Read through/cleaning of Appendix C up to C.11.

app_approx-alg-analysis.tex

@@ -162,6 +162,7 @@ N_\linput^{k+1} + N_\rinput^{k+1}\nonumber\\
 &\leq N^{k+1}.\nonumber
 \end{align}
 In the above, the first inequality follows from the inductive hypothes and \cref{def:degree} (which implies the fact that $k_\linput,k_\rinput\le k$).  Note that the RHS of this inequality is maximized when the base and exponent of one of the terms is maximized.  The second inequality follows from this fact as well as the fact that since $\circuit$ is a tree we have $N_\linput+N_\rinput=N-1$ and, lastly, the fact that $k\ge 0$. This completes the proof.
+%\AH{I don't think that it matters whether or not \circuit is a tree.  For $N=\size\inparen{\circuit}$ it must follow that $N_L + N_R + 1 = N$ regardless of whether a gate a allowed to have more than one parent.  Not true, consider when $\circuit_R = \circuit_L$.}
 \end{proof}
 
 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):

app_onepass-analysis.tex

@@ -9,7 +9,7 @@ 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 $(X + Y)(X - Y) + Y^2$.  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}. 
+ Let $\etree$ encode the expression $(X + Y)(X - Y) + Y^2$.  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 $\circuit_{\lchild}$ of \circuit, $\stree.\lwght = \stree.\rwght = \frac{1}{2}$.  This is depicted in \Cref{fig:expr-tree-T-wght}. 
 \end{Example}
 
 \begin{figure}[h!]

app_samp-monom-analysis.tex

@@ -18,6 +18,8 @@ The efficiency gains of circuits over trees is found in the capability of circui
 We first need to show that $\sampmon$ samples a valid monomial $\encMon$ by sampling and returning a set of variables $\monom$, such that $(\monom, \coef)$ is in $\expansion{\circuit}$ and $\encMon$ is indeed a monomial of the $\rpoly\inparen{\vct{X}}$ encoded in \circuit.  We show this via induction over the depth of \circuit.
 For the base case, let the depth $d$ of $\circuit$ be $0$.  We have that the single gate is either a constant $\coef$ for which by line~\ref{alg:sample-num-return} we return $\{~\}$, or we have that $\circuit.\type = \var$ and $\circuit.\val = x$, and  by line~\ref{alg:sample-var-return} we return $\{x\}$.  By \cref{def:expand-circuit}, both cases return a valid $\monom$ for some $(\monom, \coef)$ from $\expansion{\circuit}$, and the base case is proven.
 
+\AH{I think it is slightly confusing to say that depth $= 0$ in view of the definition of depth in S.4.  To say $k = 0$ is also strange, since, for a single join, we have that $k = 2$.}
+
 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 valid monomial.
 
 For the inductive step, let us take a circuit $\circuit$ with $d = k + 1$.  Note that each input has depth $d - 1 \leq k$, and by inductive hypothesis both of them sample a valid monomial.  Then the sink can be either a $\circplus$ or $\circmult$ gate.  For the case when $\circuit.\type = \circplus$, line~\ref{alg:sample-plus-bsamp} of $\sampmon$ will choose one of the inputs of the source.  By inductive hypothesis it is the case that some valid monomial is being randomly sampled from each of the inputs.  Then it follows when $\circuit.\type = \circplus$ that a valid monomial is sampled by $\sampmon$.  When the $\circuit.\type = \circmult$, line~\ref{alg:sample-times-union} computes the set union of the monomials returned by the two inputs of the sink, and it is trivial to see by \cref{def:expand-circuit} that $\encMon$ is a valid monomial encoded by some $(\monom, \coef)$ of $\expansion{\circuit}$.
@@ -26,6 +28,8 @@ We will next prove by induction on the depth $d$ of $\circuit$ that for $(\monom
 
 For the base case $d = 0$, by definition~\ref{def:circuit} we know that the $\size\inparen{\circuit} = 1$ and \circuit.\type$=$ \tnum or \var.  For either case, the probability of the value returned is $1$ since there is only one value to sample from.  When \circuit.\val $= x$, the algorithm always return the variable set $\{x\}$.  When $\circuit.\type = \tnum$, \sampmon will always return $\emptyset$.
 
+\AH{I don't think this is technically right, since \sampmon returns a tuple of two values.}
+
 For the inductive hypothesis, assume that for $d \leq k$ and $k \geq 0$ $\sampmon$ indeed returns $\monom$ in $(\monom, \coef)$ of $\expansion{\circuit}$ with probability $\frac{|\coef|}{\abs{\circuit}\polyinput{1}{1}}$.
 
 We prove now for $d = k + 1$ the inductive step holds.  It is the case that the sink of $\circuit$ has two inputs $\circuit_\linput$ and $\circuit_\rinput$.  Since $\circuit_\linput$ and $\circuit_\rinput$ are both depth $d - 1 \leq k$, by inductive hypothesis, $\sampmon$ will return $\monom_\linput$ in $(\monom_\lchild, \coef_\lchild)$ of $\expansion{\circuit_\linput}$ and $\monom_\rinput$ 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 $\frac{|\coef_\rchild|}{\abs{\circuit_\rinput}\polyinput{1}{1}}$.
@@ -43,11 +47,12 @@ and we obtain the desired result.
 
 Lastly, we show by simple induction of the depth $d$ of \circuit that \sampmon indeed returns the correct sign value of $\coef$ in $(\monom, \coef)$.
 
-In the base case, $\circuit.\type = \tnum$ or $\var$.  For the former, \sampmon correctly returns the sign value of the gate.  For the latter, \sampmon returns the correct sign of $1$, since a variable is a neutral element, and $1$ is the multiplicative identity, whose product with another sign element will not change that sign element.
+In the base case, $\circuit.\type = \tnum$ or $\var$.  For the former by~\Cref{alg:sample-num-leaf}, \sampmon correctly returns the sign value of the gate.  For the latter by~\Cref{alg:sample-var-return}, \sampmon returns the correct sign of $1$, since a variable is a neutral element, and $1$ is the multiplicative identity, whose product with another sign element will not change that sign element.
 
 For the inductive hypothesis, we assume for a circuit of depth $d \leq k$ and $k \geq 0$ that the algorithm correctly returns the sign value of $\coef$.
 
-Similar to before, for a depth $d \leq k + 1$, it is true that $\circuit_\linput$ and $\circuit_\rinput$ both return the correct sign of $\coef$.  For the case that $\circuit.\type = \circmult$, the sign value of both inputs are multiplied, which is the correct behavior by \cref{def:expand-circuit}.  When $\circuit.\type = \circplus$, only one input of $\circuit$ is sampled, and the algorithm  returns the correct sign value of $\coef$ by inductive hyptothesis.
+Similar to before, for a depth \AH{Why do we use $d = k + 1$ for the inductive cases above?}
+$d \leq k + 1$, it is true that $\circuit_\linput$ and $\circuit_\rinput$ both return the correct sign of $\coef$.  For the case that $\circuit.\type = \circmult$, the sign value of both inputs are multiplied, which is the correct behavior by \cref{def:expand-circuit}.  When $\circuit.\type = \circplus$, only one input of $\circuit$ is sampled, and the algorithm  returns the correct sign value of $\coef$ by inductive hyptothesis.
 
 
 \paragraph*{Run-time Analysis}
@@ -74,10 +79,13 @@ We prove the following inequality holds.
 2\left(\degree(\circuit) + 1\right) \cdot \depth(\circuit) + 1 \geq \cost(\circuit)\label{eq:strict-upper-bound}
 \end{equation} 
 
-Note that \cref{eq:strict-upper-bound} implies the claimed runtime.  We prove \cref{eq:strict-upper-bound} for the number of gates traversed in \sampmon using induction over $\depth(\circuit)$.  Recall how degree is defined in \cref{def:degree}.
+Note that \cref{eq:strict-upper-bound} implies the claimed runtime.  
+\AH{If the claimed runtime is from the first paragraph, then I don't follow.}
+
+We prove \cref{eq:strict-upper-bound} for the number of gates traversed in \sampmon using induction over $\depth(\circuit)$.  Recall how degree is defined in \cref{def:degree}.
 
 For the base case $\degree(\circuit) = \inset{0, 1}, \depth(\circuit) = 0$, $\cost(\circuit) = 1$, and it is trivial to see that the inequality $2\degree(\circuit) \cdot \depth(\circuit) + 1 \geq \cost(\circuit)$ holds.
-
+\AH{Why equality here instead of inequality?  Also, it could be more obvious for why depth must be at least $1$.}
 For the inductive hypothesis, we assume the bound holds for any circuit where $\ell \geq \depth(\circuit) \geq 0$.
 Now consider the case when \sampmon has an arbitrary circuit \circuit input with $\depth(\circuit) = \ell + 1$.  By definition \circuit.\type $\in \{\circplus, \circmult\}$. Note that since $\depth(\circuit) \geq 1$, \circuit must have input(s).  Further we know that by the inductive hypothesis the inputs $\circuit_i$ for $i \in \{\linput, \rinput\}$ of the sink gate \circuit uphold the bound
 \begin{equation}
@@ -132,6 +140,7 @@ To prove  (\ref{eq:plus-middle}), \cref{eq:plus-lhs-inequality} expands to
 \begin{equation}
 2\degree_{\max}\depth_{\max} + 2\degree_{\max} + 2\depth_{\max} + 2 + 1.\label{eq:plus-lhs-expanded}
 \end{equation}
+\AH{It seems more confusing to add an extra term in the RHS of the leftmost inequality.}
 Since $\degree_{\max} \cdot \depth_{\max} \geq \degree(\circuit_i)\cdot \depth(\circuit_i),$ the following upper bound holds for the expansion of \cref{eq:plus-middle}:
 \begin{equation}
 2\degree_{\max}\depth_{\max} + 2\depth_{\max} + 2 

app_samp-monom_pseudo-code.tex

@@ -21,7 +21,7 @@
 			\EndFor
 			\State $\Return ~(\vari{vars}, \vari{sgn})$
 		\ElsIf{$\circuit.\type = \tnum$}\Comment{The leaf is a coefficient}
-			\State $\Return ~\left(\{\}, \func{sgn}(\circuit.\val)\right)$\label{alg:sample-num-return}\Comment{$\func{sgn}(\cdot)$ outputs $-1$ for \circuit.\val $\geq 1$ and $-1$ for \circuit.\val $\leq -1$}
+			\State $\Return ~\left(\{\}, \func{sgn}(\circuit.\val)\right)$\label{alg:sample-num-return}\Comment{$\func{sgn}(\cdot)$ outputs $-1$ for \circuit.\val $\geq 1$ and $-1$ for \circuit.\val $\leq -1$}\label{alg:sample-num-leaf}
 		\ElsIf{$\circuit.\type = var$}
 			\State $\Return~\left(\{\circuit.\val\}, 1\right)	$\label{alg:sample-var-return}
 		\EndIf