Burxonov uktam

Akkauntni o’chirmaslik
BURXONOV UKTAM




chiqishdan oldin tekshirish kerak
Qayta shifrlash kerak

Kiberxujumlar ko’payishi

Ishonchsiz saytdan foydalanish

Maxfiy ma’lumotlarning tarqalishi








aybdorni topish
Rasmiylik kerak
Qonuniy chora ko’rish kerak

Xodim ishdan bo’shatilsa

Ishonchli odamlarga ma’lumotlarni berish kerak kerak

Tartibsizlik ko’payadi

O’rin almashtirish usullari
O’rin almashtirish shifirlash algoritimi uch qisimdan iborat.Oddiy o’rin almashtirish, kalit yordamida o’rin almashtirish va ikki tomonlama o’rin almashtirish.
1 Oddiy o’rin almashtirish
T₀ = Kiber xafsizlig fa;

K	R	S	I
I	X	I	G
B	A	Z	F
E	F	L	A

K =5x5;

T₁ =KRSI_IXIG_BAZF_EFLA
2.kalit yordamida o’rin almashtirish
T₀ = Surxondaryo viloyat;
K= Uktami;
Matnda 18- ta kalitda 3- ta harflar bor 3x6 jadval tuzamiz.

S	X	D	Y	I	Y
U	O	A	O	L	A
R	N	R	V	O	T

U	K	T	A	M	I
1	2	3	4	5	6
S	X	D	Y	I	Y
U	O	A	O	L	A
R	N	R	V	O	T

A	I	K	M	T	U
1	2	3	4	5	6
Y	Y	X	I	D	S
O	A	O	L	A	U
V	T	N	O	R	R

T1=yyxids oaolau vtnorr
Ikki tomonlama o’rin almashtirish usuli
T=Uktambek Burxonov
K1=1342 va K2=3214

K1/K2	1	3	4	2
3	U	M	B	O
2	K	B	U	N
1	T	E	R	O
4	A	K	X	V

K1/K2	1	2	3	4
1	T	O	E	R
2	K	N	B	U
3	U	O	M	B
4	A	V	K	X

T1=TOERKNBUUOMBAVKX

Vernar usul

T₀ = Burxonov Xolbekovich
K= Uktam
<sub>B=1=00001 U=20=10100 BU=10101===V
U=20=10100 K=10=01001 UK=10101===V
R=17=10001 T=19=10011 RT=00010===C
X=23=10111 A=0=00000 XA=10111===X
0=14=01110 M=12=01100 OM=00010===C
N=13=01101 U=20=10100 NU=11001===Z
O=14=01110 K=10=01010 OK=00100===E
V=21=10101 T=19=10011 VT=00110===G
X=23=10111 A=0=00000 XA=10111===X
O=14=01110 M=12=01100 0M=00010===C
L=11=01011 U=20=10100 LU=11111*
B=1=00001 K=10=01010 BK=01011===L
E=4=00100 T=19=10011 ET=10111===X
K=10=01010 A=0=00000 KA=01010===K
O=14=01110 U=20=10100 OU=10010===S
V=21=10101 K=10=01010 VK=11111===*
I=8=01000 T=19=10011 IT=11011===!
C=2=00010 A=0=00000 CA=00010===C
H=7=00111 M=12=01100 HM=01011==L
T1=VVCXCZEGXC*LXKS*!CL

A5/1 shifrlash algoritimi</sub>

A5/1 shifrlash algoritmida dastlabki kalitning uzunligi 64 bitni tashkil etib, u quyidagi uchta registorga qiymat qilib beriladi:

• 
  X: 19 bit (x₀,x₁,x₂, …,x₁₈)
  
• 
  Y: 22 bit (y₀,y₁,y₂, …,y₂₁)
  
• 
  Z: 23 bit (z₀,z₁,z₂, …,z₂₂)
  

Har bir qadamda: m = maj(x₈, y₁₀, z₁₀) hisoblanadi

• 
  masalan: maj(0,1,0) = 0 vа maj(1,1,0) = 1
  
• 
  agar x₈ = m ga teng bo‘lsa, u holda X registor qiymatlari
  
  • 
    t = x₁₃  x₁₆  x₁₇  x₁₈
    
  • 
    x_i = x_i1 for i = 18,17,…,1 va x₀ = t
    
• 
  agar y₁₀ = m ga teng bo‘lsa, u holda Y registor qiymatlari
  
  • 
    t = y₂₀  y₂₁
    
  • 
    y_i = y_i1 for i = 21,20,…,1 and y₀ = t
    
• 
  agar z₁₀ = m ga teng bo‘lsa, u holda Z registor qiymatlari
  
  • 
    t = z₇  z₂₀  z₂₁  z₂₂
    
  • 
    z_i = z_i1 for i = 22,21,…,1 and z₀ = t
    
• 
  natijaviy kalit ketma-ketligi x₁₈  y₂₁  z₂₂ ga teng bo‘ladi.
  

masalan quyidagi ko’rsatilgan hol uchun;

m = maj(x₈, y₁₀, z₁₀) = maj(1,0,1) = 1 ga teng bo‘ladi. Natijada X registor siljiydi, Y registor siljimaydi va Z registor siljiydi. O‘ng tomondagi bitlar XOR amal bo‘yicha qo‘shiladi va 0  1  0 = 1 qiymat olinadi.
Ushbu usulda bir siklda bir bit kalit x xosil qilinadi.

rsa usuli

1. 
   p=3 q=11 UKTAM
   
2. 
   2)n=p*q=33
   
3. 
   3) =(p-1)(q-1)=20
   
4. 
   e=7 e*d*mod) =1
   
5. 
   7*d*mod20=1 d=3
   
6. 
   C1=Hmodn=20⁷