, Cycle, Bridge |
| Changed | 2 November 2026 |
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