O’zbekiston respublikasi oliy va o’rta maxsus ta’lim vazirligi muhammad al-Xorazmiy nomidagi Toshkent Axborot Texnologiya universiteti

O’ZBEKISTON RESPUBLIKASI

OLIY VA O’RTA MAXSUS TA’LIM VAZIRLIGI

Muhammad al-Xorazmiy nomidagi

Toshkent Axborot Texnologiya universiteti

2-amaliy ish

Bajardi:712_18-guruh tabasi Sharifjonov Humoyun

Qabul qildi: Mardiyev U

2-amaliy mashgulot

Mavzu: Bir qiymatli o’rniga qo’yishga asoslangan shifrlar taxlili

Affin shifrlash algoritmi;

Affin shefrlash usulida quyidagi formuladan foydalanamiz:(ax+b)mod(n) Bu yerda a va b ixtiyoriy kalit hisoblanadi, n esa alifbo uzunligi.

Matn: Sharifjonovhumoyun Kalit:Humoyun

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
0	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	15	17	18	19	20	21	22	23	24	25

Xabar	S	H	A	R	I	F	J	O	N	O	V	H	U	M	O	Y	U	N
X	18	7	0	17	8	5	9	14	20	14	21	7	20	12	14	24	20	13
3x+4	58	25	4	55	28	19	31	46	64	46	67	25	64	40	46	76	64	43
(2x+4)mod26	6	25	4	3	2	19	5	20	12	20	13	25	12	14	20	24	12	17
Shefr matn	g	z	e	d	c	t	f	u	m	u	n	z	m	o	u	y	m	r

Deshifrlash jarayoni: Deshifrlash formulasi 𝐷(𝑦)=𝑎−1(𝑦−𝑏)𝑚𝑜𝑑𝑚 гаga teng bo’lib, 𝑎−1=9 , b=4 va m=26 ga teng bo'ladi.

Shifr matn	g	z	e	d	c	t	f	u	m	u	n	z	m	o	u	y	m	r
	6	25	4	3	2	19	5	20	12	20	13	25	12	14	20	24	12	17

Deshifrning umumiy ko’rinishi

Shifr matn	g	z	e	d	c	t	f	u	m	u	n	z	m	o	u	y	m	r
Y	6	25	4	3	2	19	5	20	12	20	13	25	12	14	20	24	12	17
9(y-4)	18	189	0	-9	-18	135	9	144	72	144	81	189	72	90	144	180	72	117
9(y-4)mod26	18	7	0	17	8	5	9	14	20	14	21	7	20	12	14	24	20	13

D=SharifjonovHumoyun;

2.Birlamchi akslantirish affin shifrlash yordamida;

Formulasi M=аm+b, m=26 (alfabit uzunligi), M=a*m+b;

M=Humoyu=7 20 12 14 24 20

M1=7*26+20=202

M1=12*26+14=326

M1=24*26+20=644

E(x)=аx+b (mod)m

A=157;b=580;m^2=676;

C=157М+580 (mod m2)=157*326+580 (mod676)=101688(mod 676) = 386

C1=522; C2=386; C3=288

C1=522=20*26+2;

20=u ; 2=c;

C2=386=14*26+22;

14= 0; 22=w ;

C3=288=11*26+2;

11=l ; 2=c;

C=20,2,14,22,11,2=u, c, o, w, l, c;

Deshifrlash jarayoni;Quyidagi formula orqali ifodalanadi.

М=𝑎^−1(С−𝑏)𝑚𝑜𝑑 𝑚2

M=157(-1)(522-580) mod m2=521(522-580) mod m2= -30218mod 676 = 202

M=202=7*26+20; ‘7=h’ ;20=u’;