# Box stacking problem, and variants

You are given $$n$$ boxes and want to stack them to make a tallest possible tower, but you can only stack a box on top of another if the base is smaller in both dimensions. This is a classic dynamic programming problem, and yet I can't easily find answers online to natural questions:

1. If the boxes cannot be rotated, what is the complexity of this problem? A standard dynamic programming algorithm gives $$O(n^2)$$. Can you do $$O(n \log n)$$? Any conditional lower bounds?
2. If the boxes can be rotated, what is the complexity? Note that I do not have $$n$$ types of boxes and infinite instances of each type, I have $$n$$ boxes.
3. What is the complexity in higher dimensions?
4. What is the complexity if the dimensions are integers from $$1$$ to $$k$$?

If you model a box as a point $$(b_1,b_2, \ldots, b_d)$$, and you define the dominance relationship $$p \prec q$$ $$\iff$$ $$p_i < q_i$$, for all $$i$$, then you are looking for the longest chain in this partial order. In $$d$$ dimensions this problem can be solved in $$O(n \log^{d} n)$$ times, I believe.

I would sketch a somewhat inferior solution, because it is late.

To see how to do it, consider the two dimensional case. For a point $$p$$, let $$||p||_1 = |x(p)| + |y(p)|$$. Consider the input points, $$p_1, \ldots, p_n$$, and observe that $$p_i \prec p_j$$ only if $$||p_i||_1 < ||p_j||_1$$. As such, sort the points by their $$L_1$$ norm, in increasing ordering, and let $$p_1,\ldots, p_i$$ be the points handled so far, where for each $$p_j$$, $$j=1,\ldots, i$$, we computed $$\ell(p_j)$$ -- the length of the longest path in the relation that ends at $$p_j$$. We add the extra dimension $$\ell(p_j)$$ to the point $$p_j$$, so that this is now its third coordinate.

Now, given a new point $$p_{i+1}$$, we need to compute the highest point (in the third dimension) that it dominates (in the first two dimensions) -- this is an open quadrant question, and can be solved in $$O( \log^{O(1)} n)$$ time using orthogonal range searching techniques. Assume you get the point $$p_k$$ as returned by the query -- we have that $$\ell_{i+1} = \ell(p_{i+1}) = \ell(p_k)+1$$. We need to insert this new point $$(p_{i+1}, \ell_{i+1})$$ in the the orthogonal range searching data-structure.

Thus, you can handle each point in $$O(\log^{O(1)} n)$$ time which readily implies that you can compute the longest chain in $$O( n\log^{O(1)} n)$$ time. Being more careful about the details, the running time can be improved to $$O(n \log n)$$ in two dimensions.

I have a paper on a related problem here:

https://arxiv.org/abs/1910.07586

I think a solution for your problem should be readily implied by some previous work.

More maybe later...

• It would be interesting to abstract out the general principles in these problems. Basically we have a longest path problem in a DAG where the DAG vertices correspond to objects and edges are implicitly defined and easy to compute. One would like to avoid computing the full DAG and still do longest path computation. So a proxy ordering is defined and data structures are used to speed up the computation. Oct 20 '20 at 17:59