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[DMDB] Fix error, add short remark on projections
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@@ -8,7 +8,7 @@ We have at least four metrics for the quality of results:
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\item \bi{Cosine Distance} $D_C$: defined as $1 - S_C$, values range from $0$ to $2$ and the closer to $0$, the better (lower = better)
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\item \bi{Cosine Distance} $D_C$: defined as $1 - S_C$, values range from $0$ to $2$ and the closer to $0$, the better (lower = better)
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\item \bi{Euclidian distance}: The norm of the vector between the two vectors. Values are from $0$ to $\8$, the lower the better.
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\item \bi{Euclidian distance}: The norm of the vector between the two vectors. Values are from $0$ to $\8$, the lower the better.
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\end{itemize}
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\end{itemize}
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$S_C(A, B) = \frac{A \cdot B}{||A|| \cdot ||B||}$, a
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$S_C(A, B) = \frac{A \cdot B}{||A|| \cdot ||B||}$
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\subsection{Trees/Clustering}
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\subsection{Trees/Clustering}
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@@ -1,6 +1,6 @@
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\subsection{Hash Index}
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\subsection{Hash Index}
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To measure the quality of \acrshort{ann} in comparison to \acrshort{knn}, we use $\texttt{recall@k} = |\texttt{ANN\_result} \cap \texttt{KNN\_result}| \div k$.
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To measure the quality of \acrshort{ann} in comparison to \acrshort{knn}, we use $\texttt{recall@k} = |\texttt{ANN\_result} \cap \texttt{KNN\_result}| \div k$.
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This however is not always a good metric, as two vastly different results, one objectively worse can score equally well.
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This however is not always a good metric, as for two vastly different results, the objectively worse one can score equally well.
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Alternative methods mentioned are \texttt{RDE@k} and \texttt{TDK@k}, but not elaborated any further.
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Alternative methods mentioned are \texttt{RDE@k} and \texttt{TDK@k}, but not elaborated any further.
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@@ -24,6 +24,7 @@ Given $B$ frames of memory and $N$ records, we have (typically I/Os in pages to
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\subsubsection{Projection}
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\subsubsection{Projection}
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Remember that we need the sort-based and bash-based approach to eliminate duplicates.
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\begin{itemize}
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\begin{itemize}
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\item \bi{Partitioning}: $\texttt{Cost}_\texttt{part}(R) = \texttt{cnt}(R) + \texttt{cnt}(R')$ with
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\item \bi{Partitioning}: $\texttt{Cost}_\texttt{part}(R) = \texttt{cnt}(R) + \texttt{cnt}(R')$ with
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$\texttt{cnt}(R') = \texttt{cnt}(R) Q$, with $Q$ the fraction of selected attributes divided by total attributes
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$\texttt{cnt}(R') = \texttt{cnt}(R) Q$, with $Q$ the fraction of selected attributes divided by total attributes
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