Phase transition threshold for acyclic directed graph The 2019 Stack Overflow Developer Survey Results Are InFamily of sets. Directed acyclic graph representation.Combinatorial problem: Directed Acyclic GraphDirected Acyclic Graph Questionbipartite graph vs. directed acyclic graphDirected acyclic graph problemName for “Stratified” or “Synchronized” directed acyclic graph?Longest Path in a acyclic, directed graphDirected acyclic graph and adjacency matrixPlanarity of a “cell graph”Directed acyclic graph
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Phase transition threshold for acyclic directed graph
The 2019 Stack Overflow Developer Survey Results Are InFamily of sets. Directed acyclic graph representation.Combinatorial problem: Directed Acyclic GraphDirected Acyclic Graph Questionbipartite graph vs. directed acyclic graphDirected acyclic graph problemName for “Stratified” or “Synchronized” directed acyclic graph?Longest Path in a acyclic, directed graphDirected acyclic graph and adjacency matrixPlanarity of a “cell graph”Directed acyclic graph
$begingroup$
Let $G$ be an acyclic directed graph with $MN$ vertices arranged into $M$ generations of $N$ vertices each. We stipulate that edges may only go from generation $j$ to generation $j+1$, so there are $(M-1)N^2$ permissible edges. Each permissible edge is chosen independently with probability $p$, and the resulting subset of permissible edges is used to construct $G$.
The quantity of interest is the probability $P(p)$ that a randomly-chosen vertex $x$ in the first generation of $G$ is connected by a directed path to a random-chosen vertex $y$ in the $M$th generation of $G$. As $N$ becomes large for fixed $M$, one expects a phase transition in $P$ from near $0$ to near $1$ as $p$ crosses some threshold $p_0$.
Question: What is this threshold $p_0$ in terms of $M$ and $N$?
For example, when $M=3$ (first interesting case), it seems that the threshold is $p_0=N^-frac12$, and I would guess that $p_0=N^-fracM-2M-1$ in general.
Is this correct, and how does one prove it? Or where can I find this problem analyzed? I have found many somewhat-relevant sources on percolation, but they mostly involve lattices or Erdös-Rényi graphs.
graph-theory percolation
$endgroup$
add a comment |
$begingroup$
Let $G$ be an acyclic directed graph with $MN$ vertices arranged into $M$ generations of $N$ vertices each. We stipulate that edges may only go from generation $j$ to generation $j+1$, so there are $(M-1)N^2$ permissible edges. Each permissible edge is chosen independently with probability $p$, and the resulting subset of permissible edges is used to construct $G$.
The quantity of interest is the probability $P(p)$ that a randomly-chosen vertex $x$ in the first generation of $G$ is connected by a directed path to a random-chosen vertex $y$ in the $M$th generation of $G$. As $N$ becomes large for fixed $M$, one expects a phase transition in $P$ from near $0$ to near $1$ as $p$ crosses some threshold $p_0$.
Question: What is this threshold $p_0$ in terms of $M$ and $N$?
For example, when $M=3$ (first interesting case), it seems that the threshold is $p_0=N^-frac12$, and I would guess that $p_0=N^-fracM-2M-1$ in general.
Is this correct, and how does one prove it? Or where can I find this problem analyzed? I have found many somewhat-relevant sources on percolation, but they mostly involve lattices or Erdös-Rényi graphs.
graph-theory percolation
$endgroup$
add a comment |
$begingroup$
Let $G$ be an acyclic directed graph with $MN$ vertices arranged into $M$ generations of $N$ vertices each. We stipulate that edges may only go from generation $j$ to generation $j+1$, so there are $(M-1)N^2$ permissible edges. Each permissible edge is chosen independently with probability $p$, and the resulting subset of permissible edges is used to construct $G$.
The quantity of interest is the probability $P(p)$ that a randomly-chosen vertex $x$ in the first generation of $G$ is connected by a directed path to a random-chosen vertex $y$ in the $M$th generation of $G$. As $N$ becomes large for fixed $M$, one expects a phase transition in $P$ from near $0$ to near $1$ as $p$ crosses some threshold $p_0$.
Question: What is this threshold $p_0$ in terms of $M$ and $N$?
For example, when $M=3$ (first interesting case), it seems that the threshold is $p_0=N^-frac12$, and I would guess that $p_0=N^-fracM-2M-1$ in general.
Is this correct, and how does one prove it? Or where can I find this problem analyzed? I have found many somewhat-relevant sources on percolation, but they mostly involve lattices or Erdös-Rényi graphs.
graph-theory percolation
$endgroup$
Let $G$ be an acyclic directed graph with $MN$ vertices arranged into $M$ generations of $N$ vertices each. We stipulate that edges may only go from generation $j$ to generation $j+1$, so there are $(M-1)N^2$ permissible edges. Each permissible edge is chosen independently with probability $p$, and the resulting subset of permissible edges is used to construct $G$.
The quantity of interest is the probability $P(p)$ that a randomly-chosen vertex $x$ in the first generation of $G$ is connected by a directed path to a random-chosen vertex $y$ in the $M$th generation of $G$. As $N$ becomes large for fixed $M$, one expects a phase transition in $P$ from near $0$ to near $1$ as $p$ crosses some threshold $p_0$.
Question: What is this threshold $p_0$ in terms of $M$ and $N$?
For example, when $M=3$ (first interesting case), it seems that the threshold is $p_0=N^-frac12$, and I would guess that $p_0=N^-fracM-2M-1$ in general.
Is this correct, and how does one prove it? Or where can I find this problem analyzed? I have found many somewhat-relevant sources on percolation, but they mostly involve lattices or Erdös-Rényi graphs.
graph-theory percolation
graph-theory percolation
edited Oct 25 '18 at 17:10
physmath121
asked Oct 25 '18 at 16:02
physmath121physmath121
162
162
add a comment |
add a comment |
1 Answer
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$begingroup$
I have no final answer but this is my progress:
Here is an explicit formula for $P(p)$:
$$P(p) = 1 - (1-p^M-1)^N^M-2$$
Here $1 - p^M-1$ is the probability that a complete path from $v_1$ to $v_M$ is not completely connected and $N^M-2$ is the number of possible paths from $v_1$ to $v_M$. (Where $v_i$ is some node in generation $i$).
Inserting $M=3$ and $p_0 = N^frac-12$ gives you:
$$P(p_0) = 1 - left(fracN-1Nright)^N.$$
Your general formula for $p_0$ will give you:
$$P(p_0) = 1 - left( fracK-1K right)^K$$
where $K := N^M-2$
Increasing $p_0$ above that threshold will decrease the subtractor, thus increasing $P(p_0)$, which follows the intuition. Decreasing $p_0$ will decrease the whole expression, thus making $P$ monotonically increasing in $p_0$.
I can't really see a clear transition here. Still curious if you can use that.
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I have no final answer but this is my progress:
Here is an explicit formula for $P(p)$:
$$P(p) = 1 - (1-p^M-1)^N^M-2$$
Here $1 - p^M-1$ is the probability that a complete path from $v_1$ to $v_M$ is not completely connected and $N^M-2$ is the number of possible paths from $v_1$ to $v_M$. (Where $v_i$ is some node in generation $i$).
Inserting $M=3$ and $p_0 = N^frac-12$ gives you:
$$P(p_0) = 1 - left(fracN-1Nright)^N.$$
Your general formula for $p_0$ will give you:
$$P(p_0) = 1 - left( fracK-1K right)^K$$
where $K := N^M-2$
Increasing $p_0$ above that threshold will decrease the subtractor, thus increasing $P(p_0)$, which follows the intuition. Decreasing $p_0$ will decrease the whole expression, thus making $P$ monotonically increasing in $p_0$.
I can't really see a clear transition here. Still curious if you can use that.
$endgroup$
add a comment |
$begingroup$
I have no final answer but this is my progress:
Here is an explicit formula for $P(p)$:
$$P(p) = 1 - (1-p^M-1)^N^M-2$$
Here $1 - p^M-1$ is the probability that a complete path from $v_1$ to $v_M$ is not completely connected and $N^M-2$ is the number of possible paths from $v_1$ to $v_M$. (Where $v_i$ is some node in generation $i$).
Inserting $M=3$ and $p_0 = N^frac-12$ gives you:
$$P(p_0) = 1 - left(fracN-1Nright)^N.$$
Your general formula for $p_0$ will give you:
$$P(p_0) = 1 - left( fracK-1K right)^K$$
where $K := N^M-2$
Increasing $p_0$ above that threshold will decrease the subtractor, thus increasing $P(p_0)$, which follows the intuition. Decreasing $p_0$ will decrease the whole expression, thus making $P$ monotonically increasing in $p_0$.
I can't really see a clear transition here. Still curious if you can use that.
$endgroup$
add a comment |
$begingroup$
I have no final answer but this is my progress:
Here is an explicit formula for $P(p)$:
$$P(p) = 1 - (1-p^M-1)^N^M-2$$
Here $1 - p^M-1$ is the probability that a complete path from $v_1$ to $v_M$ is not completely connected and $N^M-2$ is the number of possible paths from $v_1$ to $v_M$. (Where $v_i$ is some node in generation $i$).
Inserting $M=3$ and $p_0 = N^frac-12$ gives you:
$$P(p_0) = 1 - left(fracN-1Nright)^N.$$
Your general formula for $p_0$ will give you:
$$P(p_0) = 1 - left( fracK-1K right)^K$$
where $K := N^M-2$
Increasing $p_0$ above that threshold will decrease the subtractor, thus increasing $P(p_0)$, which follows the intuition. Decreasing $p_0$ will decrease the whole expression, thus making $P$ monotonically increasing in $p_0$.
I can't really see a clear transition here. Still curious if you can use that.
$endgroup$
I have no final answer but this is my progress:
Here is an explicit formula for $P(p)$:
$$P(p) = 1 - (1-p^M-1)^N^M-2$$
Here $1 - p^M-1$ is the probability that a complete path from $v_1$ to $v_M$ is not completely connected and $N^M-2$ is the number of possible paths from $v_1$ to $v_M$. (Where $v_i$ is some node in generation $i$).
Inserting $M=3$ and $p_0 = N^frac-12$ gives you:
$$P(p_0) = 1 - left(fracN-1Nright)^N.$$
Your general formula for $p_0$ will give you:
$$P(p_0) = 1 - left( fracK-1K right)^K$$
where $K := N^M-2$
Increasing $p_0$ above that threshold will decrease the subtractor, thus increasing $P(p_0)$, which follows the intuition. Decreasing $p_0$ will decrease the whole expression, thus making $P$ monotonically increasing in $p_0$.
I can't really see a clear transition here. Still curious if you can use that.
edited Mar 30 at 16:52
Alan Muniz
2,61311030
2,61311030
answered Mar 30 at 16:44
Lennart ScharmannLennart Scharmann
112
112
add a comment |
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