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

(cherry picked from commit 2dfeaa9222e2ed6b6e32faaf08e5b0f77318f0a7)

Co-authored-by: Mark Dickinson <dickinsm@gmail.com>

diff --git a/Modules/mathmodule.c b/Modules/mathmodule.c
index 76d821c65b4c8b8ebd23f2325d12eae668124466..82a9a14724f5e143a1dd5ebb55952e6cfbafa5f9 100644 (file)
--- a/Modules/mathmodule.c
+++ b/Modules/mathmodule.c
@@ -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)