Rsa algoritmi Ochiq matn = akmal

RSA algoritmi

Ochiq matn = AKMAL

O’zaro tub sonlar tanlanadi.

p= 5 q=7

N= p *q

N= 5*7=35

endi µ ni topamiz

µ(n)=(p-1)*(q-1)

µ(n)=(5-1)*(7-1)=24

EKUB(µ (N),e)=1 e< µ (N) e=5 tanlab oldik

ed=1mod µ(n)

5d=1mod24 d=5

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

A=1

K=11

M=13

A=1

L=12

Endi shifrlaymiz

C=M^emodN

C= 1⁵ MOD 35=1 MOD 35 = 1

C= 11⁵ MOD 35=161051 MOD 35 = 16

C= 13⁵ MOD 35= 371293 MOD 35 = 13

C= 1⁵ MOD 35= 1 MOD 35 = 1

C= 12⁵ MOD 35= 248832 MOD 35 = 17

SHIFR MATN = APMAZ

Endi deshifrlanadi

M=C ^emodN

M = 1⁵ MOD 35 = 1

M = 16⁵ MOD 35 = 11

M= 13⁵ MOD 35 = 13

M= 1⁵ MOD 35 = 1

M= 17⁵ MOD 35 = 12

OCHIQ MATN = AKMAL hosil buldi