O’zbekiston respublikasi raqamli texnologiyalari vazirligi muhammad al-Xorazmiy nomidagi Toshkent axborot texnologiyalari universiteti

GF(2³) f(x)=x² va g(x)=x²+x+1 bo’yicha qo’shish amalini bajaramiz, tartibi 3 ga teng bo‘lgan keltirilmaydigan polinom qiymati m(x)=x³+x+1 ga teng bo‘ladi.
Qo’shish: = f(x)+g(x)= (x²) + (x²+x+1) = x+13
Ayirish : f(x) - g(x) = ?
f(x) - g(x) = f(x)+(- g(x))
b=7;
-b=-7 mod 8= 1=001₂ -->> 1 = -g(x);
f(x) - g(x) = f(x)+(- g(x))=( x² ) + 1 =x;
Ko’paytirish : f(x)*g(x)= (x²) * (x²+x+1) = x⁴+x³+x²=
-->> maydondan chiqib ketganligi uchun mod m(x) ni hisoblaymiz.
x⁴+x³+x²| x³+x+1
x⁴+x²+x | x+1
x³ +x
x³ +x+1
1
Javob : f(x)*g(x)=1
Bo’lish : f(x) : g(x)= f(x)*g(x)^-1
g(x)^-1 mod m(x)
a*g(x)+b*m(x)=1
Kengaytirilgan Evklid algoritmidan foydalanib a va b larning qiymatlarini topamiz.
(x³+x+1)=(x+1)*(x²+x+1)+x
x²+x+1=(x+1)*x+1
x=x*1
1=x²+x+1 - (x+1)*x
1=x²+x+1-((x³+x+1)-(x²+x+1)*(x+1))*(x+1)
1=x²+x+1-(x+1)(x³+x+1)-(x²+x+1)*(x+1)²
1=(x²+x+1)*(x²+x+1)-(x+1)(x³+x+1)
g(x) B A m(x)


1-x²-2x-1=-x²-2x
f(x) : g(x)= f(x)*g(x)^-1=(x+1)*(x²+x+1)=x³+x²+x+x²+x+1=x³+1
Maydondan tashqarida bo’lganligi uchun mod m(x) ni hisoblaymiz.


x³+1 | x³+x+1
x³+x+1 | 1
x
Javob : f(x):g(x)=x

2. 
   a = 105, b = 65
   

Bizda a=4; b=7;
105=2⁶+2⁵ +2³+1 ; -->> x⁶+x⁵+x³+1;
65=2⁶+1 ; -->> x⁶+1;
GF(2³) f(x)=x⁶+x⁵+x³+1 va g(x)=x⁶+1 bo’yicha qo’shish amalini bajaramiz, tartibi 3 ga teng bo‘lgan keltirilmaydigan polinom qiymati m(x)=x⁸+x⁴+x³+x+1 ga teng bo‘ladi.
Qo’shish: f(x)+g(x)= x⁶+x⁵+x³+1 + x⁶+1 Javobi: x⁵ +x
Ayirish : f(x) - g(x) = ?
f(x) - g(x) = f(x)+(- g(x))
b = -65;
-b=-65 mod 256= 191 =1011111₂ -->> x⁶+x⁴+x³ +x²+1= -g(x);
f(x) - g(x) = f(x)+(- g(x))=x⁶ + x⁴ + x³ + x⁶+ x⁴ + x³+x² +1
Javobi: x²+1
Ko’paytirish:f(x)*g(x)=( x⁶+x⁵+x³+1)*
*(x⁶+1;)= x¹²+x⁶+x¹¹+x⁵+x⁹+x³+x⁶+x⁶+1=x¹²+x¹¹+x⁹+x⁶+x⁶+x⁵+x³+1 maydondan chiqib ketganligi uchun x¹²+x¹¹+x⁹+x⁵+x³+1 mod m(x) ni hisoblaymiz.
x¹²+x⁹+x⁸+x⁵+x⁴=(x⁸+x⁴+x³+x+1)*(x⁴+x)+(x⁷+x⁵+x⁴+x²+x)
Javob : f(x)*g(x)=x⁷+x⁵+x⁴+x²+x
Bo’lish : f(x) : g(x)= f(x)*g(x)^-1
g(x)^-1 mod m(x)
a*g(x)+b*m(x)=1
Kengaytirilgan Evklid algoritmidan foydalanib a va b larning qiymatlarini topamiz.


(x⁸+x⁴+x³+x+1) = (x⁶+x⁴+x)*(x²+1)+1
1 = (x⁸+x⁴+x³+x+1) - (x²+1)*(x⁶+x⁴+x)
a=1; b= - (x²+1);
b = -(x²+1) bo`lgani uchun biz buni maydonga tushirib olamiz.
Buning uchun modlab olamiz.
-(x²+1)=-5mod256=251=(x⁷+x⁶+x⁵+x⁴+x³+x+1)
g(x)^-1=x⁷+x⁶+x⁵+x⁴+x³+x+1;
f(x):g(x)=f(x)*g(x)^-1=(x⁶+x⁴+x³)*(x⁶+x²+x)= x¹³+x¹²+x¹¹+x¹⁰+x⁹+x⁷+x⁶+x¹¹+x¹⁰+x⁹+x⁸+x⁷+x⁵+x⁴+x¹⁰+
x⁹+x⁸+x⁷+x⁶+x⁴+x³=x¹³+x¹²+x¹⁰+x⁹+x⁷+x⁵+x³


Maydondan tashqarida bo’lganligi uchun mod m(x) ni hisoblaymiz.


x¹³+x¹²+x¹⁰+x⁹+x⁷+x⁵+x³ =(x⁸+x⁴+x³+x+1)*(x⁵+x⁴+x²) + x⁴+x²


Javob : f(x):g(x)=x³+x