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Necklace splitting problem

Necklace splitting problem

Overview
In mathematics
Mathematics
Mathematics is the science and study of quantity, structure, space, and change. Mathematicians seek out patterns, formulate new conjectures, and establish truth by rigorous deduction from appropriately chosen axioms and definitions....

, and in particular combinatorics
Combinatorics
Combinatorics is a branch of pure mathematics concerning the study of discrete objects. It is related to many other areas of mathematics, such as algebra, probability theory, ergodic theory and geometry, as well as to applied subjects in computer science and statistical physics...

, the necklace splitting problem arises in a variety of contexts including exact division
Exact division
In game theory, an exact or even division is a type of fair division where all the players believe everyone received the same amount.There is no finite fair division procedure for exact division of divisible goods. However there are moving knife procedures for two players...

; its picturesque name is due to mathematicians Noga Alon
Noga Alon
Noga Alon is an Israeli mathematician noted for his contributions to combinatorics and theoretical computer science, having authored hundreds of papers.-Academic background:Alon is currently on the faculty of Tel Aviv University....

  and Douglas B. West.

Suppose a necklace
Necklace
A necklace is an article of jewellery which is worn around the neck. Necklaces are frequently formed from a metal jewellery chain, often attached to a locket or pendant...

, open at the clasp, has k ·n beads. There are k · ai beads of colour i, where 1 ≤ i ≤ t. Then the necklace splitting problem is to find a partition of the necklace into k parts (not necessarily contiguous), each of which has exactly ai beads of colour i; such a split is called a k-split.
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Encyclopedia
In mathematics
Mathematics
Mathematics is the science and study of quantity, structure, space, and change. Mathematicians seek out patterns, formulate new conjectures, and establish truth by rigorous deduction from appropriately chosen axioms and definitions....

, and in particular combinatorics
Combinatorics
Combinatorics is a branch of pure mathematics concerning the study of discrete objects. It is related to many other areas of mathematics, such as algebra, probability theory, ergodic theory and geometry, as well as to applied subjects in computer science and statistical physics...

, the necklace splitting problem arises in a variety of contexts including exact division
Exact division
In game theory, an exact or even division is a type of fair division where all the players believe everyone received the same amount.There is no finite fair division procedure for exact division of divisible goods. However there are moving knife procedures for two players...

; its picturesque name is due to mathematicians Noga Alon
Noga Alon
Noga Alon is an Israeli mathematician noted for his contributions to combinatorics and theoretical computer science, having authored hundreds of papers.-Academic background:Alon is currently on the faculty of Tel Aviv University....

  and Douglas B. West.

Suppose a necklace
Necklace
A necklace is an article of jewellery which is worn around the neck. Necklaces are frequently formed from a metal jewellery chain, often attached to a locket or pendant...

, open at the clasp, has k ·n beads. There are k · ai beads of colour i, where 1 ≤ i ≤ t. Then the necklace splitting problem is to find a partition of the necklace into k parts (not necessarily contiguous), each of which has exactly ai beads of colour i; such a split is called a k-split. The size of the split is the number of cuts that are needed to separate the parts (the opening at the clasp is not included). Inevitably, one interesting question is to find a split of minimal size.
Alon explains that
the problem of finding k-splittings of small size arises naturally when k mathematically oriented thieves steal a necklace with k · ai jewel
Gemstone
A gemstone or gem is a piece of attractive mineral, which—when cut and polished—is used to make jewelry or other adornments...

s of type i, and wish to divide it fairly between them, wasting as little as possible of the metal in the links between the jewels.


If the beads of each colour are contiguous on the open necklace, then any k splitting must contain at least k − 1 cuts, so the size is at least (k − 1)t. Alon and West use the Borsuk-Ulam theorem to prove that a k-splitting can always be achieved with this number of cuts. Alon uses these and related ideas to state and prove a generalization of the Hobby–Rice theorem
Hobby–Rice theorem
In mathematics, and in particular the necklace splitting problem, the Hobby–Rice theorem is a result that is useful in establishing the existence of certain solutions. It was proved in 1965 by C. R. Hobby and J. R. Rice; a simplified proof was given in 1976 by A...

.

One cut fewer than needed


In the case of two thieves [ie 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 , a (t − 1)-split exists such that:
  • If colour , then partition 1 has more beads of colour i than partition 2;
  • If colour , then partition 2 has more beads of colour i than partition 1;
  • If colour i is in neither partition, partition 1 has as many beads of colour i as partition 2.


(note that these three conditions are mutually exclusive because D1 and D1 have empty intersection).

The upshot of this is that however the thieves may arrange their preferences in the form of two "preference" sets D1 and D2, there exists a (t − 1)-split so that thief 1 gets more beads of types in his preference set D1 than thief 2; and thief 2 gets more beads of types in her preference set D2 than thief 1.

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.

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.