Issue #8692: Improve performance of math.factorial:
(1) use a different algorithm that roughly halves the total number of
multiplications required and results in more balanced multiplications
(2) use a lookup table for small arguments
(3) fast accumulation of products in C integer arithmetic rather than
PyLong arithmetic when possible.
Typical speedup, from unscientific testing on a 64-bit laptop, is 4.5x
to 6.5x for arguments in the range 100 - 10000.
Patch by Daniel Stutzbach; extensive reviews by Alexander Belopolsky.
This commit is contained in:
Mark Dickinson
parent
ae6265f8d0
commit
4c8a9a2df3
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@ -60,6 +60,56 @@ def ulps_check(expected, got, ulps=20):
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return "error = {} ulps; permitted error = {} ulps".format(ulps_error,
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ulps)
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# Here's a pure Python version of the math.factorial algorithm, for
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# documentation and comparison purposes.
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#
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# Formula:
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#
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# factorial(n) = factorial_odd_part(n) << (n - count_set_bits(n))
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#
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# where
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#
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# factorial_odd_part(n) = product_{i >= 0} product_{0 < j <= n >> i; j odd} j
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#
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# The outer product above is an infinite product, but once i >= n.bit_length,
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# (n >> i) < 1 and the corresponding term of the product is empty. So only the
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# finitely many terms for 0 <= i < n.bit_length() contribute anything.
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#
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# We iterate downwards from i == n.bit_length() - 1 to i == 0. The inner
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# product in the formula above starts at 1 for i == n.bit_length(); for each i
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# < n.bit_length() we get the inner product for i from that for i + 1 by
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# multiplying by all j in {n >> i+1 < j <= n >> i; j odd}. In Python terms,
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# this set is range((n >> i+1) + 1 | 1, (n >> i) + 1 | 1, 2).
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def count_set_bits(n):
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"""Number of '1' bits in binary expansion of a nonnnegative integer."""
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return 1 + count_set_bits(n & n - 1) if n else 0
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def partial_product(start, stop):
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"""Product of integers in range(start, stop, 2), computed recursively.
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start and stop should both be odd, with start <= stop.
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"""
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numfactors = (stop - start) >> 1
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if not numfactors:
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return 1
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elif numfactors == 1:
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return start
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else:
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mid = (start + numfactors) | 1
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return partial_product(start, mid) * partial_product(mid, stop)
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def py_factorial(n):
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"""Factorial of nonnegative integer n, via "Binary Split Factorial Formula"
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described at http://www.luschny.de/math/factorial/binarysplitfact.html
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"""
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inner = outer = 1
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for i in reversed(range(n.bit_length())):
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inner *= partial_product((n >> i + 1) + 1 | 1, (n >> i) + 1 | 1)
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outer *= inner
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return outer << (n - count_set_bits(n))
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def acc_check(expected, got, rel_err=2e-15, abs_err = 5e-323):
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"""Determine whether non-NaN floats a and b are equal to within a
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(small) rounding error. The default values for rel_err and
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@ -365,18 +415,19 @@ class MathTests(unittest.TestCase):
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self.ftest('fabs(1)', math.fabs(1), 1)
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def testFactorial(self):
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def fact(n):
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result = 1
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for i in range(1, int(n)+1):
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result *= i
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return result
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values = list(range(10)) + [50, 100, 500]
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random.shuffle(values)
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for x in values:
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for cast in (int, float):
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self.assertEqual(math.factorial(cast(x)), fact(x), (x, fact(x), math.factorial(x)))
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self.assertEqual(math.factorial(0), 1)
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self.assertEqual(math.factorial(0.0), 1)
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total = 1
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for i in range(1, 1000):
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total *= i
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self.assertEqual(math.factorial(i), total)
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self.assertEqual(math.factorial(float(i)), total)
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self.assertEqual(math.factorial(i), py_factorial(i))
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self.assertRaises(ValueError, math.factorial, -1)
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self.assertRaises(ValueError, math.factorial, -1.0)
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self.assertRaises(ValueError, math.factorial, math.pi)
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self.assertRaises(OverflowError, math.factorial, sys.maxsize+1)
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self.assertRaises(OverflowError, math.factorial, 10e100)
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def testFloor(self):
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self.assertRaises(TypeError, math.floor)
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@ -1132,6 +1132,12 @@ Library
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Extension Modules
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-----------------
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- Issue #8692: Optimize math.factorial: replace the previous naive
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algorithm with an improved 'binary-split' algorithm that uses fewer
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multiplications and allows many of the multiplications to be
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performed using plain C integer arithmetic instead of PyLong
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arithmetic. Also uses a lookup table for small arguments.
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- Issue #8674: Fixed a number of incorrect or undefined-behaviour-inducing
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overflow checks in the audioop module.
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@ -1129,11 +1129,232 @@ PyDoc_STRVAR(math_fsum_doc,
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Return an accurate floating point sum of values in the iterable.\n\
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Assumes IEEE-754 floating point arithmetic.");
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/* Return the smallest integer k such that n < 2**k, or 0 if n == 0.
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* Equivalent to floor(lg(x))+1. Also equivalent to: bitwidth_of_type -
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* count_leading_zero_bits(x)
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*/
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/* XXX: This routine does more or less the same thing as
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* bits_in_digit() in Objects/longobject.c. Someday it would be nice to
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* consolidate them. On BSD, there's a library function called fls()
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* that we could use, and GCC provides __builtin_clz().
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*/
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static unsigned long
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bit_length(unsigned long n)
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{
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unsigned long len = 0;
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while (n != 0) {
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++len;
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n >>= 1;
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}
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return len;
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}
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static unsigned long
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count_set_bits(unsigned long n)
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{
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unsigned long count = 0;
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while (n != 0) {
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++count;
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n &= n - 1; /* clear least significant bit */
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}
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return count;
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}
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/* Divide-and-conquer factorial algorithm
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*
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* Based on the formula and psuedo-code provided at:
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* http://www.luschny.de/math/factorial/binarysplitfact.html
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*
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* Faster algorithms exist, but they're more complicated and depend on
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* a fast prime factoriazation algorithm.
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*
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* Notes on the algorithm
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* ----------------------
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*
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* factorial(n) is written in the form 2**k * m, with m odd. k and m are
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* computed separately, and then combined using a left shift.
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*
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* The function factorial_odd_part computes the odd part m (i.e., the greatest
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* odd divisor) of factorial(n), using the formula:
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*
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* factorial_odd_part(n) =
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*
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* product_{i >= 0} product_{0 < j <= n / 2**i, j odd} j
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*
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* Example: factorial_odd_part(20) =
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*
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* (1) *
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* (1) *
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* (1 * 3 * 5) *
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* (1 * 3 * 5 * 7 * 9)
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* (1 * 3 * 5 * 7 * 9 * 11 * 13 * 15 * 17 * 19)
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*
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* Here i goes from large to small: the first term corresponds to i=4 (any
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* larger i gives an empty product), and the last term corresponds to i=0.
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* Each term can be computed from the last by multiplying by the extra odd
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* numbers required: e.g., to get from the penultimate term to the last one,
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* we multiply by (11 * 13 * 15 * 17 * 19).
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*
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* To see a hint of why this formula works, here are the same numbers as above
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* but with the even parts (i.e., the appropriate powers of 2) included. For
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* each subterm in the product for i, we multiply that subterm by 2**i:
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*
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* factorial(20) =
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*
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* (16) *
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* (8) *
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* (4 * 12 * 20) *
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* (2 * 6 * 10 * 14 * 18) *
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* (1 * 3 * 5 * 7 * 9 * 11 * 13 * 15 * 17 * 19)
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*
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* The factorial_partial_product function computes the product of all odd j in
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* range(start, stop) for given start and stop. It's used to compute the
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* partial products like (11 * 13 * 15 * 17 * 19) in the example above. It
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* operates recursively, repeatedly splitting the range into two roughly equal
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* pieces until the subranges are small enough to be computed using only C
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* integer arithmetic.
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*
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* The two-valuation k (i.e., the exponent of the largest power of 2 dividing
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* the factorial) is computed independently in the main math_factorial
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* function. By standard results, its value is:
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*
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* two_valuation = n//2 + n//4 + n//8 + ....
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*
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* It can be shown (e.g., by complete induction on n) that two_valuation is
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* equal to n - count_set_bits(n), where count_set_bits(n) gives the number of
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* '1'-bits in the binary expansion of n.
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*/
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/* factorial_partial_product: Compute product(range(start, stop, 2)) using
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* divide and conquer. Assumes start and stop are odd and stop > start.
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* max_bits must be >= bit_length(stop - 2). */
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static PyObject *
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factorial_partial_product(unsigned long start, unsigned long stop,
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unsigned long max_bits)
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{
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unsigned long midpoint, num_operands;
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PyObject *left = NULL, *right = NULL, *result = NULL;
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/* If the return value will fit an unsigned long, then we can
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* multiply in a tight, fast loop where each multiply is O(1).
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* Compute an upper bound on the number of bits required to store
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* the answer.
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*
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* Storing some integer z requires floor(lg(z))+1 bits, which is
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* conveniently the value returned by bit_length(z). The
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* product x*y will require at most
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* bit_length(x) + bit_length(y) bits to store, based
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* on the idea that lg product = lg x + lg y.
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*
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* We know that stop - 2 is the largest number to be multiplied. From
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* there, we have: bit_length(answer) <= num_operands *
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* bit_length(stop - 2)
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*/
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num_operands = (stop - start) / 2;
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/* The "num_operands <= 8 * SIZEOF_LONG" check guards against the
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* unlikely case of an overflow in num_operands * max_bits. */
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if (num_operands <= 8 * SIZEOF_LONG &&
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num_operands * max_bits <= 8 * SIZEOF_LONG) {
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unsigned long j, total;
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for (total = start, j = start + 2; j < stop; j += 2)
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total *= j;
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return PyLong_FromUnsignedLong(total);
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}
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/* find midpoint of range(start, stop), rounded up to next odd number. */
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midpoint = (start + num_operands) | 1;
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left = factorial_partial_product(start, midpoint,
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bit_length(midpoint - 2));
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if (left == NULL)
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goto error;
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right = factorial_partial_product(midpoint, stop, max_bits);
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if (right == NULL)
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goto error;
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result = PyNumber_Multiply(left, right);
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error:
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Py_XDECREF(left);
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Py_XDECREF(right);
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return result;
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}
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/* factorial_odd_part: compute the odd part of factorial(n). */
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static PyObject *
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factorial_odd_part(unsigned long n)
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{
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long i;
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unsigned long v, lower, upper;
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PyObject *partial, *tmp, *inner, *outer;
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inner = PyLong_FromLong(1);
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if (inner == NULL)
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return NULL;
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outer = inner;
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Py_INCREF(outer);
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upper = 3;
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for (i = bit_length(n) - 2; i >= 0; i--) {
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v = n >> i;
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if (v <= 2)
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continue;
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lower = upper;
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/* (v + 1) | 1 = least odd integer strictly larger than n / 2**i */
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upper = (v + 1) | 1;
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/* Here inner is the product of all odd integers j in the range (0,
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n/2**(i+1)]. The factorial_partial_product call below gives the
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product of all odd integers j in the range (n/2**(i+1), n/2**i]. */
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partial = factorial_partial_product(lower, upper, bit_length(upper-2));
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/* inner *= partial */
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if (partial == NULL)
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goto error;
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tmp = PyNumber_Multiply(inner, partial);
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Py_DECREF(partial);
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if (tmp == NULL)
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goto error;
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Py_DECREF(inner);
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inner = tmp;
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/* Now inner is the product of all odd integers j in the range (0,
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n/2**i], giving the inner product in the formula above. */
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/* outer *= inner; */
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tmp = PyNumber_Multiply(outer, inner);
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if (tmp == NULL)
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goto error;
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Py_DECREF(outer);
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outer = tmp;
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}
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goto done;
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error:
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Py_DECREF(outer);
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done:
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Py_DECREF(inner);
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return outer;
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}
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/* Lookup table for small factorial values */
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static const unsigned long SmallFactorials[] = {
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1, 1, 2, 6, 24, 120, 720, 5040, 40320,
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362880, 3628800, 39916800, 479001600,
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#if SIZEOF_LONG >= 8
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6227020800, 87178291200, 1307674368000,
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20922789888000, 355687428096000, 6402373705728000,
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121645100408832000, 2432902008176640000
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#endif
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};
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static PyObject *
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math_factorial(PyObject *self, PyObject *arg)
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{
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long i, x;
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PyObject *result, *iobj, *newresult;
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long x;
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PyObject *result, *odd_part, *two_valuation;
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if (PyFloat_Check(arg)) {
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PyObject *lx;
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@ -1160,25 +1381,24 @@ math_factorial(PyObject *self, PyObject *arg)
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return NULL;
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}
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result = (PyObject *)PyLong_FromLong(1);
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if (result == NULL)
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return NULL;
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for (i=1 ; i<=x ; i++) {
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iobj = (PyObject *)PyLong_FromLong(i);
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if (iobj == NULL)
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goto error;
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newresult = PyNumber_Multiply(result, iobj);
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Py_DECREF(iobj);
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if (newresult == NULL)
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goto error;
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Py_DECREF(result);
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result = newresult;
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}
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return result;
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/* use lookup table if x is small */
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if (x < (long)(sizeof(SmallFactorials)/sizeof(SmallFactorials[0])))
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return PyLong_FromUnsignedLong(SmallFactorials[x]);
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error:
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Py_DECREF(result);
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/* else express in the form odd_part * 2**two_valuation, and compute as
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odd_part << two_valuation. */
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odd_part = factorial_odd_part(x);
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if (odd_part == NULL)
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return NULL;
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two_valuation = PyLong_FromLong(x - count_set_bits(x));
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if (two_valuation == NULL) {
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Py_DECREF(odd_part);
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return NULL;
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}
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result = PyNumber_Lshift(odd_part, two_valuation);
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Py_DECREF(two_valuation);
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Py_DECREF(odd_part);
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return result;
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}
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PyDoc_STRVAR(math_factorial_doc,
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Loading…
Reference in New Issue