Re: Divide-by-multiply algorithm reference needed

Paul Khuong <pvk@pvk.ca>
Tue, 14 Sep 2010 22:27:16 -0400

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Related articles
Divide-by-multiply algorithm reference needed newcomer@flounder.com (Joseph M. Newcomer) (2010-09-13)
Re: Divide-by-multiply algorithm reference needed rsc@swtch.com (Russ Cox) (2010-09-13)
Re: Divide-by-multiply algorithm reference needed gah@ugcs.caltech.edu (glen herrmannsfeldt) (2010-09-14)
Re: Divide-by-multiply algorithm reference needed newcomer@flounder.com (Joseph M. Newcomer) (2010-09-14)
Re: Divide-by-multiply algorithm reference needed newcomer@flounder.com (Joseph M. Newcomer) (2010-09-14)
Re: Divide-by-multiply algorithm reference needed pvk@pvk.ca (Paul Khuong) (2010-09-14)
Re: Divide-by-multiply algorithm reference needed gah@ugcs.caltech.edu (glen herrmannsfeldt) (2010-09-15)
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From: Paul Khuong <pvk@pvk.ca>
Newsgroups: comp.compilers
Date: Tue, 14 Sep 2010 22:27:16 -0400
Organization: A noiseless patient Spider
References: 10-09-013 10-09-015 10-09-016
Keywords: arithmetic, code
Posted-Date: 15 Sep 2010 00:04:57 EDT

  "Joseph M. Newcomer" <newcomer@flounder.com> wrote:
> >> My problem is I am trying to find the algorithm by which I can
> >> generate the 32-bit multiplier and the shift amount, given a specific
> >> constant divisor. ...
>
> Yes, that's the easy part. Now the hard part. In the Microsoft C
> compiler, the values they use are sometimes twice the values I
> compute, but then they throw in an additional shr instruction to
> divide the result by 2. Why? Why are my selections different from
> theirs? However, the previous message gave some citations, so I'm
> going to go look at those now.


AFAICT, the main reason for doing something like this is to maximise the
range of inputs for which the pseudoinverse will truncate to the right
value. I'm pretty sure that
<http://www.discontinuity.info/~pkhuong/pseudoinverse.pdf> derives a
tight bound for the range for which a given fixed-point multiplication
(which is what division-by-multiplication implements) always
approximates a constant (unsigned) division exactly.


Paul Khuong


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