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Misc. changes; change X to U in intro example, adjust appendix for S.3 and S.4 changes, etc.

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Aaron Huber 2022-06-03 12:11:54 -04:00
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app_approx-alg-analysis.tex
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@ -111,89 +111,6 @@ Applying this bound in the runtime bound in \Cref{lem:approx-alg} gives the firs
\qed
\end{proof}
\subsection{Proof of~\Cref{lem:ctidb-gamma}}
\begin{proof}
The circuit \circuit' is built from \circuit in the following manner. For each input gate $\gate_i$ with $\gate_i.\val = X_\tup$, replace $\gate_i$ with the circuit \subcircuit encoding the sum $\sum_{j = 1}^\bound j\cdot X_{\tup, j}$. We argue that \circuit' is a valid circuit by the following facts. Let $\pdb = \inparen{\worlds, \bpd}$ be the original \abbrCTIDB \circuit was generated from. Then, by~\Cref{prop:ctidb-reduct} there exists a \abbrOneBIDB $\pdb' = \inparen{\onebidbworlds{\tupset'}, \bpd'}$, with $\tupset' = \inset{\intuple{\tup, j}~|~\tup\in\tupset, j\in\pbox{\bound}}$, from which the conversion from \circuit to \circuit' follows. Both $\polyf\inparen{\circuit}$ and $\polyf\inparen{\circuit'}$ have the same expected multiplicity since (by~\Cref{prop:ctidb-reduct}) the distributions $\bpd$ and $\bpd'$ are equivalent and $\sum_{j=1}^\bound j\cdot\worldvec'_{\tup, j} = \worldvec_\tup$ for $\worldvec'\in\inset{0, 1}^{\bound\numvar}$ and $\worldvec\in\worlds$ such that $\worldvec_\tup\equiv\worldvec'_\tup$. Finally, note that because there exists a (sub) circuit encoding $\sum_{j = 1}^\bound j\cdot X_{\tup, j}$ that is a \emph{balanced} binary tree, the above conversion implies the claimed size and depth bounds of the lemma.
Next we argue the claim on $\gamma\inparen{\circuit'}$. Consider the list of expanded monomials $\expansion{\circuit}$ for \abbrCTIDB circuit \circuit. Let
$\encMon = X_{\tup_1}^{d_1}\cdots X_{\tup_\ell}^{d_\ell}$ be an arbitrary monomial with $\ell$ variables and let (abusing notation) $\encMon' = \inparen{\sum_{j = 1}^{\bound}j\cdot X_{\tup_1, j}}^{d_1}\cdots\inparen{\sum_{j = 1}^{\bound}j\cdot X_{\tup_\ell, j}}^{d_\ell}$. Then, for $f_\ell = \sum_{i = 1}^\ell d_i$, $\encMon$ induces the set of monomials $\inset{\prod_{i = 1}^{f_\ell} j_i\cdot X_{\tup_i, j_i}^{d_i}}_{j_i\in\pbox{\bound}}$ in the pure expansion of $\encMon'$.
%Denote the additional list elements (projecting out coefficient terms) \emph{induced} by $\monom$ as $\vari{E}_\monom\inparen{\circuit'}$. Then $\vari{E}_\monom\inparen{\circuit'}=\inset{\monom'^1~|~\encMon' \in \vari{S}}$%\inset{j_1^{d_1}\cdot X_{\tup, j_1}^{d_1}\times\cdots\times j_\ell^{d_\ell}\cdot X_{\tup, j_\ell}^{d_\ell}}_{j_1,\ldots, j_\ell \in \pbox{\bound}}$ in $\expansion{\circuit'}$.
Recall that a cancellation occurs in $\encMon'$ when there exists $\tup_{i, j}\neq\tup_{i, j'}$ in the same block $\block$ where variables $X_{\tup_i, j}, X_{\tup_i, j'}$ are in the set of variables $\monom_i'$ of $\monom_{\vari{m}_\vari{i}}\in\encMon'$. Observe that cancellations can only occur for each $X_{\tup}^{d_\tup}\in \encMon$, where the expansion $\inparen{\sum_{j = 1}^\bound j\cdot X_{\tup, j}}^{d_\tup}$ represents the monomial $X_\tup^{d_\tup}$ in $\tupset'$. Consider the number of cancellations for $\inparen{\sum_{j = 1}^\bound j\cdot X_{\tup, j}}^{d_t}$. Then $\gamma \leq 1 - \bound^{-\inparen{d_\tup - 1}}$, since
for each element in the set of cross products $\inset{\bigtimes_{i\in\pbox{d_\tup}, j_i\in\pbox{\bound}}X_{\tup, j_i}}$ there are \emph{exactly} $\bound$ surviving elements with $j_1=\cdots=j_{d_\tup}=j$, i.e. $X_{t,j}^{d_\tup}$ for each $j\in\pbox{\bound}$. The rest of the $\bound^{d_\tup}-c$ cross terms cancel. Regarding all of $\encMon'$, it is the case that the proportion of non-cancellations for each $\inparen{\sum_{j = 1}^{\bound}j\cdot X_{\tup_i, j }}^{d_i}\in\encMon'$ multiply because non-cancelling terms for $\inparen{\sum_{j = 1}^{\bound}j\cdot X_{\tup_i, j}}^{d_i}$ can only be joined with non-cancelling terms of $\inparen{\sum_{j=1}^{\bound}X_{\tup_{i'}, j}}^{d_{i'}}\in\encMon'$ for $\tup\neq\tup'$. This then yields the fraction of cancelled monomials $\gamma\le 1 - \prod_{i = 1}^{\ell}\bound^{-\inparen{d_i - 1}} \leq 1 - \bound^{-\inparen{k - 1}}$ where the inequalities take into account the fact that $f_\ell \leq k$.
Since this is true for arbitrary \monom, the bound follows for $\polyf\inparen{\circuit'}$.
\end{proof}
\qed
\subsection{Proof of \Cref{lem:val-ub}}\label{susec:proof-val-up}
\label{app:proof-lem-val-ub}
We will prove \Cref{lem:val-ub} by considering the two cases separately. We start by considering the case when $\circuit$ is a tree:
\begin{Lemma}
\label{lem:C-ub-tree}
Let $\circuit$ be a tree (i.e. the sub-circuits corresponding to two children of a node in $\circuit$ are completely disjoint). Then we have
\[\abs{\circuit}(1,\dots,1)\le \left(\size(\circuit)\right)^{\degree(\circuit)+1}.\]
\end{Lemma}
\begin{proof}[Proof of \Cref{lem:C-ub-tree}]
For notational simplicity define $N=\size(\circuit)$ and $k=\degree(\circuit)$.
We use induction on $\depth(\circuit)$ to show that $\abs{\circuit}(1,\ldots, 1) \leq N^{k+1 }$.
For the base case, we have that \depth(\circuit) $= 0$, and there can only be one node which must contain a coefficient or constant. In this case, $\abs{\circuit}(1,\ldots, 1) = 1$, and \size(\circuit) $= 1$, and by \Cref{def:degree} it is the case that $0 \leq k = \degree\inparen{\circuit} \leq 1$, and it is true that $\abs{\circuit}(1,\ldots, 1) = 1 \leq N^{k+1} = 1^{k + 1} = 1$ for $k \in \inset{0, 1}$.
Assume for $\ell > 0$ an arbitrary circuit \circuit of $\depth(\circuit) \leq \ell$ that it is true that $\abs{\circuit}(1,\ldots, 1) \leq N^{k+1 }$.
For the inductive step we consider a circuit \circuit such that $\depth(\circuit) = \ell + 1$. The sink can only be either a $\circmult$ or $\circplus$ gate. Let $k_\linput, k_\rinput$ denote \degree($\circuit_\linput$) and \degree($\circuit_\rinput$) respectively. Consider when sink node is $\circmult$.
Then note that
\begin{align}
\abs{\circuit}(1,\ldots, 1) &= \abs{\circuit_\linput}(1,\ldots, 1)\cdot \abs{\circuit_\rinput}(1,\ldots, 1) \nonumber\\
&\leq (N-1)^{k_\linput+1} \cdot (N - 1)^{k_\rinput+1}\nonumber\\
&= (N-1)^{k+1}\label{eq:sumcoeff-times-upper}\\
&\leq N^{k + 1}.\nonumber
\end{align}
In the above the first inequality follows from the inductive hypothesis (and the fact that the size of either subtree is at most $N-1$) and \Cref{eq:sumcoeff-times-upper} follows by \cref{def:degree} which states that for $k = \degree(\circuit)$ we have $k=k_\linput+k_\rinput+1$.
For the case when the sink gate is a $\circplus$ gate, then for $N_\linput = \size(\circuit_\linput)$ and $N_\rinput = \size(\circuit_\rinput)$ we have
\begin{align}
\abs{\circuit}(1,\ldots, 1) &= \abs{\circuit_\linput}(1,\ldots, 1) \circplus \abs{\circuit_\rinput}(1,\ldots, 1) \nonumber\\
&\leq
N_\linput^{k+1} + N_\rinput^{k+1}\nonumber\\
&\leq (N-1)^{k+1 } \label{eq:sumcoeff-plus-upper}\\
&\leq N^{k+1}.\nonumber
\end{align}
In the above, the first inequality follows from the inductive hypothes and \cref{def:degree} (which implies the fact that $k_\linput,k_\rinput\le k$). Note that the RHS of this inequality is maximized when the base and exponent of one of the terms is maximized. The second inequality follows from this fact as well as the fact that since $\circuit$ is a tree we have $N_\linput+N_\rinput=N-1$ and, lastly, the fact that $k\ge 0$. This completes the proof.
%\AH{I don't think that it matters whether or not \circuit is a tree. For $N=\size\inparen{\circuit}$ it must follow that $N_L + N_R + 1 = N$ regardless of whether a gate a allowed to have more than one parent. Not true, consider when $\circuit_R = \circuit_L$.}
\end{proof}
The upper bound in \Cref{lem:val-ub} for the general case is a simple variant of the above proof (but we present a proof sketch of the bound below for completeness):
\begin{Lemma}
\label{lem:C-ub-gen}
Let $\circuit$ be a (general) circuit.
Then we have
\[\abs{\circuit}(1,\dots,1)\le 2^{2^{\degree(\circuit)}\cdot \depth(\circuit)}.\]
\end{Lemma}
\begin{proof}[Proof Sketch of \Cref{lem:C-ub-gen}]
We use the same notation as in the proof of \Cref{lem:C-ub-tree} and further define $d=\depth(\circuit)$. We will prove by induction on $\depth(\circuit)$ that $\abs{\circuit}(1,\ldots, 1) \leq 2^{2^k\cdot d }$. The base case argument is similar to that in the proof of \Cref{lem:C-ub-tree}. In the inductive case we have that $d_\linput,d_\rinput\le d-1$.
For the case when the sink node is $\times$, we get that
\begin{align*}
\abs{\circuit}(1,\ldots, 1) &= \abs{\circuit_\linput}(1,\ldots, 1)\circmult \abs{\circuit_\rinput}(1,\ldots, 1) \\
&\leq {2^{2^{k_\linput}\cdot d_\linput}} \circmult {2^{2^{k_\rinput}\cdot d_\rinput}}\\
&\leq 2^{2\cdot 2^{k-1}\cdot (d-1)}\\
&\leq 2^{2^k d}.
\end{align*}
In the above the first inequality follows from inductive hypothesis while the second inequality follows from the fact that $k_\linput,k_\rinput\le k-1$ and $d_\linput, d_\rinput\le d-1$, where we substitute the upperbound into every respective term.
Now consider the case when the sink node is $+$, we get that
\begin{align*}
\abs{\circuit}(1,\ldots, 1) &= \abs{\circuit_\linput}(1,\ldots, 1) \circplus \abs{\circuit_\rinput}(1,\ldots, 1) \\
&\leq 2^{2^{k_\linput}\cdot d_\linput} + 2^{2^{k_\rinput}\cdot d_\rinput}\\
&\leq 2\cdot {2^{2^k(d-1)} } \\
&\leq 2^{2^kd}.
\end{align*}
In the above the first inequality follows from the inductive hypothesis while the second inequality follows from the facts that $k_\linput,k_\rinput\le k$ and $d_\linput,d_\rinput\le d-1$. The final inequality follows from the fact that $k\ge 0$.
\qed
\end{proof}
%%% Local Variables:
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%root: main.tex
%\textcolor{red}{Aaron: The stuff below needs to be integrated into this section.}
\subsection{\Cref{lem:ctidb-gamma},~\Cref{lem:val-ub},~\Cref{cor:approx-algo-punchline}, and Proof of~\Cref{cor:approx-algo-punchline-ctidb}} %, recalling from~\Cref{sec:intro} for \abbrCTIDB $\pdb = \inparen{\worlds, \bpd}$, where $\tupset$ is the set of possible tuples across all possible worlds of $\pdb$.
\begin{Lemma}
\label{lem:ctidb-gamma}
Given $\raPlus$ query $\query$ and \abbrCTIDB $\pdb$, let \circuit be the circuit computed by $\query\inparen{\tupset}$. Then, for the reduced \abbrOneBIDB $\pdb'$ there exists an equivalent circuit \circuit' obtained from $\query\inparen{\tupset'}$, such that $\gamma\inparen{\circuit'}\leq 1 - \bound^{-\inparen{k-1}}$ with $\size\inparen{\circuit'} \leq \size\inparen{\circuit} + \bigO{\numvar\bound}$
and $\depth\inparen{\circuit'} = \depth\inparen{\circuit} + \bigO{\log{\bound}}$.
\end{Lemma}
We briefly connect the runtime in \Cref{eq:approx-algo-runtime} to the algorithm outline earlier (where we ignore the dependence on $\multc{\cdot}{\cdot}$, which is needed to handle the cost of arithmetic operations over integers). The $\size(\circuit)$ comes from the time taken to run \onepass once (\onepass essentially computes $\abs{\circuit}(1,\ldots, 1)$ using the natural circuit evaluation algorithm on $\circuit$). We make $\frac{\log{\frac{1}{\conf}}}{\inparen{\error'}^2\cdot(1-\gamma)^2\cdot \prob_0^{2k}}$ many calls to \sampmon (each of which essentially traces $O(k)$ random sink to source paths in $\circuit$ all of which by definition have length at most $\depth(\circuit)$).
Finally, we address the $\multc{\log\left(\abs{\circuit}(1,\ldots, 1)\right)}{\log\left(\size(\circuit)\right)}$ term in the runtime.
\begin{Lemma}
\label{lem:val-ub}
For any \emph{\abbrOneBIDB} circuit $\circuit$ with $\degree(\circuit)=k$, we have
$\abs{\circuit}(1,\ldots, 1)\le 2^{2^k\cdot \depth(\circuit)}.$
Further, if $\circuit$ is a tree, then we have $\abs{\circuit}(1,\ldots, 1)\le \size(\circuit)^{O(k)}.$
\end{Lemma}
Note that the above implies that with the assumption $\prob_0>0$ and $\gamma<1$ are absolute constants from \Cref{cor:approx-algo-const-p}, then the runtime there simplifies to $O_k\left(\frac 1{\inparen{\error'}^2}\cdot\size(\circuit)^2\cdot \log{\frac{1}{\conf}}\right)$ for general circuits $\circuit$. If $\circuit$ is a tree, then the runtime simplifies to $O_k\left(\frac 1{\inparen{\error'}^2}\cdot\size(\circuit)\cdot \log{\frac{1}{\conf}}\right)$, which then answers \Cref{prob:intro-stmt} with yes for such circuits.
Finally, note that by \Cref{prop:circuit-depth} and \Cref{lem:circ-model-runtime} for any $\raPlus$ query $\query$, there exists a circuit $\circuit^*$ for $\apolyqdt$ such that $\depth(\circuit^*)\le O_{|Q|}(\log{n})$ and $\size(\circuit)\le O_k\inparen{\qruntime{\query, \tupset, \bound}}$. Using this along with \Cref{lem:val-ub}, \Cref{cor:approx-algo-const-p} and the fact that $n\le \qruntime{\query, \tupset, \bound}$, we have the following corollary:
\begin{Corollary}
\label{cor:approx-algo-punchline}
Let $\query$ be an $\raPlus$ query and $\pdb$ be a \emph{\abbrOneBIDB} with $p_0>0$ and $\gamma<1$, where $p_0,\gamma$ as in \Cref{cor:approx-algo-const-p}, are absolute constants. Let $\poly(\vct{X})=\apolyqdt$ for any result tuple $\tup$ with $\deg(\poly)=k$. Then one can compute an approximation satisfying \Cref{eq:approx-algo-bound-main} in time $O_{k,|Q|,\error',\conf}\inparen{\qruntime{\optquery{\query}, \tupset, \bound}}$ (given $\query,\tupset$ and $p_i$ for each $i\in [n]$ that defines $\pd$).
\end{Corollary}
\subsection{Proof of~\Cref{lem:ctidb-gamma}}
\begin{proof}
The circuit \circuit' is built from \circuit in the following manner. For each input gate $\gate_i$ with $\gate_i.\val = X_\tup$, replace $\gate_i$ with the circuit \subcircuit encoding the sum $\sum_{j = 1}^\bound j\cdot X_{\tup, j}$. We argue that \circuit' is a valid circuit by the following facts. Let $\pdb = \inparen{\worlds, \bpd}$ be the original \abbrCTIDB \circuit was generated from. Then, by~\Cref{prop:ctidb-reduct} there exists a \abbrOneBIDB $\pdb' = \inparen{\onebidbworlds{\tupset'}, \bpd'}$, with $\tupset' = \inset{\intuple{\tup, j}~|~\tup\in\tupset, j\in\pbox{\bound}}$, from which the conversion from \circuit to \circuit' follows. Both $\polyf\inparen{\circuit}$ and $\polyf\inparen{\circuit'}$ have the same expected multiplicity since (by~\Cref{prop:ctidb-reduct}) the distributions $\bpd$ and $\bpd'$ are equivalent and $\sum_{j=1}^\bound j\cdot\worldvec'_{\tup, j} = \worldvec_\tup$ for $\worldvec'\in\inset{0, 1}^{\bound\numvar}$ and $\worldvec\in\worlds$ such that $\worldvec_\tup\equiv\worldvec'_\tup$. Finally, note that because there exists a (sub) circuit encoding $\sum_{j = 1}^\bound j\cdot X_{\tup, j}$ that is a \emph{balanced} binary tree, the above conversion implies the claimed size and depth bounds of the lemma.
Next we argue the claim on $\gamma\inparen{\circuit'}$. Consider the list of expanded monomials $\expansion{\circuit}$ for \abbrCTIDB circuit \circuit. Let
$\encMon = X_{\tup_1}^{d_1}\cdots X_{\tup_\ell}^{d_\ell}$ be an arbitrary monomial with $\ell$ variables and let (abusing notation) $\encMon' = \inparen{\sum_{j = 1}^{\bound}j\cdot X_{\tup_1, j}}^{d_1}\cdots\inparen{\sum_{j = 1}^{\bound}j\cdot X_{\tup_\ell, j}}^{d_\ell}$. Then, for $f_\ell = \sum_{i = 1}^\ell d_i$, $\encMon$ induces the set of monomials $\inset{\prod_{i = 1}^{f_\ell} j_i\cdot X_{\tup_i, j_i}^{d_i}}_{j_i\in\pbox{\bound}}$ in the pure expansion of $\encMon'$.
%Denote the additional list elements (projecting out coefficient terms) \emph{induced} by $\monom$ as $\vari{E}_\monom\inparen{\circuit'}$. Then $\vari{E}_\monom\inparen{\circuit'}=\inset{\monom'^1~|~\encMon' \in \vari{S}}$%\inset{j_1^{d_1}\cdot X_{\tup, j_1}^{d_1}\times\cdots\times j_\ell^{d_\ell}\cdot X_{\tup, j_\ell}^{d_\ell}}_{j_1,\ldots, j_\ell \in \pbox{\bound}}$ in $\expansion{\circuit'}$.
Recall that a cancellation occurs in $\encMon'$ when there exists $\tup_{i, j}\neq\tup_{i, j'}$ in the same block $\block$ where variables $X_{\tup_i, j}, X_{\tup_i, j'}$ are in the set of variables $\monom_i'$ of $\monom_{\vari{m}_\vari{i}}\in\encMon'$. Observe that cancellations can only occur for each $X_{\tup}^{d_\tup}\in \encMon$, where the expansion $\inparen{\sum_{j = 1}^\bound j\cdot X_{\tup, j}}^{d_\tup}$ represents the monomial $X_\tup^{d_\tup}$ in $\tupset'$. Consider the number of cancellations for $\inparen{\sum_{j = 1}^\bound j\cdot X_{\tup, j}}^{d_t}$. Then $\gamma \leq 1 - \bound^{-\inparen{d_\tup - 1}}$, since
for each element in the set of cross products $\inset{\bigtimes_{i\in\pbox{d_\tup}, j_i\in\pbox{\bound}}X_{\tup, j_i}}$ there are \emph{exactly} $\bound$ surviving elements with $j_1=\cdots=j_{d_\tup}=j$, i.e. $X_{t,j}^{d_\tup}$ for each $j\in\pbox{\bound}$. The rest of the $\bound^{d_\tup}-c$ cross terms cancel. Regarding all of $\encMon'$, it is the case that the proportion of non-cancellations for each $\inparen{\sum_{j = 1}^{\bound}j\cdot X_{\tup_i, j }}^{d_i}\in\encMon'$ multiply because non-cancelling terms for $\inparen{\sum_{j = 1}^{\bound}j\cdot X_{\tup_i, j}}^{d_i}$ can only be joined with non-cancelling terms of $\inparen{\sum_{j=1}^{\bound}X_{\tup_{i'}, j}}^{d_{i'}}\in\encMon'$ for $\tup\neq\tup'$. This then yields the fraction of cancelled monomials $\gamma\le 1 - \prod_{i = 1}^{\ell}\bound^{-\inparen{d_i - 1}} \leq 1 - \bound^{-\inparen{k - 1}}$ where the inequalities take into account the fact that $f_\ell \leq k$.
Since this is true for arbitrary \monom, the bound follows for $\polyf\inparen{\circuit'}$.
\end{proof}
\qed
\subsection{Proof of \Cref{lem:val-ub}}\label{susec:proof-val-up}
\label{app:proof-lem-val-ub}
We will prove \Cref{lem:val-ub} by considering the two cases separately. We start by considering the case when $\circuit$ is a tree:
\begin{Lemma}
\label{lem:C-ub-tree}
Let $\circuit$ be a tree (i.e. the sub-circuits corresponding to two children of a node in $\circuit$ are completely disjoint). Then we have
\[\abs{\circuit}(1,\dots,1)\le \left(\size(\circuit)\right)^{\degree(\circuit)+1}.\]
\end{Lemma}
\begin{proof}[Proof of \Cref{lem:C-ub-tree}]
For notational simplicity define $N=\size(\circuit)$ and $k=\degree(\circuit)$.
We use induction on $\depth(\circuit)$ to show that $\abs{\circuit}(1,\ldots, 1) \leq N^{k+1 }$.
For the base case, we have that \depth(\circuit) $= 0$, and there can only be one node which must contain a coefficient or constant. In this case, $\abs{\circuit}(1,\ldots, 1) = 1$, and \size(\circuit) $= 1$, and by \Cref{def:degree} it is the case that $0 \leq k = \degree\inparen{\circuit} \leq 1$, and it is true that $\abs{\circuit}(1,\ldots, 1) = 1 \leq N^{k+1} = 1^{k + 1} = 1$ for $k \in \inset{0, 1}$.
Assume for $\ell > 0$ an arbitrary circuit \circuit of $\depth(\circuit) \leq \ell$ that it is true that $\abs{\circuit}(1,\ldots, 1) \leq N^{k+1 }$.
For the inductive step we consider a circuit \circuit such that $\depth(\circuit) = \ell + 1$. The sink can only be either a $\circmult$ or $\circplus$ gate. Let $k_\linput, k_\rinput$ denote \degree($\circuit_\linput$) and \degree($\circuit_\rinput$) respectively. Consider when sink node is $\circmult$.
Then note that
\begin{align}
\abs{\circuit}(1,\ldots, 1) &= \abs{\circuit_\linput}(1,\ldots, 1)\cdot \abs{\circuit_\rinput}(1,\ldots, 1) \nonumber\\
&\leq (N-1)^{k_\linput+1} \cdot (N - 1)^{k_\rinput+1}\nonumber\\
&= (N-1)^{k+1}\label{eq:sumcoeff-times-upper}\\
&\leq N^{k + 1}.\nonumber
\end{align}
In the above the first inequality follows from the inductive hypothesis (and the fact that the size of either subtree is at most $N-1$) and \Cref{eq:sumcoeff-times-upper} follows by \cref{def:degree} which states that for $k = \degree(\circuit)$ we have $k=k_\linput+k_\rinput+1$.
For the case when the sink gate is a $\circplus$ gate, then for $N_\linput = \size(\circuit_\linput)$ and $N_\rinput = \size(\circuit_\rinput)$ we have
\begin{align}
\abs{\circuit}(1,\ldots, 1) &= \abs{\circuit_\linput}(1,\ldots, 1) \circplus \abs{\circuit_\rinput}(1,\ldots, 1) \nonumber\\
&\leq
N_\linput^{k+1} + N_\rinput^{k+1}\nonumber\\
&\leq (N-1)^{k+1 } \label{eq:sumcoeff-plus-upper}\\
&\leq N^{k+1}.\nonumber
\end{align}
In the above, the first inequality follows from the inductive hypothes and \cref{def:degree} (which implies the fact that $k_\linput,k_\rinput\le k$). Note that the RHS of this inequality is maximized when the base and exponent of one of the terms is maximized. The second inequality follows from this fact as well as the fact that since $\circuit$ is a tree we have $N_\linput+N_\rinput=N-1$ and, lastly, the fact that $k\ge 0$. This completes the proof.
%\AH{I don't think that it matters whether or not \circuit is a tree. For $N=\size\inparen{\circuit}$ it must follow that $N_L + N_R + 1 = N$ regardless of whether a gate a allowed to have more than one parent. Not true, consider when $\circuit_R = \circuit_L$.}
\end{proof}
The upper bound in \Cref{lem:val-ub} for the general case is a simple variant of the above proof (but we present a proof sketch of the bound below for completeness):
\begin{Lemma}
\label{lem:C-ub-gen}
Let $\circuit$ be a (general) circuit.
Then we have
\[\abs{\circuit}(1,\dots,1)\le 2^{2^{\degree(\circuit)}\cdot \depth(\circuit)}.\]
\end{Lemma}
\begin{proof}[Proof Sketch of \Cref{lem:C-ub-gen}]
We use the same notation as in the proof of \Cref{lem:C-ub-tree} and further define $d=\depth(\circuit)$. We will prove by induction on $\depth(\circuit)$ that $\abs{\circuit}(1,\ldots, 1) \leq 2^{2^k\cdot d }$. The base case argument is similar to that in the proof of \Cref{lem:C-ub-tree}. In the inductive case we have that $d_\linput,d_\rinput\le d-1$.
For the case when the sink node is $\times$, we get that
\begin{align*}
\abs{\circuit}(1,\ldots, 1) &= \abs{\circuit_\linput}(1,\ldots, 1)\circmult \abs{\circuit_\rinput}(1,\ldots, 1) \\
&\leq {2^{2^{k_\linput}\cdot d_\linput}} \circmult {2^{2^{k_\rinput}\cdot d_\rinput}}\\
&\leq 2^{2\cdot 2^{k-1}\cdot (d-1)}\\
&\leq 2^{2^k d}.
\end{align*}
In the above the first inequality follows from inductive hypothesis while the second inequality follows from the fact that $k_\linput,k_\rinput\le k-1$ and $d_\linput, d_\rinput\le d-1$, where we substitute the upperbound into every respective term.
Now consider the case when the sink node is $+$, we get that
\begin{align*}
\abs{\circuit}(1,\ldots, 1) &= \abs{\circuit_\linput}(1,\ldots, 1) \circplus \abs{\circuit_\rinput}(1,\ldots, 1) \\
&\leq 2^{2^{k_\linput}\cdot d_\linput} + 2^{2^{k_\rinput}\cdot d_\rinput}\\
&\leq 2\cdot {2^{2^k(d-1)} } \\
&\leq 2^{2^kd}.
\end{align*}
In the above the first inequality follows from the inductive hypothesis while the second inequality follows from the facts that $k_\linput,k_\rinput\le k$ and $d_\linput,d_\rinput\le d-1$. The final inequality follows from the fact that $k\ge 0$.
\qed
\end{proof}
%\textcolor{red}{The corollary below is a repeat of the corollary on S4}
%Next, we note that the above result along with \Cref{lem:ctidb-gamma}
%answers \Cref{prob:big-o-joint-steps} in the affirmative as follows:
%\begin{Corollary}
%\label{cor:approx-algo-punchline-ctidb}
%Let $\query$ be an $\raPlus$ query and $\pdb$ be a \abbrCTIDB with $p_0>0$, where $p_0$ as in \Cref{cor:approx-algo-const-p}, is an absolute constant. Let $\poly(\vct{X})=\apolyqdt$ for any result tuple $\tup$ with $\deg(\poly)=k$. Then one can compute an approximation satisfying \Cref{eq:approx-algo-bound-main} in time $O_{k,|Q|,\error',\conf,\bound}\inparen{\qruntime{\optquery{\query}, \tupset, \bound}}$ (given $\query,\tupset$ and $\prob_{\tup, j}$ for each $\tup\in\tupset,~j\in\pbox{\bound}$ that defines $\bpd$).
%\end{Corollary}
\begin{proof}[Proof of~\Cref{cor:approx-algo-punchline-ctidb}]
By~\Cref{lem:ctidb-gamma} and~\Cref{cor:approx-algo-punchline}, the proof follows.
\end{proof}
\qed

39
app_approx-algo-defs-and-examples.tex
View File

@ -27,43 +27,4 @@ A subcircuit of a circuit $\circuit$ is a circuit \subcircuit such that \subcirc
\end{Definition}
%%%%%%%%%
\textcolor{red}{Aaron: The stuff below needs to be integrated into this section.}
Further, we can also argue the following result.%, recalling from~\Cref{sec:intro} for \abbrCTIDB $\pdb = \inparen{\worlds, \bpd}$, where $\tupset$ is the set of possible tuples across all possible worlds of $\pdb$.
\begin{Lemma}
\label{lem:ctidb-gamma}
Given $\raPlus$ query $\query$ and \abbrCTIDB $\pdb$, let \circuit be the circuit computed by $\query\inparen{\tupset}$. Then, for the reduced \abbrOneBIDB $\pdb'$ there exists an equivalent circuit \circuit' obtained from $\query\inparen{\tupset'}$, such that $\gamma\inparen{\circuit'}\leq 1 - \bound^{-\inparen{k-1}}$ with $\size\inparen{\circuit'} \leq \size\inparen{\circuit} + \bigO{\numvar\bound}$
and $\depth\inparen{\circuit'} = \depth\inparen{\circuit} + \bigO{\log{\bound}}$.
\end{Lemma}
We briefly connect the runtime in \Cref{eq:approx-algo-runtime} to the algorithm outline earlier (where we ignore the dependence on $\multc{\cdot}{\cdot}$, which is needed to handle the cost of arithmetic operations over integers). The $\size(\circuit)$ comes from the time taken to run \onepass once (\onepass essentially computes $\abs{\circuit}(1,\ldots, 1)$ using the natural circuit evaluation algorithm on $\circuit$). We make $\frac{\log{\frac{1}{\conf}}}{\inparen{\error'}^2\cdot(1-\gamma)^2\cdot \prob_0^{2k}}$ many calls to \sampmon (each of which essentially traces $O(k)$ random sink to source paths in $\circuit$ all of which by definition have length at most $\depth(\circuit)$).
Finally, we address the $\multc{\log\left(\abs{\circuit}(1,\ldots, 1)\right)}{\log\left(\size(\circuit)\right)}$ term in the runtime.
\begin{Lemma}
\label{lem:val-ub}
For any \emph{\abbrOneBIDB} circuit $\circuit$ with $\degree(\circuit)=k$, we have
$\abs{\circuit}(1,\ldots, 1)\le 2^{2^k\cdot \depth(\circuit)}.$
Further, if $\circuit$ is a tree, then we have $\abs{\circuit}(1,\ldots, 1)\le \size(\circuit)^{O(k)}.$
\end{Lemma}
Note that the above implies that with the assumption $\prob_0>0$ and $\gamma<1$ are absolute constants from \Cref{cor:approx-algo-const-p}, then the runtime there simplifies to $O_k\left(\frac 1{\inparen{\error'}^2}\cdot\size(\circuit)^2\cdot \log{\frac{1}{\conf}}\right)$ for general circuits $\circuit$. If $\circuit$ is a tree, then the runtime simplifies to $O_k\left(\frac 1{\inparen{\error'}^2}\cdot\size(\circuit)\cdot \log{\frac{1}{\conf}}\right)$, which then answers \Cref{prob:intro-stmt} with yes for such circuits.
Finally, note that by \Cref{prop:circuit-depth} and \Cref{lem:circ-model-runtime} for any $\raPlus$ query $\query$, there exists a circuit $\circuit^*$ for $\apolyqdt$ such that $\depth(\circuit^*)\le O_{|Q|}(\log{n})$ and $\size(\circuit)\le O_k\inparen{\qruntime{\query, \tupset, \bound}}$. Using this along with \Cref{lem:val-ub}, \Cref{cor:approx-algo-const-p} and the fact that $n\le \qruntime{\query, \tupset, \bound}$, we have the following corollary:
\begin{Corollary}
\label{cor:approx-algo-punchline}
Let $\query$ be an $\raPlus$ query and $\pdb$ be a \emph{\abbrOneBIDB} with $p_0>0$ and $\gamma<1$, where $p_0,\gamma$ as in \Cref{cor:approx-algo-const-p}, are absolute constants. Let $\poly(\vct{X})=\apolyqdt$ for any result tuple $\tup$ with $\deg(\poly)=k$. Then one can compute an approximation satisfying \Cref{eq:approx-algo-bound-main} in time $O_{k,|Q|,\error',\conf}\inparen{\qruntime{\optquery{\query}, \tupset, \bound}}$ (given $\query,\tupset$ and $p_i$ for each $i\in [n]$ that defines $\pd$).
\end{Corollary}
\textcolor{red}{The corollary below is a repeat of the corollary on S4}
Next, we note that the above result along with \Cref{lem:ctidb-gamma}
answers \Cref{prob:big-o-joint-steps} in the affirmative as follows:
\begin{Corollary}
\label{cor:approx-algo-punchline-ctidb}
Let $\query$ be an $\raPlus$ query and $\pdb$ be a \abbrCTIDB with $p_0>0$, where $p_0$ as in \Cref{cor:approx-algo-const-p}, is an absolute constant. Let $\poly(\vct{X})=\apolyqdt$ for any result tuple $\tup$ with $\deg(\poly)=k$. Then one can compute an approximation satisfying \Cref{eq:approx-algo-bound-main} in time $O_{k,|Q|,\error',\conf,\bound}\inparen{\qruntime{\optquery{\query}, \tupset, \bound}}$ (given $\query,\tupset$ and $\prob_{\tup, j}$ for each $\tup\in\tupset,~j\in\pbox{\bound}$ that defines $\bpd$).
\end{Corollary}
\begin{proof}[Proof of~\Cref{cor:approx-algo-punchline-ctidb}]
By~\Cref{lem:ctidb-gamma} and~\Cref{cor:approx-algo-punchline}, the proof follows.
\end{proof}
\qed

50
app_hard_single-p-proof-defs.tex
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@ -2,16 +2,16 @@
\subsection{Tools to prove \Cref{th:single-p-hard}}
Note that $\rpoly_{G}^3(\prob,\ldots, \prob)$ as a polynomial in $\prob$ has degree at most six. Next, we figure out the exact coefficients since this would be useful in our arguments:
Note that $\rpoly_{G}^3(\prob,\ldots, \prob)$ as a polynomial in $\prob$ has degree at most six. \Cref{lem:qE3-exp} shows the exact coefficients of $\rpoly^3\inparen{\prob,\ldots,\prob}$, which we prove next.% Next, we figure out the exact coefficients since this would be useful in our arguments:
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{Lemma}\label{lem:qE3-exp}
For any $\prob$, we have:
{\small
\begin{align}
\rpoly_{G}^3(\prob,\ldots, \prob) &= \numocc{G}{\ed}\prob^2 + 6\numocc{G}{\twopath}\prob^3 + 6\numocc{G}{\twodis}\prob^4 + 6\numocc{G}{\tri}\prob^3\nonumber\\
&+ 6\numocc{G}{\oneint}\prob^4 + 6\numocc{G}{\threepath}\prob^4 + 6\numocc{G}{\twopathdis}\prob^5 + 6\numocc{G}{\threedis}\prob^6.\label{claim:four-one}
\end{align}}
\end{Lemma}
%\begin{Lemma}\label{lem:qE3-exp}
%For any $\prob$, we have:
% {\small
% \begin{align}
% \rpoly_{G}^3(\prob,\ldots, \prob) &= \numocc{G}{\ed}\prob^2 + 6\numocc{G}{\twopath}\prob^3 + 6\numocc{G}{\twodis}\prob^4 + 6\numocc{G}{\tri}\prob^3\nonumber\\
% &+ 6\numocc{G}{\oneint}\prob^4 + 6\numocc{G}{\threepath}\prob^4 + 6\numocc{G}{\twopathdis}\prob^5 + 6\numocc{G}{\threedis}\prob^6.\label{claim:four-one}
% \end{align}}
%\end{Lemma}
\subsubsection{Proof for \Cref{lem:qE3-exp}}
\begin{proof}
By definition we have that
@ -58,22 +58,22 @@ The $3$-matchings in graph $\graph{2}$ satisfy the identity:
For $\ell > 1$ and any graph $\graph{\ell}$, $\numocc{\graph{\ell}}{\tri} = 0$.
\end{Lemma}
Finally, the following result immediately implies \Cref{th:single-p}:
\begin{Lemma}\label{lem:lin-sys}
Fix $\prob\in (0,1)$. Given $\rpoly_{\graph{\ell}}^3(\prob,\dots,\prob)$ for $\ell\in [2]$, we can compute in $O(m)$ time a vector $\vct{b}\in\mathbb{R}^3$ such that
\[ \begin{pmatrix}
1 - 3p & -(3\prob^2 - \prob^3)\\
10(3\prob^2 - \prob^3) & 10(3\prob^2 - \prob^3)
\end{pmatrix}
\cdot
\begin{pmatrix}
\numocc{G}{\tri}]\\
\numocc{G}{\threedis}
\end{pmatrix}
=\vct{b},
\]
allowing us to compute $\numocc{G}{\tri}$ and $\numocc{G}{\threedis}$ in $O(1)$ time.
\end{Lemma}
%Finally, the following result immediately implies \Cref{th:single-p}:
%\begin{Lemma}\label{lem:lin-sys}
%Fix $\prob\in (0,1)$. Given $\rpoly_{\graph{\ell}}^3(\prob,\dots,\prob)$ for $\ell\in [2]$, we can compute in $O(m)$ time a vector $\vct{b}\in\mathbb{R}^3$ such that
%\[ \begin{pmatrix}
%1 - 3p & -(3\prob^2 - \prob^3)\\
%10(3\prob^2 - \prob^3) & 10(3\prob^2 - \prob^3)
%\end{pmatrix}
%\cdot
%\begin{pmatrix}
%\numocc{G}{\tri}]\\
%\numocc{G}{\threedis}
%\end{pmatrix}
%=\vct{b},
%\]
%allowing us to compute $\numocc{G}{\tri}$ and $\numocc{G}{\threedis}$ in $O(1)$ time.
%\end{Lemma}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

1
appendix.tex
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@ -39,6 +39,7 @@ Product terms in lineage arise only from join operations (\Cref{fig:nxDBSemantic
\input{app_approx-alg-analysis}
\input{app_onepass-analysis}
\input{app_samp-monom-analysis}
\input{app_approx-alg-corollaries}
\subsection{Experimental Results}\label{app:subsec:experiment}
\input{experiments}

9
approx_alg.tex
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@ -83,7 +83,8 @@ Given a lineage polynomial $\poly(\vct{X})=\polyf(\circuit)$ for circuit \circui
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{equation}
\label{eq:tilde-Q-bi}
\rpoly\inparen{p_1,\dots,p_\numvar}=\hspace*{-1mm}\sum_{(\monom,\coef)\in \expansion{\circuit}}
\rpoly\inparen{p_1,\dots,p_\numvar}=%\hspace*{-1mm}
\sum_{(\monom,\coef)\in \expansion{\circuit}}
\indicator{\isInd{\encMon}
}\cdot \coef\cdot\prod_{X_i\in \monom} p_i.
\end{equation}
@ -155,17 +156,17 @@ Further, if $\circuit$ is a tree, then we have $\abs{\circuit}(1,\ldots, 1)\le
Note that the above implies that with the assumption $\prob_0>0$ and $\gamma<1$ are absolute constants from \Cref{cor:approx-algo-const-p}, then the runtime there simplifies to $O_k\left(\frac 1{\inparen{\error'}^2}\cdot\size(\circuit)^2\cdot \log{\frac{1}{\conf}}\right)$ for general circuits $\circuit$. If $\circuit$ is a tree, then the runtime simplifies to $O_k\left(\frac 1{\inparen{\error'}^2}\cdot\size(\circuit)\cdot \log{\frac{1}{\conf}}\right)$, which then answers \Cref{prob:intro-stmt} with yes for such circuits.
Finally, note that by \Cref{prop:circuit-depth} and \Cref{lem:circ-model-runtime} for any $\raPlus$ query $\query$, there exists a circuit $\circuit^*$ for $\apolyqdt$ such that $\depth(\circuit^*)\le O_{|Q|}(\log{n})$ and $\size(\circuit)\le O_k\inparen{\qruntime{\query, \tupset, \bound}}$. Using this along with \Cref{lem:val-ub}, \Cref{cor:approx-algo-const-p} and the fact that $n\le \qruntime{\query, \tupset, \bound}$, we have the following corollary:
Finally, note that by \Cref{prop:circuit-depth} and \Cref{lem:circ-model-runtime} for any $\raPlus$ query $\query$, there exists a circuit $\circuit^*$ for $\apolyqdt$\textcolor{red}{CHANGE} such that $\depth(\circuit^*)\le O_{|Q|}(\log{n})$ and $\size(\circuit)\le O_k\inparen{\qruntime{\query, \tupset, \bound}}$. Using this along with \Cref{lem:val-ub}, \Cref{cor:approx-algo-const-p} and the fact that $n\le \qruntime{\query, \tupset, \bound}$, we have the following corollary:
\begin{Corollary}
\label{cor:approx-algo-punchline}
Let $\query$ be an $\raPlus$ query and $\pdb$ be a \emph{\abbrOneBIDB} with $p_0>0$ and $\gamma<1$, where $p_0,\gamma$ as in \Cref{cor:approx-algo-const-p}, are absolute constants. Let $\poly(\vct{X})=\apolyqdt$ for any result tuple $\tup$ with $\deg(\poly)=k$. Then one can compute an approximation satisfying \Cref{eq:approx-algo-bound-main} in time $O_{k,|Q|,\error',\conf}\inparen{\qruntime{\optquery{\query}, \tupset, \bound}}$ (given $\query,\tupset$ and $p_i$ for each $i\in [n]$ that defines $\pd$).
Let $\query$ be an $\raPlus$ query and $\pdb$ be a \emph{\abbrOneBIDB} with $p_0>0$ and $\gamma<1$, where $p_0,\gamma$ as in \Cref{cor:approx-algo-const-p}, are absolute constants. Let $\poly(\vct{X})=\apolyqdt$\textcolor{red}{CHANGE} for any result tuple $\tup$ with $\deg(\poly)=k$. Then one can compute an approximation satisfying \Cref{eq:approx-algo-bound-main} in time $O_{k,|Q|,\error',\conf}\inparen{\qruntime{\optquery{\query}, \tupset, \bound}}$ (given $\query,\tupset$ and $p_i$ for each $i\in [n]$ that defines $\pd$).
\end{Corollary}
Next, we note that the above result along with \Cref{lem:ctidb-gamma}
answers \Cref{prob:big-o-joint-steps} in the affirmative as follows:
\begin{Corollary}
\label{cor:approx-algo-punchline-ctidb}
Let $\query$ be an $\raPlus$ query and $\pdb$ be a \abbrCTIDB with $p_0>0$, where $p_0$ as in \Cref{cor:approx-algo-const-p}, is an absolute constant. Let $\poly(\vct{X})=\apolyqdt$ for any result tuple $\tup$ with $\deg(\poly)=k$. Then one can compute an approximation satisfying \Cref{eq:approx-algo-bound-main} in time $O_{k,|Q|,\error',\conf,\bound}\inparen{\qruntime{\optquery{\query}, \tupset, \bound}}$ (given $\query,\tupset$ and $\prob_{\tup, j}$ for each $\tup\in\tupset,~j\in\pbox{\bound}$ that defines $\bpd$).
Let $\query$ be an $\raPlus$ query and $\pdb$ be a \abbrCTIDB with $p_0>0$, where $p_0$ as in \Cref{cor:approx-algo-const-p}, is an absolute constant. Let $\poly(\vct{X})=\apolyqdt$ \textcolor{red}{CHANGE}for any result tuple $\tup$ with $\deg(\poly)=k$. Then one can compute an approximation satisfying \Cref{eq:approx-algo-bound-main} in time $O_{k,|Q|,\error',\conf,\bound}\inparen{\qruntime{\optquery{\query}, \tupset, \bound}}$ (given $\query,\tupset$ and $\prob_{\tup, j}$ for each $\tup\in\tupset,~j\in\pbox{\bound}$ that defines $\bpd$).
\end{Corollary}
\begin{proof}[Proof of~\Cref{cor:approx-algo-punchline-ctidb}]
By~\Cref{lem:ctidb-gamma} and~\Cref{cor:approx-algo-punchline}, the proof follows.

6
binarybidb.tex
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@ -23,7 +23,6 @@ to denote the \abbrSMB form of a polynomial $\genpoly~\inparen{\poly}$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
We call a polynomial $\poly\inparen{\vct{X}}$ a \emph{\abbrCTIDB-lineage polynomial} (or simply lineage polynomial), if it is clear from context that there exists an $\raPlus$ query $\query$, \abbrCTIDB $\pdb$, and result tuple $\tup$ such that $\poly\inparen{\vct{X}} = \apolyqdt\inparen{\vct{X}}.$
\subsection{\abbrOneBIDB}\label{subsec:one-bidb}
\label{subsec:tidbs-and-bidbs}
@ -42,8 +41,9 @@ We slightly abuse notation here, denoting a world vector as $W$ rather than $\wo
\end{Definition}
Lineage polynomials for arbitrary deterministic $\gentupset'$ are constructed in a manner analogous to $1$-\abbrTIDB\xplural (see \Cref{fig:nxDBSemantics}), differing only in the base case.
In a $1$-\abbrTIDB, each tuple contributes a multiplicity of 0 or 1, and $\polyqdt{\rel}{\gentupset}{\tup} = X_\tup$.
In a \abbrOneBIDB, each tuple $\tup\in\tupset'$ contributes its corresponding multiplicity: $\polyqdt{\rel}{\gentupset}{\tup} = c_\tup\cdot X_\tup$. These semantics are fully detailed in \Cref{fig:lin-poly-bidb}.
In a $1$-\abbrTIDB, each tuple contributes a multiplicity of 0 or 1, and $\polyqdt{\rel}{\gentupset}{\tup} = X_\tup$.%\textcolor{red}{CHANGE}
In a \abbrOneBIDB, each tuple $\tup\in\tupset'$ contributes its corresponding multiplicity: %\textcolor{red}{CHANGE}
$\polyqdt{\rel}{\gentupset}{\tup} = c_\tup\cdot X_\tup$. These semantics are fully detailed in \Cref{fig:lin-poly-bidb}.
\abbrOneBIDB are powerful enough to encode \abbrCTIDB:
\begin{Proposition}[\abbrCTIDB reduction]\label{prop:ctidb-reduct}

44
introduction.tex
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@ -139,9 +139,9 @@ We drop $\query$, $\tupset$, and $\tup$ from $\apolyqdt$ when they are clear fro
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{Problem}[Expected Multiplicity of Lineage Polynomials]\label{prob:bag-pdb-poly-expected}
Given an $\raPlus$ query $\query$, \abbrCTIDB $\pdb$ and result tuple $\tup$, compute the expected
multiplicity of the polynomial $\apolyqdt$ (i.e.,
multiplicity of the polynomial $\poly$ (i.e.,
%for $\worldvec\in\worlds$,
compute $\expct_{\vct{W}\sim \pdassign}\pbox{\apolyqdt\inparen{\worldvec}}$).
compute $\expct_{\vct{W}\sim \pdassign}\pbox{\poly\inparen{\worldvec}}$).
\end{Problem}
%We note that computing \Cref{prob:expect-mult} is equivalent (yields the same result as) to computing \Cref{prob:bag-pdb-poly-expected} (see \Cref{prop:expection-of-polynom}).
@ -150,19 +150,19 @@ All of our results rely on working with a {\em reduced} form $\inparen{\rpoly}$
Next, we motivate this reduced polynomial $\rpoly$.
Consider the query $\query_1$ defined as follows over the bag relations of \Cref{fig:two-step}:
\begin{lstlisting}
\begin{lstlisting}[frame=single,framerule=0pt]
SELECT DISTINCT 1 FROM T $t_1$, R r, T $t_2$
WHERE $t_1$.Point = r.Point$_1$ AND $t_2$.Point = r.Point$_2$
\end{lstlisting}
It can be verified that $\poly\inparen{A, B, C, E, X, Y, Z}$ for the sole result tuple of $\query_1$ is $AXB + BYE + BZC$. Now consider the product query $\query_1^2 = \query_1 \times \query_1$.
The lineage polynomial for $\query_1^2$ is given by $\poly_1^2\inparen{A, B, C, E, X, Y, Z}$
It can be verified that $\poly\inparen{A, B, C, E, U, Y, Z}$ for the sole result tuple of $\query_1$ is $AUB + BYE + BZC$. Now consider the product query $\query_1^2 = \query_1 \times \query_1$.
The lineage polynomial for $\query_1^2$ is given by $\poly_1^2\inparen{A, B, C, E, U, Y, Z}$
$$
=A^2X^2B^2 + B^2Y^2E^2 + B^2Z^2C^2 + 2AXB^2YE + 2AXB^2ZC + 2B^2YEZC.
=A^2U^2B^2 + B^2Y^2E^2 + B^2Z^2C^2 + 2AUB^2YE + 2AXB^2ZC + 2B^2YEZC.
$$
To compute $\expct\pbox{\poly_1^2}$ we can use linearity of expectation and push the expectation through each summand. To keep things simple, let us focus on the monomial $\poly_1^{\inparen{ABX}^2} = A^2X^2B^2$ as the procedure is the same for all other monomials of $\poly_1^2$. Let $\randWorld_X$ be the random variable corresponding to a lineage variable $X$. Because the distinct variables in the product are independent, we can push expectation through them yielding $\expct\pbox{\randWorld_A^2\randWorld_X^2\randWorld_B^2}=\expct\pbox{\randWorld_A^2}\expct\pbox{\randWorld_X^2}\expct\pbox{\randWorld_B^2}$. Since $\randWorld_A, \randWorld_B\in \inset{0, 1}$ we can further simplify to $\expct\pbox{\randWorld_A}\expct\pbox{\randWorld_X^2}\expct\pbox{\randWorld_B}$ by the fact that for any $W\in \inset{0, 1}$, $W^2 = W$. Observe that if $W_X\in\inset{0, 1}$, then we further would have $\expct\pbox{\randWorld_A}\expct\pbox{\randWorld_X}\expct\pbox{\randWorld_B} = \prob_A\cdot\prob_X\cdot\prob_B$ (denoting $\probOf\pbox{\randWorld_A = 1} = \prob_A$) $= \rpoly_1^{\inparen{ABX}^2}\inparen{\prob_A, \prob_X, \prob_B}$ (see $ii)$ of~\Cref{def:reduced-poly}). However, in this example, we get stuck with $\expct\pbox{\randWorld_X^2}$, since $\randWorld_X\in\inset{0, 1, 2}$ and for $\randWorld_X \gets 2$, $\randWorld_X^2 \neq \randWorld_X$.
To compute $\expct\pbox{\poly_1^2}$ we can use linearity of expectation and push the expectation through each summand. To keep things simple, let us focus on the monomial $\poly_1^{\inparen{ABU}^2} = A^2U^2B^2$ as the procedure is the same for all other monomials of $\poly_1^2$. Let $\randWorld_U$ be the random variable corresponding to a lineage variable $U$. Because the distinct variables in the product are independent, we can push expectation through them yielding $\expct\pbox{\randWorld_A^2\randWorld_U^2\randWorld_B^2}=\expct\pbox{\randWorld_A^2}\expct\pbox{\randWorld_U^2}\expct\pbox{\randWorld_B^2}$. Since $\randWorld_A, \randWorld_B\in \inset{0, 1}$ we can further simplify to $\expct\pbox{\randWorld_A}\expct\pbox{\randWorld_U^2}\expct\pbox{\randWorld_B}$ by the fact that for any $W\in \inset{0, 1}$, $W^2 = W$. Observe that if $W_U\in\inset{0, 1}$, then we further would have $\expct\pbox{\randWorld_A}\expct\pbox{\randWorld_U}\expct\pbox{\randWorld_B} = \prob_A\cdot\prob_X\cdot\prob_B$ (denoting $\probOf\pbox{\randWorld_A = 1} = \prob_A$) $= \rpoly_1^{\inparen{ABX}^2}\inparen{\prob_A, \prob_U, \prob_B}$ (see $ii)$ of~\Cref{def:reduced-poly}). However, in this example, we get stuck with $\expct\pbox{\randWorld_U^2}$, since $\randWorld_U\in\inset{0, 1, 2}$ and for $\randWorld_U \gets 2$, $\randWorld_U^2 \neq \randWorld_U$.
The simple insight to get around this issue to note that the random variables $\randWorld_X$ and $\randWorld_{X_1}+2\randWorld_{X_2}$ have exactly the same distribution, where $\randWorld_{X_1},\randWorld_{X_2}\in\inset{0,1}$ and $\probOf\pbox{\randWorld_{X_j} = 1} = \probOf\pbox{\randWorld_{X} = j}$. Thus, the idea is to replace the variable $X$ by $X_1+2X_2$ (where $X_j$ corresponds to the event that $X$ has multiplicity $j$) to obtain the following polynomial:
The simple insight to get around this issue to note that the random variables $\randWorld_U$ and $\randWorld_{U_1}+2\randWorld_{U_2}$ have exactly the same distribution, where $\randWorld_{U_1},\randWorld_{U_2}\in\inset{0,1}$ and $\probOf\pbox{\randWorld_{U_j} = 1} = \probOf\pbox{\randWorld_{U} = j}$. Thus, the idea is to replace the variable $U$ by $U_1+2U_2$ (where $U_j$ corresponds to the event that $U$ has multiplicity $j$) to obtain the following polynomial:
%
%Denote the variables of $\poly$ to be $\vars{\poly}.$ In the \abbrCTIDB setting, $\poly\inparen{\vct{X}}$ has an equivalent reformulation $\inparen{\refpoly{}\inparen{\vct{X_R}}}$ that is of use to us, where $\abs{\vct{X_R}} = \bound\cdot\abs{\vct{X}}$ . Given $X_\tup \in\vars{\poly}$ and integer valuation $X_\tup \in\inset{0,\ldots, c}$. We can replace $X_\tup$ by $\sum_{j\in\pbox{\bound}}jX_{\tup, j}$ where the variables $\inparen{X_{\tup, j}}_{j\in\pbox{\bound}}$ are disjoint with integer assignments $X_{\tup, j}\in\inset{0, 1}$. Then for any $\worldvec\in\worlds$ and corresponding reformulated world $\worldvec_{\vct{R}}\in\inset{0, 1}^{\tupset\bound}$, we set $\worldvec_{\vct{R}_{\tup, j}} = 1$ for $\worldvec_\tup = j$, while $\worldvec_{\vct{R}_{\tup, j'}} = 0$ for all $j'\neq j\in\pbox{\bound}$. By construction then $\poly\inparen{\vct{X}}\equiv\refpoly{}\inparen{\vct{X_R}}$ $\inparen{\vct{X_R} = \vars{\refpoly{}}}$ since for any integer valuation $X_\tup\in\pbox{\bound}$, $X_{\tup, j}\in\inset{0, 1}$ we have the equality $X_\tup = j = \sum_{j\in\pbox{\bound}}jX_{t, j}$.
%
@ -171,14 +171,14 @@ The simple insight to get around this issue to note that the random variables $\
%\begin{multline*}
%\refpoly{1, }^{\inparen{ABX}^2}\inparen{A, X, B} = \poly_1^{\inparen{AXB}^2}\inparen{\sum_{j_1\in\pbox{\bound}}j_1A_{j_1}, \sum_{j_2\in\pbox{\bound}}j_2X_{j_2}, \sum_{j_3\in\pbox{\bound}}j_3B_{j_3}} \\
%= \inparen{\sum_{j_1\in\pbox{\bound}}j_1A_{j_1}}^2\inparen{\sum_{j_2\in\pbox{\bound}}j_2X_{j_2}}^2\inparen{\sum_{j_3\in\pbox{\bound}}j_3B_{j_3}}^2.
\[\overline{\refpoly{1, }^{\inparen{ABX}^2}}\inparen{A, X_1, X_2 B} = \poly_1^{\inparen{AXB}^2}\inparen{A,(X_1+2X_2),B}.\]
\[\overline{\refpoly{1, }^{\inparen{ABU}^2}}\inparen{A, U_1, U_2 B} = \poly_1^{\inparen{AUB}^2}\inparen{A,(U_1+2U_2),B}.\]
%\end{multline*}
%}
%Since the set of multiplicities for tuple $\tup$ by nature are disjoint we can drop all cross terms and have $\refpoly{1, }^2 = \sum_{j_1, j_2, j_3 \in \pbox{\bound}}j_1^2A^2_{j_1}j_2^2X_{j_2}^2j_3^2B^2_{j_3}$. Since we now have that all $\randWorld_{X_j}\in\inset{0, 1}$, computing expectation yields $\expct\pbox{\refpoly{1, }^2}=\sum_{j_1,j_2,j_3\in\pbox{\bound}}j_1^2j_2^2j_3^2$ \allowbreak $\expct\pbox{\randWorld_{A_{j_1}}}\expct\pbox{\randWorld_{X_{j_2}}}\expct\pbox{\randWorld_{B_{j_3}}}$.
Given that $X$ can only have multiplicity of $1$ or $2$ but not both, we drop the monomials with the term $X_1X_2$ to get
$\refpoly{1, }^{\inparen{ABX}^2}\inparen{A, X_1, X_2, B} = A^2X_1^2B^2+2^2\cdot A^2 X_2^2B^2.$
Now that all the world vectors $(\randWorld_A,\randWorld_{X_1},\randWorld_{X_2},\randWorld_A)\in\inset{0,1}^4$, we have $\expct\pbox{\refpoly{1, }^2}=\expct\pbox{\randWorld_{A}}\expct\pbox{\randWorld_{X_1}}\expct\pbox{\randWorld_{B}}+$ \\ $4\expct\pbox{\randWorld_{A}}\expct\pbox{\randWorld_{X_2}}\expct\pbox{\randWorld_{B}}\stackrel{\text{def}}{=}\rpoly_1^2\inparen{p_A,\probOf\inparen{X=1},\probOf\inparen{X=2},p_B}$. We only did the argument for a single monomial but by linearity of expectation we can apply the same argument to all monomials in $\poly_1^2$. Generalizing this argument to general $\poly$ leads to consider its follownig `reduced' version:
Given that $U$ can only have multiplicity of $1$ or $2$ but not both, we drop the monomials with the term $U_1U_2$ to get
$\refpoly{1, }^{\inparen{ABU}^2}\inparen{A, U_1, U_2, B} = A^2U_1^2B^2+2^2\cdot A^2 U_2^2B^2.$
Now that all the world vectors $(\randWorld_A,\randWorld_{U_1},\randWorld_{U_2},\randWorld_A)\in\inset{0,1}^4$, we have $\expct\pbox{\refpoly{1, }^2}=\expct\pbox{\randWorld_{A}}\expct\pbox{\randWorld_{U_1}}\expct\pbox{\randWorld_{B}}+$ \\ $4\expct\pbox{\randWorld_{A}}\expct\pbox{\randWorld_{U_2}}\expct\pbox{\randWorld_{B}}\stackrel{\text{def}}{=}\rpoly_1^2\inparen{p_A,\probOf\inparen{U=1},\probOf\inparen{U=2},p_B}$. We only did the argument for a single monomial but by linearity of expectation we can apply the same argument to all monomials in $\poly_1^2$. Generalizing this argument to general $\poly$ leads to consider its follownig `reduced' version:
\begin{Definition}\label{def:reduced-poly}
For any polynomial $\poly\inparen{\inparen{X_\tup}_{\tup\in\tupset}}$ define the reformulated polynomial $\refpoly{}\inparen{\inparen{X_{\tup, j}}_{\tup\in\tupset, j\in\pbox{\bound}}}
@ -197,11 +197,12 @@ removing all monomials containing the term $X_{\tup, j}X_{\tup, j'}$ for $\tup\i
%&= ABX_1 + AB\inparen{2}^2X_2+ BYE + BZC + 2AX_1BYE+ 2A\inparen{2}^2X_2BYE\\
%&\qquad + 2AX_1BZC + 2A\inparen{2}^2X_2BZC + 2BYEZC.
%\end{align*}
As we have essentially argued earlier, for our specific example the expectation that we want is $\rpoly_1^2(\probOf\inparen{A=1},$\allowbreak$\probOf\inparen{B=1}, \probOf\inparen{C=1}$,\allowbreak $\probOf\inparen{E=1},$\allowbreak $\probOf\inparen{X_1=1}, \probOf\inparen{X_2=1}, \probOf\inparen{Y=1}, \probOf\inparen{Z=1})$.
As we have essentially argued earlier, for our specific example the expectation that we want is $\rpoly_1^2(\probOf\inparen{A=1},$\allowbreak$\probOf\inparen{B=1}, \probOf\inparen{C=1}$,\allowbreak $\probOf\inparen{E=1},$\allowbreak $\probOf\inparen{U_1=1}, \probOf\inparen{U_2=1}, \probOf\inparen{Y=1}, \probOf\inparen{Z=1})$.
\Cref{lem:tidb-reduce-poly} generalizes the equivalence to {\em all} $\raPlus$ queries on \abbrCTIDB\xplural (proof in \Cref{subsec:proof-exp-poly-rpoly}):
\begin{Lemma}\label{lem:tidb-reduce-poly}
For any \abbrCTIDB $\pdb$, $\raPlus$ query $\query$, and lineage polynomial
$\poly\inparen{\vct{X}}=\poly\pbox{\query,\tupset,\tup}\inparen{\vct{X}}$, it holds that $
$\poly\inparen{\vct{X}}$%=\textcolor{red}{CHANGE}\poly\pbox{\query,\tupset,\tup}\inparen{\vct{X}}$
, it holds that $
\expct_{\vct{W} \sim \pdassign}\pbox{\poly\inparen{\vct{W}}} = \rpoly\inparen{\probAllTup}
$, where $\probAllTup = \inparen{\prob_{\tup,j}}_{\tup\in\tupset,j\in\pbox{\bound}}.$
\end{Lemma}
@ -222,7 +223,8 @@ Our negative results (\Cref{tab:lbs}) indicate that \abbrCTIDB{}s (even for $\bo
\input{two-step-model}
We adopt a two-step intensional model of query evaluation used in set-\abbrPDB\xplural, as illustrated in \Cref{fig:two-step}:
(i) \termStepOne (\abbrStepOne): Given input $\tupset$ and $\query$, output every tuple $\tup$ that possibly satisfies $\query$, annotated with its lineage polynomial ($\poly(\vct{X})=\apolyqdt\inparen{\vct{X}}$);
(i) \termStepOne (\abbrStepOne): Given input $\tupset$ and $\query$, output every tuple $\tup$ that possibly satisfies $\query$, annotated with its lineage polynomial $\poly(\vct{X})%=\textcolor{red}{CHANGE}\apolyqdt\inparen{\vct{X}}$
$;
(ii) \termStepTwo (\abbrStepTwo): Given $\poly(\vct{X})$ for each tuple, compute $\expct_{\randWorld\sim\bpd}\pbox{\poly(\vct{\randWorld})}$.
Let $\timeOf{\abbrStepOne}(\query,\tupset,\circuit)$ denote the runtime of \abbrStepOne when it outputs $\circuit$ (a representation of $\poly$ as an arithmetic circuit --- more on this representation in~\Cref{sec:expression-trees}).
Denote by $\timeOf{\abbrStepTwo}(\circuit, \epsilon)$ (recall $\circuit$ is the output of \abbrStepOne) the runtime of \abbrStepTwo, which we can leverage~\Cref{def:reduced-poly} and~\Cref{lem:tidb-reduce-poly} to address the next formal objective:
@ -247,28 +249,28 @@ As we also show in \Cref{sec:circuit-runtime}, the size is also bounded by $\qru
%Thus, the question of approximation can be stated as the following stronger (since~\Cref{prob:big-o-joint-steps} has access to \emph{all} equivalent \circuit representing $\query\inparen{\vct{W}}\inparen{\tup}$), but sufficient condition:
Given such a $\circuit^*$, to solve \Cref{prob:big-o-joint-steps}, it is \emph{sufficient} to solve: % the following problem:
\begin{Problem}\label{prob:intro-stmt}
Given one circuit $\circuit$ that encodes $\apolyqdt$ for all result tuples $\tup$ (one sink per $\tup$) for \abbrCTIDB $\pdb$ and $\raPlus$ query $\query$, does there exist an algorithm that computes a $(1\pm\epsilon)$-approximation of $\expct_{\rvworld\sim\bpd}\pbox{\query\inparen{\rvworld}\inparen{\tup}}$ (for all result tuples $\tup$) in $\bigO{|\circuit|}$ time?
Given one circuit $\circuit$ that encodes $\Phi\inparen{\vct{X}}$ for all result tuples $\tup$ (one sink per $\tup$) for \abbrCTIDB $\pdb$ and $\raPlus$ query $\query$, does there exist an algorithm that computes a $(1\pm\epsilon)$-approximation of $\expct_{\rvworld\sim\bpd}\pbox{\query\inparen{\rvworld}\inparen{\tup}}$ (for all result tuples $\tup$) in $\bigO{|\circuit|}$ time?
\end{Problem}
We will formalize the notions of circuits and hence, \Cref{prob:intro-stmt} in \cref{sec:expression-trees}. For an upper bound on approximating the expected count, it is easy to check that if all the probabilties are constant then (with an additive adjustment) $\rpoly\left(\prob_1,\dots, \prob_n\right)$ (recall \cref{def:reduced-poly}) is a constant factor approximation.
This is illustrated in the following example using $\query_1^2$ from earlier. To aid in presentation we again limit our focus to $\refpoly{1, }^{\inparen{ABX}^2}$, assume $\bound = 2$ for variable $X$ and $\bound = 1$ for all other variables. Let $\prob_A$ denote $\probOf\pbox{A = 1}$.
This is illustrated in the following example using $\query_1^2$ from earlier. To aid in presentation we again limit our focus to $\refpoly{1, }^{\inparen{ABU}^2}$, assume $\bound = 2$ for variable $U$ and $\bound = 1$ for all other variables. Let $\prob_A$ denote $\probOf\pbox{A = 1}$.
%In computing $\rpoly$, we have some cancellations to deal with:
Then we have:
\begin{footnotesize}
\begin{equation*}
\refpoly{1, }^{\inparen{ABX}^2}\inparen{\vct{X}} = A^2\inparen{X_1^2 + 4X_1X_2 + 4X_2^2}B^2 =A^2X_1^2B^2 + 4A^2X_1X_2B^2+4A^2X_2^2B^2
\refpoly{1, }^{\inparen{ABU}^2}\inparen{\vct{X}} = A^2\inparen{U_1^2 + 4U_1U_2 + 4U_2^2}B^2 =A^2U_1^2B^2 + 4A^2U_1U_2B^2+4A^2U_2^2B^2
%&\qquad+ 2AX_2B^2YE + 2AX_1B^2ZC + 2AX_2B^2ZC + 2B^2YEZC\\
\end{equation*}
\end{footnotesize}
Recall that
%\begin{footnotesize}
%\begin{equation*}
$\rpoly_1^{\inparen{ABX}^2}\inparen{\vct{X}} = AX_1B+4AX_2B$,
$\rpoly_1^{\inparen{ABU}^2}\inparen{\vct{X}} = AU_1B+4AU_2B$,
%\end{equation*}
%\end{footnotesize}
which implies:
\[ \refpoly{1, }^{\inparen{ABX}^2}\inparen{\probAllTup} -4\prob_A^2\prob_{X_1}\prob_{X_2}\prob_B^2=\prob_A^2\prob_{X_1}^2\prob_B^2 + 4\prob_A^2\prob_{X_2}^2\prob_B^2.\]
\[ \refpoly{1, }^{\inparen{ABU}^2}\inparen{\probAllTup} -4\prob_A^2\prob_{U_1}\prob_{U_2}\prob_B^2=\prob_A^2\prob_{U_1}^2\prob_B^2 + 4\prob_A^2\prob_{U_2}^2\prob_B^2.\]
%Substituting $\vct{\prob}$ for $\vct{X}$,
%\begin{footnotesize}
%\begin{align*}
@ -282,7 +284,7 @@ which implies:
%\end{footnotesize}
If we assume that all probability values are in $[p_0,1]$ for some $p_0>0$,
%then given access to $\refpoly{1, }^{\inparen{ABX}^2}\inparen{\vct{\prob}} - 4\prob_A^2\prob_{X_1}\prob_{X_2}\prob_B^2$
we get that $\refpoly{1, }^{\inparen{ABX}^2}\inparen{\vct{\prob}} - 4\prob_A^2\prob_{X_1}\prob_{X_2}\prob_B^2$ is in the range $\pbox{p_0^3\cdot\rpoly^{\inparen{ABX}^2}_1\inparen{\vct{\prob}}, \rpoly_1^{\inparen{ABX}^2}\inparen{\vct{\prob}}}$.
we get that $\refpoly{1, }^{\inparen{ABU}^2}\inparen{\vct{\prob}} - 4\prob_A^2\prob_{U_1}\prob_{U_2}\prob_B^2$ is in the range $\pbox{p_0^3\cdot\rpoly^{\inparen{ABU}^2}_1\inparen{\vct{\prob}}, \rpoly_1^{\inparen{ABU}^2}\inparen{\vct{\prob}}}$.
%We can simulate sampling from $\refpoly{1, }^2\inparen{\vct{X}}$ by sampling monomials from $\refpoly{1, }^2$ while ignoring any samples $A^2X_1X_2B^2$.
Note however, that this is \emph{not a tight approximation}.
In~\cref{sec:algo} we demonstrate that a $(1\pm\epsilon)$ (multiplicative) approximation with competitive performance is achievable.

1
main.tex
View File

@ -55,6 +55,7 @@
\usepackage{mdframed}
\lstdefinestyle{psql}
{
backgroundcolor=\color{black!15!white},
tabsize=2,
basicstyle=\small\upshape\ttfamily,
language=SQL,

4
prob-def.tex
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@ -2,7 +2,7 @@
%!TEX root=./main.tex
\subsection{Formalizing \Cref{prob:intro-stmt}}\label{sec:expression-trees}
We focus on the problem of computing $\expct_{\worldvec\sim\pdassign}\pbox{\apolyqdt\inparen{\vct{\randWorld}}}$ from now on, assume implicit $\query, \tupset, \tup$, and drop them from $\apolyqdt$ (i.e., $\poly\inparen{\vct{X}}$ will denote a polynomial).
We focus on the problem of computing $\expct_{\worldvec\sim\pdassign}\pbox{\poly\inparen{\vct{\randWorld}}}$ from now on.%, assume implicit $\query, \tupset, \tup$, and drop them from $\apolyqdt$ (i.e., $\poly\inparen{\vct{X}}$ will denote a polynomial).
\Cref{prob:intro-stmt} asks if there exists a linear time approximation algorithm in the size of a given circuit \circuit which encodes $\poly\inparen{\vct{X}}$. Recall that in this work we
represent lineage polynomials via {\em arithmetic circuits}~\cite{arith-complexity}, a standard way to represent polynomials over fields (particularly in the field of algebraic complexity) that we use for polynomials over $\mathbb N$ in the obvious way. Since we are specifically using circuits to model lineage polynomials, we can refer to these circuits as lineage circuits. However, when the meaning is clear, we will drop the term lineage and only refer to them as circuits.
@ -85,7 +85,7 @@ The circuit of \Cref{fig:circuit} is an element of $\circuitset{2X^2+3XY-2Y^2}$.
Let $\pdb$ be an arbitrary \abbrCTIDB and $\vct{X}$ be the set of variables annotating tuples in $\tupset$. Fix an $\raPlus$ query $\query$ and a result tuple $\tup$.
The \expectProblem is defined as follows:%\\[-7mm]
\begin{flalign*}
&\textbf{Input}: \circuit \in \circuitset{\polyX} \text{ for }\poly\inparen{\vct{X}} = \poly\pbox{\query,\tupset,\tup}&\\
&\textbf{Input}: \circuit \in \circuitset{\polyX}\\% \text{ for }\poly\inparen{\vct{X}} = \textcolor{red}{CHANGE}\poly\pbox{\query,\tupset,\tup}&\\
&\textbf{Output}: \expct_{\vct{W} \sim \bpd}\pbox{\poly\pbox{\query, \tupset, \tup}\inparen{\vct{W}}}.&
\end{flalign*}
\end{Definition}

8
two-step-model.tex
View File

@ -31,7 +31,7 @@
%\toprule
$\text{Point}_1$ & $\text{Point}_2$ & $\Phi$\\% & $\semN$ \\
\midrule
$e_1$ & $e_2$ & $X$\\% & 2 \\
$e_1$ & $e_2$ & $U$\\% & 2 \\
$e_2$ & $e_4$ & $Y$\\% & 4 \\
%& $\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ \\
$e_2$ & $e_3$ & $Z$\\% & 3 \\
@ -59,11 +59,11 @@
\midrule
%\hline
%\\\\[-3.5\medskipamount]
$e_1$ & $AX$ &\adjustbox{valign=b}{\resizebox{!}{9mm}{
$e_1$ & $AU$ &\adjustbox{valign=b}{\resizebox{!}{9mm}{
\begin{tikzpicture}[thick]
\node[gen_tree_node](sink) at (0.5, 0.8){$\boldsymbol{\circmult}$};
\node[gen_tree_node](source1) at (0, 0){$A$};
\node[gen_tree_node](source2) at (1, 0){$X$};
\node[gen_tree_node](source2) at (1, 0){$U$};
\draw[->](source1)--(sink);
\draw[->] (source2)--(sink);
\end{tikzpicture}$\inparen{1}$% & $0.5 \cdot 1.0 + 0.5 \cdot 1.0 = 1.0$
@ -124,7 +124,7 @@
%\toprule
Point & $\mathbb{E}[\poly(\vct{X})]$\\
\midrule%[0.05pt]
$e_1$ & $\inparen{\prob_{A_1} +\prob_{A_2}}\cdot\left(\prob_{X_1} + 2\prob_{X_2}\right)$\\%$1.0 \cdot 0.9 = 0.9$\\[3mm]
$e_1$ & $\inparen{\prob_{A_1} +\prob_{A_2}}\cdot\left(\prob_{U_1} + 2\prob_{U_2}\right)$\\%$1.0 \cdot 0.9 = 0.9$\\[3mm]
$e_2$ & $\inparen{\prob_{B_1} + \prob_{B_2}}\inparen{\prob_{Y_1}+2\prob_{Y_2} + \prob_{Z_1} + 2\prob_{Z_2}}$\\%$(0.5 \cdot 1.0) + $\newline $\hspace{0.2cm}(0.5 \cdot 1.0)$\newline $= 1.0$\\
\end{tabular}
};