about com (was: Motion 42: NO)
On 2013-02-06 22:07:15 +0100, Guillaume Melquiond wrote:
> - The motion states that "the final result is decorated com if and only
> if the evaluation of the whole expression was common as defined in 5.4"
> in Section 6.3. I understand the "only if" direction, but not the "if"
> direction. Indeed, Section 5.4 deals with level 1 while 6.3 deals with
> level 2, where overflow, precision, and so on, matter. I guess that the
> fact the sentence uses "f" instead of "phi" is not unrelated to my
> confusion.
As I understand it, this means that one takes the same definition
of common at Level 2: instead of considering the Level 1 evaluation,
one considers the Level 2 evaluation.
> - I do not agree with com requiring the computed interval to be bounded
> at level 2. I feel that the boundedness should only be required at level
> 1. In particular, I do not see what is gained from stripping com in case
> of a harmless overflow. What is the point of com if an unbounded
> interval from the point of view of the interval type is necessarily
> unbounded from the point of view of the decorations? Any information
> about what the mathematical function actually computes is lost.
The primary goal of com is not to give a property of the function
but to record whether the evaluation does not depend on the flavor
(assuming identical rounding and some form of reproducibility). As
some flavors (e.g. Kaucher) do not have unbounded intervals, it was
necessary to reject overflows.
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Vincent Lefèvre <vincent@xxxxxxxxxx> - Web: <http://www.vinc17.net/>
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Work: CR INRIA - computer arithmetic / AriC project (LIP, ENS-Lyon)