Thread: Goldbach's Conjecture problem

Some optimizations step by step:

Code:

for (a=2; a<=x; a++)
{
    for (b=2; b<=x; b++)
    {
        if ((x>2) && ((x%2)==0) && (a<=b) && (a+b==x))
        {
            cout << a << " and " << b << endl; 
        }
    }
}

The red condition is unnecessary, since the input has already been verified. When it comes to the orange test, it can be moved to the inner loop's condition:

Code:

for (b=2; b<=x; b++)
{
    for (a=2; a<=b; a++)
    {
        if (a+b==x)
        {
            cout << a << " and " << b << endl; 
        }
    }
}

Note that I reordered the loops, now the 'b' is outer.
Going this way you can do what iMalc said eliminating the last condition.
Given 'a' in set: <2..b> there can be only one (or zero) 'a', such that (a + b = x), and this is (a = x - b):

Code:

for (b=2; b<=x; b++)
{
    cout << x - b << " and " << b << endl; 
}

If (b == x) then (a == 0), if (b == x - 1) then (a == 1). Both 0 and 1 are not primes:

Code:

for (b=2; b < x - 1; b++)
{
    cout << x - b << " and " << b << endl; 
}

The code is much cleaner now, I think you will manage to implement IsPrime function. Just in case you get stuck:

Code:

for (b=2; b < x - 1; b++)
{
    if (IsPrime(b) && IsPrime(x - b))
    {
        cout << x - b << " and " << b << endl;
    }
}