modular-arithmetic

I'm confused about how one might supposedly bypass the final subtraction of the modulus in radix-2 montgomery modular multiplication , when used in a modular exponentiation algorithm. The following two papers put forward the conditions for bypassing the subtraction. Montgomery Exponentiation with no Final Subtractions: Improved Results Montgomery Multiplication Needs no Final Subtractions I don't understand exactly what is required in terms of the "preprocessing and postprocessing" to eliminate the need for the repetitive subtraction of the modulus at the end of the montgomery multiplication.

I have been studying this link from wikipedia of modulo of a large number, Here is the pseudocode. function modular_pow(base, exponent, modulus) result := 1 while exponent > 0 if (exponent mod 2 == 1): result := (result * base) mod modulus exponent := exponent >> 1 base = (base * base) mod modulus return result I don't understand the explanation given in wiki.Why I have to check if exp%2 is even or odd. also why I am doing the three operations? This algorithm is a combination of the Exponentiation by Squaring algorithm and modulo arithmetic. To understand what's going on, first consider a

I understand the concept of a Residual Number System and the concept of a Mixed Radix system , but I'm having difficulty getting any of the conversion methods I find to work in a simple case study. I started at Knuth's Art of Computer Programming but that had a bit too much on the theory of the conversion, and once Euler was mentioned I was lost. Wikipedia has a nice section on the subject, which I tried here and here but both times I couldn't get back to the number where I started. I found a good article here (PDF) , which I condensed the relevant sections here , but I don't understand the

How can I calculate (a ^ b) % c , where 0 <= a, b, c <= 10^18 . Here, (a ^ b) means a to the power b , not a xor b . My current code for the problem is: unsigned long long bigMod(unsigned long long b, unsigned long long p, unsigned long long m){ if(b == 1) return b; if(p == 0) return 1; if(p == 1) return b; if(p % 2 == 0){ unsigned long long temp = bigMod(b, p / 2ll, m); return ((temp) * (temp) )% m; }else return (b * bigMod(b, p-1, m)) % m; } For this input: a = 12345 b = 123456789 and c = 123456789012345 the expected output should be: 59212459031520 You have a problem with temp*temp (long

This question already has an answer here: Calculate a*a mod n without overflow 5 answers I have 3 large 64 bit numbers: A, B and C. I want to compute: (A x B) mod C considering my registers are 64 bits, i.e. writing a * b actually yields (A x B) mod 2⁶⁴. What is the best way to do it? I am coding in C, but don't think the language is relevant in this case. After getting upvotes on the comment pointing to this solution: (a * b) % c == ((a % c) * (b % c)) % c let me be specific: this isn't a solution, because ((a % c) * (b % c)) may still be bigger than 2⁶⁴, and the register would still overflow