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On 2017-08-04 15:51, Steve D'Aprano wrote: > This is a challenge for which I don't have a complete answer, only a partial > answer. > > Here are two functions for calculating the integer square root of a non-negative > int argument. The first is known to be exact but may be a bit slow: > > def isqrt_newton(n): > """Integer sqrt using Newton's Method.""" > if n == 0: > return 0 > bits = n.bit_length() > a, b = divmod(bits, 2) > x = 2**(a+b) > while True: > y = (x + n//x)//2 > if y >= x: > return x > x = y > > > The second is only exact for some values of n, and for sufficiently large values > of n, is will fail altogether: > > > import math > > def isqrt_float(n): > """Integer square root using floating point sqrt.""" > return int(math.sqrt(n)) > > > > We know that: > > - for n <= 2**53, isqrt_float(n) is exact; > > - for n >= 2**1024, isqrt_float(n) will raise OverflowError; > > - between those two values, 2**53 < n < 2**1024, isqrt_float(n) > will sometimes be exact, and sometimes not exact; > > - there is some value, let's call it M, which is the smallest > integer where isqrt_float is not exact. > > > Your mission, should you choose to accept it, is to find M. > > Hint: a linear search starting at 2**53 will find it -- eventually. But it might > take a long time. Personally I gave up after five minutes, but for all I know > if I had a faster computer I'd already have the answer. > > (But probably not.) > > > Another hint: if you run this code: > > > for i in range(53, 1024): > n = 2**i > if isqrt_newton(n) != isqrt_float(n): > print(n) > break > > > you can find a much better upper bound for M: > > 2**53 < M <= 2**105 > > which is considerable smaller that the earlier upper bound of 2**1024. But even > so, that's still 40564819207303331840695247831040 values to be tested. > > On the assumption that we want a solution before the return of Halley's Comet, > how would you go about finding M? > Why would isqrt_float not give the correct answer? Probably because of truncation (or rounding up?) of the floating point. I'd expect it to fail first near a square.

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