Fastest way to determine if an integer's square root is an integer

Question

I\'m looking for the fastest way to determine if a long value is a perfect square (i.e. its square root is another integer):

1. I\'ve done it the ea

Answer 1

You'll have to do some benchmarking. The best algorithm will depend on the distribution of your inputs.

Your algorithm may be nearly optimal, but you might want to do a quick check to rule out some possibilities before calling your square root routine. For example, look at the last digit of your number in hex by doing a bit-wise "and." Perfect squares can only end in 0, 1, 4, or 9 in base 16, So for 75% of your inputs (assuming they are uniformly distributed) you can avoid a call to the square root in exchange for some very fast bit twiddling.

Kip benchmarked the following code implementing the hex trick. When testing numbers 1 through 100,000,000, this code ran twice as fast as the original.

public final static boolean isPerfectSquare(long n)
{
    if (n < 0)
        return false;

    switch((int)(n & 0xF))
    {
    case 0: case 1: case 4: case 9:
        long tst = (long)Math.sqrt(n);
        return tst*tst == n;

    default:
        return false;
    }
}

When I tested the analogous code in C++, it actually ran slower than the original. However, when I eliminated the switch statement, the hex trick once again make the code twice as fast.

int isPerfectSquare(int n)
{
    int h = n & 0xF;  // h is the last hex "digit"
    if (h > 9)
        return 0;
    // Use lazy evaluation to jump out of the if statement as soon as possible
    if (h != 2 && h != 3 && h != 5 && h != 6 && h != 7 && h != 8)
    {
        int t = (int) floor( sqrt((double) n) + 0.5 );
        return t*t == n;
    }
    return 0;
}

Eliminating the switch statement had little effect on the C# code.