long double vs long int

Question

I'm doing a program that calculates the probability of lotteries. Specification is choose 5 numbers out of 47 and 1 out of 27

So I did the following:

#include <iostream>

long int choose(unsigned n, unsigned k);
long int factorial(unsigned n);

int main(){
    using namespace std;
    long int regularProb, megaProb;
    regularProb = choose(47, 5);
    megaProb = choose(27, 1);
    cout << "The probability of the correct number is 1 out of " << (regularProb * megaProb) << endl;

    return 0;
}

long int choose(unsigned n, unsigned k){    
    return factorial(n) / (factorial(k) * factorial(n-k));
}

long int factorial(unsigned n){
    long int result = 1;
    for (int i=2;i<=n;i++) result *= i;
    return result;
}

However the program doesn't work. The program calculates for 30 seconds, then gives me Process 4 exited with code -1,073,741,676 I have to change all the long int to long double, but that loses precision. Is it because long int is too short for the big values? Though I thought long int nowadays are 64bit? My compiler is g++ win32 (64bit host).

Answer 1

Whether long is 64-bit or not depends on the model. Windows uses a 32-bit long. Use int64_t from <stdint.h> if you need to ensure it is 64-bit.

But even if long is 64-bit it is still too small to hold factorial(47).

47!  == 2.58623242e+59
2^64 == 1.84467441e+19

although ₄₇C₅ is way smaller than that.

You should never use _nC_r == n!/(r! (n-r)!) directly do the calculation as it overflows easily. Instead, factor out the n!/(n-r)! to get:

       47 * 46 * 45 * 44 * 43
  C  = ----------------------
47 5    5 *  4 *  3 *  2 *  1

this can be managed even by a 32-bit integer.

---

BTW, for @Coffee's question: a double only has 53-bits of precision, where 47! requires 154 bits. 47! and 42! represented in double would be

47! = (0b10100100110011011110001010000100011110111001100100100 << 145) ± (1 << 144)
42! = (0b11110000010101100000011101010010010001101100101001000 << 117) ± (1 << 116)

so 47! / (42! × 5!)'s possible range of value will be

      0b101110110011111110011 = 1533939                       53 bits
                                                              v
max = 0b101110110011111110011.000000000000000000000000000000001001111...
val = 0b101110110011111110010.111111111111111111111111111111111010100...
min = 0b101110110011111110010.111111111111111111111111111111101011010...

that's enough to get the exact value ₄₇C₅.