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Merged
rhettinger merged 10 commits into from
Aug 16, 2020
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bpo-41513: Improve speed and accuracy of math.hypot() #21803

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e2b1b38
frexp/ldexp variant
rhettinger Aug 8, 2020
7cb895d
Multiply by power of two variant
rhettinger Aug 8, 2020
2d586b1
Strengthen assertions
rhettinger Aug 9, 2020
c15ce79
Added notes and commented out assertions to reach a stable resting point
rhettinger Aug 9, 2020
c853597
Two code paths. All tests pass.
rhettinger Aug 9, 2020
03ec58b
Clean-up comments and ranges
rhettinger Aug 10, 2020
ebf6a31
Remove debugging code
rhettinger Aug 10, 2020
4742ef8
Clean-up comments and over-specified tests
rhettinger Aug 10, 2020
5e9f1a9
Add blurb
rhettinger Aug 10, 2020
5c6478f
Improve the comments.
rhettinger Aug 15, 2020
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7 changes: 4 additions & 3 deletions Lib/test/test_math.py
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Original file line number Diff line number Diff line change
Expand Up @@ -795,7 +795,8 @@ def testHypot(self):
# Verify scaling for extremely large values
fourthmax = FLOAT_MAX / 4.0
for n in range(32):
self.assertEqual(hypot(*([fourthmax]*n)), fourthmax * math.sqrt(n))
self.assertTrue(math.isclose(hypot(*([fourthmax]*n)),
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fourthmax * math.sqrt(n)))

# Verify scaling for extremely small values
for exp in range(32):
Expand Down Expand Up @@ -904,8 +905,8 @@ class T(tuple):
for n in range(32):
p = (fourthmax,) * n
q = (0.0,) * n
self.assertEqual(dist(p, q), fourthmax * math.sqrt(n))
self.assertEqual(dist(q, p), fourthmax * math.sqrt(n))
self.assertTrue(math.isclose(dist(p, q), fourthmax * math.sqrt(n)))
self.assertTrue(math.isclose(dist(q, p), fourthmax * math.sqrt(n)))

# Verify scaling for extremely small values
for exp in range(32):
Expand Down
2 changes: 2 additions & 0 deletions Misc/NEWS.d/next/Library/2020-08-09-18-16-05.bpo-41513.e6K6EK.rst
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@@ -0,0 +1,2 @@
Minor algorithmic improvement to math.hypot() and math.dist() giving small
gains in speed and accuracy.
35 changes: 33 additions & 2 deletions Modules/mathmodule.c
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Expand Up @@ -2406,6 +2406,13 @@ math_fmod_impl(PyObject *module, double x, double y)
/*
Given an *n* length *vec* of values and a value *max*, compute:

sqrt(sum((x * scale) ** 2 for x in vec)) / scale

where scale is the first power of two
greater than max.

or compute:

max * sqrt(sum((x / max) ** 2 for x in vec))

The value of the *max* variable must be non-negative and
Expand Down Expand Up @@ -2437,7 +2444,8 @@ fractional digits to be dropped from *csum*.
static inline double
vector_norm(Py_ssize_t n, double *vec, double max, int found_nan)
{
double x, csum = 1.0, oldcsum, frac = 0.0;
double x, csum = 1.0, oldcsum, frac = 0.0, scale;
int max_e;
Py_ssize_t i;

if (Py_IS_INFINITY(max)) {
Expand All @@ -2449,14 +2457,37 @@ vector_norm(Py_ssize_t n, double *vec, double max, int found_nan)
if (max == 0.0 || n <= 1) {
return max;
}
frexp(max, &max_e);
if (max_e >= -1023) {
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scale = ldexp(1.0, -max_e);
assert(max * scale >= 0.5);
assert(max * scale < 1.0);
for (i=0 ; i < n ; i++) {
x = vec[i];
assert(Py_IS_FINITE(x) && fabs(x) <= max);
x *= scale;
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@serhiy-storchaka serhiy-storchaka Aug 10, 2020

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What is faster, x *= scale or x = ldexp(x, -max_e)?

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@rhettinger rhettinger Aug 10, 2020

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  • The test as written was over-specified. It should have been written this way from the start.

  • The x *= scale is faster than x = ldexp(x, -max_e). The former is a single, fast in-line instruction and the latter is an external library call.

Here's the generated code for the loop:

L284:
    movsd   (%r12,%rax,8), %xmm0
    addq    $1, %rax
    cmpq    %rax, %rbp
    mulsd   %xmm4, %xmm0           <-- x *= scale
    movapd  %xmm0, %xmm1
    mulsd   %xmm0, %xmm1           <-- x *= x
    movapd  %xmm2, %xmm0
    addsd   %xmm1, %xmm2            <-- csum += x
    subsd   %xmm2, %xmm0
    addsd   %xmm1, %xmm0
    addsd   %xmm0, %xmm3
    jg  L284
    subsd   %xmm

x = x*x;
assert(x <= 1.0);
assert(csum >= x);
oldcsum = csum;
csum += x;
frac += (oldcsum - csum) + x;
}
return sqrt(csum - 1.0 + frac) / scale;
}
/* When max_e < -1023, a scale computed given by ldexp(1.0, -max_e)
would be subnormal and break range invariants for max*scale.
Instead, we just divide by *max*.
*/
for (i=0 ; i < n ; i++) {
x = vec[i];
assert(Py_IS_FINITE(x) && fabs(x) <= max);
x /= max;
x = x*x;
assert(x <= 1.0);
assert(csum >= x);
oldcsum = csum;
csum += x;
assert(csum >= x);
frac += (oldcsum - csum) + x;
}
return max * sqrt(csum - 1.0 + frac);
Expand Down