Tim Peters <t...@python.org> added the comment:
> Is > > i, rem = isqrt_rem(n) > i + (rem != 0) > > better than > > (isqrt(n<<2) + 1) >> 1 > > or > > n and isqrt(n-1) + 1 > > ? Define "better"? The first way is by far the most obvious of the three, and the second way the least obvious. The first way also "wins" on being a variation of a uniform pattern that can deliver the floor, ceiling, or rounded result, depending on which simple comparison result is added. It's not "a trick" - it's the opposite of clever ;-) The first way is also unique in being the only one of the three that does _not_ do any Python-level arithmetic on integers as wide as `n`. `i` and `rem` are no more than about half the bit length of `n`. `n << 2` and `n - 1` in the others have to create new int objects at least as wide as `n`. ---------- _______________________________________ Python tracker <rep...@bugs.python.org> <https://bugs.python.org/issue46187> _______________________________________ _______________________________________________ Python-bugs-list mailing list Unsubscribe: https://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
