Perfect matching in random bipartite graph - with fixed probability












2












$begingroup$


as a follow up from this question :



Suppose that we have a simpler problem, where the probability $p$ is fixed. Of course we could use the above result to proove that almost every graph in the model $mathcal{G}_{n,n,p}$ contains a perfect matching, saying for instance that $$ p > frac{sqrt{n} + log{n}}{n}$$ for big enough $n$.



However, couldn't we find a more direct result?
I've been looking at the probability of findind a set $S$ (size $k$) giving a contradiction for Hall's mariage theorem. With $q=1-p$, we need each element of $S$ not connected to the $n-k-1$ elements not in $N(S)$, therefore :
$$ Pr[ |N(S)|<|S| ] leq q^{nk-k(k-1)}$$



And
$$Pr[nexists text{ a perfect matching}] leq sum_{k=1}^n binom{n}{k}q^{nk-k(k-1)}$$



But from there, trying to proove that this sum tends to 0, simple binomial approximation $sim n^k/k!$ gives too large a bound. Would you recommand any other calculation?
Thanks






graph-theory random-graphs matching-theory







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    as a follow up from this question :



    Suppose that we have a simpler problem, where the probability $p$ is fixed. Of course we could use the above result to proove that almost every graph in the model $mathcal{G}_{n,n,p}$ contains a perfect matching, saying for instance that $$ p > frac{sqrt{n} + log{n}}{n}$$ for big enough $n$.



    However, couldn't we find a more direct result?
    I've been looking at the probability of findind a set $S$ (size $k$) giving a contradiction for Hall's mariage theorem. With $q=1-p$, we need each element of $S$ not connected to the $n-k-1$ elements not in $N(S)$, therefore :
    $$ Pr[ |N(S)|<|S| ] leq q^{nk-k(k-1)}$$



    And
    $$Pr[nexists text{ a perfect matching}] leq sum_{k=1}^n binom{n}{k}q^{nk-k(k-1)}$$



    But from there, trying to proove that this sum tends to 0, simple binomial approximation $sim n^k/k!$ gives too large a bound. Would you recommand any other calculation?
    Thanks






    graph-theory random-graphs matching-theory







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      as a follow up from this question :



      Suppose that we have a simpler problem, where the probability $p$ is fixed. Of course we could use the above result to proove that almost every graph in the model $mathcal{G}_{n,n,p}$ contains a perfect matching, saying for instance that $$ p > frac{sqrt{n} + log{n}}{n}$$ for big enough $n$.



      However, couldn't we find a more direct result?
      I've been looking at the probability of findind a set $S$ (size $k$) giving a contradiction for Hall's mariage theorem. With $q=1-p$, we need each element of $S$ not connected to the $n-k-1$ elements not in $N(S)$, therefore :
      $$ Pr[ |N(S)|<|S| ] leq q^{nk-k(k-1)}$$



      And
      $$Pr[nexists text{ a perfect matching}] leq sum_{k=1}^n binom{n}{k}q^{nk-k(k-1)}$$



      But from there, trying to proove that this sum tends to 0, simple binomial approximation $sim n^k/k!$ gives too large a bound. Would you recommand any other calculation?
      Thanks






      graph-theory random-graphs matching-theory







      share|cite|improve this question









      $endgroup$




      as a follow up from this question :



      Suppose that we have a simpler problem, where the probability $p$ is fixed. Of course we could use the above result to proove that almost every graph in the model $mathcal{G}_{n,n,p}$ contains a perfect matching, saying for instance that $$ p > frac{sqrt{n} + log{n}}{n}$$ for big enough $n$.



      However, couldn't we find a more direct result?
      I've been looking at the probability of findind a set $S$ (size $k$) giving a contradiction for Hall's mariage theorem. With $q=1-p$, we need each element of $S$ not connected to the $n-k-1$ elements not in $N(S)$, therefore :
      $$ Pr[ |N(S)|<|S| ] leq q^{nk-k(k-1)}$$



      And
      $$Pr[nexists text{ a perfect matching}] leq sum_{k=1}^n binom{n}{k}q^{nk-k(k-1)}$$



      But from there, trying to proove that this sum tends to 0, simple binomial approximation $sim n^k/k!$ gives too large a bound. Would you recommand any other calculation?
      Thanks






      graph-theory random-graphs matching-theory




      graph-theory random-graphs matching-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 11 '18 at 10:03









      Thomas LesgourguesThomas Lesgourgues

      845117




      845117






















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          $begingroup$

          Many approximations to the binomial $binom nk$ are not symmetric in $k$ and $n-k$, so they perform badly when $k$ is close to $n$. (For example, $frac{n^k}{k!}$ is somewhere around $e^n$ when $k=n$, even though actually $binom nn = 1$.)



          Here, the sum is almost but not quite symmetric: we can write the power of $q$ as $q^{k(n+1-k)} < q^{k(n-k)}$, and the latter is symmetric. So we can bound the sum by
          $$
          2 sum_{k=1}^{n/2} binom nk q^{k(n-k)}
          $$
          and avoid the problematic values of $k$, getting a sum that's easy to bound using your approximation. It's even enough to write $binom nk le n^k$ and get $binom nk q^{k(n-k)} le (n cdot q^{n-k})^k le (n cdot q^{n/2})^k$.






          share|cite|improve this answer











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            1 Answer
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            $begingroup$

            Many approximations to the binomial $binom nk$ are not symmetric in $k$ and $n-k$, so they perform badly when $k$ is close to $n$. (For example, $frac{n^k}{k!}$ is somewhere around $e^n$ when $k=n$, even though actually $binom nn = 1$.)



            Here, the sum is almost but not quite symmetric: we can write the power of $q$ as $q^{k(n+1-k)} < q^{k(n-k)}$, and the latter is symmetric. So we can bound the sum by
            $$
            2 sum_{k=1}^{n/2} binom nk q^{k(n-k)}
            $$
            and avoid the problematic values of $k$, getting a sum that's easy to bound using your approximation. It's even enough to write $binom nk le n^k$ and get $binom nk q^{k(n-k)} le (n cdot q^{n-k})^k le (n cdot q^{n/2})^k$.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              Many approximations to the binomial $binom nk$ are not symmetric in $k$ and $n-k$, so they perform badly when $k$ is close to $n$. (For example, $frac{n^k}{k!}$ is somewhere around $e^n$ when $k=n$, even though actually $binom nn = 1$.)



              Here, the sum is almost but not quite symmetric: we can write the power of $q$ as $q^{k(n+1-k)} < q^{k(n-k)}$, and the latter is symmetric. So we can bound the sum by
              $$
              2 sum_{k=1}^{n/2} binom nk q^{k(n-k)}
              $$
              and avoid the problematic values of $k$, getting a sum that's easy to bound using your approximation. It's even enough to write $binom nk le n^k$ and get $binom nk q^{k(n-k)} le (n cdot q^{n-k})^k le (n cdot q^{n/2})^k$.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                Many approximations to the binomial $binom nk$ are not symmetric in $k$ and $n-k$, so they perform badly when $k$ is close to $n$. (For example, $frac{n^k}{k!}$ is somewhere around $e^n$ when $k=n$, even though actually $binom nn = 1$.)



                Here, the sum is almost but not quite symmetric: we can write the power of $q$ as $q^{k(n+1-k)} < q^{k(n-k)}$, and the latter is symmetric. So we can bound the sum by
                $$
                2 sum_{k=1}^{n/2} binom nk q^{k(n-k)}
                $$
                and avoid the problematic values of $k$, getting a sum that's easy to bound using your approximation. It's even enough to write $binom nk le n^k$ and get $binom nk q^{k(n-k)} le (n cdot q^{n-k})^k le (n cdot q^{n/2})^k$.






                share|cite|improve this answer











                $endgroup$



                Many approximations to the binomial $binom nk$ are not symmetric in $k$ and $n-k$, so they perform badly when $k$ is close to $n$. (For example, $frac{n^k}{k!}$ is somewhere around $e^n$ when $k=n$, even though actually $binom nn = 1$.)



                Here, the sum is almost but not quite symmetric: we can write the power of $q$ as $q^{k(n+1-k)} < q^{k(n-k)}$, and the latter is symmetric. So we can bound the sum by
                $$
                2 sum_{k=1}^{n/2} binom nk q^{k(n-k)}
                $$
                and avoid the problematic values of $k$, getting a sum that's easy to bound using your approximation. It's even enough to write $binom nk le n^k$ and get $binom nk q^{k(n-k)} le (n cdot q^{n-k})^k le (n cdot q^{n/2})^k$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 11 '18 at 22:28

























                answered Dec 11 '18 at 22:22









                Misha LavrovMisha Lavrov

                46.9k657107




                46.9k657107






























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