|
按照mit18_900s23_lec29.pdf的公式:$\begin{aligned} & b_0=n_0-\operatorname{rank}\left(D_1\right) \\ & b_1=n_1-\operatorname{rank}\left(D_1\right)-\operatorname{rank}\left(D_2\right) \\ & b_2=n_2-\operatorname{rank}\left(D_2\right)\end{aligned}$
对于Simplex[{1, 2, 3}]有
$D_1=\pmatrix{1&1\\-1&&1\\&-1&-1}$
$D_2=\pmatrix{1\\1\\-1}$
$n_0=3$
$n_1=3$
$\operatorname{rank}D_1=2$
$n_2=1$
$\operatorname{rank}D_2=1$
$b_0=n_0-\operatorname{rank}D_1=1$
$b_1=n_1-\operatorname{rank}D_1-\operatorname{rank}D_2=0$
$b_2=n_2-\operatorname{rank}D_2=0$
即$\{1,0,0\}$
那这个公式又是為什麼呢 |
|