> From: "Gabriel Bouvigne" <[EMAIL PROTECTED]>
> Date: Sat, 27 Nov 1999 14:03:02 +0100
>
> > The midpoint (in terms of the error) is: .5 * (1 + 2.52) = 1.76.
> > So if x<1.76, we should take i=1 and if x>1.76 we take i=2.
>
> I think that for this kind of test, something like:
> i=(int)(x^.75+.5)
> would be faster than a comparison, and produces the same output.
>
> Btw do you have any explanation about how .4054 could be optimal for any
> value?
>
.4054 is optimal when choosing between i=0 and i=1.
0^4/3 = 0
1^4/3 = 1
So x<.5 should quanitze to 0, and x>.5 should quantize to 1.
Let x = .5000. The code actually quantizes x^3/4
x^3/4 = .5946
So we want: x^3/4 > .5946 then i=1
x^3/4 < .5946 then i=0
And in this case the ISO formula is correct: i=floor( x^3/4 + .4054 )
Taking i=floor(x^.75+.5) would be suboptimal.
The problem with the ISO formula is that it is only optimal when
choosing between i=0 and i=1. When choosing between i=1 and i=2,
the number should be .4720. As i increases, this number approaches
.5, so for large values of i, the nearest int is about right.
This number is given by Takehiro's formula for adj45[].
Segher had the same formula, but I think he had a typo and was
using 3/8 instead of 3/4?
Mark
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