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

Calculating square roots by Newton's method is horrendously fast ... provided that the starting value is reasonable. However there is no reasonable starting value, and in practice we end with bisection and log(2^64) behaviour.
To be really fast we need a fast way to get at a reasonable starting value, and that means we need to descend into machine language. If a processor provides an instruction like POPCNT in the Pentium, that counts the leading zeroes we can use that to have a starting value with half the significant bits. With care we can find a a fixed number of Newton steps that will always suffice. (Thus foregoing the need to loop and have very fast execution.)

A second solution is going via the floating point facility, which may have a fast sqrt calculation (like the i87 coprocessor.) Even an excursion via exp() and log() may be faster than Newton degenerated into a binary search. There is a tricky aspect to this, a processor dependant analysis of what and if refinement afterwards is necessary.

A third solution solves a slightly different problem, but is well worth mentionning because the situation is described in the question. If you want to calculate a great many square roots for numbers that differ slightly, you can use Newton iteration, if you never reinitialise the starting value, but just leave it where the previous calculation left off. I've used this with success in at least one Euler problem.