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.






graph-theory







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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.






    graph-theory







    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.






      graph-theory







      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




      graph-theory






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      share|cite|improve this question













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      edited 13 hours ago

























      asked 15 hours ago









      cherry aldi

      183




      183






















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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 $y in A$. It's obvious that $j$ is not an edge, so the current notation is pretty confusing.



          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.






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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 $y in A$. It's obvious that $j$ is not an edge, so the current notation is pretty confusing.



            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


























              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 $y in A$. It's obvious that $j$ is not an edge, so the current notation is pretty confusing.



              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
























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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 $y in A$. It's obvious that $j$ is not an edge, so the current notation is pretty confusing.



                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 $y in A$. It's obvious that $j$ is not an edge, so the current notation is pretty confusing.



                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










                answered 8 hours ago









                Larry B.

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                2,761728






























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