## 4 Experimental Results, Analysis, and Evaluation
### Security Evaluation  
We analyzed the length of the secret key and the entropy of the encrypted 3D triangle mesh to evaluate the security of the proposed algorithm. If the secret key length is long, it is difficult to attack the encrypted 3D triangle mesh. If the entropy of the encrypted 3D triangle mesh is high, the security of the encrypted 3D triangle mesh will be high. In order to decrypt the encrypted 3D triangle mesh, a pirate must decrypt all of the encrypted facets of the 3D triangle mesh without the secret keys. In the proposed algorithm, the SHA-512 hashing function with a 512 bit salt is used to obtain random keys. Thus, if a user inputs a word with length \(L_{k}\) as his/her password, an attacker has to compute \(L_{k}\times 2^{512}\) keys to access the encrypted 3D triangle mesh. The entropy \(H(x)\) of a discrete random variable \(x\) with a possible value \(\{x_{1},x_{2},\ldots,x_{n}\}\) is defined as  
\[H(x)=-\sum_{\tau=1}^{n}p(x_{\tau})\cdot log_{2}p(x_{\tau}) \tag{15}\]  
where \(p(x_{\tau})\) is the probability density function of \(x\) on range \(\{x_{1},x_{2},\ldots,x_{n}\}\). Based on the equations in Section 3, the entropy of the encrypted 3D triangle mesh is dependent on both the secret key \(\mathbf{K}\), the interpolating vector \(\mathbf{T}_{i}\), the matrix of curvature coefficients \(\mathbf{A}_{i}\), three control points \(\mathbf{C}_{i}\), and the number of facets \(|\mathbf{M}|\). However, \(\mathbf{K}\), \(\mathbf{T}_{i}\), \(\mathbf{A}_{i}\), \(\mathbf{C}_{i}\), and \(|\mathbf{M}|\) are random independent variables. Therefore,  
Figure 6: Perceptual encryption results of our proposed method and [PERSON]’s method. (**a,c**) Results of [PERSON]’s method; (**b,d**) results of the proposed method. the entropy of the encrypted 3D triangle mesh \(H_{\mathbf{M}}\) is the sum of the entropies of variables \(\mathbf{K}\), \(\mathbf{T}_{i}\), \(\mathbf{A}_{i}\), \(\mathbf{C}_{i}\), and \(\mid\!\mathbf{M}\!\mid\) and is determined by Equation (16):  
\[ \begin{array}{ll}H_{\mathbf{M}}&=H(\mathbf{K})+H(\mathbf{T}_{i})+H(\mathbf{ A}_{i})+H(\mathbf{C}_{i})+H(\mid\!\mathbf{M}\!\mid)\\ &=\left|\mathbf{K}\!\right|\cdot\log_{2}\!\left|\mathbf{K}\right|)+\left| \mathbf{T}_{i}\!\right|\cdot\log_{2}\!\left|\mathbf{T}_{i}\right|+\left| \mathbf{A}_{i}\right|\cdot\log_{2}\!\left|\mathbf{A}_{i}\right|+\left| \mathbf{C}_{i}\right|\cdot\log_{2}\!\left|\mathbf{C}_{i}\right|+\left| \mathbf{M}\!\right|\cdot\log_{2}\!\left|\mathbf{M}\right|).\end{array}  \tag{16}\]  
\(\left|\mathbf{T}_{i}\right|=\left|\mathbf{C}_{i}\right|=3\) and \(\left|\mathbf{A}_{i}\right|=9\). In summary, the total entropy \(H_{\mathbf{M}}\) is dependent on \(\mathbf{K}\) and the number of facets \(\mid\!\mathbf{M}\!\mid\) in the 3D triangle mesh. If the secret key \(\mathbf{K}\) is fixed, we can calculate the entropy of the encrypted 3D triangle mesh according to the number of facets \(\mid\!\mathbf{M}\!\mid\) as shown in Table 1. The entropy of the encrypted 3D triangle mesh is formed from 4981 to 1.3 \(\times\) 10\({}^{6}\) dB with \(\mid\!\mathbf{M}\!\mid\in\) [544, 79162]. From Equation (16) and Table 1, we can see that, if \(\mid\!\mathbf{M}\!\mid\) is high, the entropy will be high. [PERSON] [8] divided the faces of a 3D model in \(N\) share hyper-planes before sharing. This mean a 3D model is divided into \(N\) parts before sharing. Thus, \(N\) is the secret key, and the entropy of [PERSON]'s method is always fixed. In [PERSON]'s method, \(N=4\), so the entropy of [PERSON]'s method is always 8 dB. Moreover, in the 3D space of the 3D model, the maximum number of hyper-planes is 360. Therefore, the maximum entropy of [PERSON]'s method is 3057 dB. [PERSON] used the secret key \(\mathbf{K}\) to encrypt and change the location of the vertices of the 3D mesh in OXYZ space. Clearly, [PERSON]'s method encrypted the vertices of the 3D mesh with a secret key \(\mathbf{K}\). However, the number of vertices in a 3D mesh is always smaller than the number of facets. Thus, the entropy of this method is always lower than the proposed method.