Adds some notes in the report
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@ -51,6 +51,8 @@ The wavelet transform now consists of multiplying the above matrices in a recurs
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\subsection{In place}
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\subsection{In place}
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When implementing this transform, we don't have to perform the even-odd sort. Instead, we can simply do all calculations in place and use a stride to do the recursion on the even part. This will permute the original result.
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When implementing this transform, we don't have to perform the even-odd sort. Instead, we can simply do all calculations in place and use a stride to do the recursion on the even part. This will permute the original result.
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\todo{Tell a bit more?}
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\todo{Add images to show the difference}
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\subsection{Costs}
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\subsection{Costs}
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@ -59,6 +61,40 @@ We will briefly analyze the cost of the transform by counting the number of \emp
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Using the geometric series $\sum_{i=0}^\infty 2^{-i} = 2$ we can bound the number of flops by $14n$.
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Using the geometric series $\sum_{i=0}^\infty 2^{-i} = 2$ we can bound the number of flops by $14n$.
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Compared to the FFT this is a big improvement in terms of scalability, as this wavelet transform has a linear complexity $\BigO{n}$, but the FFT has a complexity of $\BigO{n \log n}$. This is however also a burden, as it leaves not much room for overhead induced by parallelizing the algorithm. We will see an precies analysis of communication costs in section~\ref{sec:par}.
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Compared to the FFT this is a big improvement in terms of scalability, as this wavelet transform has a linear complexity $\BigO{n}$, but the FFT has a complexity of $\BigO{n \log n}$. This is however also a burden, as it leaves not much room for overhead induced by parallelizing the algorithm. We will see an precies analysis of communication costs in section~\ref{sec:par}.
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\todo{Do a simple upperbound of communication here?}
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\subsection{The inverse}
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\subsection{The inverse}
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The wavelet transform is invertible. We will proof this by first showing that $S_n$ and $W_n P_n$ are invertible. In fact, they are orthogonal, which means that the inverse is given by the transpose.
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\begin{lemma}
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The matrices $S_n$ and $W_n P_n$ are orthogonal.
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\end{lemma}
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\begin{proof}
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For $S_n$ it is clear, as it is an permutation matrix.
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For $W_n$ we should calculate the inner products of all pairs of columns.
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\todo{Calculate}
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\end{proof}
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\begin{theorem}
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The matrix $W$ is invertible with $W^{-1} = W^T$.
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\end{theorem}
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\begin{proof}
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First note that $\diag(S_m W_m P_m, I_m, \ldots, I_m)$ is orthogonal by the above lemma. Now using the fact that the multiplications of two orthogonal matrices is again orthogonal we see that $W$ is orthogonal.
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\end{proof}
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\todo{Note that I didnt parallelize the inverse}
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\subsection{Higher dimensional wavelet transform}
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Our final goal is to apply the wavelet transform to images. Of course we could simply put all the pixels of an image in a row and apply $W$. But if we do this, we don't use the spatial information of the image at all! In order to use the spatial information we have to apply $W$ in both directions. To be precise: we will apply $W$ to every row and then apply $W$ to all of the resulting columns. We can also do this the other way around, but this does not matter:
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\begin{lemma}
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Given a matrix $F$ and \todo{think of nice formulation}
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\end{lemma}
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\begin{proof}
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\todo{Give the simple calculation}
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\end{proof}
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This lemma expresses some sort of commutativity and generalises to higher dimensions by apply this commutativity recursively. As we don't need the general statement (i.e. we will only apply $W$ to images) we won't spell out the proof.
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@ -19,4 +19,8 @@
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\newcommand{\todo}[1]{
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\newcommand{\todo}[1]{
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\addcontentsline{tdo}{todo}{\protect{#1}}
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\addcontentsline{tdo}{todo}{\protect{#1}}
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$\ast$ \marginpar{\tiny $\ast$ #1}
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$\ast$ \marginpar{\tiny $\ast$ #1}
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}
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}
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\theoremstyle{plain}
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\newtheorem{theorem}{Theorem}[section]
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\newtheorem{lemma}[theorem]{Lemma}
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