# Algorithm for Finding a Conic Combination Giving the Zero Vector

Suppose I have a set of vectors $\{\mathbf s_1, \dots, \mathbf s_N\} \subset \mathbb Z^n$, and I want to find all conic combinations that give the zero vector. In other words, I want to find all vectors $\mathbf x \in \mathbb R^N$ such that $\mathbf x \geq \mathbf 0$ and

\begin{equation} \sum_{i=1}^N x_i \mathbf s_i = \mathbf 0. \end{equation}

Is there an algorithm for solving this problem?

Update: I was a little imprecise in saying that I wanted to find "all vectors...". To clarify, the set of $\mathbf x$ vectors that satisfy $\mathbf x \geq \mathbf 0$ and \begin{equation} \sum_{i=1}^N x_i \mathbf s_i = \mathbf 0. \end{equation} is a cone. I want to find the extreme rays of this cone.

In general, enumerating the extreme points or extreme rays of a polytope is a hard problem. If the size of the problem were not too large, and I had to write my own code to do it, I'd probably do something along the lines of:

1. Add the constraint $x_1 + \dots + x_N = 1$, which turns conic into convex combinations and (relevantly) turns all extreme rays into extreme points.
2. Use phase 1 of the simplex method to find one vertex.
3. Explore all the other extreme points by a standard breadth-first or depth-first search, trying all possible pivoting steps to go from a vertex to its adjacent vertices.

But there's also existing software that solves the problem for you, e.g. lrs which is just the first one I found on Google. Your task is exactly converting an H-representation of the polytope (the $2n+N$ linear inequalities that $\mathbf x$ must satisfy) into a V-representation (the extreme points and extreme rays).

Ordinary Gaussian elimination will give you a parametric equation for the subspace of all $x$s that satisfy your equation. The requirement that $\mathbf x\ge 0$ now easily translates to $N$ simultaneous linear inequalities in the parameter space.

Arguably those $N$ simultaneous linear inequalities is the best you can hope for to "find all vectors" -- in that such a system of linear inequalities is the standard way to describe a convex polytope in $\mathbb R^k$ in in the first place, such as when giving input to a linear programming solver.

If the parameter space is $2$- or perhaps $3$-dimensional you can probably understand the solution space "by eye". For higher dimensions it may not be clear whether there are solutions at all; a practical approach could be to feed all the inequalities to an off-the-shelf linear programming solver and ask it to minimize, say, $x_1$.

• I agree that there could easily be infinitely many solutions. In fact, it's straightforward to show that the set of solutions is a cone. So I suppose what I am really after is the extreme rays of this cone. I will update the question accordingly. – Garrett Aug 23 '18 at 21:47