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Given $k$ numbers $A_1 \leq A_2 \leq ... \leq A_k$ such that $\sum\limits_{i=1}^k A_i = k(2k + 1)$ is there an assignment of numbers $i_1, i_2, ... , i_{2k}$ which is a permutation of $1, 2, ... , 2k$ such that

$i_1 + i_2 \leq A_1\\ i_3 + i_4 \leq A_2\\ .\\.\\.\\ i_{2k-1} + i_{2k} \leq A_k$

?

I cannot find an efficient algorithm and that solves this problem. It seems to be a combinatorial problem. I was unable to find a similar NP-Complete problem. Does this problem look like a known NP-Complete problem or can it be solved with a polynomial algorithm?

gprime
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3 Answers3

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This problem is strongly NP-complete.

Suppose all the $A_j$ are odd. Then we know that since $i_{2j-1} + i_{2j} = A_j$ is odd, one of $i_{2j-1}$ and $i_{2j}$ is even and the other is odd. We can assume that $i_{2j-1}$ is odd and $i_{2j}$ is even. By letting $\pi_j = \frac{1}{2}(i_{2j-1}+1)$ and $\sigma_j = \frac{1}{2}(i_{2j})$, we can show that this is equivalent to asking for two permutations, $\pi$ and $\sigma$, of the numbers $1 \ldots n$ such that $\pi_j + \sigma_j = \frac{1}{2}(A_j+1)$.

This problem is known to be NP-complete; see this cstheory.se problem and this paper of W. Yu, H. Hoogeveen, and J. K. Lenstra referenced in the answer.

Community
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Peter Shor
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Here is a hint to get you started: since the sum of all numbers from $1$ to $2k$ is exactly $k(2k+1)$, a solution is possible only if in fact $i_1 + i_2 = A_1$, $i_3 + i_4 = A_2$ and so on. So given $i_1$ we know $i_2$, and so on. Also, $3 \leq A_j \leq 4k-1$.

Yuval Filmus
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It's a matching problem, and so can be solved using Edmond's algorithm. See wikipedia

Will
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