ODIn/paper-BagRelationalPDBsAreHard
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Aaron Huber 2020-12-19 12:57:27 -05:00
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ra-to-poly.tex
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@ -104,26 +104,26 @@ tree, whose internal nodes are from the set $\{+, \times\}$, with leaf nodes bei
We ignore the remaining fields (\vari{partial} and \vari{weight}) until \Cref{sec:algo}. Note that $\etree$ need not encode an expression in standard monomial basis. For instance, $\etree$ could represent a compressed form of the running example, 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 in \abbrSMB. We define $poly(\cdot)$ recursively on $\etree$ as follows, where $\etree_\lchild$ and $\etree_\rchild$ denote the left/right child of $\etree$ respectively.
%
% \begin{align*}
% &\etree.\type = +\mapsto&& \polyf(\etree_\lchild) + \polyf(\etree_\rchild)\\
% &\etree.\type = \times\mapsto&& \polyf(\etree_\lchild) \cdot \polyf(\etree_\rchild)\\
% &\etree.\type = \var \text{ OR } \tnum\mapsto&& \etree.\val
% \end{align*}
%
\begin{equation*}
\polyf(\etree) = \begin{cases}
\polyf(\etree_\lchild) + \polyf(\etree_\rchild) &\text{ if \etree.\type } = +\\
\polyf(\etree_\lchild) \cdot \polyf(\etree_\rchild) &\text{ if \etree.\type } = \times\\
\etree.\val &\text{ if \etree.\type } = \var \text{ OR } \tnum.
\end{cases}
\end{equation*}
\end{Definition}
%\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 in \abbrSMB. We define $poly(\cdot)$ recursively on $\etree$ as follows, where $\etree_\lchild$ and $\etree_\rchild$ denote the left/right child of $\etree$ respectively.
%%
%% \begin{align*}
%% &\etree.\type = +\mapsto&& \polyf(\etree_\lchild) + \polyf(\etree_\rchild)\\
%% &\etree.\type = \times\mapsto&& \polyf(\etree_\lchild) \cdot \polyf(\etree_\rchild)\\
%% &\etree.\type = \var \text{ OR } \tnum\mapsto&& \etree.\val
%% \end{align*}
%%
%\begin{equation*}
% \polyf(\etree) = \begin{cases}
% \polyf(\etree_\lchild) + \polyf(\etree_\rchild) &\text{ if \etree.\type } = +\\
% \polyf(\etree_\lchild) \cdot \polyf(\etree_\rchild) &\text{ if \etree.\type } = \times\\
% \etree.\val &\text{ if \etree.\type } = \var \text{ OR } \tnum.
% \end{cases}
%\end{equation*}
%\end{Definition}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Note that addition and multiplication above follow the standard interpretation over polynomials.
%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.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
@ -134,9 +134,8 @@ Note that addition and multiplication above follow the standard interpretation o
For our running example, $\etreeset{\smb} = \{2X^2 + 3XY - 2Y^2, (X + 2Y)(2X - Y), X(2X - Y) + 2Y(2X - Y), 2X(X + 2Y) - Y(X + 2Y)\}$. Note that \cref{def:express-tree-set} implies that $\etree \in \etreeset{poly(\etree)}$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Problem Definition}\label{sec:problem-definition}
We are now ready to formally state our main problem.
\medskip
\noindent We are now ready to formally state our \textbf{main problem}.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{Definition}[The Expected Result Multiplicity Problem]\label{def:the-expected-multipl}