3-amaliy topshiriq. 3- amaliy topshiriqlar

Axborot xavfsizligi yo’nalishi 130-20 guruh talabasi Eshmamatov Faxriddin Xayriddin o’g’li. Kriptologiyadan 3-amaliy topshiriq.


3- Amaliy topshiriqlar.
Keltirilgan shifrlash usullaridan foydalanib har bir talaba tomonidan o‘z ism – shariflari shifrlansin va deshifrlansin (kalitlar ixtiyoriy tanlansin). Shifrlash jarayoni dasturlash tillari yordamida amalga oshirilsin. Dastur interfeysi va kodi pdf formatga o’tkazilib hemis tizimiga yuklansin.


Mavzular:
1. RSA algoritmi
2. A5/1 shifrlash algoritmi

1. 
   RSA algoritmiga oid misol: Familiya va Ism so‘zlarini shifrlaymiz:
   

RSA – (Rivest, Shamir Adleman)

1. 
   Ikki xil tasodifiy tub sonlar tanlanadi: p=3 va q=11
   
2. 
   Ularning n qiymati hisoblanadi. n=p*q=3*11=33 bu modul deb ataladi.
   
3. 
   Sondan Eyler funksiyaning qiymati hisoblanadi n:
   

µ(n)=(p-1)*(q-1)= (3-1)*(11-1)=2*10=20

4. 
   Butun son tanlangan e (1
5. 
   Raqam hisoblanadi d songa ko‘patma teskari e modul µ(n) ya’ni taqqoslashni qanoatlantiradigan raqam: d*e≡1*(mod µ(n)) raqam d maxfiy ko‘rsatkich deb ataladi. Odatda kengaytirilgan Evklid algoritmi yordamida hisoblab chiqiladi.
   
6. 
   Juftlik (e,n) RSA ochiq kalit: {e;n}={7;33}
   
7. 
   Juftlik (d,n) RSA shaxsiy kalit: {d;n}={3;33}
   

A	B	C	D	E	F	G	H	I	J	K	L	M	N	O	P	Q	R	S	T	U	V	W	X	Y	Z
1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20	21	22	23	24	25	26

• 
  U holda ma’lumot Faxriddin (6,1,24,18,9,4,4,9,14) Eshmamatov (5,19,8,13,1,13,1,20,15,22) ko‘rinishda bo‘ladi va uni {e;n}={7;33} ochiq kalit bilan  bir tomonli funksiya bilan shifrlaymiz:
  

Faxriddin so’zini shifrlaymiz:
x=6da SHM₁=(6⁷)(mod33)=279936(mod33)=30,
x=1da SHM₂=(1⁷) (mod33)=1,
x=24da SHM₃=(24⁷) (mod33)=4586471424(mod33)=18, x=18da SHM₄=(18⁷)(mod33)=612220032 (mod33)=6,
x=9da SHM₅=(9⁷) (mod33)= 4782969(mod33)=15,
x=4da SHM₆=(4⁷) (mod33)=16384(mod33)=16, x=4da SHM₇=(4⁷)(mod33)= 16384(mod33)=16,
x=9da SHM₈=(9⁷) (mod33)= 4782969(mod33)=15,
x=14da SHM₉=(14⁷) (mod33)=105413504 (mod33)=20.
Eshmamatov so’zini shifrlaymiz:
x=5da SHM₁=(5⁷)(mod33)=78125(mod33)=14,
x=19da SHM₂=(19⁷) (mod33)=893871739(mod33)=13,
x=8da SHM₃=(8⁷) (mod33)=2097152(mod33)=2, x=13da SHM₄=(13⁷)(mod33)=62748517(mod33)=7,
x=1da SHM₅=(1⁷) (mod33)=1,
x=13da SHM₆=(13⁷) (mod33)= 62748517(mod33)=7, x=1da SHM₇=(1⁷) (mod33)=1,
x=20da SHM₈=(20⁷) (mod33)= 1280000000(mod33)=26,
x=15da SHM₉=(15⁷) (mod33)=170859375(mod33)=27, x=22da SHM₁₀=(22⁷) (mod33)=2494357888(mod33)=22.

• 
  Bu olingan shifrlangan (30,1,18,6,15,16,16,15,2) (14,13,2,7,1,7,1,26,27,22)ma’lumotni maxfiy {d;n}={3;33} kalit bilan ifoda orqali deshifrlaymiz: 
  

Faxriddin so’zini deshifrlaymiz:

x=30da OchM₁=(30³)(mod33)=27000(mod33)=6,
x=1da OchM₂=(1³) (mod33)=1,
x=18da OchM₃=(18³) (mod33)=5832(mod33)=24, x=6da OchM₄=(6³)(mod33)=216(mod33)=18,
x=15da OchM₅=(15³) (mod33)= 3375(mod33)=9,
x=16da OchM₆=(16³) (mod33)= 4096(mod33)=4, x=16da OchM₇=(16³)(mod33)=4096(mod33)=4,
x=15da OchM₈=(15³) (mod33)=3375(mod33)=9,
x=20da OchM₉=(20³) (mod33)=8000(mod33)=14.
Eshmamatov so’zini deshifrlaymiz:
x=14da OchM₁=(14³)(mod33)=2744(mod33)=5,
x=13da OchM₂=(13³) (mod33)=2197(mod33)=19,
x=2da OchM₃=(2³) (mod33)=8, x=7da OchM₄=(7³) (mod33)= 343(mod33)=13,
x=1da OchM₅=(1³) (mod33)=1,
x=7da OchM₆=(7³) (mod33)= 343(mod33)=13, x=1da OchM₇=(1³) (mod33)=1,
x=26da OchM₈=(26³) (mod33)= 17576(mod33)=20,
x=27da OchM₉=(27³) (mod33)=19683(mod33)=15, x=22da OchM₁₀=(22³) (mod33)=10648(mod33)=22.

Shifrlanadigan matn: Faxriddin (6,1,24,18,9,4,4,9,14)

Eshmamatov (5,19,8,13,1,13,1,20,15,22)

Shifrlangan matn: Faxriddin (30,1,18,6,15,16,16,15,2)
Eshmamatov (14,13,2,7,1,7,1,26,27,22)

Shifrlangan matnni deshifrlaymiz: Faxriddin (6,1,24,18,9,4,4,9,14) Eshmamatov (5,19,8,13,1,13,1,20,15,22)
Shifrlangan matnni deshifrlaganimizda yana shifrlanadigan matnimizni o’zi hosil bo’ldi.