Dividing a Sq. into 7 Related Rectangles
It is a continuation of my submit Dividing a square into similar rectangles, through which I mentioned this puzzle: in case you partition a sq. into n comparable rectangles, what proportions can these rectangles have?
Some individuals on Mathstodon put plenty of work into this and made some good progress. However there’s been a stunning new twist, and I’m not even speaking about the truth that the New York Occasions ran an article about this puzzle:
• Siobhan Roberts, The quest to find rectangles in a square, New York Occasions, February 7, 2023. Open-access model here.
Bear in mind the story to date. When n = 3 there are simply 3 choices for what the proportions of the rectangles could be:
I requested people on Mathstodon concerning the case n = 4 they usually discovered 11 choices, as drawn right here by Dan Piker:
From then on the quantity will increase steeply. The most important case anybody has been in a position to deal with is n = 7. Ian Henderson discovered 1371 choices in that case:
And that was nice… however now the story has gotten extra fascinating, as a result of Daniel Gerbet appears to have discovered yet another! Sure: not 1371, however 1372.
He did the computation twice, in two other ways. He despatched me a table itemizing all 1372 allowed proportions in growing order, along with the polynomial equations they obey and footage of the subdivided sq.. He wrote:
After operating the computation once more, I used to be in a position to establish the presumably lacking man precisely: the 1055th ratio. The photographs of the partitions look all the identical for smaller and bigger ratios within the tables from Ian Henderson and the one I computed. You discover it at web page 106 within the tabledesk for partitions with 7 rectangles.
The ratio is the one actual root of the polynomial
You will discover an image of a partition with this ratio within the recordsdata connected. Be aware that Ian Henderson has computed partitions with different ratios, however the identical topology as this one, e.g. the 1022th ratio.
Certainly it’s an fascinating reality, first famous by Lisanne Taams in an easier instance, that we will get two partitions with the identical topology the place the rectangles have totally different proportions!
Here’s a image of the apparently new partition that Daniel Gerbet has discovered:
It’s also possible to see it because the 1055th partition on this listing, the place the 1372 partitions are conveniently grouped into 25 teams of 54, plus 22 extra:
I mentioned “conveniently”, however in fact I used to be kidding—it’s simpler to cope with these footage in a PDF file. Daniel Gerbet has kindly allowed me to share his PDFs for all of the fascinating circumstances as much as n = 7:
• All 3 allowed proportions for 3 similar rectangles that subdivide the square.
• All 11 allowed proportions for 4 similar rectangles that subdivide the square.
• All 51 allowed proportions for 5 similar rectangles that subdivide the square.
• All 245 allowed proportions for 6 similar rectangles that subdivide the square.
• All 1372 allowed proportions for 7 similar rectangles that subdivide the square. Additionally, don’t neglect the more detailed table listing the polynomials obeyed by these proportions.
It’s also possible to get a replica of the Python program that Daniel Gerbet used to acquire his outcomes.
By the way in which, once I talked about this information to my pal Todd Trimble he immediately observed that 1372 is divisible by 7. Coincidence? Most likely.
Some excellent news right here is that Daniel Gerbet’s work confirms the earlier calculations for n ≤ 6. And that’s particularly good as a result of beforehand just one individual, Ian Henderson, had performed the n = 6 case. See my previous article for particulars).
However we want some individuals to assist settle the case n = 7!
Addendum: Greg Egan has confirmed this new resolution is actual. Ian Henderson has confirmed that it’s new, and he’s discovered the bug in his program that made him overlook it:
It’s actual! I made a mistaken assumption in my code to attempt to save computation time — as an alternative of making an attempt each method of orienting the N rectangles, it at all times orients the final rectangle the identical method, underneath the belief that the identical association however flipped shall be generated anyway.
The problem is that “final” is within the order the rectangles are visited, which can change when the sq. is flipped! (See connected picture for the association right here—the final rectangle is #6:
Both altering it to at all times orient the first rectangle in the identical method (because it’s at all times within the prime left) or simply disabling that code totally does give a complete depend of 1372.