What is the simplest way of implementing bigint in C? - c

Question

I am trying to calculate 100! (that is, the factorial of 100).
I am looking for the simplest way to accomplish this using C. I have read around but have not found a concrete answer.
If you must know, I program in Xcode in Mac os X.

Answer 1

If you're looking for a simple library, libtommath (from libtomcrypt) is probably what you want.
If you're looking to write a simple implementation yourself (either as a learning exercise or because you only need a very limited subset of bigint functionality and don't want to tack on a dependency to a large library, namespace pollution, etc.), then I might suggest the following for your problem:
Since you can bound the size of the result based on n, simply pre-allocate an array of uint32_t of the required size to hold the result. I'm guessing you'll want to print the result, so it makes sense to use a base that's a power of 10 (i.e. base 1000000000) rather than a power of 2. That is to say, each element of your array is allowed to hold a value between 0 and 999999999.
To multiply this number by a (normal, non-big) integer n, do something like:
uint32_t carry=0;
for(i=0; i<len; i++) {
uint64_t tmp = n*(uint64_t)big[i] + carry;
big[i] = tmp % 1000000000;
carry = tmp / 1000000000;
}
if (carry) big[len++] = carry;
If you know n will never be bigger than 100 (or some other small number) and want to avoid going into the 64-bit range (or if you're on a 64-bit platform and want to use uint64_t for your bigint array), then make the base a smaller power of 10 so that the multiplication result will always fit in the type.
Now, printing the result is just something like:
printf("%lu", (long)big[len-1]);
for(i=len-1; i; i--) printf("%.9lu", (long)big[i-1]);
putchar('\n');
If you want to use a power of 2 as the base, rather than a power of 10, the multiplication becomes much faster:
uint32_t carry=0;
for(i=0; i<len; i++) {
uint64_t tmp = n*(uint64_t)big[i] + carry;
big[i] = tmp;
carry = tmp >> 32;
}
if (carry) big[len++] = carry;
However, printing your result in decimal will not be so pleasant... :-) Of course if you want the result in hex, then it's easy:
printf("%lx", (long)big[len-1]);
for(i=len-1; i; i--) printf("%.8lx", (long)big[i-1]);
putchar('\n');
Hope this helps! I'll leave implementing other things (like addition, multiplication of 2 bigints, etc) as an exercise for you. Just think back to how you learned to do base-10 addition, multiplication, division, etc. in grade school and teach the computer how to do that (but in base-10^9 or base-2^32 instead) and you should have no problem.

Answer 2

If you're willing to use a library implementation the standard one seems to be GMP
mpz_t out;
mpz_init(out);
mpz_fac_ui(out,100);
mpz_out_str(stdout,10,out);
should calculate 100! from looking at the docs.

Answer 3

You asked for the simplest way to do this. So, here you go:
#include <gmp.h>
#include <stdio.h>
int main(int argc, char** argv) {
mpz_t mynum;
mpz_init(mynum);
mpz_add_ui(mynum, 100);
int i;
for (i = 99; i > 1; i--) {
mpz_mul_si(mynum, mynum, (long)i);
}
mpz_out_str(stdout, 10, mynum);
return 0;
}
I tested this code and it gives the correct answer.

Answer 4

You can also use OpenSSL bn; it is already installed in Mac OS X.

Answer 5

You can print factorial 1000 in C with just 30 lines of code, <stdio.h> and char type :
#include <stdio.h>
#define B_SIZE 3000 // number of buffered digits
struct buffer {
size_t index;
char data[B_SIZE];
};
void init_buffer(struct buffer *buffer, int n) {
for (buffer->index = B_SIZE; n; buffer->data[--buffer->index] = (char) (n % 10), n /= 10);
}
void print_buffer(const struct buffer *buffer) {
for (size_t i = buffer->index; i < B_SIZE; ++i) putchar('0' + buffer->data[i]);
}
void natural_mul_buffer(struct buffer *buffer, const int n) {
int a, b = 0;
for (size_t i = (B_SIZE - 1); i >= buffer->index; --i) {
a = n * buffer->data[i] + b;
buffer->data[i] = (char) (a % 10);
b = a / 10;
}
for (; b; buffer->data[--buffer->index] = (char) (b % 10), b /= 10);
}
int main() {
struct buffer number_1 = {0};
init_buffer(&number_1, 1);
for (int i = 2; i <= 100; ++i)
natural_mul_buffer(&number_1, i);
print_buffer(&number_1);
}
You will find faster but the “little” factorial(10000) is here computed ≈ instantly.
You can put it into a fact.c file then compile + execute :
gcc -O3 -std=c99 -Wall -pedantic fact.c ; ./a.out ;
If you want to execute some base conversion there is a solution, see also Fibonacci(10000), Thank You.