Designing function f(f(n)) == -n

Question

A question I got on my last interview:

Design a function f, such that:

f(f(n)) == -n

Where n<

Answer 1

How about:

f(n) = sign(n) - (-1)n * n

In Python:

def f(n): 
    if n == 0: return 0
    if n >= 0:
        if n % 2 == 1: 
            return n + 1
        else: 
            return -1 * (n - 1)
    else:
        if n % 2 == 1:
            return n - 1
        else:
            return -1 * (n + 1)

Python automatically promotes integers to arbitrary length longs. In other languages the largest positive integer will overflow, so it will work for all integers except that one.

---

To make it work for real numbers you need to replace the n in (-1)ⁿ with { ceiling(n) if n>0; floor(n) if n<0 }.

In C# (works for any double, except in overflow situations):

static double F(double n)
{
    if (n == 0) return 0;

    if (n < 0)
        return ((long)Math.Ceiling(n) % 2 == 0) ? (n + 1) : (-1 * (n - 1));
    else
        return ((long)Math.Floor(n) % 2 == 0) ? (n - 1) : (-1 * (n + 1));
}