## 2 Code Length Estimation and Blind Synchronization
### Principles of the Proposed Recognition Algorithm in Soft Decision Situations
#### 2.2.2 Adaptive Processing of MPCM
Because the transformation is elementary, the relationship \(Hb_{\min\_a}\times C_{j}=0\), in the hard decision situations still exists if \(C_{j}\) is a valid codeword, so we can calculate the probability \(P_{r}(S_{j,k}=0)\) defined in Equation (14) according to \(Hb_{\min_{-}a}(\alpha^{i})\). This replacement reduces the influences of the unreliable decision bits. If \(rank(Hb_{\min_{-}a}(\alpha^{i}))=m\), the left \(m\times m\) area of \(Hb_{\min_{-}a}\) forms an identity matrix. But if \(rank(Hb_{\min_{-}a}(\alpha^{i}))<m\), \(Hb_{\min_{-}a}\) becomes the following style after the elementary transformations:  
\[Hb_{\min_{-}a}(\alpha^{i})= \begin{bmatrix}1&0&0&\cdots&x&x\\ 0&1&0&\cdots&x&x\\ 0&0&1&\cdots&x&x\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\cdots&0&0\\ 0&0&0&\cdots&0&0\\ 0&0&0&\cdots&0&0\end{bmatrix}  \tag{19}\]  
In this case, only the first \(p\) rows are non-zeros, where \(p=rank(Hb_{\min_{-}a}(\alpha^{i}))\). The last \(m-p\) rows have no contribution for the calculation of \(P_{r}(\mathbf{S}_{j,i}=0)\) in Equation (11). So we modify Equation (11) and Equation (20):  
\[p_{j,i}=P_{r}(\mathbf{S}_{j,i}=0)=\frac{1}{rank(Hb_{\min_{-}a}(\alpha^{i}))} \sum_{k=1}^{rank(Hb_{\min_{-}a}(\alpha^{i}))}P_{r}(S_{j,k}=0) \tag{20}\]  
Now based on Equation (20), Equation (14) and Equation (7), we can calculate the RIDEF defined in Equation (3). However, there is still a problem in the algorithm. Note that the element \(\alpha^{2^{m-1}}=1\) is a root of a codeword polynomial only if the code weight is even. This fact severely affects the assumption of uniformity of \(p_{i}\). As shown in Figure 1, in the case of recognition of BCH (63, 51), when we assume the code length is 31, the value of \(p_{31}\) is obviously larger than other \(p\) values, although the real code length is not 31 and \(\alpha^{31}=1\) is not a root of the code. Figure 1: The problem of \(p_{31}\). To avoid that, we drop the calculation of the probability \(P_{r}(\mathbf{S}_{j,i}=0)\) on the element \(\alpha^{2^{n-1}}=1\) when recognizing the code length. So we modify Equation (3) and Equation (7) to Equation (21) and Equation (22) respectively as follows:  
\[\Delta H=\sum_{i=1}^{2^{n-2}}p_{i}\log p_{i}+\log\left(2^{n}-2\right) \tag{21}\]  
\[p_{i}=\frac{\sum_{j=1}^{M}p_{j,i}}{\sum_{i=1}^{2^{n-2}}\sum_{j=1}^{M}p_{j,i}} \tag{22}\]  
Considering that not all the rows in the parity check matrix \(Hb_{\min_{a}a}(\alpha^{i})\) have the same value of \(n_{e}\) [in Equation (14)], and this affects the comparability of \(P_{r}(S_{j,k}=0)\) among different rows. For the uniformization reasons, so we modify Equation (14) to:  
\[ \begin{split} P_{r}[S_{j,k}=0]&=\frac{1}{2}+\frac{1 }{2}\operatorname{sign}\left[\prod_{n=1}^{n_{e}}\tanh(r_{u}\ /\ \sigma^{2})\right]\left[\prod_{n=1}^{n_{e}}\tanh(r_{u}\ /\ \sigma^{2})\right]^{\frac{1}{n_{e}}}\\ &=\frac{1}{2}+\frac{1}{2}\prod_{n=1}^{n_{e}}\operatorname{sign}(r _{u})\times\left[\prod_{n=1}^{n_{e}}\tanh(r_{u}\ /\ \sigma^{2})\right]^{\frac{1}{n_{e}}}\end{split}  \tag{23}\]