Assume an $n$-dimensional discrete manifold $M$ ($n$-dimensional simplicial complex in which each ($n-1$)-simplex has exactly two adjacent $n$-simplices or only one if it is on a boundary of $M$) comprising a large number of $n$-simplices.
What is the minimum number of $1$-simplices one needs to remove from $M$ in order to create $n$-dimensional void within $M$ which void would be homeomorphic to $n$-cube or to $n$-orthoplex?
For $2$-manifold a void homeomorphic both to $2$-cube and $2$-orthoplex can be created by removing one $1$-simplex not on the boundary of $M$.
A void homeomorphic to $3$-cube can be created in $3$-manifold by removing one $3$-simplex, or specifically by removing six $1$-simplices on each $2$-face of this newly formed cube that defined this removed tetrahedron (tet in cube). This also result in creation of six quadrilateral pyramids on each face of the cube.
How about $4$-cube?
Are there any general formulas for this problem?