Fixed ~\ref in appendix.

app_approx-alg-analysis.tex

@@ -77,7 +77,7 @@ Hoeffding's inequality states that if we know that each $\randvar_i$ (which are
 \probOf\left(\left|\empmean - \expct\pbox{\empmean}\right| \geq \error\right) \leq 2\exp{\left(-\frac{2\samplesize^2\error^2}{\sum_{i = 1}^{\samplesize}(b_i -a_i)^2}\right)}.
 \end{equation*}
 
-Line ~\ref{alg:mon-sam-sample} shows that $\vari{sgn}_\vari{i}$ has a value in $\{-1, 1\}$ that is multiplied with $O(k)$ $\prob_i\in [0, 1]$, which implies the range for each $\randvar_i$ is $[-1, 1]$.
+Line~\ref{alg:mon-sam-sample} shows that $\vari{sgn}_\vari{i}$ has a value in $\{-1, 1\}$ that is multiplied with $O(k)$ $\prob_i\in [0, 1]$, which implies the range for each $\randvar_i$ is $[-1, 1]$.
 Using Hoeffding's inequality, we then get:
 \begin{equation*}
 \probOf\left(~\left| \empmean - \expct\pbox{\empmean} ~\right| \geq \error\right) \leq 2\exp{\left(-\frac{2\samplesize^2\error^2}{2^2 \samplesize}\right)} = 2\exp{\left(-\frac{\samplesize\error^2}{2 }\right)}\leq \conf,
@@ -86,7 +86,7 @@ where the last inequality follows from our choice of $\samplesize$ in \Cref{alg:
 
 For the claimed probability bound of $\probOf\left(\left|\vari{acc} - \rpoly(\prob_1,\ldots, \prob_\numvar)\right|> \error \cdot \abs{\circuit}(1,\ldots, 1)\right) \leq \conf$, note that in the algorithm, \vari{acc} is exactly $\empmean \cdot \abs{\circuit}(1,\ldots, 1)$.  Multiplying the rest of the terms by the same factor yields the said bound.
 
-This concludes the proof for the first claim of theorem ~\ref{lem:mon-samp}. We prove the claim  on the runtime next.
+This concludes the proof for the first claim of theorem~\ref{lem:mon-samp}. We prove the claim  on the runtime next.
 
 \paragraph*{Run-time Analysis}
 The runtime of the algorithm is dominated 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 the $\samplesize$ iterations of the loop in \Cref{alg:sampling-loop}. Each iteration's run time is dominated by the call to \Cref{alg:mon-sam-sample} (which by \Cref{lem:sample} takes $O\left(\log{k} \cdot k \cdot {\depth(\circuit)}\cdot \multc{\log\left(\abs{\circuit}^2(1,\ldots, 1)\right)}{\log\left(\size(\circuit)\right)}\right)$

app_notation-background.tex

@@ -131,7 +131,7 @@ Follows by the construction of $\rpoly$ in \cref{def:reduced-bi-poly}.
 \end{proof}
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 
-\subsection{Proposition ~\ref{proposition:q-qtilde}}\label{app:subsec-prop-q-qtilde}
+\subsection{Proposition~\ref{proposition:q-qtilde}}\label{app:subsec-prop-q-qtilde}
 \noindent Note the following fact:
 
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
@@ -148,7 +148,7 @@ Note that any $\poly$ in factorized form is equivalent to its \abbrSMB expansion
 
 
 
-\subsection{Proof for Lemma ~\ref{lem:exp-poly-rpoly}}
+\subsection{Proof for Lemma~\ref{lem:exp-poly-rpoly}}
 \begin{proof}
 Let $\poly$ be the generalized polynomial, i.e., the polynomial of $\numvar$ variables with highest degree $= B$: %, in which every possible monomial permutation appears,
 \[\poly(X_1,\ldots, X_\numvar) = \sum_{\vct{d} \in \{0,\ldots, B\}^\numvar}q_{\vct{d}}\cdot \prod_{\substack{i = 1\\s.t. d_i \geq 1}}^\numvar X_i^{d_i}\].
@@ -167,7 +167,7 @@ Finally, observe \Cref{p1-s5} by construction in \Cref{lem:pre-poly-rpoly}, that
 \end{proof}
 
 
-\subsection{Proof For Corollary ~\ref{cor:expct-sop}}
+\subsection{Proof For Corollary~\ref{cor:expct-sop}}
 \begin{proof}
 Note that \cref{lem:exp-poly-rpoly} shows that $\expct\pbox{\poly} =$ $\rpoly(\prob_1,\ldots, \prob_\numvar)$.  Therefore, if $\poly$ is already in \abbrSMB form, one only needs to compute $\poly(\prob_1,\ldots, \prob_\numvar)$ ignoring exponent terms (note that such a polynomial is $\rpoly(\prob_1,\ldots, \prob_\numvar)$), which indeed has $O(\smbOf{|\poly|})$ computations.
 \qed

app_one-pass-analysis.tex

@@ -13,7 +13,7 @@ Please note that it is \textit{assumed} that the original call to \onepass consi
 %\end{Definition}
 
 
-%For technical reasons, we require the invariant that every subcircuit \subcircuit corresponding to an internal gate of \circuit has $\degree\left(\subcircuit\right) \geq 1$.  \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 $\circuit \equiv \circuit'$.
+%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 $\degree(\subcircuit) \geq 1$.
@@ -171,9 +171,9 @@ For the inductive hypothesis, assume that \onepass correctly computes \subcircui
 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 $\gate_\vari{i}$.  It is also the case that $\gate_{k+1}$ has  two inputs.  Finally, 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 $\gate_\vari{s}.\type = \circmult$.
 
-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 $\gate_{{k+1}}.\rwght$.  Note that 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 is correct, as per \Cref{eq:T-all-ones}.
+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*{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.

app_samp-monom-analysis.tex

@@ -22,9 +22,9 @@ For the base case, let the depth $d$ of $\circuit$ be $0$.  We have that the roo
 
 For the inductive hypothesis, assume that for $d \leq k$ for some $k \geq 0$, that it is indeed the case that $\sampmon$ returns a monomial.
 
-For the inductive step, let us take a circuit $\circuit$ with $d = k + 1$.  Note that each input has depth $d \leq k$, and by inductive hypothesis both of them return a valid monomial.  Then the root can be either a $\circplus$ or $\circmult$ node.  For the case of a $\circplus$ root node, line~\ref{alg:sample-plus-bsamp} of $\sampmon$ will choose one of the inputs of the root.  By inductive hypothesis it is the case that a monomial in \expansion{\circuit} is being returned from either input.  Then it follows that for the case of $+$ root node a valid monomial is returned by $\sampmon$.  When the root is a $\circmult$ node, line~\ref{alg:sample-times-union} %and ~\ref{alg:sample-times-product} multiply
+For the inductive step, let us take a circuit $\circuit$ with $d = k + 1$.  Note that each input has depth $d \leq k$, and by inductive hypothesis both of them return a valid monomial.  Then the root can be either a $\circplus$ or $\circmult$ node.  For the case of a $\circplus$ root node, line~\ref{alg:sample-plus-bsamp} of $\sampmon$ will choose one of the inputs of the root.  By inductive hypothesis it is the case that a monomial in \expansion{\circuit} is being returned from either input.  Then it follows that for the case of $+$ root node a valid monomial is returned by $\sampmon$.  When the root is a $\circmult$ node, line~\ref{alg:sample-times-union} %and~\ref{alg:sample-times-product} multiply
 computes the set union of the monomials returned by the two inputs of the root, and it is trivial to see
-%by definition ~\ref{def:monomial}
+%by definition~\ref{def:monomial}
 %the product of two monomials is also a monomial, and
 by \Cref{def:expand-circuit} that \monom is a valid monomial in some $(\monom, \coef) \in \expansion{\circuit}$.
 
@@ -33,7 +33,7 @@ We will next prove by induction on the depth $d$ of $\circuit$ that the $(\monom
 
 For the base case $d = 0$, by definition~\ref{def:circuit} we know that the root has to be either a coefficient or a variable.  For either case, the probability of the value returned is $1$ since there is only one value to sample from.  When the root is a variable $x$ the algorithm correctly returns $(\{x\}, 1 )$.  When the root is a coefficient, \sampmon ~correctly returns $(\{~\}, sign(\coef_i))$.
 
-For the inductive hypothesis, assume that for $d \leq k$ and $k \geq 0$ $\sampmon$ indeed samples $\monom$ in $(\monom, \coef)$ in $\expansion{\circuit}$ with probability $\frac{|\coef|}{\abs{\circuit}\polyinput{1}{1}}$.%bove is true.%lemma ~\ref{lem:sample} is true.
+For the inductive hypothesis, assume that for $d \leq k$ and $k \geq 0$ $\sampmon$ indeed samples $\monom$ in $(\monom, \coef)$ in $\expansion{\circuit}$ with probability $\frac{|\coef|}{\abs{\circuit}\polyinput{1}{1}}$.%bove is true.%lemma~\ref{lem:sample} is true.
 
 We prove now for $d = k + 1$ the inductive step holds.  It is the case that the root of $\circuit$ has up to two inputs $\circuit_\linput$ and $\circuit_\rinput$.  Since $\circuit_\linput$ and $\circuit_\rinput$ are both depth $d \leq k$, by inductive hypothesis, $\sampmon$ will sample both monomials $\monom_\lchild$ in $(\monom_\lchild, \coef_\lchild)$ of $\expansion{\circuit_\linput}$ and $\monom_\rchild$ in $(\monom_\rchild, \coef_\rchild)$ of $\expansion{\circuit_\rinput}$, from $\circuit_\linput$ and $\circuit_\rinput$ with probability $\frac{|\coef_\lchild|}{\abs{\circuit_\linput}\polyinput{1}{1}}$ and $\frac{|\coef_\rchild|}{\abs{\circuit_\rinput}\polyinput{1}{1}}$.
 
@@ -70,7 +70,7 @@ Let \cost$(\cdot)$ be a function that models an upper bound on the number of gat
 		\end{cases}\label{eq:cost-sampmon}
 \end{equation}
 
-First note that the number of gates visited in \sampmon is $\leq\cost(\circuit)$.  To show that \Cref{eq:cost-sampmon} upper bounds the number of nodes visited by \sampmon, note that when \sampmon visits a gate such that \circuit.\type $ =\circmult$, line ~\ref{alg:sample-times-for-loop} visits each input of \circuit, as defined in (\ref{eq:cost-sampmon}).  For the case when \circuit.\type $= \circplus$, line ~\ref{alg:sample-plus-bsamp} visits exactly one of the input gates, which may or may not be the subcircuit with the maximum number of gates traversed, which makes \cost$(\cdot)$ an upperbound.  Finally, it is trivial to see that when \circuit.\type $\in \{\var, \tnum\}$, i.e., a source gate, that only one gate is visited.
+First note that the number of gates visited in \sampmon is $\leq\cost(\circuit)$.  To show that \Cref{eq:cost-sampmon} upper bounds the number of nodes visited by \sampmon, note that when \sampmon visits a gate such that \circuit.\type $ =\circmult$, line~\ref{alg:sample-times-for-loop} visits each input of \circuit, as defined in (\ref{eq:cost-sampmon}).  For the case when \circuit.\type $= \circplus$, line~\ref{alg:sample-plus-bsamp} visits exactly one of the input gates, which may or may not be the subcircuit with the maximum number of gates traversed, which makes \cost$(\cdot)$ an upperbound.  Finally, it is trivial to see that when \circuit.\type $\in \{\var, \tnum\}$, i.e., a source gate, that only one gate is visited.
 
 We prove the following inequality holds.
 \begin{equation}