Hi Daniel, As far as I know, no such method is implemented in the polyhedron class.
If I understood your situation, you want to know the set of coordinates "x_i" such that the projection along that axis is an injective function i.e. given a point in the image of the projection, you can lift it back uniquely using equations. Already here, I see that this can be non-empty if and only if you have equations, which is what you want I guess. So, to compute the coordinates for which this property is true, I would compute a basis for the affine hull of the polyhedron (essentially, being lazy, I would just compute the polyhedron by dropping the inequalities and take the lines as my basis) and if that basis has a canonical vector e_i, then that vector is in the complement of the set you're looking for. Best, J-P Le lundi 25 février 2019 11:10:24 UTC+1, Daniel Krenn a écrit : > > The H-representation consists of equations and inequalities and the > equations seem to be in some canonical form. Is there a method that > returns the non-free variables (or indices), i.e. that are the variables > completely determined by the equations meaning once a value for the > other variables is fixed, then one can compute them and no further > restrictions apply. > > E.g. for > > sage: polytopes.simplex(2).Hrepresentation() > (An equation (1, 1, 1) x - 1 == 0, > An inequality (0, -1, -1) x + 1 >= 0, > An inequality (0, 1, 0) x + 0 >= 0, > An inequality (0, 0, 1) x + 0 >= 0) > > I simply want to get the list of indices [0], as the first component is > only determined by (1, 1, 1) x - 1 and does not appear in any of the > inequalities. > -- You received this message because you are subscribed to the Google Groups "sage-support" group. To unsubscribe from this group and stop receiving emails from it, send an email to sage-support+unsubscr...@googlegroups.com. To view this discussion on the web visit https://groups.google.com/d/msgid/sage-support/c927ad01-a402-46b7-b070-8256aa5f294c%40googlegroups.com.
