Table 3 Execution steps and execution results for each step of edge sequence algorithm with \(M_{1}^{Kw}\)acting as an example

Execution steps

Results for each step

k = 1, k + 1 = 2 < (8/2 = 4)

\(e_{in} = e_{2}\),\(e_{out} = e_{16}\), thus \(S_{1 - set}^{in} = \{ \{ e_{2} \} \}\),\(S_{1 - set}^{out} = \{ \{ e_{16} \} \}\), then \(S_{2 - set}^{in} = \{ \{ e_{2} ,e_{6} \} ,\{ e_{2} ,e_{7} \} \}\), \(S_{2 - set}^{out} = \{ \{ e_{7} ,e_{16} \} \}\)

k = 2, (k + 1 = 3) < (8/2 = 4)

\(S_{3 - set}^{in} = \{ \{ e_{2} ,e_{6} ,e_{9} \} ,\{ e_{2} ,e_{6} ,e_{10} \} \}\), \(S_{3 - set}^{out} = \{ \{ e_{9} ,e_{7} ,e_{16} \} ,\{ e_{13} ,e_{7} ,e_{16} \} \}\)

k = 3, (k + 1 = 4) = (8/2 = 4)

\(S_{4 - set}^{in} = \{ \{ e_{2} ,e_{6} ,e_{9} ,e_{6} \} ,\{ e_{2} ,e_{6} ,e_{10} ,e_{13} \} ,\) \(\{ e_{2} ,e_{6} ,e_{9} ,e_{7} \} \}\),

\(S_{4 - set}^{out} = \{ \{ e_{6} ,e_{9} ,e_{7} ,e_{16} \} ,\{ e_{10} ,e_{13} ,e_{7} ,e_{16} \} \} .\)

Since

\(Ter(\{ e_{2} ,e_{6} ,e_{9} ,e_{6} \} ) = Init(\{ e_{10} ,e_{13} ,e_{7} ,e_{16} \} )\)

 = \(CR_{1}^{SR}\), and \(Ter(\{ e_{2} ,e_{6} ,e_{10} ,e_{13} \} ) = Init(\{ e_{6} ,e_{9} ,e_{7} ,e_{16} \} )\)  = \(CR_{1}^{RS}\),

\(ES_{1} (M_{1}^{Kw} )\)

 = \(\{ e_{2} ,e_{6} ,e_{9} ,e_{6} \} \cup \{ e_{10} ,e_{13} ,e_{7} ,e_{16} \}\)

 = \(\{ e_{2} ,e_{6} ,e_{9} ,e_{6} ,e_{10} ,e_{13} ,e_{7} ,e_{16} \},\)

\(ES_{2} (M_{1}^{Kw} )\)

 = \(\{ e_{2} ,e_{6} ,e_{10} ,e_{13} \} \cup \{ e_{6} ,e_{9} ,e_{7} ,e_{16} \}\)

 = \(\{ e_{2} ,e_{6} ,e_{10} ,e_{13} ,e_{6} ,e_{9} ,e_{7} ,e_{16} \}\)