Dominating sets in tournaments; is $2^{n+1}-2$ tight?












6












$begingroup$


A tournement is a directed graph such that for every pair of distinct vertices ${x,y}$, there is either an edge from $x$ to $y$ or from $y$ to $x$, but not both. I will use "$xto y$" to mean "there is an edge from $x$ to $y$."



A dominating set of a directed graph is a subset $S$ of vertices such that for every $tnotin S$, there exists $sin S$ so $sto t$.



It can be shown$^*$ that every tournament on $2^{n+1}-2$ vertices has a dominating set of size $n$. My question is whether this result is tight.




Does there exist a tournament on $2^{n+1}-1$ vertices with no dominating set of size $n$?



If not, what is the smallest tournament with no dominating set of size $n$?




My thoughts:




  • A necessary condition for a graph on $2^{n+1}-1$ vertices with no dominating set of size $n$ is that every vertex must have an out-degree of exactly $2^n-1$, so exactly half of its edges are outgoing.



  • The answer is yes when $n=1,2$.




    • The "rock-paper-scissors" graph on three vertices has no dominating set of size $1$.

    • The graph on $mathbb Z/7mathbb Z$ where each $x$ has directed edges to $x+1,x+2$ and $x+4pmod7$ has no dominating set of size $2$.




For $nge 3$, the possibilities get too large, and I cannot come up with a clever solution. Can anyone see a pattern?



I came up with this problem while thinking about this puzzle.




$^*$Consider a vertex $s$ with maximal out-degree. By the hand-shaking lemma, this degree must be at least $(2^{n+1}-3)/2$, so at least $2^n-1$. Include $s$ in $S$, then ignore $s$ and the vertices $t$ for which $sto t$. What remains is tournament of size $(2^{n+1}-2)-1-(2^{n}-1)=2^n-2$. Proceed by induction.






combinatorics discrete-mathematics graph-theory directed-graphs extremal-graph-theory







share|cite|improve this question











$endgroup$

















    6












    $begingroup$


    A tournement is a directed graph such that for every pair of distinct vertices ${x,y}$, there is either an edge from $x$ to $y$ or from $y$ to $x$, but not both. I will use "$xto y$" to mean "there is an edge from $x$ to $y$."



    A dominating set of a directed graph is a subset $S$ of vertices such that for every $tnotin S$, there exists $sin S$ so $sto t$.



    It can be shown$^*$ that every tournament on $2^{n+1}-2$ vertices has a dominating set of size $n$. My question is whether this result is tight.




    Does there exist a tournament on $2^{n+1}-1$ vertices with no dominating set of size $n$?



    If not, what is the smallest tournament with no dominating set of size $n$?




    My thoughts:




    • A necessary condition for a graph on $2^{n+1}-1$ vertices with no dominating set of size $n$ is that every vertex must have an out-degree of exactly $2^n-1$, so exactly half of its edges are outgoing.



    • The answer is yes when $n=1,2$.




      • The "rock-paper-scissors" graph on three vertices has no dominating set of size $1$.

      • The graph on $mathbb Z/7mathbb Z$ where each $x$ has directed edges to $x+1,x+2$ and $x+4pmod7$ has no dominating set of size $2$.




    For $nge 3$, the possibilities get too large, and I cannot come up with a clever solution. Can anyone see a pattern?



    I came up with this problem while thinking about this puzzle.




    $^*$Consider a vertex $s$ with maximal out-degree. By the hand-shaking lemma, this degree must be at least $(2^{n+1}-3)/2$, so at least $2^n-1$. Include $s$ in $S$, then ignore $s$ and the vertices $t$ for which $sto t$. What remains is tournament of size $(2^{n+1}-2)-1-(2^{n}-1)=2^n-2$. Proceed by induction.






    combinatorics discrete-mathematics graph-theory directed-graphs extremal-graph-theory







    share|cite|improve this question











    $endgroup$















      6












      6








      6


      3



      $begingroup$


      A tournement is a directed graph such that for every pair of distinct vertices ${x,y}$, there is either an edge from $x$ to $y$ or from $y$ to $x$, but not both. I will use "$xto y$" to mean "there is an edge from $x$ to $y$."



      A dominating set of a directed graph is a subset $S$ of vertices such that for every $tnotin S$, there exists $sin S$ so $sto t$.



      It can be shown$^*$ that every tournament on $2^{n+1}-2$ vertices has a dominating set of size $n$. My question is whether this result is tight.




      Does there exist a tournament on $2^{n+1}-1$ vertices with no dominating set of size $n$?



      If not, what is the smallest tournament with no dominating set of size $n$?




      My thoughts:




      • A necessary condition for a graph on $2^{n+1}-1$ vertices with no dominating set of size $n$ is that every vertex must have an out-degree of exactly $2^n-1$, so exactly half of its edges are outgoing.



      • The answer is yes when $n=1,2$.




        • The "rock-paper-scissors" graph on three vertices has no dominating set of size $1$.

        • The graph on $mathbb Z/7mathbb Z$ where each $x$ has directed edges to $x+1,x+2$ and $x+4pmod7$ has no dominating set of size $2$.




      For $nge 3$, the possibilities get too large, and I cannot come up with a clever solution. Can anyone see a pattern?



      I came up with this problem while thinking about this puzzle.




      $^*$Consider a vertex $s$ with maximal out-degree. By the hand-shaking lemma, this degree must be at least $(2^{n+1}-3)/2$, so at least $2^n-1$. Include $s$ in $S$, then ignore $s$ and the vertices $t$ for which $sto t$. What remains is tournament of size $(2^{n+1}-2)-1-(2^{n}-1)=2^n-2$. Proceed by induction.






      combinatorics discrete-mathematics graph-theory directed-graphs extremal-graph-theory







      share|cite|improve this question











      $endgroup$




      A tournement is a directed graph such that for every pair of distinct vertices ${x,y}$, there is either an edge from $x$ to $y$ or from $y$ to $x$, but not both. I will use "$xto y$" to mean "there is an edge from $x$ to $y$."



      A dominating set of a directed graph is a subset $S$ of vertices such that for every $tnotin S$, there exists $sin S$ so $sto t$.



      It can be shown$^*$ that every tournament on $2^{n+1}-2$ vertices has a dominating set of size $n$. My question is whether this result is tight.




      Does there exist a tournament on $2^{n+1}-1$ vertices with no dominating set of size $n$?



      If not, what is the smallest tournament with no dominating set of size $n$?




      My thoughts:




      • A necessary condition for a graph on $2^{n+1}-1$ vertices with no dominating set of size $n$ is that every vertex must have an out-degree of exactly $2^n-1$, so exactly half of its edges are outgoing.



      • The answer is yes when $n=1,2$.




        • The "rock-paper-scissors" graph on three vertices has no dominating set of size $1$.

        • The graph on $mathbb Z/7mathbb Z$ where each $x$ has directed edges to $x+1,x+2$ and $x+4pmod7$ has no dominating set of size $2$.




      For $nge 3$, the possibilities get too large, and I cannot come up with a clever solution. Can anyone see a pattern?



      I came up with this problem while thinking about this puzzle.




      $^*$Consider a vertex $s$ with maximal out-degree. By the hand-shaking lemma, this degree must be at least $(2^{n+1}-3)/2$, so at least $2^n-1$. Include $s$ in $S$, then ignore $s$ and the vertices $t$ for which $sto t$. What remains is tournament of size $(2^{n+1}-2)-1-(2^{n}-1)=2^n-2$. Proceed by induction.






      combinatorics discrete-mathematics graph-theory directed-graphs extremal-graph-theory




      combinatorics discrete-mathematics graph-theory directed-graphs extremal-graph-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 8 at 23:00







      Mike Earnest

















      asked Jan 8 at 22:06









      Mike EarnestMike Earnest

      26.7k22151




      26.7k22151






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          The answer to your question is yes: if $f(n)$ is the least number of vertices in a tournament with no $n$-vertex dominating set, then $f(n) > 2^{n+1}-1$ for large $n$ and in general we know
          $$
          (n+2) 2^{n-1} - 1 le f(n) le C cdot n^2 cdot 2^n
          $$
          for some constant $C$. Already for $n=3$ there is a $19$-vertex tournament with no dominating set of size $3$: the quadratic residue tournament mod $19$. (Here, we have an edge from $i$ to $j$ if there is some $k$ such that $i + k^2 equiv j pmod{19}$; it is a generalization of your $7$-vertex construction.)



          (For more details, see for example this paper by Szekeres and Szekeres.)






          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%2f3066793%2fdominating-sets-in-tournaments-is-2n1-2-tight%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









            2












            $begingroup$

            The answer to your question is yes: if $f(n)$ is the least number of vertices in a tournament with no $n$-vertex dominating set, then $f(n) > 2^{n+1}-1$ for large $n$ and in general we know
            $$
            (n+2) 2^{n-1} - 1 le f(n) le C cdot n^2 cdot 2^n
            $$
            for some constant $C$. Already for $n=3$ there is a $19$-vertex tournament with no dominating set of size $3$: the quadratic residue tournament mod $19$. (Here, we have an edge from $i$ to $j$ if there is some $k$ such that $i + k^2 equiv j pmod{19}$; it is a generalization of your $7$-vertex construction.)



            (For more details, see for example this paper by Szekeres and Szekeres.)






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              The answer to your question is yes: if $f(n)$ is the least number of vertices in a tournament with no $n$-vertex dominating set, then $f(n) > 2^{n+1}-1$ for large $n$ and in general we know
              $$
              (n+2) 2^{n-1} - 1 le f(n) le C cdot n^2 cdot 2^n
              $$
              for some constant $C$. Already for $n=3$ there is a $19$-vertex tournament with no dominating set of size $3$: the quadratic residue tournament mod $19$. (Here, we have an edge from $i$ to $j$ if there is some $k$ such that $i + k^2 equiv j pmod{19}$; it is a generalization of your $7$-vertex construction.)



              (For more details, see for example this paper by Szekeres and Szekeres.)






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                The answer to your question is yes: if $f(n)$ is the least number of vertices in a tournament with no $n$-vertex dominating set, then $f(n) > 2^{n+1}-1$ for large $n$ and in general we know
                $$
                (n+2) 2^{n-1} - 1 le f(n) le C cdot n^2 cdot 2^n
                $$
                for some constant $C$. Already for $n=3$ there is a $19$-vertex tournament with no dominating set of size $3$: the quadratic residue tournament mod $19$. (Here, we have an edge from $i$ to $j$ if there is some $k$ such that $i + k^2 equiv j pmod{19}$; it is a generalization of your $7$-vertex construction.)



                (For more details, see for example this paper by Szekeres and Szekeres.)






                share|cite|improve this answer









                $endgroup$



                The answer to your question is yes: if $f(n)$ is the least number of vertices in a tournament with no $n$-vertex dominating set, then $f(n) > 2^{n+1}-1$ for large $n$ and in general we know
                $$
                (n+2) 2^{n-1} - 1 le f(n) le C cdot n^2 cdot 2^n
                $$
                for some constant $C$. Already for $n=3$ there is a $19$-vertex tournament with no dominating set of size $3$: the quadratic residue tournament mod $19$. (Here, we have an edge from $i$ to $j$ if there is some $k$ such that $i + k^2 equiv j pmod{19}$; it is a generalization of your $7$-vertex construction.)



                (For more details, see for example this paper by Szekeres and Szekeres.)







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 8 at 23:37









                Misha LavrovMisha Lavrov

                48.1k657107




                48.1k657107






























                    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%2f3066793%2fdominating-sets-in-tournaments-is-2n1-2-tight%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

                    Bressuire

                    Image
                    .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                    Read more

                    Cabo Verde

                    Image
                      Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
                    Read more

                    Gyllenstierna

                    Image
                    .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                    Read more