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I have a set of $m$ linear inequalities in $R^n$, of the form $$ A x \leq b $$ These are automatically generated from the specification of my problem. Many of them could be removed because they are implied by the others.

I would like to find the minimal set of inequalities, $$ A' x \leq b' $$ such that a solution to the first problem is also a solution to the second, and vice versa.

One way to do this is to apply Fourier-Motzkin Elimination: I pick one inequality, $a_1 x \leq b_1$, I negate it as $-a_1 x < -b'$, I add it back to the system, and apply FME: if the result is an empty space, then the original inequality can be safely eliminated. Unfortunately, FME is a complex algorithm, and in the worst case it must be applied $m$ times.

Another possibility is to use linear programming: again, I remove one inequality $a_1 x \leq b$, and I call the remaining set of inequalities $A'' x \leq b''$. Then, I find an optimum for the problem $$ \max \;\; a_1 x \\ A'' x \leq b'' $$ using e.g. simplex, and finally I compare the result with $b$: if $a_1 x' < b$ then the inequality can be safely removed, otherwise it must be kept. Again, this has the same complexity of the simplex applied $m$ times (and $m$ is potentially very large).

Anyone knows a better algorithm?

Rodrigo de Azevedo
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knulp
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2 Answers2

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Using Farkas' lemma, an halfspace $\{x \mid a' x \leq b' \}$ contains the polyhedron $P= \{x |A x \leq b \}$ if and only if there exists $\lambda \in \mathbb{R}^m$ with $\lambda \geq 0$ (understood component-wise) such that: \begin{align*} a' &= \lambda^T A \\ b' &\geq \lambda^T b \end{align*} The set \begin{align*} N= \{ (a',b') | \exists \lambda \geq 0, a' = \lambda^T A, b' \geq \lambda^T b \} \end{align*} is called the polar of the polyhedron $P$. The facets of $P$ are given by the extreme rays of $N$.

An inequality $a_i x \leq b_i$ will be non-redundant (i.e is not implied by the other) if and only if it is an extreme ray of $N$.

So all you need to do is to compute the conic hull of the $m$ points $(a_1,b_1),...,(a_m,b_m)$ which is more or less equivalent (some gap to fill here) to computing a convex hull. Convex hulls can be computed in $O(m^2)$ using quickhull http://www.cse.unsw.edu.au/~lambert/java/3d/quickhull.html or any other convex hull algorithm.

Pascal
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Here is a proof of the first paragraph of Pascal's answer. It uses directly the separating hyperplane theorem, which states the following.

Given a matrix $A$ and column matrix $b$, \begin{align} Ax=b, &\text{ for some column matrix }x\ge0 \\ &\iff \\ y^TA\ge0 &\text{ for some column matrix }y \implies y^Tb\ge0 \end{align}

We put the required proposition into the above form. The premise of the proposition is \begin{align} &\begin{bmatrix} A & b \\ 0 & 1 \end{bmatrix}\begin{bmatrix} -xx_1 \\ x_1 \end{bmatrix}\ge 0 \\ &\iff \\ &\begin{cases} Ax\le b \implies a'^Tx\le b' \\ x_1\ge0 \end{cases} \\ &\implies \\ &\begin{bmatrix}a'^T &b'\end{bmatrix} \begin{bmatrix}-xx_1 \\x_1 \end{bmatrix}\ge0 \end{align}

The conclusion of the proposition is \begin{align} &\begin{bmatrix}\lambda^T & \lambda_1\end{bmatrix}\begin{bmatrix} A & b \\ 0 & 1 \end{bmatrix}=\begin{bmatrix}a'^T & b'\end{bmatrix}, \text{ for some} \begin{bmatrix}\lambda^T & \lambda_1\end{bmatrix}\ge0 \\ &\iff \\ &\exists \lambda\ge0 \ni \begin{cases} \lambda^TA=a'^T \\ \lambda^Tb\le b' \end{cases} \end{align}

The premise and the conclusion are seen to be equivalent in view of the separating hyperplane theorem as stated above. We have now proved the desired proposition.

Hans
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