Ilia,
Here's a swing at a full implementation:
http://cgit.collabora.com/git/user/fedke.m/piglit.git/log/?h=complex_tolerances_ia_v2
The __lt__ logic may not be sufficient yet. No support yet for the new
range checks in the GLSL 4.5 spec, either.
Still no idea what to do with a*b+c in complex functions - don't really
want to evaluate all the permutations if at all possible.
-mf
On 03/06/2015 04:01 PM, Ilia Mirkin wrote:
On Fri, Mar 6, 2015 at 4:50 PM, Micah Fedke <micah.fe...@collabora.co.uk> wrote:
So use the "max/min of all permutations" method for all ops? e.g.:
def __mul__(self, other):
a = self.high * other.high
b = self.high * other.low
c = self.low * other.high
d = self.low * other.low
high = numpy.float32(numpy.amax([a, b, c, d]))
low = numpy.float32(numpy.amin([a, b, c, d]))
return ValueInterval(high, low)
And tack on the tolerance at the end like this, for ops that have a
tolerance? Things should move in the right direction after high and low
have been determined, if I'm not mistaken.
def __truediv__(self, other):
tol = numpy.float32(2.5)
a = self.high / other.high
b = self.high / other.low
c = self.low / other.high
d = self.low / other.low
self.high = numpy.float32(numpy.amax([a, b, c, d]))
self.low = numpy.float32(numpy.amin([a, b, c, d]))
self.high += _ulpsize(self.high) * tol
self.low -= _ulpsize(self.low) * tol
return self
Yes, I think that's right.
As for manual fma's, that should work. I wonder, though - a double-round
manual fma has the potential to produce more error than a single-round, and
the spec allows either method, so don't we want to evaluate the more
error-ful option?
Yes and no. Both a * b + c and fma(a, b, c) have exact right answers
as defined by the spec. However for a particular a * b + c that
happens, the implementation is allowed to use either one. You could
define it as a range, but... how do you detect the a * b + c case?
Let's say I'm doing dot(x, x), which becomes
a * a + b * b + c * c + d * d.
An implementation is perfectly within its right to rewrite this as
fma(a, a, fma(b, b, fma(c, c, d * d)))
or even
fma(a, a, b * b + fma(c, c, d * d))
Since this approach only considers one op at a time, I don't see an
easy way to handle it, unfortunately...
--
Micah Fedke
Collabora Ltd.
+44 1223 362967
https://www.collabora.com/
https://twitter.com/collaboraltd
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