Hi John, On 2015-06-09, John H Palmieri <[email protected]> wrote: > sage: list(X) > [((1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0,= >=20 > 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0), > 1)] > > It looks like the tuple is a concatenation of 6 (one for each factor of X)= >=20 > tuples of length 6 (one for each generator): (1,0,0,0,0,0), (0,1,0,0,0,0),= >=20 > etc., representing a, b, a, c, c, b. So you could take that tuple, break it= >=20 > into smaller tuples of length 6 to get each factor.=20 > > Actually, after looking at further examples, the length of the tuple is=20 > unpredictable: if you first evaluate a**9, then any for element defined=20 > after that, the tuple will have length 9*6. Strange. Anyway, you should be= >=20 > able to break the tuple into length 6 chunks to get each factor, treating= >=20 > (0,0,0,0,0,0) as 1.
Sorry for answering so late: At SageDays 65, slrn was blocked by Loyola University, so, I didn't read sage-devel for a week. That's simply how the letterplace implementation works. You inject the set of all free algebra elements of degree bounded by d on n generators into the set of *commutative* polynomials over d*n generators. Hence, if you increase the degree bound for your free algebra elements during computation, the underlying commutative ring has to be increased, too. Thus, the computations will henceforth be done in a different, larger, polynomial ring. I think I have documented that principle in my wrapper of Singular's letterplace. In any case, if I understand correctly. Viviane has solved the problem during SageDays 65. Best regards, Simon -- You received this message because you are subscribed to the Google Groups "sage-devel" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To post to this group, send email to [email protected]. Visit this group at http://groups.google.com/group/sage-devel. For more options, visit https://groups.google.com/d/optout.
