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Find k compatible objects with minimum total penalty

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Problem Detail: 

Assume we have a set of $n$ objects $X=\{x_1,x_2,\ldots,x_n\}$, where each object $x_i$ has a penalty $p_i$. Additionally, we have a set of incompatibility constraints $C=\{(x_i,x_j),\ldots\}$, where a tuple $(x_i,x_j)$ says that object $x_i$ is incompatible with object $x_j$. The problem is to find a subset $Y$ of $k<n$ compatible objects that minimize the sum of penalties, i.e. $\min_{Y} \sum_{x_i \in Y} p_i$. The objects in $Y$ need to be compatible, i.e. $\forall x_i,x_j \in Y: (x_i,x_j) \not\in C$.

Let me make an example. Assume we have 4 objects $X=\{x_1,\ldots,x_4\}$ with penalties $p_1 = 2,\ p_2 = 0.1,\ p_3=3,\ p_4=100$. Furthermore the following incompatibilities are given: $C=\{(x_1,x_2),\ (x_2,x_3),\ (x_3,x_4)\}$. The $k=2$ compatible objects that minimize the function are $Y = \{x_1,x_3\}$ with a total penalty of $5$. Object $x_2$ with the least penalty is not part of the solution, because the only compatible object is $x_4$ with a penalty $p_4=100$.

I have two questions:

  • Is this problem already known under some name or a variation of a known problem?
  • Is there an efficient (polynomial time) algorithm to solve it?
Asked By : k13n
Answered By : Shreesh

First of all you have to find if there exist independent sets of size $k$ and then select the one with the minimum penalty. We have maximum weighted independent problem (size of independent set is unconstrained), but I am not aware of any optimization problem which select exactly $k$-sized independent set and minimizes/maximizes the total weight.

The decision version of the optimization problem that you describe will be: Does there exist an independent set of size $k$ and penalty less than or equal to $P$.

We can reduce standard independent set problem to this problem by specifying the penalty of each vertex as 1 and $P = k$. Thus the decision version of the described optimization problem is NP-complete.

So there won't be polynomial time algorithm for the problem unless $P = NP$.

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Question Source : http://cs.stackexchange.com/questions/56587

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