graph theory and set notation - calculating the flow in a graph












1














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?



enter image description here




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.










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    1














    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?



    enter image description here




    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.










    share|cite|improve this question



























      1












      1








      1







      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?



      enter image description here




      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.










      share|cite|improve this question















      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?



      enter image description here




      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






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 4 at 13:28







      cherry aldi

















      asked Jan 4 at 11:23









      cherry aldicherry aldi

      203




      203






















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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.






          share|cite|improve this answer























          • Thanks - do you say that it is obvious that j is not an edge because j does not follow s in the alphabet?
            – cherry aldi
            2 days ago










          • @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.
            – Larry B.
            2 days ago











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          0














          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.






          share|cite|improve this answer























          • Thanks - do you say that it is obvious that j is not an edge because j does not follow s in the alphabet?
            – cherry aldi
            2 days ago










          • @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.
            – Larry B.
            2 days ago
















          0














          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.






          share|cite|improve this answer























          • Thanks - do you say that it is obvious that j is not an edge because j does not follow s in the alphabet?
            – cherry aldi
            2 days ago










          • @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.
            – Larry B.
            2 days ago














          0












          0








          0






          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.






          share|cite|improve this answer














          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 days ago

























          answered Jan 4 at 18:16









          Larry B.Larry B.

          2,776728




          2,776728












          • Thanks - do you say that it is obvious that j is not an edge because j does not follow s in the alphabet?
            – cherry aldi
            2 days ago










          • @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.
            – Larry B.
            2 days ago


















          • Thanks - do you say that it is obvious that j is not an edge because j does not follow s in the alphabet?
            – cherry aldi
            2 days ago










          • @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.
            – Larry B.
            2 days ago
















          Thanks - do you say that it is obvious that j is not an edge because j does not follow s in the alphabet?
          – cherry aldi
          2 days ago




          Thanks - do you say that it is obvious that j is not an edge because j does not follow s in the alphabet?
          – cherry aldi
          2 days ago












          @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.
          – Larry B.
          2 days ago




          @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.
          – Larry B.
          2 days ago


















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