Bilindi\u011fi gibi asal say\u0131lar 1 ve kendinden ba\u015fka b\u00f6leni olmayan say\u0131lard\u0131r ve binlerce y\u0131ld\u0131r insano\u011flunun merak\u0131n\u0131 celbetmi\u015ftir. Bir \u00e7ok d\u00fc\u015f\u00fcn\u00fcr ve matematik\u00e7i asal say\u0131lar\u0131 tespit edecek bir form\u00fcl bulma \u00e7abas\u0131 i\u00e7ine girmi\u015f ve fakat bu \u00e7abalar h\u00fcsranla sonu\u00e7lanm\u0131\u015ft\u0131r. \u00c7\u00fcnk\u00fc asal say\u0131lar\u0131 tespit etmenin herhangi bir form\u00fcl\u00fc maalesef yoktur. Baz\u0131 yakla\u015f\u0131m metotlar\u0131 belli say\u0131lara kadar do\u011fru sonu\u00e7 verse de maalesef t\u00fcm say\u0131lar i\u00e7in genel bir yakla\u015f\u0131m ortaya konulamam\u0131\u015ft\u0131r.<\/p>\n
Asal bir\u00a0a\u00a0<\/em>i\u00e7in 2a\u00a0<\/sup><\/em>\u2013 1 bi\u00e7iminde yaz\u0131lan say\u0131lara\u00a0Mersenne<\/strong>\u00a0say\u0131lar\u0131<\/strong>\u00a0denir. Peki,\u00a0a\u00a0<\/em>asalsa,\u00a0Ma<\/sub><\/em>\u00a0= 2a<\/sup><\/em>\u00a0\u2013 1 olarak tan\u0131mlanan say\u0131 da asal m\u0131d\u0131r?<\/p>\n \u0130lk Mersenne\u00a0say\u0131lar\u0131na bakal\u0131m:<\/p>\n M<\/em>2<\/sub>\u00a0= 3<\/p>\n M<\/em>3<\/sub>\u00a0= 7<\/p>\n M<\/em>5<\/sub>\u00a0= 31<\/p>\n M<\/em>7<\/sub>\u00a0= 127<\/p>\n Bu say\u0131lar\u0131n herbiri asal. Ama bundan sonraki ilk Mersenne\u00a0say\u0131s\u0131, yani\u00a0M<\/em>11<\/sub>, asal de\u011fil:\u00a0M<\/em>11<\/sub>\u00a0= 23 x 89.<\/p>\n Hangi asallar i\u00e7in\u00a0Ma<\/sub><\/em>\u00a0asald\u0131r? Yan\u0131t bilinmiyor.<\/p>\n Fermat,\u00a0Fn<\/sub><\/em>\u00a0=(2\u00b2)^n +1 bi\u00e7iminde yaz\u0131lan b\u00fct\u00fcn say\u0131lar\u0131n asal olduklar\u0131n\u0131 san\u0131yordu. Bu y\u00fczden bu say\u0131lara\u00a0Fermat say\u0131lar\u0131<\/strong>\u00a0denir. Ger\u00e7ekten de ilk be\u015f Fermat say\u0131s\u0131,<\/p>\n F<\/em>o<\/sub>\u00a0= 3<\/p>\n F<\/em>1<\/sub>\u00a0= 5<\/p>\n F<\/em>2<\/sub>\u00a0= 17<\/p>\n F<\/em>3<\/sub>\u00a0= 257<\/p>\n F<\/em>4<\/sub>\u00a0= 65537 asald\u0131r.<\/p>\n Fermat, b\u00fct\u00fcn Fermat say\u0131lar\u0131n\u0131n asal olduklar\u0131n\u0131 kan\u0131tlamaya u\u011fra\u015ft\u0131 ama ba\u015faramad\u0131. Ba\u015far\u0131s\u0131zl\u0131\u011f\u0131n\u0131n nedeni vard\u0131: San\u0131s\u0131 do\u011fru de\u011fildi.\u00a0F<\/em>5<\/sub>\u00a0asal de\u011fildir.\u00a0F<\/em>5<\/sub>\u00a0on basamakl\u0131 bir say\u0131 oldu\u011fundan asall\u0131\u011f\u0131n\u0131 kan\u0131tlamak kolay de\u011fildi. Euler\u00a0(1707-1783),\u00a0F<\/em>5<\/sub>\u2019in 641\u2019e b\u00f6l\u00fcnd\u00fc\u011f\u00fcn\u00fc g\u00f6sterdi:<\/p>\n F<\/em>5<\/sub>\u00a0= 641 x 6700417.<\/p>\n Peki, anla\u015f\u0131lan o ki asal say\u0131lar\u0131 tespit etmek san\u0131ld\u0131\u011f\u0131n\u0131n aksine hi\u00e7te kolay de\u011fil. Ancak bir say\u0131n\u0131n asal olup olmad\u0131\u011f\u0131n\u0131 nas\u0131l anlar\u0131z sorusunun cevab\u0131n\u0131 bilgisayar yard\u0131m\u0131 ile \u00e7ok h\u0131zl\u0131 bir \u015fekilde tespit edebiliriz. Nas\u0131l m\u0131?<\/p>\n Say\u0131m\u0131za\u00a0n\u00a0<\/em>diyelim.\u00a0n<\/em>\u2019yi\u00a0n<\/em>\u2019den k\u00fc\u00e7\u00fck say\u0131lara b\u00f6lmeye \u00e7al\u0131\u015fal\u0131m. E\u011fer\u00a0n<\/em>\u2019den k\u00fc\u00e7\u00fck, 1\u2019den b\u00fcy\u00fck bir say\u0131\u00a0n<\/em>\u2019yi tam b\u00f6l\u00fcyorsa,\u00a0n<\/em>, tan\u0131m\u0131 gere\u011fi, asal olamaz. \u00d6yle bir say\u0131 bulamazsak,\u00a0n\u00a0<\/em>asald\u0131r deriz.<\/p>\n Ne var ki bu y\u00f6ntemle b\u00fcy\u00fck say\u0131lar\u0131n asall\u0131\u011f\u0131na karar vermek \u00e7ok zaman al\u0131r. Bu y\u00f6ntem ve \u00e7e\u015fitlemeleri d\u0131\u015f\u0131nda bir say\u0131n\u0131n asall\u0131\u011f\u0131na karar verebilecek genel bir y\u00f6ntem de bilinmemektedir. \u00d6rne\u011fin, \u015fu \u00e7e\u015fitleme d\u00fc\u015f\u00fcn\u00fclebilir:\u00a0n<\/em>\u2019yi\u00a0n<\/em>\u2019den k\u00fc\u00e7\u00fck her say\u0131ya b\u00f6lece\u011fimize,\u00a0n<\/em>\u2019yi \u221an<\/em>\u2019den k\u00fc\u00e7\u00fck say\u0131lara b\u00f6lmeye \u00e7al\u0131\u015fabiliriz. \u00c7\u00fcnk\u00fc\u00a0n = ab\u00a0<\/em>ve\u00a0a \u2265\u221a<\/em>n<\/em>\u00a0ise,\u00a0b\u2264\u221a<\/em>n<\/em>\u2019dir. Dolay\u0131s\u0131yla\u00a0n\u00a0<\/em>asal de\u011filse, \u221an<\/em>\u2019den k\u00fc\u00e7\u00fck bir say\u0131ya b\u00f6l\u00fcn\u00fcr. B\u00f6ylece yapmam\u0131z gereken b\u00f6lme say\u0131s\u0131 azal\u0131r.<\/p>\n Peki, bunu yapacak bir algoritma yazal\u0131m. Bunun i\u00e7in\u00a0Flowgorithm<\/a>\u00a0program\u0131 ile bir flowchart olu\u015ftural\u0131m. A\u015fa\u011f\u0131da olu\u015fturulmu\u015f bir flowchart bulunmaktad\u0131r. Bu y\u00f6ntemle \u00e7ok h\u0131zl\u0131 bir \u015fekilde herhangi bir say\u0131n\u0131n asal olup olmad\u0131\u011f\u0131n\u0131 bulmak m\u00fcmk\u00fcn.<\/p>\n