solved at
https://trac.sagemath.org/ticket/26593
Le 26/10/2018 à 16:48, 'Paul Mercat' via sage-devel a écrit :
Thank you. So the bug is with the test of equality in QQbar, not with
NumberField.
Do you know how to solve this problem with QQbar ?
I tried to look at where is the error, and it looks like there is an
infinite loop:
/Users/mercatp/sage-8.2/local/lib/python2.7/site-packages/sage/rings/polynomial/polynomial_compiled.pyx
in sage.rings.polynomial.polynomial_compiled.abc_pd.eval
(build/cythonized/sage/rings/polynomial/polynomial_compiled.c:6599)() 506
507 cdef int eval(abc_pd self, object vars, object coeffs) except -2:-->
508 pd_eval(self.left, vars, coeffs) 509 pd_eval(self.right,
vars, coeffs) 510 self.value = self.left.value * self.right.value +
coeffs[self.index]
/Users/mercatp/sage-8.2/local/lib/python2.7/site-packages/sage/rings/polynomial/polynomial_compiled.pyx
in sage.rings.polynomial.polynomial_compiled.pd_eval
(build/cythonized/sage/rings/polynomial/polynomial_compiled.c:3574)() 352 cdef
inline int pd_eval(generic_pd pd, object vars, object coeffs) except -2: 353
if pd.value is None:--> 354 pd.eval(vars, coeffs) 355 pd.hits
+= 1 356
But I don't know how to avoid this loop, because I don't know what do this
code...
Paul
Le vendredi 26 octobre 2018 16:04:57 UTC+2, John Cremona a écrit :
On Fri, 26 Oct 2018 at 13:07, 'Paul Mercat' via sage-devel <
sage-...@googlegroups.com <javascript:>> wrote:
Hi !
I have a strange bug with NumberField: when I do
sage: pi = x^7 - 2*x^6 + x^3 - 2*x^2 + 2*x - 1
sage: b = pi.roots(ring=QQbar)[3][0]
sage: pi = b.minpoly()
sage: K = NumberField(pi, 'b', embedding=b)
it works well. But if I execute it a second time, then it never
terminates!
Do you have a explanation and a correction to this problem ?
How strange. The first 3 lines are fine the second time. So is the
number field construction when you leave out the embedding.
When a number field is constructed it is cached so that a new construction
returns the same object, and not a copy, where possible. In your case that
means checking that the embeddings are the same, which means checking that
two elements of QQbar are the same. That can be expensive.
If you do
b1 = pi.roots(ring=QQbar)[3][0]
b2 = pi.roots(ring=QQbar)[3][0]
and then
b1==b2
it also takes ages. I cannot remember the algorithm here but it does not
look optimal; it may be working effectively in a field of degree 7*6 or
possibly 7!. I note that
d=b1-b2
d==0
yields True immediately.
John
Thank,
Paul
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