path: root/results.tex

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\newtheorem{proposition}{Proposition}
\newtheorem{corollary}{Corollary}
\newtheorem{problem}{Problem}

\title{Learn and/or Optimize}
\author{}
\date{}

\begin{document}
\maketitle

\section{Preliminary Results}

\subsection{Generic Random Matrix Theory}
We cite the following result from \cite{vershynin2010introduction} (Remark 5.40):

\begin{proposition}
\label{prop:non_isotropic_isometry}
Assume that $A$ is an $N \times n$ matrix whose rows $A_i$ are independent sub-gaussian random vectors in $\mathbb{R}^n$ with second moment matrix $\Sigma$. Then for every $t \geq 0$, the following inequality holds with probability at least: $1- 2 \exp(-ct^2)$:
\begin{equation}
\|\frac{1}{N} A^T A - \Sigma \| \leq \max(\delta, \delta^2) \text{ where } \delta = C\sqrt{\frac{n}{N}} + \frac{t}{\sqrt{N}}
\end{equation}
where $C$, $c$ depend only on the subgaussian norm of the rows $K \equiv \max_i \| A_i\|_{\psi_2}$.
\end{proposition}

The following result is a simple corollary of Proposition~\ref{prop:non_isotropic_isometry}:

\begin{corollary}
\label{cor:number_of_rows_needed}
Let $n \in \mathbb{N}$ and $k \in ]0,n[$. Assume that $A$ is an $N \times n$ 
matrix whose rows $A_i$ are independent Bernoulli variable vectors in 
$\mathbb{R}^n$ such that $\mathbb{P}(A_{i,j} = 1) = \frac{k}{n} = 1 - 
\mathbb{P}(A_{i,j} = 0)$ and let $\sigma \equiv \frac{k}{n}(1- \frac{k}{n}) 
\neq 0$, then if $N > (C+1)^2 n/\sigma^2$, the matrix A has full row rank 
with probability at least $1 - e^{-cn}$, where $C, c$ are constant depending only on  the subgaussian norm of the rows $K \equiv \sup_{p \geq 1} \frac{1}{\sqrt{p}}(\mathbb{E}|A_1|^p)^\frac{1}{p} = k$\footnote{Is this true? And how do the constants behave?}
\end{corollary}

\begin{proof}
It is easy to compare the kernels of $A$ and $A^TA$ and notice that $rank(A) = rank(A^T A)$. Since $A^TA$ is an $n \times n$ matrix, it follows that if $A^TA$ is invertible, then $A$ has full row rank. In other words, if $A$ has smallest singular value $\sigma_{\min}(A) > 0$, then $A$ has full row rank. 
Consider Prop.~\ref{prop:non_isotropic_isometry} with $t \equiv \sqrt{n}$ and with $N > (C+1)^2 n$, then with probability $1 - 2\exp(-cn)$:
\begin{equation}
\|\frac{1}{N} A^T A - \sigma I \| \leq (C+1)\sqrt{\frac{n}{N}}
\end{equation}

It follows that for any vector $x \in \mathbb{R}^n$, $|\frac{1}{N}\|A x\|^2_2 -\sigma | \leq (C+1)\sqrt{\frac{n}{N}} \implies \|A x\|^2_2 \geq N\left(\sigma - (C+1)\sqrt{\frac{n}{N}}\right)$. If $N > (C+1)^2n/\sigma^2$, then $\sigma_{\min}(A) > 0$ with probability at least $1 - e^{-cn}$
\end{proof}

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
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.

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.

\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 $k$?
\end{problem}


\subsection{Inverting the system}
\label{subsec:invert_the_system}

Note that for observations $(S_j, f(S_j))_{j \in N}$ we can write the system $A w = b$, where each row of $A_j$ corresponds to the indicator vector of the set $S_j$ and $b_j \equiv f(S_j)$. From Corollary~\ref{cor:number_of_rows_needed}, it is easy to see that $(C_K+1)^2 \frac{n^3}{k(n - k)}$ rows are sufficient for $A$ to have full row rank with probability $1-e^{-c_k n}$. If $A$ has full row rank, then it is easy to invert the system in polynomial time, and both learn and optimize $f$ for any cardinality constraint.

\begin{proposition}
Assume that $N$ pairs of set-value $(S_j, f(S_j))_{j \in N}$ are observed, where $S_j \sim D_p$ and $p \equiv \frac{k}{n}$. If $N > (C_k +1)^2 \frac{n}{\sigma^2}$, then we can learn $f$ exactly, and therefore optimize it under any cardinality constraint, with probability $1-e^{-cn}$.
\end{proposition}

\subsection{Average weight method}

Another method is to compute for every node $i$ the \emph{average weight of every set containing element $i$}. Let $w_i$ be the weight of element $i$, $w \equiv f(\Omega) = \sum_{i \in \Omega} w_i$ and $p \equiv \frac{k}{n}$, then
$$\forall i \in \Omega, \mathbb{E}_{S \sim D_p}\left[f(S)| i \in S\right] = pw + (1 -p)w_i$$

Note that the average weight of every set containing element $i$ preserves the ranking of the weights of the elements. For observations $(S_j, f(S_j))_{j \in N}$ and for $N_i \equiv |\{S : i \in S\}|$, we define the following estimator:

\begin{equation}
\forall i \in \Omega, w_i^{N_i} \equiv \frac{1}{N_i}\sum_{S | i \in S} \frac{f(S) - pw}{1-p}
\end{equation}

As shown above, $w_i^{N_i} \rightarrow w_i$ as $N_i \rightarrow +\infty$. We can obtain a concentration bound of $w_i^{N_i}$ around $w_i$, using Hoeffding's lemma:

\begin{equation}
\mathbb{P}\left(\middle|w_i^{N_i} - w_i \middle|  \geq \epsilon w_i \right) \leq 2e^{-2(1-p)N_i\frac{\epsilon^2 w_i^2}{w^2}}
\end{equation}

\emph{TODO: multiplicative boudns are very bad for zero weights...Need to look at additive bounds for these zeros.}

For $N_i$ sufficiently large for each element $i$, we have $\forall i \in \Omega, (1-\epsilon) w_i \leq w_i^{N_i} \leq (1 + \epsilon) w_i$. Under this assumption, if we choose the $k$ elements with largest estimated weight $W_i^{N_i}$, we obtain a $\frac{1-\epsilon}{1+\epsilon}$-approximation to OPT, where OPT is the value of the maximum weight set of $k$ elements for the function $f$. To ensure that $N_i$ is sufficiently large for each element, we note that $\mathbb{E}(N_i) = pN$ and use a Chernoff bound coupled with a classic union bound:

\begin{equation}
\mathbb{P}\left(\bigcup_{i \in \Omega} \left[N_i \leq \frac{pN}{2}\right]\right) \leq \sum_{i\in \Omega} \mathbb{P}\left(N_i \leq \frac{pN}{2}\right) \leq n e^{-\frac{pN}{8}}
\end{equation}

As such, for $C > 0$ and $N \geq (C+1)\frac{8}{p}\log n$, we have $\forall i \in \Omega, N_i \geq \frac{pN}{2}$ with probability at least $1-\frac{1}{n^C}$

\begin{proposition}
Assume that $N$ pairs of set-value $(S_j, f(S_j))_{j \in N}$ are observed, where $S_j \sim D_p$ and $p \equiv \frac{k}{n}$. If $N > XXX$, then we can $\epsilon$-learn $f$ and optimize it to a $(1+\epsilon)/(1-\epsilon)$ factor for any cardinality constraint with probability $XXX$.
\end{proposition}

\section{Passive Optimization of Submodular functions}

\subsection{Inverting the system}

A result by \cite{balcan2011learning} states that:

\begin{proposition}
\label{prop:balcan_learned}
Let ${\cal F}$ be the class of non-negative, monotone, submodular functions over $\Omega$. There is an algorithm that PMAC-learns ${\cal F}$ with approximation factor $\sqrt{n+1}$, by outputting a function $\tilde f: S\mapsto \sqrt{\frac{1}{(n+1)z}w^T \chi(S)}$.
\end{proposition}

By adapting Section~\ref{subsec:invert_the_system}, we get the following result:

\begin{proposition}
Any non-negative monotone submodular function can be passively optimized under cardinality constraint $k \in \mathbb{N}$ from $D_p$ for \emph{any} $p \in ]0, 1[$ with polynomially many samples up to a factor $1/\sqrt{n+1}$. 
\end{proposition}

\begin{proof}
We optimize $\tilde f$ from Proposition~\ref{prop:balcan_learned}, which can easily be done by solving for the system $Ax = b^2$ where $(b^2)_i = (b_i)^2$. TODO: complete proof
\end{proof}



\subsection{Average weight method}

Same thing can probably be said for the average weight method.

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\end{document}