public-keyencryptionanddigitalsignatures-資料下載頁
【導讀】Communications,7thEdition,Consideredtobe“asymmetric”,sincetwokeysare. Public-keyEncryption(Fig.). encryptionkey.key.Public-keyAuthentication(Fig.). privatekey.key.message.DigitalSignatures. canbeused.content,andsequencing.forpublic-keyencryption.LenAdleman(MIT).Encryption. –PublicKeyisKU={e,n}.integers).Decryption. –PrivateKeyisKR={d,n}.–M=Cdmodn.integersM<n.eandn.pandq.–?(n)=(p-1)x(q-1).4.Selectintegere.–gcd(?(n),e)=1;1<e<?5.Calculated.–de=mod?(n)=1.6.PublicKeyisKU={e,n}.7.PrivateKeyisKR={d,n}.–1.Selectprimes,p=17andq=11.–2.Calculaten=pxq=17x11=187.–3.Calculate?(n)=(17-1)x(11-1)=160.–4.Selecte.(n)=32x5=25x5.Possiblee’s:3,7,11,...,<?(n)=160. Choose7.–5.Determined.Solve(dx7)mod160=1,whered<160.Thismeans(dx7)=(kx160)+1forkan. integer.d=23,since23x7=161=10x160+1.–6.ThePublicKeyisKU={7,187}.EncryptionExample. –LetM=88.–C=887mod187.Xa+bmodn={(Xamodn)(Xbmodn)}modn.C={(884mod187)(882mod187)(881mod187)}mod187. C={132x77x88}mod187. C=
【正文】 = 10 x 160 +1. – 6. The Public Key is KU={7,187}. – 7. The Private Key is KR={23,187}. RSA Example () ? Encryption Example – Let M = 88. – C = 887 mod 187. – Now, consider the following property of modular arithmetic: ? Xa+b mod n={(Xa mod n)(Xb mod n)}mod n. ? C={(884mod187)(882mod187)(881mod187)}mod187 ? C={132 x 77 x 88} mod 187 ? C = 11. RSA Example () ? Decryption Example – Let C = 11. – M = 1123 mod 187. – M={(111mod187)(112mod187)(114mod187) (118mod187) (118mod187)}mod187 ? M={11 x 121 x 55 x 33x 33} mod 187 ? M = 88. Attacks on RSA ? Brute forcetry all possible keys. – This means large keys need to be used, but implementations will have longer putation time. ? Factor n, into its prime factors (p and q.) – For n large, this is a hard problem.
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