# Euclidean minimum spanning trees intersecting each unit square

The recent question "Euclidean Minimum Spanning Trees Restricted to One Vertex Per Grid Cell" can be restated in terms of "minimum spanning trees intersecting each (closed) lattice square of an $n\times n$ lattice".

I am wondering whether there is a substantial change if we require the trees to intersect every axis-parallel unit square contained in the big $n\times n$ square, not only lattice squares. Note that in both examples of the other thread, much bigger squares can be squeezed between the branches without intersecting them.

If we denote by $a(n)$ the minimal length of such a tree in the original question and by $b(n)$ the minimal length in the modified question, we have obviously $a(n)\le b(n)$. For $n=2^k+1$, we have $b(n)\le\dfrac{4^k-1}3(\sqrt{3}+1)$, and I would think intuitively that this inequality is sharp.

Can it be shown that $b(2^k+1)=\dfrac{4^k-1}3(\sqrt{3}+1)$?

The construction achieving that has a 'base tree' (i.e. one of the two minimal Steiner trees connecting the 4 unit square corners) in each square $(i,j)$ which has $\nu_2(i)=\nu_2(j)$. Here $0<i,j<2^k$ and $\nu_2(\cdot)$ is the 2-adic exponent, e.g. for $k=3$, the pattern in the $9\times9$ square is

X   X   X   X
X       X
X   X   X   X
X
X   X   X   X
X       X
X   X   X   X


where each 'X' denotes a base tree, so diagonally adjacent such cells have a common vertice. (Think of the space-filling "X-fractal" obtained by iterating this pattern in the obvious way.) Such a tree contains each lattice point, i.e. all corners of the $n\times n$ lattice squares, thus it intersects each unit square.

Likewise: Can it be shown that $a(2^{k+1})=2\dfrac{4^k-1}3(\sqrt{3}+1)$, using the same pattern but with the base trees twice as large?

I have no idea if for $k=2$ the $b(5)$ tree has a bigger length than the $n=5$ "candidate" given in the other thread. Probably it has, so:

Is it true that for all $n\ge4$, $a(n)< b(n)$? What about $\lim\limits_{n\to\infty}\dfrac{a(n)}{b(n)}$?