Upper bounds for number of $3$-flags in a $(2k, k^2)$-graph $G$
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Let $G$ be an arbitrary simple graph on $2k$ vertices with $k^2$ edges, where $k geq 2$. Let $F$ be a $3$-flag, i.e., three triangles that share a single edge (this graph has 5 vertices and 7 edges). I want to find an upper bound for the number of $3$-flags $F$ in $G$.
One trivial upper bound is $(2k)^5$, but this is too crude and I want a smaller upper bound than this. Any ideas? Thank you in advance!
combinatorics discrete-mathematics graph-theory upper-lower-bounds
asked Jan 7 at 1:16
SarahSarah
824821
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add a comment |
$begingroup$
Let $G$ be an arbitrary simple graph on $2k$ vertices with $k^2$ edges, where $k geq 2$. Let $F$ be a $3$-flag, i.e., three triangles that share a single edge (this graph has 5 vertices and 7 edges). I want to find an upper bound for the number of $3$-flags $F$ in $G$.
One trivial upper bound is $(2k)^5$, but this is too crude and I want a smaller upper bound than this. Any ideas? Thank you in advance!
combinatorics discrete-mathematics graph-theory upper-lower-bounds
asked Jan 7 at 1:16
SarahSarah
824821
$endgroup$
add a comment |
$begingroup$
Let $G$ be an arbitrary simple graph on $2k$ vertices with $k^2$ edges, where $k geq 2$. Let $F$ be a $3$-flag, i.e., three triangles that share a single edge (this graph has 5 vertices and 7 edges). I want to find an upper bound for the number of $3$-flags $F$ in $G$.
One trivial upper bound is $(2k)^5$, but this is too crude and I want a smaller upper bound than this. Any ideas? Thank you in advance!
combinatorics discrete-mathematics graph-theory upper-lower-bounds
asked Jan 7 at 1:16
SarahSarah
824821
$endgroup$
Let $G$ be an arbitrary simple graph on $2k$ vertices with $k^2$ edges, where $k geq 2$. Let $F$ be a $3$-flag, i.e., three triangles that share a single edge (this graph has 5 vertices and 7 edges). I want to find an upper bound for the number of $3$-flags $F$ in $G$.
One trivial upper bound is $(2k)^5$, but this is too crude and I want a smaller upper bound than this. Any ideas? Thank you in advance!
combinatorics discrete-mathematics graph-theory upper-lower-bounds
combinatorics discrete-mathematics graph-theory upper-lower-bounds
asked Jan 7 at 1:16
SarahSarah
824821
asked Jan 7 at 1:16
SarahSarah
824821
asked Jan 7 at 1:16
SarahSarah
824821
asked Jan 7 at 1:16
SarahSarah
824821
asked Jan 7 at 1:16
SarahSarah
824821
824821
add a comment |
add a comment |
1 Answer
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The most compact number of 3-flags would occur in a complete graph. There, any edge forms a three flag with any other vertex in the complete graph. In $K_n$, there are $binom{n}{2}$ edges. Multiply this by the $binom{n-2}{3}$ possible vertices which create a 3-flag with given edge.
So, for a quick upper bound, given $e$ edges, find the smallest $n$ for which constructing $K_n$ is impossible, and then $binom{n}{2}binom{n-2}{3}$ is your upper bound.
answered Jan 7 at 1:48
Zachary HunterZachary Hunter
1,040314
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1 Answer
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1 Answer
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active
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$begingroup$
The most compact number of 3-flags would occur in a complete graph. There, any edge forms a three flag with any other vertex in the complete graph. In $K_n$, there are $binom{n}{2}$ edges. Multiply this by the $binom{n-2}{3}$ possible vertices which create a 3-flag with given edge.
So, for a quick upper bound, given $e$ edges, find the smallest $n$ for which constructing $K_n$ is impossible, and then $binom{n}{2}binom{n-2}{3}$ is your upper bound.
answered Jan 7 at 1:48
Zachary HunterZachary Hunter
1,040314
$endgroup$
add a comment |
$begingroup$
The most compact number of 3-flags would occur in a complete graph. There, any edge forms a three flag with any other vertex in the complete graph. In $K_n$, there are $binom{n}{2}$ edges. Multiply this by the $binom{n-2}{3}$ possible vertices which create a 3-flag with given edge.
So, for a quick upper bound, given $e$ edges, find the smallest $n$ for which constructing $K_n$ is impossible, and then $binom{n}{2}binom{n-2}{3}$ is your upper bound.
answered Jan 7 at 1:48
Zachary HunterZachary Hunter
1,040314
$endgroup$
add a comment |
$begingroup$
The most compact number of 3-flags would occur in a complete graph. There, any edge forms a three flag with any other vertex in the complete graph. In $K_n$, there are $binom{n}{2}$ edges. Multiply this by the $binom{n-2}{3}$ possible vertices which create a 3-flag with given edge.
So, for a quick upper bound, given $e$ edges, find the smallest $n$ for which constructing $K_n$ is impossible, and then $binom{n}{2}binom{n-2}{3}$ is your upper bound.
answered Jan 7 at 1:48
Zachary HunterZachary Hunter
1,040314
$endgroup$
The most compact number of 3-flags would occur in a complete graph. There, any edge forms a three flag with any other vertex in the complete graph. In $K_n$, there are $binom{n}{2}$ edges. Multiply this by the $binom{n-2}{3}$ possible vertices which create a 3-flag with given edge.
So, for a quick upper bound, given $e$ edges, find the smallest $n$ for which constructing $K_n$ is impossible, and then $binom{n}{2}binom{n-2}{3}$ is your upper bound.
answered Jan 7 at 1:48
Zachary HunterZachary Hunter
1,040314
answered Jan 7 at 1:48
Zachary HunterZachary Hunter
1,040314
answered Jan 7 at 1:48
Zachary HunterZachary Hunter
1,040314
answered Jan 7 at 1:48
Zachary HunterZachary Hunter
1,040314
1,040314
add a comment |
add a comment |
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