Triangulate rectangular parallelepiped in $\mathbb{R}^{n}$ I need to triangulate the n-dimensioned rectangular parallelepiped in $\mathbb{R}^{n}$ into a set of $n$-simplices.
Could you suggest me any known algorithm for that or maybe an extension of Delaunay triangulation for $n > 2$ cases?
 A: The construction outlined by Ross suffices to dissect an $n$-cube into $n!$ congruent $n$-simplices known 
as Schläfli orthoschemes.
One might ask if $n!$ is the smallest number of simplexes needed to dissect an $n$-cube, and it is not in general.  For example the $3$-cube may be dissected into as few as five tetrahedra (not all congruent).  See Lower Bounds for the Simplexity of the $n$-Cube.  The exact minimum number needed to dissect a $4$-cube is known to be 16 rather than 24 = 4!.
A: We can consider cubes, as you can stretch each dimension as you wish.  In $\mathbb R^3$, consider the unit cube in the first octant.  Triangulate each face that touches the origin, then connect the three corners of each triangle to $(1,1,1)$.  This gives a decomposition into six tetrahedra.  One of them has corners $(0,0,0), (0,1,0), (1,1,0), (1,1,1)$
I think the same thing works in higher dimensions.  Decompose the faces that touch the origin as $n-1$ boxes, then connect all the corners to $(1,1,\ldots ,1)$  The corresponding one in $\mathbb R^4$ to the above example would be $(0,0,0,0), (1,0,0,0), (1,1,0,0), (1,1,1,0), (1,1,1,1)$
