## 3 Candidate Generation in Parallel Decoding of Rate-1 Nodes
### Candidate Generation Based on Partially Ordered Candidate Sets
An example of a partially ordered set of candidates for \(P=5\) and a constant number of flipped bits \(x=\mathrm{hw}(j)=2\) is given in Figure 4. We chose these parameters because the resulting graph depicts the correlation between the partial order and the binary expansion of \(j\) concretely. We introduce the stage \(z\) as the index in a chain of the partially ordered set of all candidates with constant \(x=\mathrm{hw}(j)\). For such a candidate, we write \(j_{x,z}\). The smallest candidate index for a given \(x\) is  
\[\operatorname{PM}_{p^{a}01q^{b}}^{l}\leq\operatorname{PM}_{p^{a}10q^{b}}^{l}. \tag{22}\]0  
From \(j_{x,z}\) within stage \(z\), \(j_{x,z+1}\) is obtained by swapping a more significant 0 with the next less significant 1 in the binary expansion of \(j_{x,z}\). We redefine the smallest possible candidate index originally given in (13) as  
\[\operatorname{PM}_{p^{a}01q^{b}}^{l}\leq\operatorname{PM}_{p^{a}10q^{b}}^{l}. \tag{22}\]1  
The values of the redefined \(m_{l,j}\) for \(L=8\) and \(P=4\) are shown in \[\operatorname{PM}_{p^{a}01q^{b}}^{l}\leq\operatorname{PM}_{p^{a}10q^{b}}^{l}. \tag{22}\]2 In addition, the smallest possible index \(m_{l,j}\) is displayed above the directed graph (Figure 4), to show how it is increased for each stage \(z\). \[\operatorname{PM}_{p^{a}01q^{b}}^{l}\leq\operatorname{PM}_{p^{a}10q^{b}}^{l}. \tag{22}\]3 Directed graph of the partially ordered set of candidates \(j_{2}\) (blue) with corresponding binary expansion (black) for \(P=5\) and \(x=2\). Based on the observation that \(m_{l,j_{x,z}}<m_{l,j_{x,z+1}}\) when \(j_{x,z+1}\) is obtained by swapping a more significant 0 with the next less significant 1 in the binary expansion of \(j_{x,z}\), we formulate Algorithm 1. It uses the PO to select the required candidates. Algorithm 1 only generates the candidates with \(m_{l,j}<L\), i.e., those candidates whose \(m_{l,j}\) are not colored in red in \[\operatorname{PM}_{p^{a}01q^{b}}^{l}\leq\operatorname{PM}_{p^{a}10q^{b}}^{l}. \tag{22}\]4  
```
1:procedureGenerateCandidates(\(L\), \(P\))
2:for\(l\gets 0,L-1\)do
3:\(candidates[l]\leftarrow\{0\}\)
4:for\(x\gets 1,P\)do\(\triangleright\) loop over the number of flipped bits, HDD (\(x=0\))
5:\(z\gets 0\)\(\triangleright\) stage in directed graph
6:\(j_{x}[z]\leftarrow\{2^{x}-1\}\)\(\triangleright\)\(j_{x,0}\) according to (27)
7:\(m_{l,j_{x}}\gets l+2^{x}-1\)\(\triangleright\)\(m_{l,j_{x,0}}\) according to (13)
8:while\(m_{l,j_{x}}<L\)do\(\triangleright\) Update candidates list
9:\(candidates[l]\leftarrow\)\(candidates[l]\cup j_{x}[z]\)
10:for\(j_{x,z}\in j_{x}[z]\)do\(\triangleright\) Generate new candidates
11:for\(c\gets 0,x-1\)do\(\triangleright\)\(x\) possible occurrences of 01
12:\(j_{x,z+1}[c]\leftarrow\)Repl_01_10(\(j_{x,z}\), \(x-1-c\))
13:endfor
15:\(j_{x}[z+1]\gets j_{x}[z+1]\cup j_{x,z+1}\)
16:endfor
17:\(z\gets z+1\)\(\triangleright\) move to next stage in directed graph
18:\(m_{l,j_{x}}\gets m_{l,j_{x}}+1\)
19:endwhile
20:endfor
21:endfor
22:returncandidates
23:endprocedure
24:procedureRepl_01_10(\(cand\), \(o\))
25:\(result\leftarrow\)Replace \(o\)-th occurrence of 01 by 10 in binary representation of \(cand\)
26:return\(result\)
27:endprocedure

\[\operatorname{PM}_{p^{a}01q^{b}}^{l}\leq\operatorname{PM}_{p^{a}10q^{b}}^{l}. \tag{22}\]5 Smallest possible candidate indices \(m_{l,j}\) for \(L=8\) and \(P=4\) with candidates \(j\in[0,2^{p}-1]\) and list index \(l\in[0,L-1]\) calculated by the PO-based method. **Theorem 4**. : _In Fast-SSCL decoding with list size \(L\), the number of required candidates originating from input path 1 to a Rate-1 node of length \(N_{v}\) is_

\[L^{l}_{\text{out,m}}=\sum_{ \begin{subarray}{c}j=0\\ m_{l,j}<L\end{subarray} }^{2^{p}-1}1. \tag{29}\]

Proof. : To prove Theorem 4, we need to verify that Equation (28) represents the smallest possible index of a candidate after sorting. Since we have proven (13) in Theorem 2, it is open to prove that the increment by \(z\) is valid. This, however, follows directly from Theorem 3. Thus, Theorem 4 is also proven.