User:TNTErick/項鍊分割

k = 2（即分為兩集合）和 t = 2（即八紅六綠兩種珠子）項鍊分割的釋例。在圖例中，其中一個集合不是相連續的，它被拆為左上和右上部分。

項鍊分割問題（Necklace Splitting Problem）是一類組合學及測度學問題。這個名字取自於它的一個較生動的類比。在複雜度上，一個將t色項鍊項鍊分成k分的分割屬於NP完全問題。該問題及其主要的解法、證明，皆由以色列數學家諾嘉亞隆及道格拉斯威斯特提出。

一般的項鍊分割問題將 t 種珠子分成若干段，而將這些段重組成 k 個部分，其中每一段中任一種珠子的顆數須相同，因此又稱為k塊 t相項鍊分割(英語：k-wise t-way necklace splitting)。再者，分出的段數愈小，即在項鍊上動的刀數愈少愈好，如此一來可以「浪費最少的金屬鍊」。

定義

在原論文中敘述及了以下幾種項鍊分割問題：

1. 離散分割：項鍊中有 $t$ 種顏色的珠子共 $k\cdot n$ 顆。而顏色 $i\in {1,2,\ldots s}$ 具有 $k\cdot a_{i}$ 顆，其中 $a_{i},n,k$ 為正整數。在 $(k-1)t$ 刀以內，將其分為 $k$ 個不一定連續的部分，使得每一個部分皆具有 $a_{i}$ 顆第 $i$ 色的珠子。考慮最差的狀況：所有顏色皆連續，則每一個顏色都須要切 $k-1$ 刀，因此該解已為最優解。
2. 連續分割:這條項鍊對應到實數區間 $[0,k\cdot n]$。區間上的每一個點都有$t$色中的一種顏色。對每一個顏色 $i\in [t]$，上了色的集合都要是勒貝格可測集而且該集合長度$k\cdot a_{i}$($a_{i}$是非負實數)。Partition the interval to $k$ parts (not necessarily contiguous), such that in each part, the total length of color $i$ is exactly $a_{i}$. Use at most $(k-1)t$ cuts.
3. Measure splitting: The necklace is a real interval. There are $t$ different measures on the interval, all absolutely continuous with respect to length. The measure of the entire necklace, according to measure $i$, is $k\cdot a_{i}$. Partition the interval to $k$ parts (not necessarily contiguous), such that the measure of each part, according to measure $i$, is exactly $a_{i}$. Use at most $(k-1)t$ cuts. This is a generalization of the Hobby–Rice theorem, and it is used to get an exact division of a cake.

前兩個情況都分別是下一個情況的特例，因此也可以以下一個問題來解決。如下：

• Discrete splitting can be solved by continuous splitting, since a discrete necklace can be converted to a coloring of the real interval $[0,k\cdot n]$ in which each interval of length 1 is colored by the color of the corresponding bead. In case the continuous splitting tries to cut inside beads, the cuts can be slid gradually such that they are made only between beads.
• Continuous splitting can be solved by measure splitting, since a coloring of an interval in $t$ colors can be converted to a set $t$ measures, such that measure $i$ measures the total length of color $i$. The opposite is also true: measure splitting can be solved by continuous splitting, using a more sophisticated reduction.

Proof

$k=2$時的情況可以用博蘇克烏蘭定理證明。當$k$是其他質數，也可以用推廣了的該定理證明。

而$k$是合成數時， the proof is as follows (demonstrated for the measure-splitting variant). Suppose $k=p\cdot q$. There are $t$ measures, each of which values the entire necklace as $p\cdot q\cdot a_{i}$. Using $(p-1)\cdot t$ cuts, divide the necklace to $p$ parts such that measure $i$ of each part is exactly $q\cdot a_{i}$. Using $(q-1)\cdot t$ cuts, divide each part to $q$ parts such that measure $i$ of each part is exactly $a_{i}$. All in all, there are now $pq$ parts such that measure $i$ of each part is exactly $a_{i}$. The total number of cuts is $(p-1)\cdot t$ plus $p(q-1)\cdot t$ which is exactly $(pq-1)\cdot t$.

Further results

One cut fewer than needed

In the case of two thieves [i.e. k = 2] and t colours, a fair split would require at most t cuts. If, however, only t − 1 cuts are available, Hungarian mathematician Gábor Simonyi shows that the two thieves can achieve an almost fair division in the following sense.

If the necklace is arranged so that no t-split is possible, then for any two subsets D₁ and D₂ of { 1, 2, ...,  t }, not both empty, such that $D_{1}\cap D_{2}=\varnothing$, a (t − 1)-split exists such that:

• If colour $i\in D_{1}$, then partition 1 has more beads of colour i than partition 2;
• If colour $i\in D_{2}$, then partition 2 has more beads of colour i than partition 1;
• If colour i is in neither partition, both partitions have equally many beads of colour i.

This means that if the thieves have preferences in the form of two "preference" sets D₁ and D₂, not both empty, there exists a (t − 1)-split so that thief 1 gets more beads of types in his preference set D₁ than thief 2; thief 2 gets more beads of types in her preference set D₂ than thief 1; and the rest are equal.

Simonyi credits Gábor Tardos with noticing that the result above is a direct generalization of Alon's original necklace theorem in the case k = 2. Either the necklace has a (t − 1)-split, or it does not. If it does, there is nothing to prove. If it does not, we may add beads of a fictitious colour to the necklace, and make D₁ consist of the fictitious colour and D₂ empty. Then Simonyi's result shows that there is a t-split with equal numbers of each real colour.

Negative result

For every $k\geq 1$ there is a measurable $(k+3)$-coloring of the real line such that no interval can be fairly split using at most $k$ cuts.

Splitting multidimensional necklaces

The result can be generalized to n probability measures defined on a d dimensional cube with any combination of n(k − 1) hyperplanes parallel to the sides for k thieves.

Approximation algorithm

An approximation algorithm for splitting a necklace can be derived from an algorithm for consensus halving.

參見

• 集合劃分
• Combinatorial necklaces
• Necklace problem

外部鏈接

• YouTube上的"Sneaky Topology", an introductory video presenting the problem with its topological solution.