Replace _nbits() with int.bit_length().
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@ -5526,23 +5526,7 @@ def _normalize(op1, op2, prec = 0):
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##### Integer arithmetic functions used by ln, log10, exp and __pow__ #####
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# This function from Tim Peters was taken from here:
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# http://mail.python.org/pipermail/python-list/1999-July/007758.html
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# The correction being in the function definition is for speed, and
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# the whole function is not resolved with math.log because of avoiding
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# the use of floats.
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def _nbits(n, correction = {
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'0': 4, '1': 3, '2': 2, '3': 2,
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'4': 1, '5': 1, '6': 1, '7': 1,
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'8': 0, '9': 0, 'a': 0, 'b': 0,
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'c': 0, 'd': 0, 'e': 0, 'f': 0}):
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"""Number of bits in binary representation of the positive integer n,
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or 0 if n == 0.
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"""
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if n < 0:
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raise ValueError("The argument to _nbits should be nonnegative.")
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hex_n = "%x" % n
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return 4*len(hex_n) - correction[hex_n[0]]
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_nbits = int.bit_length
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def _sqrt_nearest(n, a):
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"""Closest integer to the square root of the positive integer n. a is
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