Could somebody suggest a way to determine if a parallelogram contains another parallelogram? I thought of one way to do this.
Using the algorithm which determines if a point is inside a parallelogram,
one can determine if the polygon contains the point within $2N$ steps ($N=2$ for parallelogram)
I want to do this for $n$-dimensional parallelotope but I guess it will take $2^N$ check for every vertex of the $n$-parallelotope.
Could somebody suggest a better way, which is computationally faster (in a polynomial time, not exponential)?
 A: Here is a simple polynomial algorithm.
Transform your outer parallelepiped to the cube $[-1,1]^N$
by the affine map
$$A: x \mapsto 2 V^{-1}(x-x_0) - \mathbf{1},$$
where all components of $\mathbf{1}$ are 1 and
$$V = (v_1,v_2,\ldots,v_N).$$
$A$ maps the base $x_1$ of your second parallelepiped  to $\xi_1$ and its spanning vectors $w_1,\ldots,w_N$ become
$$ \omega_k=2 V^{-1} w_k.$$
The second parallelepiped  is inside the first iff for the maximum norm
$$\|\xi_1 + t_1 \omega_1 +\ldots + t_N \omega_N  \|  \leq 1 $$
for all $t_k \in [0,1]$. Because in the extremal cases $t_k \in \{ 0,1 \},$ the condition can  be easily checked componentwise.
For the $m$-th component one just has to check whether the sum of all
positive   entries of $\xi_{1,m}$  and $\omega_{k,m}$ is $\leq  1$, and
the sum of all
negative   entries of $\xi_{1,m}$  and $\omega_{k,m}$ is  $\geq -1$.
Update: This means for all $1\leq m \leq N$:
$$ \xi_{1,m}+\sum \limits_{k=1}^N \max(0,\omega_{k,m})~\leq 1,$$
$$ \xi_{1,m}+\sum \limits_{k=1}^N \min(0,\omega_{k,m})~\geq -1.$$
