■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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