Mixing
Dominating sets in tournaments; is $2^{n+1}-2$ tight?
$begingroup$
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
asked Jan 8 at 22:06
Mike EarnestMike Earnest
22.1k12051
$endgroup$
add a comment |
$begingroup$
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
asked Jan 8 at 22:06
Mike EarnestMike Earnest
22.1k12051
$endgroup$
add a comment |
$begingroup$
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
asked Jan 8 at 22:06
Mike EarnestMike Earnest
22.1k12051
$endgroup$
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
asked Jan 8 at 22:06
Mike EarnestMike Earnest
22.1k12051
asked Jan 8 at 22:06
Mike EarnestMike Earnest
22.1k12051
edited Jan 8 at 23:00
Mike Earnest
asked Jan 8 at 22:06
Mike EarnestMike Earnest
22.1k12051
asked Jan 8 at 22:06
Mike EarnestMike Earnest
22.1k12051
asked Jan 8 at 22:06
Mike EarnestMike Earnest
22.1k12051
22.1k12051
add a comment |
add a comment |
1 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.)
answered Jan 8 at 23:37
Misha LavrovMisha Lavrov
45.9k656107
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add a comment |
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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.)
answered Jan 8 at 23:37
Misha LavrovMisha Lavrov
45.9k656107
$endgroup$
add a comment |
$begingroup$
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.)
answered Jan 8 at 23:37
Misha LavrovMisha Lavrov
45.9k656107
$endgroup$
add a comment |
$begingroup$
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.)
answered Jan 8 at 23:37
Misha LavrovMisha Lavrov
45.9k656107
$endgroup$
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.)
answered Jan 8 at 23:37
Misha LavrovMisha Lavrov
45.9k656107
answered Jan 8 at 23:37
Misha LavrovMisha Lavrov
45.9k656107
answered Jan 8 at 23:37
Misha LavrovMisha Lavrov
45.9k656107
answered Jan 8 at 23:37
Misha LavrovMisha Lavrov
45.9k656107
45.9k656107
add a comment |
add a comment |
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