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Appendix (formerly C) D pass...almost complete.

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Aaron Huber 2021-09-20 12:25:29 -04:00
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23
app_approx-alg-analysis.tex
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@ -18,6 +18,7 @@ such that
\end{equation}
\end{Theorem}
\AR{\textbf{Aaron:}Just copied over from S4. The above text might need smoothening into the appendix.}
The slight abuse of notation seen in $\abs{\circuit}\inparen{1,\ldots,1}$ is explained after \Cref{def:positive-circuit} and an example is given in \Cref{ex:def-pos-circ}. The only difference in the use of this notation in \Cref{lem:approx-alg} is that we include an additional exponent to square the quantity.
\subsection{Proof of Theorem \ref{lem:approx-alg}}\label{sec:proof-lem-approx-alg}
\input{app_approx-alg-pseudo-code}
@ -28,7 +29,7 @@ The $\onepass$ function completes in time:
$$O\left(\size(\circuit) \cdot \multc{\log\left(\abs{\circuit(1\ldots, 1)}\right)}{\log{\size(\circuit}}\right)$$
$\onepass$ 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 ($\circuit_{\vari{mod}}$), $\circuit_{\vari{mod}}.\vari{partial}=\abs{\circuit}(1,\dots,1)$.
To prove correctness of \Cref{alg:mon-sam}, we only use the following fact that follows from the above lemma: for the modified circuit ($\circuit_{\vari{mod}}$) output by \onepass, $\circuit_{\vari{mod}}.\vari{partial}=\abs{\circuit}(1,\dots,1)$.
\begin{Lemma}\label{lem:sample}
The function $\sampmon$ completes in time
@ -38,7 +39,9 @@ $$O(\log{k} \cdot k \cdot \depth(\circuit)\cdot\multc{\log\left(\abs{\circuit}(1
With the above two lemmas, we are ready to argue the following result: % (proof in \Cref{sec:proofs-approx-alg}):
\begin{Theorem}\label{lem:mon-samp}
For any $\circuit$ with $\degree(poly(|\circuit|)) = k$, algorithm \ref{alg:mon-sam} outputs an estimate $\vari{acc}$ of $\rpoly(\prob_1,\ldots, \prob_\numvar)$ such that
For any $\circuit$ with
\AH{Pretty sure this is $\degree\inparen{\abs{\circuit}}$. Have to read on to be sure.}
$\degree(poly(|\circuit|)) = k$, algorithm \ref{alg:mon-sam} outputs an estimate $\vari{acc}$ of $\rpoly(\prob_1,\ldots, \prob_\numvar)$ such that
\[\probOf\left(\left|\vari{acc} - \rpoly(\prob_1,\ldots, \prob_\numvar)\right|> \error \cdot \abs{\circuit}(1,\ldots, 1)\right) \leq \conf,\]
in $O\left(\left(\size(\circuit)+\frac{\log{\frac{1}{\conf}}}{\error^2} \cdot k \cdot\log{k} \cdot \depth(\circuit)\right)\cdot \multc{\log\left(\abs{\circuit}(1,\ldots, 1)\right)}{\log{\size(\circuit)}}\right)$ time.
\end{Theorem}
@ -63,12 +66,12 @@ The claim on the runtime follows from \Cref{lem:mon-samp} since
Let us now prove \Cref{lem:mon-samp}:
\subsection{Proof of Theorem \ref{lem:mon-samp}}\label{app:subsec-th-mon-samp}
\begin{proof}
Consider now the random variables $\randvar_1,\dots,\randvar_\numsamp$, where each $\randvar_\vari{i}$ is the value of $\vari{Y}_{\vari{i}}$ in \cref{alg:mon-sam} after \cref{alg:mon-sam-product} is executed. In particular, note that we have
\[\randvar_\vari{i}= \onesymbol\inparen{\monom\mod{\mathcal{B}}\not\equiv 0}\cdot \prod_{X_i\in \var\inparen{v}} p_i,\]
Consider now the random variables $\randvar_1,\dots,\randvar_\numsamp$, where each $\randvar_\vari{i}$ is the value of $\vari{Y}_{\vari{i}}$ in \cref{alg:mon-sam} after \cref{alg:mon-sam-product} is executed. Overloading $\isInd{\cdot}$ to receive monomial input (recall $\encMon$ is the monomial composed of the variables in the set $\monom$), we have
\[\randvar_\vari{i}= \indicator{\inparen{\isInd{\encMon}}}\cdot \prod_{X_i\in \var\inparen{v}} p_i,\]
where the indicator variable handles the check in \Cref{alg:check-duplicate-block}
Then for random variable $\randvar_i$, it is the case that
\begin{align*}
\expct\pbox{\randvar_\vari{i}} &= \sum\limits_{(\monom, \coef) \in \expansion{{\circuit}} }\frac{\onesymbol\inparen{\encMon\mod{\mathcal{B}}\not\equiv 0}\cdot c\cdot\prod_{X_i\in \var\inparen{v}} p_i }{\abs{{\circuit}}(1,\dots,1)} \\
\expct\pbox{\randvar_\vari{i}} &= \sum\limits_{(\monom, \coef) \in \expansion{{\circuit}} }\frac{\indicator{\inparen{\isInd{\encMon}}}\cdot c\cdot\prod_{X_i\in \var\inparen{v}} p_i }{\abs{{\circuit}}(1,\dots,1)} \\
&= \frac{\rpoly(\prob_1,\ldots, \prob_\numvar)}{\abs{{\circuit}}(1,\ldots, 1)},
\end{align*}
where in the first equality we use the fact that $\vari{sgn}_{\vari{i}}\cdot \abs{\coef}=\coef$ and the second equality follows from \Cref{eq:tilde-Q-bi} with $X_i$ substituted by $\prob_i$.
@ -96,7 +99,7 @@ For the claimed probability bound of $\probOf\left(\left|\vari{acc} - \rpoly(\pr
This concludes the proof for the first claim of theorem~\ref{lem:mon-samp}. Next, we prove the claim on the runtime.
\paragraph*{Run-time Analysis}
The runtime of the algorithm is dominated first by \Cref{alg:mon-sam-onepass} (which by \Cref{lem:one-pass} takes time $O\left({\size(\circuit)}\cdot \multc{\log\left(\abs{\circuit}^2(1,\ldots, 1)\right)}{\log\left(\size(\circuit)\right)}\right)$) and then by $\samplesize$ iterations of the loop in \Cref{alg:sampling-loop}. Each iteration's run time is dominated by the call to \sampmon in \Cref{alg:mon-sam-sample} (which by \Cref{lem:sample} takes $O\left(\log{k} \cdot k \cdot {\depth(\circuit)}\cdot \multc{\log\left(\abs{\circuit}^2(1,\ldots, 1)\right)}{\log\left(\size(\circuit)\right)}\right)$
The runtime of the algorithm is dominated first by \Cref{alg:mon-sam-onepass} (which by \Cref{lem:one-pass} takes time $O\left({\size(\circuit)}\cdot \multc{\log\left(\abs{\circuit}(1,\ldots, 1)\right)}{\log\left(\size(\circuit)\right)}\right)$) and then by $\samplesize$ iterations of the loop in \Cref{alg:sampling-loop}. Each iteration's run time is dominated by the call to \sampmon in \Cref{alg:mon-sam-sample} (which by \Cref{lem:sample} takes $O\left(\log{k} \cdot k \cdot {\depth(\circuit)}\cdot \multc{\log\left(\abs{\circuit}(1,\ldots, 1)\right)}{\log\left(\size(\circuit)\right)}\right)$
) and the check \Cref{alg:check-duplicate-block}, which by the subsequent argument takes $O(k\log{k})$ time. We sort the $O(k)$ variables by their block IDs and then check if there is a duplicate block ID or not. Combining all the times discussed here gives us the desired overall runtime.
\qed
\end{proof}
@ -108,7 +111,7 @@ The result follows by first noting that by definition of $\gamma$, we have
Further, since each $\prob_i\ge \prob_0$ and $\poly(\vct{X})$ (and hence $\rpoly(\vct{X})$) has degree at most $k$, we have that
\[ \rpoly(1,\dots,1) \ge \prob_0^k\cdot \rpoly(1,\dots,1).\]
The above two inequalities implies $\rpoly(1,\dots,1) \ge \prob_0^k\cdot (1-\gamma)\cdot \abs{{\circuit}}(1,\dots,1)$.
Applying this bound in the runtime bound in \Cref{lem:approx-alg} gives the first claimed runtime. The final runtime of $O_k\left(\frac 1{\eps^2}\cdot\size(\circuit)\cdot \log{\frac{1}{\conf}}\cdot \multc{\log\left(\abs{\circuit}^2(1,\ldots, 1)\right)}{\log\left(\size(\circuit)\right)}\right)$ follows by noting that $\depth({\circuit})\le \size({\circuit})$ and absorbing all factors that just depend on $k$.
Applying this bound in the runtime bound in \Cref{lem:approx-alg} gives the first claimed runtime. The final runtime of $O_k\left(\frac 1{\eps^2}\cdot\size(\circuit)\cdot \log{\frac{1}{\conf}}\cdot \multc{\log\left(\abs{\circuit}(1,\ldots, 1)\right)}{\log\left(\size(\circuit)\right)}\right)$ follows by noting that $\depth({\circuit})\le \size({\circuit})$ and absorbing all factors that just depend on $k$.
\qed
\end{proof}
@ -127,9 +130,9 @@ Let $\circuit$ be a tree (i.e. the sub-circuits corresponding to two children of
\begin{proof}[Proof of \Cref{lem:C-ub-tree}]%[Proof of $\abs{\circuit}(1,\ldots, 1)$ is size $O(N)$]
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) of $1$. In this case, $\abs{\circuit}(1,\ldots, 1) = 1$, and \size(\circuit) $= 1$, and it is true that $\abs{\circuit}(1,\ldots, 1) = 1 \leq N^{k+1} = 1^{1} = 1$.
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 $\abs{\circuit}(1,\ldots, 1) \leq N^{\deg(\circuit)+1 }$.% 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 $\abs{\circuit}(1,\ldots, 1) \leq N^{k+1 }$.% 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. Let $k_\linput, k_\rinput$ denote \degree($\circuit_\linput$) and \degree($\circuit_\rinput$) respectively. Consider when sink node is $\circmult$.
Then note that
@ -140,7 +143,7 @@ For the inductive step we consider a circuit \circuit such that $\depth(\circuit
&\leq N^{k + 1}.\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 the fact that the size of either subtree is at most $N-1$) and \Cref{eq:sumcoeff-times-upper} follows by noting by \cref{def:degree} for $k = \degree(\circuit)$ we have $k=k_\linput+k_\rinput+1$.
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}

9
app_approx-alg-defs-and-examples.tex
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@ -10,20 +10,21 @@ The pure expansion of a polynomial $\poly$ is formed by computing all product of
\end{Definition}
Note that similar in spirit to \Cref{def:reduced-bi-poly}, $\expansion{\circuit}$ \Cref{def:expand-circuit} reduces all variable exponents $e > 1$ to $e = 1$. Further, it is true that $\expansion{\circuit}$ is the pure expansion of $\circuit$.
In the following, recall by \cref{def:expand-circuit} that we use $\encMon$ to denote the monomial composed of the variables in $\monom$.
%In the following, recall by \cref{def:expand-circuit} that we use $\encMon$ to denote the monomial composed of the variables in $\monom$.
\begin{Example}[Example for \Cref{def:expand-circuit}]\label{example:expr-tree-T}
\begin{Example}[Example of Pure Expansion]\label{example:expr-tree-T}
Consider the factorized representation $(X+ 2Y)(2X - Y)$ of the polynomial in \Cref{eq:poly-eg}.
Its circuit $\circuit$ is illustrated in \Cref{fig:circuit}.
The pure expansion of the product is $2X^2 - XY + 4XY - 2Y^2$ and the $\expansion{\circuit}$ is $[(X, 2), (XY, -1), (XY, 4), (Y, -2)]$.
The pure expansion of the product is $2X^2 - XY + 4XY - 2Y^2$. As an additional example of \Cref{def:expand-circuit}, $\expansion{\circuit}=[(X, 2), (XY, -1), (XY, 4), (Y, -2)]$.
\end{Example}
$\expansion{\circuit}$ effectively\footnote{The minor difference here is that $\expansion{\circuit}$ encodes the \emph{reduced} form over the SOP pure expansion of the compressed representation, as opposed to the \abbrSMB representation} encodes the \emph{reduced} form of $\polyf\inparen{\circuit}$, decoupling each monomial into a set of variables $\monom$ and a real coefficient $\coef$.
However, unlike the constraint on the input $\poly$ to compute $\rpoly$, the input circuit $\circuit$ does not need to be in \abbrSMB/SOP form.
\begin{Example}[Example for \Cref{def:positive-circuit}]\label{ex:def-pos-circ}
Using the same factorization from \Cref{example:expr-tree-T}, $\polyf(\abs{\circuit}) = (X + 2Y)(2X + Y) = 2X^2 +XY +4XY + 2Y^2 = 2X^2 + 5XY + 2Y^2$. Note that this \textit{is not} the same as the polynomial from \Cref{eq:poly-eg}.
Using the same factorization from \Cref{example:expr-tree-T}, $\polyf(\abs{\circuit}) = (X + 2Y)(2X + Y) = 2X^2 +XY +4XY + 2Y^2 = 2X^2 + 5XY + 2Y^2$. Note that this \textit{is not} the same as the polynomial from \Cref{eq:poly-eg}. As an example of the slight abuse of notation we alluded to, $\polyf\inparen{\abs{\circuit}\inparen{1,\ldots, 1}} =2\inparen{1}^2 + 5\inparen{1}\inparen{1} + 2\inparen{1}^2 = 9$.
\end{Example}
\AH{Verify whether we need pure expansion or not.}
%\begin{Definition}[Evaluation]\label{def:exp-poly-eval}
%Given a circuit $\circuit$ and a valuation $\vct{a} \in \mathbb{R}^\numvar$, we define the evaluation of $\circuit$ on $\vct{a}$ as $\circuit(\vct{a}) = \polyf(\circuit)(\vct{a})$.
%\end{Definition}

21
app_one-pass-analysis.tex
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@ -39,7 +39,7 @@ Please note that it is \textit{assumed} that the original call to \onepass consi
\subsection{$\onepass$ Example}
\begin{Example}\label{example:one-pass}
Let $\etree$ encode the expression $(X_1 + X_2)(X_1 - X_2) + X_2^2$. After one pass, \Cref{alg:one-pass-iter} would have computed the following weight distribution. For the two inputs of the root $+$ node $\etree$, $\etree.\lwght = \frac{4}{5}$ and $\etree.\rwght = \frac{1}{5}$. Similarly, let $\stree$ denote the left-subtree of $\etree_{\lchild}$, $\stree.\lwght = \stree.\rwght = \frac{1}{2}$. This is depicted in \Cref{fig:expr-tree-T-wght}.
Let $\etree$ encode the expression $(X_1 + X_2)(X_1 - X_2) + X_2^2$. After one pass, \Cref{alg:one-pass-iter} would have computed the following weight distribution. For the two inputs of the sink gate $\circuit$, $\circuit.\lwght = \frac{4}{5}$ and $\circuit.\rwght = \frac{1}{5}$. Similarly, for $\stree$ denoting the left input of $\circuit_{\lchild}$, $\stree.\lwght = \stree.\rwght = \frac{1}{2}$. This is depicted in \Cref{fig:expr-tree-T-wght}.
\end{Example}
\begin{figure}[h!]
@ -117,10 +117,11 @@ Please note that it is \textit{assumed} that the original call to \onepass consi
\node[tree_node] (b3) at (1.6, 2.5) {$\boldsymbol{\circplus}$};
%Fourth level
\node[tree_node] (a4) at (1.5, 3.75) {$\boldsymbol{\circmult}$};
\node[tree_node] (b4) at (2.8, 3.75) {$\boldsymbol{\circplus}$};
\node[tree_node] (b4) at (2.8, 4) {$\boldsymbol{\circplus}$};
\node[above right=0.15cm of b4, inner sep=0pt, font=\bfseries](labelC) {$\circuit$};
\draw[->] (a1) edge[right] node{$\frac{1}{2}$} (a3);
\draw[->] (b1) -- (c2);
\draw[->] (b1) -- (b2);
\draw[->] (a1) -- (b2);
\draw[->] (a1) edge[bend left=15] (c2);
\draw[->] (a1) edge[bend right=15] (c2);
@ -131,6 +132,7 @@ Please note that it is \textit{assumed} that the original call to \onepass consi
\draw[->] (a3) -- (a4);
\draw[->] (b3) -- (a4);
\draw[->] (a4) edge[above] node{$\frac{4}{5}$} (b4);
\draw[black] (b4) -- (labelC);
% \draw[->] (b3) -- (a4);
% \draw[->] (a4) -- (2.25, 2.75);
\end{tikzpicture}
@ -174,21 +176,22 @@ Please note that it is \textit{assumed} that the original call to \onepass consi
\subsection{Proof of \onepass (\Cref{lem:one-pass})}\label{sec:proof-one-pass}
\begin{proof}%[Proof of \Cref{lem:one-pass}]
We prove the correct computation of \prt, \lwght, \rwght values on \circuit by induction over the number of iterations in the topological order \topord (line~\ref{alg:one-pass-loop}) of the input circuit \circuit. Note that \topord follows the standard definition of a topological ordering over the DAG structure of \circuit.
We prove the correct computation of \prt, \lwght, \rwght values on \circuit by induction over the number of iterations in the topological order \topord (line~\ref{alg:one-pass-loop}) of the input circuit \circuit. \topord follows the standard definition of a topological ordering over the DAG structure of \circuit.
For the base case, we have only one gate, which by definition is a source gate and must be either \var or \tnum. In this case, as per \cref{eq:T-all-ones}, lines~\ref{alg:one-pass-var} and~\ref{alg:one-pass-num} correctly compute \circuit.\prt as $1$ and \circuit.\val respectively.
For the base case, we have only one gate, which by definition is a source gate and must be either \var or \tnum. In this case, as per \cref{eq:T-all-ones}, lines~\ref{alg:one-pass-var} and~\ref{alg:one-pass-num} correctly compute \circuit.\prt as $1$.% and \circuit.\val respectively.
For the inductive hypothesis, assume that \onepass correctly computes \subcircuit.\prt, \subcircuit.\lwght, and \subcircuit.\rwght for all gates \gate in \circuit with $k \geq 0$ iterations over \topord.
\AH{Notes above: Algo uses Reduce, but we don't use that anymore. The figure needs to change to a circuit.}
We now prove for $k + 1$ iterations that \onepass correctly computes the \prt, \lwght, and \rwght values for each gate $\gate_\vari{i}$ in \circuit for $i \in [k + 1]$.
Note that the $\gate_\vari{k + 1}$ must be in the last ordering of all gates $\gate_\vari{i}$. Note that for \size(\circuit) > 1, if $\gate_{k+1}$ is a leaf node, we are back to the base case. Otherwise $\gate_{k + 1}$ is an internal node $\gate_\vari{s}.\type = \circplus$ or $\gate_\vari{s}.\type = \circmult$, which both require binary input.
The $\gate_\vari{k + 1}$ must be in the last ordering of all gates $\gate_\vari{i}$. When \size(\circuit) > 1, if $\gate_{k+1}$ is a leaf node, we are back to the base case. Otherwise $\gate_{k + 1}$ is an internal node %$\gate_\vari{s}.\type = \circplus$ or $\gate_\vari{s}.\type = \circmult$, which both require
which requires binary input.
When $\gate_{k+1}.\type = \circplus$, then by line~\ref{alg:one-pass-plus} $\gate_{k+1}$.\prt $= \gate_{{k+1}_\lchild}$.\prt $+ \gate_{{k+1}_\rchild}$.\prt, a correct computation, as per \cref{eq:T-all-ones}. Further, lines~\ref{alg:one-pass-lwght} and~\ref{alg:one-pass-rwght} compute $\gate_{{k+1}}.\lwght = \frac{\gate_{{k+1}_\lchild}.\prt}{\gate_{{k+1}}.\prt}$ and analogously for $\gate_{{k+1}}.\rwght$. Note that all values needed for each computation have been correctly computed by the inductive hypothesis.
When $\gate_{k+1}.\type = \circplus$, then by line~\ref{alg:one-pass-plus} $\gate_{k+1}$.\prt $= \gate_{{k+1}_\lchild}$.\prt $+ \gate_{{k+1}_\rchild}$.\prt, a correct computation, as per \cref{eq:T-all-ones}. Further, lines~\ref{alg:one-pass-lwght} and~\ref{alg:one-pass-rwght} compute $\gate_{{k+1}}.\lwght = \frac{\gate_{{k+1}_\lchild}.\prt}{\gate_{{k+1}}.\prt}$ and analogously for $\gate_{{k+1}}.\rwght$. All values needed for each computation have been correctly computed by the inductive hypothesis.
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 by \cref{eq:T-all-ones} is correct.
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 by \cref{eq:T-all-ones} is correct. This concludes the proof of correctness.
\paragraph*{Runtime Analysis}
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 each of all the numbers which 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.
It is known that $\topord(G)$ is computable in linear time. There are $\size(\circuit)$ iterations, each of which takes $O\left( \multc{\log\left(\abs{\circuit(1\ldots, 1)}\right)}{\log\inparen{\size(\circuit)}}\right)$ time. This can be seen since each of all the numbers which 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.
\qed
\end{proof}
%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 $O\left(\size(\circuit)\cdot \frac{\log{\abs{\circuit}(1,\ldots, 1)}}{\log{\size(\circuit)}}\right)$.