■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
5
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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