Mixing
Number of matchings on n nodes
$begingroup$
I am trying to find the number of matchings that are possible on graphs $G=(V,E)$ with $V = [n] = {0, ..., n-1}$
Let $a_n$ be the number of matchings on $[n]$ .
Then I plan to calculate $a_n = Sigma^{n}_{k = 0} b_k$ with $b_k$ being the number of matchings on $[n]$ with n nodes.
Then $b_k$ is $0$ for all not even k, since the number of nodes in a matching has to be even.
For even k: From the n nodes I want to select k nodes for the matching. So that there are $frac{n!}{(n-k!)}$ ways to order in the selected k nodes. (n elements in the first position, n-1 in the second, ..., n-k+1 in the k-th position)
The first 2 elements then form an edge, the next 2 as well and so on. These form the matching. But it doesn't matter whether it is {a,b} or {b,a} , so I would divide by $2^{k/2}$
Also it doesn't matter whether the edge occurs at the first position or the last, so I would again divide by $(k/2)!$ so that since there are the options of ordering the edges.
This however doesn't really make a nice compact expression for using that in an generating function. The kind of binomial coefficient makes me wonder, whether to use the binomial theorem, but it doesn't quite match this.
combinatorics graph-theory
asked Jan 24 at 14:47
jt0202jt0202
152
$endgroup$
|
show 1 more comment
$begingroup$
I am trying to find the number of matchings that are possible on graphs $G=(V,E)$ with $V = [n] = {0, ..., n-1}$
Let $a_n$ be the number of matchings on $[n]$ .
Then I plan to calculate $a_n = Sigma^{n}_{k = 0} b_k$ with $b_k$ being the number of matchings on $[n]$ with n nodes.
Then $b_k$ is $0$ for all not even k, since the number of nodes in a matching has to be even.
For even k: From the n nodes I want to select k nodes for the matching. So that there are $frac{n!}{(n-k!)}$ ways to order in the selected k nodes. (n elements in the first position, n-1 in the second, ..., n-k+1 in the k-th position)
The first 2 elements then form an edge, the next 2 as well and so on. These form the matching. But it doesn't matter whether it is {a,b} or {b,a} , so I would divide by $2^{k/2}$
Also it doesn't matter whether the edge occurs at the first position or the last, so I would again divide by $(k/2)!$ so that since there are the options of ordering the edges.
This however doesn't really make a nice compact expression for using that in an generating function. The kind of binomial coefficient makes me wonder, whether to use the binomial theorem, but it doesn't quite match this.
combinatorics graph-theory
asked Jan 24 at 14:47
jt0202jt0202
152
$endgroup$
$begingroup$
To clarify: do you want a formula for counting matchings on arbitrary graphs on $n$ nodes, or (as your discussion suggests) on the complete graph $K_n$?
$endgroup$
– Connor Harris
Jan 24 at 15:48
$begingroup$
The thing I am trying to solve just says "the number of not nessesarily perfect matchings on [n]". This looks to me like every matching I can do with these set. So probably we can savely assume the $K_n$
$endgroup$
– jt0202
Jan 24 at 15:52
$begingroup$
Do you have a specific goal later? do you need an exact number, or could we bound the number by something simpler?
$endgroup$
– Thomas Lesgourgues
Jan 24 at 16:19
$begingroup$
I am supposed to find a formula and after that use the formula as coefficients in a power series and determine their convergence radius.
$endgroup$
– jt0202
Jan 24 at 16:31
$begingroup$
Maybe I just use the formula I have and use the ratio test and split it into the series for n and the n+1 part . Maybe then a bound might be good .
$endgroup$
– jt0202
Jan 24 at 16:41
|
show 1 more comment
$begingroup$
I am trying to find the number of matchings that are possible on graphs $G=(V,E)$ with $V = [n] = {0, ..., n-1}$
Let $a_n$ be the number of matchings on $[n]$ .
Then I plan to calculate $a_n = Sigma^{n}_{k = 0} b_k$ with $b_k$ being the number of matchings on $[n]$ with n nodes.
Then $b_k$ is $0$ for all not even k, since the number of nodes in a matching has to be even.
For even k: From the n nodes I want to select k nodes for the matching. So that there are $frac{n!}{(n-k!)}$ ways to order in the selected k nodes. (n elements in the first position, n-1 in the second, ..., n-k+1 in the k-th position)
The first 2 elements then form an edge, the next 2 as well and so on. These form the matching. But it doesn't matter whether it is {a,b} or {b,a} , so I would divide by $2^{k/2}$
Also it doesn't matter whether the edge occurs at the first position or the last, so I would again divide by $(k/2)!$ so that since there are the options of ordering the edges.
This however doesn't really make a nice compact expression for using that in an generating function. The kind of binomial coefficient makes me wonder, whether to use the binomial theorem, but it doesn't quite match this.
combinatorics graph-theory
asked Jan 24 at 14:47
jt0202jt0202
152
$endgroup$
I am trying to find the number of matchings that are possible on graphs $G=(V,E)$ with $V = [n] = {0, ..., n-1}$
Let $a_n$ be the number of matchings on $[n]$ .
Then I plan to calculate $a_n = Sigma^{n}_{k = 0} b_k$ with $b_k$ being the number of matchings on $[n]$ with n nodes.
Then $b_k$ is $0$ for all not even k, since the number of nodes in a matching has to be even.
For even k: From the n nodes I want to select k nodes for the matching. So that there are $frac{n!}{(n-k!)}$ ways to order in the selected k nodes. (n elements in the first position, n-1 in the second, ..., n-k+1 in the k-th position)
The first 2 elements then form an edge, the next 2 as well and so on. These form the matching. But it doesn't matter whether it is {a,b} or {b,a} , so I would divide by $2^{k/2}$
Also it doesn't matter whether the edge occurs at the first position or the last, so I would again divide by $(k/2)!$ so that since there are the options of ordering the edges.
This however doesn't really make a nice compact expression for using that in an generating function. The kind of binomial coefficient makes me wonder, whether to use the binomial theorem, but it doesn't quite match this.
combinatorics graph-theory
combinatorics graph-theory
asked Jan 24 at 14:47
jt0202jt0202
152
asked Jan 24 at 14:47
jt0202jt0202
152
asked Jan 24 at 14:47
jt0202jt0202
152
asked Jan 24 at 14:47
jt0202jt0202
152
asked Jan 24 at 14:47
jt0202jt0202
152
152
$begingroup$
To clarify: do you want a formula for counting matchings on arbitrary graphs on $n$ nodes, or (as your discussion suggests) on the complete graph $K_n$?
$endgroup$
– Connor Harris
Jan 24 at 15:48
$begingroup$
The thing I am trying to solve just says "the number of not nessesarily perfect matchings on [n]". This looks to me like every matching I can do with these set. So probably we can savely assume the $K_n$
$endgroup$
– jt0202
Jan 24 at 15:52
$begingroup$
Do you have a specific goal later? do you need an exact number, or could we bound the number by something simpler?
$endgroup$
– Thomas Lesgourgues
Jan 24 at 16:19
$begingroup$
I am supposed to find a formula and after that use the formula as coefficients in a power series and determine their convergence radius.
$endgroup$
– jt0202
Jan 24 at 16:31
$begingroup$
Maybe I just use the formula I have and use the ratio test and split it into the series for n and the n+1 part . Maybe then a bound might be good .
$endgroup$
– jt0202
Jan 24 at 16:41
|
show 1 more comment
$begingroup$
To clarify: do you want a formula for counting matchings on arbitrary graphs on $n$ nodes, or (as your discussion suggests) on the complete graph $K_n$?
$endgroup$
– Connor Harris
Jan 24 at 15:48
$begingroup$
The thing I am trying to solve just says "the number of not nessesarily perfect matchings on [n]". This looks to me like every matching I can do with these set. So probably we can savely assume the $K_n$
$endgroup$
– jt0202
Jan 24 at 15:52
$begingroup$
Do you have a specific goal later? do you need an exact number, or could we bound the number by something simpler?
$endgroup$
– Thomas Lesgourgues
Jan 24 at 16:19
$begingroup$
I am supposed to find a formula and after that use the formula as coefficients in a power series and determine their convergence radius.
$endgroup$
– jt0202
Jan 24 at 16:31
$begingroup$
Maybe I just use the formula I have and use the ratio test and split it into the series for n and the n+1 part . Maybe then a bound might be good .
$endgroup$
– jt0202
Jan 24 at 16:41
$begingroup$
To clarify: do you want a formula for counting matchings on arbitrary graphs on $n$ nodes, or (as your discussion suggests) on the complete graph $K_n$?
$endgroup$
– Connor Harris
Jan 24 at 15:48
$begingroup$
To clarify: do you want a formula for counting matchings on arbitrary graphs on $n$ nodes, or (as your discussion suggests) on the complete graph $K_n$?
$endgroup$
– Connor Harris
Jan 24 at 15:48
$begingroup$
The thing I am trying to solve just says "the number of not nessesarily perfect matchings on [n]". This looks to me like every matching I can do with these set. So probably we can savely assume the $K_n$
$endgroup$
– jt0202
Jan 24 at 15:52
$begingroup$
The thing I am trying to solve just says "the number of not nessesarily perfect matchings on [n]". This looks to me like every matching I can do with these set. So probably we can savely assume the $K_n$
$endgroup$
– jt0202
Jan 24 at 15:52
$begingroup$
Do you have a specific goal later? do you need an exact number, or could we bound the number by something simpler?
$endgroup$
– Thomas Lesgourgues
Jan 24 at 16:19
$begingroup$
Do you have a specific goal later? do you need an exact number, or could we bound the number by something simpler?
$endgroup$
– Thomas Lesgourgues
Jan 24 at 16:19
$begingroup$
I am supposed to find a formula and after that use the formula as coefficients in a power series and determine their convergence radius.
$endgroup$
– jt0202
Jan 24 at 16:31
$begingroup$
I am supposed to find a formula and after that use the formula as coefficients in a power series and determine their convergence radius.
$endgroup$
– jt0202
Jan 24 at 16:31
$begingroup$
Maybe I just use the formula I have and use the ratio test and split it into the series for n and the n+1 part . Maybe then a bound might be good .
$endgroup$
– jt0202
Jan 24 at 16:41
$begingroup$
Maybe I just use the formula I have and use the ratio test and split it into the series for n and the n+1 part . Maybe then a bound might be good .
$endgroup$
– jt0202
Jan 24 at 16:41
|
show 1 more comment
1 Answer
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$begingroup$
I arrive to the same conclusion, using the fact that the number of perfect matchings in the complete graph $K_{2n}$ is $$frac{(2n)!}{n! 2^n}$$
This can be shown by induction. If $a_n$ is the number of perfect matching in $K_{2n}$ then, you can pick a vertex $u$, match it with (2n-1) vertices, and then find a perfect matching in $K_{2n-2}$. Hence $a_n=(2n-1)a_{n-1}$, and
$$a_n=(2n-1)(2n-3)ldots 1 = frac{(2n)!}{(2n)(2n-2)ldots 2}= frac{(2n)!}{2^n n!}$$
Then the $T_n$ total number of matching in $G$ is (with $#K_{2k}$ the number of complete subgraphs on $2k$ vertices among $n$)
begin{align*}
T_n &= sum_{k=1}^{lfloor n/2 rfloor} #K_{2k} cdot frac{(2k)!}{k! 2^k}\
&= sum_{k=1}^{lfloor n/2 rfloor} binom{n}{2k} cdot frac{(2k)!}{k! 2^k}\
&= sum_{k=1}^{lfloor n/2 rfloor} frac{n!}{2^k k! (n-2k)!}\
end{align*}
And I arrive at the conclusion then you. I don't think we can do much better though. I keep looking
answered Jan 24 at 15:59
Thomas LesgourguesThomas Lesgourgues
1,128119
$endgroup$
add a comment |
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$begingroup$
I arrive to the same conclusion, using the fact that the number of perfect matchings in the complete graph $K_{2n}$ is $$frac{(2n)!}{n! 2^n}$$
This can be shown by induction. If $a_n$ is the number of perfect matching in $K_{2n}$ then, you can pick a vertex $u$, match it with (2n-1) vertices, and then find a perfect matching in $K_{2n-2}$. Hence $a_n=(2n-1)a_{n-1}$, and
$$a_n=(2n-1)(2n-3)ldots 1 = frac{(2n)!}{(2n)(2n-2)ldots 2}= frac{(2n)!}{2^n n!}$$
Then the $T_n$ total number of matching in $G$ is (with $#K_{2k}$ the number of complete subgraphs on $2k$ vertices among $n$)
begin{align*}
T_n &= sum_{k=1}^{lfloor n/2 rfloor} #K_{2k} cdot frac{(2k)!}{k! 2^k}\
&= sum_{k=1}^{lfloor n/2 rfloor} binom{n}{2k} cdot frac{(2k)!}{k! 2^k}\
&= sum_{k=1}^{lfloor n/2 rfloor} frac{n!}{2^k k! (n-2k)!}\
end{align*}
And I arrive at the conclusion then you. I don't think we can do much better though. I keep looking
answered Jan 24 at 15:59
Thomas LesgourguesThomas Lesgourgues
1,128119
$endgroup$
add a comment |
$begingroup$
I arrive to the same conclusion, using the fact that the number of perfect matchings in the complete graph $K_{2n}$ is $$frac{(2n)!}{n! 2^n}$$
This can be shown by induction. If $a_n$ is the number of perfect matching in $K_{2n}$ then, you can pick a vertex $u$, match it with (2n-1) vertices, and then find a perfect matching in $K_{2n-2}$. Hence $a_n=(2n-1)a_{n-1}$, and
$$a_n=(2n-1)(2n-3)ldots 1 = frac{(2n)!}{(2n)(2n-2)ldots 2}= frac{(2n)!}{2^n n!}$$
Then the $T_n$ total number of matching in $G$ is (with $#K_{2k}$ the number of complete subgraphs on $2k$ vertices among $n$)
begin{align*}
T_n &= sum_{k=1}^{lfloor n/2 rfloor} #K_{2k} cdot frac{(2k)!}{k! 2^k}\
&= sum_{k=1}^{lfloor n/2 rfloor} binom{n}{2k} cdot frac{(2k)!}{k! 2^k}\
&= sum_{k=1}^{lfloor n/2 rfloor} frac{n!}{2^k k! (n-2k)!}\
end{align*}
And I arrive at the conclusion then you. I don't think we can do much better though. I keep looking
answered Jan 24 at 15:59
Thomas LesgourguesThomas Lesgourgues
1,128119
$endgroup$
add a comment |
$begingroup$
I arrive to the same conclusion, using the fact that the number of perfect matchings in the complete graph $K_{2n}$ is $$frac{(2n)!}{n! 2^n}$$
This can be shown by induction. If $a_n$ is the number of perfect matching in $K_{2n}$ then, you can pick a vertex $u$, match it with (2n-1) vertices, and then find a perfect matching in $K_{2n-2}$. Hence $a_n=(2n-1)a_{n-1}$, and
$$a_n=(2n-1)(2n-3)ldots 1 = frac{(2n)!}{(2n)(2n-2)ldots 2}= frac{(2n)!}{2^n n!}$$
Then the $T_n$ total number of matching in $G$ is (with $#K_{2k}$ the number of complete subgraphs on $2k$ vertices among $n$)
begin{align*}
T_n &= sum_{k=1}^{lfloor n/2 rfloor} #K_{2k} cdot frac{(2k)!}{k! 2^k}\
&= sum_{k=1}^{lfloor n/2 rfloor} binom{n}{2k} cdot frac{(2k)!}{k! 2^k}\
&= sum_{k=1}^{lfloor n/2 rfloor} frac{n!}{2^k k! (n-2k)!}\
end{align*}
And I arrive at the conclusion then you. I don't think we can do much better though. I keep looking
answered Jan 24 at 15:59
Thomas LesgourguesThomas Lesgourgues
1,128119
$endgroup$
I arrive to the same conclusion, using the fact that the number of perfect matchings in the complete graph $K_{2n}$ is $$frac{(2n)!}{n! 2^n}$$
This can be shown by induction. If $a_n$ is the number of perfect matching in $K_{2n}$ then, you can pick a vertex $u$, match it with (2n-1) vertices, and then find a perfect matching in $K_{2n-2}$. Hence $a_n=(2n-1)a_{n-1}$, and
$$a_n=(2n-1)(2n-3)ldots 1 = frac{(2n)!}{(2n)(2n-2)ldots 2}= frac{(2n)!}{2^n n!}$$
Then the $T_n$ total number of matching in $G$ is (with $#K_{2k}$ the number of complete subgraphs on $2k$ vertices among $n$)
begin{align*}
T_n &= sum_{k=1}^{lfloor n/2 rfloor} #K_{2k} cdot frac{(2k)!}{k! 2^k}\
&= sum_{k=1}^{lfloor n/2 rfloor} binom{n}{2k} cdot frac{(2k)!}{k! 2^k}\
&= sum_{k=1}^{lfloor n/2 rfloor} frac{n!}{2^k k! (n-2k)!}\
end{align*}
And I arrive at the conclusion then you. I don't think we can do much better though. I keep looking
answered Jan 24 at 15:59
Thomas LesgourguesThomas Lesgourgues
1,128119
edited Jan 24 at 16:09
answered Jan 24 at 15:59
Thomas LesgourguesThomas Lesgourgues
1,128119
answered Jan 24 at 15:59
Thomas LesgourguesThomas Lesgourgues
1,128119
answered Jan 24 at 15:59
Thomas LesgourguesThomas Lesgourgues
1,128119
1,128119
add a comment |
add a comment |
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$begingroup$
To clarify: do you want a formula for counting matchings on arbitrary graphs on $n$ nodes, or (as your discussion suggests) on the complete graph $K_n$?
$endgroup$
– Connor Harris
Jan 24 at 15:48
$begingroup$
The thing I am trying to solve just says "the number of not nessesarily perfect matchings on [n]". This looks to me like every matching I can do with these set. So probably we can savely assume the $K_n$
$endgroup$
– jt0202
Jan 24 at 15:52
$begingroup$
Do you have a specific goal later? do you need an exact number, or could we bound the number by something simpler?
$endgroup$
– Thomas Lesgourgues
Jan 24 at 16:19
$begingroup$
I am supposed to find a formula and after that use the formula as coefficients in a power series and determine their convergence radius.
$endgroup$
– jt0202
Jan 24 at 16:31
$begingroup$
Maybe I just use the formula I have and use the ratio test and split it into the series for n and the n+1 part . Maybe then a bound might be good .
$endgroup$
– jt0202
Jan 24 at 16:41