Mixing
graph theory and set notation - calculating the flow in a graph
$begingroup$
I'm trying to understand some graph theory and set notation. Does the statement below read like this for calculating the flow in a graph?
Given some fixed input value j, sum those elements in f whose index
(s,j) satisfies (s,j)∈ E
So loop over all the s's (j is fixed) and add the terms corresponding to those js's which satisfy (s,j)∈E.
Which effectively means sum all the flows on the edges to get the flow in a network.
graph-theory
asked Jan 4 at 11:23
cherry aldicherry aldi
203
$endgroup$
add a comment |
$begingroup$
I'm trying to understand some graph theory and set notation. Does the statement below read like this for calculating the flow in a graph?
Given some fixed input value j, sum those elements in f whose index
(s,j) satisfies (s,j)∈ E
So loop over all the s's (j is fixed) and add the terms corresponding to those js's which satisfy (s,j)∈E.
Which effectively means sum all the flows on the edges to get the flow in a network.
graph-theory
asked Jan 4 at 11:23
cherry aldicherry aldi
203
$endgroup$
add a comment |
$begingroup$
I'm trying to understand some graph theory and set notation. Does the statement below read like this for calculating the flow in a graph?
Given some fixed input value j, sum those elements in f whose index
(s,j) satisfies (s,j)∈ E
So loop over all the s's (j is fixed) and add the terms corresponding to those js's which satisfy (s,j)∈E.
Which effectively means sum all the flows on the edges to get the flow in a network.
graph-theory
asked Jan 4 at 11:23
cherry aldicherry aldi
203
$endgroup$
I'm trying to understand some graph theory and set notation. Does the statement below read like this for calculating the flow in a graph?
Given some fixed input value j, sum those elements in f whose index
(s,j) satisfies (s,j)∈ E
So loop over all the s's (j is fixed) and add the terms corresponding to those js's which satisfy (s,j)∈E.
Which effectively means sum all the flows on the edges to get the flow in a network.
graph-theory
graph-theory
asked Jan 4 at 11:23
cherry aldicherry aldi
203
asked Jan 4 at 11:23
cherry aldicherry aldi
203
edited Jan 4 at 13:28
cherry aldi
asked Jan 4 at 11:23
cherry aldicherry aldi
203
asked Jan 4 at 11:23
cherry aldicherry aldi
203
asked Jan 4 at 11:23
cherry aldicherry aldi
203
203
add a comment |
add a comment |
1 Answer
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It does not read like this. The comma symbol, in relation to set membership, usually satisfies this implication: $x, y in A implies x in A $ and $y in A$. We know $j$ cannot be an edge, since it's also used in the node pair $(i,j)$ representing an edge in $E$. It's strange to have an element $j$ be both a node and an edge, but if the comma is interpreted like this, that's what we get.
To fix a variable, usually you would say $f_j = sum_{(s,j)in E} f_{s,j}$. Since $j$ is affixed as a subscript of $f$, it is made clear that $j$ is fixed for the right side of the equation.
answered Jan 4 at 18:16
Larry B.Larry B.
2,786728
$endgroup$
$begingroup$
Thanks - do you say that it is obvious that j is not an edge because j does not follow s in the alphabet?
$endgroup$
– cherry aldi
Jan 6 at 15:23
$begingroup$
@cherryaldi It's "obvious" because $j$ is immediately used within the node pair of the edge set. It's very unusual for an element to be both a node and an edge. I'll edit my answer with this insight.
$endgroup$
– Larry B.
Jan 6 at 19:08
add a comment |
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1 Answer
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1 Answer
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$begingroup$
It does not read like this. The comma symbol, in relation to set membership, usually satisfies this implication: $x, y in A implies x in A $ and $y in A$. We know $j$ cannot be an edge, since it's also used in the node pair $(i,j)$ representing an edge in $E$. It's strange to have an element $j$ be both a node and an edge, but if the comma is interpreted like this, that's what we get.
To fix a variable, usually you would say $f_j = sum_{(s,j)in E} f_{s,j}$. Since $j$ is affixed as a subscript of $f$, it is made clear that $j$ is fixed for the right side of the equation.
answered Jan 4 at 18:16
Larry B.Larry B.
2,786728
$endgroup$
$begingroup$
Thanks - do you say that it is obvious that j is not an edge because j does not follow s in the alphabet?
$endgroup$
– cherry aldi
Jan 6 at 15:23
$begingroup$
@cherryaldi It's "obvious" because $j$ is immediately used within the node pair of the edge set. It's very unusual for an element to be both a node and an edge. I'll edit my answer with this insight.
$endgroup$
– Larry B.
Jan 6 at 19:08
add a comment |
$begingroup$
It does not read like this. The comma symbol, in relation to set membership, usually satisfies this implication: $x, y in A implies x in A $ and $y in A$. We know $j$ cannot be an edge, since it's also used in the node pair $(i,j)$ representing an edge in $E$. It's strange to have an element $j$ be both a node and an edge, but if the comma is interpreted like this, that's what we get.
To fix a variable, usually you would say $f_j = sum_{(s,j)in E} f_{s,j}$. Since $j$ is affixed as a subscript of $f$, it is made clear that $j$ is fixed for the right side of the equation.
answered Jan 4 at 18:16
Larry B.Larry B.
2,786728
$endgroup$
$begingroup$
Thanks - do you say that it is obvious that j is not an edge because j does not follow s in the alphabet?
$endgroup$
– cherry aldi
Jan 6 at 15:23
$begingroup$
@cherryaldi It's "obvious" because $j$ is immediately used within the node pair of the edge set. It's very unusual for an element to be both a node and an edge. I'll edit my answer with this insight.
$endgroup$
– Larry B.
Jan 6 at 19:08
add a comment |
$begingroup$
It does not read like this. The comma symbol, in relation to set membership, usually satisfies this implication: $x, y in A implies x in A $ and $y in A$. We know $j$ cannot be an edge, since it's also used in the node pair $(i,j)$ representing an edge in $E$. It's strange to have an element $j$ be both a node and an edge, but if the comma is interpreted like this, that's what we get.
To fix a variable, usually you would say $f_j = sum_{(s,j)in E} f_{s,j}$. Since $j$ is affixed as a subscript of $f$, it is made clear that $j$ is fixed for the right side of the equation.
answered Jan 4 at 18:16
Larry B.Larry B.
2,786728
$endgroup$
It does not read like this. The comma symbol, in relation to set membership, usually satisfies this implication: $x, y in A implies x in A $ and $y in A$. We know $j$ cannot be an edge, since it's also used in the node pair $(i,j)$ representing an edge in $E$. It's strange to have an element $j$ be both a node and an edge, but if the comma is interpreted like this, that's what we get.
To fix a variable, usually you would say $f_j = sum_{(s,j)in E} f_{s,j}$. Since $j$ is affixed as a subscript of $f$, it is made clear that $j$ is fixed for the right side of the equation.
answered Jan 4 at 18:16
Larry B.Larry B.
2,786728
edited Jan 6 at 19:12
answered Jan 4 at 18:16
Larry B.Larry B.
2,786728
answered Jan 4 at 18:16
Larry B.Larry B.
2,786728
answered Jan 4 at 18:16
Larry B.Larry B.
2,786728
2,786728
$begingroup$
Thanks - do you say that it is obvious that j is not an edge because j does not follow s in the alphabet?
$endgroup$
– cherry aldi
Jan 6 at 15:23
$begingroup$
@cherryaldi It's "obvious" because $j$ is immediately used within the node pair of the edge set. It's very unusual for an element to be both a node and an edge. I'll edit my answer with this insight.
$endgroup$
– Larry B.
Jan 6 at 19:08
add a comment |
$begingroup$
Thanks - do you say that it is obvious that j is not an edge because j does not follow s in the alphabet?
$endgroup$
– cherry aldi
Jan 6 at 15:23
$begingroup$
@cherryaldi It's "obvious" because $j$ is immediately used within the node pair of the edge set. It's very unusual for an element to be both a node and an edge. I'll edit my answer with this insight.
$endgroup$
– Larry B.
Jan 6 at 19:08
$begingroup$
Thanks - do you say that it is obvious that j is not an edge because j does not follow s in the alphabet?
$endgroup$
– cherry aldi
Jan 6 at 15:23
$begingroup$
Thanks - do you say that it is obvious that j is not an edge because j does not follow s in the alphabet?
$endgroup$
– cherry aldi
Jan 6 at 15:23
$begingroup$
@cherryaldi It's "obvious" because $j$ is immediately used within the node pair of the edge set. It's very unusual for an element to be both a node and an edge. I'll edit my answer with this insight.
$endgroup$
– Larry B.
Jan 6 at 19:08
$begingroup$
@cherryaldi It's "obvious" because $j$ is immediately used within the node pair of the edge set. It's very unusual for an element to be both a node and an edge. I'll edit my answer with this insight.
$endgroup$
– Larry B.
Jan 6 at 19:08
add a comment |
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