Split network flow function












1












$begingroup$


Let

a graph $G = (V,E)$

a network $N = (G, s, t, c)$

and an integral flow funtion $f$



The value of f $v(f)=v_1+v_2+...+v_p$, where $v_i$ is a flow leaving the source.
I must prove that there are $p$ integral flows functions so that $v(f_i)=v_i$ and
$f = f_1+f_2+...+f_p$.

So I have to split the initial flow into $p$ flows so that the sum of the new flows is the original flow and $v(fi)=vi$.



I followed those instructions manually on some examples, but I can't find an algorithm to fit on all cases.

Any ideas on how I should do it?






graph-theory network-flow network







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Let

    a graph $G = (V,E)$

    a network $N = (G, s, t, c)$

    and an integral flow funtion $f$



    The value of f $v(f)=v_1+v_2+...+v_p$, where $v_i$ is a flow leaving the source.
    I must prove that there are $p$ integral flows functions so that $v(f_i)=v_i$ and
    $f = f_1+f_2+...+f_p$.

    So I have to split the initial flow into $p$ flows so that the sum of the new flows is the original flow and $v(fi)=vi$.



    I followed those instructions manually on some examples, but I can't find an algorithm to fit on all cases.

    Any ideas on how I should do it?






    graph-theory network-flow network







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let

      a graph $G = (V,E)$

      a network $N = (G, s, t, c)$

      and an integral flow funtion $f$



      The value of f $v(f)=v_1+v_2+...+v_p$, where $v_i$ is a flow leaving the source.
      I must prove that there are $p$ integral flows functions so that $v(f_i)=v_i$ and
      $f = f_1+f_2+...+f_p$.

      So I have to split the initial flow into $p$ flows so that the sum of the new flows is the original flow and $v(fi)=vi$.



      I followed those instructions manually on some examples, but I can't find an algorithm to fit on all cases.

      Any ideas on how I should do it?






      graph-theory network-flow network







      share|cite|improve this question









      $endgroup$




      Let

      a graph $G = (V,E)$

      a network $N = (G, s, t, c)$

      and an integral flow funtion $f$



      The value of f $v(f)=v_1+v_2+...+v_p$, where $v_i$ is a flow leaving the source.
      I must prove that there are $p$ integral flows functions so that $v(f_i)=v_i$ and
      $f = f_1+f_2+...+f_p$.

      So I have to split the initial flow into $p$ flows so that the sum of the new flows is the original flow and $v(fi)=vi$.



      I followed those instructions manually on some examples, but I can't find an algorithm to fit on all cases.

      Any ideas on how I should do it?






      graph-theory network-flow network




      graph-theory network-flow network






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 8 at 22:34









      qpBlazeqpBlaze

      61




      61






















          1 Answer
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          $begingroup$

          I am not perfectly sure that I correctly interpreted all notions, not defined in the question, but I guess that the following solution is OK.



          It suffices to prove the claim when $v_i=1$ for each $i$. This will allow us to split the flow $f$ into $v(f)$ flows with value $1$ each. Then given arbitrary natural $v_i$’s with $sum v_i=v(f)$, we partition these $v(f)$ onto $p$ groups consisting of $v_1,dots, v_p$ flows respectively, and next for each $i$ we merge flows of $i$-th group into one flow with value $v_i$.



          We prove the possibility of splitting a flow $f$ into $v(f)$ flows of value $1$ by induction with respect to $v(f)>0$. If $v(f)=1$ then we put $f_1=f$. Assume that we have already proved the claim for $v(f)=p$. Let $f$ be any feasible integral $s-t$ flow on $N$ with $v(f)=p+1$ Start from $s$ and go along the flow $f$ (that is, along edges $e$ such that $f(e)>0$), subtracting $1$ from $f$ on each passed edge. Since both the graph $G$ and the number $v(f)$ are finite, we can subtract only finitely many $1$’s, so we shall stick at some step. Since $operatorname{netinflow}(f, v)= operatorname{netoutlow} (f, v)$ for all vertices $v$ in $G$ except $s$ and $t$, and $operatorname{netinflow}(f, s)=0$, we can stick only at the vertex $t$. So we went a path from $s$ to $t$. This path endowed with the values which we subtracted, naturally generate an integral flow $f’$ from $s$ to to $t$ with $v(f’)=1$. Putting $f’’=f-f’$ we obtain a feasible integral $s-t$ flow on $N$ with $v(f’)=p$, which can be split into $p$ required flows by the induction hypothesis.






          share|cite|improve this answer









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            $begingroup$

            I am not perfectly sure that I correctly interpreted all notions, not defined in the question, but I guess that the following solution is OK.



            It suffices to prove the claim when $v_i=1$ for each $i$. This will allow us to split the flow $f$ into $v(f)$ flows with value $1$ each. Then given arbitrary natural $v_i$’s with $sum v_i=v(f)$, we partition these $v(f)$ onto $p$ groups consisting of $v_1,dots, v_p$ flows respectively, and next for each $i$ we merge flows of $i$-th group into one flow with value $v_i$.



            We prove the possibility of splitting a flow $f$ into $v(f)$ flows of value $1$ by induction with respect to $v(f)>0$. If $v(f)=1$ then we put $f_1=f$. Assume that we have already proved the claim for $v(f)=p$. Let $f$ be any feasible integral $s-t$ flow on $N$ with $v(f)=p+1$ Start from $s$ and go along the flow $f$ (that is, along edges $e$ such that $f(e)>0$), subtracting $1$ from $f$ on each passed edge. Since both the graph $G$ and the number $v(f)$ are finite, we can subtract only finitely many $1$’s, so we shall stick at some step. Since $operatorname{netinflow}(f, v)= operatorname{netoutlow} (f, v)$ for all vertices $v$ in $G$ except $s$ and $t$, and $operatorname{netinflow}(f, s)=0$, we can stick only at the vertex $t$. So we went a path from $s$ to $t$. This path endowed with the values which we subtracted, naturally generate an integral flow $f’$ from $s$ to to $t$ with $v(f’)=1$. Putting $f’’=f-f’$ we obtain a feasible integral $s-t$ flow on $N$ with $v(f’)=p$, which can be split into $p$ required flows by the induction hypothesis.






            share|cite|improve this answer









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              0












              $begingroup$

              I am not perfectly sure that I correctly interpreted all notions, not defined in the question, but I guess that the following solution is OK.



              It suffices to prove the claim when $v_i=1$ for each $i$. This will allow us to split the flow $f$ into $v(f)$ flows with value $1$ each. Then given arbitrary natural $v_i$’s with $sum v_i=v(f)$, we partition these $v(f)$ onto $p$ groups consisting of $v_1,dots, v_p$ flows respectively, and next for each $i$ we merge flows of $i$-th group into one flow with value $v_i$.



              We prove the possibility of splitting a flow $f$ into $v(f)$ flows of value $1$ by induction with respect to $v(f)>0$. If $v(f)=1$ then we put $f_1=f$. Assume that we have already proved the claim for $v(f)=p$. Let $f$ be any feasible integral $s-t$ flow on $N$ with $v(f)=p+1$ Start from $s$ and go along the flow $f$ (that is, along edges $e$ such that $f(e)>0$), subtracting $1$ from $f$ on each passed edge. Since both the graph $G$ and the number $v(f)$ are finite, we can subtract only finitely many $1$’s, so we shall stick at some step. Since $operatorname{netinflow}(f, v)= operatorname{netoutlow} (f, v)$ for all vertices $v$ in $G$ except $s$ and $t$, and $operatorname{netinflow}(f, s)=0$, we can stick only at the vertex $t$. So we went a path from $s$ to $t$. This path endowed with the values which we subtracted, naturally generate an integral flow $f’$ from $s$ to to $t$ with $v(f’)=1$. Putting $f’’=f-f’$ we obtain a feasible integral $s-t$ flow on $N$ with $v(f’)=p$, which can be split into $p$ required flows by the induction hypothesis.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                I am not perfectly sure that I correctly interpreted all notions, not defined in the question, but I guess that the following solution is OK.



                It suffices to prove the claim when $v_i=1$ for each $i$. This will allow us to split the flow $f$ into $v(f)$ flows with value $1$ each. Then given arbitrary natural $v_i$’s with $sum v_i=v(f)$, we partition these $v(f)$ onto $p$ groups consisting of $v_1,dots, v_p$ flows respectively, and next for each $i$ we merge flows of $i$-th group into one flow with value $v_i$.



                We prove the possibility of splitting a flow $f$ into $v(f)$ flows of value $1$ by induction with respect to $v(f)>0$. If $v(f)=1$ then we put $f_1=f$. Assume that we have already proved the claim for $v(f)=p$. Let $f$ be any feasible integral $s-t$ flow on $N$ with $v(f)=p+1$ Start from $s$ and go along the flow $f$ (that is, along edges $e$ such that $f(e)>0$), subtracting $1$ from $f$ on each passed edge. Since both the graph $G$ and the number $v(f)$ are finite, we can subtract only finitely many $1$’s, so we shall stick at some step. Since $operatorname{netinflow}(f, v)= operatorname{netoutlow} (f, v)$ for all vertices $v$ in $G$ except $s$ and $t$, and $operatorname{netinflow}(f, s)=0$, we can stick only at the vertex $t$. So we went a path from $s$ to $t$. This path endowed with the values which we subtracted, naturally generate an integral flow $f’$ from $s$ to to $t$ with $v(f’)=1$. Putting $f’’=f-f’$ we obtain a feasible integral $s-t$ flow on $N$ with $v(f’)=p$, which can be split into $p$ required flows by the induction hypothesis.






                share|cite|improve this answer









                $endgroup$



                I am not perfectly sure that I correctly interpreted all notions, not defined in the question, but I guess that the following solution is OK.



                It suffices to prove the claim when $v_i=1$ for each $i$. This will allow us to split the flow $f$ into $v(f)$ flows with value $1$ each. Then given arbitrary natural $v_i$’s with $sum v_i=v(f)$, we partition these $v(f)$ onto $p$ groups consisting of $v_1,dots, v_p$ flows respectively, and next for each $i$ we merge flows of $i$-th group into one flow with value $v_i$.



                We prove the possibility of splitting a flow $f$ into $v(f)$ flows of value $1$ by induction with respect to $v(f)>0$. If $v(f)=1$ then we put $f_1=f$. Assume that we have already proved the claim for $v(f)=p$. Let $f$ be any feasible integral $s-t$ flow on $N$ with $v(f)=p+1$ Start from $s$ and go along the flow $f$ (that is, along edges $e$ such that $f(e)>0$), subtracting $1$ from $f$ on each passed edge. Since both the graph $G$ and the number $v(f)$ are finite, we can subtract only finitely many $1$’s, so we shall stick at some step. Since $operatorname{netinflow}(f, v)= operatorname{netoutlow} (f, v)$ for all vertices $v$ in $G$ except $s$ and $t$, and $operatorname{netinflow}(f, s)=0$, we can stick only at the vertex $t$. So we went a path from $s$ to $t$. This path endowed with the values which we subtracted, naturally generate an integral flow $f’$ from $s$ to to $t$ with $v(f’)=1$. Putting $f’’=f-f’$ we obtain a feasible integral $s-t$ flow on $N$ with $v(f’)=p$, which can be split into $p$ required flows by the induction hypothesis.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 10 at 22:56









                Alex RavskyAlex Ravsky

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                42.7k32483






























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