■置換多面体の空間充填性(その162)
F4についても,
m0=Σsjsj+1+sr・sr+1 (正軸体系で最後の要素が0の場合)
を計算してみたい.
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【1】F4のm0
{3,4,3}(1,0,0,0): 6 (NG:正解は8)
{3,4,3}(0,1,0,0): 8 (NG:正解は6)
{3,4,3}(0,0,1,0): 6 (OK)
{3,4,3}(0,0,0,1): 4 (NG:正解は8)
{3,4,3}(1,1,0,0): 5 (NG:正解は4)
{3,4,3}(1,0,1,0): 6 (OK)
{3,4,3}(1,0,0,1): 6 (NG:正解は8)
{3,4,3}(0,1,1,0): 4 (OK)
{3,4,3}(0,1,0,1): 6 (OK)
{3,4,3}(0,0,1,1): 4 (OK)
{3,4,3}(1,1,1,0): 4 (OK)
{3,4,3}(1,1,0,1): 5 (OK)
{3,4,3}(1,0,1,1): 5 (OK)
{3,4,3}(0,1,1,1): 4 (OK)
{3,4,3}(1,1,1,1): 4 (OK)
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【2】F4のm1
{3,4,3}(1,0,0,0): 4 (NG:正解は8)
{3,4,3}(0,1,0,0): 6 (OK)
{3,4,3}(0,0,1,0): 6 (OK)
{3,4,3}(0,0,0,1): 4 (NG:正解は8)
{3,4,3}(1,1,0,0): 4 (OK)
{3,4,3}(1,0,1,0): 6 (OK)
{3,4,3}(1,0,0,1): 6 (NG:正解は8)
{3,4,3}(0,1,1,0): 4 (OK)
{3,4,3}(0,1,0,1): 6 (OK)
{3,4,3}(0,0,1,1): 4 (OK)
{3,4,3}(1,1,1,0): 4 (OK)
{3,4,3}(1,1,0,1): 5 (OK)
{3,4,3}(1,0,1,1): 5 (OK)
{3,4,3}(0,1,1,1): 4 (OK)
{3,4,3}(1,1,1,1): 4 (OK)
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