Number of edges in Turán graph












1












$begingroup$


I am trying to prove that the number of edges in the Turán graph $T^r(n)$ (that is, the complete $r$-partite graph on $n$ vertices whose partition sets differ in size by at most 1) satisfy the inequality
$$|E(T^r(n))| geq (1-frac{1}{r} - o(1))frac{n^2}{2},$$
and I'm stuck. Any help is much appreciated.






graph-theory extremal-graph-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    I guess $o(1)$ means $o(1)$ as $ntoinfty$ with $r$ fixed?
    $endgroup$
    – bof
    Nov 9 '18 at 8:42










  • $begingroup$
    This is my understanding as well, but I am still stuck on how to prove it.
    $endgroup$
    – Ilefen
    Nov 9 '18 at 8:43










  • $begingroup$
    I think it follows from $|E(T^r(n))|ge(1-frac1r)binom n2$ which follow from the fact that each vertex $v$ of $T^r(n)$ has degree $ge(1-frac1r)(n-1)$.
    $endgroup$
    – bof
    Nov 9 '18 at 8:49










  • $begingroup$
    There is a result in Bollobas' Modern Graph Theory which states that the number of edges of $T^r(n)$ is $(1-frac{1}{r}+o(1))binom{n}{2}$, but the result I am trying to prove has $frac{n^2}{2}$ instead of $binom{n}{2}$, that's another thing that confuses me.
    $endgroup$
    – Ilefen
    Nov 9 '18 at 12:22






  • 1




    $begingroup$
    $binom n2=(1-frac1n)frac{n^2}2=(1-o(1))frac{n^2}2$
    $endgroup$
    – bof
    Nov 9 '18 at 12:30


















1












$begingroup$


I am trying to prove that the number of edges in the Turán graph $T^r(n)$ (that is, the complete $r$-partite graph on $n$ vertices whose partition sets differ in size by at most 1) satisfy the inequality
$$|E(T^r(n))| geq (1-frac{1}{r} - o(1))frac{n^2}{2},$$
and I'm stuck. Any help is much appreciated.






graph-theory extremal-graph-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    I guess $o(1)$ means $o(1)$ as $ntoinfty$ with $r$ fixed?
    $endgroup$
    – bof
    Nov 9 '18 at 8:42










  • $begingroup$
    This is my understanding as well, but I am still stuck on how to prove it.
    $endgroup$
    – Ilefen
    Nov 9 '18 at 8:43










  • $begingroup$
    I think it follows from $|E(T^r(n))|ge(1-frac1r)binom n2$ which follow from the fact that each vertex $v$ of $T^r(n)$ has degree $ge(1-frac1r)(n-1)$.
    $endgroup$
    – bof
    Nov 9 '18 at 8:49










  • $begingroup$
    There is a result in Bollobas' Modern Graph Theory which states that the number of edges of $T^r(n)$ is $(1-frac{1}{r}+o(1))binom{n}{2}$, but the result I am trying to prove has $frac{n^2}{2}$ instead of $binom{n}{2}$, that's another thing that confuses me.
    $endgroup$
    – Ilefen
    Nov 9 '18 at 12:22






  • 1




    $begingroup$
    $binom n2=(1-frac1n)frac{n^2}2=(1-o(1))frac{n^2}2$
    $endgroup$
    – bof
    Nov 9 '18 at 12:30
















1












1








1





$begingroup$


I am trying to prove that the number of edges in the Turán graph $T^r(n)$ (that is, the complete $r$-partite graph on $n$ vertices whose partition sets differ in size by at most 1) satisfy the inequality
$$|E(T^r(n))| geq (1-frac{1}{r} - o(1))frac{n^2}{2},$$
and I'm stuck. Any help is much appreciated.






graph-theory extremal-graph-theory







share|cite|improve this question











$endgroup$




I am trying to prove that the number of edges in the Turán graph $T^r(n)$ (that is, the complete $r$-partite graph on $n$ vertices whose partition sets differ in size by at most 1) satisfy the inequality
$$|E(T^r(n))| geq (1-frac{1}{r} - o(1))frac{n^2}{2},$$
and I'm stuck. Any help is much appreciated.






graph-theory extremal-graph-theory




graph-theory extremal-graph-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 9 '18 at 8:38









bof

51k457120




51k457120










asked Nov 9 '18 at 8:22









IlefenIlefen

485215




485215












  • $begingroup$
    I guess $o(1)$ means $o(1)$ as $ntoinfty$ with $r$ fixed?
    $endgroup$
    – bof
    Nov 9 '18 at 8:42










  • $begingroup$
    This is my understanding as well, but I am still stuck on how to prove it.
    $endgroup$
    – Ilefen
    Nov 9 '18 at 8:43










  • $begingroup$
    I think it follows from $|E(T^r(n))|ge(1-frac1r)binom n2$ which follow from the fact that each vertex $v$ of $T^r(n)$ has degree $ge(1-frac1r)(n-1)$.
    $endgroup$
    – bof
    Nov 9 '18 at 8:49










  • $begingroup$
    There is a result in Bollobas' Modern Graph Theory which states that the number of edges of $T^r(n)$ is $(1-frac{1}{r}+o(1))binom{n}{2}$, but the result I am trying to prove has $frac{n^2}{2}$ instead of $binom{n}{2}$, that's another thing that confuses me.
    $endgroup$
    – Ilefen
    Nov 9 '18 at 12:22






  • 1




    $begingroup$
    $binom n2=(1-frac1n)frac{n^2}2=(1-o(1))frac{n^2}2$
    $endgroup$
    – bof
    Nov 9 '18 at 12:30




















  • $begingroup$
    I guess $o(1)$ means $o(1)$ as $ntoinfty$ with $r$ fixed?
    $endgroup$
    – bof
    Nov 9 '18 at 8:42










  • $begingroup$
    This is my understanding as well, but I am still stuck on how to prove it.
    $endgroup$
    – Ilefen
    Nov 9 '18 at 8:43










  • $begingroup$
    I think it follows from $|E(T^r(n))|ge(1-frac1r)binom n2$ which follow from the fact that each vertex $v$ of $T^r(n)$ has degree $ge(1-frac1r)(n-1)$.
    $endgroup$
    – bof
    Nov 9 '18 at 8:49










  • $begingroup$
    There is a result in Bollobas' Modern Graph Theory which states that the number of edges of $T^r(n)$ is $(1-frac{1}{r}+o(1))binom{n}{2}$, but the result I am trying to prove has $frac{n^2}{2}$ instead of $binom{n}{2}$, that's another thing that confuses me.
    $endgroup$
    – Ilefen
    Nov 9 '18 at 12:22






  • 1




    $begingroup$
    $binom n2=(1-frac1n)frac{n^2}2=(1-o(1))frac{n^2}2$
    $endgroup$
    – bof
    Nov 9 '18 at 12:30


















$begingroup$
I guess $o(1)$ means $o(1)$ as $ntoinfty$ with $r$ fixed?
$endgroup$
– bof
Nov 9 '18 at 8:42




$begingroup$
I guess $o(1)$ means $o(1)$ as $ntoinfty$ with $r$ fixed?
$endgroup$
– bof
Nov 9 '18 at 8:42












$begingroup$
This is my understanding as well, but I am still stuck on how to prove it.
$endgroup$
– Ilefen
Nov 9 '18 at 8:43




$begingroup$
This is my understanding as well, but I am still stuck on how to prove it.
$endgroup$
– Ilefen
Nov 9 '18 at 8:43












$begingroup$
I think it follows from $|E(T^r(n))|ge(1-frac1r)binom n2$ which follow from the fact that each vertex $v$ of $T^r(n)$ has degree $ge(1-frac1r)(n-1)$.
$endgroup$
– bof
Nov 9 '18 at 8:49




$begingroup$
I think it follows from $|E(T^r(n))|ge(1-frac1r)binom n2$ which follow from the fact that each vertex $v$ of $T^r(n)$ has degree $ge(1-frac1r)(n-1)$.
$endgroup$
– bof
Nov 9 '18 at 8:49












$begingroup$
There is a result in Bollobas' Modern Graph Theory which states that the number of edges of $T^r(n)$ is $(1-frac{1}{r}+o(1))binom{n}{2}$, but the result I am trying to prove has $frac{n^2}{2}$ instead of $binom{n}{2}$, that's another thing that confuses me.
$endgroup$
– Ilefen
Nov 9 '18 at 12:22




$begingroup$
There is a result in Bollobas' Modern Graph Theory which states that the number of edges of $T^r(n)$ is $(1-frac{1}{r}+o(1))binom{n}{2}$, but the result I am trying to prove has $frac{n^2}{2}$ instead of $binom{n}{2}$, that's another thing that confuses me.
$endgroup$
– Ilefen
Nov 9 '18 at 12:22




1




1




$begingroup$
$binom n2=(1-frac1n)frac{n^2}2=(1-o(1))frac{n^2}2$
$endgroup$
– bof
Nov 9 '18 at 12:30






$begingroup$
$binom n2=(1-frac1n)frac{n^2}2=(1-o(1))frac{n^2}2$
$endgroup$
– bof
Nov 9 '18 at 12:30












1 Answer
1






active

oldest

votes


















0












$begingroup$

I recall that the number $|E(T^r(n))|$ can be easily calculated exactly. Let $n=qr+s$, where $0le s<r$. Then the $r$-partition of the vertex set of the graph $T^r(n)$ has $s$ parts of size $q+1$ and $r-s$ parts of size $q$. Thus



$$|E(T^r(n))|={s choose 2}(q+1)^2+s(r-s)q(q+1)+ {r-s choose 2}q^2=frac 12left( q^2r^2-q^2r+2sqr-2sq+s^2-sright)=frac 12left(n^2-q(n+s)-sright)= frac 12left(n^2-frac {(n-s)(n+s)}{r}-sright)= frac 12left(n^2-frac {(n^2-s^2)}{r}-sright).$$



Thus
$$left|E(T^r(n))|-frac{n^2}{2}left(1-frac 1rright)right|=frac{s(r-s)}{2r}le left(frac {s+r-s}2right)^2frac 1{2r}=frac r8.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    But how can I derive from this that $|E(T^r(n))| geq (1-frac{1}{r}-o(1)) frac{n^2}{2}$?
    $endgroup$
    – Ilefen
    Dec 8 '18 at 13:07












  • $begingroup$
    Since $r$ is fixed, we have even $|E(T^r(n))|=left (1-frac 1r+Oleft(frac 1{n^2}right)right)frac {n^2}{2}.$
    $endgroup$
    – Alex Ravsky
    Dec 8 '18 at 17:28











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1 Answer
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oldest

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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









0












$begingroup$

I recall that the number $|E(T^r(n))|$ can be easily calculated exactly. Let $n=qr+s$, where $0le s<r$. Then the $r$-partition of the vertex set of the graph $T^r(n)$ has $s$ parts of size $q+1$ and $r-s$ parts of size $q$. Thus



$$|E(T^r(n))|={s choose 2}(q+1)^2+s(r-s)q(q+1)+ {r-s choose 2}q^2=frac 12left( q^2r^2-q^2r+2sqr-2sq+s^2-sright)=frac 12left(n^2-q(n+s)-sright)= frac 12left(n^2-frac {(n-s)(n+s)}{r}-sright)= frac 12left(n^2-frac {(n^2-s^2)}{r}-sright).$$



Thus
$$left|E(T^r(n))|-frac{n^2}{2}left(1-frac 1rright)right|=frac{s(r-s)}{2r}le left(frac {s+r-s}2right)^2frac 1{2r}=frac r8.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    But how can I derive from this that $|E(T^r(n))| geq (1-frac{1}{r}-o(1)) frac{n^2}{2}$?
    $endgroup$
    – Ilefen
    Dec 8 '18 at 13:07












  • $begingroup$
    Since $r$ is fixed, we have even $|E(T^r(n))|=left (1-frac 1r+Oleft(frac 1{n^2}right)right)frac {n^2}{2}.$
    $endgroup$
    – Alex Ravsky
    Dec 8 '18 at 17:28
















0












$begingroup$

I recall that the number $|E(T^r(n))|$ can be easily calculated exactly. Let $n=qr+s$, where $0le s<r$. Then the $r$-partition of the vertex set of the graph $T^r(n)$ has $s$ parts of size $q+1$ and $r-s$ parts of size $q$. Thus



$$|E(T^r(n))|={s choose 2}(q+1)^2+s(r-s)q(q+1)+ {r-s choose 2}q^2=frac 12left( q^2r^2-q^2r+2sqr-2sq+s^2-sright)=frac 12left(n^2-q(n+s)-sright)= frac 12left(n^2-frac {(n-s)(n+s)}{r}-sright)= frac 12left(n^2-frac {(n^2-s^2)}{r}-sright).$$



Thus
$$left|E(T^r(n))|-frac{n^2}{2}left(1-frac 1rright)right|=frac{s(r-s)}{2r}le left(frac {s+r-s}2right)^2frac 1{2r}=frac r8.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    But how can I derive from this that $|E(T^r(n))| geq (1-frac{1}{r}-o(1)) frac{n^2}{2}$?
    $endgroup$
    – Ilefen
    Dec 8 '18 at 13:07












  • $begingroup$
    Since $r$ is fixed, we have even $|E(T^r(n))|=left (1-frac 1r+Oleft(frac 1{n^2}right)right)frac {n^2}{2}.$
    $endgroup$
    – Alex Ravsky
    Dec 8 '18 at 17:28














0












0








0





$begingroup$

I recall that the number $|E(T^r(n))|$ can be easily calculated exactly. Let $n=qr+s$, where $0le s<r$. Then the $r$-partition of the vertex set of the graph $T^r(n)$ has $s$ parts of size $q+1$ and $r-s$ parts of size $q$. Thus



$$|E(T^r(n))|={s choose 2}(q+1)^2+s(r-s)q(q+1)+ {r-s choose 2}q^2=frac 12left( q^2r^2-q^2r+2sqr-2sq+s^2-sright)=frac 12left(n^2-q(n+s)-sright)= frac 12left(n^2-frac {(n-s)(n+s)}{r}-sright)= frac 12left(n^2-frac {(n^2-s^2)}{r}-sright).$$



Thus
$$left|E(T^r(n))|-frac{n^2}{2}left(1-frac 1rright)right|=frac{s(r-s)}{2r}le left(frac {s+r-s}2right)^2frac 1{2r}=frac r8.$$






share|cite|improve this answer









$endgroup$



I recall that the number $|E(T^r(n))|$ can be easily calculated exactly. Let $n=qr+s$, where $0le s<r$. Then the $r$-partition of the vertex set of the graph $T^r(n)$ has $s$ parts of size $q+1$ and $r-s$ parts of size $q$. Thus



$$|E(T^r(n))|={s choose 2}(q+1)^2+s(r-s)q(q+1)+ {r-s choose 2}q^2=frac 12left( q^2r^2-q^2r+2sqr-2sq+s^2-sright)=frac 12left(n^2-q(n+s)-sright)= frac 12left(n^2-frac {(n-s)(n+s)}{r}-sright)= frac 12left(n^2-frac {(n^2-s^2)}{r}-sright).$$



Thus
$$left|E(T^r(n))|-frac{n^2}{2}left(1-frac 1rright)right|=frac{s(r-s)}{2r}le left(frac {s+r-s}2right)^2frac 1{2r}=frac r8.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 2 '18 at 12:48









Alex RavskyAlex Ravsky

39.7k32281




39.7k32281












  • $begingroup$
    But how can I derive from this that $|E(T^r(n))| geq (1-frac{1}{r}-o(1)) frac{n^2}{2}$?
    $endgroup$
    – Ilefen
    Dec 8 '18 at 13:07












  • $begingroup$
    Since $r$ is fixed, we have even $|E(T^r(n))|=left (1-frac 1r+Oleft(frac 1{n^2}right)right)frac {n^2}{2}.$
    $endgroup$
    – Alex Ravsky
    Dec 8 '18 at 17:28


















  • $begingroup$
    But how can I derive from this that $|E(T^r(n))| geq (1-frac{1}{r}-o(1)) frac{n^2}{2}$?
    $endgroup$
    – Ilefen
    Dec 8 '18 at 13:07












  • $begingroup$
    Since $r$ is fixed, we have even $|E(T^r(n))|=left (1-frac 1r+Oleft(frac 1{n^2}right)right)frac {n^2}{2}.$
    $endgroup$
    – Alex Ravsky
    Dec 8 '18 at 17:28
















$begingroup$
But how can I derive from this that $|E(T^r(n))| geq (1-frac{1}{r}-o(1)) frac{n^2}{2}$?
$endgroup$
– Ilefen
Dec 8 '18 at 13:07






$begingroup$
But how can I derive from this that $|E(T^r(n))| geq (1-frac{1}{r}-o(1)) frac{n^2}{2}$?
$endgroup$
– Ilefen
Dec 8 '18 at 13:07














$begingroup$
Since $r$ is fixed, we have even $|E(T^r(n))|=left (1-frac 1r+Oleft(frac 1{n^2}right)right)frac {n^2}{2}.$
$endgroup$
– Alex Ravsky
Dec 8 '18 at 17:28




$begingroup$
Since $r$ is fixed, we have even $|E(T^r(n))|=left (1-frac 1r+Oleft(frac 1{n^2}right)right)frac {n^2}{2}.$
$endgroup$
– Alex Ravsky
Dec 8 '18 at 17:28


















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