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Fayzullo
- Bu sahifa navigatsiya:
- Modul arithmetic
- RSA algorithms.
Modul arithmetic
• Cryptography aa sonni bb songa bўlgandagy koldik rr ga teng bўlsa, u ҳolda kuyidagicha belgilanadi: aaaaaaaabb≡rr. - Dusturlash tillarida esa aaabb kabi belgilanadi. • Cryptography modul sifatida (ya'ni, bўluvchi) faqat tub sonlardan foydalanish talab etiladi. – Ya'ni, aa aaaaaa nn tenglikdaghi nn ҳar doim tub bўlishi talab etiladi. • Misollar: • 7𝑎𝑎𝑎𝑎𝑎𝑎𝑚 ≡(𝑚∗2)𝑎𝑎𝑎𝑎𝑎𝑎𝑚+1𝑎𝑎𝑎𝑎𝑎𝑎𝑚≡0+1≡1 • 14𝑎𝑎𝑎𝑎𝑎𝑎𝑚 ≡(𝑚∗4)𝑎𝑎𝑎𝑎𝑎𝑎𝑚+2𝑎𝑎𝑎𝑎𝑎𝑎𝑚≡0+2≡2 • 2𝑎𝑎𝑎𝑎𝑎𝑎𝑚≡(0∗𝑚)𝑎𝑎𝑎𝑎𝑎𝑎𝑚+2𝑎𝑎𝑎𝑎𝑎𝑎𝑚≡2 • 5𝑎𝑎𝑎𝑎𝑎𝑎7≡5 • −2𝑎𝑎𝑎𝑎𝑎𝑎5≡(−2+5)𝑎𝑎𝑎𝑎𝑎𝑎5≡𝑚𝑎𝑎𝑎𝑎𝑎𝑎5≡𝑚 • −7𝑎𝑎𝑎𝑎𝑎𝑎𝑚≡(−7+𝑚)𝑎𝑎𝑎𝑎𝑎𝑎𝑚≡−4𝑎𝑎𝑎𝑎𝑎𝑎𝑚≡(−4+𝑚)𝑎𝑎𝑎𝑎𝑎𝑎𝑚≡ −1𝑎𝑎𝑎𝑎𝑎𝑎𝑚≡(−1+𝑚)𝑎𝑎𝑎𝑎𝑎𝑎𝑚≡2 Modul arithmetic • 𝟑𝟑-1 𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎≡1/33 𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎≡? • Ushbu muammoni echishda Euclidning kengaitirilgan algorithmidan foydalanish mumkin. • Agar aa−1aaaaaann≡bb bўlsa,u ҳolda (aa∗bb)aaaaaann≡1 tenglik ўrinli bўladi. • 𝑚−1𝑎𝑎𝑎𝑎𝑎𝑎7 ≡ 𝑥𝑥 • 7 = 𝑚 ∗ 2 + 1 • 𝑚 = 1 ∗ 𝑚 + 0 • Demak, 1 sony 3 va 7 sonlari uchun ECUB bўladi. • Ohirgi tenglikdan boshlab kuyidagicha teskari amalga oshiriladi: • 1=7−𝑚∗2=7+(−2∗𝑚)=7∗1+(−2∗𝑚) Modul arithmetic • Tenglikni ikki tomonini modulga (aaaaaa7) olinadi: ((7∗1)aaaaaa7+(−2∗m)aaaaaa7)aaaaaa7≡1aaaaaa 7 Yoki (−2∗𝑚)𝑎𝑎𝑎𝑎𝑎𝑎7≡1𝑎𝑎𝑎𝑎𝑎𝑎7≡1 • Ushbu tenglikni (m∗xx)aaaaaa7≡1 hectares xx=−2 ha tengligini yoki −2aaaaaa7=5 ligani topish mumkin. Ya'ni, (m∗5)aaaaa7≡1 tenglik ўrinli. RSA algorithms. • RSA ochiq kalitli cipherlash algorithms mualliflar bўlgan uchta olimlar, Rivest, Shamir va Adleman, sharafiga қўyilgan. • RSA algorithms katta sonlarni factorlash muammosiga asoslanadi. • RSA algorithmida kuyidagi zharayonlar mavzhud: – Kalitni generationalash; – Ciphrlash; - Decipher. Download 225.98 Kb. Do'stlaringiz bilan baham: |
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