Generalized lemmas/proofs for OnePass and SampMonomial

approx_alg.tex

@@ -175,12 +175,12 @@ we first state the lemmas that summarize the relevant properties of $\onepass$ a
 
 
 \begin{Lemma}\label{lem:one-pass}
-The $\onepass$ function completes in $O(size(\circuit))$ time.  $\onepass$ guarantees two post-conditions:  First, for each subcircuit $\vari{S}$ of $\circuit$, we have that $\vari{S}.\vari{partial}$ is set to $\abs{\vari{S}}(1,\ldots, 1)$.  Second, when $\vari{S}.\type  = \circplus$, \subcircuit.\lwght $= \frac{\abs{\subcircuit_\linput}(1,\ldots, 1)}{\abs{\subcircuit}(1,\ldots, 1)}$ and likewise for \subcircuit.\rwght.
+The $\onepass$ function completes in $O(size(\circuit) \cdot \frac{\log{\abs{\circuit(1\ldots, 1)}}}{\log{N}})$ time, where $N = \size(\circuit)$.\footnote{In the appendix we give a sufficient condition when $\abs{\circuit}(1,\ldots, 1)$ is indeed $O(1)$ in arithmetic computations.  Most notably, WCOJ and FAQ results are not affected by the general runtime of arithmetic computations, a point which we also address in the appendix.}  $\onepass$ guarantees two post-conditions:  First, for each subcircuit $\vari{S}$ of $\circuit$, we have that $\vari{S}.\vari{partial}$ is set to $\abs{\vari{S}}(1,\ldots, 1)$.  Second, when $\vari{S}.\type  = \circplus$, \subcircuit.\lwght $= \frac{\abs{\subcircuit_\linput}(1,\ldots, 1)}{\abs{\subcircuit}(1,\ldots, 1)}$ and likewise for \subcircuit.\rwght.
 \end{Lemma}
 To prove correctness of~\Cref{alg:mon-sam}, we only use the following fact that follows from the above lemma: for the modified circuit ($\circuit_{\vari{mod}}$), $\circuit_{\vari{mod}}$, $\circuit_{\vari{mod}}.\vari{partial}=\abs{\circuit}(1,\dots,1)$.
 
 \begin{Lemma}\label{lem:sample}
-The function $\sampmon$ completes in $O(\log{k} \cdot k \cdot \depth(\circuit))$ time, where $k = \degree(\circuit)$.  Upon completion, every $\left(\monom, sign(\coef)\right)\in \expansion{\abs{\circuit}}$ is returned with probability $\frac{|\coef|}{\abs{\circuit}(1,\ldots, 1)}$. 
+The function $\sampmon$ completes in $O(\log{k} \cdot k \cdot \depth(\circuit)\cdot\frac{\log{\abs{\circuit}(1,\ldots, 1)}}{\log{\size(\circuit)}})$ time\footnote{Note that the same sufficient condition on \circuit to guarentee $O(1)$ arithmetic computations applies here, and when this condition is met, the runtime loses the $\frac{\log{\abs{\circuit}(1,\ldots, 1)}}{\log{\size(\circuit)}}$ factor}, where $k = \degree(\circuit)$.  Upon completion, every $\left(\monom, sign(\coef)\right)\in \expansion{\abs{\circuit}}$ is returned with probability $\frac{|\coef|}{\abs{\circuit}(1,\ldots, 1)}$. 
 \end{Lemma}
 
 Armed with the above two lemmas, we are ready to argue the following result (proof in~\Cref{sec:proofs-approx-alg}):

hardness-app.tex

@@ -622,12 +622,12 @@ When $\gate_{k+1}.\type = \circplus$, then by line ~\ref{alg:one-pass-plus} $\ga
 When $\gate_{k+1}.\type = \circmult$, then line ~\ref{alg:one-pass-mult} computes $\gate_{k+1}.\prt = \gate_{{k+1}_\lchild.\prt} \circmult \gate_{{k+1}_\rchild}.\prt$, which indeed is correct, as per \Cref{eq:T-all-ones}.
 
 \paragraph{\onepass Runtime}
-It is known that $\topord(G)$ is computable in linear time.  Next, each of the $\numvar$ iterations of the loop in ~\Cref{alg:one-pass-loop} take $O(1)$ time, thus yielding a runtime of $O\left(\size(\circuit)\right)$.
+It is known that $\topord(G)$ is computable in linear time.  Next, each of the $\numvar$ iterations of the loop in ~\Cref{alg:one-pass-loop} take $O(1)$ time.  In general it is known that an arithmetic computation which requires $M$ bits takes $O(\frac{\log{M}}{\log{N}})$ time for an input size $N$.  Since each of the arithmetic operations at a given gate has a bit size of $O(\log{\abs{\circuit}(1,\ldots, 1)})$,  thus, we obtain the general runtime of $O\left(\size(\circuit)\cdot \frac{\log{\abs{\circuit}(1,\ldots, 1)}}{\log{\size(\circuit)}}\right)$.
 
-\paragraph{$\abs{\circuit}(1,\ldots, 1)$ is size $O(N)$}
-For our runtime results to be relevant, it must be the case that the sum of the coefficients computed by \onepass is indeed size $O(N)$ since there are $O(\log{N})$ bits in the RAM model where $N$ is the size of the input.  The size of the input here is \size(\circuit).  We show that when \size$(\circuit_\linput) = N_\linput = N - c$, \size$(\circuit_\rinput) = N_\rinput \leq c$, where $N > N_\linput \geq N_\rinput$ for some constant $c$, this is indeed the case. %with the constraint that when \depth(\circuit) $\geq 1$ then \degree(\circuit) $\geq 1$, that $\abs{\circuit}(1,\ldots, 1)$ is indeed size $O(N)$ by induction on \depth(\circuit).
+\paragraph{Sufficient condition for $\abs{\circuit}(1,\ldots, 1)$ to be size $O(N)$}
+For our runtime results to be relevant, it must be the case that the sum of the coefficients computed by \onepass is indeed size $O(N)$ since there are $O(\log{N})$ bits in the RAM model where $N$ is the size of the input.  The size of the input here is \size(\circuit).  We show that when \size$(\circuit_\linput) = N_\linput$, \size$(\circuit_\rinput) = N_\rinput$, where $N_\linput + N_\rinput \leq N$, this is indeed the case.
 
-\begin{proof}[Proof of $\abs{\circuit}(1,\ldots, 1)$ is size $O(N)$]
+\begin{proof}%[Proof of $\abs{\circuit}(1,\ldots, 1)$ is size $O(N)$]
 To prove this result, we start by proving that $\abs{\circuit}(1,\ldots, 1) \leq N^{2^k }$ for \degree(\circuit) $= k$.  
 For the base case, we have that \depth(\circuit) $= 0$, and there can only be one node which must contain a coefficient (or constant) of $1$.  In this case, $\abs{\circuit}(1,\ldots, 1) = 1$, and \size(\circuit) $= 1$, and it is true that $\abs{\circuit}(1,\ldots, 1) = 1 \leq N^{2^k} = 1^{2^0} = 1$.
 
@@ -648,19 +648,15 @@ N_\linput^{2^{k_\linput}} + N_\rinput^{2^{k_\rinput}}\nonumber\\
 &\leq N_\linput^{2^k } + N_\rinput\label{eq:sumcoeff-plus-upper}\\
 &\leq N^{2^k}.\nonumber
 \end{align}
-Similar to the $\circmult$ case, \cref{eq:sumcoeff-plus-upper} upperbounds its LHS by the fact that the maximum base and exponent combination is always greater than or equal to the sum of lower base/exponent combinations.  The final equality is true given the constraint over the inputs.  Note that a symmetrical argument applies for the case when $N_\rinput > N_\linput$.
+Similar to the $\circmult$ case, \cref{eq:sumcoeff-plus-upper} upperbounds its LHS by the fact that the maximum base and exponent combination is always greater than or equal to the sum of lower base/exponent combinations.  The final equality is true given the constraint over the inputs.  
 
-Since $\abs{\circuit}(1,\ldots, 1) \leq N^{2^k}$ for all queries with the exception of subquery elimnation in recursive datalog, then $\log{N^{2^k}} = 2^k \cdot \log{N}$ which for fixed $k$ yields the desired $O(\log{N})$ bits for $O(1)$ arithmetic operations for the given query class.
+Since $\abs{\circuit}(1,\ldots, 1) \leq N^{2^k}$ for all circuits such that all $\circplus$ gates share at most one gate with their sibling (across their respective subcircuits), then $\log{N^{2^k}} = 2^k \cdot \log{N}$ which for fixed $k$ yields the desired $O(\log{N})$ bits for $O(1)$ arithmetic operations.% for the given query class.
 \end{proof} 
 
 \subsection{\sampmon Notes}
-\revision{
 While we would like to take advantage of the space efficiency gained in using a circuit \circuit instead an expression tree \etree, we do not know that such a method exists when computing a sample of the input polynomial representation.  
 
 The efficiency gains of circuits over trees is found in the capability of circuits to only require space for each \emph{distinct} term in the compressed representation.  This saves space in such polynomials containing non-distinct terms multiplied or added to each other, e.g., $x^4$.  However, to avoid biased sampling, it is imperative to sample from both inputs of a multiplication gate, independently, which is indeed the approach of \sampmon.  
-%When we perform separate, independent sampling of both inputs, the result is the same as performing the sampling computation over the space inefficient expression tree.  (Note that an expression tree \etree is a special case of a circuit with the restriction of one output per node.)  However, this doesn't harm us, since as we show, that the bounded run time is not dependent on the size of the equivalent expression tree of the input circuit \circuit, but rather on the expression tree's depth, which is the same as the depth of \circuit.
-
-}
 
 \subsection{Proof of~\Cref{lem:sample}}\label{sec:proof-sample-monom}
 We first need to show that $\sampmon$ indeed returns a monomial $\monom$,\footnote{Technically it returns $\var(\monom)$ but for less cumbersome notation we will refer to $\var(\monom)$ simply by $\monom$ in this proof.} such that $(\monom, \coef)$ is in $\expansion{\circuit}$, which we do by induction on the depth of $\circuit$.
@@ -701,7 +697,7 @@ and we obtain the desired result.
 
 
 \paragraph{Run-time Analysis}
-It is easy to check that except for~\Cref{alg:sample-times-union}, all lines take $O(1)$ time.  Now consider an execution of~\Cref{alg:sample-times-union}. We note that we will be adding a given set of variables to some set at most once: since the sum of the sizes of the sets at a given level is at most $\degree(\circuit)$, each gate visited takes $O(\log{\degree(\circuit)})$.  Deote \cost(\circuit) (\Cref{eq:cost-sampmon}) to be an upper bound of the number of nodes visited by \sampmon.  Then the runtime is $O\left(\cost(\circuit)\cdot \log{\degree(\circuit)}\right)$.
+It is easy to check that except for lines~\ref{alg:sample-times-union} and~\ref{alg:sample-plus-bsamp}, all lines take $O(1)$ time.  For \cref{alg:sample-times-uinon}, consider an execution of~\Cref{alg:sample-times-union}. We note that we will be adding a given set of variables to some set at most once: since the sum of the sizes of the sets at a given level is at most $\degree(\circuit)$, each gate visited takes $O(\log{\degree(\circuit)})$.  For \cref{alg:sample-plus-bsamp} we have $> O(1)$ time when $\abs{\circuit}(1,\ldots, 1) > \size(\circuit)$.  when this is the case that for each sample, we have $\frac{\log{\abs{\circuit}(1,\ldots, 1)}}{\log{\size(\circuit)}}$ operations, since we need to read in and then compare numbers of of $\log{{\abs{\circuit}(1,\ldots, 1)}}$ bits.  Denote \cost(\circuit) (\Cref{eq:cost-sampmon}) to be an upper bound of the number of nodes visited by \sampmon.  Then the runtime is $O\left(\cost(\circuit)\cdot \log{\degree(\circuit)}\cdot\frac{\log{\abs{\circuit}(1,\ldots, 1)}}{\log{\size(\circuit)}}\right)$.
 
 We now bound the number of recursive calls in $\sampmon$ by $O\left((\degree(\circuit) + 1)\right.$$\left.\cdot\right.$ $\left.\depth(\circuit)\right)$.  
 
@@ -787,7 +783,7 @@ As in the $\circmult$ case the \emph{reduced} invariant of \reduce implies that
 
 Similar to the case of $\circuit.\type = \circmult$, (\ref{eq:plus-rhs}) follows by equations $(\ref{eq:cost-sampmon})$ and $(\ref{eq:ih-bound-cost})$.
 
-This proves (\ref{eq:strict-upper-bound}) for the $\circplus$ case and thus the claimed $O(k\log{k}\cdot\depth(\circuit))$ runtime for $k = \degree(\circuit)$ follows.
+This proves (\ref{eq:strict-upper-bound}) for the $\circplus$ case and thus the claimed $O(k\log{k}\cdot \frac{\log{\abs{\circuit}(1,\ldots, 1)}}{\size(\circuit)}\cdot\depth(\circuit))$ runtime for $k = \degree(\circuit)$ follows.
 
 
 \subsection{Experimental Results}\label{app:subsec:experiment}

main.vtc

@@ -1 +1 @@
-\contitem\title{Standard Operating Procedure in Bag PDBs Queries Considered Harmful}\author{Su Feng, Boris Glavic, Aaron Huber, Oliver Kennedy, and Atri Rudra}\page{23:1--23:46}
+\contitem\title{Standard Operating Procedure in Bag PDBs Queries Considered Harmful}\author{Su Feng, Boris Glavic, Aaron Huber, Oliver Kennedy, and Atri Rudra}\page{23:1--23:47}