# An Untraceable Data Sharing Scheme in Wireless Sensor Networks
## 5 Secure Data Sharing between the Source Node and Sink Node
### Light-Weight Secret Sharing Scheme
\begin{array}{cccccc}0&\cdots&1&\cdots &0&\cdots&0\\ \vdots&&\vdots&&\vdots&&\vdots\\ 0&\cdots&0&\cdots&1&\cdots&0\\ 1&\cdots&k_{1}^{k_{i+1}-1}&\cdots&k_{i}^{k_{i+1}-1}&\cdots&t^{k_{i+1}-1}\\ \vdots&&\vdots&&\vdots&&\vdots\\ 1&\cdots&k_{1}^{k_{t}-1}&\cdots&k_{i}^{k_{t}-1}&\cdots&t^{k-1}\end{array}  \right)\left( \begin{array}{c}s_{1}\\ \vdots\\ s_{i}\\ s_{i+1}\\ \vdots\\ s_{t}\end{array} \right)=D\left( \begin{array}{c}s_{1}\\ \vdots\\ \vdots\\ s_{i+1}\\ \vdots\\ s_{t}\end{array} \right). \tag{6}\]  
We can compute determinant of \(D\) and find that  
\[|D|=\underset{u>u,u,v\
eq k_{1},\cdots,k_{i}}{\Pi}(u-v)\underset{v\
eq k_{1}, \cdots,k_{i}}{\Pi}v^{k_{i+1}-1}\
eq 0. \tag{7}\]  
Therefore, the matrix \(D\) is invertible and \(s_{1}\), \(s_{2},\ldots,\ s_{t}\) can be linearly expressed by \(s_{t+k_{i+1}},s_{t+k_{i+2}},\cdots,s_{t+k_{t}}\). Based on Case 1, we prove that the equations in Case 2 also have a unique solution. By combining Case 1 and Case 2, Theorem 1 is proved. Based on Theorem 1, we can guarantee that the sink node can recover the original message \(X\) once at least \(t\) shares are received. Now, we analyze the security of the secret sharing scheme and prove its security in Theorem 2. In the network, the adversary may eavesdrop and decrypt the sharesto recover \(X\). However, we can prove that even the adversary successfully decrypts a set of the shares, they cannot recover \(X\). **Theorem 2**: _Any \(T(T<t)\) shares of \(X\) cannot recover message \(X\)._  
Assume that the adversary successfully gets the plaintexts, \(s^{\prime}_{1},~{}s^{\prime}_{2},\cdots s^{\prime}_{T}\), of \(T(T<t)\) shares. Based on Equation (1), the adversary can construct a set of equations with \(t\) variables \(x_{1},x_{2},\cdots,x_{t}\) as shown in Equation (8). \[A=H_{T\times t}B, \tag{8}\]  
where \(A=\left(s^{\prime}_{1},~{}s^{\prime}_{2},\cdots s^{\prime}_{T}\right)^{\prime}\), \(B=\left(x_{1},x_{2},\cdots,x_{t}\right)^{\prime}\). Let \(F\) be a field and \(H_{T\times t}\) be a matrix over \(F\). We further assume that all the numbers in \(A\) and \(B\) are over filed \(F\). Consider the augmented matrix, \(HA_{T\times(t+1)}\), of \(H_{T\times t}\) and we can in infer that \(HA_{T\times(t+1)}=\left(H_{T\times t}~{}A\right)\). Based on the ranks of \(H_{T\times t}\) and \(A\), we prove Theorem 2 by the following two cases:  
Case 1: The rank of \(H_{T\times t}\) is smaller than that of \(HA_{T\times(t+1)}\). In this case, the equation set has no solution and the adversary cannot recover message \(X\). This may happen when the adversary gets the wrong numbers of \(s^{\prime}_{1},~{}s^{\prime}_{2},\cdots s^{\prime}_{T}\). Case 2: The rank of \(H_{T\times t}\) equals the rank of \(HA_{T\times(t+1)}\). In this case, there are \(T\) equations and \(t\) variables. Therefore, the adversary cannot accurately recover message \(X\). We consider the worst case and assume that \(T=t-1\), i.e., the adversary gets \(t-1\) shares. In this case, the adversary can get \(|F|\) lawful solutions. By combining Case 1 and Case 2, we can infer that it is impractical for the adversary to recover message \(X\) by \(T(T<t)\) shares for a field of a large size. Theorem 2 is proved. In our scheme, the shares are delivered to the sink node independently and it is extremely difficult for the adversary to capture more than \(t\) shares at the same time. Consequently, our secret sharing scheme is safe.