“…Since ≥ a 1, we confirm that there is only the following possible combination of a b c ( , , ): (1,0,0), (2, 0, 0), (3, 0, 0), (1,1,0). By (11), the only possibility is a b c ( , , ) = (3, 0, 0). In this case, deleting one vertex as S in G′ and resulting subgraph K…”