[Om3] OpenMath Symbols for Symbolic Computation

jhd at cs.bath.ac.uk jhd at cs.bath.ac.uk
Thu Sep 18 00:58:31 CEST 2008


On Wed, September 17, 2008 1:55 pm, Sebastian Freundt wrote:
> Professor James Davenport <jhd-sBK8fsN9CKk2EctHIo1CcQ at public.gmane.org>
> writes:
>> I'd be happy to look at what you have: I have Peter Horn's matrix1, but
>> don't quite see, for instance, what the point of entry_domain is?
>
> Take this one for instance:
>
> <OMOBJ xmlns="http://www.openmath.org/OpenMath" version="2.0"
> cdbase="http://www.openmath.org/cd">
>   <OMA>
>     <OMS cd="linalg2" name="matrix"/>
>     <OMA>
>       <OMS cd="linalg2" name="matrixrow"/>
>       <OMI> 1 </OMI>
>       <OMI> 0 </OMI>
>     </OMA>
>     <OMA>
>       <OMS cd="linalg2" name="matrixrow"/>
>       <OMI> 0 </OMI>
>       <OMA>
>         <OMS cd="nums1" name="rational"/>
>         <OMI> 1 </OMI>
>         <OMI> 1 </OMI>
>       </OMA>
>     </OMA>
>   </OMA>
> </OMOBJ>
>
> Whilst mathematically dodgy, with a bit of context this matrix seems to
I don't see what's dodgy about it. We'd all be happy to write
$$\pmatrix{
1&0\cr
0&\frac12\cr}$$
and the only objection to yours is the redundancy in \frac11.
> make sense.  The question if this matrix is invertible is also not too
> hard to answer, but a computer algebra system sees things differently.
> 2 cases arise:
> 1. the matrix was meant to be over Z, the (2,2)-entry is in fact bogus but
>    could be `interpreted' as rational integer (seeing as though Z is the
>    `subset' of Q whose denominators are 1).
> 2. the matrix was meant to be over Q, all but the (2,2)-entry are bogus
>    but could be (quite simply) embedded in Q properly.
>
> Now depending on that choice the invertibility question again is
> ambiguous:
> - in the simplest case and given 1. the CAS would decide not to try to
>   find an inverse since Z is not a field
> - given 1. and the more advanced concept of a unit group the CAS could try
>   an algorithm to determine whether the matrix is unimodular
> - given 2. the matrix is naturally invertible, so no problem here
This assumes a strongly typed CAS (my personal favourite, of course).
But if we take Axiom, such a matrix would be in MATRIX FRAC INT (your 2),
and invertible. If one coerced it down to MATRIX INT (your 1) it is still
invertible, since its determinant is a unit.
> However, doing the right thing here (automatically) is near to impossible
> and also extremely expensive.  The intuitive approach of choosing the
> smallest domain that `contains' all elements is doomed (though may be an
> interesting play ground for machine intelligence people), so we simply
> demand the ground domain and thus avoid all kinds of complications.
>
> Once a suitable algorithm has been chosen (or type clashes have been
> detected), the matrix elements are coerced to this entry domain, and this
> is the very first time we look at them really.
>
> Hope this clarifies our ansatz a little.
I think so - I am NOT objecting to there being a (family of) CD that does
for matrices what polyd/r/... do for polynomials, merely pointing out that
you don't NEED them to do simple things, any more than you need
polyd/r/... to make
<OMA>
  <OMS name="plus" cd="arth12/>
  <OMV name="x"/>
  <OMI>1</OMI>
</OMA>.

James Davenport
Hebron & Medlock Professor of Information Technology
Formerly RAE Coordinator and Undergraduate Director of Studies, CS Dept
Currently (thankfully briefly) Acting Head, CS Dept
Lecturer on CM30070, 30078, 50209, 50123
Chairman, Powerful Computing WP, University of Bath
OpenMath Content Dictionary Editor
IMU Committee on Electronic Information and Communication



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