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


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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

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t

u

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w

x

y

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0

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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

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f

u

m

u

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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

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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

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d

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t

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u

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u

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m

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u

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m

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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’;
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