Question

Here is the problem we should solve with C++:

GCD ( 2m , 2n )         =  2 * GCD( m , n )
GCD ( 2m , 2n+1 )       = GCD ( m , 2n+1 )
GCD ( 2m+1,  2n+1 ) = GCD ( n-m , 2m+1 )  (m<n)
GCD ( m , m )       = m

and here is the function i wrote:

int GCD(int n1, int n2)
{
    bool n1Zoj, n2Zoj;
    n1Zoj = (n1%2 == 0);
    n2Zoj = (n2%2 == 0);

    if(n1Zoj && n2Zoj)
        return 2 * GCD(n1/2, n2/2);

    if(n1Zoj && !n2Zoj)
        return GCD(n1/2, n2);

    if(!n1Zoj && !n2Zoj)
        return GCD((n2-n1)/2, n1);

    if(n1 == n2)
        return n1;
}

(*”Zoj” means “Even” in my language(persian) )

when i pass 5 as the second argument, program crashes and prints this message:

Segmentation fault (core dumped)

Exit code is 139. I’m using Code::Blocks on ubuntu 12.04 that uses g++ as compiler.

UPDATE: Program crashes with 5,10,15,20,25,…

UPDATE: i think the correct form of function is:

int GCD(int n1, int n2)
{
    if (n1 > n2)
        std::swap(n1, n2);

    //std::cout<<"GCD is called with params: "<<n1<<" & "<<n2<<std::endl;
    bool n1Zoj, n2Zoj;

    n1Zoj = (n1%2 == 0);
    n2Zoj = (n2%2 == 0);

    if(n1 == n2)
        return n1;

    if(n1Zoj && n2Zoj)
        return 2 * GCD(n1/2, n2/2);

    if(n1Zoj && !n2Zoj)
        return GCD(n1/2, n2);

    if(!n1Zoj && n2Zoj)
        return GCD(n2/2, n1);

    if(!n1Zoj && !n2Zoj)
        return GCD((n2-n1)/2, n1);
}

Answer 1

When (n1Zoj && n2Zoj) evaluates to true, what do you do? You call

return 2 * GCD(n1, n2);

which calls the function with the exact same parameters, resulting in infinite recursion, a blown out stack, and a stack overflow (segmentation fault).

Protip – learn to debug – I can’t emphasize how extremely important this is.