http://www-personal.umich.edu/~romanv/papers/non-asymptotic-rmt-plain.pdf Theorem 5.39 (page 23) gives a non-asymptotic upper bound on the largest singular value

Actually your context tells us you are interested in a much simpler problem. Rather than finding the longest directed simple path in the graph, you are only interested in the longest directed simple ...

EDIT: Woops just realized the vectors are drawn from the ball, not the sphere. But the measure of the ball is almost entirely concentrated at the sphere so the same result should apply with just ...

If $\mathcal{X}$ is the Minkowski sum of line segments $l_1,\ldots,l_k$, then $\mathcal{Y}$ is the Minkowski sum of line segments $Al_1,\ldots, Al_k$ (and hence a zonotope). As far as computing $C_y$ ...

I think the answer is yes. First observe that $CH(P) \subset CH(S)\cap H$: if $x\in CH(P)$ then $x$ is written as a convex combination of things which are convex combinations of vertices of $CH(S)$, ...

Proving this is actually problem 3 on page 164 of Integer Points in Polyhedra by Alexander Barvinok - the number of integer points is a polynomial in $t_1,...,t_k$ as long as they are non-negative ...

The nearest neighbor algorithm as I understand it (repeatedly select a neighboring vertex that hasn't been visited yet and travel to that vertex) does not guarantee that you will find a circuit even ...

A first pass to get continuity: for $||s||\geq \epsilon$, $f(s) = s/||s||$. For $||s||<\epsilon$, $f(s) = \frac{s}{||s||} (e^{1/\epsilon^2-1/||s||^2})$ If you want differentiability you just ...

Let $A\subset \mathbb{R}^d$ be (a)closed, (b) convex, and (c) contains the origin. Then $\left( A^{o} \right)^{o} = A$ where $o$ denotes the polar of $A$

Some observations which are too long for a comment to simplify the problem. The edges that are do not connect to G' are irrelevant, as is every 0 vertex that is outside of G'. So we can throw those ...

Gama doesn't have to be negative, in fact if delta is smaller than 1/2 gamma will be positive. It's known for Hadamard codes that arbitrarily large codes exist, and it seems intuitive that if the ...