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Sparse Representation Based Fisher Discrimination Dictionary Learning for Image Classification

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Abstract

The employed dictionary plays an important role in sparse representation or sparse coding based image reconstruction and classification, while learning dictionaries from the training data has led to state-of-the-art results in image classification tasks. However, many dictionary learning models exploit only the discriminative information in either the representation coefficients or the representation residual, which limits their performance. In this paper we present a novel dictionary learning method based on the Fisher discrimination criterion. A structured dictionary, whose atoms have correspondences to the subject class labels, is learned, with which not only the representation residual can be used to distinguish different classes, but also the representation coefficients have small within-class scatter and big between-class scatter. The classification scheme associated with the proposed Fisher discrimination dictionary learning (FDDL) model is consequently presented by exploiting the discriminative information in both the representation residual and the representation coefficients. The proposed FDDL model is extensively evaluated on various image datasets, and it shows superior performance to many state-of-the-art dictionary learning methods in a variety of classification tasks.

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Notes

  1. Illuminations {0,1,3,4, 6,7,8,11,13,14,16,17,18,19}.

  2. Illuminations {0,2,4,6,8,10,12,14,16,18}.

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Authors and Affiliations

  1. College of Computer Science & Software Engineering, Shenzhen University, Shenzhen, China

    Meng Yang

  2. Department of Computing, The Hong Kong Polytechnic University, Hong Kong, China

    Lei Zhang & David Zhang

  3. Department of Applied Mathematics, Xidian University, Xi’an, China

    Xiangchu Feng

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  1. Meng Yang
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  2. Lei Zhang
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  3. Xiangchu Feng
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  4. David Zhang
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Correspondence to Lei Zhang.

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Communicated by K. Ikeuchi.

Appendices

Appendix 1: \({{\varvec{tr}}}({{\varvec{S}}}_{B}({{\varvec{X}}}))\) when \(X_{i}^j =0,\;j\ne i\)

Denote by \({{\varvec{m}}}_{i}^{i}, {{\varvec{m}}}_{i}\) and \({{\varvec{m}}}\) the mean vectors of \(\varvec{X}_{i}^i , {{\varvec{X}}}_{i}\) and \({{\varvec{X}}}\), respectively. Because \(\varvec{X}_{i}^j =0\) for \(j\ne i\), we can rewrite \(\varvec{m}_{i} =\left[ {\mathbf{0};\ldots ;\varvec{m}_{i}^i ;\ldots ;\mathbf{0}} \right] \) and \(\varvec{m}={\left[ {n_1 \varvec{m}_1^1 ;\ldots ;n_{i} \varvec{m}_{i}^i ;\ldots ;n_K \varvec{m}_K^K } \right] }\Big /n\). Therefore, the between-class scatter, i.e. \(tr\left( {\varvec{S}_B \left( \varvec{X} \right) } \right) =\sum _{i=1}^K {n_{i} \left\| {\varvec{m}_{i} -\varvec{m}} \right\| _2^2 } ,\) becomes

$$\begin{aligned}&\varvec{S}_B \left( \varvec{X} \right) =\sum \nolimits _{i=1}^K {n_{i} }/{n^{2}}\left[ -n_1 \varvec{m}_1^1 ;\ldots ;\left( {n-n_{i} } \right) \varvec{m}_{i}^i ;\right. \\&\left. \quad \ldots ;-n_K \varvec{m}_K^K \right] \\&\quad \left[ -n_1 \varvec{m}_1^1 ;\ldots ;\left( {n-n_{i} } \right) \varvec{m}_{i}^i ;\ldots ;-n_K \varvec{m}_K^K \right] ^{T}. \end{aligned}$$

Denote by \(\varvec{\kappa }_{i}=1-n_{i}/n\). After some derivation, the trace of \({{\varvec{S}}}_{B}({{\varvec{X}}})\) becomes

$$\begin{aligned}&tr\left( {\varvec{S}_B \left( \varvec{X} \right) } \right) =\sum _{i=1}^K {n_{i} }\Big /{n^{2}}\\&\left\| \left[ -n_1 \varvec{m}_1^1 ;\ldots ;\left( {n-n_{i} } \right) \varvec{m}_{i}^i ;\ldots ;-n_K \varvec{m}_K^K \right] \right\| _2^2\\&=\sum _{i=1}^K {\varvec{\kappa }_{i} n_{i} \left\| {\varvec{m}_{i}^i } \right\| _2^2 } . \end{aligned}$$

Because \(\varvec{m}_{i}^i \) is the mean representation vector of the samples from the same class, which will generally have non-neglected values, the trace of between-class scatter will have big energy in general.

Appendix 2: The Derivation of Simplified FDDL Model

Denote by \(\varvec{m}_{i}^i \) and \({{\varvec{m}}}_{i}\) the mean vector of \(\varvec{X}_{i}^i \) and \({{\varvec{X}}}_{i}\), respectively. Because \(\varvec{X}_{i}^j =0\) for \(j\ne i\), we can rewrite \(\varvec{m}_{i} =\left[ {\mathbf{0};\ldots ;\varvec{m}_{i}^i ;\ldots ;\mathbf{0}} \right] \). So the within-class scatter changes to

$$\begin{aligned} \varvec{S}_W \left( \varvec{X} \right) =\sum \nolimits _{i=1}^K {\sum \nolimits _{{\varvec{x}}_k \in {\varvec{X}}_{i} } {\left( {{{\varvec{x}}}_k^i -\varvec{m}_{i}^i } \right) \left( {{{\varvec{x}}}_k^i -\varvec{m}_{i}^i } \right) ^{T}} } . \end{aligned}$$

The trace of within-class scatter is

$$\begin{aligned} tr\left( {\varvec{S}_W \left( \varvec{X} \right) } \right) =\sum \nolimits _{i=1}^K {\sum \nolimits _{{\varvec{x}}_k \in {\varvec{X}}_{i} } {\left\| {\varvec{x}_k^i -\varvec{m}_{i}^i } \right\| _2^2 } }. \end{aligned}$$

Based on Appendix 1, the trace of between-class scatter is \(tr\left( {\varvec{S}_B \left( \varvec{X} \right) } \right) =\sum \nolimits _{i=1}^K {\varvec{\kappa }_{i} n_{i} \left\| {\varvec{m}_{i}^i } \right\| _2^2 } \), where \(\kappa _{i}=1-n_{i}/n\). Therefore the discriminative coefficient term, i.e. \(f\left( \varvec{X} \right) =\left( {\varvec{S}_W \left( \varvec{X} \right) -\varvec{S}_B \left( \varvec{X} \right) } \right) +\eta \left\| \varvec{X} \right\| _F^2 \), could be simplified to

$$\begin{aligned} f\left( \varvec{X} \right)&= \sum \nolimits _{i=1}^K \left( \sum \nolimits _{{\varvec{x}}_k \in {\varvec{X}}_{i} } {\left\| {\varvec{x}_k^i -\varvec{m}_{i}^i } \right\| _2^2 }\right. \\&\quad \left. +\,\kappa _{i} \left( {\left\| {\varvec{X}_{i}^i } \right\| _F^2 -n_{i} \left\| {\varvec{m}_{i}^i } \right\| _2^2 } \right) +\left( {\eta -\kappa _{i} } \right) \left\| {\varvec{X}_{i}^i } \right\| _F^2 \right) . \end{aligned}$$

Denote by \(\varvec{E}_{i}^j =\left[ 1 \right] _{n_{i} \times n_j } \) the matrix of size \(n_{i}\times n_{j}\) with all entries being 1, then \(\varvec{M}_{i}^i =\left[ {\varvec{m}_{i}^i } \right] _{1\times n_{i} } ={\varvec{X}_{i}^i \varvec{E}_{i}^i }/{n_{i} }\). Because \({{\varvec{I}}}-{\varvec{E}_{i}^i }/{n_{i} }\left( {{\varvec{E}_{i}^i }/{n_{i} }} \right) ^{T}= ({{\varvec{I}}}-{\varvec{E}_{i}^i }/{n_{i} })({{\varvec{I}}}-{\varvec{E}_{i}^i }/{n_{i} })^{T}\), we have

$$\begin{aligned}&\left\| {\varvec{X}_{i}^i } \right\| _F^2 -n_{i} \left\| {\varvec{m}_{i}^i } \right\| _2^2 =\left\| {\varvec{X}_{i}^i } \right\| _F^2 -\left\| {\left[ {\varvec{m}_{i}^i } \right] _{1\times n_{i} } } \right\| _F^2\\&\quad =tr\left( {\varvec{X}_{i}^i \left( {\varvec{I}-{\varvec{E}_{i}^i }/{n_{i} }\left( {{\varvec{E}_{i}^i }/{n_{i} }} \right) ^{T}} \right) \left( {\varvec{X}_{i}^i } \right) ^{T}} \right) \\&\quad =tr\left( {\varvec{X}_{i}^i \left( {\varvec{I}-{\varvec{E}_{i}^i }/{n_{i} }} \right) \left( {\varvec{I}-{\varvec{E}_{i}^i }/{n_{i} }} \right) ^{T}\left( {\varvec{X}_{i}^i } \right) ^{T}} \right) \\&\quad =\left\| {\varvec{X}_{i}^i -\left[ {\varvec{m}_{i}^i } \right] _{1\times n_{i} } } \right\| _F^2 =\left\| {\varvec{X}_{i}^i -\varvec{M}_{i}^i } \right\| _F^2 \end{aligned}$$

Then the discriminative coefficient term could be written as

$$\begin{aligned} f\left( \varvec{X} \right)&=\sum \nolimits _{i=1}^K \left( {\sum \nolimits _{{\varvec{x}}_k \in {\varvec{X}}_{i} } {\left\| {{{\varvec{x}}}_k^i -\varvec{m}_{i}^i } \right\| _2^2 } +\kappa _{i} \left\| {\varvec{X}_{i}^i -\varvec{M}_{i}^i } \right\| _F^2}\right. \\&\quad \left. {+\left( {\eta -\kappa _{i} } \right) \left\| {\varvec{X}_{i}^i } \right\| _F^2 } \right) \\&=\sum \nolimits _{i=1}^K {\left( {\left( {1\!+\!\kappa _{i} } \right) \left\| {\varvec{X}_{i}^i -\varvec{M}_{i}^i } \right\| _F^2 +\left( {\eta -\kappa _{i} } \right) \left\| {\varvec{X}_{i}^i } \right\| _F^2 } \right) } \end{aligned}$$
(21)

With the constraint that \(\varvec{X}_{i}^j =0\) for \(j\ne i\) in Eq. (10), we have

$$\begin{aligned} \left\| {\varvec{A}_{i} -\varvec{DX}_{i} } \right\| _F^2 =\left\| {\varvec{A}_{i} -\varvec{D}_{i} \varvec{X}_{i}^i } \right\| _F^2 \end{aligned}$$
(22)

With Eqs. (21) and (22), the model of simplified FDDL [i.e. Eq. (10] could be written as

where \({\lambda }'_1 ={\lambda _1 }/2,{\lambda }'_2 ={\lambda _2 \left( {1+\kappa _{i} } \right) }/2\), and \({\lambda }'_3 ={\lambda _2 \left( {\eta -\kappa _{i} } \right) }/2\).

Appendix 3: The convexity of \({{\varvec{f}}}_{i}({{\varvec{X}}})\)

Let \(\varvec{E}_{i}^j =\left[ 1 \right] _{n_{i} \times n_j } \) be a matrix of size \(n_{i} \times n_{j}\) with all entries being 1, and let \(\varvec{N}_{i} =\varvec{I}_{n_{i} \times n_{i} } -{\varvec{E}_{i}^i }/{n_{i} },\,\varvec{P}_{i} ={\varvec{E}_{i}^i }/{n_{i} }-{\varvec{E}_{i}^i }/n,\,\varvec{C}_{i}^j ={\varvec{E}_{i}^j }/n\), where \(\varvec{I}_{n_{i} \times n_{i} } \) is an identity matrix of size \(n_{i}\times n_{i}\).

From \(f_{i} \left( {\varvec{X}_{i} } \right) =\left\| {\varvec{X}_{i} -\varvec{M}_{i} } \right\| _F^2 -\sum \nolimits _{k=1}^K {\left\| {\varvec{M}_k -M} \right\| _F^2 } +\eta \left\| {\varvec{X}_{i} } \right\| _F^2 \), we can derive that

$$\begin{aligned} f_{i} \left( {\varvec{X}_{i} } \right)&=\left\| {\varvec{X}_{i} \varvec{N}_{i} } \right\| _F^2 -\left\| {\varvec{X}_{i} \varvec{P}_{i} -\varvec{G}} \right\| _F^2\\&\quad -\sum \nolimits _{k=1,k\ne i}^K {\left\| {\varvec{Z}_k -\varvec{X}_{i} \varvec{C}_i^k } \right\| _F^2 } +\eta \left\| {\varvec{X}_{i} } \right\| _F^2 \end{aligned}$$
(24)

where \(\varvec{G}=\sum \nolimits _{k=1,k\ne i}^K {\varvec{X}_k \varvec{C}_k^i } ,\,\varvec{Z}_k ={\varvec{X}_k \varvec{E}_k^k }/{n_k }-\sum \nolimits _{j=1,j\ne i}^K \varvec{X}_j \varvec{C}_j^k\).

Rewrite \({{\varvec{X}}}_{i}\) as a column vector, \(\varvec{\chi }_{i} \!=\!\left[ {\varvec{r}_{i,1} ,\varvec{r}_{i,2} ,\ldots ,\varvec{r}_{i,d} } \right] ^{T}\), where \({{\varvec{r}}}_{i,j}\) is the \(j^\mathrm{th}\) row vector of \({{\varvec{X}}}_{i}\), and \(d\) is the total number of row vectors in \({{\varvec{X}}}_{i}\). Then \(f_{i}({{\varvec{X}}}_{i})\) equals to

$$\begin{aligned}&\left\| {\hbox {diag}\left( {\varvec{N}_{i} ^{T}} \right) \varvec{\chi }_{i} } \right\| _2^2 -\left\| \hbox {diag}\left( {\varvec{P}_{i} ^{T}} \right) \varvec{\chi }_{i}\right. \\&-\left. \hbox {vec}\left( {\varvec{G}^{T}} \right) \right\| _2^2 -\sum \nolimits _{k=1,k\ne i}^K \left\| \hbox {diag}\left( {\left( {\varvec{C}_{i}^k } \right) ^{T}} \right) \varvec{\chi }_{i}\right. \\&-\left. \hbox {vec}\left( {\varvec{Z}_k^T } \right) \right\| _2^2 +\eta \left\| {\varvec{\chi }_{i} } \right\| _2^2 \end{aligned}$$

where diag \(({{\varvec{T}}})\) is to construct a block diagonal matrix with each block on the diagonal being matrix \({{\varvec{T}}}\), and vec(\({{\varvec{T}}})\) is to construct a column vector by concatenating all the column vectors of \({{\varvec{T}}}\).

The convexity of \(f_{i}(\varvec{\chi }_{i})\) depends on whether its Hessian matrix \(\nabla ^{2}f_{i}(\varvec{\chi }_{i})\) is positive definite or not (Boyd and Vandenberghe 2004). We could write the Hessian matrix of \(f_{i}(\varvec{\chi }_{i})\) as

$$\begin{aligned} \nabla ^{2}f_{i} \left( {\varvec{\chi }_{i} } \right)&= 2\hbox {diag}\left( {\varvec{N}_{i} \varvec{N}_{i} ^{T}} \right) -2\hbox {diag}\left( {\varvec{P}_{i} \varvec{P}_{i} ^{T}} \right) \\&\quad -\sum \nolimits _{k=1,k\ne i}^K {\hbox {2diag}\left( {\varvec{C}_{i}^k \left( {\varvec{C}_{i}^k } \right) ^{T}} \right) } +2\eta \varvec{I}. \end{aligned}$$

\(\nabla ^{2}f_{i}(\varvec{\chi }_{i})\) will be positive definite if the following matrix \({{\varvec{S}}}\) is positive definite:

$$\begin{aligned} \varvec{S}=\varvec{N}_{i} \varvec{N}_{i}^T -\left( {\varvec{P}_{i} \varvec{P}_{i}^T +\sum _{k=1,k\ne i}^K {\varvec{C}_{i}^k \left( {\varvec{C}_{i}^k } \right) ^{T}} } \right) +\eta \varvec{I}. \end{aligned}$$

After some derivations, we have

$$\begin{aligned} \varvec{S}=\left( {1+\eta } \right) \varvec{I}-\varvec{E}_{i}^i \left( {2/{n_{i} -2/n}+\sum \nolimits _{k=1}^K {{n_k }/{n^{2}}} } \right) . \end{aligned}$$

In order to make \({{\varvec{S}}}\) positive define, each eigenvalue of \({{\varvec{S}}}\) should be greater than 0. Because the maximal eigenvalue of \({{\varvec{E}}}_{i}^{i}\) is \(n_{i}\), we should ensure

$$\begin{aligned} \left( {1+\eta } \right) -n_{i} \left( {2/{n_{i} -2/n}+\sum _{k=1}^K {{n_k }/{n^{2}}} } \right) >0 \end{aligned}$$

For \(n=n_{1}+n_{2}+{\ldots }+n_{K}\), we have \(\eta >\kappa _{i}\), which could guarantee that \(f_{i}({{\varvec{X}}}_{i})\) is convex to \({{\varvec{X}}}_{i}\). Here \(\kappa _{i}=1-n_{i}/n\).

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Yang, M., Zhang, L., Feng, X. et al. Sparse Representation Based Fisher Discrimination Dictionary Learning for Image Classification. Int J Comput Vis 109, 209–232 (2014). https://doi.org/10.1007/s11263-014-0722-8

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