Not easily. There is this internal method whose output includes which
points saturate which equation. Though you probably should just iterate
over all points and pick the ones you like.
sage: from sage.geometry.polyhedron.ppl_lattice_polytope import
LatticePolytope_PPL
sage: P = LatticePolytope_PPL([0,0], [3,0], [0,3])
sage: P.constraints()
Constraint_System {x0>=0, x1>=0, -x0-x1+3>=0}
sage: P._integral_points_saturating()
(((0, 0), frozenset([0, 1])),
((0, 1), frozenset([0])),
((0, 2), frozenset([0])),
((0, 3), frozenset([0, 2])),
((1, 0), frozenset([1])),
((1, 1), frozenset([])),
((1, 2), frozenset([2])),
((2, 0), frozenset([1])),
((2, 1), frozenset([2])),
((3, 0), frozenset([1, 2])))
On Tuesday, March 11, 2014 9:14:04 PM UTC, Soli vishkautsan wrote:
>
> Is there a quick way to calculate the number of *internal* lattice points
> of a lattice polytope? LatticePolytope object has npoints method, which
> returns the total number of lattice points, including the border points.
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