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Learning DSD in data and model domains - synthesis and analysis models

Provided that the learning-based dictionary is a tight frame such that $ (\mathbf{W}^T)^T\mathbf{W}^T=\mathbf{W}\mathbf{W}^T=\mathbf{I}$ , DSD can be learned by considering the following problem:

\begin{displaymath}\begin{split}&\min_{\mathbf{W,m}} \frac{1}{2}\parallel \mathb...
...quad and \quad \mathbf{W}\mathbf{W}^T = \mathbf{I}, \end{split}\end{displaymath} (3)

where $ t$ denotes the sparsity constrained coefficient.

Equation 3 can be solved alternatively by minimizing

\begin{displaymath}\begin{split}\min_{\mathbf{W},\mathbf{m}} \frac{1}{2}&\parall...
...el_0, \ &s.t. \mathbf{W}\mathbf{W}^T = \mathbf{I}, \end{split}\end{displaymath} (4)

where $ \lambda$ denotes the damping factor, which is connected with the sparsity constrained coefficient $ t$ . Equation 4 is called synthesis model for DSD.

Assuming the base dictionary $ \mathbf{B}$ is invertible, the synthesis-based DSD model as shown in equation 3 is equivalent to:

\begin{displaymath}\begin{split}\min_{\mathbf{W,m}} \frac{1}{2}&\parallel \mathb...
...lel_0 \ &s.t. \mathbf{W}\mathbf{W}^T = \mathbf{I}, \end{split}\end{displaymath} (5)

where $ \mathbf{B}^{-1}$ denotes the forward base transform. Equation 5 suggests that DSD can also be learned in the model domain of the multi-scale decomposition operator $ \mathbf{B}$ instead of in the data domain. Equation 5 is called analysis model for DSD.

The synthesis model (equation 4) offers more flexibility for learning DSD because there are many sparsity-promoting transforms that are not exactly invertible. The analysis model, on the other hand, offers more convenience for constructing DSD when an invertible sparsity-promoting base transform is available.


next up previous [pdf]

Next: Solving the analysis model Up: Theory Previous: Double sparsity dictionary

2016-02-27