How to prove Tait's theorem about planar cubic bridgeless graph being 3-edge-colorable?
How can be proved, that
The four-color theorem is equivalent to the claim that every planar cubic bridgeless graph is 3-edge-colorable.
I can't figure out or find any prove of this theorem.
Thanks.
graph-theory coloring
asked Jan 1 at 14:19
Přemysl Šťastný
1595
add a comment |
How can be proved, that
The four-color theorem is equivalent to the claim that every planar cubic bridgeless graph is 3-edge-colorable.
I can't figure out or find any prove of this theorem.
Thanks.
graph-theory coloring
asked Jan 1 at 14:19
Přemysl Šťastný
1595
add a comment |
How can be proved, that
The four-color theorem is equivalent to the claim that every planar cubic bridgeless graph is 3-edge-colorable.
I can't figure out or find any prove of this theorem.
Thanks.
graph-theory coloring
asked Jan 1 at 14:19
Přemysl Šťastný
1595
How can be proved, that
The four-color theorem is equivalent to the claim that every planar cubic bridgeless graph is 3-edge-colorable.
I can't figure out or find any prove of this theorem.
Thanks.
graph-theory coloring
graph-theory coloring
asked Jan 1 at 14:19
Přemysl Šťastný
1595
asked Jan 1 at 14:19
Přemysl Šťastný
1595
asked Jan 1 at 14:19
Přemysl Šťastný
1595
asked Jan 1 at 14:19
Přemysl Šťastný
1595
asked Jan 1 at 14:19
Přemysl Šťastný
1595
1595
add a comment |
add a comment |
2 Answers
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The two problems of three-edge-coloring and four-face-coloring for the same map are equivalent.
A proof of this equivalency can be round here: http://www.mathpuzzle.com/4Dec2001.htm. Search for "material added 19 November 2001" within the page.
Since the four color problem has been already proved, also the three edge coloring is true.
answered Jan 2 at 23:59
Mario Stefanutti
607613
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I'd wondered about this before, but never worked it out until now. If $G$ is a cubic bridge-less graph, it is polyhedral and its dual is a maximal planar graph. Consider a maximal planar dual, $D(G)$ colored by the colors $(a, b, c, d$). Now color each edge, $e_{ij}$, according to $(i,j)$ like so:
$$(a, b) equiv (c,d) equiv 1$$
$$(a, c) equiv (b,d) equiv 2$$
$$(a, d) equiv (b,c) equiv 3$$
In each triangle, no two edges can possible have the same color, because that require all four vertex colors. Thus, in $G$, the edges all have different colors around every vertex. Since the edge coloring is valid around every vertex, it is valid for the whole graph $G$.
answered 2 days ago
Zachary Hunter
6019
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Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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2 Answers
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2 Answers
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The two problems of three-edge-coloring and four-face-coloring for the same map are equivalent.
A proof of this equivalency can be round here: http://www.mathpuzzle.com/4Dec2001.htm. Search for "material added 19 November 2001" within the page.
Since the four color problem has been already proved, also the three edge coloring is true.
answered Jan 2 at 23:59
Mario Stefanutti
607613
add a comment |
The two problems of three-edge-coloring and four-face-coloring for the same map are equivalent.
A proof of this equivalency can be round here: http://www.mathpuzzle.com/4Dec2001.htm. Search for "material added 19 November 2001" within the page.
Since the four color problem has been already proved, also the three edge coloring is true.
answered Jan 2 at 23:59
Mario Stefanutti
607613
add a comment |
The two problems of three-edge-coloring and four-face-coloring for the same map are equivalent.
A proof of this equivalency can be round here: http://www.mathpuzzle.com/4Dec2001.htm. Search for "material added 19 November 2001" within the page.
Since the four color problem has been already proved, also the three edge coloring is true.
answered Jan 2 at 23:59
Mario Stefanutti
607613
The two problems of three-edge-coloring and four-face-coloring for the same map are equivalent.
A proof of this equivalency can be round here: http://www.mathpuzzle.com/4Dec2001.htm. Search for "material added 19 November 2001" within the page.
Since the four color problem has been already proved, also the three edge coloring is true.
answered Jan 2 at 23:59
Mario Stefanutti
607613
answered Jan 2 at 23:59
Mario Stefanutti
607613
answered Jan 2 at 23:59
Mario Stefanutti
607613
answered Jan 2 at 23:59
Mario Stefanutti
607613
607613
add a comment |
add a comment |
I'd wondered about this before, but never worked it out until now. If $G$ is a cubic bridge-less graph, it is polyhedral and its dual is a maximal planar graph. Consider a maximal planar dual, $D(G)$ colored by the colors $(a, b, c, d$). Now color each edge, $e_{ij}$, according to $(i,j)$ like so:
$$(a, b) equiv (c,d) equiv 1$$
$$(a, c) equiv (b,d) equiv 2$$
$$(a, d) equiv (b,c) equiv 3$$
In each triangle, no two edges can possible have the same color, because that require all four vertex colors. Thus, in $G$, the edges all have different colors around every vertex. Since the edge coloring is valid around every vertex, it is valid for the whole graph $G$.
answered 2 days ago
Zachary Hunter
6019
New contributor
Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I'd wondered about this before, but never worked it out until now. If $G$ is a cubic bridge-less graph, it is polyhedral and its dual is a maximal planar graph. Consider a maximal planar dual, $D(G)$ colored by the colors $(a, b, c, d$). Now color each edge, $e_{ij}$, according to $(i,j)$ like so:
$$(a, b) equiv (c,d) equiv 1$$
$$(a, c) equiv (b,d) equiv 2$$
$$(a, d) equiv (b,c) equiv 3$$
In each triangle, no two edges can possible have the same color, because that require all four vertex colors. Thus, in $G$, the edges all have different colors around every vertex. Since the edge coloring is valid around every vertex, it is valid for the whole graph $G$.
answered 2 days ago
Zachary Hunter
6019
New contributor
Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I'd wondered about this before, but never worked it out until now. If $G$ is a cubic bridge-less graph, it is polyhedral and its dual is a maximal planar graph. Consider a maximal planar dual, $D(G)$ colored by the colors $(a, b, c, d$). Now color each edge, $e_{ij}$, according to $(i,j)$ like so:
$$(a, b) equiv (c,d) equiv 1$$
$$(a, c) equiv (b,d) equiv 2$$
$$(a, d) equiv (b,c) equiv 3$$
In each triangle, no two edges can possible have the same color, because that require all four vertex colors. Thus, in $G$, the edges all have different colors around every vertex. Since the edge coloring is valid around every vertex, it is valid for the whole graph $G$.
answered 2 days ago
Zachary Hunter
6019
New contributor
Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'd wondered about this before, but never worked it out until now. If $G$ is a cubic bridge-less graph, it is polyhedral and its dual is a maximal planar graph. Consider a maximal planar dual, $D(G)$ colored by the colors $(a, b, c, d$). Now color each edge, $e_{ij}$, according to $(i,j)$ like so:
$$(a, b) equiv (c,d) equiv 1$$
$$(a, c) equiv (b,d) equiv 2$$
$$(a, d) equiv (b,c) equiv 3$$
In each triangle, no two edges can possible have the same color, because that require all four vertex colors. Thus, in $G$, the edges all have different colors around every vertex. Since the edge coloring is valid around every vertex, it is valid for the whole graph $G$.
answered 2 days ago
Zachary Hunter
6019
New contributor
Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
Zachary Hunter
6019
New contributor
Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
Zachary Hunter
6019
answered 2 days ago
Zachary Hunter
6019
6019
New contributor
Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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