On Sat, 28 Jun 2008, Matt Barber wrote:
a1 = 0.5f * (c - a); a3 = 0.5f * (d - a) + 1.5f * (b - c); a2 = a - b + a1 - a3;*out++ = ((a3 * frac + a2) * frac + a1) * frac + b; 10 +'s 6 *'s
If you compute twice the value of a1,a2,a3 and later multiply by 0.5, you end up with a multiplication of (a-b) by 2 that you can optimise by turning it into a single addition. In that case, 11 +'s 5 *'s, or 12 +'s 4 *'s if you also do the same with the multiplication by 3.0. It matters only if the CPU computes addition faster (not sure if that's still the case), or if redefining Pd samples to some weird type (floats with way too many bits, and such).
I wonder what's the lowest possible number of operations. Some possible formulas wouldn't even have ((a3 * frac + a2) * frac + a1) * frac, but I wonder whether they can be any shorter. In any case, you need at least three multiplications ;)
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