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I want to prove the NP-hardness of my problem P in scheduling. I was trying with Partition, 3-Partition and Subset product, But neither was successful.

Now, I can reduce a problem, say PRODUCT-2-PARTITION to P, but I am not sure whether such problem is $NP$-complete or not?

The problem PRODUCT-2-PARTITION is as follows:

Given $n$ integers. Can we partition them into $n/2$ subsets, where each subset contains exactly $2$ integers, and the product of the elements of any subset is equal to a given value $B$?

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  • $\begingroup$ This can be solved by blossom algorithm for general matching. $\endgroup$ – Thinh D. Nguyen Oct 18 '18 at 8:07
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Denote these integers by $x_1\ge x_2\ge\cdots\ge x_n$. We assume these integers are all positive (it is not hard to revise the following algorithm for instances including both positive and negative integers).

In a valid solution, the greatest element must match the smallest element. Otherwise, say $x_1$ matches $x_i$ where $x_i>x_n$, and $x_n$ matches $x_j$, then $B=x_1x_i>x_1x_n\ge x_jx_n=B$, a contradiction. Hence a valid solution can only be $\{x_1,x_n\},\{x_2,x_{n-1}\},\ldots$ We only need to check whether this partition is valid, i.e. whether $x_1x_n=x_2x_{n-1}=\cdots=B$.

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  • $\begingroup$ Yes, due to number-theoretic nature, this is much simpler to blossom algorithm. $\endgroup$ – Thinh D. Nguyen Oct 18 '18 at 8:41
  • $\begingroup$ Great. That is right. What if the partitions have three elements? Similar to 3-Partition problem? $\endgroup$ – Mostafa Oct 18 '18 at 9:07
  • $\begingroup$ @Mostafa Yes, you can reduce it from 3-Partition by transforming each integer $x$ to $2^x$. $\endgroup$ – xskxzr Oct 18 '18 at 9:10

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