[Solved]: What is the minimum square partition of an almost-square rectangle?

Problem Detail: 

This question is motivated by an older question about tiling an orthogonal polygon with squares. It is a generalisation of my former question about how to prove that the minimum square partition of a 3X2 rectangle has 3 squares).

Let:

• An almost-square-rectangle be a rectangle that has a width $w$ and height $h=w-1$.
• A square partitioning be a covering by non-overlapping squares; the entire rectangle must be covered, all the squares must be disjoint.
• A minimum-square-partitioning be a square partitioning, for which is no square partitioning that is made of a lesser number of squares.

Illustration:

^{Top row: The almost-square-rectangles of widths $3$, $4$ and $5$. Bottom row: Are these miminum-square-partitions of their corresponding rectangles?}

My question is now:

What is the minimum-square-partitioning of an almost-square-rectangle?

Can we prove ${\rm M{\small IN}S{\small QUARES}}(R_{w,h=w-1})=w$?

_{Note a follow-up question, Minimum square partitions for 4x3 and 5x4 rectangles.}

Asked By : Realz Slaw

Answered By : Yuval Filmus

A similar question was asked on Mathoverflow. The commenters mentioned a paper of Kenyon, which shows that the minimum number of squares required to tile a $w \times (w-1)$ rectangle is $\Theta(\log w)$. See also a related paper of Walters.

---

You can tile a $(4t+7) \times (4t+6)$ rectangle using only $t+5$ squares (for $t \geq 0$).

Best Answer from StackOverflow

Question Source : http://cs.stackexchange.com/questions/16826

 Ask a Question

 Download Related Notes/Documents