Integer square root

For computing ${\displaystyle \lfloor {\sqrt {n}}\rfloor }$ 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

${\displaystyle {x}_{k+1}=\left\lfloor {\frac {1}{2}}\left(x_{k}+\left\lfloor {\frac {n}{x_{k}}}\right\rfloor \right)\right\rfloor ,\quad k\geq 0,\quad x_{0}>0,\quad x_{0}\in \mathbb {Z} .}$

By using the fact that

${\displaystyle \left\lfloor {\frac {1}{2}}\left(x_{k}+\left\lfloor {\frac {n}{x_{k}}}\right\rfloor \right)\right\rfloor =\left\lfloor {\frac {1}{2}}\left(x_{k}+{\frac {n}{x_{k}}}\right)\right\rfloor ,}$

one can show that this will reach ${\displaystyle \lfloor {\sqrt {n}}\rfloor }$ within a finite number of iterations.

However, ${\displaystyle \lfloor {\sqrt {n}}\rfloor }$ is not necessarily a fixed point of the above iterative formula. Indeed, it can be shown that ${\displaystyle \lfloor {\sqrt {n}}\rfloor }$ is a fixed point if and only if ${\displaystyle n+1}$ is not a perfect square. If ${\displaystyle n+1}$ is a perfect square, the sequence ends up in a period-two cycle between ${\displaystyle \lfloor {\sqrt {n}}\rfloor }$ and ${\displaystyle \lfloor {\sqrt {n}}\rfloor +1}$ instead of converging.

Although ${\displaystyle {\sqrt {n}}}$ is irrational for many ${\displaystyle n}$, the sequence ${\displaystyle \{x_{k}\}}$ contains only rational terms when ${\displaystyle x_{0}}$ is rational. Thus, with this method it is unnecessary to exit the field of rational numbers in order to calculate ${\displaystyle {\mbox{isqrt}}(n)}$, a fact which has some theoretical advantages.

One can prove that ${\displaystyle c=1}$ is the largest possible number for which the stopping criterion

${\displaystyle |x_{k+1}-x_{k}|<c}$

ensures ${\displaystyle \lfloor x_{k+1}\rfloor =\lfloor {\sqrt {n}}\rfloor }$ 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.

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:

def integer_sqrt(n: int) -> int:
    assert n >= 0, "sqrt works for only non-negative inputs"
    if n < 2:
        return n

    # Recursive call:
    small_cand = integer_sqrt(n >> 2) << 1
    large_cand = small_cand + 1
    if large_cand * large_cand > n:
        return small_cand
    else:
        return large_cand


# equivalently:
def integer_sqrt_iter(n: int) -> int:
    assert n >= 0, "sqrt works for only non-negative inputs"
    if n < 2:
        return n

    # Find the shift amount. See also [[find first set]],
    # shift = ceil(log2(n) * 0.5) * 2 = ceil(ffs(n) * 0.5) * 2
    shift = 2
    while (n >> shift) != 0:
        shift += 2

    # Unroll the bit-setting loop.
    result = 0
    while shift >= 0:
        result = result << 1
        large_cand = (
            result + 1
        )  # Same as result ^ 1 (xor), because the last bit is always 0.
        if large_cand * large_cand <= n >> shift:
            result = large_cand
        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. See Methods of computing square roots § Woo abacus for an example.[1]