Problem Statement

Sometimes a function is needed to find the integer square root of   X,   where   X   can be a real non─negative number. Often   X   is actually a non─negative integer. For the purposes of this task,   X   can be an integer or a real number,   but if it simplifies things in your computer programming language,   assume it's an integer.

One of the most common uses of   Isqrt   is in the division of an integer by all factors   (or primes)   up to the   √ X    of that integer,   either to find the factors of that integer,   or to determine primality.

An alternative method for finding the   Isqrt   of a number is to calculate:       floor( sqrt(X) )

If the hardware supports the computation of (real) square roots,   the above method might be a faster method for small numbers that don't have very many significant (decimal) digits. However, floating point arithmetic is limited in the number of   (binary or decimal)   digits that it can support.

For this task, the integer square root of a non─negative number will be computed using a version of   quadratic residue,   which has the advantage that no   floating point   calculations are used,   only integer arithmetic. Furthermore, the two divisions can be performed by bit shifting,   and the one multiplication can also be be performed by bit shifting or additions. The disadvantage is the limitation of the size of the largest integer that a particular computer programming language can support.

Pseudo─code of a procedure for finding the integer square root of   X       (all variables are integers): Another version for the (above)   1st   perform   is:

Integer square roots of some values:

Compute and show all output here   (on this page)   for:

You can show more numbers for the 2nd requirement if the displays fits on one screen on Rosetta Code. If your computer programming language only supports smaller integers,   show what you can.

Let's start with the solution:

comment
  return integer square root of n using quadratic
  residue algorithm. WARNING: the function will fail 
  for x > 16,383.
end
function isqrt(x = integer) = integer
  var q, r, t = integer
  q = 1
  while q <= x do  
    q = q * 4       rem overflow may occur here!
  r = 0
  while q > 1 do
    begin
      q = q / 4
      t = x - r - q
      r = r / 2
      if t >= 0 then
        begin
          x = t
          r = r + q
        end
    end
end = r

rem - Exercise the function

var n, pow7 = integer
print "Integer square root of first 65 numbers"
for n=1 to 65
  print using "#####";isqrt(n);
next n
print
print "Integer square root of odd powers of 7"
print "  n    7^n   isqrt"
print "------------------"
for n=1 to 3 step 2
  pow7 = 7^n
  print using "###  ####  ####";n; pow7; isqrt(pow7)
next n

end


function isqrt(x = integer) = integer
   var x0, x1 = integer
   x1 = x
   repeat
      begin
         x0 = x1
         x1 = (x0 + x / x0) / 2
      end
   until x1 >= x0
end = x0