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

I like the idea to use an almost correct method on some of the input. Here is a version with a higher "offset". The code seems to work and passes my simple test case.

Just replace your:

if(n < 410881L){...}

code with this one:

if (n < 11043908100L) {
    //John Carmack hack, converted to Java.
    // See: http://www.codemaestro.com/reviews/9
    int i;
    float x2, y;

    x2 = n * 0.5F;
    y = n;
    i = Float.floatToRawIntBits(y);
    //using the magic number from 
    //http://www.lomont.org/Math/Papers/2003/InvSqrt.pdf
    //since it more accurate
    i = 0x5f375a86 - (i >> 1);
    y = Float.intBitsToFloat(i);
    y = y * (1.5F - (x2 * y * y));
    y = y * (1.5F - (x2 * y * y)); //Newton iteration, more accurate

    sqrt = Math.round(1.0F / y);
} else {
    //Carmack hack gives incorrect answer for n >= 11043908100.
    sqrt = (long) Math.sqrt(n);
}