Question

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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).

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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).

Answer 1

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…