Toshkent axborot texnologiyalari universiteti

MUXAMMAD AL- XORAZMIY NOMIDAGI

TOSHKENT AXBOROT TEXNOLOGIYALARI

UNIVERSITETI

Kriptografiya kafedrasi

Kiberxavfsizlik fani

Amaliy ish

Mavzu: Kriptografik himoyalash.

Bajardi: Ergashev Jasurbek

Tekshirdi: Shirinov Laziz

Toshkent – 2019

Topshiriq: Quyida keltirilgan usullar yordamida shifrlash va dishifrlash;

1. Sezir;

2. Polibiya kvadrati;

3. Vernam usuli;

4. Gamelton marshruti;

5. RSA;

1. Sezir usuli:

T = ERGASHEV K=8.


E 1 2 3 4 5 6 7 8=M
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


R 1 2 3 4 5 6 7 8=Z
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


G 1 2 3 4 5 6 7 8=O
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


A 1 2 3 4 5 6 7 8=I
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


S 1 2 3 4 5 6 7 8 = A
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 A


H 1 2 3 4 5 6 7 8=P
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


E 1 2 3 4 5 6 7 8=M
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


V 1 2 3 4 5 6 7 8 = D
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 A B C D

Javob: ERGASHEV  MZOIAPMD.

2. Polibiya kvadrati: T=ERGASHEV K=8.


Bu usulda olingan harf bo’yicha berilgan kalit orqali pastga birin ketin sanab harfga tegishli bo’lgan shifr harf topiladi.

	1	2	3	4	5
1	A	B	C	D	E
2	F	G	H	I/J	K
3	L	M	N	O	P
4	Q	R	S	T	U
5	V	W	X	Y	Z

ERGASHEV  UGWQHXUL.

3. Vernam usuli: T=ERGASHEV K=SALOM.

17	Q	10000
18	R	10001
19	S	10010
20	T	10011
21	U	10100
22	V	10101
23	W	10110
24	X	10111
25	Y	11000
26	Z	11001
27	#	11010
28	!	11011
29	-	11100
30	@	11101
31	?	11110
32	*	11111

1	A	00000
2	B	00001
3	C	00010
4	D	00011
5	E	00100
6	F	00101
7	G	00110
8	H	00111
9	I	01000
10	J	01001
11	K	01010
12	L	01011
13	M	01100
14	N	01101
15	O	01110
16	P	01111

R	+	A	=		R
1	+	0	=	1
0	+	0	=	0
0	+	0	=	0
0	+	0	=	0
1	+	0	=	1

G	+	L	=		N
0	+	0	=	0
0	+	1	=	1
1	+	0	=	1
1	+	1	=	0
0	+	1	=	1

E	+	S	=	W
0	+	1	=	1
0	+	0	=	0
1	+	0	=	1
0	+	1	=	1
0	+	0	=	0

S	+	M	=	?
1	+	0	=	1
0	+	1	=	1
0	+	1	=	1
1	+	0	=	1
0	+	0	=	0

A

+

O

=

O

H	+	S	=		V
0	+	1	=	1
0	+	0	=	0
1	+	0	=	1
1	+	1	=	0
1	+	0	=	1

E

+

A

=

A

V	+	L		=		?
1	+	0	=		1
0	+	1	=		1
1	+	0	=		1
0	+	1	=		1
1	+	1	=		0

Javob : ERGASHEV  WRNO?VE?.

4. Gamelton marshruti: T = ERGASHEV.


S

5

6



H



1

2



E

R



3

4



A

G

V

E

7

8

C = H V E G A R E S

K = 6 8 7 3 4 2 1 5

4. RSA algoritmi T = ERGASHEV

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

1. 
   Тub bo’lgan р=3 vа q=11 sonlarini tanlab olamiz.
   
2. 
   Ushbu n=pq=3*11=33 sonini aniqlaymiz.
   

So’ngra, sonini topamiz, hamda bu son bilan 1 dan farqli biror umumiy bo;luvchiga ega bo’lmagan e sonini, misol uchun e=3 sonini, olamiz.

3. 
   Yuqoridfa keltirilgan e*d(mod)=1 shartni qanoatlantiruvchi d sonini 3d=1 (mod 20) tenglikdan topamiz. Bu son d=7.
   
4. 
   U holda ma’lumot {ERGASHEV}{5,18,7,1,19,8,5,22} ko’rinishda bo’ladi va uni {e;n}={3;33} ochiq kalit bilan bir tomonli funksiya bilan shifrlaymiz:
   

х=5 da ShM1=(5³)(mod33)=26,

х=18 da ShM2=(18³)(mod33)=24,

х=7 da ShM3=(7³)(mod33)=13,

х=1 da ShM4=(1³)(mod33)=1,

х=19 da ShM5=(19³)(mod33)=28,

х=8 da ShM6=(8³)(mod33)=17,

х=5 da ShM7=(5³)(mod33)=26,

х=22 da ShM8=(22³)(mod33)=22,
C = {26, 24, 13, 1, 28, 17, 26, 22}

5. 
   Bu olingan shifrlangan {26, 24, 13, 1, 28, 17, 26, 22} ma’lumotni maxfiy {d;n}={7;33} kalit bilan ifoda orqali deshifrlaymiz:
   

у=26 dа ОМ1=(26⁷) (mod33)=5,

у=24 dа ОМ1=(24⁷) (mod33)=18, {5, 18, 7, 1, 19, 8, 5, 22}

у=13 dа ОМ1=(26⁷) (mod33)=7,

у=1 dа ОМ1=(26⁷) (mod33)=1,

у=28 dа ОМ1=(26⁷) (mod33)=19,

у=17 dа ОМ1=(26⁷) (mod33)=8,

у=26 dа ОМ1=(26⁷) (mod33)=5,

у=22 da ОМ1=(26⁷) (mod33)=22,