If you want to know the geometry of the solution set then you should work 
with cones directly:

sage: c1 = Cone([(1,0), (0,1)])
sage: c2 = Cone([(0,1), (-1,0)])
sage: c1.intersection(c2).polyhedron().Hrepresentation()
(An equation (1, 0) x + 0 == 0, An inequality (0, 1) x + 0 >= 0)

If you want to maximize a linear function on the solution set then you 
should use the MixedIntegerLinearProgram code.




On Friday, November 2, 2012 10:11:04 PM UTC, Ryan Davis wrote:
>
> I've been trying to do calculations on polytopes, and the associated 
> rational polyhedral cones, cone lattice, and so on.  One of the 
> computations I've been trying to implement computes a list of inequalities 
> that must hold true in terms of symbolic variables within the polytope, and 
> I would like to complete the complete solution set to these inequalities.
>
> For some of the relatively simple polytopes, the list of inequalities is 
> rather small, and I'm able to use solve([list of inequalities], [list of 
> relevant variables]) to solve for the complete solution set, which takes a 
> minute or two.  As the complexity of the polytope ramps up, though, very 
> large lists of inequalities are generated, and solve() takes an immense 
> amount of time.
>
> Is there a faster way to compute the complete set of solutions to a system 
> of linear inequalities than solve()?  From what I can see of Mixed Integer 
> Linear Programming, I'm not sure it computes the full solution set.  I'm 
> not sure how I would go about expressing these inequalities in terms of 
> matrices, or if that's even a valid approach for a linear inequality of 
> several variables.  Is there a different way of solving linear 
> inequalities, or alternatively, is my understanding of MILP incorrect?
>
> Thanks,
> Ryan Davis
>

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