“…When v = 6, there are 4 possibilities for (v, e, f ); Steinitz [42] showed that a polytope in R 3 has v vertices, e edges and f facets if and only if the vector (v, e, f ) satisfies the Euler relation v−e+f = 2 and the inequalities v ≤ 2f − 4 and f ≤ 2v − 4. Hence, if v = 6 then 5 ≤ f ≤ 8 and e = f + 4, so the possible face vectors (v, e, f ) are (6,9,5), (6,10,6), (6,11,7) and (6,12,8). Checking cases, we find that G(6, 12, 8) has the maximum value among the four possibilities.…”