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approx_alg.tex
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@ -1,40 +1,8 @@
%root: main.tex
\section{$1 \pm \epsilon$ Approximation Algorithm}
%\AH{I am attempting to rewrite this section mostly from scratch. This will involve taking 'baby' steps towards the goals we spoke of on Friday 080720 as well as throughout the following week on chat channel.}
%
%\AH{\textbf{BEGIN}: Old stuff.}
%
%
%\begin{proof}
%
%Let us now show a sampling scheme which can run in $O\left(|\poly|\cdot k\right)$ per sample.
%
%First, consider when $\poly$ is already an SOP of pure products. In this case, sampling is trivial, and one would sample from the $\setsize$ terms with probability proportional to the product of probabilitites for each variable in the sampled monomial.
%
%Second, consider when $\poly$ has a POS form with a product width of $k$. In this case, we can view $\poly$ as an expression tree, where the leaves represent the individual values of each factor. The leaves are joined together by either a $\times$ or $+$ internal node, and so on, until we reach the root, which is joining the $k$-$\times$ nodes.
%
%Then for each $\times$ node, we multiply its subtree values, while for each $+$ node, we pick one of its children with probability proportional to the product of probabilities across its variables.
%
%\AH{I think I mean to say a probability proportional to the number of elements in it's given subtree.}
%
%The above sampling scheme is in $O\left(|\poly|\cdot k\right)$ time then, since we have for either case, that at most the scheme would perform within a factor of the $|\poly|$ operations, and those operations are repeated the product width of $k$ times.
%
%Thus, it is the case, that we can approximate $\rpoly(\prob_1,\ldots, \prob_n)$ within the claimed confidence bounds and computation time, thus proving the lemma.\AH{State why.}
%
%\AH{Discuss how we have that $\rpoly \geq O(\setsize)$. Discuss that we need $b-a$ to be small.}
%\end{proof}
%
%\qed
%\AH{{\bf END:} Old Stuff}
Since it is the case that computing the expected multiplicity of a compressed representation of a bag polynomial is hard, it is then desirable to have an algorithm to approximate the multiplicity in linear time, which is what we describe next.
%\begin{Definition}[Polynomial]\label{def:polynomial}
%The expression $\poly(\vct{X})$ is a polynomial if it satisfies the standard mathematical definition of polynomial, and additionally is in the standard monomial basis.
%\end{Definition}
%To clarify defintion ~\ref{def:polynomial}, a polynomial in the standard monomial basis is one whose monomials are in SOP form, and whose non-distinct monomials have been collapsed into one distinct monomial, with its corresponding coefficient accurately reflecting the number of monomials combined.
Now, some useful definitions and notation. For illustrative purposes in the definitions below, let us consider when $\poly(\vct{X}) = 2x^2 + 3xy - 2y^2$.
First, let us introduce some useful definitions and notation. For illustrative purposes in the definitions below, let us consider when $\poly(\vct{X}) = 2x^2 + 3xy - 2y^2$.
\begin{Definition}[Degree]\label{def:degree}
The degree of polynomial $\poly(\vct{X})$ is the maximum sum of the exponents of a monomial, over all monomials.
@ -42,23 +10,15 @@ The degree of polynomial $\poly(\vct{X})$ is the maximum sum of the exponents of
The degree of $\poly(\vct{X})$ in the above example is $2$. In this paper we consider only finite degree polynomials.
For example, the expression $xy$ is a monomial from the term $3xy$ of $\poly(\vct{X})$, produced from the set of variables $\vct{X} = \{x, y\}$.
%\begin{Definition}[$|\vct{X}|$]\label{def:num-vars}
%Denote the number of variables in $\poly(\vct{X})$ as $|\vct{X}|$.
%\end{Definition}
%
%In the running example, $|\vct{X}| = 2$.
\begin{Definition}[Expression Tree]\label{def:express-tree}
An expression tree $\etree$ is a binary %an ADT logically viewed as an n-ary
tree, whose internal nodes are from the set $\{+, \times\}$, with leaf nodes being either from the set $\mathbb{R}$ $(\tnum)$ or from the set of monomials $(\var)$. The members of $\etree$ are \type, \val, \vari{partial}, \vari{children}, and \vari{weight}, where \type is the type of value stored in the node $\etree$ (i.e. one of $\{+, \times, \var, \tnum\}$, \val is the value stored, and \vari{children} is the list of $\etree$'s children where $\etree_\lchild$ is the left child and $\etree_\rchild$ the right child. Remaining fields hold values whose semantics we will fix later. When $\etree$ is used as input of ~\cref{alg:mon-sam} and ~\cref{alg:one-pass}, the values of \vari{partial} and \vari{weight} will not be set. %SEMANTICS FOR \etree: \vari{partial} is the sum of $\etree$'s coefficients , n, and \vari{weight} is the probability of $\etree$ being sampled.
\end{Definition}
Note that $\etree$ need not encode an expression in the standard monomial basis, for example, when $\etree$ represents the expression $(x + 2y)(2x - y)$.
Note that $\etree$ need not encode an expression in the standard monomial basis. For example, instead of our running example, $\etree$ could represent a compressed form such as $(x + 2y)(2x - y)$.
\begin{Definition}[poly$(\cdot)$]\label{def:poly-func}
Denote $poly(\etree)$ to be the function that takes as input expression tree $\etree$ and outputs its corresponding polynomial. Recursively defined on $\etree$ as follows, where $\etree_\lchild$ and $\etree_\rchild$ denote the left and right child of $\etree$ respectively.
Denote $poly(\etree)$ to be the function that takes as input expression tree $\etree$ and outputs its corresponding polynomial. $poly(\cdot)$ is recursively defined on $\etree$ as follows, where $\etree_\lchild$ and $\etree_\rchild$ denote the left and right child of $\etree$ respectively.
% \begin{align*}
% &\etree.\type = +\mapsto&& \polyf(\etree_\lchild) + \polyf(\etree_\rchild)\\
@ -76,8 +36,6 @@ Denote $poly(\etree)$ to be the function that takes as input expression tree $\e
\end{equation*}
\end{Definition}
\AH{
\par2) Below seems like over-defining to me. Is this really necessary? The first sentence I think is \textit{enough}.}
Note that addition and multiplication above follow the standard interpretation over polynomials.
%Specifically, when adding two monomials whose variables and respective exponents agree, the coefficients corresponding to the monomials are added and their sum is multiplied to the monomial. Multiplication here is denoted by concatenation of the monomial and coefficient. When two monomials are multiplied, the product of each corresponding coefficient is computed, and the variables in each monomial are multiplied, i.e., the exponents of like variables are added. Again we notate this by the direct product of coefficient product and all disitinct variables in the two monomials, with newly computed exponents.
@ -102,7 +60,7 @@ $\expandtree{\etree}$ is the pure sum of products expansion of $\etree$. The lo
&\expandtree{\etree} = \\
&\begin{cases}
\expandtree{\etree_\lchild} \circ \expandtree{\etree_\rchild} &\textbf{ if }\etree.\type = +\\
\left\{(\monom_\lchild \cup \monom_\rchild, \coef_\lchild \cdot \coef_\rchild) ~|~ (\monom_\lchild, \coef_\lchild) \in \expandtree{\etree_\lchild}, (\monom_\rchild, \coef_\rchild) \in \expandtree{\etree_\rchild}\right\} &\textbf{ if }\etree.\type = \times\\
\left\{(\monom_\lchild \cup \monom_\rchild, \coef_\lchild \cdot \coef_\rchild) ~|~\right.&\\ \left.(\monom_\lchild, \coef_\lchild) \in \expandtree{\etree_\lchild}, (\monom_\rchild, \coef_\rchild) \in \expandtree{\etree_\rchild}\right\} &\textbf{ if }\etree.\type = \times\\
\elist{(\emptyset, \etree.\val)} &\textbf{ if }\etree.\type = \tnum\\
\elist{(\{\etree.\val\}, 1)} &\textbf{ if }\etree.\type = \var.\\
\end{cases}
@ -173,9 +131,9 @@ For any query polynomial $\poly(\vct{X})$, an approximation of $\rpoly(\prob_1,\
\subsection{Approximating $\rpoly$}
\subsubsection{Description}
Algorithm ~\ref{alg:mon-sam} approximates $\rpoly$ using the following steps. First, a call to $\onepass$ on its input $\etree$ produces a non-biased weight distribution over the monomials of $\expandtree{\etree}$ and a correct count of $|\etree|(1,\ldots, 1)$, i.e., the number of monomials in $\expandtree{\etree}$. Next, ~\cref{alg:mon-sam} calls $\sampmon$ to sample one monomial and its sign from $\expandtree{\etree}$. The sampling is repeated $\ceil{\frac{2\log{\frac{2}{\delta}}}{\epsilon^2}}$ times, where each of the samples are evaluated over $\vct{p}$, multiplied by $1 \times sign$, and summed. The final result is scaled accordingly returning an estimate of $\rpoly$ with the claimed $(\error, \conf)$-bound of ~\cref{lem:mon-samp}.
Algorithm ~\ref{alg:mon-sam} approximates $\rpoly$ using the following steps. First, a call to $\onepass$ on its input $\etree$ produces a non-biased weight distribution over the monomials of $\expandtree{\etree}$ and a correct count of $|\etree|(1,\ldots, 1)$, i.e., the number of monomials in $\expandtree{\etree}$. Next, ~\cref{alg:mon-sam} calls $\sampmon$ to sample one monomial and its sign from $\expandtree{\etree}$. The sampling is repeated $\ceil{\frac{2\log{\frac{2}{\delta}}}{\epsilon^2}}$ times, where each of the samples are evaluated with input $\vct{p}$, multiplied by $1 \times sign$, and summed. The final result is scaled accordingly returning an estimate of $\rpoly$ with the claimed $(\error, \conf)$-bound of ~\cref{lem:mon-samp}.
Kindly recall that the notaion $[x, y]$ denotes the range of values between $x$ and $y$ inclusive. The notation $\{x, y\}$ denotes the set of values consisting of $x$ and $y$.
Recall that the notation $[x, y]$ denotes the range of values between $x$ and $y$ inclusive. The notation $\{x, y\}$ denotes the set of values consisting of $x$ and $y$.
\subsubsection{Psuedo Code}
\begin{algorithm}[H]
@ -212,10 +170,10 @@ We state the lemmas for $\onepass$ and $\sampmon$, the auxiliary algorithms on w
The $\onepass$ function completes in $O(size(\etree))$ time. After $\onepass$ returns the following post conditions hold. First, that $\abs{\vari{S}}(1,\ldots, 1)$ is correctly computed for each subtree $\vari{S}$ of $\etree$. Second, when $\vari{S}.\val = +$, the weighted distribution $\frac{\abs{\vari{S}_{\vari{child}}}(1,\ldots, 1)}{\abs{\vari{S}}(1,\ldots, 1)}$ is correctly computed for each child of $\vari{S}.$
\end{Lemma}
At the conclusion of $\onepass$, $\etree.\vari{partial}$ will hold sum of all coefficients in $\expandtree{\abs{\etree}}$, i.e., $\sum\limits_{(\monom, \coef) \in \expandtree{\abs{\etree}}}\coef$. $\etree.\vari{weight}$ will hold the weighted probability that $\etree$ is sampled from from its parent $+$ node.
At the conclusion of $\onepass$, $\etree.\vari{partial}$ will hold the sum of all coefficients in $\expandtree{\abs{\etree}}$, i.e., $\sum\limits_{(\monom, \coef) \in \expandtree{\abs{\etree}}}\coef$. $\etree.\vari{weight}$ will hold the weighted probability that $\etree$ is sampled from from its parent $+$ node.
\begin{Lemma}\label{lem:sample}
The function $\sampmon$ complete in $O(\log{k} \cdot k \cdot depth(\etree))$ time, where $k = \degree(poly(\abs{\etree})$. Upon completion, with probability $\frac{|\coef|}{\abs{\etree}(1,\ldots, 1)}$, $\sampmon$ returns the sampled term $\left(\monom, sign(\coef)\right)$ from $\expandtree{\abs{\etree}}$.
The function $\sampmon$ completes in $O(\log{k} \cdot k \cdot depth(\etree))$ time, where $k = \degree(poly(\abs{\etree})$. Upon completion, with probability $\frac{|\coef|}{\abs{\etree}(1,\ldots, 1)}$, $\sampmon$ returns the sampled term $\left(\monom, sign(\coef)\right)$ from $\expandtree{\abs{\etree}}$.
\end{Lemma}
\begin{Theorem}\label{lem:mon-samp}
@ -238,7 +196,10 @@ Consider now a set of $\samplesize$ random variables $\vct{\randvar}$, where eac
$\expct\pbox{\randvar_i} = \sum\limits_{(\monom, \coef) \in \expandtree{\etree}}\frac{\coef \cdot \evalmp(\monom, p)}{\sum\limits_{(\monom, \coef) \in \expandtree{\etree}}|\coef|} = \frac{\rpoly(\prob_1,\ldots, \prob_\numvar)}{\abs{\etree}(1,\ldots, 1)}$. Let $\empmean = \frac{1}{\samplesize}\sum_{i = 1}^{\samplesize}\randvar_i$. It is also true that
\[\expct\pbox{\empmean} = \expct\pbox{ \frac{1}{\samplesize}\sum_{i = 1}^{\samplesize}\randvar_i} = \frac{1}{\samplesize}\sum_{i = 1}^{\samplesize}\expct\pbox{\randvar_i} = \frac{1}{\samplesize}\sum_{i = 1}^{\samplesize}\sum\limits_{(\monom, \coef) \in \expandtree{\etree}}\frac{\coef \cdot \evalmp(\monom, \vct{p})}{\sum\limits_{(\monom, \coef) \in \expandtree{\etree}}|\coef|} = \frac{\rpoly(\prob_1,\ldots, \prob_\numvar)}{\abs{\etree}(1,\ldots, 1)}.\]
\begin{align*}
&\expct\pbox{\empmean} = \expct\pbox{ \frac{1}{\samplesize}\sum_{i = 1}^{\samplesize}\randvar_i} = \frac{1}{\samplesize}\sum_{i = 1}^{\samplesize}\expct\pbox{\randvar_i}\nonumber\\
&= \frac{1}{\samplesize}\sum_{i = 1}^{\samplesize}\sum\limits_{(\monom, \coef) \in \expandtree{\etree}}\frac{\coef \cdot \evalmp(\monom, \vct{p})}{\sum\limits_{(\monom, \coef) \in \expandtree{\etree}}|\coef|} = \frac{\rpoly(\prob_1,\ldots, \prob_\numvar)}{\abs{\etree}(1,\ldots, 1)}.
\end{align*}
Hoeffding' inequality can be used to compute an upper bound on the number of samples $\samplesize$ needed to establish the $(\error, \conf)$-bound. The inequality states that if we know that each $\randvar_i$ is strictly bounded by the intervals $[a_i, b_i]$, then it is true that
\begin{equation*}
@ -322,7 +283,9 @@ The evaluation of $\abs{\etree}(1,\ldots, 1)$ can be defined recursively, as fol
In the same fashion the weighted distribution can be described as above with the following modification for the case when $\etree.\type = +$:
\begin{align*}
&\abs{\etree_\lchild}(1,\ldots, 1) + \abs{\etree_\rchild}(1,\ldots, 1); \etree_\lchild.\vari{weight} \gets \frac{\abs{\etree_\lchild}(1,\ldots, 1)}{\abs{\etree_\lchild}(1,\ldots, 1) + \abs{\etree_\rchild}(1,\ldots, 1)}, \etree_\rchild.\vari{weight} \gets \frac{\abs{\etree_\rchild}(1,\ldots, 1)}{\abs{\etree_\lchild}(1,\ldots, 1)+ \abs{\etree_\rchild}(1,\ldots, 1)} &\textbf{if }\etree.\type = +
&\abs{\etree_\lchild}(1,\ldots, 1) + \abs{\etree_\rchild}(1,\ldots, 1); &\textbf{if }\etree.\type = + \\
&\etree_\lchild.\vari{weight} \gets \frac{\abs{\etree_\lchild}(1,\ldots, 1)}{\abs{\etree_\lchild}(1,\ldots, 1) + \abs{\etree_\rchild}(1,\ldots, 1)};\\
&\etree_\rchild.\vari{weight} \gets \frac{\abs{\etree_\rchild}(1,\ldots, 1)}{\abs{\etree_\lchild}(1,\ldots, 1)+ \abs{\etree_\rchild}(1,\ldots, 1)}
\end{align*}
%\begin{align*}
@ -372,13 +335,13 @@ See algorithm ~\ref{alg:one-pass} for details.
\end{algorithm}
\begin{Example}\label{example:one-pass}
Consider the when $\etree$ is $+\left(\times\left(+\left(\times\left(1, x_1\right), \times\left(1, x_2\right)\right), +\left(\times\left(1, x_1\right) as seen in ~\cref{fig:expr-tree-T-wght}, \times\left(-1, x_2\right)\right)\right), \times\left(\times\left(1, x_2\right), \times\left(1, x_2\right)\right)\right)$, which encodes the expression $(x_1 + x_2)(x_1 - x_2) + x_2^2$. After one pass, \cref{alg:one-pass} would have computed the following weight distribution. For the two children of the root $+$ node $\etree$, $\etree_\lchild.\wght = \frac{4}{5}$ and $\etree_\rchild.\wght = \frac{1}{5}$. Similarly, $\stree \gets \etree_\lchild$, $\stree_\lchild.\wght = \stree_\rchild.\wght = \frac{1}{2}$. Note that in this example, the sampling probabilities for the children of each inner $+$ node of $\stree$ are equal to one another because both parents have the same number of children, and, in each case, the children of each parent $+$ node share the same $|\coef_i|$.
Consider when $\etree$ encodes the expression $(x_1 + x_2)(x_1 - x_2) + x_2^2$. After one pass, \cref{alg:one-pass} would have computed the following weight distribution. For the two children of the root $+$ node $\etree$, $\etree_\lchild.\wght = \frac{4}{5}$ and $\etree_\rchild.\wght = \frac{1}{5}$. Similarly, $\stree \gets \etree_{\lchild_\lchild}$, $\stree_\lchild.\wght = \stree_\rchild.\wght = \frac{1}{2}$. Note that in this example, the sampling probabilities for the children of each inner $+$ node of $\stree$ are equal to one another because both parents have the same number of children, and, in each case, the children of each parent $+$ node share the same $|\coef_i|$.
\end{Example}
\begin{figure}[h!]
\begin{tikzpicture}[thick, every tree node/.style={default_node, thick, draw=black, black, circle, text width=0.3cm, font=\bfseries, minimum size=0.65cm}, every child/.style={black}, edge from parent/.style={draw, thick},
level 1/.style={sibling distance=2.5cm},
level 2/.style={sibling distance=1.25cm},
level 1/.style={sibling distance=1.25cm},
level 2/.style={sibling distance=1.0cm},
%level 2+/.style={sibling distance=0.625cm}
%level distance = 1.25cm,
%sibling distance = 1cm,

9
lin_sys.tex
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@ -218,3 +218,12 @@ Putting \cref{eq:det-1}, \cref{eq:det-2}, \cref{eq:det-3} together, we have,
Thus, by ~\cref{lem:lin-sys} we have proved ~\cref{th:single-p} for fixed $p \in (0, 1)$.
\end{proof}
\qed
\begin{Corollary}\label{cor:single-p-gen-k}
For every value $\kElem \geq 3$, there exists a query with $\kElem$ product width that is hard.
\end{Corollary}
\begin{proof}[Proof of Corollary ~\cref{cor:single-p-gen-k}]
Consider $\poly^3_{G}$ and $\poly' = 1$ such that $\poly'' = \poly^3_{G} \cdot \poly'$. By ~\cref{th:single-p}, query $\poly''$ with $\kElem = 4$ is hard.
\end{proof}
\qed