2-Амалий иш Мавзу: Криптографик ҳимоялаш

Ochiq matn va kalitning harfarini ikkilik sanoq sistemasdagi mos sonlarini xorlab chiqamiz,shu bilan bir qatorda hosil bo’lgan songa mos harflarni ham aniqlab olamiz:

(Xorlash amalini “/” qilib kiritdim)

S/S=10010/10010=00000-A

A/H=00000/00111=00111-H

F/E=00101/00100=00001-B

A/R=00000/10001=10001-R

R/A=10001/00000=10001-R

O/L=01110/01011=00101-F

V/I=10101/01000=11101-/

R/S=10001/10010=00011-D

U/H=10100/00111=10011-T

S/E=10010/00100=10110-W

T/R=10011/10001=00010-C

A/A=00000/00000=00000-A

M/L=01100/01011=00111-H

O/I=01110/01000=00110-G

V/S=10101/10010=00111-H

I/H=01000/00111=01111-P

C/E=10010/00100=10110-W

H/R=00111/10001=10110-W

Hosil bo’lgan harflarni mos ravishda yozib chiqsak shifr matn hosil bo’ladi:

S

A

F

A

R

O

V

R

U

S

T

A

M

O

V

I

C

H

Ochiq matn

S

H

E

R

A

L

I

S

H

E

R

A

L

I

S

H

E

R

Kalit

A

H

B

R

R

F

/

D

T

W

C

A

H

G

H

P

W

W

Xorlash natijasi

C=AHBRRF/DTWCAHGHPWW ;

Deshifrlash jarayonida shifr matn bilan bizga ma’lum bo’lgan kalit xorlanadi.

Shifr matn

A

H

B

R

R

F

/

D

T

W

C

A

H

G

H

P

W

W

Kalit

S

H

E

R

A

L

I

S

H

E

R

A

L

I

S

H

E

R

Shifr matn

S

A

F

A

R

O

V

R

U

S

T

A

M

O

V

I

C

H

M=SafarovRustamovich; ochiq matn hosil qilamiz.

A5/1 ma’lumotni shifrlash algoritmi

Ochiq matn sifatida M=”UM”; kalit sifatida K=”tojiakbarova” larni olindi.

0	A	00000
1	B	00001
2	C	00010
3	D	00011
4	E	00100
5	F	00101
6	G	00110
7	H	00111
8	I	01000
9	J	01001
10	K	01010
11	L	01011
12	M	01100
13	N	01101
14	O	01110
15	P	01111
16	Q	10000
17	R	10001
18	S	10010
19	T	10011
20	U	10100
21	V	10101
22	W	10110
23	X	10111
24	Y	11000
25	Z	11001
26	#	11010
27	!	11011
28	*	11100
29	/	11101
30	\	11110
31	?	11111

t	o	J	i	a	k	b	a	r	o	v	a
10011	01110	01001	01000	00000	01010	00001	00000	10001	01110	10101	00000

u	m
10100	01100

(Xorlash belgisi “^” qilib kiritildi)

Bu usulda kalit 64 bitni tashkil qilib, 3ta registorga qiymat qilib belgilanadi:

X=19 bit(x0,x1,x2,…,x18);

Y=22 bit(y0,y1,y2,..,y21);

Z= 23 bit(z0,z1,z2,…z22);

Shunga ko’ra:

x=1001101110010010100;

y=0000000101000001000001 ;

z=00010111010101000000000 ;

ekanligi kelib chiqadi.

№	0	1	2	3	4	4	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20	21	22
x	1	0	0	1	1	0	1	1	1	0	0	1	0	0	1	0	1	0	0
y	0	0	0	0	0	0	0	1	0	1	0	0	0	0	0	1	0	0	0	0	0	1
z	0	0	0	1	0	1	1	1	0	1	0	1	0	1	0	0	0	0	0	0	0	0	0

Har bir qadamda m=maj(X₈,Y₁₀, Z₁₀) hisoblanadi,ya’ni:
m=maj(1,0,0)=0;
Demak Y Z registorlar siljiydi:

Y₁=0^1=1;

Z₁=1^0^0^0=1;

Y₀ va Z₀ qiymatlari Y va Z registorlarning o’ng tomoniga qo’shiladi, hamda bu registorlarning oxirgi qiymatlari o’chiriladi:

№	0	1	2	3	4	4	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20	21	22
x	1	0	0	1	1	0	1	1	1	0	0	1	0	0	1	0	1	0	0
y	1	0	0	0	0	0	0	0	1	0	1	0	0	0	0	0	1	0	0	0	0	0
z	1	0	0	0	1	0	1	1	1	0	1	0	1	0	1	0	0	0	0	0	0	0	0

Bunda kalitimiz 0^1^1=0; orqali topiladi , ya’ni kalit 1-K₁=0