numara ataması

Verilen numaraları şekilde bir sayı atama olduğunu bir permütasyon olan bu şekilde $k$ $A_1 \leq A_2 \leq ... \leq A_k$ $\sum\limits_{i=1}^k A_i = k(2k + 1)$ $i_1, i_2, ... , i_{2k}$ $1, 2, ... , 2k$

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

Etkili bir algoritma bulamıyorum ve bu sorunu çözüyor. Kombinatoryal bir problem gibi görünüyor. Benzer bir NP-Complete sorununu bulamadım. Bu problem bilinen bir NP-Complete problemine benziyor mu yoksa polinom algoritmasıyla çözülebilir mi?

np-complete decision-problem

— Gprime
kaynak

Sorun konusunda ilerleme kaydettiniz mi?

— Yuval Filmus

Ben belirtmeyi unutmuşum

A_{1} \leq A_{2} \leq . . . \leq A_{k}

$A_1 \leq A_2 \leq ... \leq A_k$

— Gprime

İlgili problem , ayrıca tatmin edici bir cevap olmadan. (İlk bakışta nasıl ilişkili oldukları net olmayabilir, ancak

ise, problem

permütasyon bulmaya eşdeğerdir,böylece

K = 2 N

$K=2N$

1 \dots 2 N

$1 \ldots 2N$

i_{2 a - 1} - i_{2 a} = A_{i}

$i_{2a-1} - i_{2a} = A_i$

— Peter Shor

Yanıtlar:

Bu sorun kesinlikle NP-tamamlanmıştır.

Tüm garip olduğunu varsayalım . O zaman biliyoruz ki tek, ve biri çift diğeri tek. Biz varsayabiliriz garip ve bile olduğunu. İzin vererek ve , bu eşdeğer olduğunu gösterebilir iki permütasyon istemek, ve $A_j$ $i_{2j-1} + i_{2j} = A_j$ $i_{2j-1}$ $i_{2j}$ $i_{2j-1}$ $i_{2j}$ $\pi_j = \frac{1}{2}(i_{2j-1}+1)$ $\sigma_j = \frac{1}{2}(i_{2j})$ $\pi$ $\sigma$ , 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.

— Peter Shor
kaynak

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
kaynak

So how should I choose

i_{1}

$i_1$ to begin with? Im not seeing the solution. But thanks for the property

3 \leq A_{j} \leq 4 k - 1

$3 \leq A_j \leq 4k -1$

— gprime

If the

A_{i}

$A_i$ are sorted, we know

3 \leq A_{1}

$3 \leq A_1$ ,

10 \leq A_{1} + A_{2}

$10 \leq A_1 + A_2$ ,

21 \leq A_{1} + A_{2} + A_{3}

$21 \leq A_1 + A_2 + A_3$ , and so forth. Are these criteria, together with

\sum_{i} A_{i} = k (2 k + 1)

$\sum_i A_i = k(2k+1)$ , enough? If they are, there might possibly be a simple algorithm for this problem.

— Peter Shor

Yeah, they are sorted. I will try to use this...

— gprime

@PeterShor You must also consider limits from the opposite direction, i.e.

4 n - 1 \geq A_{n}, 8 n - 6 \geq A_{n - 1} + A_{n}

$4n-1 \geq A_n, 8n-6 \geq A_{n-1}+A_{n}$ , and so on and so forth. Looking at the problem anecdotally, it appears a simple greedy algorithm should discover solutions when they exist, and fail precisely when they don't - but I'm having trouble proving it.

— torquestomp

@torquestomp: You're raising a good point. In fact, the limits from one direction also imply the limits from the other, but that's not at all obvious at first sight. I looked at a similar problem, and couldn't figure out a simple algorithm (but it also looked to me like the analog of these criteria was indeed enough).

— Peter Shor

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

— Will
kaynak

The Stackexchange idea is to have a Q&A that's as complete as reasonably possible. Would you be able to expand you answer to be more than just a link to wikipedia?

— Luke Mathieson

Can you elaborate? I am failing to see how i can use that algorithms to solve my question.

— gprime

Aslında bana göre NP-complete olan 3 eşlemeli özel bir durum gibi görünüyor. Bu, OP probleminin NP-tam olduğu anlamına gelmez.

— Peter Shor

Could it be maybe a bipartite matching? I will look into the 3-matching to see if i can figure it out. Thanks!

— gprime