Integer square root
For example, because and .
Algorithm using Newton's method
One way of calculating and is to use Newton's method to find a solution for the equation , giving the iterative formula
to ensure that
Using only integer division
For computing for very large integers n, one can use the quotient of Euclidean division for both of the division operations. This has the advantage of only using integers for each intermediate value, thus making the use of floating point representations of large numbers unnecessary. It is equivalent to using the iterative formula
By using the fact that
one can show that this will reach within a finite number of iterations.
However, is not necessarily a fixed point of the above iterative formula. Indeed, it can be shown that is a fixed point if and only if is not a perfect square. If is a perfect square, the sequence ends up in a period-two cycle between and instead of converging. For termination, it suffices to check that either the number has converged or it has increased by exactly one from the previous step, in which case the new result is discarded.
Domain of computation
Although is irrational for many , the sequence contains only rational terms when is rational. Thus, with this method it is unnecessary to exit the field of rational numbers in order to calculate , a fact which has some theoretical advantages.
One can prove that is the largest possible number for which the stopping criterion
ensures in the algorithm above.
In implementations which use number formats that cannot represent all rational numbers exactly (for example, floating point), a stopping constant less than one should be used to protect against roundoff errors.
The traditional pen-and-paper algorithm for computing the square root is based on working from higher digit places to lower, and as each new digit pick the largest that will still yield a square . If stopping after the one's place, the result computed will be the integer square root.
Using bitwise operations
If working in base 2, the choice of digit is simplified to that between 0 (the "small candidate") and 1 (the "large candidate"), and digit manipulations can be expressed in terms of binary shift operations. With
* being multiplication,
<< being left shift, and
>> being logical right shift, a recursive algorithm to find the integer square root of any natural number is:
function integerSqrt(n): if n < 0: error "integerSqrt works for only nonnegative inputs" else if n < 2: return n else: # Recursive call: smallCandidate = integerSqrt(n >> 2) << 1 largeCandidate = smallCandidate + 1 if largeCandidate*largeCandidate > n: return smallCandidate else: return largeCandidate
Or, iteratively instead of recursively:
function integerSqrt(n): if n < 0: error "integerSqrt works for only nonnegative inputs" # Find greatest shift. shift = 2 nShifted = n >> shift # We check for nShifted being n, since some implementations # perform shift operations modulo the word size. while nShifted ≠ 0 and nShifted ≠ n: shift = shift + 2 nShifted = n >> shift shift = shift - 2 # Find digits of result. result = 0 while shift ≥ 0: result = result << 1 candidateResult = result + 1 if candidateResult*candidateResult ≤ n >> shift: result = candidateResult shift = shift - 2 return result
Traditional pen-and-paper presentations of the digit-by-digit algorithm include various optimisations not present in the code above, in particular the trick of presubtracting the square of the previous digits which makes a general multiplication step unnecessary.