Challenge

[helios]

Make this iterative:

typedef std::function<int()> N;

int A(int k, const N &x1, const N &x2, const N &x3, const N &x4, const N &x5){
    if (k <= 0)
        return x4() + x5();
    N B = [&](){
        k--;
        return A(k, B, x1, x2, x3, x4);
    };
    return B();
}

N F(int n){
    return [=](){ return n; };
}

int A(int k, int x1, int x2, int x3, int x4, int x5){
    return A(k, F(x1), F(x2), F(x3), F(x4), F(x5));
}

A(16, 1, -1, -1, 1, 0);

[kbw]

Tricky.

BTW, it never terminates.

[zapshe]

Plot Twist: Helios gives us his work in disguise of a challenge.

[againtry]

Plot Twist: Helios gives us his her work in disguise of a challenge.

[helios]

What do you mean, it never terminates?

[dutch]

helios wrote:

What do you mean, it never terminates?

It probably just blew his stack, as it did mine before I enlarged it (with ulimit). I hope you post the answer at some point (but not yet!).

[Ganado]

Oh, I didn't see this until now. Looks interesting. I assume the constraint is that we have to keep the interface to A(6x int) unchanged?

[helios]

Err... Sure, although it doesn't make it much easier if you don't.

[dutch]

A(50, 1, -1, -1, 1, 0) = -4357661238744040

A(100, 1, -1, -1, 1, 0) = -920829370818316202730797688469299

[helios]

Did you find a stack-free solution? My computer ran out of memory at around k=27.

[dutch]

Yes, but I basically cheated.
I never would've figured it out myself.
I'll PM you the solution and how I found it.