How to find number of simple non-isomorphic graphs for a given sequence?
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Suppose a sequence of degree of vertices of a graph is given.How to find total number of graphs for the given sequence.For example if sequence (2,2,2,2,2,2) is given , then two possible graphs are - 1.Hexagon 2.Two triangles not connected to each other.So i can tell that given sequence has at least two distinct graphs.
But without drawing graphs randomly , how to know total number of distinct graphs for given sequence.
combinatorics discrete-mathematics graph-theory
asked Jan 17 at 17:34
BhowmickBhowmick
1438
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add a comment |
$begingroup$
Suppose a sequence of degree of vertices of a graph is given.How to find total number of graphs for the given sequence.For example if sequence (2,2,2,2,2,2) is given , then two possible graphs are - 1.Hexagon 2.Two triangles not connected to each other.So i can tell that given sequence has at least two distinct graphs.
But without drawing graphs randomly , how to know total number of distinct graphs for given sequence.
combinatorics discrete-mathematics graph-theory
asked Jan 17 at 17:34
BhowmickBhowmick
1438
$endgroup$
$begingroup$
{ 2,2,2,2,2,2} = {2} is not a sequence. It is a set. The ordered six-tuple (2,2,2,2,2,2) is like a sequence.
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– William Elliot
Jan 18 at 0:56
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@WilliamElliot Sorry .That was typing error .Now i have edited the answer.
$endgroup$
– Bhowmick
Jan 18 at 16:26
1
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It is an even an open problem to count $p$-regular graphs on $n$ vertices (graphs with a degree sequence of $p$ repeated $n$ times). See mathoverflow.net/questions/77730/… for some asymptotic results.
$endgroup$
– Mike Earnest
Jan 18 at 23:59
add a comment |
$begingroup$
Suppose a sequence of degree of vertices of a graph is given.How to find total number of graphs for the given sequence.For example if sequence (2,2,2,2,2,2) is given , then two possible graphs are - 1.Hexagon 2.Two triangles not connected to each other.So i can tell that given sequence has at least two distinct graphs.
But without drawing graphs randomly , how to know total number of distinct graphs for given sequence.
combinatorics discrete-mathematics graph-theory
asked Jan 17 at 17:34
BhowmickBhowmick
1438
$endgroup$
Suppose a sequence of degree of vertices of a graph is given.How to find total number of graphs for the given sequence.For example if sequence (2,2,2,2,2,2) is given , then two possible graphs are - 1.Hexagon 2.Two triangles not connected to each other.So i can tell that given sequence has at least two distinct graphs.
But without drawing graphs randomly , how to know total number of distinct graphs for given sequence.
combinatorics discrete-mathematics graph-theory
combinatorics discrete-mathematics graph-theory
asked Jan 17 at 17:34
BhowmickBhowmick
1438
asked Jan 17 at 17:34
BhowmickBhowmick
1438
edited Jan 18 at 16:26
Bhowmick
asked Jan 17 at 17:34
BhowmickBhowmick
1438
asked Jan 17 at 17:34
BhowmickBhowmick
1438
asked Jan 17 at 17:34
BhowmickBhowmick
1438
1438
$begingroup$
{ 2,2,2,2,2,2} = {2} is not a sequence. It is a set. The ordered six-tuple (2,2,2,2,2,2) is like a sequence.
$endgroup$
– William Elliot
Jan 18 at 0:56
$begingroup$
@WilliamElliot Sorry .That was typing error .Now i have edited the answer.
$endgroup$
– Bhowmick
Jan 18 at 16:26
1
$begingroup$
It is an even an open problem to count $p$-regular graphs on $n$ vertices (graphs with a degree sequence of $p$ repeated $n$ times). See mathoverflow.net/questions/77730/… for some asymptotic results.
$endgroup$
– Mike Earnest
Jan 18 at 23:59
add a comment |
$begingroup$
{ 2,2,2,2,2,2} = {2} is not a sequence. It is a set. The ordered six-tuple (2,2,2,2,2,2) is like a sequence.
$endgroup$
– William Elliot
Jan 18 at 0:56
$begingroup$
@WilliamElliot Sorry .That was typing error .Now i have edited the answer.
$endgroup$
– Bhowmick
Jan 18 at 16:26
1
$begingroup$
It is an even an open problem to count $p$-regular graphs on $n$ vertices (graphs with a degree sequence of $p$ repeated $n$ times). See mathoverflow.net/questions/77730/… for some asymptotic results.
$endgroup$
– Mike Earnest
Jan 18 at 23:59
$begingroup$
{ 2,2,2,2,2,2} = {2} is not a sequence. It is a set. The ordered six-tuple (2,2,2,2,2,2) is like a sequence.
$endgroup$
– William Elliot
Jan 18 at 0:56
$begingroup$
{ 2,2,2,2,2,2} = {2} is not a sequence. It is a set. The ordered six-tuple (2,2,2,2,2,2) is like a sequence.
$endgroup$
– William Elliot
Jan 18 at 0:56
$begingroup$
@WilliamElliot Sorry .That was typing error .Now i have edited the answer.
$endgroup$
– Bhowmick
Jan 18 at 16:26
$begingroup$
@WilliamElliot Sorry .That was typing error .Now i have edited the answer.
$endgroup$
– Bhowmick
Jan 18 at 16:26
1
1
$begingroup$
It is an even an open problem to count $p$-regular graphs on $n$ vertices (graphs with a degree sequence of $p$ repeated $n$ times). See mathoverflow.net/questions/77730/… for some asymptotic results.
$endgroup$
– Mike Earnest
Jan 18 at 23:59
$begingroup$
It is an even an open problem to count $p$-regular graphs on $n$ vertices (graphs with a degree sequence of $p$ repeated $n$ times). See mathoverflow.net/questions/77730/… for some asymptotic results.
$endgroup$
– Mike Earnest
Jan 18 at 23:59
add a comment |
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$begingroup$
{ 2,2,2,2,2,2} = {2} is not a sequence. It is a set. The ordered six-tuple (2,2,2,2,2,2) is like a sequence.
$endgroup$
– William Elliot
Jan 18 at 0:56
$begingroup$
@WilliamElliot Sorry .That was typing error .Now i have edited the answer.
$endgroup$
– Bhowmick
Jan 18 at 16:26
1
$begingroup$
It is an even an open problem to count $p$-regular graphs on $n$ vertices (graphs with a degree sequence of $p$ repeated $n$ times). See mathoverflow.net/questions/77730/… for some asymptotic results.
$endgroup$
– Mike Earnest
Jan 18 at 23:59