## 3 The Proposed Algorithm
### Features of the Interpolating Spline Curve in 3D Space  
With each \(\mathbf{F}_{i}\) we have three vertices \(\{v_{i,j}|j\in[1,3]\}\). Each facet \(\mathbf{F}_{i}\) is used to interpolate a spline curve of degree 2 in 3D space. This spline curve passes three vertices of the facet and is controlled by three control points (see Figure 3) [15; 16]. However, these control points are determined by an interpolating vector and curvature coefficients. Thus, we could conclude that an interpolating spline curve of degree 2 in 3D space is determined by three control points, curvature coefficients, and an interpolating vector. The interpolating vector \(\mathbf{T}_{i}=\{t_{i,j}|j\in[1,3]\}\) is a set of numbers therein \(t_{i,j}<t_{i,j+1}\). The interpolating vector \(\mathbf{T}_{i}\) is then used to calculate the curvature coefficients. In order to respond to the condition \(t_{i,j}<t_{i,j+1}\), we calculate \(t_{i,j}\) by using indexes \(i,j\), the value of \(\,|\,\mathbf{M}\,|\,\) and \(\mathbf{K}\) as described in Equation (2). The SHA-512 hashing function [17] is used to generate the key value \(\mathbf{K}\) with a user's key input. Each key value has a length of 512 bits. \[t_{i,j}=\frac{i\cdot(j+1)}{|\mathbf{M}|}\times\mathbf{K}. \tag{2}\]  
With the three vertices of the facet and \(\mathbf{T}_{i}\), we can compute the curvature coefficients and three control points of spline curve of degree 2. The relationship between three control points \(\mathbf{C}_{i}=\{c_{i,k}|k\in[1,\,3]\}\), \(v_{i,j}\) and \(t_{i,j}\) is defined as follows:  
\[v_{i,j}=\sum_{k=1}^{3}B_{k,d}(t_{i,j})\cdot c_{i,k\[B_{k,0}\big{(}t_{ij}\big{)}=\left\{ \begin{array}{ll}1&\mbox{if $t_{i,k}\leq t_{i,j}<t_{i,k+1}$}\\ 0&\mbox{otherwise}\end{array} \right.. \tag{5}\]  
From Equation (3), we can see that there are three curvature coefficients \(\{B_{1,2}(t_{i,j})\), \(B_{2,2}(t_{i,j})\), \(B_{3,2}(t_{i,j})\}\) with each \(t_{i,j}\). Therefore, we have 3 \(\times\) 3 coefficients with the interpolating vector \(\mathbf{T}_{i}\). These coefficients can be organized into a 3 \(\times\) 3 matrix \(\mathbf{A}_{i}\) in which the \(j^{th}\) row contains \(B_{1,2}(t_{i,j})\), \(B_{2,2}(t_{i,j})\), \(B_{3,d}(t_{i,j})\):  
\[\mathbf{A}_{i} = \left[ \begin{array}{ll}B_{1,2}\big{(}t_{i,1}\big{)}&B_{2,2} \big{(}t_{i,1}\big{)}&B_{3,2}\big{(}t_{i,1}\big{)}\\ B_{1,2}\big{(}t_{i,2}\big{)}&B_{2,2}\big{(}t_{i,2}\big{)}&B_{3,2}\big{(}t_{i,2 }\big{)}\\ B_{1,2}\big{(}t_{i,3}\big{)}&B_{2,2}\big{(}t_{i,3}\big{)}&B_{3,2}\big{(}t_{i,3 }\big{)}\end{array} \right] \tag{6}\] \[= \left[ \begin{array}{ll}a_{1,1}&a_{1,2}&a_{1,3}\\ a_{2,1}&a_{2,2}&a_{2,3}\\ a_{3,1}&a_{3,2}&a_{3,3}\end{array} \right]\] \[= \left\{\begin{array}{ll}a_{j,k}\Big{|}j,k\in[1,\ 3]\ \big{\}}\mbox{ with }a_{j,k}=B_{k,2}\big{(}t_{i,j}\big{)}.\]  
Now, we apply three vertices of \(\mathbf{F}_{i}\) and \(t_{ij}|j\in[1,3]\) to Equation (3), the relationship between \(\mathbf{F}_{i}\), \(\mathbf{T}_{i}\), and \(\mathbf{C}_{i}\) is described as follows:  
\[\left[ \begin{array}{c}v_{i,1}\\ v_{i,2}\\ v_{i,3}\end{array} \right] = \left[ \begin{array}{ll}B_{1,2}\big{(}t_{i,1}\big{)}&B_{2,2} \big{(}t_{i,1}\big{)}&B_{3,2}\big{(}t_{i,1}\big{)}\\ B_{1,2}\big{(}t_{i,2}\big{)}&B_{2,2}\big{(}t_{i,2}\big{)}&B_{3,2}\big{(}t_{i,2 }\big{)}\\ B_{1,2}\big{(}t_{i,3}\big{)}&B_{2,2}\big{(}t_{i,3}\big{)}&B_{3,2}\big{(}t_{i,3 }\big{)}\end{array} \right]\cdot\left[ \begin{array}{c}c_{i,1}\\ c_{i,2}\\ c_{i,3}\end{array} \right]\] \[= \left[ \begin{array}{ll}a_{1,1}&a_{1,2}&a_{1,3}\\ a_{2,1}&a_{2,2}&a_{2,3}\\ a_{3,1}&a_{3,2}&a_{3,3}\end{array} \right]\cdot\left[ \begin{array}{c}c_{i,1} \\ c_{i,2}\\ c_{i,3}\end{array} \right]\] \[\Leftrightarrow\mathbf{F}_{i}=\mathbf{A}_{i}\cdot\mathbf{C}_{i}\] \[\mathbf{C}_{i}=\left(\mathbf{A}_{i}\right)^{-1}\cdot\mathbf{F}_{i}\] \[\Leftrightarrow \left[ \begin{array}{c}c_{i,1}\\ c_{i,2}\\ c_{i,3}\end{array} \right]=\left(\left[ \begin{array}{ll}a_{1,1}&a_{1,2}&a_{1,3 }\\ a_{2,1}&a_{2,2}&a_{2,3}\\ a_{3,1}&a_{3,2}&a_{3,3}\end{array} \right]\right)^{-1}\cdot\left[