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Challenge: find the first value where two functions differ


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?




-- 
Steve
?Cheer up,? they said, ?things could be worse.? So I cheered up, and sure
enough, things got worse.