■置換多面体の空間充填性(その461)

 3次元正単体のデータをもとにして,4次元正単体の場合を調べたい.

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[1]{3,3}(1000)

  {3,3}(000)→(1,0,0,0),5個

  {}(00)×{}(1)→(1,0,0,0),10個

  {}(0)×{3}(10)→(1,0,0,0),10個

  {3,3}(100)→(1,0,0,0),5個

0,10

0,0,10

0,0,0,5

 3列目より,正三角形10

 4列目より,正四面体5

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

[2]{3,3}(0100)

  {3,3}(100)→(4,6,4,1),5個

  {}(00)×{}(0)→(1,0,0,0),10個

  {}(0)×{3}(01)→(1,0,0,0),10個

  {3,3}(010)→(1,0,0,0),5個

20,−10

30,0

20,0,10

5,0,0,5

 1列目より正三角形20,3列目より正三角形10

 1列目より四面体5,4列目より八面体5

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

[3]{3,3}(0010)

  {3,3}(010)→(6,12,8,1),5個

  {}(10)×{}(0)→(3,3,1,0),10個

  {}(0)×{3}(00)→(1,0,0,0),10個

  {3,3}(001)→(1,0,0,0),5個

30,−30,10

60,−30,0

40,−10,0

5,0,0,5

 1列目より正三角形40,2列目より正三角形−10

 1列目より八面体5,4列目より四面体5→[2]と一致

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

[4]{3,3}(0001)

  {3,3}(001)→(4,6,4,1),5個

  {}(01)×{}(0)→(3,3,1,0),10個

  {}(1)×{3}(00)→(2,1,0,0),10個

  {3,3}(000)→(1,0,0,0),5個

20,−30,20,−5

30,−30,10,0

20,−10,0

5,0,0,0

 1列目より正三角形20,2列目より正三角形−10

 1列目より四面体5,4列目より四面体5→[1]と一致

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