Okay so here is the deal...
I have a CLOSED convex polyhedron $Ax \le b$ (where $x$ is in $R^n$)
and it has i vertices denoted $V_i$ such that $V_i = (x_{i1}, x_{i2}, \ldots, x_{iN})$
where $0 \le x_{ij} \le 1 $ for all components $x_{ij}$ of all vertices $V_i$
I am told ahead of time that there definitely exists a finite number of integral points on this polyhedron with components greater than 0. Is there a way to find them in polynomial of $\log(i)$ time? (obviously I can use $O(i)$ time and check them all but that is not practical for what I want to do).
My first idea was to notice if $(0,0,\ldots)$ and $(1,1,1,\ldots)$ are not solutions then I can add constraints $1 <=$ sum of all $x_{ij} \le 7$ and then consider solutions with either only 2 1's or 2 0's to tighten these bounds even further but this gets very expensive quickly and so this method can only be used to shave off time and won't be the bulk of any useful algorithm
My next idea was to basically sort the inequalities into an N-dimensional graph which gives an idea of where each inequality resides. (basically if each of the (N-1) dimensional faces of the polyhedron was transformed into a node and faces that are adjacent are connected by and edge then my linked list should be topologically equivalent to the this polyhedron graph). (This should take $m^3\log_2(m)^3$ time where M is the number of faces ($m = 2\log_2(i)$) and therefore $8\log_2(i)^3$)
Now that the inequalities are sorted I came up with the following stratagy:
http://www.maths.ed.ac.uk/~aar/papers/barvpomm.pdf demonstrates a formula for calculating the number of lattice points contained in the polyhedron.
I already know my polyhedron has a finite number of them and they only exist as vertices. So I can basically cut my topological graph (the linked list) across an axis and calculate for both halves the number of integer points respectively. For the half that reported a value greater than 0 I can slice it again and repeat this until I figure out around which edge there exists a lattice point. This takes about $log_2(i)$ steps.
Is determining the number of lattice points in this specific type of polyhedron doable in polynomial time?
Keep in mind my polyhedron is enclosed in an n-cube in $R^n$ with side-length 1
Another stratagy to shave off some more time is to determine the points with the maximum value of $x_{i}$ (for example biggest x value, biggest y value, etc...) and use that to help make the polyhedron a bit smaller and therefore easier to work with.