mf's profile - activity

thanks for the explanation!

I wonder whether there is a way to get back the short name of the function of the various rings and fields like QQ, RDF, AA, RLF, RR, etc. as a string.

For example

sage: a=QQ
sage: str(a)
'Rational Field'

But I am looking for something like:

sage: function_I_want(a)
'QQ'

I guess I could predefine a dictionary like this:

shortnames={eval(name):name for name in ['QQ', 'RDF', 'AA', 'RLF', 'RR']}

and then have a function

def function_I_want(a):
    return shortnames[a]

But this seems a bit clumsy. Is there a better way to do this; or is there somewhere in the sage code such a dictionary already defined?

The reason I am thinking about this is the following tiny bug: https://trac.sagemath.org/ticket/22132

Here is a result that I found surprising and I don't understand completely what corercian is causing it.

sage: int(.9999999999999999)
0
sage: int(.99999999999999999)
1
sage: int(0.99999999999999999)
0
sage: int(0.9999999999999999)
0
sage: int(0.999999999999999999999999999999999)
0

What is going on?

More to play with:

sage: a=.99999999999999999; b=0.999999999999999999999999999999999
sage: a<b
False
sage: int(a)<int(b)
False

and

sage: a=.999999999999999990; b=0.999999999999999999999999999999999
sage: int(a)<int(b)
False
sage: a<b
True
sage: a=.999999999999999997
sage: a<b
False
sage: a=.999999999999999996
sage: a<b
True

I am trying to get sage working with planar embedded graphs. The answer here was quite helpful, but I have a few more questions.

1. Is there a way to get represention of the embedding into sage? The graph6 format forgets the embedding (see the documentation of plantri).

2. Can I presribe the outer face of my embedding when I plot a Graph in sage using .plot(layout='planar')?

Ideally, for an embedded graph, I would like to obtain something like an oriented outer face and a list of oriented inner faces. For example for the graph below I want to have

inner_faces=[[2,0,3],[0,1,4,5,3],[1,2,4],[2,3,5],[2,5,4]]
outer_face=[0,1,2]

Maybe relating to this question: Access output from previous cells

Assume I defined a function that takes a string and prints it:

def f(s):
    print s
    return

Now the output of the cell

f('hello')

is just the word 'hello'.

Is there any way to access this output, for example read it as a variable?

The problem occurs when I have a MixedIntegerLinearProgram p and access it with p.show(). I then want to do something with the printed output of this command, e.g. write it to a file.

I am having trouble understanding an error I get, when i try the following:

ngon=polytopes.regular_polygon(122).dilation(18.1)
ngon.plot()

It says: KeyError: (frozenset([60]), frozenset([60])) The following works:

ngon=polytopes.regular_polygon(122).dilation(17.9)
ngon.plot()

What is going on? Is this a bug?

I am trying to find a numerical approximation with arbitrary precision to a real solution to a system of multivariate polynomial equations. I start out with an approximation which is somewhat close to solving the system, up to an precision of about 1e-05. (Meaning that the equations that I try to evaluate are not zero, but smaller than 1e-05 for my starting value)

In this question is it recommended to use scipy's fmin_tnc method, which is what I did. This works out very nicely and it quickly gave a new solution which now solves my system with precision 1e-07. In the Scipy doc it is stated that one can set the "epsilon" parameter, but not smaller than machine precision. So it seems like I can't get much more precision with this method?!

Let's say I want to solve my system with precision 1e-250. My questions are:

1. Can I use the fmin_tnc function to find solutions with higher precision?
2. I there another way in sage to find real solutions to polynomial systems locally (e.g. with the pari/gp)?

perfect, thanks! I don't know how i missed this..

Is there an easy way to check whether two polyhedra are combinatorial equivalent, i.e. have have isomorphic face lattices.

this does not work:

Poly1=Polyhedron(vertices=[[0,1],[1,0],[1,1]], base_ring=QQ)
Poly2=Polyhedron(vertices=[[2,0],[2,2],[0,2]], base_ring=QQ)
Poly1.face_lattice()==Poly2.face_lattice()

In this case this would work:

str(Poly1.faces(1))==str(Poly2.faces(1))

but what would be a good way to check this in general?

@vdelecroix: will do as soon as I have enough karma to do so..

@vdelecroix: I figured out how to do this, see my answer below