Question about uniquely colorable graphs
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Let $G$ be a uniquely $k$-colorable graph (for a definition, see
https://en.wikipedia.org/wiki/Uniquely_colorable_graph).
Is it true that we can always find a vertex $v$ such that $G-v$ is again uniquely $k$-colorable?
The answer is certainly 'yes' when $k=2$ and for the family of uniquely $4$-colorable Apollonian networks, but I have not yet been able to find a general proof, or a counter-example.
graph-theory
asked Dec 11 '18 at 17:57
Leen DroogendijkLeen Droogendijk
6,1701716
$endgroup$
add a comment |
$begingroup$
Let $G$ be a uniquely $k$-colorable graph (for a definition, see
https://en.wikipedia.org/wiki/Uniquely_colorable_graph).
Is it true that we can always find a vertex $v$ such that $G-v$ is again uniquely $k$-colorable?
The answer is certainly 'yes' when $k=2$ and for the family of uniquely $4$-colorable Apollonian networks, but I have not yet been able to find a general proof, or a counter-example.
graph-theory
asked Dec 11 '18 at 17:57
Leen DroogendijkLeen Droogendijk
6,1701716
$endgroup$
$begingroup$
A 1-vertex graph is still uniquely $k$-colorable for all $k$. The essence here is the partition in color classes.
$endgroup$
– Leen Droogendijk
Dec 11 '18 at 18:13
$begingroup$
$K_1$ is uniquely $2$-colorable: every $2$-coloring induces the same partition in color classes.
$endgroup$
– Leen Droogendijk
Dec 11 '18 at 18:15
add a comment |
$begingroup$
Let $G$ be a uniquely $k$-colorable graph (for a definition, see
https://en.wikipedia.org/wiki/Uniquely_colorable_graph).
Is it true that we can always find a vertex $v$ such that $G-v$ is again uniquely $k$-colorable?
The answer is certainly 'yes' when $k=2$ and for the family of uniquely $4$-colorable Apollonian networks, but I have not yet been able to find a general proof, or a counter-example.
graph-theory
asked Dec 11 '18 at 17:57
Leen DroogendijkLeen Droogendijk
6,1701716
$endgroup$
Let $G$ be a uniquely $k$-colorable graph (for a definition, see
https://en.wikipedia.org/wiki/Uniquely_colorable_graph).
Is it true that we can always find a vertex $v$ such that $G-v$ is again uniquely $k$-colorable?
The answer is certainly 'yes' when $k=2$ and for the family of uniquely $4$-colorable Apollonian networks, but I have not yet been able to find a general proof, or a counter-example.
graph-theory
graph-theory
asked Dec 11 '18 at 17:57
Leen DroogendijkLeen Droogendijk
6,1701716
asked Dec 11 '18 at 17:57
Leen DroogendijkLeen Droogendijk
6,1701716
asked Dec 11 '18 at 17:57
Leen DroogendijkLeen Droogendijk
6,1701716
asked Dec 11 '18 at 17:57
Leen DroogendijkLeen Droogendijk
6,1701716
asked Dec 11 '18 at 17:57
Leen DroogendijkLeen Droogendijk
6,1701716
6,1701716
$begingroup$
A 1-vertex graph is still uniquely $k$-colorable for all $k$. The essence here is the partition in color classes.
$endgroup$
– Leen Droogendijk
Dec 11 '18 at 18:13
$begingroup$
$K_1$ is uniquely $2$-colorable: every $2$-coloring induces the same partition in color classes.
$endgroup$
– Leen Droogendijk
Dec 11 '18 at 18:15
add a comment |
$begingroup$
A 1-vertex graph is still uniquely $k$-colorable for all $k$. The essence here is the partition in color classes.
$endgroup$
– Leen Droogendijk
Dec 11 '18 at 18:13
$begingroup$
$K_1$ is uniquely $2$-colorable: every $2$-coloring induces the same partition in color classes.
$endgroup$
– Leen Droogendijk
Dec 11 '18 at 18:15
$begingroup$
A 1-vertex graph is still uniquely $k$-colorable for all $k$. The essence here is the partition in color classes.
$endgroup$
– Leen Droogendijk
Dec 11 '18 at 18:13
$begingroup$
A 1-vertex graph is still uniquely $k$-colorable for all $k$. The essence here is the partition in color classes.
$endgroup$
– Leen Droogendijk
Dec 11 '18 at 18:13
$begingroup$
$K_1$ is uniquely $2$-colorable: every $2$-coloring induces the same partition in color classes.
$endgroup$
– Leen Droogendijk
Dec 11 '18 at 18:15
$begingroup$
$K_1$ is uniquely $2$-colorable: every $2$-coloring induces the same partition in color classes.
$endgroup$
– Leen Droogendijk
Dec 11 '18 at 18:15
add a comment |
1 Answer
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$begingroup$
The answer turns out to be "no".
There exist uniquely 3-colorable graphs without triangles.
The one on Mathworld (http://mathworld.wolfram.com/Uniquelyk-ColorableGraph.html) is incorrect (it has two different colorings as you can easily see by evaluating its chromatic polynomial at $k=3$ and dividing by $3!$), but there are others.
If the answer would be "yes" you can keep removing single vertices until you have three vertices left. Since there still is no triangle, the result cannot be uniquely $3$-colorable.
answered Dec 12 '18 at 9:33
Leen DroogendijkLeen Droogendijk
6,1701716
$endgroup$
add a comment |
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1 Answer
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$begingroup$
The answer turns out to be "no".
There exist uniquely 3-colorable graphs without triangles.
The one on Mathworld (http://mathworld.wolfram.com/Uniquelyk-ColorableGraph.html) is incorrect (it has two different colorings as you can easily see by evaluating its chromatic polynomial at $k=3$ and dividing by $3!$), but there are others.
If the answer would be "yes" you can keep removing single vertices until you have three vertices left. Since there still is no triangle, the result cannot be uniquely $3$-colorable.
answered Dec 12 '18 at 9:33
Leen DroogendijkLeen Droogendijk
6,1701716
$endgroup$
add a comment |
$begingroup$
The answer turns out to be "no".
There exist uniquely 3-colorable graphs without triangles.
The one on Mathworld (http://mathworld.wolfram.com/Uniquelyk-ColorableGraph.html) is incorrect (it has two different colorings as you can easily see by evaluating its chromatic polynomial at $k=3$ and dividing by $3!$), but there are others.
If the answer would be "yes" you can keep removing single vertices until you have three vertices left. Since there still is no triangle, the result cannot be uniquely $3$-colorable.
answered Dec 12 '18 at 9:33
Leen DroogendijkLeen Droogendijk
6,1701716
$endgroup$
add a comment |
$begingroup$
The answer turns out to be "no".
There exist uniquely 3-colorable graphs without triangles.
The one on Mathworld (http://mathworld.wolfram.com/Uniquelyk-ColorableGraph.html) is incorrect (it has two different colorings as you can easily see by evaluating its chromatic polynomial at $k=3$ and dividing by $3!$), but there are others.
If the answer would be "yes" you can keep removing single vertices until you have three vertices left. Since there still is no triangle, the result cannot be uniquely $3$-colorable.
answered Dec 12 '18 at 9:33
Leen DroogendijkLeen Droogendijk
6,1701716
$endgroup$
The answer turns out to be "no".
There exist uniquely 3-colorable graphs without triangles.
The one on Mathworld (http://mathworld.wolfram.com/Uniquelyk-ColorableGraph.html) is incorrect (it has two different colorings as you can easily see by evaluating its chromatic polynomial at $k=3$ and dividing by $3!$), but there are others.
If the answer would be "yes" you can keep removing single vertices until you have three vertices left. Since there still is no triangle, the result cannot be uniquely $3$-colorable.
answered Dec 12 '18 at 9:33
Leen DroogendijkLeen Droogendijk
6,1701716
answered Dec 12 '18 at 9:33
Leen DroogendijkLeen Droogendijk
6,1701716
answered Dec 12 '18 at 9:33
Leen DroogendijkLeen Droogendijk
6,1701716
answered Dec 12 '18 at 9:33
Leen DroogendijkLeen Droogendijk
6,1701716
6,1701716
add a comment |
add a comment |
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$begingroup$
A 1-vertex graph is still uniquely $k$-colorable for all $k$. The essence here is the partition in color classes.
$endgroup$
– Leen Droogendijk
Dec 11 '18 at 18:13
$begingroup$
$K_1$ is uniquely $2$-colorable: every $2$-coloring induces the same partition in color classes.
$endgroup$
– Leen Droogendijk
Dec 11 '18 at 18:15