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

      53516




      53516






















          1 Answer
          1






          active

          oldest

          votes


















          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











          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3035133%2fperfect-matching-in-random-bipartite-graph-with-fixed-probability%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            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












              $begingroup$

              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

                45.9k656107




                45.9k656107






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3035133%2fperfect-matching-in-random-bipartite-graph-with-fixed-probability%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Wiesbaden

                    Image
                    Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
                    Read more

                    Marschland

                    Image
                    Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
                    Read more

                    Dieringhausen

                    Image
                    Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
                    Read more