■多面体的組み合わせ論(その11)

 4次元の場合もやってみよう.

  (y0−y1)=(y1−y2)/2=(y2−y3)/3=(y3−y4)/4

  y0=1,y4=0

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[1]形状ベクトル(1,0,0,0)の場合,

  (y1−y2)/2=(y2−y3)/3=(y3−y4)/4=0

  y1=y2=y3=y4=0

[2]形状ベクトル(0,1,0,0)の場合,

  (y0−y1)=(y2−y3)/3=(y3−y4)/4=0

  y0=y1=1,y2=y3=y4=0

[3]形状ベクトル(0,0,1,0)の場合

  (y0−y1)=(y1−y2)/2=(y3−y4)/4=0

  y0=y1=y2=1,y3=y4=0

[4]形状ベクトル(0,0,0,1)の場合

  (y0−y1)=(y1−y2)/2=(y2−y3)/3=0

  y0=y1=y2=y3=1,y3=y4=0

[5]形状ベクトル(1,1,0,0)の場合

  (y0−y1)=(y1−y2)/2

  (y2−y3)/3=(y3−y4)/4=0,y2=y3=y4=0

  2(1−y1)=y1→y1=2/3

[6]形状ベクトル(1,0,1,0)の場合

  (y0−y1)=(y2−y3)/3

  (y1−y2)/2=(y3−y4)/4=0→y1=y2,y3=y4=0

  3(1−y1)=y1→y1=y2=3/4

[7]形状ベクトル(1,0,0,1)の場合

  (y0−y1)=(y3−y4)/4

  (y1−y2)/2=(y2−y3)/3=0,y1=y2=y3

  4(1−y1)=y1→y1=y2=y3=4/5

[8]形状ベクトル(0,1,1,0)の場合

  (y0−y1)=(y3−y4)/4=0→y1=1,y3=y4=0

  (y1−y2)/2=(y2−y3)/3

  3(1−y2)=y2→y2=3/4

[9]形状ベクトル(0,1,0,1)の場合

  (y0−y1)=(y2−y3)/3=0→y1=1,y2=y3

  (y1−y2)/2=(y3−y4)/4

  4(1−y2)=2y2→y2=y3=2/3

[10]形状ベクトル(0,0,1,1)の場合

  (y0−y1)=(y1−y2)/2=0→y1=y2=1

  (y2−y3)/3=(y3−y4)/4

  4(1−y3)=3y3→y3=4/7

[11]形状ベクトル(1,1,1,0)の場合

  (y0−y1)=(y1−y2)/2=(y2−y3)/3

  (y3−y4)/4=0→y3=y4=0

  2(1−y1)=(y1−y2)

  3(y1−y2)=2y2→y1=5/6,y2=1/2

[12]形状ベクトル(1,1,0,1)の場合

  (y0−y1)=(y1−y2)/2=(y3−y4)/4

  (y2−y3)/3=0→y2=y3

  2(1−y1)=y1−y2

  4(y1−y2)=2y2→y1=6/7,y2=y3=4/7

[13]形状ベクトル(1,0,1,1)の場合

  (y0−y1)=(y2−y3)/3=(y3−y4)/4

  (y1−y2)/2=0→y1=y2

  3(1−y2)=(y2−y3)

  4(y2−y3)=3y3→y1=y2=3/4,y3=y4=0

[14]形状ベクトル(0,1,1,1)の場合

  (y0−y1)=0→y1=1

  (y1−y2)/2=(y2−y3)/3=(y3−y4)/4

  3(1−y2)=2(y2−y3)

  4(y2−y3)=3y3→y2=7/9,y3=4/9

[15]形状ベクトル(1,1,1,1)の場合

  y1=1−1・2/n(n+1)

  y2=1−2・3/n(n+1)

  y3=1−3・4/n(n+1)

  yn=1−n(n+1)/n(n+1)=0

  →y1=9/10,y2=7/10,y3=4/10

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