Lower bound of chromatic number of some graph
$begingroup$
Here is the statement I'm trying to prove:
For $n = 2^m+1$, consider the graph on the pairs of integers $(i, j)$ with $1< i < j < n$ and edges between $(i, j)$ and $(k, l)$ if $j =k$.
Prove that the chromatic number is at least $m$.
I tried to do it by induction by finding some vertex that is connected to all (or enough) vertices of the graph for $m-1$, but didn't succeed.
I tried to find some structure, but I really don't know what to look for.
Let me know if you have some ideas!
Thank you
graph-theory
asked Dec 7 '18 at 11:24
SerwynSerwyn
62
$endgroup$
|
show 2 more comments
$begingroup$
Here is the statement I'm trying to prove:
For $n = 2^m+1$, consider the graph on the pairs of integers $(i, j)$ with $1< i < j < n$ and edges between $(i, j)$ and $(k, l)$ if $j =k$.
Prove that the chromatic number is at least $m$.
I tried to do it by induction by finding some vertex that is connected to all (or enough) vertices of the graph for $m-1$, but didn't succeed.
I tried to find some structure, but I really don't know what to look for.
Let me know if you have some ideas!
Thank you
graph-theory
asked Dec 7 '18 at 11:24
SerwynSerwyn
62
$endgroup$
$begingroup$
Is there any restriction on the pairs $(i,j)$ besides the inequality $ilt j$? Can $i$ and $j$ be any real numbers, or are they restricted to integers? In any case, it seems to be an infinite graph, and the chromatic number seems to be infinite.
$endgroup$
– bof
Dec 7 '18 at 12:00
$begingroup$
Was $ilt j$ a typo for $1le ilt jle n$? That will make it a finite graph. But what is $M$? Is $M=m$?
$endgroup$
– bof
Dec 7 '18 at 12:02
$begingroup$
Perhaps the answer to this question will help you: math.stackexchange.com/questions/579892/…
$endgroup$
– bof
Dec 7 '18 at 12:05
$begingroup$
Yes, n and j are integers smaller n, i edited it, thanks.
$endgroup$
– Serwyn
Dec 7 '18 at 13:30
$begingroup$
That's the same construction, thank you very much !
$endgroup$
– Serwyn
Dec 7 '18 at 13:47
|
show 2 more comments
$begingroup$
Here is the statement I'm trying to prove:
For $n = 2^m+1$, consider the graph on the pairs of integers $(i, j)$ with $1< i < j < n$ and edges between $(i, j)$ and $(k, l)$ if $j =k$.
Prove that the chromatic number is at least $m$.
I tried to do it by induction by finding some vertex that is connected to all (or enough) vertices of the graph for $m-1$, but didn't succeed.
I tried to find some structure, but I really don't know what to look for.
Let me know if you have some ideas!
Thank you
graph-theory
asked Dec 7 '18 at 11:24
SerwynSerwyn
62
$endgroup$
Here is the statement I'm trying to prove:
For $n = 2^m+1$, consider the graph on the pairs of integers $(i, j)$ with $1< i < j < n$ and edges between $(i, j)$ and $(k, l)$ if $j =k$.
Prove that the chromatic number is at least $m$.
I tried to do it by induction by finding some vertex that is connected to all (or enough) vertices of the graph for $m-1$, but didn't succeed.
I tried to find some structure, but I really don't know what to look for.
Let me know if you have some ideas!
Thank you
graph-theory
graph-theory
asked Dec 7 '18 at 11:24
SerwynSerwyn
62
asked Dec 7 '18 at 11:24
SerwynSerwyn
62
edited Dec 7 '18 at 13:29
Serwyn
asked Dec 7 '18 at 11:24
SerwynSerwyn
62
asked Dec 7 '18 at 11:24
SerwynSerwyn
62
asked Dec 7 '18 at 11:24
SerwynSerwyn
62
62
$begingroup$
Is there any restriction on the pairs $(i,j)$ besides the inequality $ilt j$? Can $i$ and $j$ be any real numbers, or are they restricted to integers? In any case, it seems to be an infinite graph, and the chromatic number seems to be infinite.
$endgroup$
– bof
Dec 7 '18 at 12:00
$begingroup$
Was $ilt j$ a typo for $1le ilt jle n$? That will make it a finite graph. But what is $M$? Is $M=m$?
$endgroup$
– bof
Dec 7 '18 at 12:02
$begingroup$
Perhaps the answer to this question will help you: math.stackexchange.com/questions/579892/…
$endgroup$
– bof
Dec 7 '18 at 12:05
$begingroup$
Yes, n and j are integers smaller n, i edited it, thanks.
$endgroup$
– Serwyn
Dec 7 '18 at 13:30
$begingroup$
That's the same construction, thank you very much !
$endgroup$
– Serwyn
Dec 7 '18 at 13:47
|
show 2 more comments
$begingroup$
Is there any restriction on the pairs $(i,j)$ besides the inequality $ilt j$? Can $i$ and $j$ be any real numbers, or are they restricted to integers? In any case, it seems to be an infinite graph, and the chromatic number seems to be infinite.
$endgroup$
– bof
Dec 7 '18 at 12:00
$begingroup$
Was $ilt j$ a typo for $1le ilt jle n$? That will make it a finite graph. But what is $M$? Is $M=m$?
$endgroup$
– bof
Dec 7 '18 at 12:02
$begingroup$
Perhaps the answer to this question will help you: math.stackexchange.com/questions/579892/…
$endgroup$
– bof
Dec 7 '18 at 12:05
$begingroup$
Yes, n and j are integers smaller n, i edited it, thanks.
$endgroup$
– Serwyn
Dec 7 '18 at 13:30
$begingroup$
That's the same construction, thank you very much !
$endgroup$
– Serwyn
Dec 7 '18 at 13:47
$begingroup$
Is there any restriction on the pairs $(i,j)$ besides the inequality $ilt j$? Can $i$ and $j$ be any real numbers, or are they restricted to integers? In any case, it seems to be an infinite graph, and the chromatic number seems to be infinite.
$endgroup$
– bof
Dec 7 '18 at 12:00
$begingroup$
Is there any restriction on the pairs $(i,j)$ besides the inequality $ilt j$? Can $i$ and $j$ be any real numbers, or are they restricted to integers? In any case, it seems to be an infinite graph, and the chromatic number seems to be infinite.
$endgroup$
– bof
Dec 7 '18 at 12:00
$begingroup$
Was $ilt j$ a typo for $1le ilt jle n$? That will make it a finite graph. But what is $M$? Is $M=m$?
$endgroup$
– bof
Dec 7 '18 at 12:02
$begingroup$
Was $ilt j$ a typo for $1le ilt jle n$? That will make it a finite graph. But what is $M$? Is $M=m$?
$endgroup$
– bof
Dec 7 '18 at 12:02
$begingroup$
Perhaps the answer to this question will help you: math.stackexchange.com/questions/579892/…
$endgroup$
– bof
Dec 7 '18 at 12:05
$begingroup$
Perhaps the answer to this question will help you: math.stackexchange.com/questions/579892/…
$endgroup$
– bof
Dec 7 '18 at 12:05
$begingroup$
Yes, n and j are integers smaller n, i edited it, thanks.
$endgroup$
– Serwyn
Dec 7 '18 at 13:30
$begingroup$
Yes, n and j are integers smaller n, i edited it, thanks.
$endgroup$
– Serwyn
Dec 7 '18 at 13:30
$begingroup$
That's the same construction, thank you very much !
$endgroup$
– Serwyn
Dec 7 '18 at 13:47
$begingroup$
That's the same construction, thank you very much !
$endgroup$
– Serwyn
Dec 7 '18 at 13:47
|
show 2 more comments
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$begingroup$
Is there any restriction on the pairs $(i,j)$ besides the inequality $ilt j$? Can $i$ and $j$ be any real numbers, or are they restricted to integers? In any case, it seems to be an infinite graph, and the chromatic number seems to be infinite.
$endgroup$
– bof
Dec 7 '18 at 12:00
$begingroup$
Was $ilt j$ a typo for $1le ilt jle n$? That will make it a finite graph. But what is $M$? Is $M=m$?
$endgroup$
– bof
Dec 7 '18 at 12:02
$begingroup$
Perhaps the answer to this question will help you: math.stackexchange.com/questions/579892/…
$endgroup$
– bof
Dec 7 '18 at 12:05
$begingroup$
Yes, n and j are integers smaller n, i edited it, thanks.
$endgroup$
– Serwyn
Dec 7 '18 at 13:30
$begingroup$
That's the same construction, thank you very much !
$endgroup$
– Serwyn
Dec 7 '18 at 13:47