11

So, a simple perfect squaring of a rectangle is a tiling of that rectangle by squares whose side lengths are all distinct integers. Additionally, not subset of the squares must form a smaller rectangle. My question is if there is a simple perfect squaring of a 1366 by 768 rectangle?

We could try reducing it to a simpler problem by splitting the rectangle into a square and a smaller rectangle, but then we need to ensure that the side lengths are different, and that the their combination is simple. So we basically back to where we started.

P.S. If you are wondering why 1366 by 768, that's the dimension of my monitor, which I am trying to artistically square (hence the tag).

Christopher King
  • 10,773
  • 3
    Just out of curiosity: are you making a wallpaper image? :) – lisyarus Mar 08 '19 at 07:18
  • 1
    @lisyarus I will. I am not much of an artist, but given the sequence of squares involved I could use a program to produce an image. I plan on making the edges black and the rest transparent, and then putting it on top of another (non-math related) wallpaper, but I'll definitely post the basic square tiling image if anyone is interested. – Christopher King Mar 08 '19 at 07:24
  • 1
    Well you can have at most $146$ squares as $\sum_{k=1}^{147}k^2>1366\times768$. A brute force search shouldn't be too hard? – Servaes Mar 08 '19 at 13:07
  • 1
    @Servaes well, $146!$ is greater than the number of atoms in the universe, so it might take a little while. :P – Christopher King Mar 08 '19 at 13:23
  • Sure, so is $146\uparrow\uparrow\uparrow\uparrow146$, but how is that relevant? Also, this shows that the largest squares is bigger than $146\times146$, and I'm sure better upper bounds than $768$ aren't hard to come by. – Servaes Mar 08 '19 at 13:50
  • @Servaes sorry, I should have been more specific. The idea is that we can't just try each of the $146$ squares, we also need to try combinations of them in different orders (hence the factorial). – Christopher King Mar 08 '19 at 13:58
  • I do not understand what you mean by "we can't just try each of the $146$ squares". Which $146$ squares? Also, a simple upper bound on the size of the largest square is $704\times704$. – Servaes Mar 08 '19 at 14:14
  • @Servaes Oh wait, nvm. However, brute force would still require more than $146$ steps. – Christopher King Mar 09 '19 at 14:59

1 Answers1

15

I looked through some of the data on Squaring.Net (actually, copies at the Internet Archive, since the site is currently down), and there is no $1366\times768$ simple perfect squared rectangle of order $17$ or less; there may be one of higher order, but that's harder to check. But there is a $1354\times764$ simple perfect squared rectangle of order $17$, so if you can accept a six-pixel border on the left and right and a two-pixel border at the top and bottom, this will do the job. It looks like this:

1354x764 squared rectangle (Click on the image for the full-size version.)

The Bouwkamp code for this rectangle dissection is $17\ 1354\ 764\ (389, 403, 311, 251)\ (60, 191)\ (83, 157, 131)\ (375, 14)\ (9, 74)\ (361, 65)\ (322)\ (296)$.

EDIT: As Servaes points out, I overlooked an even better $1366\times766$ alternative, also of order $17$:

1366x766 squared rectangle

Its Bouwkamp code is $17\ 1366\ 766\ (419, 288, 258, 401) (115, 143) (203, 85) (200) (179, 365) (347, 72) (275) (193, 7) (186)$.

FredH
  • 4,368
  • 1
    How did you search the site for that? I'm taking a look, and the navigation is confusing me. – Christopher King Mar 08 '19 at 13:25
  • @PyRulez Home -> Simple Perfect Squared Rectangles -> Simple Perfects by aspect (which gets you here); then download the .csv file. – FredH Mar 08 '19 at 13:30
  • 2
    There seems to be a $1366\times766$ rectangle in the linked file as well. – Servaes Mar 08 '19 at 15:22
  • @Servaes I missed that one. It's definitely better than the one I did notice. I'll add it. – FredH Mar 08 '19 at 15:45
  • I find the $1366\times766$ arrangement especially aesthetically pleasing. I should hope a similar configuration exists for the $1366\times768$ rectangle... – Servaes Mar 09 '19 at 01:13