# 10.4 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. 7.

Fig. 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. 10.4.3 (Bound on the Determinant Root) we show how to implement bounds involving the determinant of a PSD matrix.

## 10.4.1 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:

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

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 22 Fusion implementation of model (1). 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 = M.variable("C", Domain.inPSDCone(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())
# t <= det(C)^{1/n}
det_rootn(M, C, t)

# Objective: Maximize t
M.objective(ObjectiveSense.Maximize, t)

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. 10.4.3 (Bound on the Determinant Root).

## 10.4.2 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:

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

and then directly into Fusion code:

Listing 23 Fusion implementation of model (2). 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 = M.variable("P", Domain.inPSDCone(n))
c = M.variable("c", n, Domain.unbounded())

# (1, Px-c) in cone
M.constraint("qc",
Expr.hstack(Expr.ones(m),
Expr.sub(Expr.mul(x, P),
Var.reshape(Var.repeat(c, m), [m, n])
)
),
Domain.inQCone())

# t <= det(P)^{1/n}
det_rootn(M, P, t)

# Objective: Maximize t
M.objective(ObjectiveSense.Maximize, t)
M.solve()

P, c = P.level(), c.level()
return ([P[i:i + n] for i in range(0, n * n, n)], c)


## 10.4.3 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

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

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

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

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

(5)$t \leq (\lambda_1\cdot\ldots\cdot\lambda_n)^{1/n}.$

This is easy to implement recursively using rotated quadratic cones when $$n$$ is a power of $$2$$; otherwise we need to round $$n$$ up to the nearest power of $$2$$ as in Listing 25. For example, $$t\leq (\lambda_1\lambda_2\lambda_3\lambda_4)^{1/4}$$ is equivalent to

$\lambda_1\lambda_2\geq y_1^2,\ \lambda_3\lambda_4\geq y_2^2,\ y_1y_2\geq t^2$

while $$t\leq (\lambda_1\lambda_2\lambda_3)^{1/3}$$ can be achieved by writing $$t\leq (t\lambda_1\lambda_2\lambda_3)^{1/4}$$.

For further details and proofs see [BenTalN01] or [MOSEKApS12].

Listing 24 Approaching the determinant, see (4). Click here to download.
def det_rootn(M, X, t):
n = int(sqrt(X.size()))

# Setup variables
Y = M.variable(Domain.inPSDCone(2 * n))

# Setup Y = [X, Z; Z^T , diag(Z)]
Y11 = Y.slice([0, 0], [n, n])
Y21 = Y.slice([n, 0], [2 * n, n])
Y22 = Y.slice([n, n], [2 * n, 2 * n])

M.constraint(Expr.sub(Y21.diag(), Y22.diag()), Domain.equalsTo(0.0))
M.constraint(Expr.sub(X, Y11), Domain.equalsTo(0.0))

# t^n <= (Z11*Z22*...*Znn)
geometric_mean(M, Y22.diag(), t)

Listing 25 Bounding the geometric mean, see (5). Click here to download.
def geometric_mean(M, x, t):
def rec(x):
n = x.getShape().dim(0)
if n > 1:
y = M.variable(int(n // 2), Domain.unbounded())
M.constraint(Var.hstack(Var.reshape(
x, [n // 2, 2]), y), Domain.inRotatedQCone())
return rec(y)
else:
return x

n = x.getShape().dim(0)
l = int(ceil(log(n, 2)))
m = int(2**l) - n

# if size of x is not a power of 2 we pad it:
if m > 0:
x_padding = M.variable(m, Domain.unbounded())

# set the last m elements equal to t
Domain.equalsTo(0.0))

x = Var.vstack(x, x_padding)

M.constraint(Expr.sub(Expr.mul(2.0**(l / 2.0), t), rec(x)),
Domain.equalsTo(0.0))