Finger Vein Recognition Based on Oval Parameter-Dependent Convolutional Neural Networks

Abstract

The Gabor modulating convolutional neural network (CNN), which incorporates Gabor filter modules parallel to the convolutional layers, has made remarkable achievements in the finger vein recognition tasks. However, the Gabor module requires a bit of additional calculation and is only suitable for the shallow layers. Aiming at this problem, we proposed an oval parameter-dependent CNN (PDCNN) which is developed from the Gabor modulating CNN in two aspects but has superior performance. First, in the oval PDCNN, $3\times 3$ convolutional kernels of the first several layers are replaced by $3\times 3$ oval parameter-dependent kernels (PDKs) which are determined by 5 or fewer parameters according to a nonlinear oval function. The oval PDK can provide additional nonlinearity for feature extraction while reducing the number of parameters. In contrast to Gabor modulating modules, the oval PDKs are no longer restricted to shallow layers. Second, since the Gaussian component of the Gabor filter does not improve the network’s ability in feature extraction but rather increases the training difficulty, we remove the Gaussian component from the oval PDK to make it much easier to train. Two lightweight oval PDCNNs, with MobileNet and SqueezeNet as the basic architecture, are investigated. To illustrate the superiority of the proposed oval PDCNN, two experiments have been conducted. The first experiment compares the oval PDK with 4 other PDKs, including Gabor, cos, cross, and x, on three public finger vein datasets. The results illustrate that the oval PDCNN reduces the size of MobileNet and SqueezeNet by 0.34% and 30.36% without degrading recognition performance. Another experiment is to fit the convolutional kernels of well-trained MobileNet and SqueezeNet with PDKs to analyze their tendency. The kernels in shallow layers are not closer to any PDKs than in deep layers, which is different from the viewpoint that the property of shallow layers is close to the Gabor filter, and all kernels have not shown bias on any PDKs. It demonstrates that the advantage of oval PDK lies in its property of being easier to train than other PDKs.

Appendix A

According to the definition of kernel in Eqs. (4) and (1), the elements $K_{\text {real}}^{\Theta }$ can be re-written as below:

$$\begin{aligned} K_{11}&=e^{c_1} \cos \left( -2\pi \frac{\sin \theta +\cos \theta }{\lambda }+\psi \right) , \end{aligned}$$

(A1a)

$$\begin{aligned} K_{21}&=e^{c_2}\cos \left( -2\pi \frac{\cos \theta }{\lambda }+\psi \right) , \end{aligned}$$

(A1b)

$$\begin{aligned} K_{31}&=e^{c_3}\cos \left( 2\pi \frac{\sin \theta -\cos \theta }{\lambda }+\psi \right) , \end{aligned}$$

(A1c)

$$\begin{aligned} K_{12}&=e^{c_4}\cos \left( -2\pi \frac{\sin \theta }{\lambda }+\psi \right) , \end{aligned}$$

(A1d)

$$\begin{aligned} K_{22}&=\cos \psi , \end{aligned}$$

(A1e)

$$\begin{aligned} K_{32}&=e^{c_4}\cos \left( 2\pi \frac{\sin \theta }{\lambda }+\psi \right) , \end{aligned}$$

(A1f)

$$\begin{aligned} K_{13}&=e^{c_3}\cos \left( -2\pi \frac{\sin \theta -\cos \theta }{\lambda }+\psi \right) , \end{aligned}$$

(A1g)

$$\begin{aligned} K_{23}&=e^{c_2}\cos \left( 2\pi \frac{\cos \theta }{\lambda }+\psi \right) , \end{aligned}$$

(A1h)

$$\begin{aligned} K_{33}&=e^{c_1}\cos \left( 2\pi \frac{\sin \theta +\cos \theta }{\lambda }+\psi \right) , \end{aligned}$$

(A1i)

where

$$\begin{aligned} c_1&= {-\frac{\left( \sin \theta +\cos \theta \right) ^2+\gamma ^2 \left( \sin \theta -\cos \theta \right) ^2}{2\sigma ^2}},\\ c_2&= {-\frac{\cos ^2\theta +\gamma ^2 \sin ^2\theta }{2\sigma ^2}},\\ c_3&= {-\frac{\left( \sin \theta -\cos \theta \right) ^2+\gamma ^2 \left( \sin \theta +\cos \theta \right) ^2}{2\sigma ^2}},\\ c_4&= {-\frac{\sin ^2\theta +\gamma ^2 \cos ^2\theta }{2\sigma ^2}}. \end{aligned}$$

For a given kernel $T = [t_{ij}]_{3\times 3}$, i.e., the left of the equations are constant, we can transform the original optimization problem to an indirective one in three steps: Step1 First, normalize T according to $T=t/{ \left|t_{22} \right|}$ to make $t_{22} \in \{1,-1\}$. Then, according to Eq. (A1e), there exists:

$$\begin{aligned} \psi = \arccos t_{22}, \end{aligned}$$

(A3)

Step2 Let $x = 2\pi sin \theta / \lambda$ and $y = 2 \pi cos \theta / \lambda$, and substitute x, y, and $\psi$ into Eqs. (A1b), (A1d), (A1f), and (A1h), respectively. Then, we have

$$\begin{aligned} \begin{array}{c} \dfrac{t_{12}}{t_{32}}=\dfrac{\cos x\cos \psi +\sin x\sin \psi }{\cos x\cos \psi -\sin x\sin \psi }, \\ \dfrac{t_{21}}{t_{23}}=\dfrac{\cos y\cos \psi +\sin y\sin \psi }{\cos y\cos \psi -\sin y\sin \psi }. \end{array} \end{aligned}$$

(A4)

The solution of Eq. (A4) is

$$\begin{aligned} \begin{array}{c} x=a \pi +\arctan c_5,\\ y=b \pi +\arctan c_6. \end{array} \end{aligned}$$

(A5)

where $c_5\!=\!\dfrac{t_{12}cos\psi -t_{32}cos\psi }{t_{12}sin\psi -t_{32}sin\psi }$ and $c_6\!=\!\dfrac{t_{21}cos\psi -t_{23}cos\psi }{t_{21}sin\psi -t_{23}sin\psi }$, and $a,b \in \mathbb {Z}$ are to be determined. Thus, $\theta$ and $\lambda$ are functions of a and b, which are written as $\theta (a,b)$ and $\lambda (a,b)$, satisfying the relation below:

$$\begin{aligned} {\left\{ \begin{array}{ll} \sin \theta -\frac{1}{2\pi }\left( a\pi +\textrm{arc}\tan a \right) \lambda =0\\ \cos \theta -\frac{1}{2\pi }\left( bpi +\textrm{arc}\tan b \right) \lambda =0 \\ \end{array}\right. } \end{aligned}$$

(A6)

Step3 Given a group of (a, b), substituting $\psi$, $\theta (a,b)$, and $\lambda (a,b)$ into Eqs. (A1b), (A1d), (A1f), and (A1h), we can obtain a solution $\Theta (a,b)$ and the PDK $K_{\text {real}}^{\Theta (a,b)}$. However, $\Theta (a,b)$ may not satisfy Eqs. (A1a), (A1c), (A1g), and (A1i). Thus, the original optimal solution is transformed to search a group of (a, b) that minimizes

$$\begin{aligned} J=\sum _{i=1,3}{\sum _{j=1,2}{\Vert t_{ij}-K_{ij}^{\Theta (a,b)} \Vert _2}}, \end{aligned}$$

(A7)

Because nearly all of the elements of the kernel lie in $[-3,3]$, we can narrow the region of (a, b) to $[-10,10]$.

$\varepsilon _{\text {oval},i,j}$ and $\varepsilon _{\text {cos},i,j}$ can be calculated in the same manner.