Hi, Doing some tests with square-free factorization, I believe I have found a bug. Let e by (x+y)^2*(x-y) expanded and call sqrfree(e,x). You do not get the original expression, it remains expanded. If y is replaced by 1 it seems it works. Bernard
Hi! On Tue, 1 Aug 2000, Parisse Bernard wrote:
Doing some tests with square-free factorization, I believe I have found a bug.
Indeed. sqrfree() only works for polynomials with one indeterminate. It needs some further work. Bye, Christian -- / Coding on PowerPC and proud of it \/ http://www.uni-mainz.de/~bauec002/
Christian Bauer a écrit :
Hi!
On Tue, 1 Aug 2000, Parisse Bernard wrote:
Doing some tests with square-free factorization, I believe I have found a bug.
Indeed. sqrfree() only works for polynomials with one indeterminate. It needs some further work.
OK, don't worry about that since there is no reason to duplicate work, my own square-free works with multivariate polynomials. I have added some tricks to factor some polynomials (no Berlekamp right now, planned later), as well as partial fraction expansion, I'm beginning integration of rational fraction. You can test all this at ftp://fourier.ujf-grenoble.fr/pub/hp48/giac.tgz Bye, Bernard Parisse
On Tue, 1 Aug 2000, Parisse Bernard wrote:
Doing some tests with square-free factorization, I believe I have found a bug. Let e by (x+y)^2*(x-y) expanded and call sqrfree(e,x). You do not get the original expression, it remains expanded. If y is replaced by 1 it seems it works.
Last week I had a look at this. It was a crude copy of Yun's algorithm found in Geddes' book. That algorithm has a minor error, however. :-) I do think it's fixed in the CVS version now:
sqrfree(expand((x+y)^2*(x-y))); (x+y)^2*(x-y) sqrfree(expand((x+y)^2*(x-y)),[x]); (x+y)^2*(x-y) sqrfree(expand((x+y)^2*(x-y)),[y]); (x+y)^2*(x-y)
By the way, I downloaded giac.tgz the day before yesterday and found the remark "GiNaC import/export is now waiting for a stable interface format." What's that about? Is there anything we can do / clarify? Regards -richy. PS: You really do have factorization over Z[i]?!? Are you going to allow algrbraic extensions like RootOf and all this? -- Richard Kreckel <Richard.Kreckel@Uni-Mainz.DE> <http://wwwthep.physik.uni-mainz.de/~kreckel/>
participants (3)
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Christian Bauer
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Parisse Bernard
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Richard B. Kreckel