Possible Duplicate:
Fastest way to determine if an integer's square root is an integer
What's a way to see if a number is a perfect square?
bool IsPerfectSquare(long input)
{
// TODO
}
I'm using C# but this is language agnostic.
Bonus points for clarity and simplicity (this isn't meant to be code-golf).
Edit: This got much more complex than I expected! It turns out the problems with double precision manifest themselves a couple ways. First, Math.Sqrt takes a double which can't precisely hold a long (thanks Jon).
Second, a double's precision will lose small values ( .000...00001) when you have a huge, near perfect square. e.g., my implementation failed this test for Math.Pow(10,18)+1 (mine reported true).
bool IsPerfectSquare(long input)
{
long closestRoot = (long) Math.Sqrt(input);
return input == closestRoot * closestRoot;
}
This may get away from some of the problems of just checking "is the square root an integer" but possibly not all. You potentially need to get a little bit funkier:
bool IsPerfectSquare(long input)
{
double root = Math.Sqrt(input);
long rootBits = BitConverter.DoubleToInt64Bits(root);
long lowerBound = (long) BitConverter.Int64BitsToDouble(rootBits-1);
long upperBound = (long) BitConverter.Int64BitsToDouble(rootBits+1);
for (long candidate = lowerBound; candidate <= upperBound; candidate++)
{
if (candidate * candidate == input)
{
return true;
}
}
return false;
}
Icky, and unnecessary for anything other than really large values, but I think it should work...