Done with pass on OnePass appendix

app_one-pass-analysis.tex

@@ -9,7 +9,7 @@ Please note that it is \textit{assumed} that the original call to \onepass consi
 A circuit \circuit is equivalent to a circuit \circuit' if and only if $\polyf(\circuit) = \polyf(\circuit')$.
 \end{Definition}
 
-For technical reasons, we require the invariant that every subcircuit \subcircuit corresponding to an internal gate of \circuit has $\degree\left(\subcircuit\right) \geq 1$.  To ensure this, auxiliary algorithm ~\ref{alg:reduce} (\reduce) is called to perform any rewrites to \circuit, where an equivalent circuit \circuit' is created and returned by iteratively combining non-variable leaf nodes bottom-up until a parent node is reached which has an input \subcircuit whose subcircuit contains at least one leaf of type \var.  It is trivial to see in such a case that $\subcircuit \equiv \subcircuit'$, and this implies $\circuit \equiv \circuit'$.
+For technical reasons, we require the invariant that every subcircuit \subcircuit corresponding to an internal gate of \circuit has $\degree\left(\subcircuit\right) \geq 1$.  \revision{\textbf{AARON:} This is now trivially satisfied by the new definition of $\deg(\circuit)$ so please update this part to remove the stuff on $\reduce$. --Atri} To ensure this, auxiliary algorithm ~\ref{alg:reduce} (\reduce) is called to perform any rewrites to \circuit, where an equivalent circuit \circuit' is created and returned by iteratively combining non-variable leaf nodes bottom-up until a parent node is reached which has an input \subcircuit whose subcircuit contains at least one leaf of type \var.  It is trivial to see in such a case that $\subcircuit \equiv \subcircuit'$, and this implies $\circuit \equiv \circuit'$.
 
 \begin{Lemma}\label{lem:reduce}
 In $O(\size(\circuit))$, algorithm \reduce inspects input circuit \circuit and outputs an equivalent version \circuit' of \circuit such that all subcircuits \subcircuit of \circuit' have $\degree(\subcircuit) \geq 1$.
@@ -171,7 +171,8 @@ 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.  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 $O\left(\size(\circuit)\cdot \frac{\log{\abs{\circuit}(1,\ldots, 1)}}{\log{\size(\circuit)}}\right)$.
+It is known that $\topord(G)$ is computable in linear time.  Next, each of the $\size(\circuit)$ iterations of the loop in ~\Cref{alg:one-pass-loop} take $O\left( \multc{\log\left(\abs{\circuit(1\ldots, 1)}\right)}{\log{\size(\circuit)}}\right)$ time.  It is easy to see that all numbers that the algorithm computes is at most $\abs{\circuit}(1,\dots,1)$. Hence, by definition each such operation takes $\multc{\log\left(\abs{\circuit(1\ldots, 1)}\right)}{\log{\size(\circuit)}}$ time, which proves the claimed runtime.
+%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 $O\left(\size(\circuit)\cdot \frac{\log{\abs{\circuit}(1,\ldots, 1)}}{\log{\size(\circuit)}}\right)$.
 
 
 %%%Moved the stuff below to earlier in the appendix