# Klee-Minty cube vertices for D=5

According to Wikipedia (but this is well known), A Klee-Minty cube in $D$ dimensions is defined by

$x_1\leq 5$

$4 x_1 + x_2 \leq 25$

$8 x_1 + 4 x_2 + x_3 \leq 125$

$16 x_1 + 8 x_2 + 4 x_3 + x_4 \leq 625$

$32 x_1+ 16 x_2 + 8 x_3 + 4 x_4 + x_5 \leq 3125$

$\vdots \vdots$

$2^D x_1 + 2^{D-1} x_2 + 2^{D-2} x_3 + \ldots + 4 x_{D-1} + x_D \leq 5^D$

$x_1 \geq 0$

$x_2 \geq 0$

$x_3 \geq 0$

$x_4 \geq 0$

$x_5 \geq 0$

This polyhedron has $2^D$ vertices. How can I find them?

I have tried the following code for $D = 5$ --- that is 32 vertices.

```
A = matrix(QQ, 10, 5,
[0, 0, 0, 0, -1,
0, 0, 0, -1, -4,
0, 0, -1, -4, -8,
0, -1, -4, -8, -16,
-1, -4, -8, -16, -32,
1, 0, 0, 0, 0,
0, 1, 0, 0, 0,
0, 0, 1, 0, 0,
0, 0, 0, 1, 0,
0, 0, 0, 0, 1])
show(LatexExpr(r"\text{A = }"), A)
b = vector(QQ, [5, 254, 125, 625, 3125, 0, 0, 0, 0, 0])
show(LatexExpr(r"\text{b = }"), b)
c = vector(QQ, [2^4, 2^3, 2^2, 2^1, 2^0])
show(LatexExpr(r"\text{c = }"), c)
AA = matrix(list(list([b]) + list(transpose(A))))
show(transpose(AA))
pol = Polyhedron(ieqs=AA)
pol.Hrepresentation()
```

But I am far away from the account and some of the vertices are in the negative part of the hyperplane (which is impossible). Need some help to decipher my mistake(s). Thanks.

Une erreur ici:

Thanks to have sen that error but it doesn't change the fact that I could not find the vertices.