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












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



$$(f uparrow f')(u, v) = begin{cases} f(u,v) + f'(u, v) - f'(v, u) &
> text{if (u,v) $in$ E} \ 0 & text{otherwise} end{cases}$$




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









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    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



    $$(f uparrow f')(u, v) = begin{cases} f(u,v) + f'(u, v) - f'(v, u) &
    > text{if (u,v) $in$ E} \ 0 & text{otherwise} end{cases}$$




    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









    $endgroup$















      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



      $$(f uparrow f')(u, v) = begin{cases} f(u,v) + f'(u, v) - f'(v, u) &
      > text{if (u,v) $in$ E} \ 0 & text{otherwise} end{cases}$$




      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









      $endgroup$




      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



      $$(f uparrow f')(u, v) = begin{cases} f(u,v) + f'(u, v) - f'(v, u) &
      > text{if (u,v) $in$ E} \ 0 & text{otherwise} end{cases}$$




      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




      algorithms network-flow






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      maksadbekmaksadbek

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          $begingroup$

          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









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            $begingroup$

            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











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              $begingroup$

              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









              $endgroup$


















                4












                $begingroup$

                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









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  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 yesterday









                  D.W.♦D.W.

                  103k12129294




                  103k12129294























                      3












                      $begingroup$

                      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











                      $endgroup$


















                        3












                        $begingroup$

                        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











                        $endgroup$
















                          3












                          3








                          3





                          $begingroup$

                          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














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                          edited yesterday

























                          answered yesterday









                          Apass.JackApass.Jack

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