Is Steiner Triple System Always Regular?
$begingroup$
I know that there exists $S(2,3,n)$, when $nequiv 1,3mod 6$. $S(2,3,7)$ is the Fano plane and $S(2,3,9)$ is an affine plane. These two examples are in fact both regular hypergraphs, i.e. every vertices has the same degree. Are Steiner Triple Systems always regular?
combinatorics hypergraphs
edited Jan 9 at 11:19
draks ...
11.5k644131
asked Jun 15 '17 at 21:32
Eric ZhangEric Zhang
36029
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add a comment |
$begingroup$
I know that there exists $S(2,3,n)$, when $nequiv 1,3mod 6$. $S(2,3,7)$ is the Fano plane and $S(2,3,9)$ is an affine plane. These two examples are in fact both regular hypergraphs, i.e. every vertices has the same degree. Are Steiner Triple Systems always regular?
combinatorics hypergraphs
edited Jan 9 at 11:19
draks ...
11.5k644131
asked Jun 15 '17 at 21:32
Eric ZhangEric Zhang
36029
$endgroup$
add a comment |
$begingroup$
I know that there exists $S(2,3,n)$, when $nequiv 1,3mod 6$. $S(2,3,7)$ is the Fano plane and $S(2,3,9)$ is an affine plane. These two examples are in fact both regular hypergraphs, i.e. every vertices has the same degree. Are Steiner Triple Systems always regular?
combinatorics hypergraphs
edited Jan 9 at 11:19
draks ...
11.5k644131
asked Jun 15 '17 at 21:32
Eric ZhangEric Zhang
36029
$endgroup$
I know that there exists $S(2,3,n)$, when $nequiv 1,3mod 6$. $S(2,3,7)$ is the Fano plane and $S(2,3,9)$ is an affine plane. These two examples are in fact both regular hypergraphs, i.e. every vertices has the same degree. Are Steiner Triple Systems always regular?
combinatorics hypergraphs
combinatorics hypergraphs
edited Jan 9 at 11:19
draks ...
11.5k644131
asked Jun 15 '17 at 21:32
Eric ZhangEric Zhang
36029
edited Jan 9 at 11:19
draks ...
11.5k644131
asked Jun 15 '17 at 21:32
Eric ZhangEric Zhang
36029
edited Jan 9 at 11:19
draks ...
11.5k644131
edited Jan 9 at 11:19
draks ...
11.5k644131
edited Jan 9 at 11:19
draks ...
11.5k644131
11.5k644131
asked Jun 15 '17 at 21:32
Eric ZhangEric Zhang
36029
asked Jun 15 '17 at 21:32
Eric ZhangEric Zhang
36029
asked Jun 15 '17 at 21:32
Eric ZhangEric Zhang
36029
36029
add a comment |
add a comment |
1 Answer
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Yes, a Steiner Triple System is always regular. Consider a STS on $n$ vertices and choose any vertex $v$. We know $v$ shares exactly one edge with every other vertex, and each edge containing $v$ contains exactly 2 other vertices. For this reason, the degree of $v$ is $frac{n-1}{2}$. All vertices will have this same degree, so the STS is regular.
answered Aug 15 '17 at 18:52
Brad ElliottBrad Elliott
1016
$endgroup$
$begingroup$
To be clear, a given vertex shares an edge with any other vertex due to the fact that each block in the STS is defined by precisely two vertices. So it would appear that we're viewing every STS as a complete graph on $n$ vertices, and the $K_3$ subgraphs which correspond to blocks define our edge set?
$endgroup$
– Chickenmancer
Aug 15 '17 at 19:03
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, a Steiner Triple System is always regular. Consider a STS on $n$ vertices and choose any vertex $v$. We know $v$ shares exactly one edge with every other vertex, and each edge containing $v$ contains exactly 2 other vertices. For this reason, the degree of $v$ is $frac{n-1}{2}$. All vertices will have this same degree, so the STS is regular.
answered Aug 15 '17 at 18:52
Brad ElliottBrad Elliott
1016
$endgroup$
$begingroup$
To be clear, a given vertex shares an edge with any other vertex due to the fact that each block in the STS is defined by precisely two vertices. So it would appear that we're viewing every STS as a complete graph on $n$ vertices, and the $K_3$ subgraphs which correspond to blocks define our edge set?
$endgroup$
– Chickenmancer
Aug 15 '17 at 19:03
add a comment |
$begingroup$
Yes, a Steiner Triple System is always regular. Consider a STS on $n$ vertices and choose any vertex $v$. We know $v$ shares exactly one edge with every other vertex, and each edge containing $v$ contains exactly 2 other vertices. For this reason, the degree of $v$ is $frac{n-1}{2}$. All vertices will have this same degree, so the STS is regular.
answered Aug 15 '17 at 18:52
Brad ElliottBrad Elliott
1016
$endgroup$
$begingroup$
To be clear, a given vertex shares an edge with any other vertex due to the fact that each block in the STS is defined by precisely two vertices. So it would appear that we're viewing every STS as a complete graph on $n$ vertices, and the $K_3$ subgraphs which correspond to blocks define our edge set?
$endgroup$
– Chickenmancer
Aug 15 '17 at 19:03
add a comment |
$begingroup$
Yes, a Steiner Triple System is always regular. Consider a STS on $n$ vertices and choose any vertex $v$. We know $v$ shares exactly one edge with every other vertex, and each edge containing $v$ contains exactly 2 other vertices. For this reason, the degree of $v$ is $frac{n-1}{2}$. All vertices will have this same degree, so the STS is regular.
answered Aug 15 '17 at 18:52
Brad ElliottBrad Elliott
1016
$endgroup$
Yes, a Steiner Triple System is always regular. Consider a STS on $n$ vertices and choose any vertex $v$. We know $v$ shares exactly one edge with every other vertex, and each edge containing $v$ contains exactly 2 other vertices. For this reason, the degree of $v$ is $frac{n-1}{2}$. All vertices will have this same degree, so the STS is regular.
answered Aug 15 '17 at 18:52
Brad ElliottBrad Elliott
1016
answered Aug 15 '17 at 18:52
Brad ElliottBrad Elliott
1016
answered Aug 15 '17 at 18:52
Brad ElliottBrad Elliott
1016
answered Aug 15 '17 at 18:52
Brad ElliottBrad Elliott
1016
1016
$begingroup$
To be clear, a given vertex shares an edge with any other vertex due to the fact that each block in the STS is defined by precisely two vertices. So it would appear that we're viewing every STS as a complete graph on $n$ vertices, and the $K_3$ subgraphs which correspond to blocks define our edge set?
$endgroup$
– Chickenmancer
Aug 15 '17 at 19:03
add a comment |
$begingroup$
To be clear, a given vertex shares an edge with any other vertex due to the fact that each block in the STS is defined by precisely two vertices. So it would appear that we're viewing every STS as a complete graph on $n$ vertices, and the $K_3$ subgraphs which correspond to blocks define our edge set?
$endgroup$
– Chickenmancer
Aug 15 '17 at 19:03
$begingroup$
To be clear, a given vertex shares an edge with any other vertex due to the fact that each block in the STS is defined by precisely two vertices. So it would appear that we're viewing every STS as a complete graph on $n$ vertices, and the $K_3$ subgraphs which correspond to blocks define our edge set?
$endgroup$
– Chickenmancer
Aug 15 '17 at 19:03
$begingroup$
To be clear, a given vertex shares an edge with any other vertex due to the fact that each block in the STS is defined by precisely two vertices. So it would appear that we're viewing every STS as a complete graph on $n$ vertices, and the $K_3$ subgraphs which correspond to blocks define our edge set?
$endgroup$
– Chickenmancer
Aug 15 '17 at 19:03
add a comment |
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