## 3 Improve L1 Optimization to Correct 100% of Corrupted Propagated Errors in Network Coding
### An Example about Algorithm 1
\begin{array}{cccc}1.201&1.011&0.897&1.022\\ 0.902&0.986&0.978&1.024\\ 1.113&1.014&1.013&1.012\\ 0.967&0.972&0.997&0.977\\ \end{array} \right]\). is the estimate of \(y\prime_{B}=\left[y\prime_{\lambda+1},\ldots,y\prime_{\frac{m+n}{\mathbb{C}}}\right]\). * Reorganize the matrix \(\left(y\prime_{B}\right)^{d}\in R^{\mathbb{C}\times\left(\frac{m+n}{\mathbb{C} }-\lambda\right)}\) to the vector \(\left(y_{B}\right)^{d}\in R^{\left(m+n-\mathbb{C}\cdot\lambda\right)\times 1}\). \(\left(y_{B}\right)^{d}=y=[1.201,\ 0.902,\ 1.113,\ 0.967,\ 1.011,\ 0.986,\ 1.014,\ 0.972,\ 0.897,\ 0.978,\ 1.013,\ 0.997,\ 1.022,\ 1.024,\ 1.012,\ 0.977]^{T}(R^{16*1})\). Because \(y=\left[ \begin{array}{cccc}A\in R^{16\times 4}\\ I_{4}\end{array} \right]\times x_{0}=\left[ \begin{array}{cccc}A\times x_{0}\\ I_{4}\times x_{0}\end{array} \right]=\left[ \begin{array}{cccc}A\times x_{0}\\ I_{4}\times x_{0}\end{array} \right]=\left[ \begin{array}{cccc}A\times x_{0}\\ x_{0}\end{array} \right],y_{B}\) is the last 16 components of \(y\). Respectively, the last 4 components of \(y_{B}\) is \(x_{0}\). However, \(\left(y_{B}\right)^{d}\) is not equal to \(y_{B}\), and it is the estimate of \(y_{B}\). Thus, we cannot upload the last 4 blue components which are 1.022, 1.024, 1.012, 0.977 to 1, 1, 0, 0 where \(x_{0}=\left[1,\ 1,\ 0,\ 0\right]^{T}\). We do not know \(\kappa=\left[ \begin{array}{cccc}1\\ 1\end{array} \right]\) in advance, but we know that \(x_{0}=\left[ \begin{array}{cccc}\kappa\\ 0\\ 0\end{array} \right]\). Thus, we know 1.022, 1.024, 1.012, 0.977 will be 1.022, 1.024, 0, 0. Update \(\left(y_{B}\right)^{d}=y=[1.201,\ 0.902,\ 1.113,\ 0.967,\ 1.011,\ 0.986,\ 1.014,\ 0.972,\ 0.897,\ 0.978,\ 1.013,\ 0.997,\ 1.022,\ 1.024,\ 0,\ 0]^{T}(R^{16*1})\) The numbers with blue are components which correspond to the known numbers in 408 which can be considered as the prior knowledge. The numbers with pinkish red are 409 components which correspond to the unknown numbers in \(x_{0}\). Step 8: In the sink, receive matrix \(y^{\prime}{}_{A}=\left[ \begin{array}{c}1.012\\ 0.986\\ 1.101\\ 1.014\end{array} \right]\) through the secret channel. Denote the vector \(y_{A}=\left[ \begin{array}{c}1.012\\ 0.986\\ 1.101\\ 1.014\end{array} \right]\). * Reorganize the vector \(y_{A}\in R^{(\lambda\cdot C)\times 1}=R^{4\times 1}\) and the vector \((y_{B})^{d}\in R^{(m+n-C\cdot\lambda)\times 1}=R^{16\times 1}\) into \(y^{d}\in R^{20\times 1}=\left[ \begin{array}{c}y_{A}\\ (y_{B})^{d}\end{array} \right]\). The first 4 red components are sent by the secret channel, and the last 2 components of \(y^{d}\) are trapped by the trap matrix \(\left[ \begin{array}{c}A\in R^{16\times 4}\\ I_{4}\end{array} \right]\) which is constructed specially. \(y^{d}=\left[ \begin{array}{c}1.012,\ 0.986,\ 1.101,\ 1.014,\ 1.201,\ 0.902,\ 1.113,\ 0.967,\ 1.011,\ 0.986,\ 1.014,\ 0.972,\ 0.897,\ 0.978,\ 1.013,\ 0.997,\ 1.022,\ 1.024,\ 0. 0\end{array} \right]^{T}y=\left[ \begin{array}{c}1.012,\ 0.986,\ 1.101,\ 1.014,\ 1.052,\ 1.022,\ 1.024,\ 0.991,\ 0.957,\ 0.994,\ 1.003 0.994,\ 1.017,\ 0.990,\ 1.006,\ 1.007,\ 1,\ 1,\ 0,\ 0\end{array} \right]^{T}\). Thus, the green components are polluted. Up on \(y^{d}\), in the real field \(R\), perform L1 optimization of [PERSON] and [PERSON]'s scheme to get \(y\). Because we already know \(1.012,\ 0.986,\ 1.101,\ 1.014\) and \(0,\ 0\) of \(y\), we can certainly decoding \(y\) with [PERSON] and [PERSON]'s scheme which can recover \(100\%\) of the corrupted observations where the corrupted ratio is \((20-6)/20=70\%\). \(x_{0}=\left[ \begin{array}{c}\kappa\\ 0\\ 0\end{array} \right]=\left[ \begin{array}{c}1\\ 1\\ 0\\ 0\end{array} \right]\) is decoded successfully. The numbers with green are components which correspond to the known numbers in \(y^{d}\) which can be considered as the prior knowledge. The numbers with pinkish red are components which correspond to the unknown numbers in \(y^{d}\). * Select the first \(k\) symbols of \(y\) as \(\kappa\), that is \(\kappa=\left[ \begin{array}{c}1\\ 1\end{array} \right]\). We recover the original message successfully.