# 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 $$Ter(\{ e_{2} ,e_{6} ,e_{10} ,e_{13} \} ) = Init(\{ e_{6} ,e_{9} ,e_{7} ,e_{16} \} )$$  = $$CR_{1}^{RS}$$,$$ES_{1} (M_{1}^{Kw} )$$ = $$\{ e_{2} ,e_{6} ,e_{9} ,e_{6} \} \cup \{ e_{10} ,e_{13} ,e_{7} ,e_{16} \}$$ = $$\{ e_{2} ,e_{6} ,e_{9} ,e_{6} ,e_{10} ,e_{13} ,e_{7} ,e_{16} \},$$$$ES_{2} (M_{1}^{Kw} )$$ = $$\{ e_{2} ,e_{6} ,e_{10} ,e_{13} \} \cup \{ e_{6} ,e_{9} ,e_{7} ,e_{16} \}$$ = $$\{ e_{2} ,e_{6} ,e_{10} ,e_{13} ,e_{6} ,e_{9} ,e_{7} ,e_{16} \}$$