Reworked sampmon example over figure 3.

approx_alg.tex

@@ -94,9 +94,12 @@ Given the above, the algorithm is a sampling based algorithm for the above sum:
  to $\abs{\coef}$ and compute $\vari{Y}=\indicator{\isInd{\encMon}}
  \cdot \prod_{X_i\in \monom} p_i$. 
 Repeating the sampling an appropriate number of times
-and computing the average of $\vari{Y}$ gives us our final estimate. To illustrate the sampling procedure using~\Cref{fig:circuit}, \onepass recursively visits each gate \circuit exactly once, computing $\abs{\circuit}\inparen{1,\ldots, 1}$.  %for each gate \circuit.  
-If we consider the leftmost source gates, \onepass computes $\abs{\circuit}\inparen{1,\ldots,1} = 1$ for $\circuit.\val = X$ and $\abs{\circuit}\inparen{1,\ldots, 1} = 2$ for $\circuit.\val = 2$. For the leftmost $\circmult$ gate, \onepass computes $\abs{\circuit}\inparen{1,\ldots, 1} = 2\circmult 1$, i.e. $\abs{\circuit_\linput}\inparen{1,\ldots, 1} \times \abs{\circuit_\rinput}\inparen{1,\ldots, 1}$ for children $\circuit_\linput$ and $\circuit_\rinput$.  A level higher, the leftmost $\circplus$ gate recursively adds the values of its two children deriving $\abs{\circuit}\inparen{1,\ldots, 1} = 2 \circplus 1$, while using the expression $\frac{\abs{\circuit_i}\inparen{1,\ldots, 1}}{\abs{\circuit}\inparen{1,\ldots, 1}}$ for $i\in\inset{\linput, \rinput}$ to simultaneously compute the weights $\frac{1}{3}$ and $\frac{2}{3}$ for its children.  The final sum value is then computed in similar fashion.  % yielding $\abs{\circuit}\inparen{1,\ldots, 1} = 3 \circmult 3 = 9$.
-\sampmon then uses the weights provided by \onepass to randomly select a monomial from $\expansion{\circuit}$.  To sample the monomial $X$, \sampmon recursively traverses both children of the sink gate.  Since both children are $\circplus$ gates, \sampmon then randomly selects the left child of each $\circplus$ gate with probability $\frac{1}{3}$ and $\frac{2}{3}$ respectively.  For the fomer, the leaf node with variable $X$ and sign value $1$ is returned, while the latter visits both children of the selected child $\circmult$ gate, recursively returning $X$ with a sign value of $1$.  Finally, \sampmon performs the operation $X\cup X = X$, effectively reducing the monomial to its $\rpoly$ valuation.
+and computing the average of $\vari{Y}$ gives us our final estimate. To illustrate the sampling procedure using~\Cref{fig:circuit}, \onepass recursively computes $\abs{\circuit}\inparen{1,\ldots, 1}$, where the gray values are the intermediate values.
+% visits each gate \circuit exactly once, computing $\abs{\circuit}\inparen{1,\ldots, 1}$.  %for each gate \circuit.  
+%If we consider the leftmost source gates, \onepass computes $\abs{\circuit}\inparen{1,\ldots,1} = 1$ for $\circuit.\val = X$ and $\abs{\circuit}\inparen{1,\ldots, 1} = 2$ for $\circuit.\val = 2$. For the leftmost $\circmult$ gate, \onepass computes $\abs{\circuit}\inparen{1,\ldots, 1} = 2\circmult 1$, i.e. $\abs{\circuit_\linput}\inparen{1,\ldots, 1} \times \abs{\circuit_\rinput}\inparen{1,\ldots, 1}$ for children $\circuit_\linput$ and $\circuit_\rinput$.  A level higher, the leftmost $\circplus$ gate recursively adds the values of its two children deriving $\abs{\circuit}\inparen{1,\ldots, 1} = 2 \circplus 1$, while using the expression $\frac{\abs{\circuit_i}\inparen{1,\ldots, 1}}{\abs{\circuit}\inparen{1,\ldots, 1}}$ for $i\in\inset{\linput, \rinput}$ to simultaneously compute the weights $\frac{1}{3}$ and $\frac{2}{3}$ for its children.  The final sum value is then computed in similar fashion.  % yielding $\abs{\circuit}\inparen{1,\ldots, 1} = 3 \circmult 3 = 9$.
+Given the computed values of \onepass, \sampmon picks a sampling path by traversing both children of a $\circmult$ gate and randomly choosing a child from a $\circplus$ gate according to the weight $\frac{\abs{\circuit_i}\inparen{1,\ldots, 1}}{\abs{\circuit}\inparen{1,\ldots, 1}}$ for $i\in\inset{\linput, \rinput}$.  
+%then uses the weights provided by \onepass to randomly select a monomial from $\expansion{\circuit}$.  
+To sample the monomial $X$ in~\Cref{fig:circuit}, \sampmon recursively traverses both children of the sink gate.  Since both children are $\circplus$ gates, \sampmon then randomly selects the left child of each $\circplus$ gate with probability $\frac{1}{3}$ and $\frac{2}{3}$ respectively.  Note that the probabilty for choosing both said children is $\frac{2}{9}$, which is indeed the ratio of the number of $X$ terms to the total number of terms in $\expansion{\circuit}$, thus a correct sampling probability.  For the fomer $\circplus$ gate, the leaf gate with variable $X$ and sign value $1$ is returned, while the latter visits both children of the selected child $\circmult$ gate, recursively returning $X$ with a sign value of $1$.  Finally, \sampmon performs the operation $X\cup X = X$, effectively reducing the monomial to its $\rpoly$ valuation.  Suppose \sampmon also randomly samples $XY$ and $-Y$.  To estimate $\rpoly\inparen{\vct{\prob}}$, \approxq computes $\prob_X + \prob_X\prob_Y - \prob_Y$ and scales the accumulation accordingly.
 %such that a source gate \circuit has $\abs{\circuit}\inparen{1,\ldots, 1} = \circuit.\val$ when \circuit.\type $=$ \num and $\abs{\circuit}\inparen{1,\ldots,1} = 1$ otherwise.  For every gate \circuit, \onepass computes $\abs{\circuit}\inparen{1,\ldots, 1}$ as seen in the lighter font of~\Cref{fig:circuit}.  \onepass further weights each child $\circuit_i$ for $i\in\inset{\linput, \rinput}$, by the expression $\frac{\abs{\circuit_i}\inparen{1,\ldots, 1}}{\abs{\circuit}\inparen{1,\ldots, 1}}$.  These weight are the basis for the sampling performed by \sampmon. 
  All the algorithms details are in \Cref{sec:proofs-approx-alg}.
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