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[Powered by Google Translate] [RSA]

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[Rob Bowden] [Tommy MacWilliam] [Inivèsite Harvard]

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[Sa a se CS50.] [CS50.TV]

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Se pou nou pran yon gade nan RSA, yon algorithm lajman itilize pou chifreman done.

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Algoritm chifreman tankou Seza tande kòz ak Vigenère Algorithms yo pa twò an sekirite.

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Avèk Seza tande kòz kalkile a, yon atakè sèlman bezwen eseye 25 kle diferan

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jwenn plenn tèks mesaj la.

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Pandan ke kalkile nan Vigenère se plis sekirite pase Seza tande kòz kalkile nan

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paske nan espas ki la rechèch pi gwo pou kle, yon fwa yon atakè

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konnen longè nan kle a nan yon kalkile Vigenère,

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ki ka detèmine atravè yon analiz de pwensip ki nan tèks la chiffres,

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kalkile nan Vigenère se pa sa ki pi plis an sekirite pase Seza tande kòz kalkile a.

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RSA, sou la lòt bò, se pa vilnerab a atak tankou sa a.

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Kalkile nan Seza tande kòz ak Vigenère kalkile itilize kle a menm

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a tou de ankripte ak dekriptaj yon mesaj.

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Pwopriyete sa a fè sa yo algoritm kle Algorithms simetrik.

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Yon pwoblèm fondamantal ak algoritm simetrik kle

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se yo ke yo konte sou yon sèl nan chifreman ak voye mesaj la

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ak yon sèl la ap resevwa ak dechifre mesaj la

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te deja te dakò inicio sou kle a yo pral tou de itilize.

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Men, nou gen yon ti jan nan yon pwoblèm demaraj isit la.

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Kijan 2 òdinatè yo ki vle kominike etabli yon kle sekrè ant yo?

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Si kle a dwe sekrè a, Lè sa a, nou bezwen yon fason yo ankripte ak dekriptaj kle a.

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Si tout sa nou genyen se pou chifreman simetrik kle

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Lè sa a, nou te jis tounen vin jwenn menm pwoblèm nan.

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RSA, sou lòt men an, sèvi ak yon pè kle,

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yonn pou chifreman e yon lòt pou dekripte.

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Youn nan yo rele kle piblik la, ak lòt la se kle a prive.

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Se kle a piblik itilize yo ankripte mesaj.

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Kòm ou ta ka devine pa non li yo, nou kapab pataje kle piblik nou yo avèk

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nenpòt moun nou vle san yo pa konpwomèt sekirite a nan yon mesaj chiffres.

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Messages chiffres lè l sèvi avèk yon kle piblik

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kapab fèt sèlman dechifre ak kle korespondan li yo prive.

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Pandan ke ou ka pataje kle piblik ou, ou ta dwe toujou kenbe prive ou kle sekrè.

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epi li ka sèlman kle nan prive ap itilize yo dekriptaj mesaj

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si 2 itilizatè vle voye mesaj chiffres ak RSA

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retounen ak lide tou de itilizatè bezwen gen pwòp yo pè piblik ak prive kle.

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Mesaj soti nan itilizatè 1 a itilizatè 2

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sèlman itilize kle pè 2 itilizatè a, ak mesaj soti nan itilizatè 2 a itilizatè 1

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sèlman itilize kle pè 1 user a.

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Reyalite a ke gen 2 kle apa ankripte ak dekriptaj mesaj

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fè RSA yon asimetri algorithm kle.

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Nou pa bezwen ankripte kle piblik la yo nan lòd yo voye li nan yon lòt òdinatè

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depi kle a se piblik de tout fason.

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Sa vle di ke RSA pa gen menm pwoblèm nan demaraj

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kòm algoritm yo simetrik kle.

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Se konsa, si mwen vle voye yon mesaj lè l sèvi avèk RSA chifreman

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Rob, mwen pral premye bezwen piblik kle Rob a.

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Jenere yon pè kle, Rob bezwen chwazi 2 gwo nonb premie.

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Nimewo sa yo pral itilize nan tou de kle yo piblik ak prive,

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men kle piblik la pral sèlman itilize pwodwi a nan 2 nimewo sa yo,

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pa chif yo tèt yo.

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Yon fwa mwen te chiffres mesaj la lè l sèvi avèk kle piblik Rob a

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Mwen ka voye voye mesaj la nan Rob.

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Pou yon òdinatè, nimewo faktorizasyon se yon pwoblèm difisil.

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Kle a piblik, sonje, yo itilize pwodwi a nan 2 nonb premie.

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Sa a pwodui dwe fèt Lè sa a, gen sèlman 2 faktè,

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ki rive yo nimewo yo ki fè moute kle a prive.

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Yo nan lòd yo mesaj la dekriptaj, RSA ap itilize sa kle prive

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oswa nimewo yo miltipliye ansanm nan pwosesis pou kreye kle a piblik la.

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Paske li nan kalkil difisil yo faktè nimewo a

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itilize nan yon kle piblik nan 2 chif yo itilize nan kle a prive

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li difisil pou yon atakè figi l kle a prive

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ki pral nesesè yo dekriptaj mesaj la.

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Koulye a, kite a ale nan kèk detay pi ba nivo nan RSA.

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Se pou nou premye wè kijan nou ka jenere yon pè kle.

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Premyèman, nou pral bezwen 2 nonb premie.

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Nou pral rele nimewo sa yo 2 P ak Q.

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Yo nan lòd yo chwazi P ak Q, nan pratik nou ta pseudorandomly jenere

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nimewo gwo ak Lè sa a, sèvi ak yon tès pou detèmine si wi ou non

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moun ki nimewo yo se pwobableman premye.

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Nou kapab kenbe génération nimewo o aza sou yo ak sou ankò

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jiskaske nou gen 2 primes ke nou ka itilize.

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Isit la kite a chwazi p = 23 epi q = 43.

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Sonje byen, nan pratik, P ak Q yo ta dwe anpil pi gwo nonb.

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Osi lwen ke nou konnen, pi gwo a nimewo yo, pi rèd nan li se

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krak yon mesaj chiffres.

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Men, li la tou plis chè ankripte ak dekriptaj mesaj.

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Jodi a li nan souvan rekòmande pou P ak Q gen omwen 1024 Bits,

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ki mete chak nimewo nan plis pase 300 chif desimal.

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Men, nou pral chwazi nimewo sa yo ti pou egzanp sa a.

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Koulye a, nou pral anpil anpil pitit P ak Q ansanm yo ka resevwa yon nimewo 3rd,

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ki nou ap rele n.

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Nan ka nou an, n = 23 * 43, ki = 989.

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Nou te n = 989.

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Next nou pral anpil anpil pitit p - 1 ak q - 1

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jwenn yon nimewo 4yèm, ki nou ap rele m.

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Nan ka nou an, m = 22 * ​​42, ki = 924.

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Nou gen m = 924.

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Koulye a, nou pral bezwen yon nimewo e ke se relativman pwemye m

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ak mwens pase m.

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De nonb yo relativman pwemye oswa coprime

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si nonm antye a sèlman pozitif ki divize yo tou de respire se 1.

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Nan lòt mo, pi gran divizè a komen nan e ak m

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dwe 1.

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An pratik, li a ki komen pou e yo dwe nimewo a pwemye 65537

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osi lontan ke nimewo sa a pa rive yo dwe yon faktè de m.

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Pou kle nou yo, nou pral chwazi e = 5

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depi 5 se relativman pwemye 924.

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Finalman, nou pral bezwen yon sèl plis nimewo, ki nou ap rele d.

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D dwe gen kèk valè ki satisfè ekwasyon an de = 1 (Mod m).

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Sa a m Mod vle nou pral sèvi ak yon bagay yo rele modilè aritmetik.

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Nan aritmetik modilè, yon fwa yon nimewo vin pi wo pase kèk mare anwo

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li pral vlope tounen otou a 0.

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Yon revèy, pou egzanp, itilize aritmetik modilè.

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Yon minit apre 1:59, pou egzanp, se 2:00,

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pa 1:60.

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Te men nan minit mare nan ren a 0

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lè l genyen yon anwo mare l 'nan 60.

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Se konsa, nou ka di 60 se ekivalan 0 (Mod 60)

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ak 125 ki ekivalan a 65 ki ekivalan a 5 (Mod 60).

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Kle piblik nou yo ap e la pè ak n

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kote nan ka sa a e se 5 epi n se 989.

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Kle prive nou yo ap d la pè ak n,

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ki nan ka nou an, se 185 ak 989.

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Remake orijinal nou primes P ak Q pa parèt

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nenpòt kote nan kle prive oswa piblik nou yo.

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Kounye a ke nou gen pè nou an kle, kite a pran yon gade nan ki jan nou ka ankripte

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ak dekriptaj yon mesaj.

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Mwen vle voye yon mesaj bay Rob,

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pou l 'ka fè yon sèl la yo kreye sa a pè kle.

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Lè sa a, mwen pral mande Rob pou kle piblik li a, ki m ap sèvi ak

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ankripte yon mesaj pou voye bay l 'yo.

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Sonje byen, li la totalman oke pou Rob yo pataje kle piblik li avè m '.

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Men, li pa ta dwe oke yo pataje kle prive l 'yo.

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Mwen pa gen okenn lide ki sa kle prive li ye.

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Nou ka kraze m mesaj nou an moute nan sègman plizyè

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tout pi piti pase n ak Lè sa a, ankripte chak nan sa yo fragman.

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Nou pral ankripte CS50 nan fisèl, ki nou ka kraze moute nan 4 sègman,

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youn pou chak lèt.

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Yo nan lòd yo ankripte mesaj mwen, mwen pral bezwen konvèti li an

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kèk kalite reprezantasyon nimerik.

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Se pou nou anchene valè yo ASCII ak karaktè yo ki nan mesaj mwen an.

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Yo nan lòd yo ankripte yon m mesaj bay

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Mwen pral bezwen kalkile c = m e an (n Mod).

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Men, m dwe pi piti pase n,

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oswa lòt moun mesaj la plen pa kapab eksprime modulo n.

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Nou ka kraze m moute nan sègman plizyè, nan tout ki se pi piti pase n,

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ak ankripte chak nan sa yo fragman.

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Chifreman chak nan sa yo fragman, nou jwenn c1 = 67 a 5 an (Mod 989)

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ki = 658 yo.

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Pou moso dezyèm nou an, nou gen 83 a 5 an (Mod 989)

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ki = 15.

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Pou moso twazyèm nou an, nou gen 53 a 5 an (Mod 989)

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ki = 799.

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Epi finalman, pou moso dènye nou an, nou gen 48 a 5 an (Mod 989)

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ki = 975.

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Koulye a, nou ka voye voye sou valè sa yo chiffres Rob.

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La a ou ale, Rob.

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Pandan ke mesaj nou an se nan vòl, kite a pran yon lòt gade

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nan ki jan nou te resevwa ki valè pou d.

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D nimewo nou an bezwen satisfè 5D = 1 (Mod 924).

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Sa fè d Envès miltiplikatif nan 5 modulo 924.

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Bay 2 nonm antye yo, A ak B yo, pwolonje algorithm nan eklidyen

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yo ka itilize yo jwenn pi gran divizè a komen nan 2 sa yo nonm antye relatif.

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Li pral tou ba nou 2 nimewo lòt, x ak y,

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ki satisfè rach la ekwasyon + By = pi gran divizè a komen nan yon b ak.

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Kijan sa a ede nou?

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Oke, rakorde nan e = 5 pou yon

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ak m = 924 pou b

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nou deja konnen ke nimewo sa yo se coprime.

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Pi gran yo divizè komen se 1.

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Sa ban-nou 5x + 924y = 1

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oswa 5x = 1 - 924y.

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Men, si nou sèlman pran swen sou tout bagay modulo 924

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Lè sa a, nou ka lage - 924y la.

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Panse tounen nan revèy la.

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Si se men nan minit se sou 1 ak Lè sa a, egzakteman 10 èdtan pase,

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nou konnen men an minit ap toujou gen sou 1 an.

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Isit la nou kòmanse nan 1 ak Lè sa a, vlope toutotou fwa egzakteman y,

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se konsa nou pral toujou gen 1.

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Nou gen 5x = 1 (Mod 924).

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Ak isit la sa a x se menm bagay la kòm d nan nou te kap chèche anvan,

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Se konsa, si nou itilize pwolonje algorithm nan eklidyen

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jwenn sa a x nimewo, sa se nimewo a nou ta dwe itilize kòm d nou an.

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Koulye a, kite a kouri pwolonje algorithm nan eklidyen pou yon 5 =

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epi b = 924.

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Nou pral sèvi ak yon metòd ki rele metòd la tab.

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Tab nou yo ap gen 4 kolòn, x, y, d, ak k.

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Tab nou an kòmanse koupe ak 2 ranje.

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Nan premye ranje a nou gen 1, 0, donk valè nou an yon, ki se 5,

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ak ranje dezyèm nou an se 0, 1, ak valè nou yo pou b, ki se 924.

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Valè a nan kolòn nan 4yèm, k, yo pral rezilta nan

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nan divize valè a nan d nan ranje ki anwo a li ak valè a nan d

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sou ranje a menm.

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Nou gen 5 divize pa 924 se 0 ak kèk rès.

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Sa vle di nou gen k = 0.

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Koulye a, valè a nan chak selil lòt yo pral valè a nan 2 ranje yo selil pi wo pase l

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mwens valè a nan ranje a pi wo pase l fwa k.

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Kite la kòmanse avèk d nan ranje a 3yèm.

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Nou gen 5 - 924 * 0 = 5.

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Next nou gen 0 - 1 * 0 ki se 0

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ak 1 - 0 * 0 ki se 1.

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Pa twò move, kidonk kite yo deplase sou ranje a kap vini an.

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Premye nou bezwen valè nou an k.

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924 divize pa 5 184 = ak kèk rès,

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00:13:43,000 --> 00:13:46,000
se konsa valè nou an pou k se 184.

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

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

226
00:14:05,000 --> 00:14:07,000
Tout dwa, kite a fè ranje a kap vini an.

227
00:14:07,000 --> 00:14:10,000
Valè nou yo ki genyen k ap 1 paske

228
00:14:10,000 --> 00:14:15,000
5 divize pa 4 1 = ak kèk rès.

229
00:14:15,000 --> 00:14:17,000
Se pou nou ranpli nan kolòn yo ak lòt.

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
Ak 1 - 184 * 1 se 185.

233
00:14:33,000 --> 00:14:35,000
Ann wè sa ki valè pwochen nou yo nan k ta dwe.

234
00:14:35,000 --> 00:14:40,000
Oke, li sanble nou gen 4 divize pa 1, ki se 4.

235
00:14:40,000 --> 00:14:43,000
Nan ka sa a kote n ap divize pa 1 sa yo ki k se egal a

236
00:14:43,000 --> 00:14:50,000
valè a nan d nan ranje ki anwo a vle di ke nou ap fè ak algorithm nou an.

237
00:14:50,000 --> 00:14:58,000
Nou ka wè isit la ke nou gen x = 185 epi y = -1 nan ranje ki sot pase yo.

238
00:14:58,000 --> 00:15:00,000
Se pou nou kounye a tounen vin jwenn objektif orijinal nou an.

239
00:15:00,000 --> 00:15:04,000
Nou te di ke valè x kòm yon konsekans kouri sa a algorithm

240
00:15:04,000 --> 00:15:08,000
ta dwe envès la miltiplikatif yon (Mod b).

241
00:15:08,000 --> 00:15:15,000
Sa vle di ke 185 se Envès miltiplikatif nan 5 (Mod 924)

242
00:15:15,000 --> 00:15:20,000
ki vle di ke nou gen yon valè de 185 pou d.

243
00:15:20,000 --> 00:15:23,000
Lefèt ke d = 1 nan ranje nan dènye

244
00:15:23,000 --> 00:15:26,000
verifye te ke e coprime m.

245
00:15:26,000 --> 00:15:30,000
Si se pa t '1 Lè sa a, nou ta gen yo chwazi yon e nouvo.

246
00:15:30,000 --> 00:15:33,000
Koulye a, kite a wè si Rob te resevwa mesaj mwen an.

247
00:15:33,000 --> 00:15:35,000
Lè yon moun ap voye m 'yon mesaj chiffres

248
00:15:35,000 --> 00:15:38,000
osi lontan ke mwen te kenbe kle prive mwen yon sekrè

249
00:15:38,000 --> 00:15:41,000
Mwen yon sèl la sèlman ki moun ki ka dekriptaj mesaj la.

250
00:15:41,000 --> 00:15:46,000
Dekriptaj yon moso c mwen ka kalkile mesaj orijinal la

251
00:15:46,000 --> 00:15:53,000
ki egal a moso nan d pouvwa (Mod n).

252
00:15:53,000 --> 00:15:57,000
Sonje ke d ak n yo soti nan kle prive mwen an.

253
00:15:57,000 --> 00:16:01,000
Pou jwenn yon mesaj plen soti nan sègman li nou dekriptaj chak moso

254
00:16:01,000 --> 00:16:04,000
ak anchene rezilta yo.

255
00:16:04,000 --> 00:16:08,000
Egzakteman kouman sekirite sa, se RSA?

256
00:16:08,000 --> 00:16:10,000
Verite a se, nou pa konnen.

257
00:16:10,000 --> 00:16:14,000
Sekirite a baze sou konbyen tan li ta pran yon atakè krak yon mesaj

258
00:16:14,000 --> 00:16:16,000
chiffres ak RSA.

259
00:16:16,000 --> 00:16:19,000
Sonje ke yon atakè gen aksè a kle piblik ou a,

260
00:16:19,000 --> 00:16:21,000
ki gen tou de e ak n.

261
00:16:21,000 --> 00:16:26,000
Si atakè a jere faktè n nan 2 primes li yo, P ak Q,

262
00:16:26,000 --> 00:16:30,000
Lè sa a, li te kapab kalkile d lè l sèvi avèk pwolonje algorithm nan eklidyen.

263
00:16:30,000 --> 00:16:35,000
Sa a ba l 'kle nan prive, ki ka itilize yo dekriptaj nenpòt mesaj.

264
00:16:35,000 --> 00:16:38,000
Men, ki jan nou ka byen vit faktè nonb antye relatif?

265
00:16:38,000 --> 00:16:41,000
Yon fwa ankò, nou pa konnen.

266
00:16:41,000 --> 00:16:43,000
Okenn moun pa te jwenn yon fason vit nan fè li,

267
00:16:43,000 --> 00:16:46,000
ki vle di ke yo bay gwo ase n

268
00:16:46,000 --> 00:16:49,000
li ta pran yon atakè irealist long

269
00:16:49,000 --> 00:16:51,000
faktè nimewo a.

270
00:16:51,000 --> 00:16:54,000
Si yon moun devwale yon fason vit nan nonm antye relatif faktorizasyon

271
00:16:54,000 --> 00:16:57,000
RSA ta gen pou kase.

272
00:16:57,000 --> 00:17:01,000
Men, menm si nonb antye relatif faktorizasyon se natirèlman ralanti

273
00:17:01,000 --> 00:17:04,000
algorithm nan RSA te ka toujou gen kèk defo nan li

274
00:17:04,000 --> 00:17:07,000
ki pèmèt pou dekripte fasil nan mesaj.

275
00:17:07,000 --> 00:17:10,000
Okenn moun pa te jwenn ak devwale tankou yon defo ankò,

276
00:17:10,000 --> 00:17:12,000
men sa pa vle di yon sèl pa egziste.

277
00:17:12,000 --> 00:17:17,000
Nan teyori, yon moun ta ka soti la lekti tout done chiffres ak RSA.

278
00:17:17,000 --> 00:17:19,000
Genyen yon lòt ti jan nan yon pwoblèm konfidansyalite.

279
00:17:19,000 --> 00:17:23,000
Si Tommy chifre kèk mesaj lè l sèvi avèk kle piblik m '

280
00:17:23,000 --> 00:17:26,000
ak yon atakè chifre menm mesaj la lè l sèvi avèk kle piblik m '

281
00:17:26,000 --> 00:17:29,000
atakè a pral wè ke mesaj yo 2 se idantik

282
00:17:29,000 --> 00:17:32,000
e konsa konnen ki sa Tommy chiffres.

283
00:17:32,000 --> 00:17:36,000
Yo nan lòd yo anpeche sa a, mesaj yo, se tipikman Matlasye ak Bits o aza

284
00:17:36,000 --> 00:17:39,000
avan ke yo te chiffres pou ke menm mesaj la chiffres

285
00:17:39,000 --> 00:17:44,000
plizyè fwa pral gade diferan osi lontan ke kouvèti a sou mesaj la se diferan.

286
00:17:44,000 --> 00:17:47,000
Men, sonje ki jan nou dwe divize mesaj nan sègman

287
00:17:47,000 --> 00:17:50,000
pou ke chak moso se ki pi piti pase n?

288
00:17:50,000 --> 00:17:52,000
Kouvèti sègman yo vle di ke nou ka gen yo fann bagay yo

289
00:17:52,000 --> 00:17:57,000
nan menm plis sègman depi moso nan Matlasye dwe pi piti pase n.

290
00:17:57,000 --> 00:18:01,000
Chifreman ak dekripte yo relativman chè ak RSA,

291
00:18:01,000 --> 00:18:05,000
ak sa ki bezwen kraze moute yon mesaj nan sègman anpil ka trè koute chè.

292
00:18:05,000 --> 00:18:09,000
Si yon volim gwo done bezwen yo dwe chiffres ak dechifre

293
00:18:09,000 --> 00:18:12,000
nou ka konbine benefis ki genyen nan algoritm simetrik kle

294
00:18:12,000 --> 00:18:16,000
ak sa yo ki nan RSA jwenn tou de sekirite ak efikasite.

295
00:18:16,000 --> 00:18:18,000
Malgre ke nou pa pral antre li isit la,

296
00:18:18,000 --> 00:18:23,000
AES se yon algorithm simetrik kle tankou Vigenère la ak Seza tande kòz algorithms

297
00:18:23,000 --> 00:18:25,000
men pi difisil krak.

298
00:18:25,000 --> 00:18:30,000
Natirèlman, nou pa ka sèvi ak AES san etabli yon pataje kle sekrè

299
00:18:30,000 --> 00:18:34,000
ant 2 sistèm yo, epi nou te wè pwoblèm nan ak ke anvan.

300
00:18:34,000 --> 00:18:40,000
Men koulye a, nou kapab itilize RSA etabli pataje kle a sekrè ant 2 sistèm yo.

301
00:18:40,000 --> 00:18:43,000
Nou pral rele òdinatè a voye done yo moun k la

302
00:18:43,000 --> 00:18:46,000
ak òdinatè a ap resevwa done yo reseptè la.

303
00:18:46,000 --> 00:18:49,000
Reseptè a gen yon pè RSA kle epi voye

304
00:18:49,000 --> 00:18:51,000
kle a piblik la moun k la.

305
00:18:51,000 --> 00:18:54,000
Moun k nan jenere yon kle AES,

306
00:18:54,000 --> 00:18:57,000
chifre li ak RSA kle reseptè a nan piblik la,

307
00:18:57,000 --> 00:19:00,000
epi voye kle a AES reseptè la.

308
00:19:00,000 --> 00:19:04,000
Reseptè a dechifr mesaj la ak kle RSA li yo prive.

309
00:19:04,000 --> 00:19:09,000
Tou de moun k la ak reseptè a koulye a gen yon pataje AES kle ant yo.

310
00:19:09,000 --> 00:19:14,000
AES, ki se anpil pi vit nan chifreman ak dekripte pase RSA,

311
00:19:14,000 --> 00:19:18,000
kapab kounye a ka itilize ankripte komèsan ki gwo done epi voye yo bay reseptè a,

312
00:19:18,000 --> 00:19:21,000
ki moun ki ka dekriptaj lè l sèvi avèk kle a menm.

313
00:19:21,000 --> 00:19:26,000
AES, ki se anpil pi vit nan chifreman ak dekripte pase RSA,

314
00:19:26,000 --> 00:19:30,000
kapab kounye a ka itilize ankripte komèsan ki gwo done epi voye yo bay reseptè a,

315
00:19:30,000 --> 00:19:32,000
ki moun ki ka dekriptaj lè l sèvi avèk kle a menm.

316
00:19:32,000 --> 00:19:36,000
Nou jis bezwen RSA transfere kle a pataje.

317
00:19:36,000 --> 00:19:40,000
Nou pa bezwen sèvi ak RSA nan tout.

318
00:19:40,000 --> 00:19:46,000
Li sanble ke mwen te gen yon mesaj.

319
00:19:46,000 --> 00:19:49,000
Li pa enpòtan si nenpòt moun ki li sa ki la sou avyon an papye anvan m 'kenbe li

320
00:19:49,000 --> 00:19:55,000
paske mwen yon sèl la sèlman ki gen kle a prive.

321
00:19:55,000 --> 00:19:57,000
Se pou nou dekriptaj chak nan sègman yo ki nan mesaj la.

322
00:19:57,000 --> 00:20:07,000
Moso a an premye, 658, nou leve sou pouvwa a d, ki se 185,

323
00:20:07,000 --> 00:20:18,000
Mod n, ki se 989, ki egal a 67,

324
00:20:18,000 --> 00:20:24,000
ki se C a lèt nan ASCII.

325
00:20:24,000 --> 00:20:31,000
Koulye a, sou moso, dezyèm lan.

326
00:20:31,000 --> 00:20:35,000
Moso nan dezyèm gen valè 15,

327
00:20:35,000 --> 00:20:41,000
ki nou leve sou pouvwa a 185th,

328
00:20:41,000 --> 00:20:51,000
989 Mod, e sa se egal a 83

329
00:20:51,000 --> 00:20:57,000
ki se S nan lèt nan ASCII.

330
00:20:57,000 --> 00:21:06,000
Koulye a, pou moso nan twazyèm, ki te gen valè 799, nou leve a 185,

331
00:21:06,000 --> 00:21:17,000
989 Mod, e sa se egal a 53,

332
00:21:17,000 --> 00:21:24,000
ki se valè a an karaktè a 5 nan ASCII.

333
00:21:24,000 --> 00:21:30,000
Koulye a, pou moso nan dènye, ki te gen valè 975,

334
00:21:30,000 --> 00:21:41,000
nou leve a 185, Mod 989,

335
00:21:41,000 --> 00:21:51,000
ak sa a se egal a 48, ki se valè de 0 a karaktè nan ASCII.

336
00:21:51,000 --> 00:21:57,000
Non mwen se Rob Bowden, ak sa a se CS50.

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

338
00:22:06,000 --> 00:22:08,000
RSA nan tout.

339
00:22:08,000 --> 00:22:14,000
RSA nan tout. [Ri]

340
00:22:14,000 --> 00:22:17,000
Nan tout.