Mixing
Tournament with the property $S_k$
$begingroup$
I'm wondering what's the matter with this property on oriented graphs :
$$ T= (V(T),E(T)) text{ has prop } S_k text{ if for any k vertices } v_1, cdots, v_k in V(T) \ text{ there exist a vertex } u in V(T) text{ such that } uv_1, uv_2, cdots, uv_k in E(T) $$
I think that $S_1$ is the necessary and sufficient condition for having a connected graph (is this even true? I don't know if we consider the direction of the edges), but what about this property for any k ? what is it saying?
combinatorics graph-theory
asked Jan 17 at 14:27
Marine GalantinMarine Galantin
860316
$endgroup$
add a comment |
$begingroup$
I'm wondering what's the matter with this property on oriented graphs :
$$ T= (V(T),E(T)) text{ has prop } S_k text{ if for any k vertices } v_1, cdots, v_k in V(T) \ text{ there exist a vertex } u in V(T) text{ such that } uv_1, uv_2, cdots, uv_k in E(T) $$
I think that $S_1$ is the necessary and sufficient condition for having a connected graph (is this even true? I don't know if we consider the direction of the edges), but what about this property for any k ? what is it saying?
combinatorics graph-theory
asked Jan 17 at 14:27
Marine GalantinMarine Galantin
860316
$endgroup$
add a comment |
$begingroup$
I'm wondering what's the matter with this property on oriented graphs :
$$ T= (V(T),E(T)) text{ has prop } S_k text{ if for any k vertices } v_1, cdots, v_k in V(T) \ text{ there exist a vertex } u in V(T) text{ such that } uv_1, uv_2, cdots, uv_k in E(T) $$
I think that $S_1$ is the necessary and sufficient condition for having a connected graph (is this even true? I don't know if we consider the direction of the edges), but what about this property for any k ? what is it saying?
combinatorics graph-theory
asked Jan 17 at 14:27
Marine GalantinMarine Galantin
860316
$endgroup$
I'm wondering what's the matter with this property on oriented graphs :
$$ T= (V(T),E(T)) text{ has prop } S_k text{ if for any k vertices } v_1, cdots, v_k in V(T) \ text{ there exist a vertex } u in V(T) text{ such that } uv_1, uv_2, cdots, uv_k in E(T) $$
I think that $S_1$ is the necessary and sufficient condition for having a connected graph (is this even true? I don't know if we consider the direction of the edges), but what about this property for any k ? what is it saying?
combinatorics graph-theory
combinatorics graph-theory
asked Jan 17 at 14:27
Marine GalantinMarine Galantin
860316
asked Jan 17 at 14:27
Marine GalantinMarine Galantin
860316
asked Jan 17 at 14:27
Marine GalantinMarine Galantin
860316
asked Jan 17 at 14:27
Marine GalantinMarine Galantin
860316
asked Jan 17 at 14:27
Marine GalantinMarine Galantin
860316
860316
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$S_k$ says that every $k$ vertices have a common (direct) ancestor (predecessor may be the better, or more familiar term).
This does not say that the graph is connected for $k=1$.
For $k=1$ (and $T$ finite!) it does say that there is a directed cycle. The disjoint union of two directed cycles however satisfies $S_1$ but is not connected.
For $k>1$ and $T$ finite it does imply that the underlying simple graph is connected, but $T$ still does not need to be strongly connected.
For infinite graphs all bets are off: there need not even be cycles.
EDIT: above is for general digraphs (overlooking the fact that the question title clearly states tournaments).
In a tournament you can show that property $S_k$ implies minimum in-degree $k$ (provided $k$ is smaller than the size of the tournament). You can prove this as follows:
Suppose a vertex $v$ has in-degree $l<k$. Let $v_1u,ldots,v_lu$ be the $l$ in-edges of $u$. Let $v_{l+1},ldots,v_{k-1}$ other vertices (possibly zero if $l=k-1$), different from $u$, and all $v_i$ different from each other. By definition $u,v_1,v_2,ldots,v_{k-1}$ have a common predecessor $w$ that must be different from all $v_i$ and therefore is another in-edge for $u$. Contradiction.
answered Jan 17 at 15:32
Leen DroogendijkLeen Droogendijk
6,1351716
$endgroup$
$begingroup$
what's an ancestor please. I m only studying finite graphs
$endgroup$
– Marine Galantin
Jan 17 at 15:42
$begingroup$
In a directed graph $u$ is a direct ancestor of $v$ if there is a directed edge $uv$. I only mention finiteness, because if you want to prove that there is a directed cycle, this finiteness is explicitly required.
$endgroup$
– Leen Droogendijk
Jan 17 at 15:47
1
$begingroup$
I'm also familiar with to notion of degre of a vertex. Does S_k imply that every vertex is of at least degree k?
$endgroup$
– Marine Galantin
Jan 17 at 15:58
1
$begingroup$
Unfortunately not. If you have a graph $T$ satisfying $S_2$, you can add four new vertices: $u$, $v$, $w$, $x$, such that $u$, $v$ and $w$ point to every other vertex, and only $u$ points to $x$. The new graph satisfies $S_2$ again, but $x$ has in-degree $1$ and outdegree $0$. This does require that you allow allow both edge $xy$ and directed edge $yx$ at the same time in your graph.
$endgroup$
– Leen Droogendijk
Jan 17 at 16:14
1
$begingroup$
Your intuition about the degree was correct. Answer has been edited to reflect that.
$endgroup$
– Leen Droogendijk
Jan 17 at 22:41
|
show 5 more comments
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1 Answer
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$begingroup$
$S_k$ says that every $k$ vertices have a common (direct) ancestor (predecessor may be the better, or more familiar term).
This does not say that the graph is connected for $k=1$.
For $k=1$ (and $T$ finite!) it does say that there is a directed cycle. The disjoint union of two directed cycles however satisfies $S_1$ but is not connected.
For $k>1$ and $T$ finite it does imply that the underlying simple graph is connected, but $T$ still does not need to be strongly connected.
For infinite graphs all bets are off: there need not even be cycles.
EDIT: above is for general digraphs (overlooking the fact that the question title clearly states tournaments).
In a tournament you can show that property $S_k$ implies minimum in-degree $k$ (provided $k$ is smaller than the size of the tournament). You can prove this as follows:
Suppose a vertex $v$ has in-degree $l<k$. Let $v_1u,ldots,v_lu$ be the $l$ in-edges of $u$. Let $v_{l+1},ldots,v_{k-1}$ other vertices (possibly zero if $l=k-1$), different from $u$, and all $v_i$ different from each other. By definition $u,v_1,v_2,ldots,v_{k-1}$ have a common predecessor $w$ that must be different from all $v_i$ and therefore is another in-edge for $u$. Contradiction.
answered Jan 17 at 15:32
Leen DroogendijkLeen Droogendijk
6,1351716
$endgroup$
$begingroup$
what's an ancestor please. I m only studying finite graphs
$endgroup$
– Marine Galantin
Jan 17 at 15:42
$begingroup$
In a directed graph $u$ is a direct ancestor of $v$ if there is a directed edge $uv$. I only mention finiteness, because if you want to prove that there is a directed cycle, this finiteness is explicitly required.
$endgroup$
– Leen Droogendijk
Jan 17 at 15:47
1
$begingroup$
I'm also familiar with to notion of degre of a vertex. Does S_k imply that every vertex is of at least degree k?
$endgroup$
– Marine Galantin
Jan 17 at 15:58
1
$begingroup$
Unfortunately not. If you have a graph $T$ satisfying $S_2$, you can add four new vertices: $u$, $v$, $w$, $x$, such that $u$, $v$ and $w$ point to every other vertex, and only $u$ points to $x$. The new graph satisfies $S_2$ again, but $x$ has in-degree $1$ and outdegree $0$. This does require that you allow allow both edge $xy$ and directed edge $yx$ at the same time in your graph.
$endgroup$
– Leen Droogendijk
Jan 17 at 16:14
1
$begingroup$
Your intuition about the degree was correct. Answer has been edited to reflect that.
$endgroup$
– Leen Droogendijk
Jan 17 at 22:41
|
show 5 more comments
$begingroup$
$S_k$ says that every $k$ vertices have a common (direct) ancestor (predecessor may be the better, or more familiar term).
This does not say that the graph is connected for $k=1$.
For $k=1$ (and $T$ finite!) it does say that there is a directed cycle. The disjoint union of two directed cycles however satisfies $S_1$ but is not connected.
For $k>1$ and $T$ finite it does imply that the underlying simple graph is connected, but $T$ still does not need to be strongly connected.
For infinite graphs all bets are off: there need not even be cycles.
EDIT: above is for general digraphs (overlooking the fact that the question title clearly states tournaments).
In a tournament you can show that property $S_k$ implies minimum in-degree $k$ (provided $k$ is smaller than the size of the tournament). You can prove this as follows:
Suppose a vertex $v$ has in-degree $l<k$. Let $v_1u,ldots,v_lu$ be the $l$ in-edges of $u$. Let $v_{l+1},ldots,v_{k-1}$ other vertices (possibly zero if $l=k-1$), different from $u$, and all $v_i$ different from each other. By definition $u,v_1,v_2,ldots,v_{k-1}$ have a common predecessor $w$ that must be different from all $v_i$ and therefore is another in-edge for $u$. Contradiction.
answered Jan 17 at 15:32
Leen DroogendijkLeen Droogendijk
6,1351716
$endgroup$
$begingroup$
what's an ancestor please. I m only studying finite graphs
$endgroup$
– Marine Galantin
Jan 17 at 15:42
$begingroup$
In a directed graph $u$ is a direct ancestor of $v$ if there is a directed edge $uv$. I only mention finiteness, because if you want to prove that there is a directed cycle, this finiteness is explicitly required.
$endgroup$
– Leen Droogendijk
Jan 17 at 15:47
1
$begingroup$
I'm also familiar with to notion of degre of a vertex. Does S_k imply that every vertex is of at least degree k?
$endgroup$
– Marine Galantin
Jan 17 at 15:58
1
$begingroup$
Unfortunately not. If you have a graph $T$ satisfying $S_2$, you can add four new vertices: $u$, $v$, $w$, $x$, such that $u$, $v$ and $w$ point to every other vertex, and only $u$ points to $x$. The new graph satisfies $S_2$ again, but $x$ has in-degree $1$ and outdegree $0$. This does require that you allow allow both edge $xy$ and directed edge $yx$ at the same time in your graph.
$endgroup$
– Leen Droogendijk
Jan 17 at 16:14
1
$begingroup$
Your intuition about the degree was correct. Answer has been edited to reflect that.
$endgroup$
– Leen Droogendijk
Jan 17 at 22:41
|
show 5 more comments
$begingroup$
$S_k$ says that every $k$ vertices have a common (direct) ancestor (predecessor may be the better, or more familiar term).
This does not say that the graph is connected for $k=1$.
For $k=1$ (and $T$ finite!) it does say that there is a directed cycle. The disjoint union of two directed cycles however satisfies $S_1$ but is not connected.
For $k>1$ and $T$ finite it does imply that the underlying simple graph is connected, but $T$ still does not need to be strongly connected.
For infinite graphs all bets are off: there need not even be cycles.
EDIT: above is for general digraphs (overlooking the fact that the question title clearly states tournaments).
In a tournament you can show that property $S_k$ implies minimum in-degree $k$ (provided $k$ is smaller than the size of the tournament). You can prove this as follows:
Suppose a vertex $v$ has in-degree $l<k$. Let $v_1u,ldots,v_lu$ be the $l$ in-edges of $u$. Let $v_{l+1},ldots,v_{k-1}$ other vertices (possibly zero if $l=k-1$), different from $u$, and all $v_i$ different from each other. By definition $u,v_1,v_2,ldots,v_{k-1}$ have a common predecessor $w$ that must be different from all $v_i$ and therefore is another in-edge for $u$. Contradiction.
answered Jan 17 at 15:32
Leen DroogendijkLeen Droogendijk
6,1351716
$endgroup$
$S_k$ says that every $k$ vertices have a common (direct) ancestor (predecessor may be the better, or more familiar term).
This does not say that the graph is connected for $k=1$.
For $k=1$ (and $T$ finite!) it does say that there is a directed cycle. The disjoint union of two directed cycles however satisfies $S_1$ but is not connected.
For $k>1$ and $T$ finite it does imply that the underlying simple graph is connected, but $T$ still does not need to be strongly connected.
For infinite graphs all bets are off: there need not even be cycles.
EDIT: above is for general digraphs (overlooking the fact that the question title clearly states tournaments).
In a tournament you can show that property $S_k$ implies minimum in-degree $k$ (provided $k$ is smaller than the size of the tournament). You can prove this as follows:
Suppose a vertex $v$ has in-degree $l<k$. Let $v_1u,ldots,v_lu$ be the $l$ in-edges of $u$. Let $v_{l+1},ldots,v_{k-1}$ other vertices (possibly zero if $l=k-1$), different from $u$, and all $v_i$ different from each other. By definition $u,v_1,v_2,ldots,v_{k-1}$ have a common predecessor $w$ that must be different from all $v_i$ and therefore is another in-edge for $u$. Contradiction.
answered Jan 17 at 15:32
Leen DroogendijkLeen Droogendijk
6,1351716
edited Jan 17 at 22:39
answered Jan 17 at 15:32
Leen DroogendijkLeen Droogendijk
6,1351716
answered Jan 17 at 15:32
Leen DroogendijkLeen Droogendijk
6,1351716
answered Jan 17 at 15:32
Leen DroogendijkLeen Droogendijk
6,1351716
6,1351716
$begingroup$
what's an ancestor please. I m only studying finite graphs
$endgroup$
– Marine Galantin
Jan 17 at 15:42
$begingroup$
In a directed graph $u$ is a direct ancestor of $v$ if there is a directed edge $uv$. I only mention finiteness, because if you want to prove that there is a directed cycle, this finiteness is explicitly required.
$endgroup$
– Leen Droogendijk
Jan 17 at 15:47
1
$begingroup$
I'm also familiar with to notion of degre of a vertex. Does S_k imply that every vertex is of at least degree k?
$endgroup$
– Marine Galantin
Jan 17 at 15:58
1
$begingroup$
Unfortunately not. If you have a graph $T$ satisfying $S_2$, you can add four new vertices: $u$, $v$, $w$, $x$, such that $u$, $v$ and $w$ point to every other vertex, and only $u$ points to $x$. The new graph satisfies $S_2$ again, but $x$ has in-degree $1$ and outdegree $0$. This does require that you allow allow both edge $xy$ and directed edge $yx$ at the same time in your graph.
$endgroup$
– Leen Droogendijk
Jan 17 at 16:14
1
$begingroup$
Your intuition about the degree was correct. Answer has been edited to reflect that.
$endgroup$
– Leen Droogendijk
Jan 17 at 22:41
|
show 5 more comments
$begingroup$
what's an ancestor please. I m only studying finite graphs
$endgroup$
– Marine Galantin
Jan 17 at 15:42
$begingroup$
In a directed graph $u$ is a direct ancestor of $v$ if there is a directed edge $uv$. I only mention finiteness, because if you want to prove that there is a directed cycle, this finiteness is explicitly required.
$endgroup$
– Leen Droogendijk
Jan 17 at 15:47
1
$begingroup$
I'm also familiar with to notion of degre of a vertex. Does S_k imply that every vertex is of at least degree k?
$endgroup$
– Marine Galantin
Jan 17 at 15:58
1
$begingroup$
Unfortunately not. If you have a graph $T$ satisfying $S_2$, you can add four new vertices: $u$, $v$, $w$, $x$, such that $u$, $v$ and $w$ point to every other vertex, and only $u$ points to $x$. The new graph satisfies $S_2$ again, but $x$ has in-degree $1$ and outdegree $0$. This does require that you allow allow both edge $xy$ and directed edge $yx$ at the same time in your graph.
$endgroup$
– Leen Droogendijk
Jan 17 at 16:14
1
$begingroup$
Your intuition about the degree was correct. Answer has been edited to reflect that.
$endgroup$
– Leen Droogendijk
Jan 17 at 22:41
$begingroup$
what's an ancestor please. I m only studying finite graphs
$endgroup$
– Marine Galantin
Jan 17 at 15:42
$begingroup$
what's an ancestor please. I m only studying finite graphs
$endgroup$
– Marine Galantin
Jan 17 at 15:42
$begingroup$
In a directed graph $u$ is a direct ancestor of $v$ if there is a directed edge $uv$. I only mention finiteness, because if you want to prove that there is a directed cycle, this finiteness is explicitly required.
$endgroup$
– Leen Droogendijk
Jan 17 at 15:47
$begingroup$
In a directed graph $u$ is a direct ancestor of $v$ if there is a directed edge $uv$. I only mention finiteness, because if you want to prove that there is a directed cycle, this finiteness is explicitly required.
$endgroup$
– Leen Droogendijk
Jan 17 at 15:47
1
1
$begingroup$
I'm also familiar with to notion of degre of a vertex. Does S_k imply that every vertex is of at least degree k?
$endgroup$
– Marine Galantin
Jan 17 at 15:58
$begingroup$
I'm also familiar with to notion of degre of a vertex. Does S_k imply that every vertex is of at least degree k?
$endgroup$
– Marine Galantin
Jan 17 at 15:58
1
1
$begingroup$
Unfortunately not. If you have a graph $T$ satisfying $S_2$, you can add four new vertices: $u$, $v$, $w$, $x$, such that $u$, $v$ and $w$ point to every other vertex, and only $u$ points to $x$. The new graph satisfies $S_2$ again, but $x$ has in-degree $1$ and outdegree $0$. This does require that you allow allow both edge $xy$ and directed edge $yx$ at the same time in your graph.
$endgroup$
– Leen Droogendijk
Jan 17 at 16:14
$begingroup$
Unfortunately not. If you have a graph $T$ satisfying $S_2$, you can add four new vertices: $u$, $v$, $w$, $x$, such that $u$, $v$ and $w$ point to every other vertex, and only $u$ points to $x$. The new graph satisfies $S_2$ again, but $x$ has in-degree $1$ and outdegree $0$. This does require that you allow allow both edge $xy$ and directed edge $yx$ at the same time in your graph.
$endgroup$
– Leen Droogendijk
Jan 17 at 16:14
1
1
$begingroup$
Your intuition about the degree was correct. Answer has been edited to reflect that.
$endgroup$
– Leen Droogendijk
Jan 17 at 22:41
$begingroup$
Your intuition about the degree was correct. Answer has been edited to reflect that.
$endgroup$
– Leen Droogendijk
Jan 17 at 22:41
|
show 5 more comments
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