Cycle double cover conjecture for complete graphs?
$begingroup$
The cycle double cover conjecture states that for every bridgeless finite $G$ there is a collection of cycles $mathcal{C}$ in $G$ such that every edges on $G$ occurs in precisely two of the cycles in $mathcal{C}$. Is the cycle double cover conjecture known for complete graphs?
I would actually be interested in knowing about stronger results pertaining to the various stronger versions of the double cover conjecture - for example does every complete graph admit a strong embedding into some orientable surface (i.e. every face is a cycle in the graph)? Can we take it to be 5-face colorable?
I am also interested in the same question for complete bipartite graphs. References would be welcome!
graph-theory reference-request topological-graph-theory
asked Jan 8 at 3:14
user101010user101010
1,972516
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add a comment |
$begingroup$
The cycle double cover conjecture states that for every bridgeless finite $G$ there is a collection of cycles $mathcal{C}$ in $G$ such that every edges on $G$ occurs in precisely two of the cycles in $mathcal{C}$. Is the cycle double cover conjecture known for complete graphs?
I would actually be interested in knowing about stronger results pertaining to the various stronger versions of the double cover conjecture - for example does every complete graph admit a strong embedding into some orientable surface (i.e. every face is a cycle in the graph)? Can we take it to be 5-face colorable?
I am also interested in the same question for complete bipartite graphs. References would be welcome!
graph-theory reference-request topological-graph-theory
asked Jan 8 at 3:14
user101010user101010
1,972516
$endgroup$
add a comment |
$begingroup$
The cycle double cover conjecture states that for every bridgeless finite $G$ there is a collection of cycles $mathcal{C}$ in $G$ such that every edges on $G$ occurs in precisely two of the cycles in $mathcal{C}$. Is the cycle double cover conjecture known for complete graphs?
I would actually be interested in knowing about stronger results pertaining to the various stronger versions of the double cover conjecture - for example does every complete graph admit a strong embedding into some orientable surface (i.e. every face is a cycle in the graph)? Can we take it to be 5-face colorable?
I am also interested in the same question for complete bipartite graphs. References would be welcome!
graph-theory reference-request topological-graph-theory
asked Jan 8 at 3:14
user101010user101010
1,972516
$endgroup$
The cycle double cover conjecture states that for every bridgeless finite $G$ there is a collection of cycles $mathcal{C}$ in $G$ such that every edges on $G$ occurs in precisely two of the cycles in $mathcal{C}$. Is the cycle double cover conjecture known for complete graphs?
I would actually be interested in knowing about stronger results pertaining to the various stronger versions of the double cover conjecture - for example does every complete graph admit a strong embedding into some orientable surface (i.e. every face is a cycle in the graph)? Can we take it to be 5-face colorable?
I am also interested in the same question for complete bipartite graphs. References would be welcome!
graph-theory reference-request topological-graph-theory
graph-theory reference-request topological-graph-theory
asked Jan 8 at 3:14
user101010user101010
1,972516
asked Jan 8 at 3:14
user101010user101010
1,972516
asked Jan 8 at 3:14
user101010user101010
1,972516
asked Jan 8 at 3:14
user101010user101010
1,972516
asked Jan 8 at 3:14
user101010user101010
1,972516
1,972516
add a comment |
add a comment |
1 Answer
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A complete graph with an odd number of vertices has a single cycle cover: Arrange the vertices in a regular polygon and consider each possible length of diagonals separately. Now just take each of those cycles twice.
If you have an even number of vertices, then set one of them aside and split the rest of the graph into cycles as above. Then take one of the resulting outer cycles and disassemble it into single edges; let each of those edges make a triangle with the vertex you set aside. This again gives a cycle double cover.
answered Jan 8 at 3:30
Henning MakholmHenning Makholm
244k17312556
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1 Answer
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1 Answer
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$begingroup$
A complete graph with an odd number of vertices has a single cycle cover: Arrange the vertices in a regular polygon and consider each possible length of diagonals separately. Now just take each of those cycles twice.
If you have an even number of vertices, then set one of them aside and split the rest of the graph into cycles as above. Then take one of the resulting outer cycles and disassemble it into single edges; let each of those edges make a triangle with the vertex you set aside. This again gives a cycle double cover.
answered Jan 8 at 3:30
Henning MakholmHenning Makholm
244k17312556
$endgroup$
add a comment |
$begingroup$
A complete graph with an odd number of vertices has a single cycle cover: Arrange the vertices in a regular polygon and consider each possible length of diagonals separately. Now just take each of those cycles twice.
If you have an even number of vertices, then set one of them aside and split the rest of the graph into cycles as above. Then take one of the resulting outer cycles and disassemble it into single edges; let each of those edges make a triangle with the vertex you set aside. This again gives a cycle double cover.
answered Jan 8 at 3:30
Henning MakholmHenning Makholm
244k17312556
$endgroup$
add a comment |
$begingroup$
A complete graph with an odd number of vertices has a single cycle cover: Arrange the vertices in a regular polygon and consider each possible length of diagonals separately. Now just take each of those cycles twice.
If you have an even number of vertices, then set one of them aside and split the rest of the graph into cycles as above. Then take one of the resulting outer cycles and disassemble it into single edges; let each of those edges make a triangle with the vertex you set aside. This again gives a cycle double cover.
answered Jan 8 at 3:30
Henning MakholmHenning Makholm
244k17312556
$endgroup$
A complete graph with an odd number of vertices has a single cycle cover: Arrange the vertices in a regular polygon and consider each possible length of diagonals separately. Now just take each of those cycles twice.
If you have an even number of vertices, then set one of them aside and split the rest of the graph into cycles as above. Then take one of the resulting outer cycles and disassemble it into single edges; let each of those edges make a triangle with the vertex you set aside. This again gives a cycle double cover.
answered Jan 8 at 3:30
Henning MakholmHenning Makholm
244k17312556
answered Jan 8 at 3:30
Henning MakholmHenning Makholm
244k17312556
answered Jan 8 at 3:30
Henning MakholmHenning Makholm
244k17312556
answered Jan 8 at 3:30
Henning MakholmHenning Makholm
244k17312556
244k17312556
add a comment |
add a comment |
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