I'm sorry, it may be simple and stupid but I didn't find any relative solutions on the Internet.
Given the unit hypercube $C$ in the Euclidean space $R^n$, how to divide (or we can say "partition", I hope they have identical meanings in this aspect) $C$ into $m$ equilateral $n$-simplices, where $m$ is a user predefined parameter that can be an any large number?
Note that:
(1) We should divide $C$ as much as possible, which means the left "offcut" space (that cannot be fitly divided) should be minimal. So we need to carefully choose the edge length $l$.
(2) Any two closest $n$-simplices must be joint. They have joint vertexes, joint edges, joint faces and other joint high dimensional elements. In addition, the joint faces and joint high dimensional elements must also be simplices, e.g., the joint face with 3 joint vertexes is a 2-simplex with edge length $l$, the joint tetrahedron with 4 joint vertexes is a 3-simplex with edge length $l$, and so on in higher dimensional space.
(3) As we describes it above, we try to make all the vertices equally distributed. It requires all the simplices try to have the same edge length and volume. In this way, the distances from all the points to their nearest neighbors are uniform.
(4) The number of divided simplices $m$ is large enough. Its exact constraints may be related to the number of nearest neighbors that is discussed below. Let's just simply assume $m$ is a very large number. It's predefined and can be changed in different trial, which means the final algorithm should hold any large $m$ instead of only working on some specific values of $m$.
I have a naive idea for it. Since we can calculate the volume $V$ of each simplex (I suppose $V$ is a function $V(l)$ in terms of $l$), we can build the equation $\frac{1}{m}=V(l)$ and solve the $l$. Then we can start from the geometry center $O$ of cube $C$, around $O$ we can add vertexes layer by layer. For example, for $n=2$, there are 6 nearest vertexes and 6 second nearest vertexes and so on. For $n=3$, there may be 20 nearest vertexes and 20 second nearest vertexes and so on. I can't imagine the case for $n>3$. Therefore, if the idea works, the new question arises here is:
(#) Given vertex $O$, how to calculate its number of nearest neighbor vertexes for any $n$ in such case? Is there any theory indicating it?
To explain "6 nearest vertexes and 6 second nearest vertexes" for $n=2$, I give an example in the following figure. For the center red point, it has 6 nearest blue vertexes, 6 second nearest green vertexes and 6 third nearest yellow vertexes, and we can do it continuously if $m$ is large enough.
I found a question with similar description but different purpose A question about partitioning the unit cube into simplexes.
I'm looking for a feasible algorithm to implement it. I guess it may be related into surface triangulation and Simplicial Complex. But I'm not familiar with $n$-dimensional topology and I don't know the naive idea is right or not. Any help is highly appreciated.