Probability of an edge in directed random configuration graph
$begingroup$
I am considering Bollobas' directed random configuration graph of size $N$, constructed by the following random algorithm:
Draw a sequence of $N$ node-degree pairs $(j_1,k_1),...,(j_N,k_N)$ independently from the degree distribution $P$.
Accept the draw only if it is feasible, i.e. only if $sum_{n in [N]}(j_n-k_n)=0$.For every node $n$, create $j_n$ in-stubs and $k_n$ out-stubs, where in-/out-stubs are open ended half-edges with an in-/out-arrow.
- For any unpaired out-stub, select iteratively uniformly at random an unpaired in-stub and connect them.
Each such resulting pair of stubs forms a directed edge of the graph.
I am interested in the probabilities of an edge. My suggestion is:
Under this random matching approach, the probability that there is an edge from a node $i$ to a node $l$ is, with $m$ being the number of all edges:
$$p_{il}=frac{k_i cdot j_l}{(2m-1)},$$
as node $i$ has $k_i$ out-stubs and $j_l$ in-stubs out of $2m-1$ (excluding of node $i$) attached to $l$ to which it could connect.
And similarly the probability that there is an edge from a node $l$ to $i$ is:$$p_{il}=frac{j_i cdot k_l}{(2m-1)}$$
Is this correct, or am I missing something?
probability probability-theory graph-theory random-graphs
asked Nov 9 '18 at 15:09
AlisatAlisat
639
$endgroup$
add a comment |
$begingroup$
I am considering Bollobas' directed random configuration graph of size $N$, constructed by the following random algorithm:
Draw a sequence of $N$ node-degree pairs $(j_1,k_1),...,(j_N,k_N)$ independently from the degree distribution $P$.
Accept the draw only if it is feasible, i.e. only if $sum_{n in [N]}(j_n-k_n)=0$.For every node $n$, create $j_n$ in-stubs and $k_n$ out-stubs, where in-/out-stubs are open ended half-edges with an in-/out-arrow.
- For any unpaired out-stub, select iteratively uniformly at random an unpaired in-stub and connect them.
Each such resulting pair of stubs forms a directed edge of the graph.
I am interested in the probabilities of an edge. My suggestion is:
Under this random matching approach, the probability that there is an edge from a node $i$ to a node $l$ is, with $m$ being the number of all edges:
$$p_{il}=frac{k_i cdot j_l}{(2m-1)},$$
as node $i$ has $k_i$ out-stubs and $j_l$ in-stubs out of $2m-1$ (excluding of node $i$) attached to $l$ to which it could connect.
And similarly the probability that there is an edge from a node $l$ to $i$ is:$$p_{il}=frac{j_i cdot k_l}{(2m-1)}$$
Is this correct, or am I missing something?
probability probability-theory graph-theory random-graphs
asked Nov 9 '18 at 15:09
AlisatAlisat
639
$endgroup$
add a comment |
$begingroup$
I am considering Bollobas' directed random configuration graph of size $N$, constructed by the following random algorithm:
Draw a sequence of $N$ node-degree pairs $(j_1,k_1),...,(j_N,k_N)$ independently from the degree distribution $P$.
Accept the draw only if it is feasible, i.e. only if $sum_{n in [N]}(j_n-k_n)=0$.For every node $n$, create $j_n$ in-stubs and $k_n$ out-stubs, where in-/out-stubs are open ended half-edges with an in-/out-arrow.
- For any unpaired out-stub, select iteratively uniformly at random an unpaired in-stub and connect them.
Each such resulting pair of stubs forms a directed edge of the graph.
I am interested in the probabilities of an edge. My suggestion is:
Under this random matching approach, the probability that there is an edge from a node $i$ to a node $l$ is, with $m$ being the number of all edges:
$$p_{il}=frac{k_i cdot j_l}{(2m-1)},$$
as node $i$ has $k_i$ out-stubs and $j_l$ in-stubs out of $2m-1$ (excluding of node $i$) attached to $l$ to which it could connect.
And similarly the probability that there is an edge from a node $l$ to $i$ is:$$p_{il}=frac{j_i cdot k_l}{(2m-1)}$$
Is this correct, or am I missing something?
probability probability-theory graph-theory random-graphs
asked Nov 9 '18 at 15:09
AlisatAlisat
639
$endgroup$
I am considering Bollobas' directed random configuration graph of size $N$, constructed by the following random algorithm:
Draw a sequence of $N$ node-degree pairs $(j_1,k_1),...,(j_N,k_N)$ independently from the degree distribution $P$.
Accept the draw only if it is feasible, i.e. only if $sum_{n in [N]}(j_n-k_n)=0$.For every node $n$, create $j_n$ in-stubs and $k_n$ out-stubs, where in-/out-stubs are open ended half-edges with an in-/out-arrow.
- For any unpaired out-stub, select iteratively uniformly at random an unpaired in-stub and connect them.
Each such resulting pair of stubs forms a directed edge of the graph.
I am interested in the probabilities of an edge. My suggestion is:
Under this random matching approach, the probability that there is an edge from a node $i$ to a node $l$ is, with $m$ being the number of all edges:
$$p_{il}=frac{k_i cdot j_l}{(2m-1)},$$
as node $i$ has $k_i$ out-stubs and $j_l$ in-stubs out of $2m-1$ (excluding of node $i$) attached to $l$ to which it could connect.
And similarly the probability that there is an edge from a node $l$ to $i$ is:$$p_{il}=frac{j_i cdot k_l}{(2m-1)}$$
Is this correct, or am I missing something?
probability probability-theory graph-theory random-graphs
probability probability-theory graph-theory random-graphs
asked Nov 9 '18 at 15:09
AlisatAlisat
639
asked Nov 9 '18 at 15:09
AlisatAlisat
639
edited Jan 6 at 19:23
Alisat
asked Nov 9 '18 at 15:09
AlisatAlisat
639
asked Nov 9 '18 at 15:09
AlisatAlisat
639
asked Nov 9 '18 at 15:09
AlisatAlisat
639
639
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We can choose a uniformly random matching in any order. So let's choose the $k_i$ edges out of node $i$ one at a time and see the probability none of them go to node $l$.
The first edge we choose has probability $frac{j_l}{m}$ of going to $l$. If it doesn't, then the second edge has probability $frac{j_l}{m-1}$. If that also doesn't happen, then the third edge has probability $frac{j_l}{m-2}$, and so on. So the overall probability that none of the edges go to node $l$ is
$$
1-p_{il} = left(1 - frac{j_l}{m}right)left(1 - frac{j_l}{m-1}right) dotsb left(1 - frac{j_l}{m-k_i+1}right)
$$
and the probability you want is the complement of this one.
Asymptotically, however, assuming $k_i$ and $j_l$ are small relative to $m$, the answer is in fact close to $frac{k_i cdot j_l}{m}$ which is what you have, except that $m$ and not $2m-1$ is the number of options for the first edge. (Unlike the undirected configuration model, here there are $m$ out-stubs and $m$ in-stubs, which we distinguish.)
Additionally, because there are $k_i$ edges with probability $frac{j_l}{m}$ of going to $l$, then $frac{k_i j_l}{m}$ edges is the expected number of edges from $i$ to $l$.
answered Jan 6 at 17:16
Misha LavrovMisha Lavrov
49k757107
$endgroup$
$begingroup$
So I guess there is no "neat" form for the probability as in the undirected one, where it is of the form $p_{ij} =frac{ k_i k_j} {2m−1}$. ? (Only for the large limits as you stated $p_{il} =frac{ k_i j_l} {m}$)
$endgroup$
– Alisat
Jan 6 at 18:35
1
$begingroup$
The "neat" form is not the probability but the expected number of edges from $i$ to $l$.
$endgroup$
– Misha Lavrov
Jan 6 at 20:10
$begingroup$
And for the undirected case, the formula you give is also the expectation, not the probability.
$endgroup$
– Misha Lavrov
Jan 6 at 20:15
1
$begingroup$
The paper does not use the configuration mode. Instead, they define $p_{ij} = frac{k_i k_j}{2m}$ then add an edge $ij$ with that probability. As they mention on p. 3 (second paragraph) the actual degree of vertex $i$ is not guaranteed to be $k_i$ in this model.
$endgroup$
– Misha Lavrov
Jan 6 at 21:19
1
$begingroup$
Upon further reading, the paper does seem to claim that $frac{k_i k_j}{2m-1}$ is the exact edge probability in the random matching model, which is incorrect.
$endgroup$
– Misha Lavrov
Jan 6 at 21:48
|
show 1 more comment
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1 Answer
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1 Answer
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$begingroup$
We can choose a uniformly random matching in any order. So let's choose the $k_i$ edges out of node $i$ one at a time and see the probability none of them go to node $l$.
The first edge we choose has probability $frac{j_l}{m}$ of going to $l$. If it doesn't, then the second edge has probability $frac{j_l}{m-1}$. If that also doesn't happen, then the third edge has probability $frac{j_l}{m-2}$, and so on. So the overall probability that none of the edges go to node $l$ is
$$
1-p_{il} = left(1 - frac{j_l}{m}right)left(1 - frac{j_l}{m-1}right) dotsb left(1 - frac{j_l}{m-k_i+1}right)
$$
and the probability you want is the complement of this one.
Asymptotically, however, assuming $k_i$ and $j_l$ are small relative to $m$, the answer is in fact close to $frac{k_i cdot j_l}{m}$ which is what you have, except that $m$ and not $2m-1$ is the number of options for the first edge. (Unlike the undirected configuration model, here there are $m$ out-stubs and $m$ in-stubs, which we distinguish.)
Additionally, because there are $k_i$ edges with probability $frac{j_l}{m}$ of going to $l$, then $frac{k_i j_l}{m}$ edges is the expected number of edges from $i$ to $l$.
answered Jan 6 at 17:16
Misha LavrovMisha Lavrov
49k757107
$endgroup$
$begingroup$
So I guess there is no "neat" form for the probability as in the undirected one, where it is of the form $p_{ij} =frac{ k_i k_j} {2m−1}$. ? (Only for the large limits as you stated $p_{il} =frac{ k_i j_l} {m}$)
$endgroup$
– Alisat
Jan 6 at 18:35
1
$begingroup$
The "neat" form is not the probability but the expected number of edges from $i$ to $l$.
$endgroup$
– Misha Lavrov
Jan 6 at 20:10
$begingroup$
And for the undirected case, the formula you give is also the expectation, not the probability.
$endgroup$
– Misha Lavrov
Jan 6 at 20:15
1
$begingroup$
The paper does not use the configuration mode. Instead, they define $p_{ij} = frac{k_i k_j}{2m}$ then add an edge $ij$ with that probability. As they mention on p. 3 (second paragraph) the actual degree of vertex $i$ is not guaranteed to be $k_i$ in this model.
$endgroup$
– Misha Lavrov
Jan 6 at 21:19
1
$begingroup$
Upon further reading, the paper does seem to claim that $frac{k_i k_j}{2m-1}$ is the exact edge probability in the random matching model, which is incorrect.
$endgroup$
– Misha Lavrov
Jan 6 at 21:48
|
show 1 more comment
$begingroup$
We can choose a uniformly random matching in any order. So let's choose the $k_i$ edges out of node $i$ one at a time and see the probability none of them go to node $l$.
The first edge we choose has probability $frac{j_l}{m}$ of going to $l$. If it doesn't, then the second edge has probability $frac{j_l}{m-1}$. If that also doesn't happen, then the third edge has probability $frac{j_l}{m-2}$, and so on. So the overall probability that none of the edges go to node $l$ is
$$
1-p_{il} = left(1 - frac{j_l}{m}right)left(1 - frac{j_l}{m-1}right) dotsb left(1 - frac{j_l}{m-k_i+1}right)
$$
and the probability you want is the complement of this one.
Asymptotically, however, assuming $k_i$ and $j_l$ are small relative to $m$, the answer is in fact close to $frac{k_i cdot j_l}{m}$ which is what you have, except that $m$ and not $2m-1$ is the number of options for the first edge. (Unlike the undirected configuration model, here there are $m$ out-stubs and $m$ in-stubs, which we distinguish.)
Additionally, because there are $k_i$ edges with probability $frac{j_l}{m}$ of going to $l$, then $frac{k_i j_l}{m}$ edges is the expected number of edges from $i$ to $l$.
answered Jan 6 at 17:16
Misha LavrovMisha Lavrov
49k757107
$endgroup$
$begingroup$
So I guess there is no "neat" form for the probability as in the undirected one, where it is of the form $p_{ij} =frac{ k_i k_j} {2m−1}$. ? (Only for the large limits as you stated $p_{il} =frac{ k_i j_l} {m}$)
$endgroup$
– Alisat
Jan 6 at 18:35
1
$begingroup$
The "neat" form is not the probability but the expected number of edges from $i$ to $l$.
$endgroup$
– Misha Lavrov
Jan 6 at 20:10
$begingroup$
And for the undirected case, the formula you give is also the expectation, not the probability.
$endgroup$
– Misha Lavrov
Jan 6 at 20:15
1
$begingroup$
The paper does not use the configuration mode. Instead, they define $p_{ij} = frac{k_i k_j}{2m}$ then add an edge $ij$ with that probability. As they mention on p. 3 (second paragraph) the actual degree of vertex $i$ is not guaranteed to be $k_i$ in this model.
$endgroup$
– Misha Lavrov
Jan 6 at 21:19
1
$begingroup$
Upon further reading, the paper does seem to claim that $frac{k_i k_j}{2m-1}$ is the exact edge probability in the random matching model, which is incorrect.
$endgroup$
– Misha Lavrov
Jan 6 at 21:48
|
show 1 more comment
$begingroup$
We can choose a uniformly random matching in any order. So let's choose the $k_i$ edges out of node $i$ one at a time and see the probability none of them go to node $l$.
The first edge we choose has probability $frac{j_l}{m}$ of going to $l$. If it doesn't, then the second edge has probability $frac{j_l}{m-1}$. If that also doesn't happen, then the third edge has probability $frac{j_l}{m-2}$, and so on. So the overall probability that none of the edges go to node $l$ is
$$
1-p_{il} = left(1 - frac{j_l}{m}right)left(1 - frac{j_l}{m-1}right) dotsb left(1 - frac{j_l}{m-k_i+1}right)
$$
and the probability you want is the complement of this one.
Asymptotically, however, assuming $k_i$ and $j_l$ are small relative to $m$, the answer is in fact close to $frac{k_i cdot j_l}{m}$ which is what you have, except that $m$ and not $2m-1$ is the number of options for the first edge. (Unlike the undirected configuration model, here there are $m$ out-stubs and $m$ in-stubs, which we distinguish.)
Additionally, because there are $k_i$ edges with probability $frac{j_l}{m}$ of going to $l$, then $frac{k_i j_l}{m}$ edges is the expected number of edges from $i$ to $l$.
answered Jan 6 at 17:16
Misha LavrovMisha Lavrov
49k757107
$endgroup$
We can choose a uniformly random matching in any order. So let's choose the $k_i$ edges out of node $i$ one at a time and see the probability none of them go to node $l$.
The first edge we choose has probability $frac{j_l}{m}$ of going to $l$. If it doesn't, then the second edge has probability $frac{j_l}{m-1}$. If that also doesn't happen, then the third edge has probability $frac{j_l}{m-2}$, and so on. So the overall probability that none of the edges go to node $l$ is
$$
1-p_{il} = left(1 - frac{j_l}{m}right)left(1 - frac{j_l}{m-1}right) dotsb left(1 - frac{j_l}{m-k_i+1}right)
$$
and the probability you want is the complement of this one.
Asymptotically, however, assuming $k_i$ and $j_l$ are small relative to $m$, the answer is in fact close to $frac{k_i cdot j_l}{m}$ which is what you have, except that $m$ and not $2m-1$ is the number of options for the first edge. (Unlike the undirected configuration model, here there are $m$ out-stubs and $m$ in-stubs, which we distinguish.)
Additionally, because there are $k_i$ edges with probability $frac{j_l}{m}$ of going to $l$, then $frac{k_i j_l}{m}$ edges is the expected number of edges from $i$ to $l$.
answered Jan 6 at 17:16
Misha LavrovMisha Lavrov
49k757107
edited Jan 6 at 20:09
answered Jan 6 at 17:16
Misha LavrovMisha Lavrov
49k757107
answered Jan 6 at 17:16
Misha LavrovMisha Lavrov
49k757107
answered Jan 6 at 17:16
Misha LavrovMisha Lavrov
49k757107
49k757107
$begingroup$
So I guess there is no "neat" form for the probability as in the undirected one, where it is of the form $p_{ij} =frac{ k_i k_j} {2m−1}$. ? (Only for the large limits as you stated $p_{il} =frac{ k_i j_l} {m}$)
$endgroup$
– Alisat
Jan 6 at 18:35
1
$begingroup$
The "neat" form is not the probability but the expected number of edges from $i$ to $l$.
$endgroup$
– Misha Lavrov
Jan 6 at 20:10
$begingroup$
And for the undirected case, the formula you give is also the expectation, not the probability.
$endgroup$
– Misha Lavrov
Jan 6 at 20:15
1
$begingroup$
The paper does not use the configuration mode. Instead, they define $p_{ij} = frac{k_i k_j}{2m}$ then add an edge $ij$ with that probability. As they mention on p. 3 (second paragraph) the actual degree of vertex $i$ is not guaranteed to be $k_i$ in this model.
$endgroup$
– Misha Lavrov
Jan 6 at 21:19
1
$begingroup$
Upon further reading, the paper does seem to claim that $frac{k_i k_j}{2m-1}$ is the exact edge probability in the random matching model, which is incorrect.
$endgroup$
– Misha Lavrov
Jan 6 at 21:48
|
show 1 more comment
$begingroup$
So I guess there is no "neat" form for the probability as in the undirected one, where it is of the form $p_{ij} =frac{ k_i k_j} {2m−1}$. ? (Only for the large limits as you stated $p_{il} =frac{ k_i j_l} {m}$)
$endgroup$
– Alisat
Jan 6 at 18:35
1
$begingroup$
The "neat" form is not the probability but the expected number of edges from $i$ to $l$.
$endgroup$
– Misha Lavrov
Jan 6 at 20:10
$begingroup$
And for the undirected case, the formula you give is also the expectation, not the probability.
$endgroup$
– Misha Lavrov
Jan 6 at 20:15
1
$begingroup$
The paper does not use the configuration mode. Instead, they define $p_{ij} = frac{k_i k_j}{2m}$ then add an edge $ij$ with that probability. As they mention on p. 3 (second paragraph) the actual degree of vertex $i$ is not guaranteed to be $k_i$ in this model.
$endgroup$
– Misha Lavrov
Jan 6 at 21:19
1
$begingroup$
Upon further reading, the paper does seem to claim that $frac{k_i k_j}{2m-1}$ is the exact edge probability in the random matching model, which is incorrect.
$endgroup$
– Misha Lavrov
Jan 6 at 21:48
$begingroup$
So I guess there is no "neat" form for the probability as in the undirected one, where it is of the form $p_{ij} =frac{ k_i k_j} {2m−1}$. ? (Only for the large limits as you stated $p_{il} =frac{ k_i j_l} {m}$)
$endgroup$
– Alisat
Jan 6 at 18:35
$begingroup$
So I guess there is no "neat" form for the probability as in the undirected one, where it is of the form $p_{ij} =frac{ k_i k_j} {2m−1}$. ? (Only for the large limits as you stated $p_{il} =frac{ k_i j_l} {m}$)
$endgroup$
– Alisat
Jan 6 at 18:35
1
1
$begingroup$
The "neat" form is not the probability but the expected number of edges from $i$ to $l$.
$endgroup$
– Misha Lavrov
Jan 6 at 20:10
$begingroup$
The "neat" form is not the probability but the expected number of edges from $i$ to $l$.
$endgroup$
– Misha Lavrov
Jan 6 at 20:10
$begingroup$
And for the undirected case, the formula you give is also the expectation, not the probability.
$endgroup$
– Misha Lavrov
Jan 6 at 20:15
$begingroup$
And for the undirected case, the formula you give is also the expectation, not the probability.
$endgroup$
– Misha Lavrov
Jan 6 at 20:15
1
1
$begingroup$
The paper does not use the configuration mode. Instead, they define $p_{ij} = frac{k_i k_j}{2m}$ then add an edge $ij$ with that probability. As they mention on p. 3 (second paragraph) the actual degree of vertex $i$ is not guaranteed to be $k_i$ in this model.
$endgroup$
– Misha Lavrov
Jan 6 at 21:19
$begingroup$
The paper does not use the configuration mode. Instead, they define $p_{ij} = frac{k_i k_j}{2m}$ then add an edge $ij$ with that probability. As they mention on p. 3 (second paragraph) the actual degree of vertex $i$ is not guaranteed to be $k_i$ in this model.
$endgroup$
– Misha Lavrov
Jan 6 at 21:19
1
1
$begingroup$
Upon further reading, the paper does seem to claim that $frac{k_i k_j}{2m-1}$ is the exact edge probability in the random matching model, which is incorrect.
$endgroup$
– Misha Lavrov
Jan 6 at 21:48
$begingroup$
Upon further reading, the paper does seem to claim that $frac{k_i k_j}{2m-1}$ is the exact edge probability in the random matching model, which is incorrect.
$endgroup$
– Misha Lavrov
Jan 6 at 21:48
|
show 1 more comment
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