Extension of Mantel's theorem












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How would one extend the proof of Mantel's Theorem to show that if $G$ is a graph with $n$ vertices and $lfloor n^2/4rfloor - t$ edges then $G$ contains a bipartite subgraph with at least $lfloor n^2/4rfloor - 2t$ edges?










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  • 1




    $begingroup$
    Since this is intended to be an extension of Mantel's Theorem, do you assume that $G$ is triangle-free?
    $endgroup$
    – Andrew Uzzell
    Apr 23 '13 at 14:58










  • $begingroup$
    Yes, we assume G has no K3.
    $endgroup$
    – Musegirl
    Apr 23 '13 at 15:12
















4












$begingroup$


How would one extend the proof of Mantel's Theorem to show that if $G$ is a graph with $n$ vertices and $lfloor n^2/4rfloor - t$ edges then $G$ contains a bipartite subgraph with at least $lfloor n^2/4rfloor - 2t$ edges?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Since this is intended to be an extension of Mantel's Theorem, do you assume that $G$ is triangle-free?
    $endgroup$
    – Andrew Uzzell
    Apr 23 '13 at 14:58










  • $begingroup$
    Yes, we assume G has no K3.
    $endgroup$
    – Musegirl
    Apr 23 '13 at 15:12














4












4








4





$begingroup$


How would one extend the proof of Mantel's Theorem to show that if $G$ is a graph with $n$ vertices and $lfloor n^2/4rfloor - t$ edges then $G$ contains a bipartite subgraph with at least $lfloor n^2/4rfloor - 2t$ edges?










share|cite|improve this question











$endgroup$




How would one extend the proof of Mantel's Theorem to show that if $G$ is a graph with $n$ vertices and $lfloor n^2/4rfloor - t$ edges then $G$ contains a bipartite subgraph with at least $lfloor n^2/4rfloor - 2t$ edges?







graph-theory






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edited Apr 23 '13 at 14:58









Kundor

5,47621338




5,47621338










asked Apr 23 '13 at 14:43









MusegirlMusegirl

241




241








  • 1




    $begingroup$
    Since this is intended to be an extension of Mantel's Theorem, do you assume that $G$ is triangle-free?
    $endgroup$
    – Andrew Uzzell
    Apr 23 '13 at 14:58










  • $begingroup$
    Yes, we assume G has no K3.
    $endgroup$
    – Musegirl
    Apr 23 '13 at 15:12














  • 1




    $begingroup$
    Since this is intended to be an extension of Mantel's Theorem, do you assume that $G$ is triangle-free?
    $endgroup$
    – Andrew Uzzell
    Apr 23 '13 at 14:58










  • $begingroup$
    Yes, we assume G has no K3.
    $endgroup$
    – Musegirl
    Apr 23 '13 at 15:12








1




1




$begingroup$
Since this is intended to be an extension of Mantel's Theorem, do you assume that $G$ is triangle-free?
$endgroup$
– Andrew Uzzell
Apr 23 '13 at 14:58




$begingroup$
Since this is intended to be an extension of Mantel's Theorem, do you assume that $G$ is triangle-free?
$endgroup$
– Andrew Uzzell
Apr 23 '13 at 14:58












$begingroup$
Yes, we assume G has no K3.
$endgroup$
– Musegirl
Apr 23 '13 at 15:12




$begingroup$
Yes, we assume G has no K3.
$endgroup$
– Musegirl
Apr 23 '13 at 15:12










2 Answers
2






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0












$begingroup$

Here is an attempt: Consider a bipartition $(A,B)$ of the vertices of the graph $G$ such that the number of edges between the two bipartition is maximized. So then we can see that this bipartite subgraph $H$ has the property: $v in V(H)$ then $deg(v)_{H} geq frac{deg(v)_G}{2}$. So $2|E(H)|=sum_{vin V(H)}deg(v) geq sum_{vin V(G)}frac{deg(v)}{2}=frac{2|E(G)|}{2}=lfloor frac{n^2}{4} rfloor -t$. so then $|E(H)|geq frac{lfloor frac{n^2}{4} rfloor -t}{2}$.






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  • $begingroup$
    can anyone complete the proof or give a solution?
    $endgroup$
    – nafhgood
    Dec 6 '18 at 20:29





















0





+100







$begingroup$

For every vertex $x$, we can let $A_x$ be all the neighbors of $x$ and $B_x$ be all the non-neighbors of $x$. Because $G$ is triangle-free, $A_x$ is an independent set, so if we delete all the edges in $E(B_x)$, we get a bipartite graph with bipartition $(A_x, B_x cup {x})$.



It remains to choose a vertex $x$ for which $|E(B_x)|$ is small. What follows is just an elaboration on the one-line proof of Lemma 2.1 in "How to make a graph bipartite" by Erdős, Faudree, Pach, and Spencer.



We have
$$
sum_{x in V(G)} |E(B_x)| = sum_{vw in E(G)} (n - deg(v) - deg(w)).
$$

The second sum counts for every edge $vw$ all the vertices not adjacent to $v$ or $w$: here we are using the fact that $G$ is triangle-free and therefore no vertex is adjacent to both. The vertices not adjacent to $v$ or $w$ are precisely the vertices $x$ for which $vw in E(B_x)$. Therefore the two sums count pairs $(x, vw)$ where $vw in E(B_x)$ in two different ways, so they're equal.



Expanding out the second sum, we get
$$
sum_{x in V(G)} |E(B_x)| = sum_{vw in E(G)} (n - deg(v) - deg(w)) = n|E(G)| - sum_{v in V(G)} deg(v)^2
$$

because $deg(v)$ is subtracted once for each of the $deg(v)$ edges out of $v$. By the convexity of $x mapsto x^2$, or by Cauchy-Schwarz, the sum of squares is minimized when $deg(v) = frac{2|E(G)|}{n}$ for all $v$, and therefore
$$
sum_{x in V(G)} |E(B_x)| = n|E(G)| - sum_{v in V(G)} deg(v)^2 le n|E(G)| - n cdot frac{4|E(G)|^2}{n^2}.
$$

If we substitute $|E(G)| = frac{n^2}{4} - t$ into this inequality and simplify, the left-hand side becomes $n(frac{n^2}{4}-t) - frac{4}{n}(frac{n^2}{4}-t)^2 = frac{n^3}{4} - nt - frac{n^3}{4} + 2nt - frac{4t^2}{n} = nt - frac{4t^2}{n}$, so we get
$$
sum_{x in V(G)} |E(B_x)| le tn - frac{4t^2}{n} iff frac1n sum_{x in V(G)} |E(B_x)| le t - frac{4t^2}{n^2}
$$

and therefore there is some choice of $x$ for which $|E(B_x)| le t - frac{4t^2}{n^2}$. By deleting the edges of $B_x$ from $G$, we get a bipartite subgraph containing all but at most $ t - frac{4t^2}{n^2}$ edges of $G$: therefore we have at least
$$
frac{n^2}{4} - t - left(t - frac{4t^2}{n^2}right) = frac{n^2}{4} - 2t + frac{4t^2}{n^2} ge frac{n^2}{4} - 2t
$$

edges in the bipartite subgraph.






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    2 Answers
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    2 Answers
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    active

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    0












    $begingroup$

    Here is an attempt: Consider a bipartition $(A,B)$ of the vertices of the graph $G$ such that the number of edges between the two bipartition is maximized. So then we can see that this bipartite subgraph $H$ has the property: $v in V(H)$ then $deg(v)_{H} geq frac{deg(v)_G}{2}$. So $2|E(H)|=sum_{vin V(H)}deg(v) geq sum_{vin V(G)}frac{deg(v)}{2}=frac{2|E(G)|}{2}=lfloor frac{n^2}{4} rfloor -t$. so then $|E(H)|geq frac{lfloor frac{n^2}{4} rfloor -t}{2}$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      can anyone complete the proof or give a solution?
      $endgroup$
      – nafhgood
      Dec 6 '18 at 20:29


















    0












    $begingroup$

    Here is an attempt: Consider a bipartition $(A,B)$ of the vertices of the graph $G$ such that the number of edges between the two bipartition is maximized. So then we can see that this bipartite subgraph $H$ has the property: $v in V(H)$ then $deg(v)_{H} geq frac{deg(v)_G}{2}$. So $2|E(H)|=sum_{vin V(H)}deg(v) geq sum_{vin V(G)}frac{deg(v)}{2}=frac{2|E(G)|}{2}=lfloor frac{n^2}{4} rfloor -t$. so then $|E(H)|geq frac{lfloor frac{n^2}{4} rfloor -t}{2}$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      can anyone complete the proof or give a solution?
      $endgroup$
      – nafhgood
      Dec 6 '18 at 20:29
















    0












    0








    0





    $begingroup$

    Here is an attempt: Consider a bipartition $(A,B)$ of the vertices of the graph $G$ such that the number of edges between the two bipartition is maximized. So then we can see that this bipartite subgraph $H$ has the property: $v in V(H)$ then $deg(v)_{H} geq frac{deg(v)_G}{2}$. So $2|E(H)|=sum_{vin V(H)}deg(v) geq sum_{vin V(G)}frac{deg(v)}{2}=frac{2|E(G)|}{2}=lfloor frac{n^2}{4} rfloor -t$. so then $|E(H)|geq frac{lfloor frac{n^2}{4} rfloor -t}{2}$.






    share|cite|improve this answer









    $endgroup$



    Here is an attempt: Consider a bipartition $(A,B)$ of the vertices of the graph $G$ such that the number of edges between the two bipartition is maximized. So then we can see that this bipartite subgraph $H$ has the property: $v in V(H)$ then $deg(v)_{H} geq frac{deg(v)_G}{2}$. So $2|E(H)|=sum_{vin V(H)}deg(v) geq sum_{vin V(G)}frac{deg(v)}{2}=frac{2|E(G)|}{2}=lfloor frac{n^2}{4} rfloor -t$. so then $|E(H)|geq frac{lfloor frac{n^2}{4} rfloor -t}{2}$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 4 '18 at 17:30









    nafhgoodnafhgood

    1,801422




    1,801422












    • $begingroup$
      can anyone complete the proof or give a solution?
      $endgroup$
      – nafhgood
      Dec 6 '18 at 20:29




















    • $begingroup$
      can anyone complete the proof or give a solution?
      $endgroup$
      – nafhgood
      Dec 6 '18 at 20:29


















    $begingroup$
    can anyone complete the proof or give a solution?
    $endgroup$
    – nafhgood
    Dec 6 '18 at 20:29






    $begingroup$
    can anyone complete the proof or give a solution?
    $endgroup$
    – nafhgood
    Dec 6 '18 at 20:29













    0





    +100







    $begingroup$

    For every vertex $x$, we can let $A_x$ be all the neighbors of $x$ and $B_x$ be all the non-neighbors of $x$. Because $G$ is triangle-free, $A_x$ is an independent set, so if we delete all the edges in $E(B_x)$, we get a bipartite graph with bipartition $(A_x, B_x cup {x})$.



    It remains to choose a vertex $x$ for which $|E(B_x)|$ is small. What follows is just an elaboration on the one-line proof of Lemma 2.1 in "How to make a graph bipartite" by Erdős, Faudree, Pach, and Spencer.



    We have
    $$
    sum_{x in V(G)} |E(B_x)| = sum_{vw in E(G)} (n - deg(v) - deg(w)).
    $$

    The second sum counts for every edge $vw$ all the vertices not adjacent to $v$ or $w$: here we are using the fact that $G$ is triangle-free and therefore no vertex is adjacent to both. The vertices not adjacent to $v$ or $w$ are precisely the vertices $x$ for which $vw in E(B_x)$. Therefore the two sums count pairs $(x, vw)$ where $vw in E(B_x)$ in two different ways, so they're equal.



    Expanding out the second sum, we get
    $$
    sum_{x in V(G)} |E(B_x)| = sum_{vw in E(G)} (n - deg(v) - deg(w)) = n|E(G)| - sum_{v in V(G)} deg(v)^2
    $$

    because $deg(v)$ is subtracted once for each of the $deg(v)$ edges out of $v$. By the convexity of $x mapsto x^2$, or by Cauchy-Schwarz, the sum of squares is minimized when $deg(v) = frac{2|E(G)|}{n}$ for all $v$, and therefore
    $$
    sum_{x in V(G)} |E(B_x)| = n|E(G)| - sum_{v in V(G)} deg(v)^2 le n|E(G)| - n cdot frac{4|E(G)|^2}{n^2}.
    $$

    If we substitute $|E(G)| = frac{n^2}{4} - t$ into this inequality and simplify, the left-hand side becomes $n(frac{n^2}{4}-t) - frac{4}{n}(frac{n^2}{4}-t)^2 = frac{n^3}{4} - nt - frac{n^3}{4} + 2nt - frac{4t^2}{n} = nt - frac{4t^2}{n}$, so we get
    $$
    sum_{x in V(G)} |E(B_x)| le tn - frac{4t^2}{n} iff frac1n sum_{x in V(G)} |E(B_x)| le t - frac{4t^2}{n^2}
    $$

    and therefore there is some choice of $x$ for which $|E(B_x)| le t - frac{4t^2}{n^2}$. By deleting the edges of $B_x$ from $G$, we get a bipartite subgraph containing all but at most $ t - frac{4t^2}{n^2}$ edges of $G$: therefore we have at least
    $$
    frac{n^2}{4} - t - left(t - frac{4t^2}{n^2}right) = frac{n^2}{4} - 2t + frac{4t^2}{n^2} ge frac{n^2}{4} - 2t
    $$

    edges in the bipartite subgraph.






    share|cite|improve this answer











    $endgroup$


















      0





      +100







      $begingroup$

      For every vertex $x$, we can let $A_x$ be all the neighbors of $x$ and $B_x$ be all the non-neighbors of $x$. Because $G$ is triangle-free, $A_x$ is an independent set, so if we delete all the edges in $E(B_x)$, we get a bipartite graph with bipartition $(A_x, B_x cup {x})$.



      It remains to choose a vertex $x$ for which $|E(B_x)|$ is small. What follows is just an elaboration on the one-line proof of Lemma 2.1 in "How to make a graph bipartite" by Erdős, Faudree, Pach, and Spencer.



      We have
      $$
      sum_{x in V(G)} |E(B_x)| = sum_{vw in E(G)} (n - deg(v) - deg(w)).
      $$

      The second sum counts for every edge $vw$ all the vertices not adjacent to $v$ or $w$: here we are using the fact that $G$ is triangle-free and therefore no vertex is adjacent to both. The vertices not adjacent to $v$ or $w$ are precisely the vertices $x$ for which $vw in E(B_x)$. Therefore the two sums count pairs $(x, vw)$ where $vw in E(B_x)$ in two different ways, so they're equal.



      Expanding out the second sum, we get
      $$
      sum_{x in V(G)} |E(B_x)| = sum_{vw in E(G)} (n - deg(v) - deg(w)) = n|E(G)| - sum_{v in V(G)} deg(v)^2
      $$

      because $deg(v)$ is subtracted once for each of the $deg(v)$ edges out of $v$. By the convexity of $x mapsto x^2$, or by Cauchy-Schwarz, the sum of squares is minimized when $deg(v) = frac{2|E(G)|}{n}$ for all $v$, and therefore
      $$
      sum_{x in V(G)} |E(B_x)| = n|E(G)| - sum_{v in V(G)} deg(v)^2 le n|E(G)| - n cdot frac{4|E(G)|^2}{n^2}.
      $$

      If we substitute $|E(G)| = frac{n^2}{4} - t$ into this inequality and simplify, the left-hand side becomes $n(frac{n^2}{4}-t) - frac{4}{n}(frac{n^2}{4}-t)^2 = frac{n^3}{4} - nt - frac{n^3}{4} + 2nt - frac{4t^2}{n} = nt - frac{4t^2}{n}$, so we get
      $$
      sum_{x in V(G)} |E(B_x)| le tn - frac{4t^2}{n} iff frac1n sum_{x in V(G)} |E(B_x)| le t - frac{4t^2}{n^2}
      $$

      and therefore there is some choice of $x$ for which $|E(B_x)| le t - frac{4t^2}{n^2}$. By deleting the edges of $B_x$ from $G$, we get a bipartite subgraph containing all but at most $ t - frac{4t^2}{n^2}$ edges of $G$: therefore we have at least
      $$
      frac{n^2}{4} - t - left(t - frac{4t^2}{n^2}right) = frac{n^2}{4} - 2t + frac{4t^2}{n^2} ge frac{n^2}{4} - 2t
      $$

      edges in the bipartite subgraph.






      share|cite|improve this answer











      $endgroup$
















        0





        +100







        0





        +100



        0




        +100



        $begingroup$

        For every vertex $x$, we can let $A_x$ be all the neighbors of $x$ and $B_x$ be all the non-neighbors of $x$. Because $G$ is triangle-free, $A_x$ is an independent set, so if we delete all the edges in $E(B_x)$, we get a bipartite graph with bipartition $(A_x, B_x cup {x})$.



        It remains to choose a vertex $x$ for which $|E(B_x)|$ is small. What follows is just an elaboration on the one-line proof of Lemma 2.1 in "How to make a graph bipartite" by Erdős, Faudree, Pach, and Spencer.



        We have
        $$
        sum_{x in V(G)} |E(B_x)| = sum_{vw in E(G)} (n - deg(v) - deg(w)).
        $$

        The second sum counts for every edge $vw$ all the vertices not adjacent to $v$ or $w$: here we are using the fact that $G$ is triangle-free and therefore no vertex is adjacent to both. The vertices not adjacent to $v$ or $w$ are precisely the vertices $x$ for which $vw in E(B_x)$. Therefore the two sums count pairs $(x, vw)$ where $vw in E(B_x)$ in two different ways, so they're equal.



        Expanding out the second sum, we get
        $$
        sum_{x in V(G)} |E(B_x)| = sum_{vw in E(G)} (n - deg(v) - deg(w)) = n|E(G)| - sum_{v in V(G)} deg(v)^2
        $$

        because $deg(v)$ is subtracted once for each of the $deg(v)$ edges out of $v$. By the convexity of $x mapsto x^2$, or by Cauchy-Schwarz, the sum of squares is minimized when $deg(v) = frac{2|E(G)|}{n}$ for all $v$, and therefore
        $$
        sum_{x in V(G)} |E(B_x)| = n|E(G)| - sum_{v in V(G)} deg(v)^2 le n|E(G)| - n cdot frac{4|E(G)|^2}{n^2}.
        $$

        If we substitute $|E(G)| = frac{n^2}{4} - t$ into this inequality and simplify, the left-hand side becomes $n(frac{n^2}{4}-t) - frac{4}{n}(frac{n^2}{4}-t)^2 = frac{n^3}{4} - nt - frac{n^3}{4} + 2nt - frac{4t^2}{n} = nt - frac{4t^2}{n}$, so we get
        $$
        sum_{x in V(G)} |E(B_x)| le tn - frac{4t^2}{n} iff frac1n sum_{x in V(G)} |E(B_x)| le t - frac{4t^2}{n^2}
        $$

        and therefore there is some choice of $x$ for which $|E(B_x)| le t - frac{4t^2}{n^2}$. By deleting the edges of $B_x$ from $G$, we get a bipartite subgraph containing all but at most $ t - frac{4t^2}{n^2}$ edges of $G$: therefore we have at least
        $$
        frac{n^2}{4} - t - left(t - frac{4t^2}{n^2}right) = frac{n^2}{4} - 2t + frac{4t^2}{n^2} ge frac{n^2}{4} - 2t
        $$

        edges in the bipartite subgraph.






        share|cite|improve this answer











        $endgroup$



        For every vertex $x$, we can let $A_x$ be all the neighbors of $x$ and $B_x$ be all the non-neighbors of $x$. Because $G$ is triangle-free, $A_x$ is an independent set, so if we delete all the edges in $E(B_x)$, we get a bipartite graph with bipartition $(A_x, B_x cup {x})$.



        It remains to choose a vertex $x$ for which $|E(B_x)|$ is small. What follows is just an elaboration on the one-line proof of Lemma 2.1 in "How to make a graph bipartite" by Erdős, Faudree, Pach, and Spencer.



        We have
        $$
        sum_{x in V(G)} |E(B_x)| = sum_{vw in E(G)} (n - deg(v) - deg(w)).
        $$

        The second sum counts for every edge $vw$ all the vertices not adjacent to $v$ or $w$: here we are using the fact that $G$ is triangle-free and therefore no vertex is adjacent to both. The vertices not adjacent to $v$ or $w$ are precisely the vertices $x$ for which $vw in E(B_x)$. Therefore the two sums count pairs $(x, vw)$ where $vw in E(B_x)$ in two different ways, so they're equal.



        Expanding out the second sum, we get
        $$
        sum_{x in V(G)} |E(B_x)| = sum_{vw in E(G)} (n - deg(v) - deg(w)) = n|E(G)| - sum_{v in V(G)} deg(v)^2
        $$

        because $deg(v)$ is subtracted once for each of the $deg(v)$ edges out of $v$. By the convexity of $x mapsto x^2$, or by Cauchy-Schwarz, the sum of squares is minimized when $deg(v) = frac{2|E(G)|}{n}$ for all $v$, and therefore
        $$
        sum_{x in V(G)} |E(B_x)| = n|E(G)| - sum_{v in V(G)} deg(v)^2 le n|E(G)| - n cdot frac{4|E(G)|^2}{n^2}.
        $$

        If we substitute $|E(G)| = frac{n^2}{4} - t$ into this inequality and simplify, the left-hand side becomes $n(frac{n^2}{4}-t) - frac{4}{n}(frac{n^2}{4}-t)^2 = frac{n^3}{4} - nt - frac{n^3}{4} + 2nt - frac{4t^2}{n} = nt - frac{4t^2}{n}$, so we get
        $$
        sum_{x in V(G)} |E(B_x)| le tn - frac{4t^2}{n} iff frac1n sum_{x in V(G)} |E(B_x)| le t - frac{4t^2}{n^2}
        $$

        and therefore there is some choice of $x$ for which $|E(B_x)| le t - frac{4t^2}{n^2}$. By deleting the edges of $B_x$ from $G$, we get a bipartite subgraph containing all but at most $ t - frac{4t^2}{n^2}$ edges of $G$: therefore we have at least
        $$
        frac{n^2}{4} - t - left(t - frac{4t^2}{n^2}right) = frac{n^2}{4} - 2t + frac{4t^2}{n^2} ge frac{n^2}{4} - 2t
        $$

        edges in the bipartite subgraph.







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        edited Dec 9 '18 at 19:14

























        answered Dec 9 '18 at 17:17









        Misha LavrovMisha Lavrov

        45.1k656107




        45.1k656107






























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