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

This a rework from decimal to binary of the old Marchant calculator algorithm (sorry, I don't have a reference), in Ruby, adapted specifically for this question:

def isexactsqrt(v)
    value = v.abs
    residue = value
    root = 0
    onebit = 1
    onebit <<= 8 while (onebit < residue)
    onebit >>= 2 while (onebit > residue)
    while (onebit > 0)
        x = root + onebit
        if (residue >= x) then
            residue -= x
            root = x + onebit
        end
        root >>= 1
        onebit >>= 2
    end
    return (residue == 0)
end

Here's a workup of something similar (please don't vote me down for coding style/smells or clunky O/O - it's the algorithm that counts, and C++ is not my home language). In this case, we're looking for residue == 0:

#include   

using namespace std;  
typedef unsigned long long int llint;

class ISqrt {           // Integer Square Root
    llint value;        // Integer whose square root is required
    llint root;         // Result: floor(sqrt(value))
    llint residue;      // Result: value-root*root
    llint onebit, x;    // Working bit, working value

public:

    ISqrt(llint v = 2) {    // Constructor
        Root(v);            // Take the root 
    };

    llint Root(llint r) {   // Resets and calculates new square root
        value = r;          // Store input
        residue = value;    // Initialise for subtracting down
        root = 0;           // Clear root accumulator

        onebit = 1;                 // Calculate start value of counter
        onebit <<= (8*sizeof(llint)-2);         // Set up counter bit as greatest odd power of 2 
        while (onebit > residue) {onebit >>= 2; };  // Shift down until just < value

        while (onebit > 0) {
            x = root ^ onebit;          // Will check root+1bit (root bit corresponding to onebit is always zero)
            if (residue >= x) {         // Room to subtract?
                residue -= x;           // Yes - deduct from residue
                root = x + onebit;      // and step root
            };
            root >>= 1;
            onebit >>= 2;
        };
        return root;                    
    };
    llint Residue() {           // Returns residue from last calculation
        return residue;                 
    };
};

int main() {
    llint big, i, q, r, v, delta;
    big = 0; big = (big-1);         // Kludge for "big number"
    ISqrt b;                            // Make q sqrt generator
    for ( i = big; i > 0 ; i /= 7 ) {   // for several numbers
        q = b.Root(i);                  // Get the square root
        r = b.Residue();                // Get the residue
        v = q*q+r;                      // Recalc original value
        delta = v-i;                    // And diff, hopefully 0
        cout << i << ": " << q << " ++ " << r << " V: " << v << " Delta: " << delta << "\n";
    };
    return 0;
};