Bir\u00e7ok sihirbazl\u0131k g\u00f6sterisinin arka plan\u0131nda matematiksel d\u00fc\u015f\u00fcnce vard\u0131r; hatta baz\u0131 oyunlar\u0131n haz\u0131rlan\u0131\u015f\u0131nda do\u011frudan matematik kullan\u0131l\u0131r. Persi Diaconis, Ron Graham gibi bir\u00e7ok \u00fcnl\u00fc matematik\u00e7inin bu konuyla ilgili \u00e7al\u0131\u015fmalar yapt\u0131\u011f\u0131n\u0131 biliyoruz.<\/p>\n
Bu yaz\u0131da matematiksel d\u00fc\u015f\u00fcnce \u00fczerinden kurgulanm\u0131\u015f, el \u00e7abuklu\u011fu, hile gibi y\u00f6ntemlerin bulunmad\u0131\u011f\u0131 oyun k\u00e2\u011f\u0131tlar\u0131yla yap\u0131lan iki ilgin\u00e7 sihirbazl\u0131k g\u00f6sterisini ele alaca\u011f\u0131z.<\/p>\n
25\u2019lik desteyle sihirbazl\u0131k. <\/strong>25 oyun k\u00e2\u011f\u0131d\u0131n\u0131 5\u00d75 boyutlar\u0131nda bir kare olu\u015facak \u015fekilde diziyor, kar\u015f\u0131m\u0131zdaki ki\u015fiden bu k\u00e2\u011f\u0131tlardan birini akl\u0131ndan tutmas\u0131n\u0131 ve tuttu\u011fu kart\u0131n hangi sat\u0131rda bulundu\u011funu s\u00f6ylemesini istiyoruz. Daha sonra kartlar\u0131 bir kurala g\u00f6re toplay\u0131p, tekrar 5\u00d75\u2019lik bir kare olu\u015facak \u015fekilde, bir kurala g\u00f6re diziyor ve yine tutulan kart\u0131n hangi sat\u0131rda oldu\u011funu soruyoruz. Art\u0131k ald\u0131\u011f\u0131m\u0131z yan\u0131ta g\u00f6re tutulan kart\u0131 bilebiliriz. Bunun i\u00e7in kartlar\u0131 toplarken ve dizerken yukar\u0131da s\u00f6z\u00fc edilen kurallar ne olmal\u0131?(1)<\/p>\n Yan\u0131t. <\/strong>Basit, ama ho\u015f bir ak\u0131l y\u00fcr\u00fctmeyle bu sihirbazl\u0131\u011f\u0131n kurgusunu \u00e7\u00f6zebiliriz: Ba\u015flang\u0131\u00e7taki dizili\u015fte tutulan kart\u0131n hangi sat\u0131rda oldu\u011funu biliyoruz. Kartlar\u0131 toplarken kartlar\u0131n ve sat\u0131rlar\u0131n s\u0131ras\u0131n\u0131 bozmamal\u0131y\u0131z. \u015e\u00f6yle ki; sa\u011f alt k\u00f6\u015fedeki kart\u0131 en alta, yan\u0131ndakini onun \u00fcst\u00fcne ve bu \u015fekilde son sat\u0131rdaki (5. sat\u0131r) kartlar\u0131 sa\u011fdan sola do\u011fru \u00fcst \u00fcste koyarak toplar\u0131z. Sonras\u0131nda 4\u2019\u00fcnc\u00fc sat\u0131rdaki ve s\u0131ras\u0131yla di\u011fer sat\u0131rlardaki (3, 2 ve 1) kartlar\u0131 ayn\u0131 \u015fekilde sa\u011fdan sola do\u011fru sat\u0131rlar\u0131n s\u0131ras\u0131n\u0131 bozmadan toplamal\u0131y\u0131z. B\u00f6ylece elimizdeki destenin en \u00fcst\u00fcndeki 5 kart 1\u2019inci sat\u0131rdaki kartlardan, sonraki 5 kart 2\u2019inci sat\u0131rdaki kartlardan olu\u015fuyor ve di\u011fer sat\u0131rlardaki (3, 4 ve 5) kartlar da bu \u015fekilde s\u0131ralan\u0131yor.<\/p>\n \u015eimdi, desteyi s\u00fctunlara g\u00f6re dizelim, yani en \u00fcstteki 5 karttan 1\u2019inci s\u00fctunu, sonraki 5 karttan 2\u2019inci s\u00fctunu ve bu \u015fekilde devam ederek di\u011fer s\u00fctunlar\u0131 da olu\u015ftural\u0131m. Kar\u015f\u0131m\u0131zdaki ki\u015fiden tuttu\u011fu kart\u0131n yeni dizili\u015fte hangi sat\u0131rda oldu\u011funu \u00f6\u011frendikten sonra art\u0131k o kart\u0131 kolayl\u0131kla s\u00f6yleyebiliriz; \u00e7\u00fcnk\u00fc eski sat\u0131rlar yeni s\u00fctunlara d\u00f6n\u00fc\u015ft\u00fc\u011f\u00fcnden tutulan say\u0131n\u0131n son dizili\u015fte bulundu\u011fu sat\u0131r\u0131n numaras\u0131 ilk dizili\u015fteki s\u00fctun numaras\u0131na e\u015fit olacakt\u0131r. B\u00f6ylece arad\u0131\u011f\u0131m\u0131z kart\u0131n ilk dizili\u015fte hangi sat\u0131r ve s\u00fctunda bulundu\u011funu \u00f6\u011frenmi\u015f oluruz.<\/p>\n Yukar\u0131da ayr\u0131nt\u0131l\u0131 bir \u015fekilde anlatt\u0131\u011f\u0131m\u0131z uygulamalar\u0131n \u00f6zeti, eski sat\u0131rlar\u0131 yeni s\u00fctunlara d\u00f6n\u00fc\u015ft\u00fcrmek. Lineer cebir bilen okur bu i\u015fleme matrislerde transpoze (devrik) ad\u0131n\u0131n verildi\u011fini an\u0131msayacakt\u0131r. Oyun k\u00e2\u011f\u0131tlar\u0131yla ula\u015ft\u0131\u011f\u0131m\u0131z bu sonu\u00e7 lineer cebirde bir teorem olarak kan\u0131tlanabilir.<\/p>\n 52\u2019lik desteyle sihirbazl\u0131k. <\/strong>Bu sihirbazl\u0131k oyunundaki ad\u0131m say\u0131s\u0131 daha fazla, ama sab\u0131rl\u0131 okur g\u00fczel bir sonla kar\u015f\u0131la\u015facak.(2)<\/p>\n \u00d6nce kar\u015f\u0131m\u0131zdaki ki\u015fiyi kartlar\u0131n de\u011ferleri hakk\u0131nda bilgilendirmeliyiz: Her kart\u0131n \u00fczerindeki say\u0131 o kart\u0131n de\u011ferini g\u00f6sterir. Ayr\u0131ca, \u00fczerlerinde say\u0131 olmayan kartlar\u0131n de\u011ferleri: V(J) = 11, D(Q) = 12, K = 13, A = 1.<\/p>\n Kar\u015f\u0131m\u0131zdaki ki\u015fiden destenin en \u00fcst\u00fcndeki kart\u0131 y\u00fcz\u00fc yukar\u0131 bakacak \u015fekilde a\u00e7\u0131p masaya koymas\u0131n\u0131 isteyelim.<\/p>\n Bu kart\u0131n \u00fcst\u00fcne kart\u0131n de\u011ferini 13\u2019e tamamlayacak say\u0131da y\u00fczleri a\u00e7\u0131k kartlar koyal\u0131m. \u00d6rne\u011fin izleyicinin a\u00e7t\u0131\u011f\u0131 kart 8 ise 5 kart\u0131 y\u00fcz\u00fc a\u00e7\u0131k olarak 8 numaral\u0131 kart\u0131n \u00fcst\u00fcne koyal\u0131m; e\u011fer a\u00e7\u0131lan kart papaz (K) ise kart koymuyoruz. Tabii ki, izleyicinin a\u00e7\u0131lan kart\u0131n de\u011ferini 13\u2019e tamamlad\u0131\u011f\u0131m\u0131z\u0131 anlamamas\u0131 gerekir.<\/p>\n Elde etti\u011fimiz bu y\u0131\u011f\u0131n\u0131 ters \u00e7evirerek ay\u0131ral\u0131m, kalan kartlarla yukar\u0131daki ad\u0131mlar\u0131 tekrarlayal\u0131m, yani izleyici yine kalan destenin en \u00fcst\u00fcndeki kart\u0131 a\u00e7s\u0131n ve biz de bu kart\u0131n de\u011ferini 13\u2019e tamamlayacak say\u0131da kart\u0131 ekleyelim. Bu y\u0131\u011f\u0131n\u0131 da ters \u00e7evirerek masada ayr\u0131 bir yere koyal\u0131m ve destedeki kartlar bitinceye dek ayn\u0131 \u015fekilde y\u00fczleri kapal\u0131 y\u0131\u011f\u0131nlar olu\u015ftural\u0131m. E\u011fer son y\u0131\u011f\u0131nda elimizdeki kartlar\u0131n say\u0131s\u0131 a\u00e7\u0131lan kart\u0131n say\u0131s\u0131n\u0131 13\u2019e tamamlamaya yetmiyorsa o kartlar\u0131 elimizde tutal\u0131m.<\/p>\n \u0130zleyiciden masadaki y\u0131\u011f\u0131nlardan \u00fc\u00e7\u00fcn\u00fc se\u00e7mesini isteyelim. Kalan y\u0131\u011f\u0131nlar\u0131 elimizdeki kalan kartlarla birle\u015ftirelim.<\/p>\n Bu \u201cbirle\u015ftirilmi\u015f\u201d desteden 10 kart\u0131 \u00e7\u0131kar\u0131p, bir kenara koyal\u0131m. Art\u0131k bu kartlar gerekmeyecek. \u0130zleyiciden o \u00fc\u00e7 y\u0131\u011f\u0131n\u0131n herhangi ikisini se\u00e7- mesini ve bu iki y\u0131\u011f\u0131n\u0131n en \u00fcst\u00fcnde bulunan birer kart\u0131 y\u00fcz\u00fc g\u00f6r\u00fclecek \u015fekilde \u00e7evirmesini isteyelim.<\/p>\n Bu iki kart\u0131n de\u011ferini ifade eden say\u0131lar\u0131 toplayal\u0131m ve elimizdeki desteden o say\u0131da kart\u0131 \u00e7\u0131karal\u0131m. Art\u0131k bu kartlara da gerek olmayacak.<\/p>\n Son ad\u0131m: Elimizde kalan kartlar\u0131n say\u0131s\u0131 masada kapal\u0131 olarak duran destenin en \u00fcst\u00fcndeki kart\u0131n de\u011ferine e\u015fit olan say\u0131d\u0131r! Art\u0131k izleyiciye o kart\u0131n de\u011ferini ifade eden say\u0131y\u0131 s\u00f6yleyebilirsiniz.<\/p>\n Sihirbazl\u0131\u011f\u0131n s\u0131rr\u0131<\/strong><\/p>\n \u0130zleyicinin \u00fc\u00e7 y\u0131\u011f\u0131n\u0131 se\u00e7ti\u011fi ad\u0131ma bakal\u0131m. Asl\u0131nda izleyici bu ad\u0131mda \u00fc\u00e7 y\u0131\u011f\u0131n\u0131n en \u00fcst\u00fcndeki \u00fc\u00e7 kart\u0131 se\u00e7iyor ve bu kartlar daha \u00f6nce izleyicinin tek tek a\u00e7t\u0131\u011f\u0131 kartlardan herhangi \u00fc\u00e7\u00fc; \u00e7\u00fcnk\u00fc y\u0131\u011f\u0131nlar\u0131 \u00e7evirdi\u011fimiz i\u00e7in her y\u0131\u011f\u0131n\u0131n en alt\u0131ndaki kartlar en \u00fcste gelmi\u015f oluyor.<\/p>\n Se\u00e7ilen \u00fc\u00e7 y\u0131\u011f\u0131n\u0131n en \u00fcst\u00fcndeki kartlar\u0131n de\u011ferleri a<\/em>, b <\/em>ve c <\/em>olsun. De\u011feri a <\/em>olan kart\u0131n bulundu\u011fu y\u0131\u011f\u0131nda 13\u2013a<\/em>+1 tane kart var, \u00e7\u00fcnk\u00fc bu y\u0131\u011f\u0131n\u0131 olu\u015ftururken de\u011feri a <\/em>olan kart\u0131n \u00fcst\u00fcne 13\u2013a <\/em>tane kart koymu\u015ftuk, bir de a <\/em>kart\u0131 var, toplam bu y\u0131\u011f\u0131nda 14\u2013a <\/em>tane kart bulunuyor. Ayn\u0131 \u015fekilde de\u011ferleri b <\/em>ve c <\/em>olan kartlar\u0131n bulundu\u011fu y\u0131\u011f\u0131nlarda da 14\u2013b <\/em>ve 14\u2013c <\/em>tane kart var.<\/p>\n \u015eimdi, izleyicinin \u00fc\u00e7 y\u0131\u011f\u0131n\u0131 se\u00e7mesinden sonra geriye kalan b\u00fct\u00fcn kartlar\u0131 birle\u015ftirerek elde etti\u011fimiz destedeki kart say\u0131s\u0131n\u0131 bulal\u0131m: 52\u2013(14\u2013a<\/em>)\u2013(14\u2013b<\/em>)\u2013(14\u2013c<\/em>).<\/p>\n Buradan, 10+a<\/em>+b<\/em>+c <\/em>bulunur.<\/p>\n Elimizdeki desteden 10 tane kart\u0131 \u00e7\u0131karm\u0131\u015ft\u0131k. Bu durumda destede a<\/em>+b<\/em>+c <\/em>kart kal\u0131yor. Son olarak izleyicinin se\u00e7ti\u011fi iki kart\u0131n de\u011ferinin toplam\u0131 kadar kart\u0131 da \u00e7\u0131kard\u0131\u011f\u0131m\u0131zdan elimizdeki kart say\u0131s\u0131 a\u00e7\u0131lmam\u0131\u015f y\u0131\u011f\u0131n\u0131n en \u00fcst\u00fcndeki kart\u0131n de\u011ferine e\u015fit oluyor. \u00d6rne\u011fin izleyici de\u011ferleri b <\/em>ve c <\/em>olan kartlar\u0131 se\u00e7mi\u015f olsun. Bu durumda elimizdeki a<\/em>+b<\/em>+c <\/em>tane karttan b<\/em>+c <\/em>tane kart\u0131 \u00e7\u0131kar\u0131yoruz, a <\/em>tane kart kal\u0131yor ve a <\/em>say\u0131s\u0131 da a\u00e7\u0131lmayan y\u0131\u011f\u0131n\u0131n en \u00fcst\u00fcndeki kart\u0131n de\u011ferine e\u015fit oluyor.<\/p>\n Kaynaklar<\/strong><\/p>\n 1) Refail Alizade, \u201cProblemler ve \u00c7\u00f6z\u00fcmleri\u201d, Matematik D\u00fcnyas\u0131<\/em>, 2013-III.<\/p>\n 2) Asl\u0131 Nesin, \u201cAbrakadabra\u201d, Matematik D\u00fcnyas\u0131<\/em>, 2004-IV.<\/p>\n 3) Persi Diaconis, Ron Graham, Magical Mathematics<\/em>, Princeton UniversityPress, 2012. \u00a0\u00a0\u00a0<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":" Bir\u00e7ok sihirbazl\u0131k g\u00f6sterisinin arka plan\u0131nda matematiksel d\u00fc\u015f\u00fcnce vard\u0131r; hatta baz\u0131 oyunlar\u0131n haz\u0131rlan\u0131\u015f\u0131nda do\u011frudan matematik kullan\u0131l\u0131r. Persi Diaconis, Ron Graham gibi bir\u00e7ok \u00fcnl\u00fc matematik\u00e7inin bu konuyla ilgili \u00e7al\u0131\u015fmalar yapt\u0131\u011f\u0131n\u0131 biliyoruz. Bu yaz\u0131da matematiksel d\u00fc\u015f\u00fcnce \u00fczerinden kurgulanm\u0131\u015f, el \u00e7abuklu\u011fu, hile gibi y\u00f6ntemlerin bulunmad\u0131\u011f\u0131 oyun k\u00e2\u011f\u0131tlar\u0131yla yap\u0131lan iki ilgin\u00e7 sihirbazl\u0131k g\u00f6sterisini ele alaca\u011f\u0131z. 25\u2019lik desteyle sihirbazl\u0131k. 25 oyun […]<\/p>\n","protected":false},"author":375,"featured_media":23707,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[168,514],"tags":[208,2913,2914],"class_list":["post-23706","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-131-sayi","category-matematik-sohbetleri","tag-matematik","tag-oyun-kagitlari","tag-sihirbazlik"],"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