Turn math.isqrt assertion into a comment to clarify its purpose. (GH-14131)

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Mark Dickinson 2019-06-16 17:53:21 +01:00 committed by GitHub
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@ -1527,10 +1527,10 @@ Here's Python code equivalent to the C implementation below:
a = 1
d = 0
for s in reversed(range(c.bit_length())):
# Loop invariant: (a-1)**2 < (n >> 2*(c - d)) < (a+1)**2
e = d
d = c >> s
a = (a << d - e - 1) + (n >> 2*c - e - d + 1) // a
assert (a-1)**2 < n >> 2*(c - d) < (a+1)**2
return a - (a*a > n)