{"id":19573,"date":"2015-08-01T16:12:28","date_gmt":"2015-08-01T13:12:28","guid":{"rendered":"https:\/\/bilimvegelecek.com.tr\/?p=19573"},"modified":"2018-02-16T16:17:40","modified_gmt":"2018-02-16T13:17:40","slug":"sen-kes-ben-seceyim","status":"publish","type":"post","link":"https:\/\/bilimvegelecek.com.tr\/index.php\/2015\/08\/01\/sen-kes-ben-seceyim","title":{"rendered":"Sen kes, ben se\u00e7eyim!"},"content":{"rendered":"

Bir pastay\u0131 (ya da bir somun ekme\u011fi) arkada\u015f\u0131n\u0131zla payla\u015facaks\u0131n\u0131z ve elinizde sadece bir b\u0131\u00e7ak var. Her iki taraf\u0131n da g\u00f6n\u00fcl rahatl\u0131\u011f\u0131yla kabul edece\u011fi bir payla\u015f\u0131m nas\u0131l yap\u0131labilir? Pop\u00fcler matematik merakl\u0131lar\u0131nca iyi bilinen bu sorunun yan\u0131t\u0131 \u00e7ok basit: Arkada\u015f\u0131n\u0131za \u201cSen kes, ben se\u00e7eyim\u201d \u00f6nerisinde bulunursunuz, olur biter. B\u00f6ylece art\u0131k onun amac\u0131 (kendine g\u00f6re) pastay\u0131 e\u015fit iki par\u00e7aya b\u00f6lmektir, \u00e7\u00fcnk\u00fc se\u00e7imi siz yapacaks\u0131n\u0131z. E\u011fer siz par\u00e7alar\u0131n e\u015fit olmad\u0131\u011f\u0131n\u0131 d\u00fc\u015f\u00fcn\u00fcyorsan\u0131z b\u00fcy\u00fck olan\u0131 se\u00e7ersiniz. Bu durumda arkada\u015f\u0131n\u0131z sonu\u00e7tan memnun de\u011filse, pastay\u0131 daha dikkatli kesmesi gerekti\u011fini ve \u015fik\u00e2yet etme hakk\u0131n\u0131n olmad\u0131\u011f\u0131n\u0131 d\u00fc\u015f\u00fcnmelidir. E\u011fer bu payla\u015f\u0131mdan siz memnun de\u011filseniz, o zaman yanl\u0131\u015f dilimi se\u00e7mi\u015fsiniz demektir. Sonu\u00e7ta iki taraf da adil bir payla\u015f\u0131m yap\u0131lmad\u0131\u011f\u0131n\u0131 iddia edemez. E\u011fer ba\u015flang\u0131\u00e7ta arkada\u015f\u0131n\u0131z pastay\u0131 kesmeyi de\u011fil, se\u00e7meyi tercih ederse bu kez sadece roller de\u011fi\u015fir, yine ayn\u0131 y\u00f6ntemle adil bir payla\u015f\u0131m yap\u0131labilir.<\/p>\n

K\u0131saca \u201ckes ve se\u00e7\u201d olarak adland\u0131raca\u011f\u0131m\u0131z bu y\u00f6ntemin \u015fu \u00f6zellikleri sa\u011flad\u0131\u011f\u0131n\u0131 g\u00f6r\u00fcyoruz:<\/p>\n

1) K\u0131skan\u00e7l\u0131k yok<\/strong>. \u0130ki taraf da paylar\u0131na d\u00fc\u015fen par\u00e7alar\u0131n en b\u00fcy\u00fck par\u00e7adan daha k\u00fc\u00e7\u00fck olmad\u0131\u011f\u0131n\u0131 d\u00fc\u015f\u00fcn\u00fcr ve kar\u015f\u0131s\u0131ndakini k\u0131skanmaz.<\/p>\n

2) Verimli. <\/strong>Taraflar\u0131n birbirlerine zarar vermedi\u011fi bu payla\u015f\u0131mdan daha iyi bir da\u011f\u0131l\u0131m yap\u0131lamaz.<\/p>\n

Bu y\u00f6ntemin sa\u011flamad\u0131\u011f\u0131 bir \u00f6zellik e\u015fitliktir<\/strong>. \u0130ki taraf da kendi paylar\u0131n\u0131 s\u00fcbjektif \u00f6l\u00e7\u00fclerle belirlediklerinden mutlak bir e\u015fitlikten s\u00f6z edilemez.<\/p>\n

\u00c7ikolatal\u0131 ve vanilyal\u0131 pastan\u0131n payla\u015f\u0131m\u0131 <\/strong><\/p>\n

\u015eimdi, problemi biraz zenginle\u015ftirerek pastan\u0131n sol yar\u0131s\u0131n\u0131n vanilyal\u0131, sa\u011f yar\u0131s\u0131n\u0131n da \u00e7ikolatal\u0131 oldu\u011funu varsayal\u0131m. A <\/em>ve B <\/em>gibi iki ki\u015fiden A <\/em>kessin, B <\/em>de se\u00e7sin, ama ayr\u0131ca A <\/em>i\u00e7in vanilyal\u0131 pasta kakaoluya g\u00f6re iki kat de\u011ferde olsun. \u201cKes ve se\u00e7\u201d payla\u015f\u0131m\u0131 nas\u0131l yap\u0131labilir? (Kaynak 1)<\/p>\n

Bu durumda A<\/em>, tabii ki pay\u0131n\u0131 pastan\u0131n sol yar\u0131s\u0131ndan almak isteyecektir, ama se\u00e7imi B <\/em>yapaca\u011f\u0131ndan \u201cadil\u201d bir kesim yapmay\u0131 ama\u00e7lamal\u0131d\u0131r. A<\/em>, bu kesimi vanilyan\u0131n \u00e7ikolatan\u0131n 2 kat\u0131 de\u011fere sahip oldu\u011funu d\u00fc\u015f\u00fcnerek yapaca\u011f\u0131ndan pastay\u0131 tam ortadan kesmeyecektir. Onun i\u00e7in \u201cadil\u201d olan, yani onun \u00f6l\u00e7\u00fclerine g\u00f6re yar\u0131 yar\u0131ya olacak payla\u015f\u0131m pastan\u0131n vanilyal\u0131 taraf\u0131nda ve yar\u0131m porsiyondan k\u00fc\u00e7\u00fck olmal\u0131d\u0131r.<\/p>\n

Pastan\u0131n, soldan sa\u011fa do\u011fru say\u0131 do\u011frusu \u00fczerinde [0, 1] aral\u0131\u011f\u0131ndaki say\u0131larla temsil edildi\u011fini d\u00fc\u015f\u00fcnelim. [0, 1\/2] aral\u0131\u011f\u0131 vanilyal\u0131, [1\/2, 1] aral\u0131\u011f\u0131 da \u00e7ikolatal\u0131 b\u00f6l\u00fcm\u00fc g\u00f6stersin. Bu aral\u0131kta alaca\u011f\u0131m\u0131z herhangi bir reel say\u0131 A<\/em>\u2019n\u0131n pastay\u0131 hangi oranda kesece\u011fini ifade eder, ama unutmayal\u0131m A <\/em>i\u00e7in vanilyal\u0131 b\u00f6l\u00fcm \u00e7ikolatal\u0131 b\u00f6l\u00fcm\u00fcn iki kat\u0131 de\u011ferinde. Bu durumda A<\/em>\u2019n\u0131n hangi oranda pastay\u0131 kesece\u011fini g\u00f6steren say\u0131 x <\/em>ise x <\/em>\u2264 1\/2 olacakt\u0131r.<\/p>\n

\"\"<\/p>\n

Buna g\u00f6re a\u015fa\u011f\u0131daki denklemi yazabiliriz:<\/p>\n

2x=2(1\/2-x)+1-1\/2<\/p>\n

A<\/em>\u2019ya g\u00f6re pastan\u0131n ilk yar\u0131s\u0131ndan ald\u0131\u011f\u0131 3\/8\u2019lik dilim her iki taraf i\u00e7in adil ve e\u015fit bir payla\u015f\u0131m\u0131 ifade ediyor, \u00e7\u00fcnk\u00fc A <\/em>vanilyaya \u00e7ikolataya g\u00f6re 2 kat de\u011fer bi\u00e7iyor.<\/p>\n

\"\"<\/p>\n

Tabloda da g\u00f6r\u00fcld\u00fc\u011f\u00fc gibi A<\/em>, griye boyal\u0131 olan 6p<\/em>\u2019lik b\u00f6l\u00fcm\u00fc almay\u0131 ama\u00e7l\u0131yor ve B<\/em>\u2019ye de yine 6p<\/em>\u2019lik b\u00f6l\u00fcm kal\u0131yor.<\/p>\n

Bir kez daha hat\u0131rlamakta fayda var: Bu payla\u015f\u0131mda iki porsiyonun e\u015fitli\u011fi A<\/em>\u2019n\u0131n pastaya bi\u00e7ti\u011fi de\u011fere g\u00f6re belirleniyor ve A<\/em>\u2019n\u0131n bu de\u011ferlendirmesinden B<\/em>\u2019nin haberi yok. E\u011fer B<\/em>, vanilya ve \u00e7ikolataya e\u015fit de\u011fer veriyorsa, bu durumda 5\/8\u2019lik porsiyonu tercih ederek b\u00fcy\u00fck par\u00e7ay\u0131 alm\u0131\u015f olacak. Bu payla\u015f\u0131m B<\/em>\u2019yi mutlu edecek \u00e7\u00fcnk\u00fc kendini A<\/em>\u2019ya g\u00f6re daha iyi durumda g\u00f6recek. \u00d6te yandan A <\/em>da amac\u0131na ula\u015fm\u0131\u015f olacak ve kendi a\u00e7\u0131s\u0131ndan hakkaniyetli bir da\u011f\u0131l\u0131m\u0131n yap\u0131ld\u0131\u011f\u0131n\u0131 d\u00fc\u015f\u00fcnecektir. Dolay\u0131s\u0131yla bu payla\u015f\u0131m \u201ckes ve se\u00e7\u201d y\u00f6nteminin t\u00fcm \u00f6zelliklerini sa\u011fl\u0131yor.<\/p>\n

E\u011fer kesme i\u015flemini B <\/em>yapar, A <\/em>se\u00e7erse yine \u201ckes ve se\u00e7\u201d y\u00f6ntemine uygun bir da\u011f\u0131l\u0131m yap\u0131lacakt\u0131r. \u015e\u00f6yle ki: Bu kez B <\/em>pastay\u0131 tam ortadan keserek yar\u0131s\u0131 vanilyal\u0131, di\u011fer yar\u0131s\u0131 da \u00e7ikolatal\u0131 olacak \u015fekilde iki e\u015fit porsiyona ay\u0131rabilir \u00e7\u00fcnk\u00fc onun i\u00e7in her iki b\u00f6l\u00fcm de e\u015fit de\u011ferde. Bu durumda A <\/em>tabii ki vanilyal\u0131 b\u00f6l\u00fcm\u00fc se\u00e7ecek ve b\u00f6ylece her iki taraf da \u201ckes ve se\u00e7\u201d y\u00f6ntemine g\u00f6re istedi\u011fi sonuca ula\u015fm\u0131\u015f olacakt\u0131r.<\/p>\n

Bu payla\u015f\u0131m\u0131n \u201ckes ve se\u00e7\u201d y\u00f6ntemine uygun olarak yap\u0131labilmesi i\u00e7in A<\/em>\u2019n\u0131n pastay\u0131 soldan (vanilyal\u0131 b\u00f6l\u00fcmden) sa\u011fa do\u011fru kesme i\u015flemi i\u00e7in belirledi\u011fimiz x <\/em>say\u0131s\u0131 [3\/8,1\/2] aral\u0131\u011f\u0131nda bulunmal\u0131, yani 3\/8 \u2264 x <\/em>\u2264 1\/2 olmal\u0131. Bu aral\u0131ktan se\u00e7ilecek her x <\/em>de\u011feri i\u00e7in her iki taraf\u0131 da memnun eden payla\u015f\u0131mlar yap\u0131labilir.<\/p>\n

Bu problemi genelle\u015ftirmek m\u00fcmk\u00fcn, yani vanilya ile \u00e7ikolata aras\u0131ndaki de\u011ferlerin ki\u015filere g\u00f6re de\u011fi\u015fmesi durumunda \u201ckes ve se\u00e7\u201d y\u00f6nteminin nas\u0131l \u00e7al\u0131\u015faca\u011f\u0131 ele al\u0131nabilir. Ama biz b\u00f6ylesi bir incelemeyi sonraki yaz\u0131lara b\u0131rakarak, belki de bir\u00e7ok okurun merak etmi\u015f olabilece\u011fi \u201ckes ve se\u00e7\u201d y\u00f6nteminin 3 ki\u015fi aras\u0131ndaki uygulamas\u0131n\u0131 ele almaya \u00e7al\u0131\u015fal\u0131m.<\/p>\n

\u00dc\u00e7 ki\u015fi i\u00e7in \u201ckes ve se\u00e7\u201d y\u00f6ntemi <\/strong><\/p>\n

Amerika\u2019l\u0131 ve \u0130ngiliz matematik\u00e7iler John Selfridge ve Horton Conway\u2019\u0131n 1960\u2019lar\u0131n ba\u015f\u0131nda A<\/em>, B<\/em>, C <\/em>gibi \u00fc\u00e7 ki\u015fi i\u00e7in form\u00fcle ettikleri \u201ckes ve se\u00e7\u201d y\u00f6nteminde \u015fu ad\u0131mlar uygulan\u0131yor:<\/p>\n

1) A<\/em>, pastay\u0131 \u201ckes ve se\u00e7\u201d y\u00f6ntemine g\u00f6re (adil olarak) \u00fc\u00e7 par\u00e7aya b\u00f6ler.<\/p>\n

2) B<\/em>, e\u011fer en b\u00fcy\u00fck iki par\u00e7an\u0131n birbirine e\u015fit oldu\u011funu d\u00fc\u015f\u00fcn\u00fcrse ya s\u0131ras\u0131n\u0131 atlar, ya da par\u00e7alar\u0131 e\u015fitlemek amac\u0131yla b\u00fcy\u00fck olandan par\u00e7a ya da par\u00e7alar keser. Ad\u0131na \u201cart\u0131k\u201d diyece\u011fimiz bu par\u00e7alar bir yerde toplan\u0131r.<\/p>\n

3) C<\/em>, B <\/em>ve A<\/em>, kendilerince en b\u00fcy\u00fck bulduklar\u0131 par\u00e7ay\u0131 belirtilen s\u0131raya g\u00f6re se\u00e7erler. E\u011fer B <\/em>ikinci ad\u0131mda s\u0131ras\u0131n\u0131 atlamad\u0131ysa dilimleri e\u015fitlemek i\u00e7in k\u00fc\u00e7\u00fck par\u00e7alar kesti\u011fi dilimi se\u00e7mek zorunda, tabii daha \u00f6nce C<\/em>, o dilimi se\u00e7mediyse.<\/p>\n

Bu durumda pasta, B<\/em>\u2019nin kesti\u011fi k\u00fc\u00e7\u00fck par\u00e7alar hari\u00e7 \u201ckes ve se\u00e7\u201d y\u00f6nteminin \u00f6zelliklerine uygun olarak payla\u015f\u0131lm\u0131\u015f olur.<\/p>\n

4) E\u011fer B<\/em>, ikinci ad\u0131mda s\u0131ras\u0131n\u0131 atlad\u0131ysa art\u0131k par\u00e7a <\/em>bulunmayaca\u011f\u0131ndan payla\u015f\u0131m i\u015flemi tamamlanm\u0131\u015f demektir. E\u011fer atlamad\u0131ysa C <\/em>veya B art\u0131k par\u00e7alar\u0131 <\/em>birle\u015ftir. Bu oyunculardan biri art\u0131k par\u00e7a <\/em>y\u0131\u011f\u0131n\u0131n\u0131 kesecektir. Bu oyuncuyu \u201ckesen\u201d , di\u011ferini \u201ckesmeyen\u201d olarak adland\u0131ral\u0131m. \u201cKesen\u201d art\u0131k par\u00e7a <\/em>y\u0131\u011f\u0131n\u0131n\u0131 e\u015fit oldu\u011funu d\u00fc\u015f\u00fcnd\u00fc\u011f\u00fc \u00fc\u00e7 par\u00e7aya ay\u0131r\u0131r.<\/p>\n

Bu a\u015famada A<\/em>\u2019n\u0131n durumuna de\u011finelim: A<\/em>, \u201ckesmeyen\u201d oyuncuya g\u00f6re ba\u015f\u0131ndan avantajl\u0131, \u00e7\u00fcnk\u00fc A<\/em>, ba\u015flang\u0131\u00e7ta b\u00fct\u00fcn par\u00e7alar\u0131n e\u015fit oldu\u011funu d\u00fc\u015f\u00fcnd\u00fc\u011f\u00fc bir payla\u015f\u0131m yapm\u0131\u015ft\u0131 ve \u201ckesmeyen\u201d oyuncu k\u00fc\u00e7\u00fck par\u00e7alar kesildikten sonra kalan par\u00e7ay\u0131 alm\u0131\u015ft\u0131. Bu y\u00fczden A<\/em>, \u201ckesmeyen\u201d oyuncunun adil olmayan bir pay almad\u0131\u011f\u0131n\u0131 d\u00fc\u015f\u00fcnecek ve \u201cart\u0131klar\u0131n\u201d payla\u015f\u0131l\u0131yor olmas\u0131ndan dolay\u0131 \u201ckesmeyeni\u201d k\u0131skanmayacakt\u0131r.<\/p>\n

5) \u00dc\u00e7e b\u00f6l\u00fcnm\u00fc\u015f art\u0131k par\u00e7alar <\/em>oyuncular aras\u0131nda \u015fu s\u0131rayla payla\u015f\u0131l\u0131r: \u201cKesmeyen\u201d, A <\/em>ve \u201ckesen\u201d. Elbette, her oyuncu kendilerine g\u00f6re en b\u00fcy\u00fck par\u00e7ay\u0131 se\u00e7ecektir. Bu payla\u015f\u0131mda \u201ckesmeyen\u201d ilk se\u00e7imi yapt\u0131\u011f\u0131 i\u00e7in k\u0131skanaca\u011f\u0131 bir durum s\u00f6z konusu de\u011fildir. A<\/em>, \u201ckesmeyeni\u201d k\u0131skanmaz \u00e7\u00fcnk\u00fc daha ba\u015flang\u0131\u00e7tan avantajl\u0131, \u201ckesen\u201d oyuncudan \u00f6nce se\u00e7im yapt\u0131\u011f\u0131ndan onu da k\u0131skanmaz. \u201cKesen\u201d oyuncu par\u00e7alar\u0131 kendi kesti\u011fi i\u00e7in hi\u00e7 kimseyi k\u0131skanmaz.<\/p>\n

Selfridge ve Conway\u2019in \u00fc\u00e7 ki\u015fi i\u00e7in geli\u015ftirdikleri \u201ckes ve se\u00e7\u201d y\u00f6ntemi daha sonra siyaset bilimci Steven J. Brams ve Alan D. Taylor taraf\u0131ndan d\u00f6rt ki\u015fi i\u00e7in de form\u00fcle edildi. Sonras\u0131nda be\u015f ve daha fazla oyuncu i\u00e7in de kesim say\u0131s\u0131 s\u0131n\u0131rl\u0131 olmayan algoritmalar olu\u015fturuldu. Gelecek ayki yaz\u0131da bu ilgin\u00e7 konuyu incelemeyi s\u00fcrd\u00fcrece\u011fiz.<\/p>\n

Kaynaklar <\/strong><\/p>\n

1) S. J. Brams, Micheal A. Jones, Christian Klamler, Better Ways to Cut a Cake, Notices of the AMS<\/em>, Volume 53, Number 11, 2006.<\/p>\n

2) Ian Stewart, Matematik Histerisi<\/em>, \u00c7ev. A. N. Narman, Aylak Kitap, 2013.<\/p>\n","protected":false},"excerpt":{"rendered":"

Bir pastay\u0131 (ya da bir somun ekme\u011fi) arkada\u015f\u0131n\u0131zla payla\u015facaks\u0131n\u0131z ve elinizde sadece bir b\u0131\u00e7ak var. Her iki taraf\u0131n da g\u00f6n\u00fcl rahatl\u0131\u011f\u0131yla kabul edece\u011fi bir payla\u015f\u0131m nas\u0131l yap\u0131labilir? Pop\u00fcler matematik merakl\u0131lar\u0131nca iyi bilinen bu sorunun yan\u0131t\u0131 \u00e7ok basit: Arkada\u015f\u0131n\u0131za \u201cSen kes, ben se\u00e7eyim\u201d \u00f6nerisinde bulunursunuz, olur biter. B\u00f6ylece art\u0131k onun amac\u0131 (kendine g\u00f6re) pastay\u0131 e\u015fit iki […]<\/p>\n","protected":false},"author":375,"featured_media":19574,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[175,25,514],"tags":[2363,208,1530],"class_list":["post-19573","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-138-sayi","category-matematik","category-matematik-sohbetleri","tag-bolme","tag-matematik","tag-sayilar"],"acf":[],"aioseo_notices":[],"aioseo_head":"\n\t\t\n\t\n\t\n\t\n\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t