76 lines
3.7 KiB
TeX
76 lines
3.7 KiB
TeX
% -*- root: main.tex -*-
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\section{Exact Results}
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\label{sec:exact}
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We turn to computing the exact values of
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\begin{equation}
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\sum\limits_{\wVec \in \pw } \sketchJParam{\sketchHashParam{\wVec}}\sketchPolarParam{\wVec} =
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\sum\limits_{\wVec \in \pw } \kMapParam{\wVec}\sketchPolarParam{\wVec}
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\sum_{\substack{\wVecPrime \in \pw\st\\
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\sketchHashParam{\wVec} = \sketchHashParam{\wVecPrime}}} \sketchPolarParam{\wVecPrime}\label{eq:exact-results} .
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\end{equation}
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Starting with the latter term $\gIJ = \sum\limits_{\wVecPrime \in \pw}\sketchPolarParam{\wVecPrime}$, by the definition of the image of $\sketchPolar$ and the property of associativity in addition, we can break the sum into
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\begin{equation*}
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\gIJ = \sum_{\substack{\wVecPrime \in \pw \st\\
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\sketchPolarParam{\wVecPrime} = 0}} 1 + \sum_{\substack{\wVecPrime \in \pw \st\\
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\sketchPolarParam{\wVecPrime} = 1}} -1.
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\end{equation*}
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Setting the terms to $T_1 = \sum\limits_{\substack{\wVecPrime \in \pw \st\\
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\sketchPolarParam{\wVecPrime} = 0}} 1$ and $T_2 = \sum\limits_{\substack{\wVecPrime \in \pw \st\\
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\sketchPolarParam{\wVecPrime} = 1}} -1$ and fixing $\buck$ to a specific value, gives a system of linear equations for each term. It is a known result for a consistent matrix multiplication that the number of solutions are $| \kDom |^{\numTup - rank(\matrixH')}$, where $\kDom$ is the set being considered. This gives us an exact calculation for both terms,
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\begin{align*}
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T_1 = |\{\wVec \st \matrixH' \cdot \wVec = \buck^{(0)}\}|\rightarrow T_1 \in \{0, 2^{\numTup - rank(\matrixH')}\},\\
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T_2 = |\{\wVec \st \matrixH' \cdot \wVec = \buck^{(1)}\}|\rightarrow T_2 \in \{0, 2^{\numTup - rank(\matrixH')}\},
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\end{align*}
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where the notation $\jpbit{y}$ denotes the polarity bit $\lenB$ value of the $\buck$ bucket identifier, specifically $\buck(b)$, such that $\buck(b)\in \{0, 1\}$.
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\subsection{Algorithm for $\gIJ$}
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\begin{algorithmic}
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\If {$\matrixH' \cdot \wVec = j^{(0)}$ is consistent}
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\If {$\matrixH' \cdot \wVec = j^{(1)}$ is consistent}
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\State $\gIJ = 0$
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\Else
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\State $\gIJ = 2^{\numTup - computeRank(\matrixH')}$
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\EndIf
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\ElsIf{$\matrixH' \cdot \wVec = \buck^{(1)}$ is consistent}
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\State $\gIJ = 2^{\numTup - computeRank(\matrixH')}$
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\Else
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\State $\gIJ = 0$
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\EndIf.
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\end{algorithmic}
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For examining the first term of equation \eqref{eq:exact-results}, we fix $\kMap{t}$ to be defined as
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\begin{equation*}
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\kMapParam{\wVec} = \begin{cases}
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1,&\text{if } w_t = 1\\
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0, &\text{otherwise}.
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\end{cases}
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\end{equation*}
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%Therefore, by definition we have
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%\begin{equation*}
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%\sum_{\wVec \in \pw}\sketchJParam{\sketchHashParam{\wVec}} = \sum_{\wVec \in \pw}\kMapParam{\wVec}\sketchPolarParam{\wVec},
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%\end{equation*}
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Using the same argument as in $\gIJ$ yields
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\begin{equation*}
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\sum_{\wVec \in \pw \st \sketchPolarParam{\wVec} = 0}\kMapParam{\wVec} - \sum_{\wVec \in \pw \st \sketchPolarParam{\wVec} = 1}\kMapParam{\wVec}.
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\end{equation*}
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Setting $T_3 = \sum\limits_{\wVec \in \pw \st \sketchPolarParam{\wVec} = 0}\kMapParam{\wVec}$, $T_4 = \sum\limits_{\wVec \in \pw \st \sketchPolarParam{\wVec} = 1}\kMapParam{\wVec}$ gives an exact calculation for each term given a fixed $\buck$:
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\begin{equation*}
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T_3 = | \{\wVec \st \matrixH \cdot \wVec = \buck^{(0)}, \kMapParam{\wVec} = 1\}\rightarrow T_3 \in [0, 2^{\numTup - rank(\matrixH')}]
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\end{equation*}
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\begin{equation*}
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T_4 = | \{\wVec \st \matrixH \cdot \wVec = \buck^{(1)}, \kMapParam{\wVec} = 1\}\rightarrow T_4 \in [0, 2^{\numTup - rank(\matrixH') - 1}]
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\end{equation*}
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\subsection{Algorithm for Initialization}
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\begin{algorithmic}
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\ForAll{$\wVec \st \kMapParam{\wVec} = 1$}
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\State $\buck = \matrixH' \cdot \wVec$
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\If{$\buck(\lenB) = 0$}
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\State $\sketchIj += 1$
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\Else
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\State$\sketchIj -= 1$
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\EndIf
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\EndFor.
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\end{algorithmic} |