{"id":24253,"date":"2018-05-01T00:00:31","date_gmt":"2018-04-30T21:00:31","guid":{"rendered":"https:\/\/bilimvegelecek.com.tr\/?p=24253"},"modified":"2018-05-01T10:26:02","modified_gmt":"2018-05-01T07:26:02","slug":"cozum-basitse-yanlistir","status":"publish","type":"post","link":"https:\/\/bilimvegelecek.com.tr\/index.php\/2018\/05\/01\/cozum-basitse-yanlistir","title":{"rendered":"\u00c7\u00f6z\u00fcm basitse yanl\u0131\u015ft\u0131r!"},"content":{"rendered":"

– Harika bir olas\u0131l\u0131k sorusu \u00f6\u011frendim, hi\u00e7 hesaplama gerektirmeyen \u00e7ok zarif ve k\u0131sa bir \u00e7\u00f6z\u00fcm\u00fc var: Bir k\u00fcre y\u00fczeyi \u00fczerinde rastgele 4 nokta se\u00e7erek bu noktalar\u0131 k\u00f6\u015fe kabul eden bir d\u00f6rty\u00fczl\u00fc olu\u015fturuyoruz. K\u00fcrenin merkezinin bu d\u00f6rty\u00fczl\u00fc i\u00e7inde olma olas\u0131l\u0131\u011f\u0131 ka\u00e7t\u0131r?
\n– G\u00fczel bir soru, ama kolay de\u011fil.
\n– \u00c7\u00f6z\u00fcm basit bir d\u00fc\u015f\u00fcnceye dayan\u0131yor. Bu soruyu Amerika\u2019da d\u00fczenlenen Putnam Matematik Yar\u0131\u015fmas\u0131n\u0131n ilk b\u00f6l\u00fcm\u00fcn\u00fcn son sorusu olarak sormu\u015flar, yani en zor sorulardan biri. Ne diyorsun, \u00e7\u00f6zebilir misin? Hat\u0131rlarsan bir zamanlar bana sordu\u011fun bir olas\u0131l\u0131k sorusunu an\u0131nda yan\u0131tlad\u0131\u011f\u0131mda cevab\u0131m\u0131n do\u011fru ama \u00e7\u00f6z\u00fcm\u00fcm\u00fcn yanl\u0131\u015f oldu\u011funu s\u00f6ylemi\u015ftin. \u201cNas\u0131l \u00e7\u00f6zd\u00fc\u011f\u00fcm\u00fc bilmiyorsun ki, neden yanl\u0131\u015f?\u201d diye sordu\u011fumda, \u201cBasit bir \u00e7\u00f6z\u00fcm yapt\u0131\u011f\u0131n\u0131 anlad\u0131m, matematikte \u00e7\u00f6z\u00fcm basitse yanl\u0131\u015ft\u0131r\u201d diyerek dalga ge\u00e7mi\u015ftin. \u015eimdi bu kez ben soruyorum ve olduk\u00e7a basit bir \u00e7\u00f6z\u00fcm\u00fcn oldu\u011funu s\u00f6yl\u00fcyorum, bakal\u0131m bulabilecek misin? Bir saat s\u00fcre, ama internete ba\u015fvurmak yok!
\n– Anla\u015f\u0131ld\u0131, ge\u00e7mi\u015fin intikam\u0131! Peki, bir saat sonra g\u00f6r\u00fc\u015felim.<\/p>\n

Yukar\u0131daki diyalog ye\u011fenim Do\u011fucan G\u00fcn-c\u00fc\u2019yle yapt\u0131\u011f\u0131m\u0131z bir telefon g\u00f6r\u00fc\u015fmesinde ge\u00e7ti. Bir saat sonra arad\u0131. \u00c7\u00f6zememi\u015ftim. G\u00fczeli g\u00f6rmenin heyecan\u0131 ve keskin zek\u00e2s\u0131yla \u00e7\u00f6z\u00fcm\u00fc anlatt\u0131. Ger\u00e7ekten \u00e7ok ho\u015f ve yal\u0131n bir ak\u0131l y\u00fcr\u00fctmeyle ger\u00e7ekle\u015fen bu \u00e7\u00f6z\u00fcm\u00fc Bilim ve Gelecek okurlar\u0131yla payla\u015fmak istedim.<\/p>\n

Problemi bir kez daha ifade edelim: K\u00fcre y\u00fczeyinde rastgele al\u0131nan d\u00f6rt noktayla bir d\u00f6rty\u00fczl\u00fc olu\u015fturuyoruz ve k\u00fcrenin merkezinin bu d\u00f6rty\u00fczl\u00fcn\u00fcn i\u00e7inde olma olas\u0131l\u0131\u011f\u0131n\u0131 bulmak istiyoruz. \u00d6rme\u011fin a\u015fa\u011f\u0131daki \u015fekilde sa\u011fdaki d\u00f6rty\u00fczl\u00fc k\u00fcrenin merkezini kaps\u0131yorken, soldaki kapsam\u0131yor.<\/p>\n

\"\"Matematikte bir problemi \u00e7\u00f6zerken daha basit bir analog problem bularak \u00e7\u00f6z\u00fcme ula\u015fmak \u00e7ok s\u0131k kullan\u0131lan bir y\u00f6ntemdir. Biz de bu problemi \u015fimdilik bir kenara b\u0131rak\u0131p, bu sorunun \u00e7ember versiyonuna bakal\u0131m.<\/p>\n

Bir \u00e7ember \u00fczerinde rastgele \u00fc\u00e7 nokta alarak bu noktalar\u0131 k\u00f6\u015fe kabul eden bir \u00fc\u00e7gen olu\u015fturuyor ve \u015fu soruyu soruyoruz: \u00c7emberin merkezinin \u00fc\u00e7genin i\u00e7inde olma olas\u0131l\u0131\u011f\u0131 ka\u00e7t\u0131r?<\/p>\n

\u00c7\u00f6z\u00fcm olduk\u00e7a basit ve ak\u0131l dolu: \u00d6nce, \u00e7ember \u00fczerinde rastgele \u00fc\u00e7 nokta se\u00e7mek yerine rastgele bir \u015fekilde [BB’] ve [CC’] \u00e7aplar\u0131n\u0131 \u00e7izelim ve \u00e7ember \u00fczerinde sadece bir noktay\u0131 sabitleyelim. Bu noktay\u0131 A ile g\u00f6sterirsek, A noktas\u0131 [BB’] ve [CC’] \u00e7aplar\u0131n\u0131n \u00e7ember \u00fczerinde ay\u0131rd\u0131\u011f\u0131 4 farkl\u0131 yaydan birinin \u00fczerinde olacakt\u0131r, dolay\u0131s\u0131yla A\u2019y\u0131 herhangi bir yere koyabiliriz, BC (k\u0131sa) yay\u0131 \u00fczerinde olsun. Bu durumda art\u0131k di\u011fer iki noktan\u0131n sabitlenmesi i\u00e7in 4 farkl\u0131 se\u00e7im yapabiliriz, \u015f\u00f6yle ki: noktalardan birini B veya B’ di\u011ferini ise C veya C’ noktalar\u0131ndan se\u00e7meliyiz, yani, 2\u00d72=4 farkl\u0131 se\u00e7im yapabiliriz: B\u2013C, B’\u2013C, B\u2013C’, B’\u2013C’. B\u00f6ylece A d\u0131\u015f\u0131ndaki iki noktay\u0131 4 farkl\u0131 \u015fekilde sabitleyece\u011fimizi g\u00f6rm\u00fc\u015f olduk, \u015fimdi art\u0131k ABC, AB’C, ABC’ ve AB’C’ \u00fc\u00e7genlerini \u00e7izebiliriz. A\u015fa\u011f\u0131daki \u015fekilde bu 4 \u00fc\u00e7gen g\u00f6r\u00fclmektedir. Bu \u00fc\u00e7genlerden sadece AB’C’ \u00fc\u00e7geni \u00e7emberin merkezini kaps\u0131yor. O halde \u00e7ember \u00fczerinde rastgele al\u0131nan 3 noktay\u0131 k\u00f6\u015fe kabul eden \u00fc\u00e7genin \u00e7emberin merkezini kapsamas\u0131 olas\u0131l\u0131\u011f\u0131 1\/4\u2019t\u00fcr. M\u00fckemmel bir \u00e7\u00f6z\u00fcm!<\/p>\n

\"\"As\u0131l sorumuza, k\u00fcre y\u00fczeyinde rastgele al\u0131nan 4 nokta problemine geri d\u00f6nelim ve yukar\u0131da ula\u015ft\u0131\u011f\u0131m\u0131z bulgular\u0131 bu problemin \u00e7\u00f6z\u00fcm\u00fcnde kullanal\u0131m.
\n\u00c7emberin 2 farkl\u0131 \u00e7ap\u0131n\u0131 \u00e7izmi\u015ftik, bu kez k\u00fcrenin 3 farkl\u0131 \u00e7ap\u0131n\u0131 \u00e7izerek \u00e7\u00f6z\u00fcme ba\u015flayal\u0131m. Bu \u00e7aplar\u0131 [AB], [CD], [EF] ile g\u00f6sterirsek her bir \u00e7ap\u0131 kapsayan \u00fc\u00e7 d\u00fczlemle k\u00fcre sekiz par\u00e7aya b\u00f6l\u00fcnm\u00fc\u015f olacakt\u0131r. A\u015fa\u011f\u0131daki \u015fekle bak\u0131p, bu k\u00fcre par\u00e7alar\u0131n\u0131 hayal etmemiz m\u00fcmk\u00fcn.<\/p>\n

\"\"\u015eimdi, t\u0131pk\u0131 \u00e7emberin \u00e7aplar\u0131n\u0131n u\u00e7 noktalar\u0131n\u0131 se\u00e7ti\u011fimiz gibi burada da 3 \u00e7ap i\u00e7in 2\u00d72\u00d72=8 farkl\u0131 se\u00e7im yapabiliriz, \u015f\u00f6yle ki A ya da B\u2019yi veya C ya da D\u2019yi veya E ya da F\u2019yi se\u00e7ebiliriz. B\u00f6ylece 4 noktan\u0131n 3\u2019\u00fcn\u00fc sabitlemi\u015f oluruz. Bu 8 se\u00e7ene\u011fin ortaya \u00e7\u0131kard\u0131\u011f\u0131 \u00fc\u00e7genleri \u015fu \u015fekilde s\u0131ralayabiliriz: ACE, ACF, ADE, ADF, BCE, BCF, BDE, BDF. D\u00f6rd\u00fcnc\u00fc nokta sekiz k\u00fcre par\u00e7as\u0131ndan herhangi biri \u00fczerinde olabilir. Bu noktay\u0131 P ile g\u00f6sterip, AE yay\u0131n\u0131n arka taraf\u0131ndaki k\u00fcre par\u00e7as\u0131n\u0131n \u00fczerine koydu\u011fumuzu varsayarsak (P, BCF) d\u00f6rty\u00fczl\u00fcs\u00fc k\u00fcrenin merkezini kapsarken di\u011fer d\u00f6rty\u00fczl\u00fcler merkezi kapsamayacakt\u0131r. \u00c7\u00fcnk\u00fc P noktas\u0131n\u0131 BCF \u00fc\u00e7geninin tam kar\u015f\u0131s\u0131ndaki b\u00f6lgeye koymu\u015f olduk; bu durum di\u011fer \u00fc\u00e7genler i\u00e7in ge\u00e7erli olmad\u0131\u011f\u0131 i\u00e7in sadece (P, BCF) d\u00f6rty\u00fczl\u00fcs\u00fc merkezi kaps\u0131yor. B\u00f6ylece k\u00fcre \u00fczerinde rastgele se\u00e7ilen 4 noktay\u0131 k\u00f6\u015fe kabul eden d\u00f6rty\u00fczl\u00fcn\u00fcn k\u00fcrenin merkezini kapsama olas\u0131l\u0131\u011f\u0131 1\/8 olarak bulunur.<\/p>\n

Ne kadar ince ve zekice bir yol, de\u011fil mi? \u00c7\u00f6z\u00fcm \u00e7ok basit ve yanl\u0131\u015f de\u011fil! \u00d6te yandan, problemi \u00f6nce iki boyutta ele al\u0131p, oradaki y\u00f6ntemi \u00fc\u00e7 boyuta ta\u015f\u0131mak da m\u00fckemmel bir hamle! Burada ister istemez \u015fu soru geliyor insan\u0131n akl\u0131na: \u0130ki boyutta olas\u0131l\u0131k 1\/4, \u00fc\u00e7 boyutta 1\/8 oldu\u011funa g\u00f6re d\u00f6rt boyutta(!) 1\/16 m\u0131 olur? B\u00f6ylesi bir problemi ortaya koyup \u00e7\u00f6zmeden \u201c1\/16 olur\u201d diyemeyiz elbette, ama problemi bir boyuta indirgersek bu kez sonu\u00e7 1\/2 oluyor, \u015f\u00f6yle ki: \u201c[0,1] aral\u0131\u011f\u0131ndan rastgele iki reel say\u0131 al\u0131rsak s\u0131f\u0131r say\u0131s\u0131n\u0131n bu say\u0131lar aras\u0131nda kalma olas\u0131l\u0131\u011f\u0131 ka\u00e7t\u0131r?\u201d diye soral\u0131m ve \u015f\u00f6yle d\u00fc\u015f\u00fcnelim: Say\u0131lardan birini 0\u2019\u0131n soluna ya da sa\u011f\u0131na koyma olas\u0131l\u0131\u011f\u0131 1\/2\u2019dir, art\u0131k di\u011fer say\u0131n\u0131n di\u011fer tarafta olma olas\u0131l\u0131\u011f\u0131 1 olaca\u011f\u0131ndan sonu\u00e7 1\/2\u2019dir. Dolay\u0131s\u0131yla \u00fc\u00e7 boyuta kadar boyut say\u0131s\u0131na ba\u011fl\u0131 olarak istenen olas\u0131l\u0131k 1\/2n form\u00fcl\u00fcyle hesaplanabiliyor. Sonras\u0131n\u0131 bilmiyoruz!<\/p>\n

Not:<\/strong> Merakl\u0131 okur www.brilliant.org adresinden bu problemin \u00e7\u00f6z\u00fcm\u00fcn\u00fcn \u00fc\u00e7 boyutlu sim\u00fclasyonunun yer ald\u0131\u011f\u0131 videoyu izleyebilir.<\/p>\n","protected":false},"excerpt":{"rendered":"

– Harika bir olas\u0131l\u0131k sorusu \u00f6\u011frendim, hi\u00e7 hesaplama gerektirmeyen \u00e7ok zarif ve k\u0131sa bir \u00e7\u00f6z\u00fcm\u00fc var: Bir k\u00fcre y\u00fczeyi \u00fczerinde rastgele 4 nokta se\u00e7erek bu noktalar\u0131 k\u00f6\u015fe kabul eden bir d\u00f6rty\u00fczl\u00fc olu\u015fturuyoruz. K\u00fcrenin merkezinin bu d\u00f6rty\u00fczl\u00fc i\u00e7inde olma olas\u0131l\u0131\u011f\u0131 ka\u00e7t\u0131r? – G\u00fczel bir soru, ama kolay de\u011fil. – \u00c7\u00f6z\u00fcm basit bir d\u00fc\u015f\u00fcnceye dayan\u0131yor. Bu soruyu […]<\/p>\n","protected":false},"author":375,"featured_media":24257,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[2985,25,514],"tags":[3023,3022,208,1572],"class_list":["post-24253","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-171-sayi","category-matematik","category-matematik-sohbetleri","tag-cember","tag-kure","tag-matematik","tag-problemler"],"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