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Why CLRS example on residual networks does not follows its formula?


Why is the complexity of negative-cycle-cancelling $O(V^2AUW)$?CLRS - Maxflow Augmented Flow Lemma 26.1 - don't understand use of def. in proofFord-Fulkerson algorithm clarificationLinear programming formulation of cheapest k-edge path between two nodesWhy is it that the flow value can increased along an augmenting path $p$ in a residual network?Maximum flow with Edmonds–Karp algorithmHow would one construct conjunctively local predicate of order k for checking if a shape is Convex?Equivalence of minimum cost circulation problem and minimum cost max flow problemGiven max-flow determine if edge is in a min-cutWhat is the intuition behind the way of reading off a dual optimal solution from simplex primal tabular in CLRS?













1












$begingroup$


I am learning algorithms to solve Maximum Flow problem by reading the CRLS book and confused by the following figure:



figure 26.4



That is:




A flow in a residual network provides a roadmap for adding flow to the
original flow network. If $f$ is a flow in $G$ and $f'$ is a flow in
the corresponding residual network $G_f$, we define $f uparrow f'$,
the augmentation of flow $f$ by $f'$, to be a function from $V times V$ to
$R$, defined by



$$(f uparrow f')(u, v) = begincases f(u,v) + f'(u, v) - f'(v, u) &
> textif (u,v) $in$ E \ 0 & textotherwise endcases$$




How the flow network in (c), for example $(s, v_2)$ got the flow 12 ?
If we follow the formula, it must have a flow 5:
$8 + 5 - 8 = 5$










share|cite|improve this question









$endgroup$
















    1












    $begingroup$


    I am learning algorithms to solve Maximum Flow problem by reading the CRLS book and confused by the following figure:



    figure 26.4



    That is:




    A flow in a residual network provides a roadmap for adding flow to the
    original flow network. If $f$ is a flow in $G$ and $f'$ is a flow in
    the corresponding residual network $G_f$, we define $f uparrow f'$,
    the augmentation of flow $f$ by $f'$, to be a function from $V times V$ to
    $R$, defined by



    $$(f uparrow f')(u, v) = begincases f(u,v) + f'(u, v) - f'(v, u) &
    > textif (u,v) $in$ E \ 0 & textotherwise endcases$$




    How the flow network in (c), for example $(s, v_2)$ got the flow 12 ?
    If we follow the formula, it must have a flow 5:
    $8 + 5 - 8 = 5$










    share|cite|improve this question









    $endgroup$














      1












      1








      1





      $begingroup$


      I am learning algorithms to solve Maximum Flow problem by reading the CRLS book and confused by the following figure:



      figure 26.4



      That is:




      A flow in a residual network provides a roadmap for adding flow to the
      original flow network. If $f$ is a flow in $G$ and $f'$ is a flow in
      the corresponding residual network $G_f$, we define $f uparrow f'$,
      the augmentation of flow $f$ by $f'$, to be a function from $V times V$ to
      $R$, defined by



      $$(f uparrow f')(u, v) = begincases f(u,v) + f'(u, v) - f'(v, u) &
      > textif (u,v) $in$ E \ 0 & textotherwise endcases$$




      How the flow network in (c), for example $(s, v_2)$ got the flow 12 ?
      If we follow the formula, it must have a flow 5:
      $8 + 5 - 8 = 5$










      share|cite|improve this question









      $endgroup$




      I am learning algorithms to solve Maximum Flow problem by reading the CRLS book and confused by the following figure:



      figure 26.4



      That is:




      A flow in a residual network provides a roadmap for adding flow to the
      original flow network. If $f$ is a flow in $G$ and $f'$ is a flow in
      the corresponding residual network $G_f$, we define $f uparrow f'$,
      the augmentation of flow $f$ by $f'$, to be a function from $V times V$ to
      $R$, defined by



      $$(f uparrow f')(u, v) = begincases f(u,v) + f'(u, v) - f'(v, u) &
      > textif (u,v) $in$ E \ 0 & textotherwise endcases$$




      How the flow network in (c), for example $(s, v_2)$ got the flow 12 ?
      If we follow the formula, it must have a flow 5:
      $8 + 5 - 8 = 5$







      algorithms network-flow






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 5 hours ago









      maksadbekmaksadbek

      1185




      1185




















          2 Answers
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          2












          $begingroup$

          That's not what the formula gives you. As the caption says, the capacity of the augmenting path in the residual network in (b) is $4$. Therefore we send 4 units of flow along the augmenting path from $s$ to $t$, namely, the path $s to v_2 to v_3 to t$. In particular, $f(s,v_2)=8$, $f'(s,v_2)=4$, and $f'(v_2,s)=0$, so the updated flow is $8+4-0=12$.






          share|cite|improve this answer









          $endgroup$




















            2












            $begingroup$

            It is explained in part (b) of the caption of Figure 26.4.




            The residual network $G_f$ with augmenting path $p$ shaded; its residual capacity is $c_f(p)=c_f(v_2,v_3)=4$.




            Since the capacity of path $p$ is 4 (not 5), we find a flow $f'$ in the residual network $G_f$ that is defined by $f'(s,v_2)=f'(v_2,v_3)=f'(v_3,t)=4$. So for the network flow $fuparrow f'$ in (c), we have
            $$ (fuparrow f')(v_2, v_3)=f(v_2,v_3)+f'(v_2,v_3) = 8+4=12.$$






            share|cite|improve this answer











            $endgroup$








            • 2




              $begingroup$
              I think you mean $=12$ not $=8$.
              $endgroup$
              – D.W.
              3 hours ago











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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

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            active

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            active

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            2












            $begingroup$

            That's not what the formula gives you. As the caption says, the capacity of the augmenting path in the residual network in (b) is $4$. Therefore we send 4 units of flow along the augmenting path from $s$ to $t$, namely, the path $s to v_2 to v_3 to t$. In particular, $f(s,v_2)=8$, $f'(s,v_2)=4$, and $f'(v_2,s)=0$, so the updated flow is $8+4-0=12$.






            share|cite|improve this answer









            $endgroup$

















              2












              $begingroup$

              That's not what the formula gives you. As the caption says, the capacity of the augmenting path in the residual network in (b) is $4$. Therefore we send 4 units of flow along the augmenting path from $s$ to $t$, namely, the path $s to v_2 to v_3 to t$. In particular, $f(s,v_2)=8$, $f'(s,v_2)=4$, and $f'(v_2,s)=0$, so the updated flow is $8+4-0=12$.






              share|cite|improve this answer









              $endgroup$















                2












                2








                2





                $begingroup$

                That's not what the formula gives you. As the caption says, the capacity of the augmenting path in the residual network in (b) is $4$. Therefore we send 4 units of flow along the augmenting path from $s$ to $t$, namely, the path $s to v_2 to v_3 to t$. In particular, $f(s,v_2)=8$, $f'(s,v_2)=4$, and $f'(v_2,s)=0$, so the updated flow is $8+4-0=12$.






                share|cite|improve this answer









                $endgroup$



                That's not what the formula gives you. As the caption says, the capacity of the augmenting path in the residual network in (b) is $4$. Therefore we send 4 units of flow along the augmenting path from $s$ to $t$, namely, the path $s to v_2 to v_3 to t$. In particular, $f(s,v_2)=8$, $f'(s,v_2)=4$, and $f'(v_2,s)=0$, so the updated flow is $8+4-0=12$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 3 hours ago









                D.W.D.W.

                103k12129294




                103k12129294





















                    2












                    $begingroup$

                    It is explained in part (b) of the caption of Figure 26.4.




                    The residual network $G_f$ with augmenting path $p$ shaded; its residual capacity is $c_f(p)=c_f(v_2,v_3)=4$.




                    Since the capacity of path $p$ is 4 (not 5), we find a flow $f'$ in the residual network $G_f$ that is defined by $f'(s,v_2)=f'(v_2,v_3)=f'(v_3,t)=4$. So for the network flow $fuparrow f'$ in (c), we have
                    $$ (fuparrow f')(v_2, v_3)=f(v_2,v_3)+f'(v_2,v_3) = 8+4=12.$$






                    share|cite|improve this answer











                    $endgroup$








                    • 2




                      $begingroup$
                      I think you mean $=12$ not $=8$.
                      $endgroup$
                      – D.W.
                      3 hours ago















                    2












                    $begingroup$

                    It is explained in part (b) of the caption of Figure 26.4.




                    The residual network $G_f$ with augmenting path $p$ shaded; its residual capacity is $c_f(p)=c_f(v_2,v_3)=4$.




                    Since the capacity of path $p$ is 4 (not 5), we find a flow $f'$ in the residual network $G_f$ that is defined by $f'(s,v_2)=f'(v_2,v_3)=f'(v_3,t)=4$. So for the network flow $fuparrow f'$ in (c), we have
                    $$ (fuparrow f')(v_2, v_3)=f(v_2,v_3)+f'(v_2,v_3) = 8+4=12.$$






                    share|cite|improve this answer











                    $endgroup$








                    • 2




                      $begingroup$
                      I think you mean $=12$ not $=8$.
                      $endgroup$
                      – D.W.
                      3 hours ago













                    2












                    2








                    2





                    $begingroup$

                    It is explained in part (b) of the caption of Figure 26.4.




                    The residual network $G_f$ with augmenting path $p$ shaded; its residual capacity is $c_f(p)=c_f(v_2,v_3)=4$.




                    Since the capacity of path $p$ is 4 (not 5), we find a flow $f'$ in the residual network $G_f$ that is defined by $f'(s,v_2)=f'(v_2,v_3)=f'(v_3,t)=4$. So for the network flow $fuparrow f'$ in (c), we have
                    $$ (fuparrow f')(v_2, v_3)=f(v_2,v_3)+f'(v_2,v_3) = 8+4=12.$$






                    share|cite|improve this answer











                    $endgroup$



                    It is explained in part (b) of the caption of Figure 26.4.




                    The residual network $G_f$ with augmenting path $p$ shaded; its residual capacity is $c_f(p)=c_f(v_2,v_3)=4$.




                    Since the capacity of path $p$ is 4 (not 5), we find a flow $f'$ in the residual network $G_f$ that is defined by $f'(s,v_2)=f'(v_2,v_3)=f'(v_3,t)=4$. So for the network flow $fuparrow f'$ in (c), we have
                    $$ (fuparrow f')(v_2, v_3)=f(v_2,v_3)+f'(v_2,v_3) = 8+4=12.$$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 3 hours ago

























                    answered 3 hours ago









                    Apass.JackApass.Jack

                    14k1940




                    14k1940







                    • 2




                      $begingroup$
                      I think you mean $=12$ not $=8$.
                      $endgroup$
                      – D.W.
                      3 hours ago












                    • 2




                      $begingroup$
                      I think you mean $=12$ not $=8$.
                      $endgroup$
                      – D.W.
                      3 hours ago







                    2




                    2




                    $begingroup$
                    I think you mean $=12$ not $=8$.
                    $endgroup$
                    – D.W.
                    3 hours ago




                    $begingroup$
                    I think you mean $=12$ not $=8$.
                    $endgroup$
                    – D.W.
                    3 hours ago

















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