Ryukzak algoritmi

Shifrlash:

1. 
   BESH so’zini ASCII jadvali yordamida bit ko’rinishga o’tkazamiz:
   

01000010 01000101 01010011 01001000.

2. 
   Bitlardan iborat matnni 8 bitdan bloklarga ajratamiz va har bir blokni o’nlik sanoq sistemasiga o’tkazamiz: M₁=66, M₂=69, M₃=83, M₄=72.
   
3. 
   C_i= mod n. formula yordamida shifrlanadi:
   

C₁= mod n = 66¹¹ mod 1517 = (66*(66⁵nod1517)²) mod 1517 = =(66*532²mod1517) mod 1517 = (66*862)mod1517=763,
C₂= mod n = 69¹¹ mod 1517 =1441,
C₃= mod n = 83¹¹ mod 1517 =821,
C₄= mod n = 72¹¹ mod 1517 =1097.
Hosil bo’lgan shifrtekst quyidagicha:C={763,1441,821,1097}
Shifrni ochish: Modul son n=p*q=1517, (p-1)*(q-1) ko’paytma bilan o’zaro tub bo’lgan e=11 ochiq kalit, matn uzunligi L=8 bit va C={763,1441,821,1097} bizga ma‘lum.

1. 
   e sonining (p-1)*(q-1)modul bo’yicha teskarisini topamiz:
   d=e^-1mod((p-1)*(q-1))=11^-1mod1440=11³⁸³mod1440=131
   
2. 
   M_i= mod n. formula yordamida shifrni ochamiz:
   

M₁= mod n=763¹³¹mod1517=66, M₂= mod n=1441¹³¹mod1517=69,
M₃= mod n=821¹³¹mod1517=83, M₄= mod n=1097¹³¹mod1517=72.

3. 
   M_i larni o’nlikdan ikkilikka o’tkazib, ASCII jadval yordamida harflarga o’tamiz va natijada M=BESH so’zi paydo bo’ladi.
   

Mustaqil ish uchun misollar.

1. 
   n=1517, e=11, C={ 413, 665, 620, 30 }, L=8 bit, M=?
   
2. 
   n=1517, e=11, C={ 593, 27, 763, 30 } L=8 bit, M=?
   
3. 
   n=1517, e=11, C={ 763, 30, 593, 27 }, L=8 bit, M=?
   
4. 
   n=1517, e=11, C={ 1270, 1408, 902, 665 }, L=8 bit, M=?
   
5. 
   n=1517, e=11, C={ 902, 1408, 620, 30 }, L=8 bit, M=?
   
6. 
   n=1517, e=11, C={ 13, 665, 1408, 1390 }, L=8 bit, M=?
   
7. 
   n=1517, e=11, C={ 1437, 1408, 902, 285 }, L=8 bit, M=?
   
8. 
   n=1147, e=11, C={162,881,767,753,162,516,881,753,139,881,894, 162,1109,894,229,881,162,894,890,162,1109}, L=8 bit, M=?
   
9. 
   n=1147, e=11, C={778,890,753,890,946,778,881,946,1109,139,881, 231,881,778,516,162,881,778}, L=8 bit, M=?
   
10. 
    n=1189, e=11,C={1071,416,529,50,368,1096,801,55,368,368,1096, 148,55,575,646,628,1096,495,50,529,352}, L=8 bit, M=?
    
11. 
    n=1189, e=11,C={220,1096,451,416,529,352,416,801,449,86,1096, 575,416,628,575,50,801,220,1096,451,368,50}, L=8 bit, M=? 12. n=341, e=13, L=6 bit, C={ 335, 161, 130, 276} M=?
    

13. 
    n=407, e=13, L=6 bit, C={ 280, 334, 235, 315} M=?
    
14. 
    n=341, e=13, L=6 bit, C={ 335, 232, 282, 226} M=?
    
15. 
    n=407, e=13, L=6 bit, C={ 280, 165, 261, 348} M=?
    
16. 
    n=341, e=13, L=6 bit, C={ 246, 112, 1, 293} M=?
    
17. 
    n=407, e=13, L=6 bit, C={ 280, 23, 185, 94} M=?
    
18. 
    n=341, e=13, L=6 bit, C={ 246, 198, 1, 226} M=?
    
19. 
    n=407, e=13, L=6 bit, C={ 288, 334, 235, 310} M=?
    
20. 
    n=341, e=13, L=6 bit, C={ 246, 161, 168, 130} M=?
    
21. 
    n=407, e=13, L=6 bit, C={ 280, 165, 312, 310} M=?
    
22. 
    n=341, e=13, L=6 bit, C={ 178, 161, 191, 151 } M=?
    
23. 
    n=407, e=13, L=6 bit, C={ 174, 23, 258, 382 } M=?
    
24. 
    n=341, e=13, L=6 bit, C={ 208, 208, 80, 132 } M=?
    
25. 
    n=187, e=13, L=6 bit, C={ 21, 21, 25, 33 } M=?
    
26. 
    n=209, e=13, L=6 bit, C={ 98, 98, 25, 154 } M=?
    
27. 
    n=209, e=7, L=6 bit, C={ 36, 19, 4, 2 } M=?
    
28. 
    n=209, e=13, L=6 bit, C={ 81, 98, 25, 204 } M=?
    
29. 
    n=187, e=13, L=6 bit, C={ 169, 21, 25, 6 } M=?
    
30. 
    n=341, e=7, L=6 bit, C={ 181, 145, 192, 271 } M=?
    
31. 
    n=187, e=7, L=6 bit, C={ 104, 47, 181, 7 } M=?
    

III. Rabin algoritmi
Bu shifrlash usuli 1979 yilda Maykl Rabin tomonidan chop etilgan. Algoritmning xavfsizligi katta tub sonlarga va ko’paytuvchilarga ajratish muammosiga asoslangan. Bunda ikkita katta tub son tanlanadi va ularning har birini to’rt soniga bo’lganda uch qoldiq chiqishi kerak. Bu sonlar yopiq kalit hisoblanadi. Ularning ko’paytmasi ochiq kalit hisoblanadi. p, q tub sonlar tanlanadi. Yuqoridagi shartga ko’ra ular quyidagilarni qanoatlantirishi kerak:
p mod4=3, q mod4=3.
Ochiq kalit n=p·q. M ochiq xabar va MShifrlash: C=M² modn;