#8095: is_primitive of WordMorphism is broken
-----------------------------+----------------------------------------------
Reporter: slabbe | Owner: slabbe
Type: defect | Status: new
Priority: major | Milestone: sage-4.3.2
Component: combinatorics | Keywords:
Author: | Upstream: N/A
Reviewer: | Merged:
Work_issues: |
-----------------------------+----------------------------------------------
Description changed by slabbe:
Old description:
> Let us define the following morphism over 3 letters:
> {{{
> sage: substitution=WordMorphism('a->b,b->ac,c->a')
> }}}
> Then we get
> {{{
> sage: substitution.is_primitive()
> False
> }}}
> but also
> {{{
> sage: (substitution^2).is_primitive()
> True
> }}}
>
> ------
>
> expected behaviour:
>
> See the description of ".is_primitive()":
> Returns True if self is primitive.
> A morphism ϕ is primitive if there exists an positive integer k such
> that for all α∈Σ, ϕk(α) contains all the letters of Σ.
>
> So, if a morphism is primitive, so are all its powers. And if there is
> a power which is primitive, so is the morphism itself. In the example
> above, both outputs should be "True".
New description:
Let us define the following morphism over 3 letters:
{{{
sage: substitution=WordMorphism('a->b,b->ac,c->a')
}}}
Then we get
{{{
sage: substitution.is_primitive()
False
}}}
but also
{{{
sage: (substitution^2).is_primitive()
True
}}}
------
expected behaviour:
See the description of ".is_primitive()":
Returns True if self is primitive.
A morphism ϕ is primitive if there exists an positive integer k such
that for all α∈Σ, ϕk(α) contains all the letters of Σ.
So, if a morphism is primitive, so are all its powers. And if there is
a power which is primitive, so is the morphism itself. In the example
above, both outputs should be "True".
This was reported here (via 'Report a problem'):
http://groups.google.com/group/sage-combinat-
devel/browse_thread/thread/5ed1186c229e7343?hl=en
--
--
Ticket URL: <http://trac.sagemath.org/sage_trac/ticket/8095#comment:1>
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