Proof of NP-completeness for the following
I have encountered a problem similar to Set Cover (and Maximum Coverage):
We have several sets in a universe with $N$ elements. What is the maximum number of sets so that the number of elements found in these sets is not greater than $frac{N}{2}$.
Can this problem be reduced to Set Cover or Maximum Coverage?
computational-complexity np-complete
asked Nov 28 '12 at 22:38
manix
83
add a comment |
I have encountered a problem similar to Set Cover (and Maximum Coverage):
We have several sets in a universe with $N$ elements. What is the maximum number of sets so that the number of elements found in these sets is not greater than $frac{N}{2}$.
Can this problem be reduced to Set Cover or Maximum Coverage?
computational-complexity np-complete
asked Nov 28 '12 at 22:38
manix
83
add a comment |
I have encountered a problem similar to Set Cover (and Maximum Coverage):
We have several sets in a universe with $N$ elements. What is the maximum number of sets so that the number of elements found in these sets is not greater than $frac{N}{2}$.
Can this problem be reduced to Set Cover or Maximum Coverage?
computational-complexity np-complete
asked Nov 28 '12 at 22:38
manix
83
I have encountered a problem similar to Set Cover (and Maximum Coverage):
We have several sets in a universe with $N$ elements. What is the maximum number of sets so that the number of elements found in these sets is not greater than $frac{N}{2}$.
Can this problem be reduced to Set Cover or Maximum Coverage?
computational-complexity np-complete
computational-complexity np-complete
asked Nov 28 '12 at 22:38
manix
83
asked Nov 28 '12 at 22:38
manix
83
edited yesterday
asked Nov 28 '12 at 22:38
manix
83
asked Nov 28 '12 at 22:38
manix
83
asked Nov 28 '12 at 22:38
manix
83
83
add a comment |
add a comment |
1 Answer
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Yes. The above problem can be reduced to a generalization of the Maximum Coverage Problem, called Budgeted Maximum Coverage.
Using the terminology of wiki page, select the weight of each object $e_j =1$ and cost $C(S_i) = |S_i|$, the size of $S_i$. Select the budget to be $frac{N}{2}$. Then your problem is exactly equal to solving the given problem.
See the wiki page and reference within. Hope it helps.
answered Nov 28 '12 at 22:53
dexter04
1,6701027
This seems right. I am not sure about C(Si)=|Si|. The same elements can be on different sets, so the cost varies.
– manix
Nov 28 '12 at 23:25
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
Yes. The above problem can be reduced to a generalization of the Maximum Coverage Problem, called Budgeted Maximum Coverage.
Using the terminology of wiki page, select the weight of each object $e_j =1$ and cost $C(S_i) = |S_i|$, the size of $S_i$. Select the budget to be $frac{N}{2}$. Then your problem is exactly equal to solving the given problem.
See the wiki page and reference within. Hope it helps.
answered Nov 28 '12 at 22:53
dexter04
1,6701027
This seems right. I am not sure about C(Si)=|Si|. The same elements can be on different sets, so the cost varies.
– manix
Nov 28 '12 at 23:25
add a comment |
Yes. The above problem can be reduced to a generalization of the Maximum Coverage Problem, called Budgeted Maximum Coverage.
Using the terminology of wiki page, select the weight of each object $e_j =1$ and cost $C(S_i) = |S_i|$, the size of $S_i$. Select the budget to be $frac{N}{2}$. Then your problem is exactly equal to solving the given problem.
See the wiki page and reference within. Hope it helps.
answered Nov 28 '12 at 22:53
dexter04
1,6701027
This seems right. I am not sure about C(Si)=|Si|. The same elements can be on different sets, so the cost varies.
– manix
Nov 28 '12 at 23:25
add a comment |
Yes. The above problem can be reduced to a generalization of the Maximum Coverage Problem, called Budgeted Maximum Coverage.
Using the terminology of wiki page, select the weight of each object $e_j =1$ and cost $C(S_i) = |S_i|$, the size of $S_i$. Select the budget to be $frac{N}{2}$. Then your problem is exactly equal to solving the given problem.
See the wiki page and reference within. Hope it helps.
answered Nov 28 '12 at 22:53
dexter04
1,6701027
Yes. The above problem can be reduced to a generalization of the Maximum Coverage Problem, called Budgeted Maximum Coverage.
Using the terminology of wiki page, select the weight of each object $e_j =1$ and cost $C(S_i) = |S_i|$, the size of $S_i$. Select the budget to be $frac{N}{2}$. Then your problem is exactly equal to solving the given problem.
See the wiki page and reference within. Hope it helps.
answered Nov 28 '12 at 22:53
dexter04
1,6701027
answered Nov 28 '12 at 22:53
dexter04
1,6701027
answered Nov 28 '12 at 22:53
dexter04
1,6701027
answered Nov 28 '12 at 22:53
dexter04
1,6701027
1,6701027
This seems right. I am not sure about C(Si)=|Si|. The same elements can be on different sets, so the cost varies.
– manix
Nov 28 '12 at 23:25
add a comment |
This seems right. I am not sure about C(Si)=|Si|. The same elements can be on different sets, so the cost varies.
– manix
Nov 28 '12 at 23:25
This seems right. I am not sure about C(Si)=|Si|. The same elements can be on different sets, so the cost varies.
– manix
Nov 28 '12 at 23:25
This seems right. I am not sure about C(Si)=|Si|. The same elements can be on different sets, so the cost varies.
– manix
Nov 28 '12 at 23:25
add a comment |
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