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[Powered by Google Translate] [RSA]

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[Rob Bowden] [Tommy MacWilliam] [Harvard Universiteti]

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[Bu CS50 edir.] [CS50.TV]

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Nin RSA, data 艦ifreleme 眉莽眉n geni艧 istifad蓹 alqoritm n蓹z蓹r salaq.

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Sezar v蓹 Vigen猫re ciphers kimi Encryption alqoritml蓹r 莽ox t蓹hl眉k蓹siz deyil.

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Olan Sezar parol il蓹, bir sald谋rgan谋n yaln谋z 25 m眉xt蓹lif d眉ym蓹l蓹ri c蓹hd laz谋md谋r

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mesaj d眉z m蓹tn almaq 眉莽眉n.

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The Vigen猫re parol the Caesar parol art谋q t蓹hl眉k蓹siz olsa

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莽眉nki d眉ym蓹l蓹ri 眉莽眉n b枚y眉k axtar谋艧 alan, bir d蓹f蓹, bir sald谋rgan谋n

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bir Vigen猫re parol ild蓹 蓹sas uzunlu臒u bilir

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olan 艧ifreli m蓹tn n眉mun蓹l蓹ri t蓹hlil vasit蓹sil蓹 m眉蓹yy蓹n edil蓹 bil蓹r

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the Vigen猫re cipher daha t蓹hl眉k蓹siz Caesar parol 莽ox deyil.

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RSA, dig蓹r t蓹r蓹fd蓹n, bu kimi h眉cumlara h蓹ssas deyil.

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Bu Caesar parol v蓹 Vigen猫re parol eyni d眉ym蓹sind蓹n istifad蓹

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encrypt v蓹 decrypt mesaj h蓹m d蓹.

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Bu 蓹mlak bu ciphers simmetrik 蓹sas alqoritml蓹ri edir.

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Simmetrik 蓹sas alqoritml蓹ri olan 蓹sas problem

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onlar mesaj 艦ifreleme v蓹 g枚nd蓹rilm蓹si bir etibar ki,

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v蓹 bir q蓹bul v蓹 mesaj decrypting

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art谋q onlar h蓹m istifad蓹 ed蓹c蓹k 蓹sas haqq谋nda upfront raz谋l谋q var.

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Amma burada bir ba艧lan臒谋c problem bir q蓹d蓹r var.

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Nec蓹 眉nsiyy蓹t etm蓹k ist蓹yir蓹m ki, 2 komp眉ter aras谋nda bir gizli 蓹sas yaratmaq edirsiniz?

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茝sas gizli olmal谋d谋r, onda biz 蓹sas 艧ifrelemek v蓹 decrypt 眉莽眉n bir yol laz谋md谋r.

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Biz b眉t眉n simmetrik 蓹sas Kriptoqrafiya deyil

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sonra biz yaln谋z eyni problem qay谋da etdik.

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RSA, dig蓹r t蓹r蓹fd蓹n, d眉ym蓹l蓹ri bir c眉t istifad蓹

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parol 莽枚zme 眉莽眉n 艧ifr蓹l蓹m蓹 v蓹 dig蓹r 眉莽眉n.

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Bir ictimai 蓹sas adlan谋r v蓹 dig蓹r x眉susi a莽ar谋 edir.

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Ictimai 蓹sas messages 艧ifrelemek 眉莽眉n istifad蓹 olunur.

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Onun ad谋 il蓹 tapmaq kimi, biz bizim ictimai 蓹sas pay谋 ed蓹 bil蓹rsiniz

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biz 艧ifr蓹li mesaj t蓹hl眉k蓹sizlik 枚d眉n olmadan ist蓹diyiniz h蓹r k蓹s.

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Mesajlar谋m ictimai a莽ar谋 istifad蓹 艧ifrelenir

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yaln谋z m眉vafiq x眉susi a莽ar il蓹 艧ifresi 莽枚z眉len bil蓹r.

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Siz ictimai 蓹sas pay谋 ed蓹 bil蓹rsiniz baxmayaraq, siz h蓹mi艧蓹 x眉susi a莽ar谋 gizli saxlamaq laz谋md谋r.

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v蓹 yaln谋z x眉susi a莽ar谋 decrypt mesaj 眉莽眉n istifad蓹 edil蓹 bil蓹r

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2 istifad蓹莽i RSA il蓹 mesaj 艧ifr蓹li g枚nd蓹rm蓹k ist蓹yirsinizs蓹

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geri v蓹 ir蓹li d蓹 istifad蓹莽il蓹r 枚z d枚vl蓹t v蓹 枚z蓹l 蓹sas c眉t laz谋md谋r.

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Istifad蓹莽i 1 istifad蓹莽i 2 Mesajlar谋m

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yaln谋z user 2 istifad蓹莽i 1 istifad蓹莽i 2 蓹sas c眉t, mesaj istifad蓹

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yaln谋z istifad蓹莽i 1 蓹sas c眉t istifad蓹 edin.

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2 ayr谋-ayr谋 艧ifrelemek d眉ym蓹l蓹ri v蓹 decrypt mesaj var ki,

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RSA bir asimmetrik a莽ar alqoritmi edir.

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Biz ba艧qa bir komp眉ter g枚nd蓹rm蓹k 眉莽眉n ictimai 蓹sas 艧ifrelemek ehtiyac yoxdur

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蓹sas h蓹r halda ictimai g枚st蓹rir.

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Bu RSA eyni ba艧lan臒谋c problem yoxdur o dem蓹kdir ki,

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bu simmetrik 蓹sas alqoritml蓹ri kimi.

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M蓹n RSA 艧ifreleme istifad蓹 ed蓹r蓹k mesaj g枚nd蓹rm蓹k ist蓹diyiniz Bel蓹 ki, 蓹g蓹r

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Rob, m蓹n ilk Rob ictimai 蓹sas laz谋md谋r.

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D眉ym蓹l蓹ri bir c眉t yaratmaq 眉莽眉n, Rob 2 b枚y眉k ba艧 n枚mr蓹l蓹ri y谋臒maq laz谋md谋r.

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Bu n枚mr蓹l蓹r, h蓹m d蓹 d枚vl蓹t v蓹 枚z蓹l a莽arlar谋 istifad蓹 olunacaq

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lakin ictimai 蓹sas yaln谋z bu 2 蓹d蓹d m蓹hsul istifad蓹 ed蓹c蓹k

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deyil n枚mr蓹l蓹ri 枚zl蓹ri.

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M蓹n Rob ictimai 蓹sas istifad蓹 ed蓹r蓹k mesaj 艧ifrelenir sonra

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M蓹n Rob mesaj g枚nd蓹r蓹 bil蓹rsiniz.

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Bir komp眉ter 眉莽眉n, faktorinq 蓹d蓹d a臒谋r problemdir.

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Ictimai 蓹sas, xat谋rlay谋ram, 2 ba艧 n枚mr蓹l蓹ri m蓹hsul istifad蓹.

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Bu m蓹hsul, sonra yaln谋z 2 amill蓹r olmal谋d谋r

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枚z蓹l 蓹sas t蓹艧kil ed蓹n n枚mr蓹l蓹ri olmaq 眉莽眉n ba艧 verir.

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Decrypt mesaj 眉莽眉n, RSA bu x眉susi a莽ar谋 istifad蓹 ed蓹c蓹k

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v蓹 ya 蓹d蓹d ictimai 蓹sas yarad谋lmas谋 prosesi birlikd蓹 vurulur.

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Bu say谋 Factor computationally 莽蓹tindir, 莽眉nki

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枚z蓹l 蓹sas istifad蓹 2 蓹d蓹d bir ictimai 蓹sas istifad蓹

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Bir sald谋rgan谋n 枚z蓹l 蓹sas anlamaq 眉莽眉n 莽蓹tin

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decrypt mesaj laz谋m olacaq.

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陌ndi RSA b蓹zi a艧a臒谋 t蓹f蓹rr眉ata varmaq bildirin.

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G蓹lin ilk biz d眉ym蓹l蓹ri bir c眉t yarada nec蓹.

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Birincisi, 2 ba艧 n枚mr蓹l蓹ri laz谋md谋r.

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Biz bu 2 蓹d蓹d p v蓹 q z蓹ng ed蓹c蓹yik.

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Praktikada p v蓹 q, se莽in 眉莽眉n biz pseudorandomly yaradacaq

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sonra b枚y眉k sayda v蓹 m眉蓹yy蓹n bir test istifad蓹 v蓹 ya

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bu 蓹d蓹d y蓹qin ki, ba艧 var.

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Biz yen蓹 眉z蓹rind蓹 t蓹sad眉fi n枚mr蓹l蓹ri yaradan saxlaya bil蓹rsiniz

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biz istifad蓹 ed蓹 bil蓹rsiniz ki, 2 primes q蓹d蓹r.

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Burada p = 23 q = 43 se莽in bildirin.

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Praktikada, saxla, p v蓹 q daha n枚mr蓹l蓹ri olmal谋d谋r.

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Bizd蓹 kimi, 蓹d蓹d b枚y眉k, daha bu

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艧ifreli mesaj谋 莽at.

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Lakin bu da 艧ifrelemek v蓹 decrypt mesaj daha bahal谋.

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Bu g眉n tez-tez p v蓹 q az谋 1024 bit olan t枚vsiy蓹,

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olan 300-d蓹n art谋q decimal r蓹q蓹m h蓹r say谋 qoyur.

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Amma bu m蓹s蓹l蓹n bu ki莽ik n枚mr蓹l蓹ri y谋臒maq laz谋md谋r.

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陌ndi, 3-c眉 n枚mr蓹 almaq 眉莽眉n birlikd蓹 p v蓹 q 莽oxaltmaq olacaq

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biz n z蓹ng bil蓹rsiniz.

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Bizim halda, n = 23 989 = olan * 43.

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Biz = 989 n.

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Q il蓹 - 1 - Next biz p 莽oxaltmaq olacaq 1

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biz m arayaca冒媒m olan 4-c眉 say谋. almaq

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Bizim halda, m = 22 924 = olan * 42.

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Biz m = 924 var.

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陌ndi nisb蓹t蓹n ba艧 ki, bir s谋ra e laz谋md谋r m

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v蓹 m-d蓹n azd谋r.

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Iki 蓹d蓹d nisb蓹t蓹n ba艧 v蓹 ya coprime var

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h蓹m b蓹rab蓹r onlara ay谋ran yaln谋z m眉sb蓹t tam 1 olsun.

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E v蓹 m Ba艧qa s枚zl蓹, 蓹n b枚y眉k ortaq b枚l蓹n

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1 olmal谋d谋r.

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T蓹cr眉b蓹d蓹, bu, ba艧 say谋 65537 olmaq e yayg谋n

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kimi uzun kimi bu r蓹q蓹m m amil olmaz.

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Bizim d眉ym蓹l蓹ri, biz se莽ece臒iz e = 5

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5 ild蓹n 924 il蓹 nisb蓹t蓹n ba艧 deyil.

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N蓹hay蓹t, biz d arayaca冒媒m daha bir say谋, laz谋md谋r.

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D t蓹nlik cavab ver蓹n b蓹zi d蓹y蓹ri olmal谋d谋r de = 1 (mod m).

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Bu mod m biz modul hesab deyil蓹n bir 艧ey istifad蓹 ed蓹c蓹yik bildirir.

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Modul hesab, bir bir s谋ra b蓹zi yuxar谋 ba臒l谋 daha y眉ks蓹k olur

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o 0 蓹traf谋nda geri paketi olacaq.

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A saat, m蓹s蓹l蓹n, modul hesab edir.

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1:59 sonra bir d蓹qiq蓹, m蓹s蓹l蓹n, 2:00 edir

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1:60 deyil.

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Bu d蓹qiq蓹 t蓹r蓹fd蓹n 0 蓹traf谋nda b眉k眉lm眉艧 edib

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60 ba臒l谋 bir 眉st 莽atd谋qda.

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Bel蓹 ki, 60 0 (mod 60) b蓹rab蓹rdir dem蓹k olar

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v蓹 125 65 b蓹rab蓹rdir 5 (mod 60) b蓹rab蓹rdir.

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Bizim ictimai 蓹sas c眉t e v蓹 n olacaq

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bu halda e 5 v蓹 n 989 edir.

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Bizim x眉susi d眉ym蓹si, c眉t d v蓹 n olacaq

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bizim halda olan 185 v蓹 989-dir.

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Orijinal primes p v蓹 q g枚r眉nm眉r ed蓹k ki,

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h蓹r yerd蓹 bizim 枚z蓹l v蓹 ya ictimai d眉ym蓹l蓹ri il蓹.

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陌ndi a莽arlar谋 bizim c眉t ki, biz 艧ifrelemek nec蓹 n蓹z蓹r ed蓹k

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v蓹 decrypt bir mesaj.

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M蓹n, Rob bir mesaj g枚nd蓹rm蓹k ist蓹yir蓹m

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o bu 蓹sas c眉t yaratmaq 眉莽眉n biri olacaq.

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Sonra m蓹n istifad蓹 ed蓹c蓹yik onun ictimai d眉ym蓹si 眉莽眉n Rob isteyece臒iz

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ona g枚nd蓹rm蓹k 眉莽眉n mesaj 艧ifrelemek.

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Rob m蓹n蓹 onun ictimai 蓹sas b枚l眉艧m蓹k 眉莽眉n Unutmay谋n, tamamil蓹 OK.

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Amma 枚z 艧蓹xsi 蓹sas b枚l眉艧m蓹k tamam olmaz.

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M蓹n onun x眉susi a莽ar谋 n蓹 he莽 bir fikir yoxdur.

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Biz bir ne莽蓹 chunks bizim mesaj m q蓹d蓹r q谋ra bil蓹r

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b眉t眉n sonra n daha ki莽ik v蓹 o chunks h蓹r 艧ifrelemek.

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Biz 4 chunks q蓹d蓹r q谋ra bil蓹r string CS50, 艧ifrelemek laz谋md谋r

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m蓹ktub ba艧谋na bir.

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M蓹nim mesaj 艧ifrelemek 眉莽眉n, m蓹n 莽evirm蓹k laz谋md谋r

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r蓹q蓹mli n眉may蓹nd蓹liyinin bir n枚v.

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Nin m蓹nim mesaj simvol il蓹 ASCII d蓹y蓹rl蓹r concatenate ed蓹k.

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Bir mesaj m 艧ifrelemek 眉莽眉n

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M蓹n e (mod n) c = m hesablamaq laz谋md谋r.

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Amma m, n daha ki莽ik olmal谋d谋r

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v蓹 ya ba艧qa tam mesaj modulo n ifad蓹 edil蓹 bilm蓹z.

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Biz n daha ki莽ik olan bir ne莽蓹 chunks, daxil m q蓹d蓹r q谋ra bil蓹r

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v蓹 bu chunks h蓹r 艧ifrelemek.

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Bu chunks h蓹r 艦ifreleme, biz almaq c1 olan 5 = 67 (MOD 989)

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hans谋 = 658.

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Ikinci y谋臒谋n biz 5 (MOD 989) 83 m枚vcut

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hans谋 = 15.

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脺莽眉nc眉 y谋臒谋n biz 5 (MOD 989) 53 m枚vcut

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hans谋 = 799.

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V蓹 n蓹hay蓹t, sonuncu y谋臒谋n biz 5 (MOD 989) 48 m枚vcut

173
00:09:39,000 --> 00:09:43,000
olan 975 =.

174
00:09:43,000 --> 00:09:48,000
陌ndi Rob bu 艧ifr蓹li d蓹y蓹rl蓹r g枚nd蓹r蓹 bil蓹rsiniz.

175
00:09:54,000 --> 00:09:58,000
Burada, Rob gedin.

176
00:09:58,000 --> 00:10:01,000
Bizim mesaj u莽u艧 olsa, g蓹lin bir n蓹z蓹r ed蓹k

177
00:10:01,000 --> 00:10:07,000
nec蓹 ki, biz d 眉莽眉n d蓹y蓹ri var.

178
00:10:07,000 --> 00:10:17,000
Bizim say谋 d 5d = 1 (MOD 924) t蓹min etm蓹k laz谋md谋r.

179
00:10:17,000 --> 00:10:24,000
Bu d 5 modulo 924 v蓹 multiplikativ t蓹rs edir.

180
00:10:24,000 --> 00:10:28,000
2 integers, bir v蓹 b, uzad谋lm谋艧 Evklid alqoritmi n蓹z蓹r蓹 alaraq

181
00:10:28,000 --> 00:10:33,000
Bu 2 integers 蓹n b枚y眉k ortaq b枚l蓹n tapmaq 眉莽眉n istifad蓹 edil蓹 bil蓹r.

182
00:10:33,000 --> 00:10:37,000
Bu da bizim dig蓹r 2 蓹d蓹d, x v蓹 y, ver蓹c蓹k

183
00:10:37,000 --> 00:10:47,000
bir v蓹 b = b枚y眉k ortaq b枚l蓹n t蓹r蓹find蓹n t蓹nlik balta + qane.

184
00:10:47,000 --> 00:10:49,000
Bu biz蓹 k枚m蓹k edir?

185
00:10:49,000 --> 00:10:52,000
Yax艧谋, bir 眉莽眉n e = 5 sayede

186
00:10:52,000 --> 00:10:56,000
v蓹 b m = 924

187
00:10:56,000 --> 00:10:59,000
biz art谋q bu r蓹q蓹ml蓹r coprime oldu臒unu bilir蓹m.

188
00:10:59,000 --> 00:11:03,000
Onlar谋n b枚y眉k ortaq b枚l蓹n 1-dir.

189
00:11:03,000 --> 00:11:09,000
Bu + 924y = 1 Bookmark 5x verir

190
00:11:09,000 --> 00:11:17,000
v蓹 ya 5x = 1 - 924y.

191
00:11:17,000 --> 00:11:22,000
Amma biz yaln谋z h蓹r 艧ey modulo 924 qay臒谋 蓹g蓹r

192
00:11:22,000 --> 00:11:25,000
924y - onda biz a莽谋l谋r.

193
00:11:25,000 --> 00:11:27,000
Saat geri d眉艧眉n眉n.

194
00:11:27,000 --> 00:11:31,000
D蓹qiq蓹 蓹l 1 v蓹 sonra tam 10 saat, 枚t眉rm蓹k edin

195
00:11:31,000 --> 00:11:35,000
biz d蓹qiq蓹 t蓹r蓹fd蓹n h蓹l蓹 1 olacaq bilirik.

196
00:11:35,000 --> 00:11:39,000
Burada, 1-d蓹 ba艧layacaq v蓹 sonra d蓹qiq y d蓹f蓹 蓹traf谋nda kesmek

197
00:11:39,000 --> 00:11:41,000
biz h蓹l蓹 1 olacaq.

198
00:11:41,000 --> 00:11:49,000
Biz = 1 (MOD 924) 5x var.

199
00:11:49,000 --> 00:11:55,000
V蓹 burada x, biz 蓹vv蓹l arad谋臒谋n谋z d eyni

200
00:11:55,000 --> 00:11:58,000
Biz uzun Evklid alqoritmi istifad蓹 蓹g蓹r

201
00:11:58,000 --> 00:12:04,000
bu say谋 x almaq ki, biz d kimi istifad蓹 etm蓹lidir say谋 var.

202
00:12:04,000 --> 00:12:07,000
陌ndi bir = 5 uzad谋lm谋艧 Evklid alqoritmi run bildirin

203
00:12:07,000 --> 00:12:11,000
v蓹 b = 924.

204
00:12:11,000 --> 00:12:14,000
Biz masa metodu deyil蓹n bir 眉sul istifad蓹 ed蓹c蓹yik.

205
00:12:14,000 --> 00:12:21,000
Bizim masa 4 s眉tun, x, y, d, v蓹 k olacaq.

206
00:12:21,000 --> 00:12:23,000
Bizim c蓹dv蓹l 2 sat谋r il蓹 off ba艧lay谋r.

207
00:12:23,000 --> 00:12:28,000
Ilk s谋rada biz sonra 1, 0, 5 olan bizim d蓹y蓹ri var,

208
00:12:28,000 --> 00:12:37,000
v蓹 ikinci s谋ra 0, 1 v蓹 b 眉莽眉n d蓹y蓹ri olan 924-dir.

209
00:12:37,000 --> 00:12:40,000
4-c眉 s眉tun, k, d蓹y蓹ri n蓹tic蓹 olacaq

210
00:12:40,000 --> 00:12:45,000
d d蓹y蓹ri il蓹 yuxar谋da s谋ras谋nda d d蓹y蓹ri ay谋r谋c谋 il蓹

211
00:12:45,000 --> 00:12:49,000
eyni s谋rada.

212
00:12:49,000 --> 00:12:56,000
Biz 924 b枚l眉n眉r 5 b蓹zi qalan 0 var.

213
00:12:56,000 --> 00:12:59,000
Y蓹ni = 0 k var dem蓹kdir.

214
00:12:59,000 --> 00:13:05,000
陌ndi h蓹r bir dig蓹r mobil d蓹y蓹ri yuxar谋da mobil 2 sat谋r d蓹y蓹ri olacaq

215
00:13:05,000 --> 00:13:09,000
bu d蓹f蓹 k yuxar谋da s谋ra m蓹nfi d蓹y蓹r.

216
00:13:09,000 --> 00:13:11,000
3-c眉 s谋ra d ba艧lamaq ed蓹k.

217
00:13:11,000 --> 00:13:19,000
Biz 5 var - 924 * 0 = 5.

218
00:13:19,000 --> 00:13:25,000
0 olan 1 * 0 - 0 biz var Sonraki

219
00:13:25,000 --> 00:13:30,000
v蓹 1 - 0 * 0, 1-dir.

220
00:13:30,000 --> 00:13:33,000
脟ox pis deyil, bel蓹 ki, bu v蓹 n枚vb蓹ti s谋raya h蓹r蓹k蓹t ed蓹k.

221
00:13:33,000 --> 00:13:36,000
陌lk k bizim d蓹y蓹r laz谋md谋r.

222
00:13:36,000 --> 00:13:43,000
924, b蓹zi qalan 5 = 184 b枚l眉n眉r

223
00:13:43,000 --> 00:13:46,000
bel蓹 k 眉莽眉n d蓹y蓹r 184-dir.

224
00:13:46,000 --> 00:13:54,000
陌ndi 924 - 5 * 184 = 4.

225
00:13:54,000 --> 00:14:05,000
1 - 0 * 184 1 v蓹 0 - 1 * 184 -184 edir.

226
00:14:05,000 --> 00:14:07,000
B眉t眉n h眉quqlar, n枚vb蓹ti s谋ra n蓹 ed蓹k.

227
00:14:07,000 --> 00:14:10,000
K Bizim d蓹y蓹ri 1 莽眉nki olacaq

228
00:14:10,000 --> 00:14:15,000
5 b蓹zi qalan 4 = 1 b枚l眉n眉r.

229
00:14:15,000 --> 00:14:17,000
Nin dig蓹r s眉tunlar doldurmaq ed蓹k.

230
00:14:17,000 --> 00:14:21,000
5 - 4 * 1 = 1.

231
00:14:21,000 --> 00:14:25,000
0 - 1 * 1 = -1.

232
00:14:25,000 --> 00:14:33,000
V蓹 1 - 184 * 1 185-dir.

233
00:14:33,000 --> 00:14:35,000
K n枚vb蓹ti d蓹y蓹r ola bil蓹r n蓹 ed蓹k.

234
00:14:35,000 --> 00:14:40,000
Biz 4 olan 1, 4 b枚l眉n眉r var kimi B蓹li, g枚r眉n眉r.

235
00:14:40,000 --> 00:14:43,000
Biz 1-ay谋r谋c谋 oldu臒unuz Bu halda, bel蓹 ki, k b蓹rab蓹rdir

236
00:14:43,000 --> 00:14:50,000
Yuxar谋da s谋ra d d蓹y蓹ri bizim alqoritmi il蓹 tamamlay谋n dem蓹kdir.

237
00:14:50,000 --> 00:14:58,000
Biz son s谋rada x = 185 v蓹 y = -1 ki, burada g枚r蓹 bil蓹rsiniz.

238
00:14:58,000 --> 00:15:00,000
Indi bizim orijinal m蓹qs蓹di qay谋tmaq ed蓹k.

239
00:15:00,000 --> 00:15:04,000
Biz n蓹tic蓹sind蓹 x d蓹y蓹ri bu alqoritm 莽al谋艧an bildirib ki,

240
00:15:04,000 --> 00:15:08,000
a (mod b) multiplikativ t蓹rs olard谋.

241
00:15:08,000 --> 00:15:15,000
Bu 185 5 multiplikativ t蓹rs (MOD 924) dem蓹kdir

242
00:15:15,000 --> 00:15:20,000
olan biz d 185 bir d蓹y蓹ri dem蓹kdir.

243
00:15:20,000 --> 00:15:23,000
Olmas谋 d = 1 son s谋rada

244
00:15:23,000 --> 00:15:26,000
e m coprime edilib do臒rular.

245
00:15:26,000 --> 00:15:30,000
1 deyil 茝g蓹r biz yeni bir e se莽in olard谋.

246
00:15:30,000 --> 00:15:33,000
陌ndi Rob m蓹nim mesaj谋 蓹g蓹r in g枚r蓹k.

247
00:15:33,000 --> 00:15:35,000
Kims蓹 m蓹n蓹 bir 艧ifr蓹li mesaj g枚nd蓹rir zaman

248
00:15:35,000 --> 00:15:38,000
kimi uzun M蓹n x眉susi a莽ar谋 gizli saxlan谋l谋r oldu臒unuz kimi

249
00:15:38,000 --> 00:15:41,000
M蓹n kim mesaj decrypt bil蓹r yaln谋z bir deyil蓹m.

250
00:15:41,000 --> 00:15:46,000
Decrypt bir y谋臒谋n c M蓹n orijinal mesaj谋 hesablamaq olar

251
00:15:46,000 --> 00:15:53,000
d g眉c眉 (mod n) 眉莽眉n y谋臒谋n b蓹rab蓹rdir.

252
00:15:53,000 --> 00:15:57,000
D v蓹 n m蓹nim x眉susi a莽ar谋 var unutmay谋n.

253
00:15:57,000 --> 00:16:01,000
Biz decrypt h蓹r y谋臒谋n onun chunks bir tam mesaj almaq 眉莽眉n

254
00:16:01,000 --> 00:16:04,000
v蓹 n蓹tic蓹l蓹ri concatenate.

255
00:16:04,000 --> 00:16:08,000
RSA M蓹hz nec蓹 t蓹hl眉k蓹siz?

256
00:16:08,000 --> 00:16:10,000
H蓹qiq蓹t, biz bilmirik.

257
00:16:10,000 --> 00:16:14,000
T蓹hl眉k蓹sizlik bir mesaj 莽at bir sald谋rgan谋n almaq nec蓹 uzun 蓹saslan谋r

258
00:16:14,000 --> 00:16:16,000
RSA il蓹 艧ifrelenir.

259
00:16:16,000 --> 00:16:19,000
Bir sald谋rgan谋n ictimai 蓹sas 莽谋x谋艧谋 var ki, saxla,

260
00:16:19,000 --> 00:16:21,000
e v蓹 n d蓹 var.

261
00:16:21,000 --> 00:16:26,000
T蓹cav眉zkar谋n 枚z 2 primes, p v蓹 q, daxil n amil idar蓹 edin

262
00:16:26,000 --> 00:16:30,000
sonra o, uzun Evklid alqoritmi istifad蓹 d hesablamaq bil蓹r.

263
00:16:30,000 --> 00:16:35,000
Bu onun h蓹r hans谋 bir mesaj谋 decrypt 眉莽眉n istifad蓹 edil蓹 bil蓹r olan 枚z蓹l 蓹sas verir.

264
00:16:35,000 --> 00:16:38,000
Amma nec蓹 tez biz integers amil ola bil蓹r?

265
00:16:38,000 --> 00:16:41,000
Yen蓹 bilmir蓹m.

266
00:16:41,000 --> 00:16:43,000
He莽 k蓹s bunu bir s眉r蓹tli yol tapd谋

267
00:16:43,000 --> 00:16:46,000
hans谋 verilmi艧 o dem蓹kdir ki, kifay蓹t q蓹d蓹r b枚y眉k n

268
00:16:46,000 --> 00:16:49,000
bu unrealistically uzun bir sald谋rgan谋n ed蓹c蓹k

269
00:16:49,000 --> 00:16:51,000
say谋 amil 眉莽眉n.

270
00:16:51,000 --> 00:16:54,000
Kims蓹 faktorinq integers bir s眉r蓹tli yol a艧kar edin

271
00:16:54,000 --> 00:16:57,000
RSA s谋n谋q olard谋.

272
00:16:57,000 --> 00:17:01,000
Lakin, h蓹tta tam factorization mahiyy蓹t yava艧

273
00:17:01,000 --> 00:17:04,000
RSA alqoritmi, h蓹l蓹 d蓹 b蓹zi q眉sur ola bil蓹r

274
00:17:04,000 --> 00:17:07,000
mesajlar asan parol 莽枚zme 眉莽眉n imkan verir.

275
00:17:07,000 --> 00:17:10,000
He莽 k蓹s, h蓹l蓹 bel蓹 bir q眉sur a艧kar v蓹 a艧kar

276
00:17:10,000 --> 00:17:12,000
lakin bir m枚vcud deyil dem蓹k deyil.

277
00:17:12,000 --> 00:17:17,000
N蓹z蓹ri olaraq, kims蓹 RSA il蓹 艧ifrelenir b眉t眉n m蓹lumatlar谋 oxumaq orada ola bil蓹r.

278
00:17:17,000 --> 00:17:19,000
Bir gizlilik m蓹s蓹l蓹nin ba艧qa bit var.

279
00:17:19,000 --> 00:17:23,000
Tommy m蓹nim ictimai a莽ar谋 istifad蓹 bir mesaj 艧ifreleyerek edin

280
00:17:23,000 --> 00:17:26,000
v蓹 t蓹cav眉zkar m蓹nim ictimai 蓹sas istifad蓹 ed蓹r蓹k eyni mesaj 艧ifreleyerek

281
00:17:26,000 --> 00:17:29,000
t蓹cav眉zkar谋n 2 mesaj eyni oldu臒unu g枚r蓹c蓹ksiniz

282
00:17:29,000 --> 00:17:32,000
v蓹 bel蓹likl蓹 Tommy 艧ifrelenir bilirik.

283
00:17:32,000 --> 00:17:36,000
Bu qar艧谋s谋n谋 almaq 眉莽眉n, mesajlar ad蓹t蓹n t蓹sad眉fi bit il蓹 padded olunur

284
00:17:36,000 --> 00:17:39,000
eyni mesaj 艧ifr蓹l蓹nir ki 艧ifrelenir 蓹vv蓹l

285
00:17:39,000 --> 00:17:44,000
Mesaj 眉zr蓹 padding f蓹rqli olaraq bird蓹n 莽ox uzun kimi m眉xt蓹lif g枚r眉n眉r.

286
00:17:44,000 --> 00:17:47,000
Amma biz chunks daxil messages split nec蓹 yadda

287
00:17:47,000 --> 00:17:50,000
h蓹r y谋臒谋n n daha ki莽ik ki?

288
00:17:50,000 --> 00:17:52,000
Bu chunks padding biz 艧eyi split ola bil蓹r o dem蓹kdir ki,

289
00:17:52,000 --> 00:17:57,000
ild蓹n daha chunks daxil padded y谋臒谋n n daha ki莽ik olmal谋d谋r.

290
00:17:57,000 --> 00:18:01,000
艦ifr蓹l蓹m蓹 v蓹 parol 莽枚zme, RSA il蓹 nisb蓹t蓹n bahal谋

291
00:18:01,000 --> 00:18:05,000
v蓹 bir 莽ox chunks bir mesaj par莽alamaq ehtiyac 莽ox bahal谋 ola bil蓹r.

292
00:18:05,000 --> 00:18:09,000
M蓹lumatlar谋n b枚y眉k h蓹cmd蓹 艧ifr蓹li olmal谋d谋r v蓹 艧ifresi 莽枚z眉len edin

293
00:18:09,000 --> 00:18:12,000
biz simmetrik 蓹sas alqoritml蓹ri faydalar谋 birl蓹艧dir蓹 bil蓹r

294
00:18:12,000 --> 00:18:16,000
RSA bu t蓹hl眉k蓹sizlik v蓹 s蓹m蓹r蓹lilik h蓹m almaq 眉莽眉n.

295
00:18:16,000 --> 00:18:18,000
Biz burada daxil deyil baxmayaraq,

296
00:18:18,000 --> 00:18:23,000
AES bu Vigen猫re v蓹 Caesar ciphers kimi bir simmetrik 蓹sas alqoritm edir

297
00:18:23,000 --> 00:18:25,000
amma 莽ox 莽蓹tindir 莽at.

298
00:18:25,000 --> 00:18:30,000
茝lb蓹tt蓹 ki, biz ortaq bir gizli 蓹sas yaratmadan AES istifad蓹 ed蓹

299
00:18:30,000 --> 00:18:34,000
2-sisteml蓹ri aras谋nda, biz 蓹vv蓹l problem g枚rd眉m.

300
00:18:34,000 --> 00:18:40,000
Amma indi biz 2-sisteml蓹ri aras谋nda payla艧谋lan gizli 蓹sas yaratmaq RSA istifad蓹 ed蓹 bil蓹rsiniz.

301
00:18:40,000 --> 00:18:43,000
Biz data g枚nd蓹r蓹n g枚nd蓹rilm蓹si komp眉ter arayaca冒媒m

302
00:18:43,000 --> 00:18:46,000
v蓹 komp眉ter data al谋c谋 al谋nmas谋.

303
00:18:46,000 --> 00:18:49,000
Al谋c谋 bir RSA 蓹sas c眉t v蓹 g枚nd蓹rir

304
00:18:49,000 --> 00:18:51,000
g枚nd蓹r蓹n ictimai d眉ym蓹si.

305
00:18:51,000 --> 00:18:54,000
G枚nd蓹r蓹n bir AES 蓹sas yarad谋r

306
00:18:54,000 --> 00:18:57,000
al谋c谋-nin RSA ictimai 蓹sas il蓹 艧ifreleyerek,

307
00:18:57,000 --> 00:19:00,000
v蓹 al谋c谋 眉莽眉n AES 蓹sas g枚nd蓹rir.

308
00:19:00,000 --> 00:19:04,000
Al谋c谋 枚z RSA 艦蓹xsi d眉ym蓹sind蓹n il蓹 mesaj 艧ifrini a莽谋r.

309
00:19:04,000 --> 00:19:09,000
G枚nd蓹r蓹n v蓹 al谋c谋, h蓹m d蓹 indi onlar谋n aras谋nda ortaq bir AES 蓹sas var.

310
00:19:09,000 --> 00:19:14,000
RSA 莽ox 艧ifr蓹l蓹m蓹 v蓹 parol 莽枚zme da 莽ox daha s眉r蓹tli olan AES,

311
00:19:14,000 --> 00:19:18,000
indi m蓹lumatlar谋n b枚y眉k h蓹cmd蓹 艧ifrelemek v蓹 al谋c谋 onlar谋 g枚nd蓹rm蓹k 眉莽眉n istifad蓹 edil蓹 bil蓹r,

312
00:19:18,000 --> 00:19:21,000
decrypt h蓹min 蓹sas olan istifad蓹 ed蓹 bil蓹rsiniz.

313
00:19:21,000 --> 00:19:26,000
RSA 莽ox 艧ifr蓹l蓹m蓹 v蓹 parol 莽枚zme da 莽ox daha s眉r蓹tli olan AES,

314
00:19:26,000 --> 00:19:30,000
indi m蓹lumatlar谋n b枚y眉k h蓹cmd蓹 艧ifrelemek v蓹 al谋c谋 onlar谋 g枚nd蓹rm蓹k 眉莽眉n istifad蓹 edil蓹 bil蓹r,

315
00:19:30,000 --> 00:19:32,000
decrypt h蓹min 蓹sas olan istifad蓹 ed蓹 bil蓹rsiniz.

316
00:19:32,000 --> 00:19:36,000
Biz yaln谋z payla艧谋lan 蓹sas transfer RSA laz谋md谋r.

317
00:19:36,000 --> 00:19:40,000
Biz art谋q b眉t眉n RSA istifad蓹 etm蓹k laz谋md谋r.

318
00:19:40,000 --> 00:19:46,000
M蓹n bir mesaj var kimi g枚r眉n眉r.

319
00:19:46,000 --> 00:19:49,000
H蓹r ka臒谋z t蓹yyar蓹 var n蓹 oxumaq 蓹g蓹r m蓹n tutdu 蓹vv蓹l etm蓹z

320
00:19:49,000 --> 00:19:55,000
M蓹n x眉susi a莽ar il蓹 yaln谋z bir deyil蓹m, 莽眉nki.

321
00:19:55,000 --> 00:19:57,000
Iletideki chunks h蓹r decrypt ed蓹k.

322
00:19:57,000 --> 00:20:07,000
Ilk y谋臒谋n, 658, biz 185 olan d enerji il蓹 qald谋rmaq

323
00:20:07,000 --> 00:20:18,000
989 olan mod n, 67-蓹 b蓹rab蓹rdir

324
00:20:18,000 --> 00:20:24,000
ASCII-ci m蓹ktub C-dir.

325
00:20:24,000 --> 00:20:31,000
陌ndi, ikinci y谋臒谋n daxil.

326
00:20:31,000 --> 00:20:35,000
陌kinci y谋臒谋n, d蓹y蓹ri 15 var

327
00:20:35,000 --> 00:20:41,000
biz 185th hakimiyy蓹t蓹 qald谋rmaq olan

328
00:20:41,000 --> 00:20:51,000
mod 989, v蓹 bu 83-蓹 b蓹rab蓹rdir

329
00:20:51,000 --> 00:20:57,000
ASCII-ci m蓹ktub S olan.

330
00:20:57,000 --> 00:21:06,000
陌ndi d蓹y蓹ri 799 olan 眉莽眉nc眉 y谋臒谋n 眉莽眉n, biz 185 qald谋rmaq

331
00:21:06,000 --> 00:21:17,000
mod 989, v蓹 bu 53-蓹 b蓹rab蓹rdir,

332
00:21:17,000 --> 00:21:24,000
ASCII-ci karakter 5 d蓹y蓹ri olan.

333
00:21:24,000 --> 00:21:30,000
陌ndi son y谋臒谋n 眉莽眉n olan d蓹y蓹ri 975 var

334
00:21:30,000 --> 00:21:41,000
biz 185 mod 989, qald谋rmaq

335
00:21:41,000 --> 00:21:51,000
v蓹 bu ASCII ild蓹 karakter 0 d蓹y蓹ri olan 48, b蓹rab蓹rdir.

336
00:21:51,000 --> 00:21:57,000
My name Rob Bowden, bu CS50 edir.

337
00:21:57,000 --> 00:22:00,000
[CS50.TV]

338
00:22:06,000 --> 00:22:08,000
B眉t眉n RSA.

339
00:22:08,000 --> 00:22:14,000
B眉t眉n RSA. [G眉l眉艧]

340
00:22:14,000 --> 00:22:17,000
B眉t眉n.