Mondrian rectangles

Composition by Piet Mondrian.

To begin with, a clarification/correction: last week I said that "in the sequence 1, 11, 111, 1111, 11111... there is no perfect square", and I should have added, obviously, "except for the trivial case of 1, which it is the square of itself.”

My apologies.

The problem of the comic page divided into rectangular panels is still not fully resolved (although in the last comments of the previous installment there are some interesting approximations), so the question is still open and expandable to 4x4, 5x5 grids...

And in relation to this, Manuel Amorós posed a similar problem (which connects with some of the same type seen previously): In how many different ways can a chessboard be covered with dominoes?

And, on the other hand, the problem of the vignettes refers directly to that of the Mondrian rectangles, as we will see below.

The paradox of the fly and the handlebar hasn't received a satisfactory (or unsatisfactory) answer either, so I'll pose the usual meta-question for note: How is the paradox of the fly related to Zeno's famous paradoxes about motion?

Is it identical to the paradox of the arrow?

Mondrian puzzle

The well-known geometric compositions of the Dutch painter Piet Mondrian - an extreme simplification of pictorial abstraction based on rectangular figures and primary colors - have served as inspiration for various mathematical games and pastimes.

Here is one of the most interesting:

We divide an nxn lattice into rectangles containing an integer number of cells and all of them different: it may have the same surface, but not the same shape (there may be, for example, a 2x2 one —needless to say, squares are also rectangles— and one 1x4, but not one 1x4 and one 4x1).

We call the “score” of one of these divisions the difference between the surface of the largest rectangle and that of the smallest, and we look for the division with the lowest score.

For example, in the attached figure we see a 4x4 grid divided into a 3x3 square, a 4x1 rectangle and a 1x3 rectangle;

the surfaces of the three parts are, respectively, 9, 4, and 3 square units, so the score for this division is 9 – 3 = 6. Can it be improved?

Yes, the score can be lowered to 4 (by what division?).

Obviously, the situation becomes more complicated as the size of the grid increases.

If, as we have seen in previous weeks, it is not easy to find the number of different divisions in rectangles of a simple 3x3 grid, neither is it easy to solve the problem of Mondrian rectangles for ever larger grids.

In fact, there is (as far as I know) no formula or algorithm for determining the minimum score of an nxn grid as a function of n.

And precisely the absence of such an algorithm turns Mondrian's rectangles into a fascinating puzzle that can provide my astute readers with long hours of solace and/or despair.

For now, I suggest you look up the minimum scores for the 5x5, 6x6, 7x7, and 8x8 grids.

As a hint and example, here is a division of the grid of 10x10 of minimum score, with two rectangles of maximum area (5x4 and 10x2) and one of minimum area (2x6), so the score is 20 – 12 = 8. Is this minimum score division unique?

And as for the Mondrian painting at the top of this article, could we fit it into a grid and consider it one of the divisions we just looked at?

And if this is not possible, how would it have to be modified to make it fit?

Courage, not every day they give you the opportunity to amend the page of a great contemporary painter.

Carlo Frabetti

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