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.
--
https://mail.python.org/mailman/listinfo/python-list
