## Lec #16: Ellipsoid and Center-of-Gravity

We will talk a bit more about the Ellipsoid algorithm (and separation vs optimization) on Friday, maybe start talking about the Newton method, and then move on to more details on interior point techniques on Monday.

• One point to emphasize: the ${t^{th}}$ step of Ellipsoid cuts ${\mathcal{E}_t}$ into two, and builds ${\mathcal{E}_{t+1}}$ around the relavant half. Suppose ${\mathcal{E}_t = B}$, where ${B}$ is the unit ball. Then ${\mathcal{E}_{t+1} = \mathcal{E}'}$ is some ellipsoid with volume at most ${e^{1/(2(n+1))}}$ times the volume of the unit ball. Now since any ellipsoid is a linear transformation of the ball, if ${\mathcal{E}_t = L_t B}$, where ${L_t}$ is the associated linear transformation, then ${\mathcal{E}_{t+1} = L_t \mathcal{E}'}$. But for any body ${K}$, the volume of ${L_t K}$ is ${|det(L_t)| \cdot volume(K)}$. So

$\displaystyle \frac{volume(\mathcal{E}_{t+1})}{volume(\mathcal{E}_{t})} = \frac{|det(L_t)| \cdot volume(\mathcal{E}')}{|det(L_t)| \cdot volume(B)} \leq e^{1/(2(n+1))}.$

In fact, you can imagine that each step we are just transforming the current elipsoid back to a ball, making a cut, and then transforming things back. One problem, of course, is that at each step we make the transformation more complicated. Indeed, in the notation above, if ${E' = L' B}$, then ${L_{t+1} := L_t L'}$. So the numbers involved can get larger at each step: we may need to round the numbers to control how big they get. These and other numerical issues are at the heart of Khachiyan’s proof that the algorithm runs in polynomial time.

• My apologies: the alternate version of Ellipsoid I thought used boxes, in fact, uses simplices. The analysis that it runs in polynomial time is due to Boris Yamnitsky and (Leonid) Levin. The main idea is that half of some simplex ${S}$ can be contained in another simplex of volume at most ${e^{1/(2(n+1)^2)} \approx 1 - \Theta(1/n^2)}$ times the volume of ${S}$. (A proof appears in Vasek Chvatal’s book.) Note this factor is worse than the Ellipsoid factor of ${e^{1/(2(n+1))}}$ by another factor of ${n}$, but the numerical calculations in the algorithm definition do not require the use of square roots. The original paper (also here) is not quite kosher, since it can lead to the size of the numbers blowing up: this report by Bartels gives an example, and also suggests rounding approaches to control the problem. Finally, some notes on the simplices algorithm by Yossi Azar (parts 1, 2) which I have not had a chance to go over in detail.

And a couple words about the center-of-gravity algorithm:

• The center-of-gravity definition. It’s the natural extension of the discrete case. Indeed, if we have ${N}$ objects in ${{\mathbb R}^n}$, the ${i^{th}}$ one having location ${x_i \in {\mathbb R}^n}$ and mass ${\mu_i \geq 0}$, the center of gravity (or the center of mass, or centroid) is defined as

$\displaystyle c = \frac{\sum_{i \in [N]} x_i \mu_i }{\sum_{i \in [N]} \mu_i}.$

The continuous analog of this where we have a general measure ${\mu(x)}$ over ${{\mathbb R}^n}$ (basically replacing sums by integrals), is

$\displaystyle c = \frac{\int_{x \in R^n} x \; d\mu(x) }{\int_{x \in R^n} d\mu(x)}.$

The numerator is the total measure over ${{\mathbb R}^n}$. (In class I was implcitly assuming the uniform measure over ${K \subseteq {\mathbb R}^n}$, which is given by ${d\mu(x) = \mathbf{1}_{x \in K} dx}$.

The ${1/e}$ in Grunbaum’s theorem (that each hyperplane through the centroid of a convex body contains at least ${1/e}$ fraction of the mass on either side) is best possible for convex bodies. And the proof is clever but not difficult See Grunbaum’s (very short) paper for examples and proof. Or these notes by Jon Kelner or Santosh Vempala.

What happens if we don’t have the uniform measure over a convex body, but a more general distribution? Then things change quite a bit. E.g., consider ${n+1}$ equal point masses at the vertices of an ${n}$-dimensional simplex. No matter which point you choose, you can find a hyperplane through it that contains only a single point (which is ${1/n+1}$ of the mass) on one side. Grunbaum actually shows (in the same paper) that you can find a point that ensures at least ${1/(n+1)}$ fraction of the mass on either side.