Necklace Separate is a beautiful name given to several related problems in combinatorics and measuring theory. The name and solution is due to mathematician Noga Alon and Douglas B. West.
The basic setting involves a necklace with different color beads. The necklace must be split between several partners, so each pair receives the same number of colors. In addition, the number of cuts should be as small as possible (to remove as little as possible the metal in the link between the beads).
Video Necklace splitting problem
Varian
Setiap masalah dapat diselesaikan dengan masalah berikutnya:
- Pemisahan discrit dapat dipecahkan dengan pemisahan berkelanjutan, karena kalung terpisah dapat diubah menjadi pewarna interval nyata di mana setiap interval panjang 1 diwarnai oleh warna manik yang sesuai. Jika pembelahan terus menerus mencoba memotong manik-manik, potongan dapat digeser secara bertahap sehingga hanya dibuat di antara manik-manik.
- Pembelahan berkelanjutan dapat dipecahkan dengan mengukur pemisahan, karena pewarnaan interval dalam warna dapat dikonversi that satu set ukuran, seperti itu mengukur mengukur panjang total warna . Kebalikannya juga benar: pemecahan ukuran dapat diselesaikan dengan pemecahan terus menerus, menggunakan reduksi yang lebih canggih.
Maps Necklace splitting problem
Bukti
Kasus gives Dorset Dwarf theorem Borsuk-Ulam.
Ketika adalah bilangan prima yang ganjil, buktinya melibatkan generalisasi teorema Borsuk-Ulam.
In the case of two thieves [ie. k = 2] and t colors, a fair split will require the most t cuts. However, if only t Ã-1 cuts are available, Hungarian mathematician GÃÆ'¡bor Simonyi points out that two thieves can achieve an almost fair division in the following sense.
Jika kalung itu diatur sehingga tidak ada t -split yang mungkin, maka untuk dua himpunan bagian D 1 dan D 2 dari {Ã, 1, Ã, 2, Ã,..., Ã, t Ã,}, tidak kosong , sehingga < span> , a ( t Ã, - 1) -split ada sehingga:
- Jika warna , maka partisi 1 memiliki lebih banyak manik-manik berwarna i daripada partisi 2;
- Jika warna , maka partisi 2 memiliki lebih banyak manik-manik berwarna i daripada partisi 1;
- Jika warna i tidak dalam partisi, kedua partisi memiliki banyak manik-manik berwarna yang sama. i .
This means that if the thieves have preference in the form of two "preferences" set D 1 and D 2 , do not both are empty, there is ( t Ã,-1) -split so that thief 1 gets more bead types in the preference set D 1 from thief 2; Thief 2 gets more beads of the kind in his preferences set D 2 than thief 1; and the rest are the same.
Simonyi praises GÃÆ'¡bor Tardos by noting that the above result is a direct generalization of Alon's original necklace theorem in this case k = 2. Either the necklace has ( t -Ã, 1) -split, or not. If yes, nothing needs to be proven. Otherwise we can add fictitious color beads to the necklace, and make D 1 consist of fictitious colors and D 2 is empty. So the Simonyi result shows that there is t -split with the same number of each original color.
Negative results
Untuk setiap ada -pengukuran garis nyata sehingga tidak ada selang yang dapat dibagi rata menggunakan paling banyak memotong.
Memisahkan kalung multidimensi
The result can be generalized to the n measurement probabilities defined in the dimensional cube d with the combination n ( k Ã, -Ã , 1) Hyperplanes parallel to the side for k thieves.
Approximation approximation
The approach algorithm for separating the necklace can be derived from the algorithm for a half consensus.
See also
- Combinatorial necklace
- Necklace problem
- Share exactly
References
Source of the article : Wikipedia