Independence system/ graph-theory












1












$begingroup$


$ (E,S)$ is an independence system with $ w: E rightarrow mathbb{R_+}$.



$E= { e_1,...,e_m} $ is the set of edges with $ w(e_1) geq....geq w(e_m), w(e_{m+1}):=0$. Define the set: $ E_i:= { e_1,...,e_i}$ and $ T subset E$ with $ w'(T) = sum_{ e in T} w(e)$.
I will show: $$ w'(T)= sum_{ i=1}^m |T cap E_i|big(w(e_i)-w(e_{i+1})big)$$



Can somebody give me a hint to do that?






combinatorics graph-theory independence matroids







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$endgroup$

















    1












    $begingroup$


    $ (E,S)$ is an independence system with $ w: E rightarrow mathbb{R_+}$.



    $E= { e_1,...,e_m} $ is the set of edges with $ w(e_1) geq....geq w(e_m), w(e_{m+1}):=0$. Define the set: $ E_i:= { e_1,...,e_i}$ and $ T subset E$ with $ w'(T) = sum_{ e in T} w(e)$.
    I will show: $$ w'(T)= sum_{ i=1}^m |T cap E_i|big(w(e_i)-w(e_{i+1})big)$$



    Can somebody give me a hint to do that?






    combinatorics graph-theory independence matroids







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      $ (E,S)$ is an independence system with $ w: E rightarrow mathbb{R_+}$.



      $E= { e_1,...,e_m} $ is the set of edges with $ w(e_1) geq....geq w(e_m), w(e_{m+1}):=0$. Define the set: $ E_i:= { e_1,...,e_i}$ and $ T subset E$ with $ w'(T) = sum_{ e in T} w(e)$.
      I will show: $$ w'(T)= sum_{ i=1}^m |T cap E_i|big(w(e_i)-w(e_{i+1})big)$$



      Can somebody give me a hint to do that?






      combinatorics graph-theory independence matroids







      share|cite|improve this question











      $endgroup$




      $ (E,S)$ is an independence system with $ w: E rightarrow mathbb{R_+}$.



      $E= { e_1,...,e_m} $ is the set of edges with $ w(e_1) geq....geq w(e_m), w(e_{m+1}):=0$. Define the set: $ E_i:= { e_1,...,e_i}$ and $ T subset E$ with $ w'(T) = sum_{ e in T} w(e)$.
      I will show: $$ w'(T)= sum_{ i=1}^m |T cap E_i|big(w(e_i)-w(e_{i+1})big)$$



      Can somebody give me a hint to do that?






      combinatorics graph-theory independence matroids




      combinatorics graph-theory independence matroids






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 12 at 16:01









      greedoid

      39.9k114798




      39.9k114798










      asked Jan 12 at 9:32









      Leon1998Leon1998

      266




      266






















          1 Answer
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          $begingroup$

          It is enough to think what is $w'(T_j)$ where $T_j ={e_j}$.



          Let $$w''(T_j)=sum_{ i=1}^m |T_j cap E_i|big(w(e_i)-w(e_{i+1})big)$$



          then



          begin{eqnarray}w''(T_j) &=&underbrace{|T_j cap E_1|}_{=0}big(w(e_1)-w(e_{2})big)+...+underbrace{|T_j cap E_{j-1}|}_{=0}big(w(e_{j-1})-w(e_{j})big) +\
          &+& underbrace{|T_j cap E_{j}|}_{=1}big(w(e_{j})-w(e_{j+1})big)+...+underbrace{|T_j cap E_{m}|}_{=1}big(w(e_m)-w(e_{m+1})big) \
          &=& underbrace{0+0+...+0}_{j-1}+big(w(e_{j})-w(e_{j+1})big)+...+big(w(e_m)-w(e_{m+1})big) \
          &=&w(j)-w(e_{m+1})\
          &=&w(j)-0\
          &= & w'(T_j)
          end{eqnarray}



          So $w'$ and $w''$ are the same functions on singeltons and now should not be difficult to see that they are the same functions:



          For every $T$ and every $i$ we have: $$|Tcap E_i| = sum _{jin T}|T_jcap E_i|$$ so:



          begin{eqnarray}w''(T) &=& sum _{i=1}^mcolor{red}{Big(}sum _{jin T}|T_jcap E_i|color{red}{Big)}big(w(e_i)-w(e_{i+1})big)\
          &=& sum _{jin T}color{red}{Big(}sum _{i=1}^m|T_jcap E_i|big(w(e_i)-w(e_{i+1})big)color{red}{Big)}\
          &= &sum _{jin T}w'(T_j)\
          &= & w'(T)
          end{eqnarray}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why is it enough to consider only 1 edge $e_j$? Thank you very much for your proof:) I try understand what u have done.
            $endgroup$
            – Leon1998
            Jan 12 at 13:17













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          1 Answer
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          1












          $begingroup$

          It is enough to think what is $w'(T_j)$ where $T_j ={e_j}$.



          Let $$w''(T_j)=sum_{ i=1}^m |T_j cap E_i|big(w(e_i)-w(e_{i+1})big)$$



          then



          begin{eqnarray}w''(T_j) &=&underbrace{|T_j cap E_1|}_{=0}big(w(e_1)-w(e_{2})big)+...+underbrace{|T_j cap E_{j-1}|}_{=0}big(w(e_{j-1})-w(e_{j})big) +\
          &+& underbrace{|T_j cap E_{j}|}_{=1}big(w(e_{j})-w(e_{j+1})big)+...+underbrace{|T_j cap E_{m}|}_{=1}big(w(e_m)-w(e_{m+1})big) \
          &=& underbrace{0+0+...+0}_{j-1}+big(w(e_{j})-w(e_{j+1})big)+...+big(w(e_m)-w(e_{m+1})big) \
          &=&w(j)-w(e_{m+1})\
          &=&w(j)-0\
          &= & w'(T_j)
          end{eqnarray}



          So $w'$ and $w''$ are the same functions on singeltons and now should not be difficult to see that they are the same functions:



          For every $T$ and every $i$ we have: $$|Tcap E_i| = sum _{jin T}|T_jcap E_i|$$ so:



          begin{eqnarray}w''(T) &=& sum _{i=1}^mcolor{red}{Big(}sum _{jin T}|T_jcap E_i|color{red}{Big)}big(w(e_i)-w(e_{i+1})big)\
          &=& sum _{jin T}color{red}{Big(}sum _{i=1}^m|T_jcap E_i|big(w(e_i)-w(e_{i+1})big)color{red}{Big)}\
          &= &sum _{jin T}w'(T_j)\
          &= & w'(T)
          end{eqnarray}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why is it enough to consider only 1 edge $e_j$? Thank you very much for your proof:) I try understand what u have done.
            $endgroup$
            – Leon1998
            Jan 12 at 13:17


















          1












          $begingroup$

          It is enough to think what is $w'(T_j)$ where $T_j ={e_j}$.



          Let $$w''(T_j)=sum_{ i=1}^m |T_j cap E_i|big(w(e_i)-w(e_{i+1})big)$$



          then



          begin{eqnarray}w''(T_j) &=&underbrace{|T_j cap E_1|}_{=0}big(w(e_1)-w(e_{2})big)+...+underbrace{|T_j cap E_{j-1}|}_{=0}big(w(e_{j-1})-w(e_{j})big) +\
          &+& underbrace{|T_j cap E_{j}|}_{=1}big(w(e_{j})-w(e_{j+1})big)+...+underbrace{|T_j cap E_{m}|}_{=1}big(w(e_m)-w(e_{m+1})big) \
          &=& underbrace{0+0+...+0}_{j-1}+big(w(e_{j})-w(e_{j+1})big)+...+big(w(e_m)-w(e_{m+1})big) \
          &=&w(j)-w(e_{m+1})\
          &=&w(j)-0\
          &= & w'(T_j)
          end{eqnarray}



          So $w'$ and $w''$ are the same functions on singeltons and now should not be difficult to see that they are the same functions:



          For every $T$ and every $i$ we have: $$|Tcap E_i| = sum _{jin T}|T_jcap E_i|$$ so:



          begin{eqnarray}w''(T) &=& sum _{i=1}^mcolor{red}{Big(}sum _{jin T}|T_jcap E_i|color{red}{Big)}big(w(e_i)-w(e_{i+1})big)\
          &=& sum _{jin T}color{red}{Big(}sum _{i=1}^m|T_jcap E_i|big(w(e_i)-w(e_{i+1})big)color{red}{Big)}\
          &= &sum _{jin T}w'(T_j)\
          &= & w'(T)
          end{eqnarray}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why is it enough to consider only 1 edge $e_j$? Thank you very much for your proof:) I try understand what u have done.
            $endgroup$
            – Leon1998
            Jan 12 at 13:17
















          1












          1








          1





          $begingroup$

          It is enough to think what is $w'(T_j)$ where $T_j ={e_j}$.



          Let $$w''(T_j)=sum_{ i=1}^m |T_j cap E_i|big(w(e_i)-w(e_{i+1})big)$$



          then



          begin{eqnarray}w''(T_j) &=&underbrace{|T_j cap E_1|}_{=0}big(w(e_1)-w(e_{2})big)+...+underbrace{|T_j cap E_{j-1}|}_{=0}big(w(e_{j-1})-w(e_{j})big) +\
          &+& underbrace{|T_j cap E_{j}|}_{=1}big(w(e_{j})-w(e_{j+1})big)+...+underbrace{|T_j cap E_{m}|}_{=1}big(w(e_m)-w(e_{m+1})big) \
          &=& underbrace{0+0+...+0}_{j-1}+big(w(e_{j})-w(e_{j+1})big)+...+big(w(e_m)-w(e_{m+1})big) \
          &=&w(j)-w(e_{m+1})\
          &=&w(j)-0\
          &= & w'(T_j)
          end{eqnarray}



          So $w'$ and $w''$ are the same functions on singeltons and now should not be difficult to see that they are the same functions:



          For every $T$ and every $i$ we have: $$|Tcap E_i| = sum _{jin T}|T_jcap E_i|$$ so:



          begin{eqnarray}w''(T) &=& sum _{i=1}^mcolor{red}{Big(}sum _{jin T}|T_jcap E_i|color{red}{Big)}big(w(e_i)-w(e_{i+1})big)\
          &=& sum _{jin T}color{red}{Big(}sum _{i=1}^m|T_jcap E_i|big(w(e_i)-w(e_{i+1})big)color{red}{Big)}\
          &= &sum _{jin T}w'(T_j)\
          &= & w'(T)
          end{eqnarray}






          share|cite|improve this answer











          $endgroup$



          It is enough to think what is $w'(T_j)$ where $T_j ={e_j}$.



          Let $$w''(T_j)=sum_{ i=1}^m |T_j cap E_i|big(w(e_i)-w(e_{i+1})big)$$



          then



          begin{eqnarray}w''(T_j) &=&underbrace{|T_j cap E_1|}_{=0}big(w(e_1)-w(e_{2})big)+...+underbrace{|T_j cap E_{j-1}|}_{=0}big(w(e_{j-1})-w(e_{j})big) +\
          &+& underbrace{|T_j cap E_{j}|}_{=1}big(w(e_{j})-w(e_{j+1})big)+...+underbrace{|T_j cap E_{m}|}_{=1}big(w(e_m)-w(e_{m+1})big) \
          &=& underbrace{0+0+...+0}_{j-1}+big(w(e_{j})-w(e_{j+1})big)+...+big(w(e_m)-w(e_{m+1})big) \
          &=&w(j)-w(e_{m+1})\
          &=&w(j)-0\
          &= & w'(T_j)
          end{eqnarray}



          So $w'$ and $w''$ are the same functions on singeltons and now should not be difficult to see that they are the same functions:



          For every $T$ and every $i$ we have: $$|Tcap E_i| = sum _{jin T}|T_jcap E_i|$$ so:



          begin{eqnarray}w''(T) &=& sum _{i=1}^mcolor{red}{Big(}sum _{jin T}|T_jcap E_i|color{red}{Big)}big(w(e_i)-w(e_{i+1})big)\
          &=& sum _{jin T}color{red}{Big(}sum _{i=1}^m|T_jcap E_i|big(w(e_i)-w(e_{i+1})big)color{red}{Big)}\
          &= &sum _{jin T}w'(T_j)\
          &= & w'(T)
          end{eqnarray}







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 12 at 10:56

























          answered Jan 12 at 10:29









          greedoidgreedoid

          39.9k114798




          39.9k114798












          • $begingroup$
            Why is it enough to consider only 1 edge $e_j$? Thank you very much for your proof:) I try understand what u have done.
            $endgroup$
            – Leon1998
            Jan 12 at 13:17




















          • $begingroup$
            Why is it enough to consider only 1 edge $e_j$? Thank you very much for your proof:) I try understand what u have done.
            $endgroup$
            – Leon1998
            Jan 12 at 13:17


















          $begingroup$
          Why is it enough to consider only 1 edge $e_j$? Thank you very much for your proof:) I try understand what u have done.
          $endgroup$
          – Leon1998
          Jan 12 at 13:17






          $begingroup$
          Why is it enough to consider only 1 edge $e_j$? Thank you very much for your proof:) I try understand what u have done.
          $endgroup$
          – Leon1998
          Jan 12 at 13:17




















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