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@nlw0
Last active Nov 20, 2018
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Exponential function benchmark in Julia
@inline @fastmath function myexpm1(x :: Float64)
ln2_hi = 6.93147180369123816490e-01
ln2_lo = 1.90821492927058770002e-10
one = 1.0
Q1 = -3.33333333333331316428e-02
Q2 = 1.58730158725481460165e-03
Q3 = -7.93650757867487942473e-05
Q4 = 4.00821782732936239552e-06
Q5 = -2.01099218183624371326e-07
hi = x - ln2_hi
lo = ln2_lo
k = 1
x = hi - lo;
c = (hi - x) - lo;
hfx = 0.5 * x;
hxs = x * hfx;
R1 = one + hxs * Q1;
h2 = hxs * hxs;
R2 = Q2 + hxs * Q3;
h4 = h2 * h2;
R3 = Q4 + hxs * Q5;
r1 = R1 + h2 * R2 + h4 * R3;
t = 3.0 - r1 * hfx;
e = hxs * ((r1 - t) / (6.0 - x * t));
e = (x * (e - c) - c);
e -= hxs;
one + 2.0 * (x - e);
end
@inline myexp(x) = myexpm1(x) + 1
@inline calcerr(x) = abs(myexp(x) - exp(x))
@inline function test_myexp_for(n)
err = 0.0
for i in 0:n
err = max(err, calcerr(0.4+0.6*i/n))
end
err
end
@inline function test_myexp_fold(n)
@inline errmax(acc, i) = max(acc, calcerr(0.4+0.6*i/n))
foldl(errmax, 0:n, init=0)
end
for n in reverse([1e3, 1e4, 1e5, 1e6, 1e7, 1e8, 1e9])
tt = @elapsed er=test_myexp_for(convert(Int, n))
println("($n, $tt, $er)")
end
for n in reverse([1e3, 1e4, 1e5, 1e6, 1e7, 1e8, 1e9])
tt = @elapsed er=test_myexp_fold(convert(Int, n))
println("($n, $tt, $er)")
end
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