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>> [MUSIC PLAYING]

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>> DOUG LLOYD: Siz y蓹qin ki, hesab edir蓹m ki,
code yaln谋z bir tap艧谋r谋q yerin蓹 yetirm蓹k 眉莽眉n istifad蓹 olunur.

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Siz onu yazmaq.

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Bu bir 艧ey yoxdur.

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Y蓹ni, bu, olduqca 莽ox var.

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>> Siz t蓹rtib edir.

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Siz proqram run.

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Siz getm蓹k iyi.

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>> Lakin iman v蓹 ya, 蓹g蓹r
Siz uzun m眉dd蓹t kod

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Siz, h蓹qiq蓹t蓹n, g枚rm蓹k 眉莽眉n g蓹lm蓹k bil蓹r
g枚z蓹l bir 艧ey kimi kodu.

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Bu problem h蓹ll edir
莽ox maraql谋 bir 艧蓹kild蓹,

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v蓹 ya, h蓹qiq蓹t蓹n, yaln谋z bir 艧ey var
g枚r眉n眉r yolu haqq谋nda s蓹liq蓹li.

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Siz laughing ola bil蓹r
M蓹n蓹, ancaq bu do臒ru deyil.

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V蓹 recursion bir yoldur
sort bu fikir almaq 眉莽眉n

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g枚z蓹l, z蓹rif g枚r眉n眉艧l眉 kodu.

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Bu yolla probleml蓹ri h蓹ll ki,
, g枚r眉nt眉l蓹m蓹k 眉莽眉n asan, maraql谋

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v蓹 t蓹蓹cc眉bl眉 q谋sa.

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>> yol recursion i艧l蓹ri
bir recursive funksiyas谋 var

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莽a臒谋r谋r funksiyas谋 kimi m眉蓹yy蓹n edilir
枚z眉 icra hiss蓹si kimi.

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Ki, bir az q蓹rib蓹 g枚r眉n蓹 bil蓹r
v蓹 biz bir az g枚r眉rs眉n眉z

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bu bir anda nec蓹 haqq谋nda.

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Ancaq yen蓹 d蓹, bu
recursive prosedurlar谋 var

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bel蓹 z蓹rif olacaq
onlar olacaq, 莽眉nki

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olmadan bu problemi h蓹ll etm蓹k 眉莽眉n
B眉t眉n bu dig蓹r funksiyalar谋 olan

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v蓹 ya bu uzun loops.

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Siz bu recursive ki, g枚r眉rs眉n眉z
prosedurlar谋 bel蓹 q谋sa baxmaq 眉莽眉n gedir.

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Onlar h蓹qiq蓹t蓹n etm蓹k 眉莽眉n gedir
Sizin code daha 莽ox g枚z蓹l baxmaq.

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>> M蓹n siz蓹 bir n眉mun蓹 verim
Bu nec蓹

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bir recursive proseduru m眉蓹yy蓹n edil蓹 bil蓹r.

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Bu il蓹 tan谋艧 de臒ilseniz bel蓹
ill蓹r 蓹vv蓹l math sinif

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bir 艧ey var deyirl蓹r
ad蓹t蓹n fakt枚ryel funksiyas谋,

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bir 眉nlem kimi qeydi olan
b眉t眉n m眉sb蓹t tam 蓹d蓹dl蓹r 眉z蓹rind蓹 m眉蓹yy蓹n edilir.

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V蓹 yol ki, n
fakt枚ryel hesablan谋r

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Siz b眉t眉n 莽oxaltmaq edilir
daha n枚mr蓹l蓹ri az

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v蓹 ya b蓹rab蓹r n together-- 眉莽眉n
b眉t眉n integers az

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v蓹 ya birlikd蓹 n b蓹rab蓹r.

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>> Bel蓹 ki, 5 fakt枚ryel 5 d蓹f蓹
4 d蓹f蓹 3 d蓹f蓹 2 d蓹f蓹 1.

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And 4 fakt枚ryel 4 d蓹f蓹
3 d蓹f蓹 2 d蓹f蓹 1 v蓹 s.

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Siz fikir almaq.

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>> Proqram莽谋lar kimi, biz deyil
n, 眉nlem istifad蓹 edin.

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Bel蓹likl蓹, biz fakt枚ryel m眉蓹yy蓹n ed蓹c蓹yik
n fakt kimi f蓹aliyy蓹t g枚st蓹rir.

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V蓹 biz yaratmaq 眉莽眉n fakt枚ryel istifad蓹 ed蓹c蓹yik
bir problemin recursive h蓹lli.

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M蓹n tapa bil蓹r edir蓹m
Bu daha 莽ox vizual ki,

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iterativ art谋q m眉raci蓹t
Bu versiyas谋 olan

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biz d蓹 bir anda n蓹z蓹r laz谋md谋r.

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>> Bel蓹 ki, burada bir ne莽蓹
facts-- pun intended--

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haqq谋nda factorial--
fakt枚ryel funksiyas谋.

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Dediyim kimi 1 fakt枚ryel, 1.

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2 fakt枚ryel 2 d蓹f蓹 1.

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3 fakt枚ryel 3
d蓹f蓹 2 bel蓹 d蓹f蓹 1, v蓹.

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Biz art谋q 4 v蓹 5 b蓹hs etdi.

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>> Amma bu baxaraq, bu do臒ru deyil?

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2 fakt枚ryel deyil, yaln谋z
2 d蓹f蓹 1 fakt枚ryel?

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M蓹n dem蓹k, 1 fakt枚ryel 1.

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Bel蓹 ki, niy蓹 biz yaln谋z dem蓹k olmaz ki,
2 fakt枚ryel 2 d蓹f蓹 1-ci ild蓹n,

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Bu, h蓹qiq蓹t蓹n, yaln谋z 2 d蓹f蓹 var
1 fakt枚ryel?

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>> V蓹 sonra, ki, fikir uzanan
3 fakt枚ryel deyil

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yaln谋z 3 d蓹f蓹 2 fakt枚ryel?

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V蓹 4 fakt枚ryel 4 d蓹f蓹
bel蓹 3 v蓹 fakt枚ryel?

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茝slind蓹, fakt枚ryel
h蓹r hans谋 bir say谋 yaln谋z bil蓹rsiniz

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c眉r 蓹g蓹r ifad蓹
蓹b蓹di bu h蓹yata ke莽irir.

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Biz n枚v 眉mumil蓹艧dirm蓹k bil蓹r
fakt枚ryel problem

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bu kimi n d蓹f蓹
n minus 1 fakt枚ryel.

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Bu n d蓹f蓹 m蓹hsul var
b眉t眉n n枚mr蓹l蓹ri m蓹n蓹 az.

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>> Bu fikir, bu
Problemin 眉mumil蓹艧dirilm蓹si,

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Biz蓹 recursively 眉莽眉n imkan verir
fakt枚ryel funksiyas谋 m眉蓹yy蓹n edir.

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Bir funksiyas谋 m眉蓹yy蓹n zaman
recursively var

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Bunun bir hiss蓹si olmaq 眉莽眉n laz谋m olan iki 艧ey.

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Siz bir 艧ey deyil蓹n laz谋md谋r
baza halda, hans谋, siz onu tetiklemek zaman,

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recursive prosesi dayanacaq.

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>> 茝ks halda, bir funksiyas谋 莽a臒谋r谋r
枚z眉 siz imagine-- bil蓹r

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蓹b蓹di davam ed蓹 bil蓹r.

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Function funksiyas谋n谋 莽a臒谋r谋r
funksiyas谋 z蓹ngl蓹r 莽a臒谋r谋r

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funksiyas谋 funksiyas谋 莽a臒谋r谋r.

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Bir yol yoxsa
, proqram dayand谋rmaq 眉莽眉n

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s蓹m蓹r蓹li vurulmu艧 olunacaq
sonsuz loop edir.

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Bu, n蓹tic蓹d蓹 q蓹za ed蓹c蓹k
Bu yadda艧 t枚k眉lm蓹k laz谋md谋r, 莽眉nki.

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Amma bu n枚qt蓹d蓹 ba艧qa var.

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>> Biz dayand谋rmaq 眉莽眉n b蓹zi dig蓹r yol laz谋md谋r
proqram 艧aqq谋lt谋l谋 ba艧qa 艧eyl蓹r

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Yem蓹yini bir proqramd谋r, 莽眉nki
y蓹qin ki, g枚z蓹l v蓹 ya z蓹rif deyil.

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V蓹 bel蓹 ki, biz bu 蓹sas i艧i 莽a臒谋r谋r谋q.

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Bu sad蓹 h蓹ll edir
N枚qt蓹 bir problem

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meydana g蓹l蓹n recursive prosesi.

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Bel蓹 ki, bir hiss蓹si var
bir recursive funksiyas谋.

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>> ikinci hiss蓹si recursive hald谋r.

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Bu harada recursion edir
h蓹qiq蓹t蓹n ke莽iril蓹c蓹k.

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Bu harada
funksiyas谋 枚z眉 z蓹ng ed蓹c蓹k.

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>> M蓹hz 枚z眉n眉 z蓹ng deyil
Eyni 艧蓹kild蓹 bu adlan谋rd谋.

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Bu c眉zi variasiya olacaq
ki, bu problem edir

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bir ufac谋k az ki莽ik h蓹ll etm蓹y蓹 莽al谋艧谋r谋q.

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Amma 眉mumiyy蓹tl蓹 dollar ke莽ir
h蓹lli hiss蓹sini h蓹ll

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x蓹tti a艧a臒谋 f蓹rqli z蓹ng.

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>> Bu g枚r眉n眉艧眉 hans谋
Burada 蓹sas halda kimi?

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Hans谋 kimi bu g枚r眉n眉r bir
bir problem 眉莽眉n sad蓹 h蓹ll?

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Biz factorials bir d蓹st蓹 var,
v蓹 davam ed蓹 bil蓹r

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bel蓹 Us 6, 7, 8, 9, 10, v蓹 gedir.

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>> Amma kimi bu g枚r眉n眉r bir
yax艧谋 hal baza halda olmaq.

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Bu, 莽ox sad蓹 h蓹ll edir.

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Biz x眉susi bir 艧ey yoxdur.

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>> 1 fakt枚ryel yaln谋z 1-dir.

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Biz he莽 n蓹 yoxdur
vurma b眉t眉n.

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Gedirik 蓹g蓹r kimi g枚r眉n眉r
c蓹hd v蓹 bu problemi h蓹ll etm蓹k,

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v蓹 biz dayand谋rmaq laz谋md谋r
haradasa Recursion,

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biz y蓹qin ki, dayand谋rmaq ist蓹yir蓹m
biz 1 almaq zaman.

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Biz bundan 蓹vv蓹l dayand谋rmaq ist蓹mir蓹m.

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>> Biz m眉蓹yy蓹n edirik Bel蓹 ki
Bizim fakt枚ryel funksiyas谋,

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Burada bir skelet 眉莽眉n var
biz bunu nec蓹.

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Biz o iki 艧ey plug laz谋md谋r
baza halda v蓹 recursive halda.

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Baza halda n蓹dir?

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N 1 b蓹rab蓹r deyil, qay谋tmaq 1 var ki
h蓹qiq蓹t蓹n sad蓹 problem h蓹ll etsin.

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>> 1 fakt枚ryel 1.

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Bu 1 d蓹f蓹 bir 艧ey var.

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Bu, sad蓹c蓹 1 var.

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Bu, 莽ox asan faktd谋r.

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V蓹 bel蓹 ki, bizim 蓹sas i艧i ola bil蓹r.

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Biz bu 1 ke莽di almaq
funksiyas谋, biz yaln谋z 1 qay谋tmaq laz谋md谋r.

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>> Recursive n蓹dir
hal y蓹qin ki, kimi baxmaq?

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H蓹r say谋 眉莽眉n
1 Bundan ba艧qa, model n蓹dir?

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Yax艧谋, biz q蓹bul edirsinizs蓹
n fakt枚ryel,

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bu n d蓹f蓹 n fakt枚ryel minus 1.

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>> Biz 3 fakt枚ryel alaraq edirsinizs蓹,
Bu, 3 minus 1 3 d蓹f蓹 fakt枚ryel

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v蓹 ya 2.

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V蓹 biz deyilik 蓹g蓹r
ba艧qa, 1 baxaraq

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qaytar谋lmas谋 n d蓹f蓹
n minus 1 fakt枚ryel.

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Bu olduqca sad蓹.

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>> V蓹 az olan namin蓹
t蓹miz v蓹 kodu daha z蓹rif,

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bilirik ki, biz bir-line loops varsa
v蓹 ya bir-line 艧蓹rti filiallar谋,

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Biz b眉t眉n xilas ed蓹 bil蓹rsiniz
onlar谋n 蓹traf谋nda q谋vr谋m a艧谋rma.

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Bel蓹likl蓹, biz bu bu birl蓹艧dirm蓹k olar.

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Bu eyni var
bu kimi funksionall谋q.

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>> M蓹n yaln谋z buruq 眉z alaraq al谋ram
yaln谋z bir x蓹tt var, 莽眉nki, a艧谋rma

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h蓹min 艧蓹rti filial daxilind蓹.

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Bel蓹 ki, bu eyni davran谋rlar.

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N 1 b蓹rab蓹r deyil, 1 qay谋tmaq.

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茝ks halda n d蓹f蓹 qay谋tmaq
n minus 1 fakt枚ryel.

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Bel蓹 ki, biz ki莽ik problem edirik.

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N 5 kimi ba艧lay谋r, biz olacaq
4 5 d蓹f蓹 fakt枚ryel qay谋tmaq.

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V蓹 biz dan谋艧maq zaman bir d蓹qiq蓹 g枚r眉rs眉n眉z
ba艧qa video z蓹ng y谋臒谋n谋 haqq谋nda

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biz haqq谋nda dan谋艧maq
biz 枚yr蓹nm蓹k laz谋md谋r y谋臒谋n谋 z蓹ng

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m蓹hz bu prosesi i艧l蓹ri niy蓹.

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>> Amma 5 is蓹 fakt枚ryel deyir
5 d蓹f蓹 fakt枚ryel 4 qay谋tmaq v蓹 4

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yax艧谋, OK, dem蓹k gedir, geri
4 d蓹f蓹 3 fakt枚ryel.

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G枚rd眉y眉n眉z kimi, biz ist蓹yirik
sort 1-yax谋nla艧谋r.

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Biz yax谋n 蓹ld蓹 edirik v蓹
ki, baza halda yax谋n.

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>> V蓹 biz baza halda hit bir d蓹f蓹,
蓹vv蓹lki funksiyalar谋 b眉t眉n

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onlar arad谋臒谋n谋z cavab var.

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2 fakt枚ryel qaytar谋lmas谋 deyirdi
2 d蓹f蓹 1 fakt枚ryel.

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Yax艧谋, 1 yekunlar谋 1 fakt枚ryel.

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Faktorial Bel蓹 ki z蓹ng
2, 2 d蓹f蓹 1 qay谋da bil蓹r

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00:07:10,090 --> 00:07:13,980
v蓹 fakt枚ryel ki, geri verm蓹k
Ki, n蓹tic蓹 g枚zl蓹yir 3.

151
00:07:13,980 --> 00:07:17,110
>> V蓹 sonra hesablamaq olar
onun n蓹tic蓹sind蓹 3 d蓹f蓹 2, 6

152
00:07:17,110 --> 00:07:18,907
v蓹 4 fakt枚ryel geri verir.

153
00:07:18,907 --> 00:07:20,740
V蓹 yen蓹, biz var
z蓹ng y谋臒谋n谋 video

154
00:07:20,740 --> 00:07:23,810
bu bir az t蓹svir olunur
陌ndi deyir蓹m daha 莽ox.

155
00:07:23,810 --> 00:07:25,300
Lakin bu deyil.

156
00:07:25,300 --> 00:07:29,300
Bu t蓹k h蓹ll edir
bir s谋ra fakt枚ryel hesablanmas谋.

157
00:07:29,300 --> 00:07:31,527
>> Bu kod yaln谋z d枚rd x蓹tl蓹ri var.

158
00:07:31,527 --> 00:07:32,610
Bu do臒ru, olduqca s蓹rin var?

159
00:07:32,610 --> 00:07:35,480
Bu sexy n枚v眉 var.

160
00:07:35,480 --> 00:07:38,580
>> Bel蓹 ki, 眉mumiyy蓹tl蓹, lakin
h蓹mi艧蓹 bir recursive funksiyas谋

161
00:07:38,580 --> 00:07:41,190
bir bir loop 蓹v蓹z ed蓹 bilm蓹z
qeyri-recursive funksiyas谋.

162
00:07:41,190 --> 00:07:46,100
Bel蓹 ki, burada, yan-yana, iterativ var
fakt枚ryel funksiyas谋 versiyas谋.

163
00:07:46,100 --> 00:07:49,650
Bu hesablamaq, h蓹m d蓹
eyni 艧ey.

164
00:07:49,650 --> 00:07:52,170
>> Onlar h蓹m n fakt枚ryel hesablamaq.

165
00:07:52,170 --> 00:07:54,990
sol version
bunu recursion istifad蓹 edir.

166
00:07:54,990 --> 00:07:58,320
sa臒 version
bunu iteration istifad蓹 edir.

167
00:07:58,320 --> 00:08:02,050
>> V蓹 bildiri艧, biz elan var
tam m蓹hsul d蓹yi艧蓹n.

168
00:08:02,050 --> 00:08:02,940
V蓹 sonra biz loop.

169
00:08:02,940 --> 00:08:06,790
Bel蓹 ki, uzun n kimi, daha 莽ox 0
n ki, m蓹hsul vurulmas谋 saxlamaq

170
00:08:06,790 --> 00:08:09,890
q蓹d蓹r n decrementing
biz m蓹hsul hesablamaq.

171
00:08:09,890 --> 00:08:14,600
Bel蓹 ki, bu iki funksiyalar谋, yen蓹,
eyni 艧ey.

172
00:08:14,600 --> 00:08:19,980
Amma onlar bunu etm蓹yin
tam eyni 艧蓹kild蓹.

173
00:08:19,980 --> 00:08:22,430
>> 陌ndi, bu m眉mk眉nd眉r
daha 莽ox bazas谋

174
00:08:22,430 --> 00:08:25,770
cinay蓹t i艧in蓹 v蓹 ya daha 莽ox
recursive halda, as谋l谋 olaraq

175
00:08:25,770 --> 00:08:27,670
n蓹 sizin funksiyas谋 etm蓹y蓹 莽al谋艧谋r.

176
00:08:27,670 --> 00:08:31,650
Siz m眉tl蓹q yaln谋z m蓹hdud deyil
bir baza halda v蓹 ya bir recursive

177
00:08:31,650 --> 00:08:32,370
halda.

178
00:08:32,370 --> 00:08:35,320
Bir 艧ey bel蓹 bir n眉mun蓹
脟ox bazas谋 hallarda

179
00:08:35,320 --> 00:08:37,830
ola bil蓹r 艧eyl蓹rdir
Fibonacci say谋 ard谋c谋ll谋臒谋.

180
00:08:37,830 --> 00:08:41,549
>> Siz geri bil蓹r
ibtidai m蓹kt蓹b g眉n

181
00:08:41,549 --> 00:08:45,740
Fibonacci ard谋c谋ll谋臒谋 m眉蓹yy蓹n edilir ki,
Bu kimi ilk element 0.

182
00:08:45,740 --> 00:08:46,890
陌kinci element 1.

183
00:08:46,890 --> 00:08:49,230
O h蓹m d蓹 yaln谋z m眉蓹yy蓹n olunur.

184
00:08:49,230 --> 00:08:55,920
>> Sonra h蓹r element m眉蓹yy蓹n edilir
n minus 1 v蓹 n minus 2 c蓹mi kimi.

185
00:08:55,920 --> 00:09:00,330
脺莽眉nc眉 element So
0 plus 1 1 olacaq.

186
00:09:00,330 --> 00:09:03,280
V蓹 sonra d枚rd眉nc眉 element
陌kinci element, 1 olacaq,

187
00:09:03,280 --> 00:09:06,550
plus 眉莽眉nc眉 element, 1.

188
00:09:06,550 --> 00:09:08,507
V蓹 2 olard谋.

189
00:09:08,507 --> 00:09:09,340
V蓹 s v蓹 s.

190
00:09:09,340 --> 00:09:11,680
>> Bel蓹 ki, bu halda, biz iki baza hallarda var.

191
00:09:11,680 --> 00:09:14,850
N 1 b蓹rab蓹r deyil, 0 qay谋tmaq.

192
00:09:14,850 --> 00:09:18,560
N 2 b蓹rab蓹r olduqda, 1 qay谋tmaq.

193
00:09:18,560 --> 00:09:25,930
茝ks halda, n Fibonacci qay谋tmaq
minus 1 plus n minus 2 Fibonacci.

194
00:09:25,930 --> 00:09:27,180
>> Bel蓹 ki, bir 莽ox baza hallarda var.

195
00:09:27,180 --> 00:09:29,271
N蓹 莽ox recursive bar蓹d蓹?

196
00:09:29,271 --> 00:09:31,520
Yax艧谋, bir 艧ey var
Collatz z蓹nn 莽a臒谋r谋b.

197
00:09:31,520 --> 00:09:34,630
M蓹n dem蓹k fikrind蓹 deyil蓹m
Siz ki, n蓹

198
00:09:34,630 --> 00:09:38,170
Bu, h蓹qiq蓹t蓹n, bizim son 莽眉nki
bu video 眉莽眉n problem.

199
00:09:38,170 --> 00:09:43,220
V蓹 bu, bizim h蓹yata var
birlikd蓹 i艧l蓹m蓹k 眉莽眉n.

200
00:09:43,220 --> 00:09:46,760
>> Bel蓹 ki, burada n蓹
Collatz z蓹nn is--

201
00:09:46,760 --> 00:09:48,820
h蓹r m眉sb蓹t tam t蓹tbiq edilir.

202
00:09:48,820 --> 00:09:51,500
V蓹 bu ki, speculates
h蓹mi艧蓹 m眉mk眉n geri almaq 眉莽眉n

203
00:09:51,500 --> 00:09:55,060
1 bu ad谋mlar谋 蓹g蓹r.

204
00:09:55,060 --> 00:09:57,560
N 1 deyil, dayand谋rmaq.

205
00:09:57,560 --> 00:10:00,070
N 1 蓹g蓹r biz 1 geri var.

206
00:10:00,070 --> 00:10:05,670
>> 茝ks halda, bu ke莽ir
proses yenid蓹n n 2 b枚l眉n眉r.

207
00:10:05,670 --> 00:10:08,200
Siz 1 geri ala bil蓹rsiniz, 蓹g蓹r bax谋n.

208
00:10:08,200 --> 00:10:13,260
N t蓹k 茝ks t蓹qdird蓹, ke莽m蓹k
daha 3n plus 1-d蓹 bu proses,

209
00:10:13,260 --> 00:10:15,552
v蓹 ya 3 d蓹f蓹 n plus 1.

210
00:10:15,552 --> 00:10:17,010
Bel蓹 ki, burada biz bir baza halda var.

211
00:10:17,010 --> 00:10:18,430
N 1 b蓹rab蓹r deyil, dayand谋rmaq.

212
00:10:18,430 --> 00:10:20,230
Biz bir daha recursion m蓹艧臒ul deyilik.

213
00:10:20,230 --> 00:10:23,730
>> Amma biz iki recursive hallarda var.

214
00:10:23,730 --> 00:10:28,750
N h蓹tta varsa, biz bir recursive etm蓹k
halda, n 2 b枚l眉n眉r z蓹ng.

215
00:10:28,750 --> 00:10:33,950
N t蓹k deyil, f蓹rqli bir etm蓹k
3 d蓹f蓹 n plus 1-d蓹 recursive halda.

216
00:10:33,950 --> 00:10:39,120
>> V蓹 bu video 眉莽眉n m蓹qs蓹di
, ikinci almaq video fasil蓹,

217
00:10:39,120 --> 00:10:42,440
v蓹 c蓹hd v蓹 bu yazmaq
recursive funksiyas谋 Collatz

218
00:10:42,440 --> 00:10:47,640
harada, d蓹y蓹ri n ke莽ir v蓹
Bu nec蓹 bir 莽ox add谋mlar hesablay谋r

219
00:10:47,640 --> 00:10:52,430
Siz n ba艧lamaq 蓹g蓹r 1 almaq 眉莽眉n
v蓹 yuxar谋da bu ad谋mlar谋 edin.

220
00:10:52,430 --> 00:10:56,660
N 1 olarsa, bu, 0 add谋mlar at谋r.

221
00:10:56,660 --> 00:11:00,190
茝ks halda, bu olacaq
Lakin bir add谋m plus almaq

222
00:11:00,190 --> 00:11:06,200
bu da n qal谋r 莽ox add谋mlar
2 b枚l眉n眉r n h蓹tta, v蓹 ya 3N plus 1, 蓹g蓹r

223
00:11:06,200 --> 00:11:08,100
n t蓹k deyil.

224
00:11:08,100 --> 00:11:11,190
>> 陌ndi m蓹n burada ekranda qoymaq etdik
Sizin 眉莽眉n test 艧eyi bir ne莽蓹,

225
00:11:11,190 --> 00:11:15,690
Sizin 眉莽眉n testl蓹r hallarda bir ne莽蓹 g枚rm蓹k
bu m眉xt蓹lif Collatz n枚mr蓹l蓹ri n蓹,

226
00:11:15,690 --> 00:11:17,440
v蓹 h蓹m莽inin ill眉strasiya
add谋mlar ki,

227
00:11:17,440 --> 00:11:20,390
bel蓹 ki, siz ke莽mi艧dir laz谋md谋r
sort f蓹aliyy蓹t bu prosesi g枚rm蓹k.

228
00:11:20,390 --> 00:11:24,222
N b蓹rab蓹rdir Bel蓹 ki
1, n Collatz 0.

229
00:11:24,222 --> 00:11:26,180
Siz n蓹 yoxdur
bir 艧ey 1 geri almaq 眉莽眉n.

230
00:11:26,180 --> 00:11:27,600
茝g蓹r siz art谋q orada ist蓹yirik.

231
00:11:27,600 --> 00:11:30,550
>> N 2 olarsa, bu, davam edir
bir add谋m 1 almaq 眉莽眉n.

232
00:11:30,550 --> 00:11:31,810
Siz 2 il蓹 ba艧lay谋n.

233
00:11:31,810 --> 00:11:33,100
Yax艧谋, 2 1 b蓹rab蓹r deyil.

234
00:11:33,100 --> 00:11:36,580
Bel蓹 ki, bir add谋m olacaq
plus lakin 莽ox add谋mlar onu

235
00:11:36,580 --> 00:11:38,015
qal谋r n 2 b枚l眉n眉r.

236
00:11:38,015 --> 00:11:41,280

237
00:11:41,280 --> 00:11:42,910
>> 2 b枚l眉n眉r 2 1.

238
00:11:42,910 --> 00:11:47,200
Bel蓹 ki, lakin bir add谋m plus edir
莽ox add谋mlar 1-edir.

239
00:11:47,200 --> 00:11:49,720
1 s谋f谋r add谋mlar at谋r.

240
00:11:49,720 --> 00:11:52,370
>> G枚rd眉y眉n眉z kimi 3, var
bir ne莽蓹 add谋mlar i艧tirak edir.

241
00:11:52,370 --> 00:11:53,590
Siz 3 gedin.

242
00:11:53,590 --> 00:11:56,710
V蓹 sonra getm蓹k
10, 5, 16, 8, 4, 2, 1.

243
00:11:56,710 --> 00:11:58,804
Bu 1 geri almaq 眉莽眉n yeddi add谋mlar at谋r.

244
00:11:58,804 --> 00:12:01,220
G枚rd眉y眉n眉z kimi, var bir
Burada ne莽蓹 dig蓹r test hallarda

245
00:12:01,220 --> 00:12:02,470
proqram test 眉莽眉n.

246
00:12:02,470 --> 00:12:03,970
Bel蓹 ki, yen蓹, video fasil蓹.

247
00:12:03,970 --> 00:12:09,210
M蓹n indi geri jump getm蓹k laz谋md谋r
faktiki prosesi burada n蓹,

248
00:12:09,210 --> 00:12:11,390
bu z蓹nn n蓹.

249
00:12:11,390 --> 00:12:14,140
>> Siz anlamaq bil蓹r bax谋n
n Collatz m眉蓹yy蓹n etm蓹k 眉莽眉n nec蓹

250
00:12:14,140 --> 00:12:19,967
Bu nec蓹 bir 莽ox hesablay谋r ki,
Bu 1 almaq 眉莽眉n add谋mlar.

251
00:12:19,967 --> 00:12:23,050
Bel蓹likl蓹, 眉mid edir蓹m, Siz video durdurulmu艧 var
v蓹 yaln谋z m蓹ni g枚zl蓹yir deyil

252
00:12:23,050 --> 00:12:25,820
Burada siz蓹 cavab verm蓹k.

253
00:12:25,820 --> 00:12:29,120
Amma 蓹g蓹r, yax艧谋,
Burada cavab h蓹r halda var.

254
00:12:29,120 --> 00:12:33,070
>> Bel蓹 ki, burada m眉mk眉n m眉蓹yy蓹n var
Collatz funksiyas谋.

255
00:12:33,070 --> 00:12:35,610
N, 蓹g蓹r baza case--
1 b蓹rab蓹r, biz 0 qay谋tmaq.

256
00:12:35,610 --> 00:12:38,250
H蓹r hans谋 bir da艧谋m谋r
add谋mlar 1 geri almaq 眉莽眉n.

257
00:12:38,250 --> 00:12:42,710
>> 茝ks halda, biz iki recursive cases-- var
h蓹tta n枚mr蓹l蓹ri 眉莽眉n bir v蓹 t蓹k 眉莽眉n.

258
00:12:42,710 --> 00:12:47,164
M蓹n h蓹tta n枚mr蓹l蓹ri 眉莽眉n test yolu
n mod 2 0 b蓹rab蓹rdir yoxlamaq 眉莽眉n edir.

259
00:12:47,164 --> 00:12:49,080
Bu, yen蓹 蓹sas蓹n
sual,

260
00:12:49,080 --> 00:12:54,050
n蓹 mod is-- geri 蓹g蓹r, 蓹g蓹r m蓹n
2 b枚lm蓹k n he莽 bir qal谋q var?

261
00:12:54,050 --> 00:12:55,470
Y蓹ni h蓹tta say谋 olard谋.

262
00:12:55,470 --> 00:13:01,370
>> V蓹 bel蓹 n mod 2 0 b蓹rab蓹rdir 蓹g蓹r
test Bu daha say谋.

263
00:13:01,370 --> 00:13:04,250
茝g蓹r bel蓹dirs蓹, m蓹n 1 qay谋tmaq ist蓹yir蓹m,
bu m眉tl蓹q, 莽眉nki

264
00:13:04,250 --> 00:13:09,270
bir add谋m plus Collatz alaraq
n蓹 say谋 m蓹n蓹 yar谋s谋.

265
00:13:09,270 --> 00:13:13,910
茝ks halda, m蓹n 1 qay谋tmaq ist蓹yir蓹m
plus Collatz 3 d蓹f蓹 n plus 1.

266
00:13:13,910 --> 00:13:16,060
Dig蓹r oldu
recursive add谋m ki, biz

267
00:13:16,060 --> 00:13:19,470
hesablamaq bil蓹r
Add谋mlar say谋 Collatz--

268
00:13:19,470 --> 00:13:22,610
geri almaq 眉莽眉n
1 bir s谋ra verilir.

269
00:13:22,610 --> 00:13:24,610
Bel蓹 ki, 眉mid edir蓹m ki, bu n眉mun蓹
bir az verdi

270
00:13:24,610 --> 00:13:26,620
recursive prosedurlar bir dad.

271
00:13:26,620 --> 00:13:30,220
脺mid edir蓹m ki, kodu hesab edir蓹m
az daha 蓹g蓹r g枚z蓹l h蓹yata

272
00:13:30,220 --> 00:13:32,760
z蓹rif, recursive 艧蓹kild蓹.

273
00:13:32,760 --> 00:13:35,955
H蓹tta 蓹g蓹r Lakin, recursion bir
Buna baxmayaraq, h蓹qiq蓹t蓹n g眉cl眉 bir vasit蓹dir.

274
00:13:35,955 --> 00:13:38,330
V蓹 bel蓹 ki, m眉tl蓹q bir 艧ey
蓹traf谋nda ba艧 almaq 眉莽眉n,

275
00:13:38,330 --> 00:13:41,360
yaratmaq ed蓹 bil蓹rsiniz, 莽眉nki
recursion istifad蓹 olduqca s蓹rin proqramlar谋

276
00:13:41,360 --> 00:13:45,930
ki, 蓹ks halda yazmaq 眉莽眉n m眉r蓹kk蓹b ola bil蓹r
Siz loops v蓹 iteration kullan谋yorsan谋z.

277
00:13:45,930 --> 00:13:46,980
M蓹n Doug Lloyd edir蓹m.

278
00:13:46,980 --> 00:13:48,780
Bu CS50 edir.

279
00:13:48,780 --> 00:13:50,228