B\u00fcy\u00fck veri y\u0131\u011f\u0131nlar\u0131 aras\u0131nda kaybolan bir i\u011fneyi bulmak gibi mi geliyor s\u0131ralama i\u015flemi? G\u00fcn\u00fcm\u00fcz d\u00fcnyas\u0131nda veriler, i\u015fletmelerden bilimsel ara\u015ft\u0131rmalara, ki\u015fisel cihazlar\u0131m\u0131zdan bulut tabanl\u0131 sistemlere kadar her yerde kar\u015f\u0131m\u0131za \u00e7\u0131k\u0131yor. Bu verileri anlaml\u0131 hale getirmenin en temel yollar\u0131ndan biri ise onlar\u0131 belirli bir d\u00fczene g\u00f6re s\u0131ralamakt\u0131r. Ancak milyarlarca veri noktas\u0131n\u0131 s\u0131ralamak, standart algoritmalarla bile bazen beklenenden \u00e7ok daha uzun s\u00fcrebilir. \u0130\u015fte bu noktada, lineer zamanl\u0131 s\u0131ralama algoritmalar\u0131 devreye girerek, belirli ko\u015fullar alt\u0131nda kar\u015f\u0131la\u015ft\u0131rma tabanl\u0131 algoritmalar\u0131n O(n log n) bariyerini a\u015fmam\u0131z\u0131 sa\u011flar ve muazzam performans art\u0131\u015flar\u0131 sunar.<\/p>\n

Bu makalede, veri s\u0131ralama d\u00fcnyas\u0131n\u0131n \u00fc\u00e7\u00fcnc\u00fc ve en h\u0131zl\u0131 b\u00f6l\u00fcm\u00fcne odaklanaca\u011f\u0131z: Lineer zamanl\u0131 s\u0131ralama algoritmalar\u0131. \u00d6zellikle Counting Sort<\/strong>, Radix Sort<\/strong> ve Bucket Sort<\/strong> gibi y\u00f6ntemlerin nas\u0131l \u00e7al\u0131\u015ft\u0131\u011f\u0131n\u0131, ne zaman kullan\u0131lmalar\u0131 gerekti\u011fini ve ger\u00e7ek d\u00fcnya senaryolar\u0131nda nas\u0131l bir fark yaratt\u0131klar\u0131n\u0131 ad\u0131m ad\u0131m inceleyece\u011fiz. Bu algoritmalar, geleneksel s\u0131ralama y\u00f6ntemlerinin aksine, verinin yap\u0131s\u0131ndan (\u00f6rne\u011fin, tam say\u0131 olmas\u0131 veya belirli bir aral\u0131kta bulunmas\u0131) faydalanarak \u015fa\u015f\u0131rt\u0131c\u0131 bir h\u0131zla sonu\u00e7 verirler. Gelin, bu g\u00fc\u00e7l\u00fc ara\u00e7lar\u0131 yak\u0131ndan tan\u0131yal\u0131m ve veri setlerinizi en verimli \u015fekilde nas\u0131l organize edebilece\u011finizi ke\u015ffedelim.<\/p>\n

Veri s\u0131ralama, bilgisayar bilimlerinin en temel ve s\u0131k kar\u015f\u0131la\u015f\u0131lan problemlerinden biridir. Geleneksel olarak, Quicksort, Mergesort veya Heapsort gibi kar\u015f\u0131la\u015ft\u0131rma tabanl\u0131 s\u0131ralama algoritmalar\u0131, en iyi ihtimalle O(n log n) zaman karma\u015f\u0131kl\u0131\u011f\u0131na sahiptir. Bu, n elemanl\u0131 bir veri setini s\u0131ralamak i\u00e7in yakla\u015f\u0131k n * log(n) i\u015flem gerekti\u011fi anlam\u0131na gelir. N de\u011feri b\u00fcy\u00fcd\u00fck\u00e7e, bu karma\u015f\u0131kl\u0131k ciddi bir performans darbo\u011faz\u0131 yaratabilir.<\/p>\n

Peki, daha h\u0131zl\u0131s\u0131 m\u00fcmk\u00fcn m\u00fc? Evet, e\u011fer veri setimiz belirli \u00f6zelliklere sahipse! Lineer zamanl\u0131 s\u0131ralama algoritmalar\u0131, kar\u015f\u0131la\u015ft\u0131rma yapmadan, verinin kendisindeki de\u011ferlerin da\u011f\u0131l\u0131m\u0131n\u0131 veya basamaklar\u0131n\u0131 kullanarak s\u0131ralama yapar. Bu da onlara teorik olarak O(n+k) veya O(d*(n+k)) gibi lineer bir zaman karma\u015f\u0131kl\u0131\u011f\u0131na ula\u015fma yetene\u011fi kazand\u0131r\u0131r; burada ‘n’ eleman say\u0131s\u0131, ‘k’ de\u011fer aral\u0131\u011f\u0131 ve ‘d’ basamak say\u0131s\u0131d\u0131r. Bu, n’nin \u00e7ok b\u00fcy\u00fck oldu\u011fu durumlarda, kar\u015f\u0131la\u015ft\u0131rma tabanl\u0131 algoritmalar\u0131n log(n) fakt\u00f6r\u00fcn\u00fc ortadan kald\u0131rarak devasa bir h\u0131z avantaj\u0131 sa\u011flar. \u00d6rne\u011fin, 1 milyon eleman i\u00e7in log(1 milyon) yakla\u015f\u0131k 20’dir, yani lineer algoritmalar potansiyel olarak 20 kat daha h\u0131zl\u0131 olabilir!<\/p>\n

Bu algoritmalar\u0131n temel g\u00fcc\u00fc, “kar\u015f\u0131la\u015ft\u0131rma” i\u015fleminin maliyetinden ka\u00e7\u0131nmalar\u0131d\u0131r. Bunun yerine, elemanlar\u0131n do\u011frudan dizinlerini (Counting Sort), basamak de\u011ferlerini (Radix Sort) veya belirli aral\u0131klar\u0131n\u0131 (Bucket Sort) kullanarak s\u0131ralama yaparlar. Ancak bu performans avantaj\u0131, baz\u0131 k\u0131s\u0131tlamalarla gelir. Genellikle, bu algoritmalar yaln\u0131zca tam say\u0131lar veya belirli bir aral\u0131ktaki ondal\u0131k say\u0131lar \u00fczerinde etkili bir \u015fekilde \u00e7al\u0131\u015f\u0131r. Ayr\u0131ca, “k” de\u011feri (maksimum eleman de\u011feri veya aral\u0131k geni\u015fli\u011fi) \u00e7ok b\u00fcy\u00fckse, alan veya zaman karma\u015f\u0131kl\u0131\u011f\u0131 yine de artabilir. Dolay\u0131s\u0131yla, do\u011fru algoritmay\u0131 se\u00e7mek, veri setinizin do\u011fas\u0131n\u0131 ve k\u0131s\u0131tlamalar\u0131n\u0131 anlamaktan ge\u00e7er. Bu h\u0131zl\u0131 s\u0131ralama y\u00f6ntemleri, \u00f6zellikle geni\u015f veri tabanlar\u0131nda, g\u00f6r\u00fcnt\u00fc i\u015fleme uygulamalar\u0131nda veya a\u011f paketlerinin s\u0131ralanmas\u0131 gibi ger\u00e7ek zamanl\u0131 sistemlerde kritik \u00f6neme sahiptir.<\/p>\n

\n Uzman \u0130pucu: Kar\u015f\u0131la\u015ft\u0131rma tabanl\u0131 s\u0131ralama algoritmalar\u0131n\u0131n alt s\u0131n\u0131r\u0131 O(n log n) iken, lineer zamanl\u0131 algoritmalar\u0131n bu s\u0131n\u0131r\u0131 a\u015fabilmesinin ana nedeni, elemanlar\u0131 do\u011frudan kar\u015f\u0131la\u015ft\u0131rmak yerine onlar\u0131n \u00f6zelliklerini (de\u011fer, basamak) kullanmalar\u0131d\u0131r. Bu detay, do\u011fru senaryoda inan\u0131lmaz performans art\u0131\u015f\u0131 sa\u011flayabilir.\n<\/div>\n
Counting Sort: Tam Say\u0131lar\u0131 H\u0131zla Nas\u0131l S\u0131ralars\u0131n\u0131z?<\/h2>\n
Counting Sort, belirli bir aral\u0131ktaki tam say\u0131lar\u0131, di\u011fer kar\u015f\u0131la\u015ft\u0131rma tabanl\u0131 algoritmalardan \u00e7ok daha h\u0131zl\u0131 s\u0131ralayabilen kararl\u0131 bir lineer zamanl\u0131 s\u0131ralama algoritmas\u0131d\u0131r. \u00d6zellikle s\u0131ralanacak elemanlar\u0131n pozitif tam say\u0131 oldu\u011fu ve de\u011fer aral\u0131\u011f\u0131n\u0131n (k) eleman say\u0131s\u0131na (n) g\u00f6re \u00e7ok b\u00fcy\u00fck olmad\u0131\u011f\u0131 durumlarda olduk\u00e7a etkilidir. Ad\u0131ndan da anla\u015f\u0131laca\u011f\u0131 gibi, bu algoritma elemanlar\u0131n “say\u0131s\u0131na” odaklan\u0131r.<\/p>\n
Algoritma \u00fc\u00e7 ana ad\u0131mda ilerler:<\/p>\n
\n
Frekans Say\u0131m\u0131:<\/strong> \u0130lk olarak, orijinal dizideki her eleman\u0131n ka\u00e7 kez ge\u00e7ti\u011fini saymak i\u00e7in bir frekans dizisi (veya say\u0131c\u0131 dizi) olu\u015fturulur. Bu dizi genellikle elemanlar\u0131n alabilece\u011fi en b\u00fcy\u00fck de\u011fere (k) kadar boyutland\u0131r\u0131l\u0131r. \u00d6rne\u011fin, e\u011fer en b\u00fcy\u00fck say\u0131 10 ise, dizinin boyutu 11 olur (0’dan 10’a kadar).<\/li>\n
K\u00fcm\u00fclatif Toplam (\u00d6n Eki Toplam\u0131):<\/strong> Frekans dizisi \u00fczerinde bir \u00f6n eki toplam\u0131 i\u015flemi yap\u0131l\u0131r. Bu ad\u0131mda, her eleman\u0131n de\u011feri, kendinden \u00f6nceki elemanlar\u0131n toplam\u0131 ile g\u00fcncellenir. Bu i\u015flem sonucunda frekans dizisindeki her indeks, orijinal dizideki o de\u011ferden k\u00fc\u00e7\u00fck veya ona e\u015fit ka\u00e7 eleman oldu\u011funu g\u00f6sterir. Ba\u015fka bir deyi\u015fle, bu bize s\u0131ralanm\u0131\u015f dizide her eleman\u0131n son konumunu verir.<\/li>\n
\u00c7\u0131k\u0131\u015f Dizisi Olu\u015fturma:<\/strong> Orijinal dizi tersten taran\u0131r. Her eleman i\u00e7in, k\u00fcm\u00fclatif toplam dizisinde bulunan de\u011fer (yani s\u0131ralanm\u0131\u015f dizideki konumu) belirlenir ve bu konuma eleman yerle\u015ftirilir. Ard\u0131ndan, k\u00fcm\u00fclatif toplam dizisindeki ilgili de\u011fer bir azalt\u0131l\u0131r. Bu, ayn\u0131 say\u0131dan birden fazla varsa do\u011fru kararl\u0131 s\u0131ralamay\u0131 sa\u011flamak i\u00e7indir.<\/li>\n<\/ol>\n
Counting Sort’un zaman karma\u015f\u0131kl\u0131\u011f\u0131 O(n + k) ve alan karma\u015f\u0131kl\u0131\u011f\u0131 O(n + k)’dir. Kararl\u0131 olmas\u0131 (yani e\u015fit elemanlar\u0131n orijinal s\u0131ras\u0131n\u0131 korumas\u0131) Radix Sort gibi di\u011fer algoritmalar i\u00e7in bir alt rutin olarak kullan\u0131lmas\u0131nda kritik bir avantaj sa\u011flar. Ancak, ‘k’ de\u011feri ‘n’den \u00e7ok b\u00fcy\u00fck oldu\u011funda, algoritman\u0131n hem zaman hem de alan karma\u015f\u0131kl\u0131\u011f\u0131 artar ve verimsiz hale gelebilir.<\/p>\n
Counting Sort Ad\u0131m Ad\u0131m \u00d6rnek<\/h3>\n
Dizi: [4, 2, 2, 8, 3, 3, 1]<\/code> (En b\u00fcy\u00fck de\u011fer k=8)<\/p>\n
\nFrekans Say\u0131m\u0131:<\/strong>\n\ncount = [0, 0, 0, 0, 0, 0, 0, 0, 0]<\/code> (Boyut k+1 = 9)<\/li>\n Diziyi tara:<\/li>\n 4 -> count[4]++<\/code> -> count = [0, 0, 0, 0, 1, 0, 0, 0, 0]<\/code><\/li>\n 2 -> count[2]++<\/code> -> count = [0, 0, 1, 0, 1, 0, 0, 0, 0]<\/code><\/li>\n 2 -> count[2]++<\/code> -> count = [0, 0, 2, 0, 1, 0, 0, 0, 0]<\/code><\/li>\n 8 -> count[8]++<\/code> -> count = [0, 0, 2, 0, 1, 0, 0, 0, 1]<\/code><\/li>\n 3 -> count[3]++<\/code> -> count = [0, 0, 2, 1, 1, 0, 0, 0, 1]<\/code><\/li>\n 3 -> count[3]++<\/code> -> count = [0, 0, 2, 2, 1, 0, 0, 0, 1]<\/code><\/li>\n 1 -> count[1]++<\/code> -> count = [0, 1, 2, 2, 1, 0, 0, 0, 1]<\/code><\/li>\n<\/ul>\nSonu\u00e7: count = [0, 1, 2, 2, 1, 0, 0, 0, 1]<\/code><\/p>\n<\/li>\n K\u00fcm\u00fclatif Toplam:<\/strong>\n\ncount[0] = 0<\/code><\/li>\n count[1] = count[0] + count[1] = 0 + 1 = 1<\/code><\/li>\n count[2] = count[1] + count[2] = 1 + 2 = 3<\/code><\/li>\n count[3] = count[2] + count[3] = 3 + 2 = 5<\/code><\/li>\n count[4] = count[3] + count[4] = 5 + 1 = 6<\/code><\/li>\n count[5] = count[4] + count[5] = 6 + 0 = 6<\/code><\/li>\n count[6] = count[5] + count[6] = 6 + 0 = 6<\/code><\/li>\n count[7] = count[6] + count[7] = 6 + 0 = 6<\/code><\/li>\n count[8] = count[7] + count[8] = 6 + 1 = 7<\/code><\/li>\n<\/ul>\nSonu\u00e7: count = [0, 1, 3, 5, 6, 6, 6, 6, 7]<\/code> (Bu dizideki de\u011ferler, orijinal dizideki elemanlar\u0131n s\u0131ralanm\u0131\u015f dizideki konumunu (1-tabanl\u0131) g\u00f6sterir.)<\/p>\n<\/li>\n \u00c7\u0131k\u0131\u015f Dizisi Olu\u015fturma:<\/strong> (output<\/code> dizisi boyutu n=7)\n\nOrijinal dizi: [4, 2, 2, 8, 3, 3, 1]<\/code><\/li>\n 1. eleman: 1<\/code>. count[1]<\/code> = 1. output[1-1] = output[0] = 1<\/code>. count[1]--<\/code> -> 0.<\/li>\n 2. eleman: 3<\/code>. count[3]<\/code> = 5. output[5-1] = output[4] = 3<\/code>. count[3]--<\/code> -> 4.<\/li>\n 3. eleman: 3<\/code>. count[3]<\/code> = 4. output[4-1] = output[3] = 3<\/code>. count[3]--<\/code> -> 3.<\/li>\n 4. eleman: 8<\/code>. count[8]<\/code> = 7. output[7-1] = output[6] = 8<\/code>. count[8]--<\/code> -> 6.<\/li>\n 5. eleman: 2<\/code>. count[2]<\/code> = 3. output[3-1] = output[2] = 2<\/code>. count[2]--<\/code> -> 2.<\/li>\n 6. eleman: 2<\/code>. count[2]<\/code> = 2. output[2-1] = output[1] = 2<\/code>. count[2]--<\/code> -> 1.<\/li>\n 7. eleman: 4<\/code>. count[4]<\/code> = 6. output[6-1] = output[5] = 4<\/code>. count[4]--<\/code> -> 5.<\/li>\n<\/ul>\nSonu\u00e7: output = [1, 2, 2, 3, 3, 4, 8]<\/code><\/p>\n<\/li>\n<\/ol>\n Python ile Counting Sort Kod \u00d6rne\u011fi<\/h3>\n\ndef counting_sort(arr):\n # Dizi bo\u015fsa veya tek elemanl\u0131ysa s\u0131ralamaya gerek yok\n if not arr or len(arr) <= 1:\n return arr\n\n # 1. En b\u00fcy\u00fck eleman\u0131 bul (k de\u011ferini belirlemek i\u00e7in)\n # Negatif say\u0131lar i\u00e7in offset y\u00f6netimi yap\u0131labilir, burada pozitif tam say\u0131lar varsay\u0131l\u0131yor.\n max_val = max(arr)\n\n # 2. Frekans say\u0131m dizisini olu\u015ftur ve ba\u015flat\n # Boyut max_val + 1 (0'dan max_val'a kadar olan de\u011ferler i\u00e7in)\n count = [0] * (max_val + 1)\n\n # 3. Her eleman\u0131n frekans\u0131n\u0131 say\n for num in arr:\n count[num] += 1\n\n # 4. K\u00fcm\u00fclatif toplam\u0131 hesapla\n # count[i] \u015fimdi, i de\u011ferine e\u015fit veya ondan k\u00fc\u00e7\u00fck ka\u00e7 eleman oldu\u011funu g\u00f6sterir.\n for i in range(1, len(count)):\n count[i] += count[i-1]\n\n # 5. Sonu\u00e7 dizisini olu\u015ftur\n output = [0] * len(arr)\n\n # 6. Orijinal diziyi tersten tara ve elemanlar\u0131 \u00e7\u0131k\u0131\u015f dizisine yerle\u015ftir\n # Tersten taramak, algoritman\u0131n kararl\u0131l\u0131\u011f\u0131n\u0131 (stable) sa\u011flar.\n for num in reversed(arr):\n output[count[num] - 1] = num # count[num]-1 \u00e7\u00fcnk\u00fc dizinler 0'dan ba\u015flar\n count[num] -= 1 # Ayn\u0131 say\u0131dan birden fazla varsa do\u011fru konuma yerle\u015ftirmek i\u00e7in\n\n return output\n\n# Vaka Analizi: Anket Sonu\u00e7lar\u0131\n# Bir anketten al\u0131nan 1-5 aras\u0131 puanlar\nanket_puanlari = [3, 1, 4, 2, 5, 3, 1, 4, 2, 3]\nsirali_puanlar = counting_sort(anket_puanlari)\nprint(f\"S\u0131ralanmam\u0131\u015f Anket Puanlar\u0131: {anket_puanlari}\")\nprint(f\"Counting Sort ile S\u0131ralanm\u0131\u015f Puanlar: {sirali_puanlar}\")\n\n# Vaka Analizi 2: Ya\u015f Da\u011f\u0131l\u0131m\u0131 (k\u00fc\u00e7\u00fck ya\u015f aral\u0131\u011f\u0131)\n# Bir kre\u015fteki \u00e7ocuklar\u0131n ya\u015flar\u0131\ncocuk_yaslari = [3, 2, 4, 3, 1, 2, 4, 3, 2, 1, 5]\nsirali_yaslar = counting_sort(cocuk_yaslari)\nprint(f\"S\u0131ralanmam\u0131\u015f \u00c7ocuk Ya\u015flar\u0131: {cocuk_yaslari}\")\nprint(f\"Counting Sort ile S\u0131ralanm\u0131\u015f Ya\u015flar: {sirali_yaslar}\")\n<\/pre>\n<\/code><\/p>\n Counting Sort, \u00f6zellikle veri aral\u0131\u011f\u0131n\u0131n (k) k\u00fc\u00e7\u00fck oldu\u011fu ve elemanlar\u0131n pozitif tam say\u0131lar oldu\u011fu senaryolarda inan\u0131lmaz derecede h\u0131zl\u0131d\u0131r. Anket sonu\u00e7lar\u0131, ya\u015f da\u011f\u0131l\u0131m\u0131 gibi s\u0131k\u00e7a kar\u015f\u0131la\u015f\u0131lan veri setlerinde bu algoritmay\u0131 kullanarak performanstan \u00f6nemli \u00f6l\u00e7\u00fcde kazan\u00e7 sa\u011flayabilirsiniz.<\/p>\n Radix Sort: En B\u00fcy\u00fck Say\u0131lar\u0131 Bile Nas\u0131l D\u00fczenlersiniz?<\/h2>\nRadix Sort, Counting Sort gibi lineer zamanl\u0131 bir s\u0131ralama algoritmas\u0131d\u0131r ancak \u00e7ok daha b\u00fcy\u00fck say\u0131lar\u0131 veya farkl\u0131 veri tiplerini (\u00f6rne\u011fin, metin dizeleri) s\u0131ralamak i\u00e7in kullan\u0131labilir. Temel fikir, say\u0131lar\u0131 basamak basamak (veya metinleri karakter karakter) ele alarak s\u0131ralamakt\u0131r. En az anlaml\u0131 basamaktan (Least Significant Digit - LSB) ba\u015flay\u0131p en \u00e7ok anlaml\u0131 basama\u011fa (Most Significant Digit - MSB) do\u011fru ilerleyerek, her basamak i\u00e7in kararl\u0131 bir s\u0131ralama algoritmas\u0131 (genellikle Counting Sort) kullan\u0131l\u0131r.<\/p>\n Radix Sort'un \u00e7al\u0131\u015fma prensibi \u015fu \u015fekildedir:<\/p>\n \nEn B\u00fcy\u00fck Eleman\u0131 Bulma:<\/strong> S\u0131ralanacak dizideki en b\u00fcy\u00fck eleman bulunur. Bu, ka\u00e7 basamak (d) \u00fczerinden s\u0131ralama yap\u0131laca\u011f\u0131n\u0131 belirlemek i\u00e7in gereklidir.<\/li>\n Basamak Bazl\u0131 S\u0131ralama:<\/strong> En az anlaml\u0131 basamaktan (birler basama\u011f\u0131) ba\u015flayarak, her basamak i\u00e7in dizi kararl\u0131 bir s\u0131ralama algoritmas\u0131 (\u00f6rn. Counting Sort) kullan\u0131larak s\u0131ralan\u0131r. \u00d6rne\u011fin, birler basama\u011f\u0131na g\u00f6re s\u0131ralama tamamland\u0131ktan sonra, onlar basama\u011f\u0131na ge\u00e7ilir ve bu basama\u011fa g\u00f6re s\u0131ralama yap\u0131l\u0131rken, birler basama\u011f\u0131ndaki s\u0131ralama korunur. Bu i\u015flem, en \u00e7ok anlaml\u0131 basama\u011fa ula\u015f\u0131lana kadar tekrarlan\u0131r.<\/li>\n<\/ol>\nHer basamak i\u00e7in yap\u0131lan s\u0131ralama i\u015fleminin kararl\u0131 olmas\u0131 (yani e\u015fit de\u011ferli elemanlar\u0131n orijinal g\u00f6receli s\u0131ras\u0131n\u0131 korumas\u0131) Radix Sort i\u00e7in hayati \u00f6nem ta\u015f\u0131r. Aksi takdirde, \u00f6nceki basamaklarda yap\u0131lan s\u0131ralama bozulur ve nihai sonu\u00e7 yanl\u0131\u015f olur. Counting Sort'un kararl\u0131 yap\u0131s\u0131 nedeniyle, Radix Sort i\u00e7inde genellikle Counting Sort bir alt rutin olarak kullan\u0131l\u0131r.<\/p>\n Radix Sort'un zaman karma\u015f\u0131kl\u0131\u011f\u0131 O(d * (n + k))'dir; burada 'd' basamak say\u0131s\u0131, 'n' eleman say\u0131s\u0131 ve 'k' her basamaktaki rakam aral\u0131\u011f\u0131d\u0131r (genellikle 10'luk taban i\u00e7in 0-9 aras\u0131 oldu\u011fu i\u00e7in 10'dur). Alan karma\u015f\u0131kl\u0131\u011f\u0131 ise O(n + k)'dir. Counting Sort'a k\u0131yasla 'd' \u00e7arpan\u0131 eklenmesi, \u00e7ok uzun say\u0131lar i\u00e7in performans maliyetini art\u0131rabilir, ancak pratikte 'd' genellikle log(maksimum de\u011fer) ile s\u0131n\u0131rl\u0131d\u0131r ve bu da algoritmay\u0131 hala olduk\u00e7a h\u0131zl\u0131 k\u0131lar.<\/p>\n Radix Sort Ad\u0131m Ad\u0131m \u00d6rnek<\/h3>\nDizi: [170, 45, 75, 90, 802, 24, 2, 66]<\/code><\/p>\n \nBirler Basama\u011f\u0131na G\u00f6re S\u0131ralama (digit = 0):<\/strong>\nCounting Sort kullan\u0131larak:<\/p>\n \n[170<\/u>, 90<\/u>, 802<\/u>, 2<\/u>, 24<\/u>, 45<\/u>, 75<\/u>, 66<\/u>]<\/li>\n Sonu\u00e7: [170, 90, 802, 2, 24, 45, 75, 66]<\/code> (\u00d6rnek: 2 ve 802 ayn\u0131 basama\u011fa sahip, orijinal s\u0131ras\u0131 korunur)<\/li>\n<\/ul>\n<\/li>\n Onlar Basama\u011f\u0131na G\u00f6re S\u0131ralama (digit = 1):<\/strong>\n\u00d6nceki ad\u0131m\u0131n sonucunu kullanarak Counting Sort ile:<\/p>\n \n[80<\/u>2, 2<\/u>, 24<\/u>, 45<\/u>, 66<\/u>, 17<\/u>0, 75<\/u>, 90<\/u>]<\/li>\n Sonu\u00e7: [802, 2, 24, 45, 66, 170, 75, 90]<\/code> (\u00d6rnek: 2 ve 802'deki s\u0131f\u0131rlar ayn\u0131 basama\u011fa sahiptir. \u00d6nceki ad\u0131mdan gelen 802, 2'den \u00f6nceydi, burada da \u00f6nce kal\u0131r.)<\/li>\n<\/ul>\n<\/li>\n Y\u00fczler Basama\u011f\u0131na G\u00f6re S\u0131ralama (digit = 2):<\/strong>\n\u00d6nceki ad\u0131m\u0131n sonucunu kullanarak Counting Sort ile:<\/p>\n \n[2<\/u>, 24<\/u>, 45<\/u>, 66<\/u>, 75<\/u>, 90<\/u>, 1<\/u>70, 8<\/u>02]<\/li>\n Sonu\u00e7: [2, 24, 45, 66, 75, 90, 170, 802]<\/code><\/li>\n<\/ul>\n<\/li>\n<\/ol>\nNihai Sonu\u00e7: [2, 24, 45, 66, 75, 90, 170, 802]<\/code><\/p>\n Python ile Radix Sort Kod \u00d6rne\u011fi<\/h3>\n\n# Counting Sort, Radix Sort'un alt rutini olarak kullan\u0131lacak.\n# Bu fonksiyon, belirli bir basama\u011fa g\u00f6re s\u0131ralama yapar.\ndef counting_sort_for_radix(arr, exp):\n n = len(arr)\n output = [0] * n\n count = [0] * 10 # 0-9 aras\u0131 rakamlar i\u00e7in\n\n # 1. Belirli basama\u011fa g\u00f6re frekans say\u0131m\u0131\n for i in range(n):\n index = (arr[i] \/\/ exp) % 10\n count[index] += 1\n\n # 2. K\u00fcm\u00fclatif toplam hesaplama\n for i in range(1, 10):\n count[i] += count[i-1]\n\n # 3. Elemanlar\u0131 \u00e7\u0131k\u0131\u015f dizisine yerle\u015ftirme (kararl\u0131 bir \u015fekilde)\n i = n - 1\n while i >= 0:\n index = (arr[i] \/\/ exp) % 10\n output[count[index] - 1] = arr[i]\n count[index] -= 1\n i -= 1\n\n # Orijinal diziyi s\u0131ralanm\u0131\u015f diziden kopyala\n for i in range(n):\n arr[i] = output[i]\n\ndef radix_sort(arr):\n # Dizi bo\u015fsa veya tek elemanl\u0131ysa s\u0131ralamaya gerek yok\n if not arr or len(arr) <= 1:\n return arr\n\n # 1. En b\u00fcy\u00fck eleman\u0131 bul (basamak say\u0131s\u0131n\u0131 belirlemek i\u00e7in)\n max_val = max(arr)\n\n # 2. Her basamak i\u00e7in counting_sort_for_radix'i \u00e7a\u011f\u0131r\n # exp (exponent) ile mevcut basama\u011f\u0131 kontrol ediyoruz (1, 10, 100, ...)\n exp = 1\n while max_val \/\/ exp > 0:\n counting_sort_for_radix(arr, exp)\n exp *= 10\n return arr\n\n# Vaka Analizi: Telefon Numaralar\u0131\ntelefon_numaralari = [8345678901, 1234567890, 5678901234, 9012345678, 3456789012]\nsirali_telefon_numaralari = list(telefon_numaralari) # Kopya \u00fczerinde \u00e7al\u0131\u015f\nradix_sort(sirali_telefon_numaralari)\nprint(f\"S\u0131ralanmam\u0131\u015f Telefon Numaralar\u0131: {telefon_numaralari}\")\nprint(f\"Radix Sort ile S\u0131ralanm\u0131\u015f Telefon Numaralar\u0131: {sirali_telefon_numaralari}\")\n\n# Vaka Analizi 2: IP Adresleri (tam say\u0131ya \u00e7evrilmi\u015f hali)\n# \u00d6rnek olarak IP adreslerini tam say\u0131ya \u00e7evirdi\u011fimizi varsayal\u0131m\nip_adresleri_int = [19216811, 10000001, 192168100, 172016001, 10000002]\nsirali_ip_adresleri = list(ip_adresleri_int)\nradix_sort(sirali_ip_adresleri)\nprint(f\"S\u0131ralanmam\u0131\u015f IP Adresleri (int): {ip_adresleri_int}\")\nprint(f\"Radix Sort ile S\u0131ralanm\u0131\u015f IP Adresleri (int): {sirali_ip_adresleri}\")\n<\/pre>\n<\/code><\/p>\n Radix Sort, \u00f6zellikle tam say\u0131 tabanl\u0131 veritaban\u0131 indeksleri, b\u00fcy\u00fck telefon numaras\u0131 listeleri veya IP adresleri gibi basamakl\u0131 yap\u0131ya sahip veri setlerinde olduk\u00e7a etkilidir. Di\u011fer bir deyi\u015fle, elemanlar\u0131n b\u00fcy\u00fckl\u00fc\u011f\u00fcnden ziyade, basamak say\u0131s\u0131n\u0131n belirleyici oldu\u011fu senaryolarda parlak bir \u00e7\u00f6z\u00fcm sunar.<\/p>\n Bucket Sort: Veri Da\u011f\u0131l\u0131m\u0131ndan Nas\u0131l Faydalan\u0131l\u0131r?<\/h2>\nBucket Sort (Kova S\u0131ralamas\u0131), verinin belirli bir aral\u0131kta ve m\u00fcmk\u00fcn oldu\u011funca e\u015fit bir \u015fekilde da\u011f\u0131ld\u0131\u011f\u0131 durumlarda olduk\u00e7a etkili olan, lineer zamanl\u0131 bir s\u0131ralama algoritmas\u0131d\u0131r. Temel prensip, s\u0131ralanacak elemanlar\u0131 belirli say\u0131da \"kovaya\" (bucket) da\u011f\u0131tmak ve her kovay\u0131 ayr\u0131 ayr\u0131 s\u0131ralad\u0131ktan sonra kovalar\u0131 birle\u015ftirerek nihai s\u0131ral\u0131 listeyi olu\u015fturmakt\u0131r.<\/p>\n Algoritma \u015fu ad\u0131mlar\u0131 takip eder:<\/p>\n \nKova Olu\u015fturma:<\/strong> Belirli say\u0131da bo\u015f kova (genellikle bir liste veya dizi listesi) olu\u015fturulur. Kova say\u0131s\u0131, veri setinin b\u00fcy\u00fckl\u00fc\u011f\u00fcne ve da\u011f\u0131l\u0131m\u0131na g\u00f6re belirlenebilir. Genellikle kova say\u0131s\u0131, n (eleman say\u0131s\u0131) veya n'nin karek\u00f6k\u00fc kadar al\u0131n\u0131r.<\/li>\n Elemanlar\u0131 Kovalara Da\u011f\u0131tma:<\/strong> Her eleman, belirlenen bir da\u011f\u0131t\u0131m fonksiyonu (hashing fonksiyonu) kullan\u0131larak uygun kovaya yerle\u015ftirilir. Bu fonksiyon genellikle eleman\u0131n de\u011ferini, kova say\u0131s\u0131yla \u00e7arp\u0131p tam say\u0131 k\u0131sm\u0131n\u0131 alarak hangi kovaya d\u00fc\u015fece\u011fini belirler. \u00d6rne\u011fin, 0-1 aral\u0131\u011f\u0131ndaki ondal\u0131k say\u0131lar i\u00e7in kov_index = int(eleman * kova_say\u0131s\u0131)<\/code> gibi bir form\u00fcl kullan\u0131labilir.<\/li>\n Kovalar\u0131 S\u0131ralama:<\/strong> Her kovan\u0131n i\u00e7indeki elemanlar, ayr\u0131 ayr\u0131 s\u0131ralan\u0131r. Bu ad\u0131mda, genellikle Insertion Sort gibi basit bir s\u0131ralama algoritmas\u0131 kullan\u0131l\u0131r. Kovalar\u0131n i\u00e7indeki eleman say\u0131s\u0131 nispeten k\u00fc\u00e7\u00fck oldu\u011fundan, Insertion Sort'un O(n^2) en k\u00f6t\u00fc durum karma\u015f\u0131kl\u0131\u011f\u0131 burada ciddi bir performans sorununa yol a\u00e7maz. Ancak, bir kova \u00e7ok fazla eleman i\u00e7eriyorsa, bu ad\u0131m Bucket Sort'un genel performans\u0131n\u0131 d\u00fc\u015f\u00fcrebilir.<\/li>\n Kovalar\u0131 Birle\u015ftirme:<\/strong> T\u00fcm kovalar s\u0131raland\u0131ktan sonra, s\u0131ralanm\u0131\u015f kovalar s\u0131rayla birle\u015ftirilerek nihai s\u0131ral\u0131 dizi olu\u015fturulur.<\/li>\n<\/ol>\nBucket Sort'un zaman karma\u015f\u0131kl\u0131\u011f\u0131, veri da\u011f\u0131l\u0131m\u0131na ve kova i\u00e7inde kullan\u0131lan s\u0131ralama algoritmas\u0131na b\u00fcy\u00fck \u00f6l\u00e7\u00fcde ba\u011fl\u0131d\u0131r. Ortalama durumda, e\u011fer veriler kovalar aras\u0131nda e\u015fit da\u011f\u0131lm\u0131\u015fsa, zaman karma\u015f\u0131kl\u0131\u011f\u0131 O(n + k) olabilir; burada 'n' eleman say\u0131s\u0131 ve 'k' kova say\u0131s\u0131d\u0131r. Ancak en k\u00f6t\u00fc durumda, t\u00fcm elemanlar tek bir kovaya d\u00fc\u015ferse, kova i\u00e7indeki s\u0131ralama algoritmas\u0131n\u0131n karma\u015f\u0131kl\u0131\u011f\u0131na (\u00f6rn. Insertion Sort i\u00e7in O(n^2)) d\u00fc\u015febilir. Alan karma\u015f\u0131kl\u0131\u011f\u0131 ise O(n + k)'dir.<\/p>\n Bucket Sort Ad\u0131m Ad\u0131m \u00d6rnek<\/h3>\nDizi: [0.89, 0.45, 0.65, 0.12, 0.69, 0.55, 0.23, 0.78]<\/code> (0-1 aras\u0131 ondal\u0131k say\u0131lar)<\/p>\n Kova Say\u0131s\u0131: 5 (\u00f6rne\u011fin)<\/p>\n \nKova Olu\u015fturma:<\/strong> buckets = [ [], [], [], [], [] ]<\/code> (5 bo\u015f liste)<\/li>\n Elemanlar\u0131 Kovalara Da\u011f\u0131tma:<\/strong> (kova_index = int(eleman * kova_say\u0131s\u0131)<\/code>)\n\n0.89 -> int(0.89 * 5) = 4<\/code> -> buckets[4].append(0.89)<\/code><\/li>\n 0.45 -> int(0.45 * 5) = 2<\/code> -> buckets[2].append(0.45)<\/code><\/li>\n 0.65 -> int(0.65 * 5) = 3<\/code> -> buckets[3].append(0.65)<\/code><\/li>\n 0.12 -> int(0.12 * 5) = 0<\/code> -> buckets[0].append(0.12)<\/code><\/li>\n 0.69 -> int(0.69 * 5) = 3<\/code> -> buckets[3].append(0.69)<\/code><\/li>\n 0.55 -> int(0.55 * 5) = 2<\/code> -> buckets[2].append(0.55)<\/code><\/li>\n 0.23 -> int(0.23 * 5) = 1<\/code> -> buckets[1].append(0.23)<\/code><\/li>\n 0.78 -> int(0.78 * 5) = 3<\/code> -> buckets[3].append(0.78)<\/code><\/li>\n<\/ul>\nKovalar\u0131n durumu:<\/p>\n \nbuckets[0] = [0.12]<\/code><\/li>\n buckets[1] = [0.23]<\/code><\/li>\n buckets[2] = [0.45, 0.55]<\/code><\/li>\n buckets[3] = [0.65, 0.69, 0.78]<\/code><\/li>\n buckets[4] = [0.89]<\/code><\/li>\n<\/ul>\n<\/li>\n Kovalar\u0131 S\u0131ralama:<\/strong> Her kovay\u0131 Insertion Sort ile s\u0131rala\n\nbuckets[0] = [0.12]<\/code> (zaten s\u0131ral\u0131)<\/li>\n buckets[1] = [0.23]<\/code> (zaten s\u0131ral\u0131)<\/li>\n buckets[2] = [0.45, 0.55]<\/code> (zaten s\u0131ral\u0131)<\/li>\n buckets[3] = [0.65, 0.69, 0.78]<\/code> (zaten s\u0131ral\u0131)<\/li>\n buckets[4] = [0.89]<\/code> (zaten s\u0131ral\u0131)<\/li>\n<\/ul>\n<\/li>\n Kovalar\u0131 Birle\u015ftirme:<\/strong> Kovalar\u0131 s\u0131ras\u0131yla birle\u015ftir\n[0.12, 0.23, 0.45, 0.55, 0.65, 0.69, 0.78, 0.89]<\/code><\/p>\n<\/li>\n<\/ol>\n Nihai Sonu\u00e7: [0.12, 0.23, 0.45, 0.55, 0.65, 0.69, 0.78, 0.89]<\/code><\/p>\n Python ile Bucket Sort Kod \u00d6rne\u011fi<\/h3>\n\ndef bucket_sort(arr):\n # Dizi bo\u015fsa veya tek elemanl\u0131ysa s\u0131ralamaya gerek yok\n if not arr or len(arr) <= 1:\n return arr\n\n # Kova say\u0131s\u0131n\u0131 belirle, genellikle eleman say\u0131s\u0131na yak\u0131n bir de\u011fer\n # E\u011fer elemanlar 0-1 aral\u0131\u011f\u0131nda de\u011filse, \u00f6nce normalize etmek gerekebilir.\n # Burada varsay\u0131m, elemanlar\u0131n 0 ile 1 aras\u0131nda olmas\u0131d\u0131r.\n num_buckets = len(arr)\n \n # 1. Kovalar\u0131 olu\u015ftur\n buckets = [[] for _ in range(num_buckets)]\n\n # 2. Elemanlar\u0131 kovalar\u0131na da\u011f\u0131t\n for num in arr:\n # Kovaya yerle\u015ftirme indeksini hesapla\n # num * num_buckets i\u015flemi ile 0-1 aral\u0131\u011f\u0131ndaki say\u0131y\u0131 kova indeksine \u00e7eviriyoruz\n # E\u011fer num == 1 olursa, son kovaya d\u00fc\u015fmesi i\u00e7in min(idx, num_buckets-1) kullan\u0131l\u0131r\n idx = min(int(num * num_buckets), num_buckets - 1)\n buckets[idx].append(num)\n\n # 3. Her kovay\u0131 ayr\u0131 ayr\u0131 s\u0131rala (genellikle Insertion Sort kullan\u0131l\u0131r)\n # Python'\u0131n yerle\u015fik sort'u TimSort kullan\u0131r, bu da k\u00fc\u00e7\u00fck listeler i\u00e7in h\u0131zl\u0131d\u0131r.\n for i in range(num_buckets):\n buckets[i].sort() # Kova i\u00e7indeki elemanlar\u0131 s\u0131rala\n\n # 4. Kovalar\u0131 birle\u015ftir\n sorted_arr = []\n for bucket in buckets:\n sorted_arr.extend(bucket) # Kovalar\u0131n i\u00e7eri\u011fini birle\u015ftir\n\n return sorted_arr\n\n# Vaka Analizi: \u00d6\u011frenci Notlar\u0131 (0.0 - 1.0 aral\u0131\u011f\u0131na normalize edilmi\u015f)\nogrenci_notlari = [0.85, 0.12, 0.95, 0.33, 0.71, 0.48, 0.05, 0.66, 0.29]\nsirali_notlar = bucket_sort(ogrenci_notlari)\nprint(f\"S\u0131ralanmam\u0131\u015f \u00d6\u011frenci Notlar\u0131: {ogrenci_notlari}\")\nprint(f\"Bucket Sort ile S\u0131ralanm\u0131\u015f Notlar: {sirali_notlar}\")\n\n# Vaka Analizi 2: Stok Fiyat De\u011fi\u015fimleri (\u00f6rne\u011fin -1.0 ile 1.0 aras\u0131 bir aral\u0131\u011fa normalize edilmi\u015f)\n# Negatif de\u011ferler i\u00e7in \u00f6nce offset ekleyerek pozitif aral\u0131\u011fa ta\u015f\u0131mak gerekir.\n# Bu \u00f6rnekte yine 0-1 aral\u0131\u011f\u0131 i\u00e7in \u00e7al\u0131\u015f\u0131yoruz.\nfiyat_degisimleri = [0.01, 0.99, 0.50, 0.25, 0.75, 0.11, 0.88, 0.44, 0.66, 0.33]\nsirali_fiyat_degisimleri = bucket_sort(fiyat_degisimleri)\nprint(f\"S\u0131ralanmam\u0131\u015f Fiyat De\u011fi\u015fimleri: {fiyat_degisimleri}\")\nprint(f\"Bucket Sort ile S\u0131ralanm\u0131\u015f Fiyat De\u011fi\u015fimleri: {sirali_fiyat_degisimleri}\")\n<\/pre>\n<\/code><\/p>\n Bucket Sort, \u00f6zellikle ondal\u0131k say\u0131lar veya belirli bir aral\u0131kta tekd\u00fcze da\u011f\u0131lm\u0131\u015f b\u00fcy\u00fck veri setleri i\u00e7in idealdir. \u00d6rne\u011fin, belirli bir s\u0131cakl\u0131k aral\u0131\u011f\u0131ndaki sens\u00f6r verileri veya finansal veriler gibi senaryolarda kullan\u0131ld\u0131\u011f\u0131nda y\u00fcksek performans g\u00f6sterebilir. Verinin da\u011f\u0131l\u0131m\u0131 ne kadar e\u015fitse, bu algoritma o kadar verimli \u00e7al\u0131\u015f\u0131r.<\/p>\n Lineer Zamanl\u0131 S\u0131ralama Algoritmalar\u0131n\u0131n Kar\u015f\u0131la\u015ft\u0131rmas\u0131 ve \u0130leri Seviye \u0130pu\u00e7lar\u0131<\/h2>\n\u00dc\u00e7 ana lineer zamanl\u0131 s\u0131ralama algoritmas\u0131n\u0131 inceledik: Counting Sort, Radix Sort ve Bucket Sort. Her birinin kendine \u00f6zg\u00fc g\u00fc\u00e7l\u00fc y\u00f6nleri ve k\u0131s\u0131tlamalar\u0131 bulunuyor. Peki, ne zaman hangisini tercih etmeliyiz? Bu algoritmalar aras\u0131ndaki farklar\u0131 daha iyi anlamak i\u00e7in bir kar\u015f\u0131la\u015ft\u0131rma tablosu ile ba\u015flayabiliriz:<\/p>\n \n\n\n\n\n\n\n\nAlgoritma<\/th>\n Zaman Karma\u015f\u0131kl\u0131\u011f\u0131<\/th>\n Alan Karma\u015f\u0131kl\u0131\u011f\u0131<\/th>\n Kararl\u0131l\u0131k (Stability)<\/th>\n Uygulama Alan\u0131 \/ K\u0131s\u0131tlama<\/th>\n<\/tr>\n<\/thead>\n Counting Sort<\/strong><\/td>\n O(n + k)<\/td>\n O(n + k)<\/td>\n Evet<\/td>\n K\u00fc\u00e7\u00fck aral\u0131kta (k) pozitif tam say\u0131lar i\u00e7in idealdir. k \u00e7ok b\u00fcy\u00fckse verimsizle\u015fir.<\/td>\n<\/tr>\n Radix Sort<\/strong><\/td>\n O(d * (n + k))<\/td>\n O(n + k)<\/td>\n Evet<\/td>\n B\u00fcy\u00fck tam say\u0131lar veya ayn\u0131 basamak say\u0131s\u0131na sahip veriler i\u00e7in (telefon numaras\u0131, IP adresi). Kararl\u0131 bir alt rutin gerektirir.<\/td>\n<\/tr>\n Bucket Sort<\/strong><\/td>\n Ort. O(n + k), K\u00f6t\u00fc O(n^2)<\/td>\n O(n + k)<\/td>\n Evet (kova i\u00e7i s\u0131ralama kararl\u0131ysa)<\/td>\n E\u015fit da\u011f\u0131lm\u0131\u015f, genellikle ondal\u0131k say\u0131lar i\u00e7in uygundur. Kova say\u0131s\u0131 ve da\u011f\u0131l\u0131m \u00e7ok kritik.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\nNe Zaman Hangi Algoritmay\u0131 Se\u00e7meli?<\/h3>\n\nCounting Sort'u Tercih Edin:<\/strong>\n\nE\u011fer s\u0131ralaman\u0131z gereken elemanlar pozitif tam say\u0131larsa.<\/li>\n Elemanlar\u0131n alabilece\u011fi en b\u00fcy\u00fck de\u011fer (k), eleman say\u0131s\u0131na (n) g\u00f6re \u00e7ok b\u00fcy\u00fck de\u011filse (yani k \u2248 n).<\/li>\n \u00d6rnek: Bir s\u0131navdaki \u00f6\u011frenci puanlar\u0131 (0-100), ya\u015f da\u011f\u0131l\u0131mlar\u0131 (0-150).<\/li>\n<\/ul>\n<\/li>\n Radix Sort'u Tercih Edin:<\/strong>\n\nS\u0131ralaman\u0131z gereken elemanlar tam say\u0131larsa ve de\u011ferleri \u00e7ok b\u00fcy\u00fckse (Counting Sort i\u00e7in 'k' \u00e7ok b\u00fcy\u00fck olurdu).<\/li>\n Elemanlar\u0131n basamak say\u0131s\u0131 (d) nispeten k\u00fc\u00e7\u00fckse.<\/li>\n \u00d6rnek: Uluslararas\u0131 telefon numaralar\u0131, kimlik numaralar\u0131, IP adresleri, uzun hash de\u011ferleri.<\/li>\n Counting Sort'un kararl\u0131 yap\u0131s\u0131ndan faydalanarak daha karma\u015f\u0131k veri yap\u0131lar\u0131n\u0131 (\u00f6rne\u011fin s\u00f6zl\u00fckleri) s\u0131ralayabilir.<\/li>\n<\/ul>\n<\/li>\n Bucket Sort'u Tercih Edin:<\/strong>\n\nS\u0131ralaman\u0131z gereken elemanlar, belirli bir aral\u0131kta tekd\u00fcze (uniform) da\u011f\u0131lm\u0131\u015fsa. Genellikle ondal\u0131k say\u0131lar i\u00e7in idealdir.<\/li>\n E\u015fit da\u011f\u0131l\u0131m varsay\u0131m\u0131 nedeniyle ortalama durumda \u00e7ok h\u0131zl\u0131 olabilir.<\/li>\n \u00d6rnek: 0.0 ile 1.0 aras\u0131ndaki float de\u011ferleri, fiyat aral\u0131klar\u0131, olas\u0131l\u0131k da\u011f\u0131l\u0131mlar\u0131.<\/li>\n Veri da\u011f\u0131l\u0131m\u0131 homojen de\u011filse, baz\u0131 kovalar \u00e7ok kalabal\u0131kla\u015fabilir ve performans d\u00fc\u015f\u00fc\u015f\u00fc ya\u015fanabilir (en k\u00f6t\u00fc durum O(n^2)).<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\u0130leri Seviye \u0130pu\u00e7lar\u0131 ve P\u00fcf Noktalar\u0131:<\/h3>\n\nVeri Normalizasyonu:<\/strong> Bucket Sort gibi algoritmalar genellikle belirli bir aral\u0131kta (\u00f6rn. 0-1) \u00e7al\u0131\u015f\u0131r. E\u011fer verileriniz bu aral\u0131kta de\u011filse, \u00f6nce min-max normalizasyonu gibi y\u00f6ntemlerle onlar\u0131 uygun aral\u0131\u011fa \u00e7ekmeniz gerekebilir. Bu i\u015flem, algoritman\u0131n genel etkinli\u011fini art\u0131rabilir.<\/li>\n Hibrit Yakla\u015f\u0131mlar:<\/strong> Ger\u00e7ek d\u00fcnya senaryolar\u0131nda, tek bir algoritma yerine hibrit yakla\u015f\u0131mlar tercih edilebilir. \u00d6rne\u011fin, \u00e7ok b\u00fcy\u00fck bir veri setini Bucket Sort ile \u00f6n s\u0131ralay\u0131p, her kovan\u0131n i\u00e7indeki k\u00fc\u00e7\u00fck eleman gruplar\u0131n\u0131 Quicksort veya Insertion Sort ile s\u0131ralamak performans\u0131 optimize edebilir.<\/li>\n Haf\u0131za Hiyerar\u015fisi:<\/strong> Lineer zamanl\u0131 algoritmalar genellikle ek alan kullan\u0131r. B\u00fcy\u00fck veri k\u00fcmeleriyle \u00e7al\u0131\u015f\u0131rken, bu ek belle\u011fin CPU \u00f6nbelle\u011fine s\u0131\u011f\u0131p s\u0131\u011fmad\u0131\u011f\u0131 performans a\u00e7\u0131s\u0131ndan kritik olabilir. \u00d6n belle\u011fe uygun eri\u015fim desenleri (locality of reference) olan algoritmalar daha iyi performans g\u00f6sterir. Radix Sort'un basamak basamak ilerlemesi, veri eri\u015fimini daha lokal hale getirebilir.<\/li>\n Kova Say\u0131s\u0131 ve Da\u011f\u0131t\u0131m Fonksiyonu Optimizasyonu (Bucket Sort i\u00e7in):<\/strong> Kova say\u0131s\u0131, performans \u00fczerinde do\u011frudan etkilidir. \u00c7ok az kova, her kovay\u0131 kalabal\u0131kla\u015ft\u0131r\u0131r ve i\u00e7 s\u0131ralaman\u0131n maliyetini art\u0131r\u0131r. \u00c7ok fazla kova ise, kova olu\u015fturma ve birle\u015ftirme maliyetini art\u0131r\u0131r. En uygun kova say\u0131s\u0131, genellikle veri setinin b\u00fcy\u00fckl\u00fc\u011f\u00fcne ve da\u011f\u0131l\u0131m\u0131na ba\u011fl\u0131d\u0131r. Da\u011f\u0131t\u0131m fonksiyonu da, elemanlar\u0131n kovalar aras\u0131nda m\u00fcmk\u00fcn oldu\u011funca e\u015fit da\u011f\u0131lmas\u0131n\u0131 sa\u011flamal\u0131d\u0131r.<\/li>\n<\/ul>\n\n Uzman \u0130pucu: Lineer zamanl\u0131 s\u0131ralama algoritmalar\u0131, \u00f6ncelikle CPU \u00f6nbelleklerinde yer alan verilerle \u00e7al\u0131\u015ft\u0131\u011f\u0131nda maksimum performans\u0131n\u0131 sergiler. Veri boyutunuz \u00f6nbellek limitlerini a\u015ft\u0131\u011f\u0131nda, disk G\/\u00c7 veya ana bellek eri\u015fim s\u00fcreleri, teorik zaman karma\u015f\u0131kl\u0131\u011f\u0131 avantaj\u0131n\u0131 azaltabilir. Bu durumu g\u00f6z \u00f6n\u00fcnde bulundurarak algoritma se\u00e7imi yapmak \u00f6nemlidir.\n<\/div>\nMobil Cihazlarda S\u0131ralama Performans\u0131 \u0130\u00e7in \u0130pu\u00e7lar\u0131<\/h2>\nMobil cihazlar, masa\u00fcst\u00fc bilgisayarlara g\u00f6re daha k\u0131s\u0131tl\u0131 i\u015flem g\u00fcc\u00fc, bellek ve pil \u00f6mr\u00fcne sahiptir. Bu nedenle, mobil uygulamalarda b\u00fcy\u00fck veri k\u00fcmelerini s\u0131ralarken performans optimizasyonu kritik hale gelir. Lineer zamanl\u0131 s\u0131ralama algoritmalar\u0131, do\u011fru kullan\u0131ld\u0131\u011f\u0131nda mobil cihazlarda \u00f6nemli avantajlar sa\u011flayabilir.<\/p>\n Neden Mobil Cihazlarda Performans \u00d6nemli?<\/h3>\n\nKullan\u0131c\u0131 Deneyimi:<\/strong> Yava\u015f s\u0131ralama i\u015flemleri, uygulaman\u0131n donmas\u0131na veya gecikmelere neden olarak k\u00f6t\u00fc bir kullan\u0131c\u0131 deneyimi yarat\u0131r. H\u0131zl\u0131 tepki s\u00fcreleri, kullan\u0131c\u0131 memnuniyetini art\u0131r\u0131r.<\/li>\n Pil \u00d6mr\u00fc:<\/strong> Yo\u011fun CPU kullanan s\u0131ralama i\u015flemleri, pil t\u00fcketimini art\u0131r\u0131r. Verimli algoritmalar pil \u00f6mr\u00fcn\u00fc korumaya yard\u0131mc\u0131 olur.<\/li>\n Bellek K\u0131s\u0131tlamalar\u0131:<\/strong> Mobil cihazlarda bellek s\u0131n\u0131rl\u0131 olabilir. Ekstra alan gerektiren (O(n+k)) algoritmalar\u0131n bellek t\u00fcketimi dikkatle y\u00f6netilmelidir.<\/li>\n<\/ul>\nHTML ve CSS ile S\u0131ralama Sonu\u00e7lar\u0131n\u0131n Mobil Uyumlu G\u00f6sterimi \u0130\u00e7in \u0130pu\u00e7lar\u0131<\/h3>\nS\u0131ralanm\u0131\u015f verileri kullan\u0131c\u0131ya sunarken, \u00f6zellikle tablolar gibi karma\u015f\u0131k yap\u0131lar mobil cihazlarda sorun yaratabilir. Responsive tasar\u0131m prensiplerini uygulayarak bu sorunlar\u0131n \u00fcstesinden gelebilirsiniz.<\/p>\n \nFluid Layouts (Ak\u0131\u015fkan D\u00fczenler):<\/strong> CSS %<\/code> birimleri, vw<\/code>\/vh<\/code> gibi viewport birimleri ve flexbox<\/code> veya CSS Grid<\/code> kullanarak i\u00e7eri\u011fin ekran boyutuna g\u00f6re otomatik olarak ayarlanmas\u0131n\u0131 sa\u011flay\u0131n.<\/li>\n Responsive Tablolar:<\/strong> B\u00fcy\u00fck tablolar mobil ekranlarda ta\u015fabilir. \u0130\u015fte bunu \u00f6nlemek i\u00e7in kullan\u0131lan baz\u0131 yayg\u0131n teknikler:\n\nScrollable Tablolar:<\/strong> Tablonun etraf\u0131na overflow-x: auto;<\/code> \u00f6zelli\u011fine sahip bir kapsay\u0131c\u0131 ekleyerek yatay kayd\u0131rma \u00e7ubu\u011fu olu\u015fturun.<\/li>\n Stacking Columns (S\u00fctunlar\u0131 Alt Alta Y\u0131\u011fma):<\/strong> K\u00fc\u00e7\u00fck ekranlarda tablo ba\u015fl\u0131klar\u0131n\u0131 ve verilerini alt alta y\u0131\u011fmak i\u00e7in CSS Media Queries kullan\u0131n. Her h\u00fccreyi blok element olarak g\u00f6sterip, :before<\/code> ps\u00f6do-eleman\u0131n\u0131 kullanarak s\u00fctun ba\u015fl\u0131\u011f\u0131n\u0131 ekleyebilirsiniz.<\/li>\n Less Important Columns:<\/strong> Daha az \u00f6nemli s\u00fctunlar\u0131 mobil g\u00f6r\u00fcn\u00fcmde gizleyebilir veya varsay\u0131lan olarak daraltabilirsiniz.<\/li>\n<\/ul>\n<\/li>\n Lazy Loading:<\/strong> E\u011fer s\u0131ralanacak veya g\u00f6sterilecek veri \u00e7ok b\u00fcy\u00fckse, t\u00fcm veriyi ayn\u0131 anda y\u00fcklemek yerine, kullan\u0131c\u0131 kayd\u0131rd\u0131k\u00e7a veya ihtiya\u00e7 duyduk\u00e7a veriyi par\u00e7a par\u00e7a y\u00fckleyin. Bu, ba\u015flang\u0131\u00e7 y\u00fckleme s\u00fcresini k\u0131salt\u0131r ve bellek t\u00fcketimini azalt\u0131r.<\/li>\n Minimalist Tasar\u0131m:<\/strong> Mobil cihazlarda karma\u015f\u0131k aray\u00fczler performans\u0131 d\u00fc\u015f\u00fcrebilir. Sade ve odaklanm\u0131\u015f bir tasar\u0131m, hem performans hem de kullan\u0131c\u0131 deneyimi a\u00e7\u0131s\u0131ndan daha iyidir.<\/li>\n<\/ul>\nMobil Uyumlu Responsive Tablo i\u00e7in CSS \u00d6rne\u011fi<\/h3>\n\n

Algoritma<\/th>\n	Zaman Karma\u015f\u0131kl\u0131\u011f\u0131<\/th>\n	Alan Karma\u015f\u0131kl\u0131\u011f\u0131<\/th>\n	Kararl\u0131l\u0131k (Stability)<\/th>\n	Uygulama Alan\u0131 \/ K\u0131s\u0131tlama<\/th>\n<\/tr>\n<\/thead>\n
Counting Sort<\/strong><\/td>\n	O(n + k)<\/td>\n	O(n + k)<\/td>\n	Evet<\/td>\n	K\u00fc\u00e7\u00fck aral\u0131kta (k) pozitif tam say\u0131lar i\u00e7in idealdir. k \u00e7ok b\u00fcy\u00fckse verimsizle\u015fir.<\/td>\n<\/tr>\n
Radix Sort<\/strong><\/td>\n	O(d * (n + k))<\/td>\n	O(n + k)<\/td>\n	Evet<\/td>\n	B\u00fcy\u00fck tam say\u0131lar veya ayn\u0131 basamak say\u0131s\u0131na sahip veriler i\u00e7in (telefon numaras\u0131, IP adresi). Kararl\u0131 bir alt rutin gerektirir.<\/td>\n<\/tr>\n
Bucket Sort<\/strong><\/td>\n	Ort. O(n + k), K\u00f6t\u00fc O(n^2)<\/td>\n	O(n + k)<\/td>\n	Evet (kova i\u00e7i s\u0131ralama kararl\u0131ysa)<\/td>\n	E\u015fit da\u011f\u0131lm\u0131\u015f, genellikle ondal\u0131k say\u0131lar i\u00e7in uygundur. Kova say\u0131s\u0131 ve da\u011f\u0131l\u0131m \u00e7ok kritik.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n Ne Zaman Hangi Algoritmay\u0131 Se\u00e7meli?<\/h3>\n \n Counting Sort'u Tercih Edin:<\/strong>\n \n E\u011fer s\u0131ralaman\u0131z gereken elemanlar pozitif tam say\u0131larsa.<\/li>\n Elemanlar\u0131n alabilece\u011fi en b\u00fcy\u00fck de\u011fer (k), eleman say\u0131s\u0131na (n) g\u00f6re \u00e7ok b\u00fcy\u00fck de\u011filse (yani k \u2248 n).<\/li>\n \u00d6rnek: Bir s\u0131navdaki \u00f6\u011frenci puanlar\u0131 (0-100), ya\u015f da\u011f\u0131l\u0131mlar\u0131 (0-150).<\/li>\n<\/ul>\n<\/li>\n Radix Sort'u Tercih Edin:<\/strong>\n \n S\u0131ralaman\u0131z gereken elemanlar tam say\u0131larsa ve de\u011ferleri \u00e7ok b\u00fcy\u00fckse (Counting Sort i\u00e7in 'k' \u00e7ok b\u00fcy\u00fck olurdu).<\/li>\n Elemanlar\u0131n basamak say\u0131s\u0131 (d) nispeten k\u00fc\u00e7\u00fckse.<\/li>\n \u00d6rnek: Uluslararas\u0131 telefon numaralar\u0131, kimlik numaralar\u0131, IP adresleri, uzun hash de\u011ferleri.<\/li>\n Counting Sort'un kararl\u0131 yap\u0131s\u0131ndan faydalanarak daha karma\u015f\u0131k veri yap\u0131lar\u0131n\u0131 (\u00f6rne\u011fin s\u00f6zl\u00fckleri) s\u0131ralayabilir.<\/li>\n<\/ul>\n<\/li>\n Bucket Sort'u Tercih Edin:<\/strong>\n \n S\u0131ralaman\u0131z gereken elemanlar, belirli bir aral\u0131kta tekd\u00fcze (uniform) da\u011f\u0131lm\u0131\u015fsa. Genellikle ondal\u0131k say\u0131lar i\u00e7in idealdir.<\/li>\n E\u015fit da\u011f\u0131l\u0131m varsay\u0131m\u0131 nedeniyle ortalama durumda \u00e7ok h\u0131zl\u0131 olabilir.<\/li>\n \u00d6rnek: 0.0 ile 1.0 aras\u0131ndaki float de\u011ferleri, fiyat aral\u0131klar\u0131, olas\u0131l\u0131k da\u011f\u0131l\u0131mlar\u0131.<\/li>\n Veri da\u011f\u0131l\u0131m\u0131 homojen de\u011filse, baz\u0131 kovalar \u00e7ok kalabal\u0131kla\u015fabilir ve performans d\u00fc\u015f\u00fc\u015f\u00fc ya\u015fanabilir (en k\u00f6t\u00fc durum O(n^2)).<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n \u0130leri Seviye \u0130pu\u00e7lar\u0131 ve P\u00fcf Noktalar\u0131:<\/h3>\n \n Veri Normalizasyonu:<\/strong> Bucket Sort gibi algoritmalar genellikle belirli bir aral\u0131kta (\u00f6rn. 0-1) \u00e7al\u0131\u015f\u0131r. E\u011fer verileriniz bu aral\u0131kta de\u011filse, \u00f6nce min-max normalizasyonu gibi y\u00f6ntemlerle onlar\u0131 uygun aral\u0131\u011fa \u00e7ekmeniz gerekebilir. Bu i\u015flem, algoritman\u0131n genel etkinli\u011fini art\u0131rabilir.<\/li>\n Hibrit Yakla\u015f\u0131mlar:<\/strong> Ger\u00e7ek d\u00fcnya senaryolar\u0131nda, tek bir algoritma yerine hibrit yakla\u015f\u0131mlar tercih edilebilir. \u00d6rne\u011fin, \u00e7ok b\u00fcy\u00fck bir veri setini Bucket Sort ile \u00f6n s\u0131ralay\u0131p, her kovan\u0131n i\u00e7indeki k\u00fc\u00e7\u00fck eleman gruplar\u0131n\u0131 Quicksort veya Insertion Sort ile s\u0131ralamak performans\u0131 optimize edebilir.<\/li>\n Haf\u0131za Hiyerar\u015fisi:<\/strong> Lineer zamanl\u0131 algoritmalar genellikle ek alan kullan\u0131r. B\u00fcy\u00fck veri k\u00fcmeleriyle \u00e7al\u0131\u015f\u0131rken, bu ek belle\u011fin CPU \u00f6nbelle\u011fine s\u0131\u011f\u0131p s\u0131\u011fmad\u0131\u011f\u0131 performans a\u00e7\u0131s\u0131ndan kritik olabilir. \u00d6n belle\u011fe uygun eri\u015fim desenleri (locality of reference) olan algoritmalar daha iyi performans g\u00f6sterir. Radix Sort'un basamak basamak ilerlemesi, veri eri\u015fimini daha lokal hale getirebilir.<\/li>\n Kova Say\u0131s\u0131 ve Da\u011f\u0131t\u0131m Fonksiyonu Optimizasyonu (Bucket Sort i\u00e7in):<\/strong> Kova say\u0131s\u0131, performans \u00fczerinde do\u011frudan etkilidir. \u00c7ok az kova, her kovay\u0131 kalabal\u0131kla\u015ft\u0131r\u0131r ve i\u00e7 s\u0131ralaman\u0131n maliyetini art\u0131r\u0131r. \u00c7ok fazla kova ise, kova olu\u015fturma ve birle\u015ftirme maliyetini art\u0131r\u0131r. En uygun kova say\u0131s\u0131, genellikle veri setinin b\u00fcy\u00fckl\u00fc\u011f\u00fcne ve da\u011f\u0131l\u0131m\u0131na ba\u011fl\u0131d\u0131r. Da\u011f\u0131t\u0131m fonksiyonu da, elemanlar\u0131n kovalar aras\u0131nda m\u00fcmk\u00fcn oldu\u011funca e\u015fit da\u011f\u0131lmas\u0131n\u0131 sa\u011flamal\u0131d\u0131r.<\/li>\n<\/ul>\n \n Uzman \u0130pucu: Lineer zamanl\u0131 s\u0131ralama algoritmalar\u0131, \u00f6ncelikle CPU \u00f6nbelleklerinde yer alan verilerle \u00e7al\u0131\u015ft\u0131\u011f\u0131nda maksimum performans\u0131n\u0131 sergiler. Veri boyutunuz \u00f6nbellek limitlerini a\u015ft\u0131\u011f\u0131nda, disk G\/\u00c7 veya ana bellek eri\u015fim s\u00fcreleri, teorik zaman karma\u015f\u0131kl\u0131\u011f\u0131 avantaj\u0131n\u0131 azaltabilir. Bu durumu g\u00f6z \u00f6n\u00fcnde bulundurarak algoritma se\u00e7imi yapmak \u00f6nemlidir.\n<\/div>\n Mobil Cihazlarda S\u0131ralama Performans\u0131 \u0130\u00e7in \u0130pu\u00e7lar\u0131<\/h2>\n Mobil cihazlar, masa\u00fcst\u00fc bilgisayarlara g\u00f6re daha k\u0131s\u0131tl\u0131 i\u015flem g\u00fcc\u00fc, bellek ve pil \u00f6mr\u00fcne sahiptir. Bu nedenle, mobil uygulamalarda b\u00fcy\u00fck veri k\u00fcmelerini s\u0131ralarken performans optimizasyonu kritik hale gelir. Lineer zamanl\u0131 s\u0131ralama algoritmalar\u0131, do\u011fru kullan\u0131ld\u0131\u011f\u0131nda mobil cihazlarda \u00f6nemli avantajlar sa\u011flayabilir.<\/p>\n Neden Mobil Cihazlarda Performans \u00d6nemli?<\/h3>\n \n Kullan\u0131c\u0131 Deneyimi:<\/strong> Yava\u015f s\u0131ralama i\u015flemleri, uygulaman\u0131n donmas\u0131na veya gecikmelere neden olarak k\u00f6t\u00fc bir kullan\u0131c\u0131 deneyimi yarat\u0131r. H\u0131zl\u0131 tepki s\u00fcreleri, kullan\u0131c\u0131 memnuniyetini art\u0131r\u0131r.<\/li>\n Pil \u00d6mr\u00fc:<\/strong> Yo\u011fun CPU kullanan s\u0131ralama i\u015flemleri, pil t\u00fcketimini art\u0131r\u0131r. Verimli algoritmalar pil \u00f6mr\u00fcn\u00fc korumaya yard\u0131mc\u0131 olur.<\/li>\n Bellek K\u0131s\u0131tlamalar\u0131:<\/strong> Mobil cihazlarda bellek s\u0131n\u0131rl\u0131 olabilir. Ekstra alan gerektiren (O(n+k)) algoritmalar\u0131n bellek t\u00fcketimi dikkatle y\u00f6netilmelidir.<\/li>\n<\/ul>\n HTML ve CSS ile S\u0131ralama Sonu\u00e7lar\u0131n\u0131n Mobil Uyumlu G\u00f6sterimi \u0130\u00e7in \u0130pu\u00e7lar\u0131<\/h3>\n S\u0131ralanm\u0131\u015f verileri kullan\u0131c\u0131ya sunarken, \u00f6zellikle tablolar gibi karma\u015f\u0131k yap\u0131lar mobil cihazlarda sorun yaratabilir. Responsive tasar\u0131m prensiplerini uygulayarak bu sorunlar\u0131n \u00fcstesinden gelebilirsiniz.<\/p>\n \n Fluid Layouts (Ak\u0131\u015fkan D\u00fczenler):<\/strong> CSS `%<\/code> birimleri, vw<\/code>\/vh<\/code> gibi viewport birimleri ve flexbox<\/code> veya CSS Grid<\/code> kullanarak i\u00e7eri\u011fin ekran boyutuna g\u00f6re otomatik olarak ayarlanmas\u0131n\u0131 sa\u011flay\u0131n.<\/li>\n` Responsive Tablolar:<\/strong> B\u00fcy\u00fck tablolar mobil ekranlarda ta\u015fabilir. \u0130\u015fte bunu \u00f6nlemek i\u00e7in kullan\u0131lan baz\u0131 yayg\u0131n teknikler:\n\nScrollable Tablolar:<\/strong> Tablonun etraf\u0131na overflow-x: auto;<\/code> \u00f6zelli\u011fine sahip bir kapsay\u0131c\u0131 ekleyerek yatay kayd\u0131rma \u00e7ubu\u011fu olu\u015fturun.<\/li>\n Stacking Columns (S\u00fctunlar\u0131 Alt Alta Y\u0131\u011fma):<\/strong> K\u00fc\u00e7\u00fck ekranlarda tablo ba\u015fl\u0131klar\u0131n\u0131 ve verilerini alt alta y\u0131\u011fmak i\u00e7in CSS Media Queries kullan\u0131n. Her h\u00fccreyi blok element olarak g\u00f6sterip, :before<\/code> ps\u00f6do-eleman\u0131n\u0131 kullanarak s\u00fctun ba\u015fl\u0131\u011f\u0131n\u0131 ekleyebilirsiniz.<\/li>\n Less Important Columns:<\/strong> Daha az \u00f6nemli s\u00fctunlar\u0131 mobil g\u00f6r\u00fcn\u00fcmde gizleyebilir veya varsay\u0131lan olarak daraltabilirsiniz.<\/li>\n<\/ul>\n<\/li>\n Lazy Loading:<\/strong> E\u011fer s\u0131ralanacak veya g\u00f6sterilecek veri \u00e7ok b\u00fcy\u00fckse, t\u00fcm veriyi ayn\u0131 anda y\u00fcklemek yerine, kullan\u0131c\u0131 kayd\u0131rd\u0131k\u00e7a veya ihtiya\u00e7 duyduk\u00e7a veriyi par\u00e7a par\u00e7a y\u00fckleyin. Bu, ba\u015flang\u0131\u00e7 y\u00fckleme s\u00fcresini k\u0131salt\u0131r ve bellek t\u00fcketimini azalt\u0131r.<\/li>\n Minimalist Tasar\u0131m:<\/strong> Mobil cihazlarda karma\u015f\u0131k aray\u00fczler performans\u0131 d\u00fc\u015f\u00fcrebilir. Sade ve odaklanm\u0131\u015f bir tasar\u0131m, hem performans hem de kullan\u0131c\u0131 deneyimi a\u00e7\u0131s\u0131ndan daha iyidir.<\/li>\n<\/ul>\nMobil Uyumlu Responsive Tablo i\u00e7in CSS \u00d6rne\u011fi<\/h3>\n\n