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[Powered by Google Translate] [RSA]

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[Rob Bowden] [Tommy MacWilliam] [Chuo Kikuu cha Harvard]

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[Hii ni CS50.] [CS50.TV]

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Hebu tuangalie RSA, algorithm sana kutumika kwa ajili ya encrypting data.

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Algorithms encryption kama Kaisari na ciphers Vigenère ni si salama sana.

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Kwa cipher Kaisari, mshambulizi tu mahitaji ya kujaribu 25 funguo mbalimbali

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kupata ujumbe wa wazi asilia.

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Wakati cipher Vigenère ni salama zaidi kuliko cipher Kaisari

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sababu ya kubwa tafuta nafasi kwa funguo, mara moja mshambulizi

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anajua urefu wa muhimu katika cipher Vigenère,

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ambayo inaweza kuamua kupitia uchambuzi wa ruwaza katika maandishi encrypted,

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Cipher Vigenère si kwamba salama zaidi kuliko cipher Kaisari.

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RSA, kwa upande mwingine, si katika mazingira magumu na mashambulizi kama hii.

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Cipher Kaisari na Vigenère cipher kutumia ufunguo huo

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kwa wote Simba na decrypt ujumbe.

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Hii inafanya mali hizo ciphers algorithms muhimu symmetriska.

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Tatizo la msingi na muhimu algorithms symmetriska

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ni kwamba wao wanategemea moja encrypting na kutuma ujumbe

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na moja ya kupokea na decrypting ujumbe

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kuwa tayari walikubaliana upfront juu ya ufunguo wao wote kutumia.

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Lakini tuna tatizo kidogo ya startup hapa.

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Jinsi gani 2 kompyuta kwamba unataka kuwasiliana kuanzisha muhimu siri kati yao?

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Kama muhimu lazima siri, basi tunahitaji njia encrypt na decrypt muhimu.

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Kama wote tuna ni symmetriska key cryptography

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kisha tumekuwa tu kurudi tatizo moja.

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RSA, kwa upande mwingine, anatumia jozi ya funguo,

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moja kwa encryption na mwingine kwa ajili ya decryption.

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Moja inaitwa muhimu ya umma, na nyingine ni ufunguo binafsi.

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muhimu ya umma hutumiwa encrypt ujumbe.

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Kama unaweza nadhani kwa jina lake, tunaweza kushiriki ufunguo yetu ya umma na

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mtu yeyote tunataka bila kuacha usalama wa ujumbe ambao umefungiwa.

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Ujumbe encrypted kutumia ufunguo wa umma

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inaweza tu decrypted na ufunguo wake sambamba binafsi.

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Wakati unaweza kushiriki muhimu yako ya umma, unapaswa daima kuweka ufunguo wako binafsi siri.

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na tu ufunguo binafsi inaweza kutumika kwa ujumbe decrypt

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ikiwa 2 watumiaji unataka kutuma ujumbe fiche pamoja na RSA

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na kurudi wote watumiaji haja ya kuwa na umma na binafsi zao wenyewe jozi ufunguo.

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Ujumbe kutoka kwa mtumiaji 1 kwa mtumiaji 2

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tu kutumia mtumiaji 2 muhimu ya jozi, na ujumbe kutoka kwa mtumiaji 2 kwa mtumiaji 1

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tu kutumia mtumiaji 1 muhimu ya jozi.

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ukweli kwamba kuna 2 tofauti funguo encrypt na ujumbe decrypt

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hufanya RSA Differentierade muhimu algorithm.

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Hatuna haja ya kutia fumbo muhimu ya umma ili kutuma kwa kompyuta nyingine

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tangu muhimu ni umma anyway.

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Hii ina maana kwamba RSA hana huo startup tatizo

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kama symmetriska algorithms muhimu.

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Hivyo kama mimi unataka kutuma ujumbe kwa kutumia encryption RSA

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Rob, mimi kwanza utahitaji Rob wa umma muhimu.

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Kuzalisha jozi ya funguo, Rob inahitaji kuchukua 2 idadi kubwa ya waziri mkuu.

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Namba hizi zitatumika katika wawili funguo za umma na binafsi,

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lakini muhimu ya umma tu kutumia bidhaa za namba hizi 2,

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si idadi wenyewe.

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Mara nimekuwa encrypted ujumbe kwa kutumia Rob wa umma ufunguo

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Siwezi kutuma ujumbe kwa Rob.

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Kwa ajili ya kompyuta, factoring idadi ni tatizo ngumu.

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muhimu ya umma, kumbuka, kutumika bidhaa za namba 2 ya waziri mkuu.

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Bidhaa hii lazima basi kuwa sababu tu 2,

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ambayo kutokea kwa kuwa idadi kwamba kufanya juu ya ufunguo binafsi.

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Ili decrypt ujumbe, RSA kutumia ufunguo binafsi

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au namba tele pamoja katika mchakato wa kuunda muhimu ya umma.

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Sababu ni vigumu computationally sababu idadi

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kutumika katika muhimu ya umma ndani ya namba 2 kutumika katika ufunguo binafsi

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ni vigumu kwa mshambulizi kufikiri ufunguo binafsi

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kwamba itakuwa muhimu decrypt ujumbe.

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Sasa hebu kwenda katika maelezo ya chini baadhi ngazi ya RSA.

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Hebu kwanza tuone jinsi gani tunaweza kuzalisha jozi ya funguo.

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Kwanza, sisi itabidi 2 idadi ya waziri mkuu.

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Tutamwita nambari hizi 2 p na q.

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Ili kubaini p na q, katika mazoezi tunataka pseudorandomly kuzalisha

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idadi kubwa na kisha kutumia mtihani kwa ajili ya kuamua iwapo au

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wale idadi ni pengine waziri mkuu.

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Tunaweza kuendelea kuzalisha idadi random tena na tena

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mpaka tuna primes 2 kwamba tunaweza kutumia.

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Hapa hebu pick p = 23 na q = 43.

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Kumbuka, katika mazoezi, p na q wanapaswa kuwa kubwa idadi.

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Mbali kama sisi kujua, idadi kubwa, ngumu ni

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ufa ujumbe ambao umefungiwa.

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Lakini pia ni ghali zaidi ujumbe Simba na decrypt.

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Leo ni mara nyingi ilipendekeza kwamba p na q ni angalau 1024 bits,

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ambao unaweka idadi ya kila saa digits zaidi ya 300 decimal.

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Lakini tutaweza kuchukua namba hizi ndogo kwa ajili ya mfano huu.

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Sasa tutaweza kuzidisha p na q pamoja na kupata idadi 3,

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ambayo Tutamwita n.

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Katika kesi yetu, n = 23 * 43, ambayo = 989.

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Tuna n = 989.

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Next tutaweza kuzidisha p - 1 na q - 1

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kupata idadi 4, ambayo Tutamwita m.

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Katika kesi yetu, m = 22 * ​​42, ambayo = 924.

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Tuna m = 924.

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Sasa tutaweza haja e idadi ambayo ni kiasi mkuu m

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na chini ya m.

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Namba mbili ni kiasi mkuu au coprime

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ikiwa tu integer chanya kwamba mgawanyiko wote wawili ni sawasawa 1.

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Kwa maneno mengine, kubwa kawaida kigawanyo ya e na m

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lazima 1.

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Katika mazoezi, ni kawaida kwa e kuwa idadi mkuu 65,537

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muda mrefu kama idadi hii haina kutokea kwa kuwa sababu ya m.

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Kwa funguo wetu, tutaweza kuchukua e = 5

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tangu 5 ni kiasi mkuu 924.

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Hatimaye, tutaweza haja moja zaidi ya namba, ambayo Tutamwita d.

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D lazima baadhi thamani utakaokubalika equation de = 1 (Mod m).

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Hii m Mod kunaashiria tutaweza kutumia kitu kinachoitwa byggelement hesabu.

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Katika hesabu za msimu, mara moja idadi anapata juu kuliko amefungwa baadhi ya juu

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ni litamalizika nyuma karibu na 0.

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saa, kwa mfano, hutumia hesabu byggelement.

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Dakika moja baada ya 1:59, kwa mfano, ni 2:00,

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si 1:60.

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mkono dakika ina kung'ata kwa 0

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baada ya kufikia juu amefungwa ya 60.

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Hivyo, tunaweza kusema 60 ni sawa na 0 (Mod 60)

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na 125 ni sawa na 65 ni sawa na 5 (Mod 60).

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Ufunguo yetu ya umma itakuwa e jozi na n

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ambapo katika kesi hii e ni 5 na n ni 989.

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Ufunguo yetu binafsi itakuwa d jozi na n,

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ambayo katika kesi zetu ni 185 na 989.

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Ona kwamba primes yetu ya awali na p q wala kuonekana

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mahali popote katika funguo yetu binafsi au ya umma.

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Sasa kwa kuwa tuna jozi yetu ya funguo, hebu tuangalie ni jinsi gani tunaweza encrypt

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na decrypt ujumbe.

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Nataka kupeleka ujumbe kwa Rob,

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hivyo yeye atakuwa mmoja wa kuzalisha hii jozi ufunguo.

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Basi mimi itabidi kuuliza Rob kwa ufunguo wake wa umma, ambayo mimi itabidi kutumia

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encrypt ujumbe kutuma kwake.

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Kumbuka, ni kabisa sawa kwa Rob kushiriki ufunguo wake wa umma na mimi.

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Lakini itakuwa ni sawa na kushiriki ufunguo wake binafsi.

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Mimi si kuwa na wazo lolote nini ufunguo wake binafsi ni.

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Tunaweza kuvunja ujumbe wetu m juu katika chunks kadhaa

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ndogo kuliko zote n na kisha encrypt kila chunks hizo.

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Tutaweza encrypt CS50 kamba, ambayo tunaweza kuvunja katika chunks 4,

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moja kwa kila barua.

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Ili kuficha ujumbe wangu, mimi itabidi kubadili kwenye

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baadhi ya aina ya uwakilishi numeric.

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Hebu concatenate maadili ASCII na wahusika katika ujumbe wangu.

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Ili kuficha kupewa ujumbe m

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Mimi itabidi compute c = m e (Mod n).

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Lakini m lazima kuwa ndogo kuliko n,

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au mwingine ujumbe kamili hawezi kuwa walionyesha modulo n.

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Tunaweza kuvunja m juu katika chunks kadhaa, ambayo yote ni ndogo kuliko n,

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na simba kila chunks hizo.

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Encrypting kila chunks hizi, sisi kupata C1 = 67-5 (Mod 989)

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ambayo = 658.

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Kwa chunk pili yetu tuna 83-5 (Mod 989)

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ambayo = 15.

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Kwa chunk yetu ya tatu tuna 53-5 (Mod 989)

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ambayo = 799.

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Na hatimaye, kwa chunk yetu ya mwisho tuna 48-5 (Mod 989)

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ambayo = 975.

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Sasa tunaweza kutuma juu ya maadili haya fiche Rob.

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Hapa kwenda, Rob.

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Wakati ujumbe wetu ni katika ndege, hebu kuchukua kuangalia mwingine

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katika jinsi sisi got kwamba thamani kwa d.

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Idadi yetu d zinahitajika kukidhi 5d = 1 (Mod 924).

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Hii inafanya d inverse multiplicative ya 5 modulo 924.

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Kutokana 2 integers, b, na kupanuliwa Euclidean algorithm

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inaweza kutumika kwa kupata kubwa ya kawaida integers kigawanyo ya hizi 2.

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Itakuwa pia kutupa 2 idadi nyingine, x na y,

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kwamba kukidhi shoka equation + na = kigawanyo kubwa ya kawaida ya b na.

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Jinsi gani hii kutusaidia?

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Naam, kuziba katika e = 5 kwa

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na m = 924 kwa b

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sisi tayari kujua kuwa hizi namba ni coprime.

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Kawaida wao mkuu kigawanyo ni 1.

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Hii inatupa 5x + 924y = 1

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au 5x = 1 - 924y.

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Lakini kama sisi tu huduma ya juu kila kitu modulo 924

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basi tunaweza kuacha - 924y.

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Fikiria nyuma ya saa.

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Kama mkono dakika ni juu ya 1 na kisha hasa masaa 10 kupita,

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tunajua mkono dakika bado kuwa juu ya 1.

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Hapa sisi kuanza saa 1 na kisha wrap kuzunguka mara hasa y,

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hivyo tutaweza bado kuwa katika 1.

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Tuna 5x = 1 (Mod 924).

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Na hapa x hii ni sawa kama sisi d walikuwa wanatafuta kabla,

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hivyo kama sisi kutumia kupanuliwa Euclidean algorithm

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kupata hii x idadi, hiyo ni idadi tunapaswa kutumia kama d wetu.

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Sasa hebu kukimbia kupanuliwa Euclidean algorithm kwa 5 =

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na b = 924.

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Tutaweza kutumia njia inayoitwa njia meza.

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Meza yetu itakuwa na nguzo 4, x, y, d, na k.

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Meza yetu kuanza mbali na safu 2.

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Katika safu ya kwanza tuna 1, 0, basi thamani yetu ya, ambayo ni 5,

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na safu ya pili yetu ni 0, 1, na thamani yetu kwa b, ambayo ni 924.

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thamani ya safu ya 4, k, itakuwa matokeo

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ya kugawa thamani ya d katika mstari hapo juu kwa thamani ya d

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juu ya mstari huo.

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Tuna 5 kugawanywa na 924 ni 0 kwa salio baadhi.

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Hiyo ina maana tuna k = 0.

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Sasa thamani ya kila kiini nyingine itakuwa thamani ya safu kiini 2 hapo juu

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bala thamani ya mstari hapo juu ni mara k.

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Hebu kuanza na d katika mstari 3.

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Tuna 5-924 * 0 = 5.

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Next tuna 0-1 * 0 ambayo ni 0

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na 1-0 * 0 ambayo ni 1.

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Si mbaya sana, hivyo basi hoja juu ya mstari unaofuata.

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Kwanza tunahitaji thamani yetu ya k.

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924 kugawanywa na 5 = 184 na salio baadhi,

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hivyo thamani yetu kwa k ni 184.

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Sasa 924-5 * 184 = 4.

225
00:13:54,000 --> 00:14:05,000
1-0 * 184 ni 1 na 0-1 * 184 ni -184.

226
00:14:05,000 --> 00:14:07,000
Haki zote, hebu kufanya mstari unaofuata.

227
00:14:07,000 --> 00:14:10,000
Thamani yetu ya k itakuwa 1 kwa sababu

228
00:14:10,000 --> 00:14:15,000
5 kugawanywa na 4 = 1 na salio baadhi.

229
00:14:15,000 --> 00:14:17,000
Hebu kujaza katika nguzo nyingine.

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
Na 1-184 * 1 ni 185.

233
00:14:33,000 --> 00:14:35,000
Hebu kuona nini thamani yetu ya pili ya k itakuwa.

234
00:14:35,000 --> 00:14:40,000
Naam, inaonekana kama tuna 4 kugawanywa na 1, ambayo ni 4.

235
00:14:40,000 --> 00:14:43,000
Katika kesi hii ambapo sisi ni kugawa na 1 vile k kwamba ni sawa na

236
00:14:43,000 --> 00:14:50,000
thamani ya d katika mstari hapo juu ina maana kwamba sisi ni kosa na kompyuta yetu.

237
00:14:50,000 --> 00:14:58,000
Tunaweza kuona hapa kwamba tuna x = 185 na y = -1 katika mstari wa mwisho.

238
00:14:58,000 --> 00:15:00,000
Hebu sasa kurudi lengo yetu ya awali.

239
00:15:00,000 --> 00:15:04,000
Sisi alisema kuwa thamani ya x kama matokeo ya kuendesha algorithm

240
00:15:04,000 --> 00:15:08,000
itakuwa inverse multiplicative ya (Mod b).

241
00:15:08,000 --> 00:15:15,000
Hiyo ina maana kwamba ni 185 inverse multiplicative ya 5 (Mod 924)

242
00:15:15,000 --> 00:15:20,000
ambayo ina maana kwamba tuna thamani ya 185 kwa ajili ya d.

243
00:15:20,000 --> 00:15:23,000
ukweli kwamba d = 1 katika mstari wa mwisho

244
00:15:23,000 --> 00:15:26,000
inathibitisha e kwamba alikuwa coprime kwa m.

245
00:15:26,000 --> 00:15:30,000
Kama si 1 kisha sisi ingekuwa kuchukua e mpya.

246
00:15:30,000 --> 00:15:33,000
Sasa hebu angalia kama Rob imepokea ujumbe wangu.

247
00:15:33,000 --> 00:15:35,000
Wakati mtu anatuma ujumbe ambao umefungiwa yangu

248
00:15:35,000 --> 00:15:38,000
muda mrefu kama mimi ve naendelea ufunguo wangu binafsi siri

249
00:15:38,000 --> 00:15:41,000
Mimi ni mmoja tu ambao wanaweza decrypt ujumbe.

250
00:15:41,000 --> 00:15:46,000
Ili decrypt c chunk naweza mahesabu ya ujumbe halisi

251
00:15:46,000 --> 00:15:53,000
ni sawa na chunk d nguvu (Mod n).

252
00:15:53,000 --> 00:15:57,000
Kumbuka kwamba d na n ni kutoka ufunguo wangu binafsi.

253
00:15:57,000 --> 00:16:01,000
Ili kupata ujumbe kamili kutoka chunks yake sisi decrypt kila chunk

254
00:16:01,000 --> 00:16:04,000
na concatenate matokeo.

255
00:16:04,000 --> 00:16:08,000
Hasa jinsi salama ni RSA?

256
00:16:08,000 --> 00:16:10,000
Ukweli ni kwamba, sisi hatujui.

257
00:16:10,000 --> 00:16:14,000
Usalama ni msingi kwa muda gani itachukua mshambulizi kwa ufa ujumbe

258
00:16:14,000 --> 00:16:16,000
umesimbwa kwa RSA.

259
00:16:16,000 --> 00:16:19,000
Kumbuka kwamba mshambulizi ina kupata ufunguo yako ya umma,

260
00:16:19,000 --> 00:16:21,000
ambayo ina wote e na n.

261
00:16:21,000 --> 00:16:26,000
Kama mshambulizi itaweza Factor n ndani yake primes 2, uk na q,

262
00:16:26,000 --> 00:16:30,000
kisha aliweza kukokotoa d kutumia kupanuliwa Euclidean algorithm.

263
00:16:30,000 --> 00:16:35,000
Hii inatoa yake muhimu binafsi, ambayo inaweza kutumika decrypt ujumbe wowote.

264
00:16:35,000 --> 00:16:38,000
Lakini jinsi ya haraka tunaweza Factor integers?

265
00:16:38,000 --> 00:16:41,000
Tena, hatujui.

266
00:16:41,000 --> 00:16:43,000
Hakuna ina kupatikana kwa njia ya haraka ya kufanya hivyo,

267
00:16:43,000 --> 00:16:46,000
ambayo ina maana kwamba aliyopewa kubwa ya kutosha n

268
00:16:46,000 --> 00:16:49,000
itachukua muda mrefu mshambulizi unrealistically

269
00:16:49,000 --> 00:16:51,000
kwa sababu idadi.

270
00:16:51,000 --> 00:16:54,000
Kama mtu umebaini njia ya haraka ya integers factoring

271
00:16:54,000 --> 00:16:57,000
RSA itakuwa kuvunjwa.

272
00:16:57,000 --> 00:17:01,000
Lakini hata kama integer factorization ni inneboende polepole

273
00:17:01,000 --> 00:17:04,000
Algorithm RSA bado anaweza kuwa na baadhi ya flaw katika hilo

274
00:17:04,000 --> 00:17:07,000
ambayo inaruhusu kwa decryption rahisi ya ujumbe.

275
00:17:07,000 --> 00:17:10,000
Hakuna mtu ambaye kupatikana na umebaini flaw vile bado,

276
00:17:10,000 --> 00:17:12,000
lakini hiyo haina maana moja haipo.

277
00:17:12,000 --> 00:17:17,000
Katika nadharia, mtu angeweza kuwa nje kuna kusoma data zote umesimbwa kwa RSA.

278
00:17:17,000 --> 00:17:19,000
Kuna mwingine kidogo ya suala faragha.

279
00:17:19,000 --> 00:17:23,000
Kama Tommy encrypts baadhi ujumbe kwa kutumia ufunguo wangu umma

280
00:17:23,000 --> 00:17:26,000
na mshambulizi encrypts ujumbe huo kutumia ufunguo wangu umma

281
00:17:26,000 --> 00:17:29,000
mshambulizi utaona kwamba ujumbe 2 ni sawa

282
00:17:29,000 --> 00:17:32,000
na hivyo kujua nini Tommy msimbo.

283
00:17:32,000 --> 00:17:36,000
Ili kuzuia hili, ujumbe ni kawaida padded na bits random

284
00:17:36,000 --> 00:17:39,000
kabla ya kuwa encrypted ili kwamba ujumbe huo encrypted

285
00:17:39,000 --> 00:17:44,000
mara nyingi utaangalia tofauti kwa muda mrefu kama padding juu ya ujumbe ni tofauti.

286
00:17:44,000 --> 00:17:47,000
Lakini kumbuka jinsi tuna pasua ujumbe katika chunks

287
00:17:47,000 --> 00:17:50,000
ili kila chunk ni ndogo kuliko n?

288
00:17:50,000 --> 00:17:52,000
Padding chunks ina maana kwamba sisi tupate kuwa na mgawanyiko mambo up

289
00:17:52,000 --> 00:17:57,000
katika chunks hata zaidi tangu chunk padded lazima kuwa ndogo kuliko n.

290
00:17:57,000 --> 00:18:01,000
Encryption na decryption ni kiasi ghali na RSA,

291
00:18:01,000 --> 00:18:05,000
na hivyo wanaohitaji kuvunja ujumbe katika chunks wengi wanaweza kuwa gharama kubwa sana.

292
00:18:05,000 --> 00:18:09,000
Kama kiasi kikubwa cha data mahitaji ya kuwa na encrypted decrypted

293
00:18:09,000 --> 00:18:12,000
tunaweza kuchanganya faida ya algorithms symmetriska ufunguo

294
00:18:12,000 --> 00:18:16,000
na wale wa RSA kupata wawili wa usalama na ufanisi.

295
00:18:16,000 --> 00:18:18,000
Ingawa sisi si kwenda ndani yake hapa,

296
00:18:18,000 --> 00:18:23,000
AES ni muhimu symmetriska algorithm kama Vigenère na Kaisari ciphers

297
00:18:23,000 --> 00:18:25,000
lakini vigumu sana ufa.

298
00:18:25,000 --> 00:18:30,000
Bila shaka, hatuwezi kutumia AES bila kuanzisha pamoja ufunguo wa siri

299
00:18:30,000 --> 00:18:34,000
kati ya mifumo ya 2, na tuliona tatizo na kwamba kabla.

300
00:18:34,000 --> 00:18:40,000
Lakini sasa tunaweza kutumia RSA kuanzisha pamoja ufunguo wa siri kati ya mifumo ya 2.

301
00:18:40,000 --> 00:18:43,000
Tutamwita kompyuta kutuma data mtumaji

302
00:18:43,000 --> 00:18:46,000
na kompyuta kupokea data receiver.

303
00:18:46,000 --> 00:18:49,000
receiver ina RSA jozi ufunguo na zituma

304
00:18:49,000 --> 00:18:51,000
muhimu ya umma kwa mtumaji.

305
00:18:51,000 --> 00:18:54,000
Sender inazalisha muhimu AES,

306
00:18:54,000 --> 00:18:57,000
encrypts kwa ufunguo receiver ya RSA umma,

307
00:18:57,000 --> 00:19:00,000
na zituma muhimu AES kwa mpokeaji.

308
00:19:00,000 --> 00:19:04,000
receiver decrypts ujumbe na ufunguo wake RSA binafsi.

309
00:19:04,000 --> 00:19:09,000
Wote mtumaji na mpokeaji sasa kuwa pamoja AES muhimu kati yao.

310
00:19:09,000 --> 00:19:14,000
AES, ambayo ni kwa kasi zaidi katika encryption na decryption kuliko RSA,

311
00:19:14,000 --> 00:19:18,000
sasa inaweza kutumika encrypt kiasi kikubwa cha data na kuwatuma receiver,

312
00:19:18,000 --> 00:19:21,000
ambao wanaweza decrypt kutumia ufunguo huo.

313
00:19:21,000 --> 00:19:26,000
AES, ambayo ni kwa kasi zaidi katika encryption na decryption kuliko RSA,

314
00:19:26,000 --> 00:19:30,000
sasa inaweza kutumika encrypt kiasi kikubwa cha data na kuwatuma receiver,

315
00:19:30,000 --> 00:19:32,000
ambao wanaweza decrypt kutumia ufunguo huo.

316
00:19:32,000 --> 00:19:36,000
Sisi tu inahitajika RSA kuhamisha muhimu pamoja.

317
00:19:36,000 --> 00:19:40,000
Sisi hakuna tena haja ya kutumia RSA wakati wote.

318
00:19:40,000 --> 00:19:46,000
Inaonekana kama mimi nimepata ujumbe.

319
00:19:46,000 --> 00:19:49,000
Haijalishi kama mtu yeyote kusoma nini juu ya ndege karatasi kabla mimi hawakupata hiyo

320
00:19:49,000 --> 00:19:55,000
kwa sababu mimi nina moja tu na ufunguo binafsi.

321
00:19:55,000 --> 00:19:57,000
Hebu decrypt kila chunks katika ujumbe.

322
00:19:57,000 --> 00:20:07,000
chunk kwanza, 658, sisi kuongeza nguvu d, ambayo ni 185,

323
00:20:07,000 --> 00:20:18,000
Mod n, ambayo ni 989, ni sawa na 67,

324
00:20:18,000 --> 00:20:24,000
ambayo ni C barua katika ASCII.

325
00:20:24,000 --> 00:20:31,000
Sasa, kwenye chunk pili.

326
00:20:31,000 --> 00:20:35,000
chunk pili ana thamani 15,

327
00:20:35,000 --> 00:20:41,000
ambayo sisi kuongeza kwa nguvu 185,

328
00:20:41,000 --> 00:20:51,000
Mod 989, na hii ni sawa na 83

329
00:20:51,000 --> 00:20:57,000
ambayo ni S barua katika ASCII.

330
00:20:57,000 --> 00:21:06,000
Sasa kwa ajili ya chunk ya tatu, ambayo ina thamani 799, sisi kuongeza kwa 185,

331
00:21:06,000 --> 00:21:17,000
Mod 989, na hii ni sawa na 53,

332
00:21:17,000 --> 00:21:24,000
ambayo ni thamani ya tabia 5 katika ASCII.

333
00:21:24,000 --> 00:21:30,000
Sasa kwa ajili ya chunk ya mwisho, ambayo ina thamani 975,

334
00:21:30,000 --> 00:21:41,000
sisi kuongeza kwa 185, 989 Mod,

335
00:21:41,000 --> 00:21:51,000
na hii ni sawa na 48, ambayo ni thamani ya 0 tabia katika ASCII.

336
00:21:51,000 --> 00:21:57,000
Jina langu ni Rob Bowden, na hii ni CS50.

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

338
00:22:06,000 --> 00:22:08,000
RSA wakati wote.

339
00:22:08,000 --> 00:22:14,000
RSA wakati wote. [Kicheko]

340
00:22:14,000 --> 00:22:17,000
Wakati wote.