From f0b3b883fd8f652fdaa245c99d9ccd29edff2dc5 Mon Sep 17 00:00:00 2001
From: Thibaut Horel <thibaut.horel@gmail.com>
Date: Wed, 18 Mar 2015 18:05:46 -0400
Subject: Attempt at a more combinatorial argument

---
 results.tex | 57 +++++++++++++++++++++++++++++++++++++++++++++++++++++++--
 1 file changed, 55 insertions(+), 2 deletions(-)

(limited to 'results.tex')

diff --git a/results.tex b/results.tex
index 2c0873d..974209b 100644
--- a/results.tex
+++ b/results.tex
@@ -1,5 +1,6 @@
 \documentclass[10pt]{article}
 \usepackage{fullpage, amsmath, amssymb, amsthm}
+\usepackage[utf8x]{inputenc}
 
 \newtheorem{proposition}{Proposition}
 \newtheorem{corollary}{Corollary}
@@ -14,6 +15,7 @@
 
 \section{Preliminary Results}
 
+\subsection{Generic Random Matrix Theory}
 We cite the following result from \cite{vershynin2010introduction} (Remark 5.40):
 
 \begin{proposition}
@@ -49,9 +51,60 @@ It follows that for any vector $x \in \mathbb{R}^n$, $|\frac{1}{N}\|A x\|^2_2 -\
 
 Note that this is clearly not optimal. For example, note that for $k=1$, the solution to the coupon collector problem states that only ${\cal O}(n\log n)$ rows are sufficient before the matrix $A$ has full row rank with high probability, and not ${\cal O}(n^2)$ as stated by Corollary~\ref{cor:number_of_rows_needed}.
 
+\subsection{More Direct Approach}
+
+Here we work on $\mathbb{F}_2$. An $n\times n$ binary matrix can be seen as
+a matrix over $\mathbb{F}_2$. Let us assume that each row of $A$ is chosen
+uniformly at random among all binary rows of length $n$. A standard counting
+arguments shows that the number of non-singular matrices over $\mathbb{F}_2$
+is:
+\begin{displaymath}
+    N_n = (2^n-1)(2^n - 2)\dots (2^n- 2^{n-1})
+    = (2^n)^n\prod_{i=1}^n\left(1-\frac{1}{2^i}\right)
+\end{displaymath}
+
+Hence the probability that our random matrix $A$ is invertible is:
+\begin{displaymath}
+    P_n  = \prod_{i=1}^n\left(1-\frac{1}{2^i}\right)
+\end{displaymath}
+It is easy to see that $P_n$ is a decreasing sequence. Its limit is
+$\phi(\frac{1}{2})$ where $\phi$ is the Euler function. We have
+$\phi(\frac{1}{2})\approx
+0.2887880950866024212788997219292307800889119048406857$ and $P_n$ converges
+exponentially fast to this constant (one way to see this is to use the
+Pentagonal number theorem).
+
+Hence, if we observe $\ell\cdot n$ uniformly random binary rows, the
+probability that they will have full column rank is at least:
+\begin{displaymath}
+    P_{\ell\cdot n}\geq 1 - \left(1-\phi\left(\frac{1}{2}\right)\right)^{\ell}
+\end{displaymath}
+
+Note that this is the probability of having full column rank over
+$\mathbb{F}_2$. A standard linear algebra argument shows that this implies full
+column rank over $\mathbb{R}$.
+
+\paragraph{TODO:} Study the case where we only observe sets of size exactly
+$k$, or at most $k$. This amounts to replacing $2^n$ in the computation above
+by:
+\begin{displaymath}
+{n\choose k}\quad\text{or}\quad\sum_{j=0}^k{n\choose j}
+\end{displaymath}
+
+Thinking about it, why do we assume that the sample sets are of size exactly
+$k$. I think it would make more sense from the learning perspective to consider
+uniformly random sets. In this case, the above approach allows us to conclude
+directly.
+
 \section{Passive Optimization of Additive Functions}
 
-Let $\Omega$ be a universe of elements of size $n$. For $k \in [n]$, let $D_k$ be the uniform probability distribution over all sets of size $k$, and let $D_p$ be the product probability distribution over sets such that $\forall j \in \Omega, \mathbb{P}_{S \sim D_p}(j \in S) = p$. Note that $D_{k}$ and $D_{p}$ for $p \equiv \frac{k}{n}$ are very similar, and will be considered almost interchangeably throughout the following notes. Consider the following problem:
+Let $\Omega$ be a universe of elements of size $n$. For $k \in [n]$, let $D_k$ be the uniform probability distribution over all sets of size $k$, and let $D_p$ be the product probability distribution over sets such that $\forall j \in \Omega, \mathbb{P}_{S \sim D_p}(j \in S) = p$. Note that $D_{k}$ and $D_{p}$ for $p \equiv \frac{k}{n}$ are very similar, and will be considered almost interchangeably throughout the following notes.
+
+\textbf{TODO:} Make the argument about independently choosing elements with
+proba $k/n$ vs choosing a set of size $k$ rigorous. The “more direct” approach
+above should allow us to consider the case of sets of size $k$ directly.
+
+Consider the following problem:
 
 \begin{problem}
 Let $f = (w_i)_{i \in \Omega}$ be an unknown additive function: $f(S) \equiv \sum_{u \in S} w_u$. How many (set, value) pairs $(S, f(S))$ where $S \sim D_k$ must be observed until we can optimize $f$ exactly or to a constant approximation factor under cardinality constraint $k$?
@@ -129,4 +182,4 @@ Same thing can probably be said for the average weight method.
 \bibliographystyle{apalike}
 
 
-\end{document}
\ No newline at end of file
+\end{document}
-- 
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From d6da106c03463113bb1356943f6b4084834868ef Mon Sep 17 00:00:00 2001
From: Thibaut Horel <thibaut.horel@gmail.com>
Date: Wed, 18 Mar 2015 18:32:02 -0400
Subject: Add small comment

---
 results.tex | 4 ++++
 1 file changed, 4 insertions(+)

(limited to 'results.tex')

diff --git a/results.tex b/results.tex
index 974209b..2a21586 100644
--- a/results.tex
+++ b/results.tex
@@ -96,6 +96,10 @@ $k$. I think it would make more sense from the learning perspective to consider
 uniformly random sets. In this case, the above approach allows us to conclude
 directly.
 
+More generally, I think the “right” way to think about this is to look at $A$
+as the adjacency matrix of a random $k$-regular graph. There are tons of
+results about the probability of the adjacency matrix to be non singular.
+
 \section{Passive Optimization of Additive Functions}
 
 Let $\Omega$ be a universe of elements of size $n$. For $k \in [n]$, let $D_k$ be the uniform probability distribution over all sets of size $k$, and let $D_p$ be the product probability distribution over sets such that $\forall j \in \Omega, \mathbb{P}_{S \sim D_p}(j \in S) = p$. Note that $D_{k}$ and $D_{p}$ for $p \equiv \frac{k}{n}$ are very similar, and will be considered almost interchangeably throughout the following notes.
-- 
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From 3c091fa90c6a4251777f8eea4fbe3751e76a785f Mon Sep 17 00:00:00 2001
From: Thibaut Horel <thibaut.horel@gmail.com>
Date: Wed, 18 Mar 2015 18:33:57 -0400
Subject: Comment on coupon collector

---
 results.tex | 5 +++++
 1 file changed, 5 insertions(+)

(limited to 'results.tex')

diff --git a/results.tex b/results.tex
index 2a21586..a9e79e1 100644
--- a/results.tex
+++ b/results.tex
@@ -51,6 +51,11 @@ It follows that for any vector $x \in \mathbb{R}^n$, $|\frac{1}{N}\|A x\|^2_2 -\
 
 Note that this is clearly not optimal. For example, note that for $k=1$, the solution to the coupon collector problem states that only ${\cal O}(n\log n)$ rows are sufficient before the matrix $A$ has full row rank with high probability, and not ${\cal O}(n^2)$ as stated by Corollary~\ref{cor:number_of_rows_needed}.
 
+\paragraph{Note:} $\Theta(n\log n)$ is in expectation for the coupon collector
+problem. If one want an exponentially small probability of failure, then
+I think results on the coupon collector problem also imply that a quadratic
+number of rows are required.
+
 \subsection{More Direct Approach}
 
 Here we work on $\mathbb{F}_2$. An $n\times n$ binary matrix can be seen as
-- 
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