[New-bugs-announce] [issue46187] Optionally support rounding for math.isqrt()

[Raymond Hettinger]

New submission from Raymond Hettinger <raymond.hettinger at gmail.com>:

By default, isqrt(n) gives the floor of the exact square of n.  It would be nice to have a flag to give a rounded result:

    y = isqrt(n, round=True)

Alternatively, set a mode argument to one of {'floor', 'round', 'ceil'}:

    y = isqrt(n, mode='round')

I would like something better than this:

    def risqrt(x):
        'Big integer version of: round(sqrt(x)).'
        y = isqrt(x)
        s = y ** 2
        return y if x <= s + y else y + 1

    def cisqrt(x):
        'Big integer version of: ceil(sqrt(x)).'
        return isqrt(x - 1) + 1

My use case arose when building a table of square roots incorporated in arbitrary precision functions implemented with scaled integer arithmetic:

    def get_root_table(base, steps, scale):
        s = []
        x = round(base * scale)
        for i in range(steps):
            x = risqrt(x * scale)
            s.append(x)
        return s

----------
assignee: mark.dickinson
components: Library (Lib)
messages: 409243
nosy: mark.dickinson, rhettinger, tim.peters
priority: normal
severity: normal
status: open
title: Optionally support rounding for math.isqrt()
type: enhancement
versions: Python 3.11

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Python tracker <report at bugs.python.org>
<https://bugs.python.org/issue46187>
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