Chromatic number on a graph












2












$begingroup$



Let $G$ be any graph with order $p$ and size $q$. Prove that $$χ(G)ge {p^2 over p^2-2q}$$




If someone could give me a hint it would be amazing!






graph-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    I guess order is number of vertices, so what is size?
    $endgroup$
    – coffeemath
    Dec 6 '18 at 3:35






  • 1




    $begingroup$
    number of edges
    $endgroup$
    – John Smith
    Dec 6 '18 at 4:05










  • $begingroup$
    You want to prove $qleq p^2cdotfrac{chi-1}{2chi}$. Take a graph with $p$ vertices and $chi$-coloring. What is the most edges it can have if all bipartite subgraphs between different color pairs are complete?
    $endgroup$
    – Michal Adamaszek
    Dec 6 '18 at 7:20










  • $begingroup$
    Wow that is really cool! I had another idea. Is it possible to get this directly, by finding a lower bound on the largest clique in a graph? Then $chi geq textrm{largest clique size}$. My idea was to use an averaging method, but no success yet.
    $endgroup$
    – Mriganka Basu Roy Chowdhury
    Dec 6 '18 at 7:30








  • 1




    $begingroup$
    Possible duplicate of Prove that if G is a simple graph, $chi geq frac{|V|^2}{|V|^2-2|E|}$
    $endgroup$
    – Kuifje
    Dec 6 '18 at 9:20
















2












$begingroup$



Let $G$ be any graph with order $p$ and size $q$. Prove that $$χ(G)ge {p^2 over p^2-2q}$$




If someone could give me a hint it would be amazing!






graph-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    I guess order is number of vertices, so what is size?
    $endgroup$
    – coffeemath
    Dec 6 '18 at 3:35






  • 1




    $begingroup$
    number of edges
    $endgroup$
    – John Smith
    Dec 6 '18 at 4:05










  • $begingroup$
    You want to prove $qleq p^2cdotfrac{chi-1}{2chi}$. Take a graph with $p$ vertices and $chi$-coloring. What is the most edges it can have if all bipartite subgraphs between different color pairs are complete?
    $endgroup$
    – Michal Adamaszek
    Dec 6 '18 at 7:20










  • $begingroup$
    Wow that is really cool! I had another idea. Is it possible to get this directly, by finding a lower bound on the largest clique in a graph? Then $chi geq textrm{largest clique size}$. My idea was to use an averaging method, but no success yet.
    $endgroup$
    – Mriganka Basu Roy Chowdhury
    Dec 6 '18 at 7:30








  • 1




    $begingroup$
    Possible duplicate of Prove that if G is a simple graph, $chi geq frac{|V|^2}{|V|^2-2|E|}$
    $endgroup$
    – Kuifje
    Dec 6 '18 at 9:20














2












2








2


2



$begingroup$



Let $G$ be any graph with order $p$ and size $q$. Prove that $$χ(G)ge {p^2 over p^2-2q}$$




If someone could give me a hint it would be amazing!






graph-theory







share|cite|improve this question











$endgroup$





Let $G$ be any graph with order $p$ and size $q$. Prove that $$χ(G)ge {p^2 over p^2-2q}$$




If someone could give me a hint it would be amazing!






graph-theory




graph-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 2:53









Chinnapparaj R

5,4331928




5,4331928










asked Dec 6 '18 at 2:48









John SmithJohn Smith

333




333












  • $begingroup$
    I guess order is number of vertices, so what is size?
    $endgroup$
    – coffeemath
    Dec 6 '18 at 3:35






  • 1




    $begingroup$
    number of edges
    $endgroup$
    – John Smith
    Dec 6 '18 at 4:05










  • $begingroup$
    You want to prove $qleq p^2cdotfrac{chi-1}{2chi}$. Take a graph with $p$ vertices and $chi$-coloring. What is the most edges it can have if all bipartite subgraphs between different color pairs are complete?
    $endgroup$
    – Michal Adamaszek
    Dec 6 '18 at 7:20










  • $begingroup$
    Wow that is really cool! I had another idea. Is it possible to get this directly, by finding a lower bound on the largest clique in a graph? Then $chi geq textrm{largest clique size}$. My idea was to use an averaging method, but no success yet.
    $endgroup$
    – Mriganka Basu Roy Chowdhury
    Dec 6 '18 at 7:30








  • 1




    $begingroup$
    Possible duplicate of Prove that if G is a simple graph, $chi geq frac{|V|^2}{|V|^2-2|E|}$
    $endgroup$
    – Kuifje
    Dec 6 '18 at 9:20


















  • $begingroup$
    I guess order is number of vertices, so what is size?
    $endgroup$
    – coffeemath
    Dec 6 '18 at 3:35






  • 1




    $begingroup$
    number of edges
    $endgroup$
    – John Smith
    Dec 6 '18 at 4:05










  • $begingroup$
    You want to prove $qleq p^2cdotfrac{chi-1}{2chi}$. Take a graph with $p$ vertices and $chi$-coloring. What is the most edges it can have if all bipartite subgraphs between different color pairs are complete?
    $endgroup$
    – Michal Adamaszek
    Dec 6 '18 at 7:20










  • $begingroup$
    Wow that is really cool! I had another idea. Is it possible to get this directly, by finding a lower bound on the largest clique in a graph? Then $chi geq textrm{largest clique size}$. My idea was to use an averaging method, but no success yet.
    $endgroup$
    – Mriganka Basu Roy Chowdhury
    Dec 6 '18 at 7:30








  • 1




    $begingroup$
    Possible duplicate of Prove that if G is a simple graph, $chi geq frac{|V|^2}{|V|^2-2|E|}$
    $endgroup$
    – Kuifje
    Dec 6 '18 at 9:20
















$begingroup$
I guess order is number of vertices, so what is size?
$endgroup$
– coffeemath
Dec 6 '18 at 3:35




$begingroup$
I guess order is number of vertices, so what is size?
$endgroup$
– coffeemath
Dec 6 '18 at 3:35




1




1




$begingroup$
number of edges
$endgroup$
– John Smith
Dec 6 '18 at 4:05




$begingroup$
number of edges
$endgroup$
– John Smith
Dec 6 '18 at 4:05












$begingroup$
You want to prove $qleq p^2cdotfrac{chi-1}{2chi}$. Take a graph with $p$ vertices and $chi$-coloring. What is the most edges it can have if all bipartite subgraphs between different color pairs are complete?
$endgroup$
– Michal Adamaszek
Dec 6 '18 at 7:20




$begingroup$
You want to prove $qleq p^2cdotfrac{chi-1}{2chi}$. Take a graph with $p$ vertices and $chi$-coloring. What is the most edges it can have if all bipartite subgraphs between different color pairs are complete?
$endgroup$
– Michal Adamaszek
Dec 6 '18 at 7:20












$begingroup$
Wow that is really cool! I had another idea. Is it possible to get this directly, by finding a lower bound on the largest clique in a graph? Then $chi geq textrm{largest clique size}$. My idea was to use an averaging method, but no success yet.
$endgroup$
– Mriganka Basu Roy Chowdhury
Dec 6 '18 at 7:30






$begingroup$
Wow that is really cool! I had another idea. Is it possible to get this directly, by finding a lower bound on the largest clique in a graph? Then $chi geq textrm{largest clique size}$. My idea was to use an averaging method, but no success yet.
$endgroup$
– Mriganka Basu Roy Chowdhury
Dec 6 '18 at 7:30






1




1




$begingroup$
Possible duplicate of Prove that if G is a simple graph, $chi geq frac{|V|^2}{|V|^2-2|E|}$
$endgroup$
– Kuifje
Dec 6 '18 at 9:20




$begingroup$
Possible duplicate of Prove that if G is a simple graph, $chi geq frac{|V|^2}{|V|^2-2|E|}$
$endgroup$
– Kuifje
Dec 6 '18 at 9:20










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$begingroup$

Suppose a graph with $v$ vertices has chromatic number $chi$, so we have a splitting of the vertex set into independent sets of sizes $v_1,ldots,v_chi$. Then the number of edges $e$ is at most
$$eleq sum_{1leq i<jleq chi}v_iv_jleq frac{chi-1}{2chi}(sum_i v_i)^2=frac{chi-1}{2chi}v^2 $$
where the middle inequality is something that holds for all real numbers as it is just $sum_{i<j}(v_i-v_j)^2geq 0$.






share|cite|improve this answer









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    $begingroup$

    Suppose a graph with $v$ vertices has chromatic number $chi$, so we have a splitting of the vertex set into independent sets of sizes $v_1,ldots,v_chi$. Then the number of edges $e$ is at most
    $$eleq sum_{1leq i<jleq chi}v_iv_jleq frac{chi-1}{2chi}(sum_i v_i)^2=frac{chi-1}{2chi}v^2 $$
    where the middle inequality is something that holds for all real numbers as it is just $sum_{i<j}(v_i-v_j)^2geq 0$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Suppose a graph with $v$ vertices has chromatic number $chi$, so we have a splitting of the vertex set into independent sets of sizes $v_1,ldots,v_chi$. Then the number of edges $e$ is at most
      $$eleq sum_{1leq i<jleq chi}v_iv_jleq frac{chi-1}{2chi}(sum_i v_i)^2=frac{chi-1}{2chi}v^2 $$
      where the middle inequality is something that holds for all real numbers as it is just $sum_{i<j}(v_i-v_j)^2geq 0$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Suppose a graph with $v$ vertices has chromatic number $chi$, so we have a splitting of the vertex set into independent sets of sizes $v_1,ldots,v_chi$. Then the number of edges $e$ is at most
        $$eleq sum_{1leq i<jleq chi}v_iv_jleq frac{chi-1}{2chi}(sum_i v_i)^2=frac{chi-1}{2chi}v^2 $$
        where the middle inequality is something that holds for all real numbers as it is just $sum_{i<j}(v_i-v_j)^2geq 0$.






        share|cite|improve this answer









        $endgroup$



        Suppose a graph with $v$ vertices has chromatic number $chi$, so we have a splitting of the vertex set into independent sets of sizes $v_1,ldots,v_chi$. Then the number of edges $e$ is at most
        $$eleq sum_{1leq i<jleq chi}v_iv_jleq frac{chi-1}{2chi}(sum_i v_i)^2=frac{chi-1}{2chi}v^2 $$
        where the middle inequality is something that holds for all real numbers as it is just $sum_{i<j}(v_i-v_j)^2geq 0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 6 '18 at 10:31









        Michal AdamaszekMichal Adamaszek

        2,08148




        2,08148






























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