Randomized Rounding for Set Cover modification












0












$begingroup$


I know that the following is the standard Randomized Rounding for Set Cover problem:



$min sum_{S in mathcal{S}} c(S)x(S)$

s.t
$sum_{S: e in S} x(S) geq 1 forall e in U$
$x(S) in [0,1] S in mathcal{S}$



The above linear program cover every element at least one time ($min sum_{S: e in S} x(S) geq 1 forall e in U$).



Set $S$ is picked with probability $p(S) = x^*(S)$



$E[ALG] = sum_{S in mathcal{S}} c(S)p(S) = sum_{S in mathcal{S}} c(S)x^*(S) = OPT^{LP} leq OPT$



Considering $a in S,
Pr[a is covered] = 1 - (1 - p(S_1)) times ... times (1 - p(S_k)) geq 1 - (1 - frac{1}{k})^k geq 1 - frac{1}{e} $

So each element $a in U$ is covered with $Prob geq 1 - frac{1}{e} $



Picking $d$ log $n$ subcollections $C^prime = C_1 cup ... cup C_{d log n}$ with $d$ such that :
$Pr[a not covered] leq (frac{1}{e})^{d log n} leq frac{1}{4n}$



We obtain:
$E[COST(C^prime)] leq d times log n OPT^{LP}$
$Pr[COST(C^prime)] geq 4d times log n OPT^{LP} leq frac{1}{4}$
$Pr[COST(C^prime) not easible] leq n times frac{1}{4n} leq frac{1}{4}$
$Pr[COST(C^prime) geq 4d times log n OPT^{LP} AND C^prime is feasible] geq frac{1}{2}$

Expected number of repetitions = 2.



Now my question is: how do thinks change if we consider the following LP?



$min sum_{S in mathcal{S}} c(S)x(S)$

s.t
$sum_{S: e in S} x(S) geq m forall e in U$
$x(S) in [0,1] S in mathcal{S}$



Considering $m = 1,2,..,n$. Of course for $m = 1$ we re-obtain the above steps but for $m geq 2$? For example for $m = 3$? Is there a generalized solution $forall m$?






algorithms







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    0












    $begingroup$


    I know that the following is the standard Randomized Rounding for Set Cover problem:



    $min sum_{S in mathcal{S}} c(S)x(S)$

    s.t
    $sum_{S: e in S} x(S) geq 1 forall e in U$
    $x(S) in [0,1] S in mathcal{S}$



    The above linear program cover every element at least one time ($min sum_{S: e in S} x(S) geq 1 forall e in U$).



    Set $S$ is picked with probability $p(S) = x^*(S)$



    $E[ALG] = sum_{S in mathcal{S}} c(S)p(S) = sum_{S in mathcal{S}} c(S)x^*(S) = OPT^{LP} leq OPT$



    Considering $a in S,
    Pr[a is covered] = 1 - (1 - p(S_1)) times ... times (1 - p(S_k)) geq 1 - (1 - frac{1}{k})^k geq 1 - frac{1}{e} $

    So each element $a in U$ is covered with $Prob geq 1 - frac{1}{e} $



    Picking $d$ log $n$ subcollections $C^prime = C_1 cup ... cup C_{d log n}$ with $d$ such that :
    $Pr[a not covered] leq (frac{1}{e})^{d log n} leq frac{1}{4n}$



    We obtain:
    $E[COST(C^prime)] leq d times log n OPT^{LP}$
    $Pr[COST(C^prime)] geq 4d times log n OPT^{LP} leq frac{1}{4}$
    $Pr[COST(C^prime) not easible] leq n times frac{1}{4n} leq frac{1}{4}$
    $Pr[COST(C^prime) geq 4d times log n OPT^{LP} AND C^prime is feasible] geq frac{1}{2}$

    Expected number of repetitions = 2.



    Now my question is: how do thinks change if we consider the following LP?



    $min sum_{S in mathcal{S}} c(S)x(S)$

    s.t
    $sum_{S: e in S} x(S) geq m forall e in U$
    $x(S) in [0,1] S in mathcal{S}$



    Considering $m = 1,2,..,n$. Of course for $m = 1$ we re-obtain the above steps but for $m geq 2$? For example for $m = 3$? Is there a generalized solution $forall m$?






    algorithms







    share|cite|improve this question









    $endgroup$



    migrated from datascience.stackexchange.com Jan 6 at 16:21


    This question came from our site for Data science professionals, Machine Learning specialists, and those interested in learning more about the field.





















      0












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      0





      $begingroup$


      I know that the following is the standard Randomized Rounding for Set Cover problem:



      $min sum_{S in mathcal{S}} c(S)x(S)$

      s.t
      $sum_{S: e in S} x(S) geq 1 forall e in U$
      $x(S) in [0,1] S in mathcal{S}$



      The above linear program cover every element at least one time ($min sum_{S: e in S} x(S) geq 1 forall e in U$).



      Set $S$ is picked with probability $p(S) = x^*(S)$



      $E[ALG] = sum_{S in mathcal{S}} c(S)p(S) = sum_{S in mathcal{S}} c(S)x^*(S) = OPT^{LP} leq OPT$



      Considering $a in S,
      Pr[a is covered] = 1 - (1 - p(S_1)) times ... times (1 - p(S_k)) geq 1 - (1 - frac{1}{k})^k geq 1 - frac{1}{e} $

      So each element $a in U$ is covered with $Prob geq 1 - frac{1}{e} $



      Picking $d$ log $n$ subcollections $C^prime = C_1 cup ... cup C_{d log n}$ with $d$ such that :
      $Pr[a not covered] leq (frac{1}{e})^{d log n} leq frac{1}{4n}$



      We obtain:
      $E[COST(C^prime)] leq d times log n OPT^{LP}$
      $Pr[COST(C^prime)] geq 4d times log n OPT^{LP} leq frac{1}{4}$
      $Pr[COST(C^prime) not easible] leq n times frac{1}{4n} leq frac{1}{4}$
      $Pr[COST(C^prime) geq 4d times log n OPT^{LP} AND C^prime is feasible] geq frac{1}{2}$

      Expected number of repetitions = 2.



      Now my question is: how do thinks change if we consider the following LP?



      $min sum_{S in mathcal{S}} c(S)x(S)$

      s.t
      $sum_{S: e in S} x(S) geq m forall e in U$
      $x(S) in [0,1] S in mathcal{S}$



      Considering $m = 1,2,..,n$. Of course for $m = 1$ we re-obtain the above steps but for $m geq 2$? For example for $m = 3$? Is there a generalized solution $forall m$?






      algorithms







      share|cite|improve this question









      $endgroup$




      I know that the following is the standard Randomized Rounding for Set Cover problem:



      $min sum_{S in mathcal{S}} c(S)x(S)$

      s.t
      $sum_{S: e in S} x(S) geq 1 forall e in U$
      $x(S) in [0,1] S in mathcal{S}$



      The above linear program cover every element at least one time ($min sum_{S: e in S} x(S) geq 1 forall e in U$).



      Set $S$ is picked with probability $p(S) = x^*(S)$



      $E[ALG] = sum_{S in mathcal{S}} c(S)p(S) = sum_{S in mathcal{S}} c(S)x^*(S) = OPT^{LP} leq OPT$



      Considering $a in S,
      Pr[a is covered] = 1 - (1 - p(S_1)) times ... times (1 - p(S_k)) geq 1 - (1 - frac{1}{k})^k geq 1 - frac{1}{e} $

      So each element $a in U$ is covered with $Prob geq 1 - frac{1}{e} $



      Picking $d$ log $n$ subcollections $C^prime = C_1 cup ... cup C_{d log n}$ with $d$ such that :
      $Pr[a not covered] leq (frac{1}{e})^{d log n} leq frac{1}{4n}$



      We obtain:
      $E[COST(C^prime)] leq d times log n OPT^{LP}$
      $Pr[COST(C^prime)] geq 4d times log n OPT^{LP} leq frac{1}{4}$
      $Pr[COST(C^prime) not easible] leq n times frac{1}{4n} leq frac{1}{4}$
      $Pr[COST(C^prime) geq 4d times log n OPT^{LP} AND C^prime is feasible] geq frac{1}{2}$

      Expected number of repetitions = 2.



      Now my question is: how do thinks change if we consider the following LP?



      $min sum_{S in mathcal{S}} c(S)x(S)$

      s.t
      $sum_{S: e in S} x(S) geq m forall e in U$
      $x(S) in [0,1] S in mathcal{S}$



      Considering $m = 1,2,..,n$. Of course for $m = 1$ we re-obtain the above steps but for $m geq 2$? For example for $m = 3$? Is there a generalized solution $forall m$?






      algorithms




      algorithms






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      asked Jan 6 at 10:10









      CuriousMindCuriousMind

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      migrated from datascience.stackexchange.com Jan 6 at 16:21


      This question came from our site for Data science professionals, Machine Learning specialists, and those interested in learning more about the field.









      migrated from datascience.stackexchange.com Jan 6 at 16:21


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          $begingroup$

          Hint: the change is in the calculation of $P[text{a is covered}]$ and its effect on the part of after that. You should consider more cases if $m > 1$ which is different choosing a subset of $S_1, ldots, S_k$ which are not covered.
          $$P[text{a is covered}] = 1 - left((1-P(S_1))times cdots times (1-P(S_k)) + binom{k}{1} (p(S_i)times text{multiplication of other $1-p(S_j)$}) + cdots + binom{k}{m} (p(S_i)times cdots times p(S_{i + m}))times text{multiplication of other $1-p(S_j)$}right) geq 1- left((1-frac{1}{k})^k + kfrac{1}{k}(1-frac{1}{k-1})^{k-1} + cdots + binom{k}{m}(frac{1}{k})^m(1-frac{1}{k-m})^{k-m}right) $$



          which $S_i, ldots, S_{i+k}$ are covered in different cases.



          As you can see, simplification of this is not that easy. You can't say it is greater than $1- frac{m}{e}$ here as the multiplier here is not simple here, in contrast with the first case.






          share|cite|improve this answer









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            $begingroup$

            Hint: the change is in the calculation of $P[text{a is covered}]$ and its effect on the part of after that. You should consider more cases if $m > 1$ which is different choosing a subset of $S_1, ldots, S_k$ which are not covered.
            $$P[text{a is covered}] = 1 - left((1-P(S_1))times cdots times (1-P(S_k)) + binom{k}{1} (p(S_i)times text{multiplication of other $1-p(S_j)$}) + cdots + binom{k}{m} (p(S_i)times cdots times p(S_{i + m}))times text{multiplication of other $1-p(S_j)$}right) geq 1- left((1-frac{1}{k})^k + kfrac{1}{k}(1-frac{1}{k-1})^{k-1} + cdots + binom{k}{m}(frac{1}{k})^m(1-frac{1}{k-m})^{k-m}right) $$



            which $S_i, ldots, S_{i+k}$ are covered in different cases.



            As you can see, simplification of this is not that easy. You can't say it is greater than $1- frac{m}{e}$ here as the multiplier here is not simple here, in contrast with the first case.






            share|cite|improve this answer









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              0












              $begingroup$

              Hint: the change is in the calculation of $P[text{a is covered}]$ and its effect on the part of after that. You should consider more cases if $m > 1$ which is different choosing a subset of $S_1, ldots, S_k$ which are not covered.
              $$P[text{a is covered}] = 1 - left((1-P(S_1))times cdots times (1-P(S_k)) + binom{k}{1} (p(S_i)times text{multiplication of other $1-p(S_j)$}) + cdots + binom{k}{m} (p(S_i)times cdots times p(S_{i + m}))times text{multiplication of other $1-p(S_j)$}right) geq 1- left((1-frac{1}{k})^k + kfrac{1}{k}(1-frac{1}{k-1})^{k-1} + cdots + binom{k}{m}(frac{1}{k})^m(1-frac{1}{k-m})^{k-m}right) $$



              which $S_i, ldots, S_{i+k}$ are covered in different cases.



              As you can see, simplification of this is not that easy. You can't say it is greater than $1- frac{m}{e}$ here as the multiplier here is not simple here, in contrast with the first case.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Hint: the change is in the calculation of $P[text{a is covered}]$ and its effect on the part of after that. You should consider more cases if $m > 1$ which is different choosing a subset of $S_1, ldots, S_k$ which are not covered.
                $$P[text{a is covered}] = 1 - left((1-P(S_1))times cdots times (1-P(S_k)) + binom{k}{1} (p(S_i)times text{multiplication of other $1-p(S_j)$}) + cdots + binom{k}{m} (p(S_i)times cdots times p(S_{i + m}))times text{multiplication of other $1-p(S_j)$}right) geq 1- left((1-frac{1}{k})^k + kfrac{1}{k}(1-frac{1}{k-1})^{k-1} + cdots + binom{k}{m}(frac{1}{k})^m(1-frac{1}{k-m})^{k-m}right) $$



                which $S_i, ldots, S_{i+k}$ are covered in different cases.



                As you can see, simplification of this is not that easy. You can't say it is greater than $1- frac{m}{e}$ here as the multiplier here is not simple here, in contrast with the first case.






                share|cite|improve this answer









                $endgroup$



                Hint: the change is in the calculation of $P[text{a is covered}]$ and its effect on the part of after that. You should consider more cases if $m > 1$ which is different choosing a subset of $S_1, ldots, S_k$ which are not covered.
                $$P[text{a is covered}] = 1 - left((1-P(S_1))times cdots times (1-P(S_k)) + binom{k}{1} (p(S_i)times text{multiplication of other $1-p(S_j)$}) + cdots + binom{k}{m} (p(S_i)times cdots times p(S_{i + m}))times text{multiplication of other $1-p(S_j)$}right) geq 1- left((1-frac{1}{k})^k + kfrac{1}{k}(1-frac{1}{k-1})^{k-1} + cdots + binom{k}{m}(frac{1}{k})^m(1-frac{1}{k-m})^{k-m}right) $$



                which $S_i, ldots, S_{i+k}$ are covered in different cases.



                As you can see, simplification of this is not that easy. You can't say it is greater than $1- frac{m}{e}$ here as the multiplier here is not simple here, in contrast with the first case.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 6 at 10:51









                OmGOmG

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