Sure: take x0=1/2.

On Wed, 21 May 2025 at 20:20, Dima Pasechnik <dimp...@gmail.com> wrote:
>
> I forgot - does Sage do the Minkowski decomposition into compact and 
> non-compact parts (so you do have a ray in your polyhedron)
> Is it even possible to define over Q a non-compact polyhedron without integer 
> points?
> (sorry, my geometry of numbers knowledge is pretty bad)
>
>
>
> On Wed, May 21, 2025, 12:52 Vincent Delecroix <20100.delecr...@gmail.com> 
> wrote:
>>
>> Note that for triangulating, sage does not help
>>
>> sage: M.polyhedron().triangulate()
>> Traceback (most recent call last):
>> ...
>> NotImplementedError: triangulation of non-compact polyhedra that are
>> not cones is not supported
>>
>> On Wed, 21 May 2025 at 18:39, Dima Pasechnik <dimp...@gmail.com> wrote:
>> >
>> > PS. You also have not set an objective function, not sure, but it could be 
>> > why you have no termination
>> >
>> > On Wed, May 21, 2025, 11:37 Dima Pasechnik <dimp...@gmail.com> wrote:
>> >>
>> >> It should be possible to construct the polyhedron determined by the 
>> >> feasible set of the LP, triangulate it, and do simplex by simplex or 
>> >> perhaps use various results relating volumes and presence of integral 
>> >> points in polyhedra.
>> >>
>> >> On Wed, May 21, 2025, 11:27 Vincent Delecroix <20100.delecr...@gmail.com> 
>> >> wrote:
>> >>>
>> >>> Dear all,
>> >>>
>> >>> I have a 7 variables 3 constraints linear program that I want to solve
>> >>> with integers
>> >>>
>> >>> x = M.new_variable(integer=True, nonnegative=True)
>> >>> M.add_constraint(x[0] - x[1] + x[2] - x[3] - x[4] - 2 * x[6] == 2)
>> >>> M.add_constraint(x[0] - x[1] + x[2] - 2 * x[4] - x[5] - x[6] == 2)
>> >>> M.add_constraint(2 * x[0] - x[3] - x[4] - x[5] - x[6] == -1)
>> >>>
>> >>> However, with both
>> >>>
>> >>> M = MixedIntegerLinearProgram(solver="PPL")
>> >>>
>> >>> and
>> >>>
>> >>> M = MixedIntegerLinearProgram(solver="GLPK")
>> >>>
>> >>> The command M.solve() does not terminate in reasonable time... I do
>> >>> not expect the system to have solutions, but I would like a proof of
>> >>> it.
>> >>>
>> >>> One subtlety of the system is that there are (infinitely many)
>> >>> positive integral solutions of the homogeneous version (ie linear
>> >>> combination == 0). I wondered if that was the reason why it is harder
>> >>> for a solver.
>> >>>
>> >>> If anyone knows of an alternative way to provide an open source
>> >>> computer assisted proof that there is no solution I would be
>> >>> interested.
>> >>>
>> >>> Best
>> >>> Vincent
>> >>>
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