Finding the smallest number $n$ such that every two-colouring of the edges of $K_n$ contains a Path on 3...












2












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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?










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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
















2












$begingroup$


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?










share|cite|improve this question











$endgroup$












  • $begingroup$
    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














2












2








2





$begingroup$


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?










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Dec 6 '18 at 23:26









user376343

3,7383827




3,7383827










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


















  • $begingroup$
    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










1 Answer
1






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2












$begingroup$

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.
enter image description here






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

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2












$begingroup$

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.
enter image description here






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















2












$begingroup$

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.
enter image description here






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














2












2








2





$begingroup$

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.
enter image description here






share|cite|improve this answer











$endgroup$



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.
enter image description here







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 6 '18 at 20:12

























answered Dec 6 '18 at 17:05









nafhgoodnafhgood

1,805422




1,805422












  • $begingroup$
    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










  • $begingroup$
    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


















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