Still working on S4 appendix

app_approx-alg-analysis.tex

@@ -70,66 +70,68 @@ We will prove~\Cref{lem:val-ub} by considering the three cases separetly. We fir
 \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
-\[\abs{\circuit}(1,\dots,1)\le \left(\size(\circuit)\right)^{2^{\depth(\circuit)}}.\]
+\[\abs{\circuit}(1,\dots,1)\le \left(\size(\circuit)\right)^{\degree(\circuit)+1}.\]
 \end{Lemma}
 \begin{proof}%[Proof of $\abs{\circuit}(1,\ldots, 1)$ is size $O(N)$]
-For notational simplcity define $N=\size(\circuit)$ and $k=\depth(\circuit)$.
-To prove this result, we by prove by induction on $k$ that $\abs{\circuit}(1,\ldots, 1) \leq N^{2^k }$.
-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 $\abs{\circuit}(1,\ldots, 1) = 1 \leq N^{2^k} = 1^{2^0} = 1$.
+For notational simplcity define $N=\size(\circuit)$ and $k=\degree(\circuit)$.
+To prove this result, we by prove by induction on $\depth(\circuit)$ 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) of $1$.  In this case, $\abs{\circuit}(1,\ldots, 1) = 1$, and \size(\circuit) $= 1$, and it is true that $\abs{\circuit}(1,\ldots, 1) = 1 \leq N^{k+1} = 1^{1} = 1$.
 
-Assume for $\ell > 0$ an arbitrary circuit \circuit of $\depth(\circuit) \leq \ell$ that it is true that $\abs{\circuit}(1,\ldots, 1) \leq N^{2^\ell }$.% for $k \geq 1$ when \depth(C) $\geq 1$.
+Assume for $\ell > 0$ an arbitrary circuit \circuit of $\depth(\circuit) \leq \ell$ that it is true that $\abs{\circuit}(1,\ldots, 1) \leq N^{\deg(\circuit)+1 }$.% for $k \geq 1$ when \depth(C) $\geq 1$.
 
 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.  Consider when sink node is $\circmult$.  Let $k_\linput, k_\rinput$ denote \degree($\circuit_\linput$) and \degree($\circuit_\rinput$) respectively.  %Note that this case does not require the constraint on $N_\linput$ or $N_\rinput$.
-In this case we do not use the fact that $\circuit$ is a tree and just assume that $N_\linput,N_\rinput\le N-1$. Then note that
+%In this case we do not use the fact that $\circuit$ is a tree and just assume that $N_\linput,N_\rinput\le N-1$. 
+ Then note that
 \begin{align}
 \abs{\circuit}(1,\ldots, 1) &= \abs{\circuit_\linput}(1,\ldots, 1)\circmult \abs{\circuit_\rinput}(1,\ldots, 1) \nonumber\\
-&\leq (N-1)^{2^{k_\linput}} \circmult (N - 1)^{2^{k_\rinput}}\nonumber\\
- &\leq (N-1)^{2^{k}}\label{eq:sumcoeff-times-upper}\\
+&\leq (N-1)^{k_\linput+1} \circmult (N - 1)^{k_\rinput+1}\nonumber\\
+ &= (N-1)^{k+1}\label{eq:sumcoeff-times-upper}\\
  &\leq N^{2^k}.\nonumber
 \end{align}
 %We derive the upperbound of \cref{eq:sumcoeff-times-upper} by noting that the maximum value of the LHS occurs when both the base and exponent are maximized.
-In the above the first inequality follows from the inductive hypothesis and \cref{eq:sumcoeff-times-upper} follows by nothing that for $\times$ node we have $k=k_\linput+k_\rinput$.
+In the above the first inequality follows from the inductive hypothesis (and the fact that $N_\linput,N_\rinput\le N-1$) and \cref{eq:sumcoeff-times-upper} follows by nothing that for $\times$ node we have $k=k_\linput+k_\rinput+1$.
 
 For the case when the sink node is a $\circplus$ node, then 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^{2^{k_\linput}} + N_\rinput^{2^{k_\rinput}}\nonumber\\
-&\leq (N-1)^{2^k } \label{eq:sumcoeff-plus-upper}\\
-&\leq N^{2^k}.\nonumber
+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 hypothesis while the second inequality follows from the fact that since $\circuit$ is a tree we have $N_\linput+N_\rinput=N-1$ and the fact that $0\le k_\linput,k_\rinput\le k$. This compeletes the proof.
+In the above, the first inequality follows from the inductive hypothesis (and the fact that $k_\linput,k_\rinput\le k$)  while the second inequality follows from the fact that since $\circuit$ is a tree we have $N_\linput+N_\rinput=N-1$ and the fact that $k\ge 0$. This compeletes the proof.
 %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 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}
 
-\revision{\textbf{THE PART BELOW NEEDS WORK. --Atri}}
+%\revision{\textbf{THE PART BELOW NEEDS WORK. --Atri}}
 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. % tree (i.e. the sub-circuits corresponding to two children of a node in $\circuit$ are completely disjoint). 
 Then we have
-\[\abs{\circuit}(1,\dots,1)\le 2^{\depth(\circuit)\cdot \size(\circuit)}.\]
+\[\abs{\circuit}(1,\dots,1)\le 2^{2^{\degree(\circuit)}\cdot \size(\circuit)}.\]
 \end{Lemma}
 \begin{proof}[Proof Sketch]
-We use the same notation as in the proof of~\Cref{lem:C-ub-tree}. We will prove by induction on $k$ that $\abs{\circuit}(1,\ldots, 1) \leq 2^{(k+1)N }$. The base case argument is similar to that in the proof of~\Cref{lem:C-ub-tree}. In the inductive case we have that $N_\linput,N_\rinput\le N-1$.
+We use the same notation as in the proof of~\Cref{lem:C-ub-tree}. We will prove by induction on $\depth(\circuit)$ that $\abs{\circuit}(1,\ldots, 1) \leq 2^{2^k\cdot N }$. The base case argument is similar to that in the proof of~\Cref{lem:C-ub-tree}. In the inductive case we have that $N_\linput,N_\rinput\le N-1$.
 
 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^{(k_\linput+1)\cdot N_\linput}} \circmult {2^{(k_\rinput+1)\cdot N_\rinput}}\\
- &\leq {2\cdot 2^{(\max(k_\linput,k_\rinput)+1)(N-1)}}\\
- &\leq 2^{(k+1) N}.
+&\leq {2^{2^{k_\linput}\cdot N_\linput}} \circmult {2^{2^{k_\rinput}\cdot N_\rinput}}\\
+ &\leq 2^{2\cdot 2^{k-1}\cdot (N-1)}\\
+ &\leq 2^{2^k N}.
 \end{align*}
-In the above the first inequality follows from inductive hypothesis while the third inequality follows from the fact that $k_\linput+k_\rinput=k$ (and hence $\max(k_\linput,k_\rinput)\le k$) as well as the fact that $k\ge 0$.
+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 $N_\linput, N_\rinput\le N-1$.
+%$k_\linput+k_\rinput=k$ (and hence $\max(k_\linput,k_\rinput)\le k$) as well as the fact that $k\ge 0$.
 
 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^{(k_\linput+1)\cdot N_\linput} + 2^{(k_\rinput+1)\cdot N_\rinput}\\
-&\leq 2\cdot {2^{(k+1)(N-1)} } \\
-&\leq 2^{(k+1)N}.
+&\leq 2^{2^{k_\linput}\cdot N_\linput} + 2^{2^{k_\rinput}\cdot N_\rinput}\\
+&\leq 2\cdot {2^{2^k(N-1)} } \\
+&\leq 2^{2^kN}.
 \end{align*}
-In the above the first inequality follows from the inductive hypothesis while the second inequality follows from the fact that $k_\linput,k_\rinput\le k$. The final inequality follows from the fact that $k\ge 0$.
+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 $N_\linput,N_\rinput\le N-1$. The final inequality follows from the fact that $k\ge 0$.
 \end{proof}

approx_alg.tex

@@ -150,7 +150,7 @@ For any circuit $\circuit$ with $\degree(\circuit)=k$, we have
 Further, under the following conditions:
 \begin{enumerate}
 \item $\circuit$ is a tree,
-\item $\circuit$ is the output of a FAQ query from algorithm in~\cite{DBLP:conf/pods/KhamisNR16},
+\item $\circuit$ encodes the run of the algorithm in~\cite{DBLP:conf/pods/KhamisNR16} on an FAQ query,
 \end{enumerate}
 we have
 \[\abs{\circuit}(1,\ldots, 1)\le  \size(\circuit)^{O(k)}.\]