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Lecture notes #5 - EECS: www
... For example, 29 and 5 are congruent modulo 12 because 12 divides 29 − 5. We can also write 22 ≡ −2 (mod 12). Notice that x and y are congruent modulo m iff they have the same remainder modulo m. What is the set of numbers that are congruent to 0 (mod 12)? These are all the multiples of 12: {. . . , ...
... For example, 29 and 5 are congruent modulo 12 because 12 divides 29 − 5. We can also write 22 ≡ −2 (mod 12). Notice that x and y are congruent modulo m iff they have the same remainder modulo m. What is the set of numbers that are congruent to 0 (mod 12)? These are all the multiples of 12: {. . . , ...
Lecture 5 1 Integer multiplication via polynomial multiplication
... bit operations. (The details are left as an exercise.) Hence, O( N log N ) multiplications by powers of ω in R takes O(N log N ) bit operations. The purpose of choosing ω, a power of 2, is to reduce multiplications by powers of ω to shift ...
... bit operations. (The details are left as an exercise.) Hence, O( N log N ) multiplications by powers of ω in R takes O(N log N ) bit operations. The purpose of choosing ω, a power of 2, is to reduce multiplications by powers of ω to shift ...
Lecture Notes: Cryptography – Part 2
... 63 ≡ 0 (mod 63) 20 ≡ 20 (mod 63) 3 ≡ −3 · 20 (mod 63) 2 ≡ 19 · 20 (mod 63) 1 ≡ −22 · 20 (mod 63) 0 ≡ 63 · 20 (mod 63) Now we must stop, because we cannot divide 1 by 0. Recall that we were looking for a multiplicative inverse of 20 modulo 63, i.e., we wanted some number b such that 20 · b ≡ 1 (mod 6 ...
... 63 ≡ 0 (mod 63) 20 ≡ 20 (mod 63) 3 ≡ −3 · 20 (mod 63) 2 ≡ 19 · 20 (mod 63) 1 ≡ −22 · 20 (mod 63) 0 ≡ 63 · 20 (mod 63) Now we must stop, because we cannot divide 1 by 0. Recall that we were looking for a multiplicative inverse of 20 modulo 63, i.e., we wanted some number b such that 20 · b ≡ 1 (mod 6 ...