A separation oracle (also called a cutting-plane oracle) is a concept in the mathematical theory of convex optimization. It is a method to describe a convex set that is given as an input to an optimization algorithm. Separation oracles are used as input to ellipsoid methods.^{[1]: 87, 96, 98}

Definition

Let K be a convex and compact set in Rⁿ. A strong separation oracle for K is an oracle (black box) that, given a vector y in Rⁿ, returns one of the following:^[1]: 48

• Assert that y is in K.
• Find a hyperplane that separates y from K: a vector a in Rⁿ, such that ${\displaystyle a\cdot y>a\cdot x}$ for all x in K.

A strong separation oracle is completely accurate, and thus may be hard to construct. For practical reasons, a weaker version is considered, which allows for small errors in the boundary of K and the inequalities. Given a small error tolerance d>0, we say that:

• A vector y is d-near K if its Euclidean distance from K is at most d;
• A vector y is d-deep in K if it is in K, and its Euclidean distance from any point in outside K is at least d.

The weak version also considers rational numbers, which have a representation of finite length, rather than arbitrary real numbers. A weak separation oracle for K is an oracle that, given a vector y in Qⁿ and a rational number d>0, returns one of the following::^[1]: 51

• Assert that y is d-near K;
• Find a vector a in Qⁿ, normalized such that its maximum element is 1, such that ${\displaystyle a\cdot y+d\geq a\cdot x}$ for all x that are d-deep in K.

Implementation

A special case of a convex set is a set represented by linear inequalities: ${\displaystyle K=\{x|Ax\leq b\))$. Such a set is called a convex polytope. A strong separation oracle for a convex polytope can be implemented, but its run-time depends on the input format.

Representation by inequalities

If the matrix A and the vector b are given as input, so that ${\displaystyle K=\{x|Ax\leq b\))$, then a strong separation oracle can be implemented as follows.^[2] Given a point y, compute ${\displaystyle Ay}$:

• If the outcome is at most ${\displaystyle b}$, then y is in K by definition;
• Otherwise, there is at least one row ${\displaystyle c}$ of A, such that ${\displaystyle c\cdot y}$ is larger than the corresponding value in ${\displaystyle b}$; this row ${\displaystyle c}$ gives us the separating hyperplane, as ${\displaystyle c\cdot y>b\geq c\cdot x}$ for all x in K.

This oracle runs in polynomial time as long as the number of constraints is polynomial.

Representation by vertices

Suppose the set of vertices of K is given as an input, so that ${\displaystyle K={\text{conv))(v_{1},\ldots ,v_{k})=}$ the convex hull of its vertices. Then, deciding whether y is in K requires to check whether y is a convex combination of the input vectors, that is, whether there exist coefficients z₁,...,z_k such that: ^[1]: 49

• ${\displaystyle z_{1}\cdot v_{1}+\cdots +z_{k}\cdot v_{k}=y}$;
• ${\displaystyle 0\leq z_{i}\leq 1}$ for all i in 1,...,k.

This is a linear program with k variables and n equality constraints (one for each element of y). If y is not in K, then the above program has no solution, and the separation oracle needs to find a vector c such that

• ${\displaystyle c\cdot y>c\cdot v_{i))$ for all i in 1,...,k.

Note that the two above representations can be very different in size: it is possible that a polytope can be represented by a small number of inequalities, but has exponentially many vertices (for example, an n-dimensional cube). Conversely, it is possible that a polytope has a small number of vertices, but requires exponentially many inequalities (for example, the convex hull of the 2n vectors of the form (0,...,±1,...,0).

Problem-specific representation

In some linear optimization problems, even though the number of constraints is exponential, one can still write a custom separation oracle that works in polynomial time. Some examples are:

• The minimum-cost arborescence problem: given a weighted directed graph and a vertex r in it, find a subgraph of minimum cost that contains a directed path from r to any other vertex. The problem can be presented as an LP with a constraint for each subset of vertices, which is an exponential number of constraints. However, a separation oracle can be implemented using n-1 applications of the minimum cut procedure.^[3]
• The maximum independent set problem. It can be approximated by an LP with a constraint for every odd-length cycle. While there are exponentially-many such cycles, a separation oracle that works in polynomial time can be implemented by just finding an odd cycle of minimum length, which can be done in polynomial time.^[3]
• The dual of the configuration linear program for the bin packing problem. It can be approximated by an LP with a constraint for each feasible configuration. While there are exponentially-many such cycles, a separation oracle that works in pseudopolynomial time can be implemented by solving a knapsack problem. This is used by the Karmarkar-Karp bin packing algorithms.

Non-linear sets

Let f be a convex function on Rⁿ. The set ${\displaystyle K=\{(x,t)|f(x)\leq t\))$ is a convex set in Rⁿ⁺¹. Given an evaluation oracle for f (a black box that returns the value of f for every given point), one can easily check whether a vector (y, t) is in K. In order to get a separation oracle, we need also an oracle to evaluate the subgradient of f.^[1]: 49 Suppose some vector (y, s) is not in K, so f(y) > s. Let g be the subgradient of f at y (g is a vector in Rⁿ). Denote ${\displaystyle c:=(g,-1)}$.Then, ${\displaystyle c\cdot (y,s)=g\cdot y-s>g\cdot y-f(y)}$, and for all (x, t) in K: ${\displaystyle c\cdot (x,t)=g\cdot x-t\leq g\cdot x-f(x)}$. By definition of a subgradient: ${\displaystyle f(x)\geq f(y)+g\cdot (x-y)}$ for all x in Rⁿ. Therefore, ${\displaystyle g\cdot y-f(y)\geq g\cdot x-f(x)}$, so ${\displaystyle c\cdot (y,s)>c\cdot (x,t)}$ , and c represents a separating hyperplane.

Usage

A strong separation oracle can be given as an input to the ellipsoid method for solving a linear program. Consider the linear program ${\displaystyle {\text{maximize))~~c\cdot x~~{\text{subject to))~~Ax\leq b,x\geq 0}$. The ellipsoid method maintains an ellipsoid that initially contains the entire feasible domain ${\displaystyle Ax\leq b}$. At each iteration t, it takes the center ${\displaystyle x_{t))$ of the current ellipsoid, and sends it to the separation oracle:

• If the oracle says that ${\displaystyle x_{t))$ is feasible (that is, contained in the set ${\displaystyle Ax\leq b}$), then we do an "optimality cut" at ${\displaystyle x_{t))$: we cut from the ellipsoid all points x for which ${\displaystyle c\cdot x<c\cdot x_{t))$. These points are definitely not optimal.
• If the oracle says that ${\displaystyle x_{t))$ is infeasible, then it typically returns a specific constraint that is violated by ${\displaystyle x_{t))$, that is, a row ${\displaystyle a_{j))$ in the matrix A, such that ${\displaystyle a_{j}\cdot x_{t}>b_{j))$. Since ${\displaystyle a_{j}\cdot x\leq b_{j))$ for all feasible x, this implies that ${\displaystyle a_{j}\cdot x_{t}>a_{j}\cdot x}$ for all feasible x. Then, we do a "feasibility cut" at ${\displaystyle x_{t))$: we cut from the ellipsoid all points y for which ${\displaystyle a_{j}\cdot y>a_{j}\cdot x_{t))$. These points are definitely not feasible.

After making a cut, we construct a new, smaller ellipsoid, that contains the remaining region. It can be shown that this process converges to an approximate solution, in time polynomial in the required accuracy.

Converting a weak oracle to a strong oracle

Given a weak separation oracle for a polyhedron, it is possible to construct a strong separation oracle by a careful method of rounding, or by diophantine approximations.^[1]: 159