Robust 3D Mesh Watermarking Scheme for an Anti-Collusion Fingerprint Code

Abstract

Collusion attack is one of the techniques used for unauthorized removal of embedded marks. In this paper, we propose a robust 3D mesh fingerprinting scheme for an anti-collusion code. In contrast to the existing robust mesh watermarking which provides unsuitable primitives for anti-collusion code, the proposed method has well-operated capacity to carry the anti-collusion fingerprint code. In order to minimize the detection error, we also modeled the response of the detector and herein present optimized thresholds for our method. Based on the experiments, the proposed method outperformed conventional robust mesh watermarking against collusion attack in all cases.

6 Appendix

In this appendix, we focus on mathematical derivation, in order to provide the threshold value $\tau _b$ illustrated in Fig. 1(c). From Eq. 14, $\tau _b$ is value that makes the derivative of $e_1+e_2$ by t to be zero.

$$\begin{aligned} \frac{de_1}{dt}= \left( \frac{1}{n}\right) ^n \cdot f ( t | 1 , \sigma ^2 ) \end{aligned}$$

(15)

$$\begin{aligned} \frac{de_2}{dt}= -\sum _{k=0}^{n-1} \left( \left( {\begin{array}{c}n\\ k\end{array}}\right) \frac{1}{n}^k (1- \frac{1}{n})^{n-k} \cdot f ( t | \frac{k}{n} , \sigma ^2 ) \right) \end{aligned}$$

(16)

Substitute $\left( {\begin{array}{c}n\\ k\end{array}}\right) \frac{1}{n}^k (1- \frac{1}{n})^{n-k}$ to $\alpha _k$ for $k=1$ to n.

$$\begin{aligned} \frac{de_1+de_2}{dt} = \alpha _n \cdot f ( t | 1 , \sigma ^2 ) - \left( \sum _{k=0}^{n-1} \alpha _k \cdot f ( t | \frac{k}{n} , \sigma ^2 ) \right) \end{aligned}$$

(17)

$$\begin{aligned} = \frac{1}{\sqrt{2\pi }\sigma } \cdot \left( \alpha _n \cdot e^{- \frac{(t-1)^2}{2\sigma ^2}} - \left( \sum _{k=0}^{n-1} \alpha _k \cdot e^{-\frac{(t-\frac{k}{n})^2}{2\sigma ^2}} \right) \right) \end{aligned}$$

(18)

$$\begin{aligned} =\frac{1}{\sqrt{2\pi }\sigma }e^{-\frac{t^2}{2\sigma ^2}} \cdot \left( \alpha _n e^{\frac{2t-1}{2\sigma ^2}} - \left( \sum _{k=0}^{n-1} \alpha _k e^{\frac{\frac{2k}{n}t-\frac{k^2}{n^2}}{2\sigma ^2}} \right) \right) \end{aligned}$$

(19)

Substitute $e^{\frac{t}{n\sigma ^2}}$ to x and $\alpha _k e^{-\frac{k^2}{2n^2\sigma ^2}}$ to $\beta _k$ for $k=1$ to n.

$$\begin{aligned} =\frac{1}{\sqrt{2\pi }\sigma } e^{-\frac{t^2}{2\sigma ^2}} \cdot \left( \beta _n x^n - \sum _{k=0}^{n-1} \beta _k x^k \right) \end{aligned}$$

(20)

Therefore, the value $\tau _b$ that minimizes $e_1+e_2$ can be derived from solving following polynomial of degree n.

$$\begin{aligned} \beta _n x^n - \sum _{k=0}^{n-1} \beta _k x^k=0 \end{aligned}$$

(21)

Finally, we can get $\tau _b$ by using solution of (Eq. 21)

$$\begin{aligned} \tau _b= n \cdot \sigma ^2 \cdot ln \left( x \right) \end{aligned}$$

(22)