Is there a way to build e.g. (853467 * 21660421200929) % 100000000000007 without BigInteger libraries (note that each number fits into a 64 bit integer but the multiplication result does not)?
This solution seems inefficient:
int64_t mulmod(int64_t a, int64_t b, int64_t m) {
if (b < a)
std::swap(a, b);
int64_t res = 0;
for (int64_t i = 0; i < a; i++) {
res += b;
res %= m;
}
return res;
}
You should use Russian Peasant multiplication. It uses repeated doubling to compute all the values (b*2^i)%m, and adds them in if the ith bit of a is set.
uint64_t mulmod(uint64_t a, uint64_t b, uint64_t m) {
int64_t res = 0;
while (a != 0) {
if (a & 1) res = (res + b) % m;
a >>= 1;
b = (b << 1) % m;
}
return res;
}
It improves upon your algorithm because it takes O(log(a)) time, not O(a) time.
Caveats: unsigned, and works only if m is 63 bits or less.
Keith Randall's answer is good, but as he said, a caveat is that it works only if m is 63 bits or less.
Here is a modification which has two advantages:
m is 64 bits.(Note that the res -= m and temp_b -= m lines rely on 64-bit unsigned integer overflow in order to give the expected results. This should be fine since unsigned integer overflow is well-defined in C and C++. For this reason it's important to use unsigned integer types.)
uint64_t mulmod(uint64_t a, uint64_t b, uint64_t m) {
uint64_t res = 0;
uint64_t temp_b;
/* Only needed if b may be >= m */
if (b >= m) {
if (m > UINT64_MAX / 2u)
b -= m;
else
b %= m;
}
while (a != 0) {
if (a & 1) {
/* Add b to res, modulo m, without overflow */
if (b >= m - res) /* Equiv to if (res + b >= m), without overflow */
res -= m;
res += b;
}
a >>= 1;
/* Double b, modulo m */
temp_b = b;
if (b >= m - b) /* Equiv to if (2 * b >= m), without overflow */
temp_b -= m;
b += temp_b;
}
return res;
}
Both methods work for me. The first one is the same as yours, but I changed your numbers to excplicit ULL. Second one uses assembler notation, which should work faster. There are also algorithms used in cryptography (RSA and RSA based cryptography mostly I guess), like already mentioned Montgomery reduction as well, but I think it will take time to implement them.
#include <algorithm>
#include <iostream>
__uint64_t mulmod1(__uint64_t a, __uint64_t b, __uint64_t m) {
if (b < a)
std::swap(a, b);
__uint64_t res = 0;
for (__uint64_t i = 0; i < a; i++) {
res += b;
res %= m;
}
return res;
}
__uint64_t mulmod2(__uint64_t a, __uint64_t b, __uint64_t m) {
__uint64_t r;
__asm__
( "mulq %2\n\t"
"divq %3"
: "=&d" (r), "+%a" (a)
: "rm" (b), "rm" (m)
: "cc"
);
return r;
}
int main() {
using namespace std;
__uint64_t a = 853467ULL;
__uint64_t b = 21660421200929ULL;
__uint64_t c = 100000000000007ULL;
cout << mulmod1(a, b, c) << endl;
cout << mulmod2(a, b, c) << endl;
return 0;
}
An improvement to the repeating doubling algorithm is to check how many bits at once can be calculated without an overflow. An early exit check can be done for both arguments -- speeding up the (unlikely?) event of N not being prime.
e.g. 100000000000007 == 0x00005af3107a4007, which allows 16 (or 17) bits to be calculated per each iteration. The actual number of iterations will be 3 with the example.
// just a conceptual routine
int get_leading_zeroes(uint64_t n)
{
int a=0;
while ((n & 0x8000000000000000) == 0) { a++; n<<=1; }
return a;
}
uint64_t mulmod(uint64_t a, uint64_t b, uint64_t n)
{
uint64_t result = 0;
int N = get_leading_zeroes(n);
uint64_t mask = (1<<N) - 1;
a %= n;
b %= n; // Make sure all values are originally in the proper range?
// n is not necessarily a prime -- so both a & b can end up being zero
while (a>0 && b>0)
{
result = (result + (b & mask) * a) % n; // no overflow
b>>=N;
a = (a << N) % n;
}
return result;
}
You could try something that breaks the multiplication up into additions:
// compute (a * b) % m:
unsigned int multmod(unsigned int a, unsigned int b, unsigned int m)
{
unsigned int result = 0;
a %= m;
b %= m;
while (b)
{
if (b % 2 != 0)
{
result = (result + a) % m;
}
a = (a * 2) % m;
b /= 2;
}
return result;
}
a * b % m equals a * b - (a * b / m) * m
Use floating point arithmetic to approximate a * b / m. The approximation leaves a value small enough for normal 64 bit integer operations, for m up to 63 bits.
This method is limited by the significand of a double, which is usually 52 bits.
uint64_t mod_mul_52(uint64_t a, uint64_t b, uint64_t m) {
uint64_t c = (double)a * b / m - 1;
uint64_t d = a * b - c * m;
return d % m;
}
This method is limited by the significand of a long double, which is usually 64 bits or larger. The integer arithmetic is limited to 63 bits.
uint64_t mod_mul_63(uint64_t a, uint64_t b, uint64_t m) {
uint64_t c = (long double)a * b / m - 1;
uint64_t d = a * b - c * m;
return d % m;
}
These methods require that a and b be less than m. To handle arbitrary a and b, add these lines before c is computed.
a = a % m;
b = b % m;
In both methods, the final % operation could be made conditional.
return d >= m ? d % m : d;
I can suggest an improvement for your algorithm.
You actually calculate a * b iteratively by adding each time b, doing modulo after each iteration. It's better to add each time b * x, whereas x is determined so that b * x won't overflow.
int64_t mulmod(int64_t a, int64_t b, int64_t m)
{
a %= m;
b %= m;
int64_t x = 1;
int64_t bx = b;
while (x < a)
{
int64_t bb = bx * 2;
if (bb <= bx)
break; // overflow
x *= 2;
bx = bb;
}
int64_t ans = 0;
for (; x < a; a -= x)
ans = (ans + bx) % m;
return (ans + a*b) % m;
}
