(mod (4) This question deals with the finite field GF(28), which can be obtained as Z[x]...

 

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(mod (4) This question deals with the finite field GF(28), which can be obtained as Z₂[x] x³+x²+x³+x+1). This is the polynomial the AES cryptosystem uses, but it is not the only polynomial that gives GF(28) (see the last part of this question). In this question, you will learn the extended Euclidean algorithm for polynomials. A brief description of this algorithm is given below. Compute the following elements of Z₂[x] (mod x³ + x² + x³ + x + 1); your polynomials must be polynomials of degree at most 7. (a) (x6 +x5 + x¹ + x²) (x³ + x + 1). (b) (x5 + x² + 1) + (x³ + x²) (c) (x² + x + 1) - (x² + x + 1). (d) (x² + x) (x³ + 1)-¹ (I could have also written this as (x² + x)/(x³ + 1)). To compute (x³ + 1)-¹ in GF(28), we proceed as in the case of integers. Step 1. First compute the gcd (x³+1, x³ +xª+x³+x+1). Here use the Euclidean algorithm for polynomials. Each step in the Euclidean algorithm is a long division of polynomials with remainder. At every step, just make sure that the remainder has a degree less than that of the divisor. Step 2. As in the case of integers, trace your steps back and find polynomials s(x) and t(x) such that (x³ + 1) s(x) + (x³ + x² + x³ + x + 1)t(x) = gcd(x³ + 1, x³ + x² + x³ + x + 1). (e) GF (28) can also be obtained as Z₂[x] (mod x³ + x²+x6 +x+1). Under this new polynomial, compute again Parts (a) and (c). 4+3+3+7+(3+2)=22 marks (mod (4) This question deals with the finite field GF(28), which can be obtained as Z₂[x] x³+x²+x³+x+1). This is the polynomial the AES cryptosystem uses, but it is not the only polynomial that gives GF(28) (see the last part of this question). In this question, you will learn the extended Euclidean algorithm for polynomials. A brief description of this algorithm is given below. Compute the following elements of Z₂[x] (mod x³ + x² + x³ + x + 1); your polynomials must be polynomials of degree at most 7. (a) (x6 +x5 + x¹ + x²) (x³ + x + 1). (b) (x5 + x² + 1) + (x³ + x²) (c) (x² + x + 1) - (x² + x + 1). (d) (x² + x) (x³ + 1)-¹ (I could have also written this as (x² + x)/(x³ + 1)). To compute (x³ + 1)-¹ in GF(28), we proceed as in the case of integers. Step 1. First compute the gcd (x³+1, x³ +xª+x³+x+1). Here use the Euclidean algorithm for polynomials. Each step in the Euclidean algorithm is a long division of polynomials with remainder. At every step, just make sure that the remainder has a degree less than that of the divisor. Step 2. As in the case of integers, trace your steps back and find polynomials s(x) and t(x) such that (x³ + 1) s(x) + (x³ + x² + x³ + x + 1)t(x) = gcd(x³ + 1, x³ + x² + x³ + x + 1). (e) GF (28) can also be obtained as Z₂[x] (mod x³ + x²+x6 +x+1). Under this new polynomial, compute again Parts (a) and (c). 4+3+3+7+(3+2)=22 marks

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