Finding the smallest number $n$ such that every two-colouring of the edges of $K_n$ contains a Path on 3...
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What is the smallest number $n$ such that every two-colouring of the edges of $K_n$ contains a (not necessarily induced) path on 3 vertices?
graph-theory coloring
edited Dec 6 '18 at 23:26
user376343
3,2982825
asked Dec 6 '18 at 16:46
ipponippon
1345
$endgroup$
add a comment |
$begingroup$
What is the smallest number $n$ such that every two-colouring of the edges of $K_n$ contains a (not necessarily induced) path on 3 vertices?
graph-theory coloring
edited Dec 6 '18 at 23:26
user376343
3,2982825
asked Dec 6 '18 at 16:46
ipponippon
1345
$endgroup$
$begingroup$
Sorry, Do you mean a path where every edges has the same color?
$endgroup$
– nafhgood
Dec 6 '18 at 16:55
$begingroup$
@mathnoob yes, all edges should have to have the same color
$endgroup$
– ippon
Dec 6 '18 at 17:01
$begingroup$
Seems like $K_3$ already does the job which makes me think that you meant to phrase the problem differently. (A path on $3$ edges, for example.)
$endgroup$
– Misha Lavrov
Dec 6 '18 at 17:02
add a comment |
$begingroup$
What is the smallest number $n$ such that every two-colouring of the edges of $K_n$ contains a (not necessarily induced) path on 3 vertices?
graph-theory coloring
edited Dec 6 '18 at 23:26
user376343
3,2982825
asked Dec 6 '18 at 16:46
ipponippon
1345
$endgroup$
What is the smallest number $n$ such that every two-colouring of the edges of $K_n$ contains a (not necessarily induced) path on 3 vertices?
graph-theory coloring
graph-theory coloring
edited Dec 6 '18 at 23:26
user376343
3,2982825
asked Dec 6 '18 at 16:46
ipponippon
1345
edited Dec 6 '18 at 23:26
user376343
3,2982825
asked Dec 6 '18 at 16:46
ipponippon
1345
edited Dec 6 '18 at 23:26
user376343
3,2982825
edited Dec 6 '18 at 23:26
user376343
3,2982825
edited Dec 6 '18 at 23:26
user376343
3,2982825
3,2982825
asked Dec 6 '18 at 16:46
ipponippon
1345
asked Dec 6 '18 at 16:46
ipponippon
1345
asked Dec 6 '18 at 16:46
ipponippon
1345
1345
$begingroup$
Sorry, Do you mean a path where every edges has the same color?
$endgroup$
– nafhgood
Dec 6 '18 at 16:55
$begingroup$
@mathnoob yes, all edges should have to have the same color
$endgroup$
– ippon
Dec 6 '18 at 17:01
$begingroup$
Seems like $K_3$ already does the job which makes me think that you meant to phrase the problem differently. (A path on $3$ edges, for example.)
$endgroup$
– Misha Lavrov
Dec 6 '18 at 17:02
add a comment |
$begingroup$
Sorry, Do you mean a path where every edges has the same color?
$endgroup$
– nafhgood
Dec 6 '18 at 16:55
$begingroup$
@mathnoob yes, all edges should have to have the same color
$endgroup$
– ippon
Dec 6 '18 at 17:01
$begingroup$
Seems like $K_3$ already does the job which makes me think that you meant to phrase the problem differently. (A path on $3$ edges, for example.)
$endgroup$
– Misha Lavrov
Dec 6 '18 at 17:02
$begingroup$
Sorry, Do you mean a path where every edges has the same color?
$endgroup$
– nafhgood
Dec 6 '18 at 16:55
$begingroup$
Sorry, Do you mean a path where every edges has the same color?
$endgroup$
– nafhgood
Dec 6 '18 at 16:55
$begingroup$
@mathnoob yes, all edges should have to have the same color
$endgroup$
– ippon
Dec 6 '18 at 17:01
$begingroup$
@mathnoob yes, all edges should have to have the same color
$endgroup$
– ippon
Dec 6 '18 at 17:01
$begingroup$
Seems like $K_3$ already does the job which makes me think that you meant to phrase the problem differently. (A path on $3$ edges, for example.)
$endgroup$
– Misha Lavrov
Dec 6 '18 at 17:02
$begingroup$
Seems like $K_3$ already does the job which makes me think that you meant to phrase the problem differently. (A path on $3$ edges, for example.)
$endgroup$
– Misha Lavrov
Dec 6 '18 at 17:02
add a comment |
1 Answer
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If it is a path on 3 vertices , $K_3$ works as it is garanted that two of the three edges have the same color, well then they form a path on three vertices. If what we want is a path of length $3$, my guess is $5$. Look at a vertex $u in V(K_n)$, there are 4 edges connect to it. It is guarantee that at least two of the edges $e_1,e_2$ connected to it have the same color. So the edges connected to $e_1$ and $e_2$ not via $u$ must be color the other color. But then that gives a path of length 3 in the other color.
Also, here is an example of coloring of $K_4$ that does not admit a path of length 3, the edges are colored using numbers 1 and 3.
answered Dec 6 '18 at 17:05
nafhgoodnafhgood
1,803422
$endgroup$
$begingroup$
Am I right????????????
$endgroup$
– nafhgood
Dec 6 '18 at 17:19
$begingroup$
Seems right to me, modulo a coloring of $K_4$ that does not contain a path of length $3$ (which is not hard to find).
$endgroup$
– Misha Lavrov
Dec 6 '18 at 18:17
$begingroup$
Great! Thank you!
$endgroup$
– nafhgood
Dec 6 '18 at 18:20
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
If it is a path on 3 vertices , $K_3$ works as it is garanted that two of the three edges have the same color, well then they form a path on three vertices. If what we want is a path of length $3$, my guess is $5$. Look at a vertex $u in V(K_n)$, there are 4 edges connect to it. It is guarantee that at least two of the edges $e_1,e_2$ connected to it have the same color. So the edges connected to $e_1$ and $e_2$ not via $u$ must be color the other color. But then that gives a path of length 3 in the other color.
Also, here is an example of coloring of $K_4$ that does not admit a path of length 3, the edges are colored using numbers 1 and 3.
answered Dec 6 '18 at 17:05
nafhgoodnafhgood
1,803422
$endgroup$
$begingroup$
Am I right????????????
$endgroup$
– nafhgood
Dec 6 '18 at 17:19
$begingroup$
Seems right to me, modulo a coloring of $K_4$ that does not contain a path of length $3$ (which is not hard to find).
$endgroup$
– Misha Lavrov
Dec 6 '18 at 18:17
$begingroup$
Great! Thank you!
$endgroup$
– nafhgood
Dec 6 '18 at 18:20
add a comment |
$begingroup$
If it is a path on 3 vertices , $K_3$ works as it is garanted that two of the three edges have the same color, well then they form a path on three vertices. If what we want is a path of length $3$, my guess is $5$. Look at a vertex $u in V(K_n)$, there are 4 edges connect to it. It is guarantee that at least two of the edges $e_1,e_2$ connected to it have the same color. So the edges connected to $e_1$ and $e_2$ not via $u$ must be color the other color. But then that gives a path of length 3 in the other color.
Also, here is an example of coloring of $K_4$ that does not admit a path of length 3, the edges are colored using numbers 1 and 3.
answered Dec 6 '18 at 17:05
nafhgoodnafhgood
1,803422
$endgroup$
$begingroup$
Am I right????????????
$endgroup$
– nafhgood
Dec 6 '18 at 17:19
$begingroup$
Seems right to me, modulo a coloring of $K_4$ that does not contain a path of length $3$ (which is not hard to find).
$endgroup$
– Misha Lavrov
Dec 6 '18 at 18:17
$begingroup$
Great! Thank you!
$endgroup$
– nafhgood
Dec 6 '18 at 18:20
add a comment |
$begingroup$
If it is a path on 3 vertices , $K_3$ works as it is garanted that two of the three edges have the same color, well then they form a path on three vertices. If what we want is a path of length $3$, my guess is $5$. Look at a vertex $u in V(K_n)$, there are 4 edges connect to it. It is guarantee that at least two of the edges $e_1,e_2$ connected to it have the same color. So the edges connected to $e_1$ and $e_2$ not via $u$ must be color the other color. But then that gives a path of length 3 in the other color.
Also, here is an example of coloring of $K_4$ that does not admit a path of length 3, the edges are colored using numbers 1 and 3.
answered Dec 6 '18 at 17:05
nafhgoodnafhgood
1,803422
$endgroup$
If it is a path on 3 vertices , $K_3$ works as it is garanted that two of the three edges have the same color, well then they form a path on three vertices. If what we want is a path of length $3$, my guess is $5$. Look at a vertex $u in V(K_n)$, there are 4 edges connect to it. It is guarantee that at least two of the edges $e_1,e_2$ connected to it have the same color. So the edges connected to $e_1$ and $e_2$ not via $u$ must be color the other color. But then that gives a path of length 3 in the other color.
Also, here is an example of coloring of $K_4$ that does not admit a path of length 3, the edges are colored using numbers 1 and 3.
answered Dec 6 '18 at 17:05
nafhgoodnafhgood
1,803422
edited Dec 6 '18 at 20:12
answered Dec 6 '18 at 17:05
nafhgoodnafhgood
1,803422
answered Dec 6 '18 at 17:05
nafhgoodnafhgood
1,803422
answered Dec 6 '18 at 17:05
nafhgoodnafhgood
1,803422
1,803422
$begingroup$
Am I right????????????
$endgroup$
– nafhgood
Dec 6 '18 at 17:19
$begingroup$
Seems right to me, modulo a coloring of $K_4$ that does not contain a path of length $3$ (which is not hard to find).
$endgroup$
– Misha Lavrov
Dec 6 '18 at 18:17
$begingroup$
Great! Thank you!
$endgroup$
– nafhgood
Dec 6 '18 at 18:20
add a comment |
$begingroup$
Am I right????????????
$endgroup$
– nafhgood
Dec 6 '18 at 17:19
$begingroup$
Seems right to me, modulo a coloring of $K_4$ that does not contain a path of length $3$ (which is not hard to find).
$endgroup$
– Misha Lavrov
Dec 6 '18 at 18:17
$begingroup$
Great! Thank you!
$endgroup$
– nafhgood
Dec 6 '18 at 18:20
$begingroup$
Am I right????????????
$endgroup$
– nafhgood
Dec 6 '18 at 17:19
$begingroup$
Am I right????????????
$endgroup$
– nafhgood
Dec 6 '18 at 17:19
$begingroup$
Seems right to me, modulo a coloring of $K_4$ that does not contain a path of length $3$ (which is not hard to find).
$endgroup$
– Misha Lavrov
Dec 6 '18 at 18:17
$begingroup$
Seems right to me, modulo a coloring of $K_4$ that does not contain a path of length $3$ (which is not hard to find).
$endgroup$
– Misha Lavrov
Dec 6 '18 at 18:17
$begingroup$
Great! Thank you!
$endgroup$
– nafhgood
Dec 6 '18 at 18:20
$begingroup$
Great! Thank you!
$endgroup$
– nafhgood
Dec 6 '18 at 18:20
add a comment |
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$begingroup$
Sorry, Do you mean a path where every edges has the same color?
$endgroup$
– nafhgood
Dec 6 '18 at 16:55
$begingroup$
@mathnoob yes, all edges should have to have the same color
$endgroup$
– ippon
Dec 6 '18 at 17:01
$begingroup$
Seems like $K_3$ already does the job which makes me think that you meant to phrase the problem differently. (A path on $3$ edges, for example.)
$endgroup$
– Misha Lavrov
Dec 6 '18 at 17:02