atakoute_CS3626_Assignment04

- 1 - CS3626 Homework 04 Spring 2024 AES Total Points: 25 Be as brief as possible and use your own words when describing concepts.   SHOW ALL WORK for Questions requiring calculations and algorithms. Q-1: Calculate the following polynomial operations over a m(x), GF(2^x) where a i ∈ GF(2) given m(x) Given these polynomials where: A = x 7 + x 6 + 0 + x 4 + 0 + x 2 + 0 + x 0 B = x 6 + x 5 + 0 + 0 + 0 + 0 + x 0 m(x) = x 8 + 0 + 0 + 0 + x 4 + x 3 + 0 + x 1 + x 0 a i operations are modulo-2 math. Calculate the operations below; show the setup and operation steps and result, remember they are over m(x):

- 2 - A mod B 2*A + 3*B + (A* B) + (A+B) = // recall the shortcut of 3*A = A + 2*A = A + (A << 1) over m(x) // you already have A*B and A+B from above

- 4 - Q-2: Build the 4x4 State box from the following 128-bit plain text input: Input = 4e 6f 20 6d 6f 72 65 20 73 65 63 72 65 74 73 2e 1 point Q-3: Take the above state box convert the state box values by applying each byte to the AES-SBOX. What is the new state box result?

- 5 - 2 points Q-4: What is the new state box result of applying the AES shift Columns function to the state box above: 2 points Q-5: Given the above state box, calculate the Mix Columns result of ONLY s’ 1,1 Recall:

- 7 - Q-6: Implement the following polynomial operation in a language (C++, C#, Python, Java): (i) Get Polynomial Degree Write a function to find the highest degree of a given polynomial: e.g. if given x 13 + x 7 + x 4 + x 1 + x 0 the function get_degree( a(x)) will return 13. A pseudocode algorithm: // returns highest exponent value of shortcut representation of a polynomial // returns -1 if a(x) == 0 get_degree( unsigned int a(x)) { int degree = -1; if( a(x) == 0) { return(degree); } while( a(x) != 0) { ++degree; // keep incrementing while there are 1’s in number a(x) = a(x) >> 1; // divide by 2 which shifts out lowest bit value } return( degree); } (ii) Polynomial Modulus over GF with m(x) Write a function to long divide two polynomials, i.e. (A(x), m(x)). Use the binary shortcut to indicate the values of A(x) = a n *x n + a n-1 *x n-1 +… a 5 *x 5 + a 4 x 4 + a 3 x 3 + a 2 x 2 + a 1 x 1 + a 0 x 0 becomes the binary number a n a n-1 … a 5 a 4 a 3 a 2 a 1 a 0 and perform the respective bitwise operations. This makes m(x) = 0x11B in hexadecimal for example. The function use modulo 2 math. A possible algorithm: Unsigned int divide_galois( A(x), m(x)) { result = A(x); while (get_degree(result) >= get_degree(m(x)) { result = result XOR ( get_degree(result) - get_degree(m(x))); } return( result); } unsigned int A(x) = multiplicand unsigned int B(x) = multiplier unsigned int m(x) = modulus = x 8 + 0 + 0 + 0 + x 4 + x 3 + 0 + x 1 + x 0 = 0x11B unsigned int multiply_galois( A(x), B(x), m(x)) { int index = 0; int result = 0;

- 8 - while( B(x) != 0) { // run until no bits left in multiplier If ( b 0 == 1) { // Test the lowest bit of b(x) result = result XOR (A(x) << index)) // index is the current degree of the multiplicand } B(x) = B(x) >> 1 // SHIFT out the current b 0 and bring new b 0 ++index; // raise the degree of the multiplicand for next loop in the line above } return( divide_galois( result, m(x)); } Using the above calculate the following and provide the results here: get_degree( 0x3CF0) = 13 get_degree( 0x10000) = 16 get_degree( 0x00) = -1 get_degree( 0x01) = 0 divide_galois( 0x1000, 0x11B) = 0xAB divide_galois( 0xE1, 0x11B) = 0xE1 divide_galois( 0x32CFE1, 0x11B) = 0xD3 divide_galois( 0xE1, 0x11B) = 0xE1 multiply_galois( 0xD5, 0x61, 0x11B) = 0xE1 multiply_galois( 0x1E3C, 0x1E3C, 0x11B) = 0xA (i) 3 points (ii) 4 points (iii) 4 points Keep source code, it will be needed for future exercises. AES KEYGEN and DECRYPTION will be in Next Assignment