## Appendix: Proof of the faultiness of the Hypothesis 1  
In this Appendix, we present that the Hypothesis 1 proposed in [22; 23; 24] is not always correct. The proof is shown below:  
**Proof**. We can consider a codeword \(C\) with length \(n\) and the binary pattern of one of the MPCM \(Hb_{\min}(\alpha^{i})\), where \(\alpha^{i}(0\leq i\leq 2^{m}-1)\) is an element in \(\mathbf{G}_{m}\). We let \(c(x)\) be the codeword polynomial of \(C\). If \(\alpha^{i}\) is a root of \(c(x)\), then we have \(c(\alpha^{i})=0\) and \(Hb_{\min}(\alpha^{i})\times C=0\). There are \(m\) rows in \(Hb_{\min}(\alpha^{i})\) and we define \(\textbf{\emph{h}}_{j}(1\leq j\leq m)\) to be the \(j^{th}\) row of \(Hb_{\min}(\alpha^{i})\). Then the equation \(Hb_{\min}(\alpha^{i})\times C=0\) means the productions of all the rows with the codeword \(C\) equal to zeros, as shown in Equation (35):  
\[Hb_{\min}(\alpha^{i})\times C=0\Leftrightarrow \begin{cases}\textbf{\emph{h}}_ {i}\times C=0\\ \textbf{\emph{h}}_{2}\times C=0\\ \vdots\\ \textbf{\emph{h}}_{m}\times C=0\end{cases}  \tag{35}\]  
So we can calculate the probability of \(\alpha^{i}\) being the root of \(c(x)\), _i.e.,_ the probability of \(Hb_{\min}(\alpha^{i})\times C=0\) as follows:  
\[P_{r}\Big{[}Hb_{\min}(\alpha^{i})\times C=0\Big{]}=P_{r}\left(\textbf{\emph{h} }_{1}\times C=0,\textbf{\emph{h}}_{2}\times C=0,...,\textbf{\emph{h}}_{m} \times C=0\right) \tag{36}\]  
Let \(h_{j^{\prime}}(1\leq l\leq n)\) and \(C_{l}\) be the \(l^{th}\) element in the vector \(\textbf{\emph{h}}_{j}\) and \(C\) and we define the _checking indexing set_\(\textbf{S}_{j}\) for \(\textbf{\emph{h}}_{j}\) and \(C_{l}\) as follows:  
\[\textbf{S}_{j}=\{C_{l}\mid h_{j^{\prime}}=1\} \tag{37}\]  
Obviously, when the number of nonzero elements in \(\textbf{S}_{j}\) is even, we have:  
\[\textbf{\emph{h}}_{j}\times C=0 \tag{38}\]  
And when the number of nonzero elements in \(\textbf{S}_{j}\) is odd, we have:  
\[\textbf{\emph{h}}_{j}\times C=1 \tag{39}\]  
When the code length and synchronization positions are not estimated correctly, the restriction among the elements in \(C\) does exist. So the elements in the codewords can be considered to appear randomly. In this case, the probabilities of the number of nonzero elements in \(\textbf{S}_{j}\) being odd and even are all about 0.5. When \(Hb_{\min}(\alpha^{i})\) is full rank, the rows of \(Hb_{\min}(\alpha^{i})\) is linearly independent, so we can calculate Equation (36) as follows:  
\[P_{r}\Big{[}Hb_{\min}(\alpha^{i})\times C=0\Big{]}=\prod_{j=1}^{m}P_{r}\left( \textbf{\emph{h}}_{j}\times C=0\right)=(0.5)^{m} \tag{40}\]  
But if \(Hb_{\min}(\alpha^{i})\) is not full rank, the calculation of Equation (36) is not correct. We define the _maximum linearly independent vector group_**MI** of the row vectors set \(\textbf{H}\)=\(\left\{\textbf{\emph{h}}_{j}\mid 1\leq j\leq m\right\}\) as follows:  
**MI** is a subset of **H** and meets the following conditions:  
(1) The vectors in **MI** is linearly independent;  
(2) Any vector in **H** can be obtained by linear combinations of the vectors in **MI**. And it is easy to prove that the number of vectors in \(\mathbf{MI}\) is the rank of \(\mathit{Hb}_{\min}(\alpha^{i})\).