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







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$endgroup$

















    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




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      edited Jan 8 at 23:00







      Mike Earnest

















      asked Jan 8 at 22:06









      Mike EarnestMike Earnest

      22.1k12051




      22.1k12051






















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          $begingroup$

          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









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            $begingroup$

            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

                45.9k656107




                45.9k656107






























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