## 2 Preliminaries and Related Works
### Network Coding and Its Fundamental Concepts
Divide the matrix \(y\prime\) into two parts: \(y\prime=\left[y\prime_{A}\;;\;y\prime_{B}\right]\) where \(y\prime_{A}=\left[y\prime_{1},\ldots,y\prime_{\lambda}\right]\) and \(y\prime_{B}=\left[y\prime_{\lambda+1},\ldots,y\prime_{\frac{m+n}{C}}\right]\). Send the matrix \(y\prime_{A}\in R^{C\times\lambda}\) through a secret channel. Send the matrix \(y\prime_{B}\in R^{C\times(\frac{m+n}{C}-\lambda)}\) through the networks where the networks coding is performed in the relay nodes. Each time, \(y\prime_{j}\in R^{C\times 1}\)\((j=\lambda+1,\lambda+2,\ldots,\frac{m+n}{C})\) which is a column of \(y\prime_{B}\), is sent with network coding method. The matrix \(y\prime_{B}\) is needed to be sent \(\frac{m+n}{C}-\lambda\) times. Step 5 In the sink, receive the matrix \(y\prime_{A}=\left[y\prime_{1},\ldots,y\prime_{\lambda}\right]\) through a secret channel and the matrix \(Y_{B}\) through the networks where the networks coding is performed. \(Y_{B}=\left[Y\prime_{\lambda+1},\ldots,Y\prime_{\frac{m+n}{C}}\right]\) responds to \(y\prime_{B}=\left[y\prime_{\lambda+1},\ldots,y\prime_{\frac{m+n}{C}}\right]\). After the network coding effect and errors effect, \(y\prime_{B}\) will become \(Y_{B}\). Based on Equation (11), we know the received message responding \(y\prime_{j}\) is \(Y_{j}=\hat{T}\cdot y\prime_{j}+T_{Z\to Y}\cdot\left(Z-L\cdot y\prime_{j} \right)\)\((j=\lambda+1,\lambda+2,\ldots,\frac{m+n}{C})\). Step 6 Perform network coding decoding algorithm to \(Y_{B}=\left[Y_{\lambda+1},\ldots,Y_{\frac{m+n}{C}}\right]\). In the finite field \(F_{q}\), perform decoding of the network coding scheme in every sink to get \(\hat{T}^{-1}\). In the real field \(R\), compute \(\left(y\prime_{j}\right)^{d}=\hat{T}^{-1}\cdot Y_{j}\)\((j=\lambda+1,\lambda+2,\ldots,\frac{m+n}{C})\) in random network, where \(\left(y\prime_{j}\right)^{d}\) is the estimate of \(y\prime_{j}\). After \(\frac{m}{C}-\lambda\) times, we get \(\left(y\prime_{B}\right)^{d}=\left[\left(y\prime_{\lambda+1}\right)^{d},\ldots,\left(y\prime_{\frac{m+n}{C}}\right)^{d}\right]\)\((j=\lambda+1,\lambda+2,\ldots,\frac{m+n}{C})\) where \(\left(y\prime_{B}\right)^{d}\) is the estimate of \(y\prime_{B}=\left[y\prime_{\lambda+1},\ldots,y\prime_{\frac{m+n}{C}}\right]\). Reorganize the matrix \(\left(y\prime_{B}\right)^{d}\in R^{C\times(\frac{m+n}{C}-\lambda)}\) to the vector \(\left(y_{B}\right)^{d}\in R^{(m+n-C\cdot\lambda)\times 1}\). Step 7 Reorganize the matrix \(\left(y\prime_{B}\right)^{d}\in R^{C\times(\frac{m+n}{C}-\lambda)}\) to the vector \(\left(y_{B}\right)^{d}\in R^{(m+n-C\cdot\lambda)\times 1}\). Divide \(\left(y_{B}\right)^{d}\) into two parts and it is \(\left(y_{B}\right)^{d}=\left[ \begin{array}{c}\left(y_{(m+n-C\cdot\lambda)- (n-k)}\right)^{d}\\ \left(y\prime_{n-k}\right)^{d}\end{array} \right]\). Set \(\left(y_{n-k}\right)^{d}=0^{(n-k)\times 1}\). Update \(\left(y_{B}\right)^{d}=\left[ \begin{array}{c}\left(y_{(m+n-C\cdot\lambda)- (n-k)}\right)^{d}\\ 0^{(n-k)\times 1}\end{array} \right]\). Step 8 In the sink, receive the matrix \(y\prime_{A}\) and then reorganize the matrix \(y\prime_{A}\in R^{C\times\lambda}\) into the vector \(y_{A}\in R^{(\lambda\cdot C)\times 1}\). Step 9 Reorganize the vector \(y_{A}\in R^{(\lambda\cdot C)\times 1}\) and the vector \(\left(y_{B}\right)^{d}\in R^{(m+n-C\cdot\lambda)\times 1}\) into \(y^{d}=\left[ \begin{array}{c}y_{A}\\ \left(y_{B}\right)^{d}\end{array} \right]\). Up on \(y^{d}\), in the real field \(R\), perform L1 optimization which is based on [PERSON] and [PERSON]'s scheme to get \(y\). Step 10 Select the first \(k\) symbols of \(y\) as \(\kappa\). ```  
**Algorithm 1**LOECNC AlgorithmDecoding for network coding: In a sink, for example, sink 6, the decoding matrix is \(\left[ \begin{array}{c}afaeik+cgjk\\ bfbeik+dgjk\\ afaeik+cgjk\\ bfbeik+dgjk\end{array} \right]\). If the vectors \(\left[ \begin{array}{c}af\\ bf\end{array} \right]\) and \(\left[ \begin{array}{c}aeik+cgjk\\ beik+dgjk\end{array} \right]\) are linearly independent or the matrix \(\left[ \begin{array}{c}afaeik+cgjk\\ bfbeik+dgjk\end{array}