Weighted Discriminative Sparse Representation for Image Classification

Abstract

Sparse representation methods based on $l _2$ norm regularization have attracted much attention due to its low computational cost and competitive performance. How to enhance the discriminability of $l _2$ norm regularization-based representation method is a meaningful work. In this paper, we put forward a novel $l _2$ norm regularization-based representation method, called Weighted Discriminative Sparse Representation for Classification (WDSRC), in which we consider the global discriminability and the local discriminability using two discriminative regularization terms of representation. The global discriminability is obtained by decorrelating the representation results stemming from all distinct classes. The local discriminability is achieved by the weighted representation in which the representation coefficient of the training images dissimilar to the test image will be reduced and the representation coefficient of the training images similar to the test image will be increased, which restrains the training images dissimilar to the test image and promotes the training images similar to the test image as much as possible in representing the test sample. By considering the global and local discriminability of representations simultaneously, the proposed WDSRC method can gain more discriminative representation for classification. Extensive experiments on benchmark datasets of object, face, action and flower demonstrate the effectiveness of the proposed WDSRC method.

Appendix

$$\begin{aligned} \min _{b } \Vert {y-Xb}\Vert _2 ^2 + \gamma \sum _{i=1}^{C}\sum _{j=1}^{C} \Vert {X_i b_i + X_j b_j} \Vert _2 ^2+\lambda \Vert {Wb}\Vert _2 ^2 \end{aligned}$$

(11)

The derivative of the objective function (11) can be computed as follows. First

$$\begin{aligned}&\frac{\mathrm {d}\Vert {y-Xb}\Vert _2 ^2}{\mathrm {d}b}= -2X^T(y-Xb) \end{aligned}$$

(16)

$$\begin{aligned}&\frac{\mathrm {d}\Vert Wb\Vert _2 ^2}{\mathrm {d}b}= 2W^TWb \end{aligned}$$

(17)

Let $f(b)=\gamma \sum _{i=1}^{C}\sum _{j=1}^{C} \Vert {X_i b_i + X_j b_j} \Vert _2 ^2$, since f(b) dose not overtly include b, we first compute partial derivatives $\frac{\partial f(b)}{\partial b_k}$ and then obtain $\frac{\mathrm {d}f}{\mathrm {d}b}$ by using all $\frac{\partial f(b)}{\partial b_k}$ (k = 1,2,...,C).

$$\begin{aligned} \begin{aligned} f(b)&= \gamma \left( \sum _{i=1,2,\ldots ,C,i\ne k} \Vert {X_i b_i + X_k b_k }\Vert _2 ^2 + \sum _{j=1,2,\ldots ,C,j\ne k} \Vert {X_k b_k + X_j b_j }\Vert _2 ^2 \right. \\&\left. \qquad \sum _{i=1,2,\ldots ,C,i\ne k}\sum _{j=1,2,\ldots ,C,j\ne k} \Vert {X_i b_i + X_j b_j}\Vert _2 ^2\right) \\&= \gamma \left( 2 \sum _{i=1,2,\ldots ,C,i\ne k} \Vert {X_i b_i + X_k b_k }\Vert _2 ^2 + \sum _{i=1,2,\ldots ,C,i\ne k}\sum _{j=1,2,\ldots ,C,j\ne k} \Vert {X_i b_i + X_j b_j}\Vert _2 ^2\right) \end{aligned} \end{aligned}$$

(18)

Accordingly, the partial derivative of f(b) with respect to $b_k$ can be computed as Equation in (19)

$$\begin{aligned} \begin{aligned} \frac{\partial f(b)}{\partial b_k}&= \frac{\partial }{\partial b_k} \left( \gamma \sum _{i=1}^{C}\sum _{j=1}^{C} \Vert {X_i b_i + X_j b_j} \Vert _2 ^2 \right) \\&= \gamma \frac{\partial }{\partial b_k}\left( 2 \sum _{i=1,2,\ldots ,C,i\ne k} \Vert {X_i b_i + X_k b_k }\Vert _2 ^2 + \sum _{i=1,2,\ldots ,C,i\ne k}\sum _{j=1,2,\ldots ,C,j\ne k} \Vert {X_i b_i + X_j b_j}\Vert _2 ^2\right) \\&= 2 \gamma \sum _{i=1,2,\ldots ,C,i\ne k} 2X^T_k(X_kb_k+X_ib_i) \\&= 4 \gamma X^T_k\left( (C-1)X_kb_k + \sum \limits _{i=1,2,\ldots ,C,i\ne k}(X_kb_k+X_ib_i)\right) \\&= 4 \gamma X^T_k\left( (C-2)X_kb_k + \sum \limits _{i=1,2,\ldots ,C}(X_kb_k+X_ib_i)\right) \\&= 4 \gamma X^T_k((C-2){{X_kb_k}} + Xb)\\&= 4 \gamma (C-2)X^T_k{{X_kb_k}} + 4 \gamma X^T_kXb \end{aligned} \end{aligned}$$

(19)

Hence, $\frac{\mathrm {d}f}{\mathrm {d}b}$ is computed as

$$\begin{aligned} \begin{aligned} \frac{\mathrm {d}f(b)}{\mathrm {d}b}&= \left( \begin{array}{c} \frac{ \partial f}{\partial b_1} \\ \ldots \\ \frac{\partial f}{\partial b_C}\\ \end{array} \right) = \left( \begin{array}{c} 4 \gamma (C-2) X^T_1{{X_1b_1}} + 4 \gamma X^T_1Xb\\ \ldots \\ 4 \gamma (C-2) X^T_C{{X_Cb_C}} + 4 \gamma X^TXb\\ \end{array} \right) \\&= 4\gamma (C-2)\left( \begin{array}{ccc} X^T_1{{X_1}} &{}\quad \ldots &{}\quad \mathbf{0 }\\ \ldots &{}\quad \ldots &{}\quad \ldots \\ \mathbf{0 } &{} \quad \ldots &{}\quad X^T_C{{X_C}}\\ \end{array} \right) \left( \begin{array}{ccc} {{b_1}}\\ \ldots \\ {{b_C}}\\ \end{array}\right) + 4\gamma \left( \begin{array}{ccc} X^T\\ \ldots \\ X^T_C\\ \end{array}\right) Xb \\&= 4\gamma (C-2){{Hb}} +4\gamma X^TXb \end{aligned} \end{aligned}$$

(20)

where $H=\left( \begin{array}{ccc} X_1^T X_1 &{}\quad \ldots &{}\quad 0 \\ \vdots &{}\quad \ddots &{}\quad \vdots \\ 0 &{}\quad \ldots &{}\quad X_C^T X_C \\ \end{array} \right)$.

Finally, let $p = \Vert {y-Xb}\Vert _2 ^2 + \gamma \sum _{i=1}^{C}\sum _{j=1}^{C} \Vert {X_i b_i + X_j b_j} \Vert _2 ^2+\lambda \Vert {Wb}\Vert _2 ^2$, we have $\frac{\mathrm {d}p}{\mathrm {d}{b}} = ((1+2\gamma )X^TX + 2\gamma (C-2)H + \lambda W^TW){{b}} - X^Ty$. The solution of (11) is obtained when $\frac{\mathrm {d}p}{\mathrm {d}b}= 0$, which leads to $((1+2\gamma )X^TX + 2\gamma (C-2)H + \lambda W^TW){{b}} = X^Ty$. In conclusion, the solution to (11) is

$$\begin{aligned} b =((1+2\gamma )X^T X+2\gamma (C-2)H+\lambda W^T W)^- X^Ty \end{aligned}$$

(13)