# Toshkent axborot texnologiyalari universiteti

 Sana 07.11.2019 Hajmi 80.86 Kb.

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;

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.

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.

1. Yuqoridfa keltirilgan e*d(mod)=1 shartni qanoatlantiruvchi d sonini 3d=1 (mod 20) tenglikdan topamiz. Bu son d=7.

2. 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}

1. 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,