Asila Elchiboyeva A5/1 ma’lumotni shifrlash algoritmi

691-19

Asila Elchiboyeva

A5/1 ma’lumotni shifrlash algoritmi

Ochiq matn sifatida M=”AS”; kalit sifatida K=”ELCHIBOYEVA” 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

E	L	C	H	I	B	O	Y	E	V	A
00100	01011	00010	00111	01000	00001	01110	11000	00100	10101	00000

A	S
00000	10010

(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=0010001011000100011;

y=1010000000101110110000;

z=01001010100000000001001;

ekanligi kelib chiqadi.

№	0	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20	21	22
x	0	0	1	0	0	0	1	0	1	1	0	0	0	1	0	0	0	1	1
y	1	0	1	0	0	0	0	0	0	1	0	1	1	1	0	1	1	0	0	0	0	0
z	0	1	0	0	1	0	1	0	1	0	0	0	0	0	0	0	0	0	0	0	0	0	1

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

X₁=1^0^1^1=1;

Z₁=0^0^0^1=1;

X₀ va Z₀ qiymatlari Y va Z registorlarning o’ng tomoniga qo’shiladi, hamda bu registorlarning oxirgi qiymatlari o’chiriladi:

№	0	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20	21	22
x	1	0	0	1	0	0	0	1	0	1	1	0	0	0	1	0	0	0	1	0
y	1	0	1	0	0	0	0	0	0	1	0	1	1	1	0	1	1	0	0	0	0	0
z	1	0	1	0	0	1	0	1	0	1	0	0	0	0	0	0	0	0	0	0	0	0	0

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

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

X₂=0^0^0^1=1;

Y₂=0^0=0

Z₂=1^0^0^0=1;

№	0	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20	21	22
x	1	1	0	0	1	0	0	0	1	0	1	1	0	0	0	1	0	0	0	1
y	0	1	0	1	0	0	0	0	0	0	1	0	1	1	1	0	1	1	0	0	0	0
z	1	1	0	1	0	0	1	0	1	0	1	0	0	0	0	0	0	0	0	0	0	0	0

Bunda kalitimiz 1^0^1=0; orqali topiladi , ya’ni kalit 2-K₂=0

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

X₃=0^0^0^0=0;

Z₃=0^0^0^0=0;

№	0	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20	21	22
x	0	1	1	0	0	1	0	0	0	1	0	1	1	0	0	0	1	0	0	0
y	0	1	0	1	0	0	0	0	0	0	1	0	1	1	1	0	1	1	0	0	0	0
z	0	1	1	0	1	0	0	1	0	1	0	1	0	0	0	0	0	0	0	0	0	0	0

Bunda kalitimiz 0^0^0=0; orqali topiladi , ya’ni kalit 3-K₃=0