Fixed typo.

master
Aaron Huber 2022-06-06 09:19:50 -04:00
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@ -101,7 +101,7 @@ Given the computed values of \onepass, \sampmon picks a sampling path by travers
%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}.
All the algorithm details are in \Cref{sec:proofs-approx-alg}.
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