11.6 Inner and outer Löwner-John Ellipsoids

In this section we show how to compute the Löwner-John inner and outer ellipsoidal approximations of a polytope. They are defined as, respectively, the largest volume ellipsoid contained inside the polytope and the smallest volume ellipsoid containing the polytope, as seen in Fig. 11.7.

_images/ellipses_polygon.png

Fig. 11.7 The inner and outer Löwner-John ellipse of a polygon.

For further mathematical details, such as uniqueness of the two ellipsoids, consult [BenTalN01]. Our solution is a mix of conic quadratic and semidefinite programming. Among other things, in Sec. 11.6.3 (Bound on the Determinant Root) we show how to implement bounds involving the determinant of a PSD matrix.

11.6.1 Inner Löwner-John Ellipsoids

Suppose we have a polytope given by an h-representation

\[\mathcal{P} = \{ x \in \real^n \mid Ax \leq b \}\]

and we wish to find the inscribed ellipsoid with maximal volume. It will be convenient to parametrize the ellipsoid as an affine transformation of the standard disk:

\[\mathcal{E} = \{ x \mid x = Cu + d,\ u\in\real^n,\ \| u \|_2 \leq 1 \}.\]

Every non-degenerate ellipsoid has a parametrization such that \(C\) is a positive definite symmetric \(n\times n\) matrix. Now the volume of \(\mathcal{E}\) is proportional to \(\mbox{det}(C)^{1/n}\). The condition \(\mathcal{E}\subseteq\mathcal{P}\) is equivalent to the inequality \(A(Cu+d)\leq b\) for all \(u\) with \(\|u\|_2\leq 1\). After a short computation we obtain the formulation:

(11.22)\[\begin{split}\begin{array}{lll} \maximize & t & \\ \st & t \leq \mbox{det}(C)^{1/n}, & \\ & (b-Ad)_i\geq \|(AC)_i\|_2, & i=1,\ldots,m,\\ & C \succeq 0, & \end{array}\end{split}\]

where \(X_i\) denotes the \(i\)-th row of the matrix \(X\). This can easily be implemented using Fusion, where the sequence of conic inequalities can be realized at once by feeding in the matrices \(b-Ad\) and \(AC\).

Listing 11.11 Fusion implementation of model (11.22). Click here to download.
def lownerjohn_inner(A, b): with Model("lownerjohn_inner") as M: M.setLogHandler(sys.stdout) m, n = len(A), len(A[0]) # Setup variables t = M.variable("t", 1, Domain.greaterThan(0.0)) C = det_rootn(M, t, n) d = M.variable("d", n, Domain.unbounded()) # (b-Ad, AC) generate cones M.constraint("qc", Expr.hstack(Expr.sub(b, Expr.mul(A, d)), Expr.mul(A, C)), Domain.inQCone()) # Objective: Maximize t M.objective(ObjectiveSense.Maximize, t) M.writeTask("lj-inner.ptf") M.solve() C, d = C.level(), d.level() return ([C[i:i + n] for i in range(0, n * n, n)], d) 

The only black box is the method det_rootn which implements the constraint \(t\leq \mbox{det}(C)^{1/n}\). It will be described in Sec. 11.6.3 (Bound on the Determinant Root).

11.6.2 Outer Löwner-John Ellipsoids

To compute the outer ellipsoidal approximation to a polytope, let us now start with a v-representation

\[\mathcal{P} = \mbox{conv}\{ x_1, x_2, \ldots , x_m \} \subseteq \real^n,\]

of the polytope as a convex hull of a set of points. We are looking for an ellipsoid given by a quadratic inequality

\[\mathcal{E} = \{ x\in\real^n \mid \| Px-c \|_2 \leq 1 \},\]

whose volume is proportional to \(\mbox{det}(P)^{-1/n}\), so we are after maximizing \(\mbox{det}(P)^{1/n}\). Again, there is always such a representation with a symmetric, positive definite matrix \(P\). The inclusion conditions \(x_i\in\mathcal{E}\) translate into a straightforward problem formulation:

(11.23)\[\begin{split}\begin{array}{lll} \maximize & t &\\ \st & t \leq \mbox{det}(P)^{1/n}, &\\ & \|Px_i - c\|_2 \leq 1, &i=1,\ldots,m,\\ & P \succeq 0, & \end{array}\end{split}\]

and then directly into Fusion code:

Listing 11.12 Fusion implementation of model (11.23). Click here to download.
def lownerjohn_outer(x): with Model("lownerjohn_outer") as M: M.setLogHandler(sys.stdout) m, n = len(x), len(x[0]) # Setup variables t = M.variable("t", 1, Domain.greaterThan(0.0)) P = det_rootn(M, t, n) c = M.variable("c", Domain.unbounded().withShape(1,n)) # (1, Px-c) in cone M.constraint("qc", Expr.hstack(Expr.ones(m), Expr.sub(Expr.mul(x, P), Expr.repeat(c,m,0))), Domain.inQCone().axis(1)) # Objective: Maximize t M.objective(ObjectiveSense.Maximize, t) M.writeTask("lj-outer.ptf") M.solve() P, c = P.level(), c.level() return ([P[i:i + n] for i in range(0, n * n, n)], c) 

11.6.3 Bound on the Determinant Root

It remains to show how to express the bounds on \(\mbox{det}(X)^{1/n}\) for a symmetric positive definite \(n\times n\) matrix \(X\) using PSD and conic quadratic variables. We want to model the set

(11.24)\[C = \lbrace (X, t) \in \PSD^n \times \real \mid t \leq \mbox{det}(X)^{1/n} \rbrace.\]

A standard approach when working with the determinant of a PSD matrix is to consider a semidefinite cone

(11.25)\[\begin{split}\left( {\begin{array}{cc}X & Z \\ Z^T & \mbox{Diag}(Z) \\ \end{array} } \right) \succeq 0\end{split}\]

where \(Z\) is a matrix of additional variables and where we intuitively identify \(\mbox{Diag}(Z)=\{\lambda_1,\ldots,\lambda_n\}\) with the eigenvalues of \(X\). With this in mind, we are left with expressing the constraint

(11.26)\[t \leq (\lambda_1\cdot\ldots\cdot\lambda_n)^{1/n}.\]

This is easy to implement recursively using rotated quadratic cones when \(n\) is a power of \(2\). In general it is convenient to express (11.26) as a composition of power cones, see [BenTalN01] or Modeling Cookbook.

Listing 11.13 Approaching the determinant, see (11.25). Click here to download.
def det_rootn(M, t, n): # Setup variables Y = M.variable(Domain.inPSDCone(2 * n)) # Setup Y = [X, Z; Z^T , diag(Z)] X = Y.slice([0, 0], [n, n]) Z = Y.slice([0, n], [n, 2 * n]) DZ = Y.slice([n, n], [2 * n, 2 * n]) # Z is lower-triangular M.constraint(Z.pick([[i,j] for i in range(n) for j in range(i+1,n)]), Domain.equalsTo(0.0)) # DZ = Diag(Z) M.constraint(Expr.sub(DZ, Expr.mulElm(Z, Matrix.eye(n))), Domain.equalsTo(0.0)) # t^n <= (Z11*Z22*...*Znn) geometric_mean(M, DZ.diag(), t) # Return an n x n PSD variable which satisfies t <= det(X)^(1/n) return X 
Listing 11.14 Bounding the geometric mean, see (11.26). Click here to download.
def geometric_mean(M, x, t): n = int(x.getSize()) if n==1: M.constraint(Expr.sub(t, x), Domain.lessThan(0.0)) else: t2 = M.variable() M.constraint(Var.hstack(t2, x.index(n-1), t), Domain.inPPowerCone(1-1.0/n)) geometric_mean(M, x.slice(0,n-1), t2)