The following are explicit non-decomposable tilings for the lowest three N values.
In order to prove that a non-decomposable tiling is not possible for any other N, let us consider the rectangles in terms of the coordinates of their top, bottom, left and right. Consider only the tops and bottoms. Let E be the number of distinct values taken by the tops and bottoms, not including the top and bottom of the original rectangle. We'll call these values "internal edges". The top and bottom of the original rectangle, we'll call "external edges".
Note the following:
On the other hand, in a non-decomposable tiling with N>2 no rectangle can go all the way from the bottom to the top, so if E=1, this internal edge separates the tiling into two sub-tilings, again refuting non-decomposability.
The only remaining case is E=2 with N=6.
Repeating the same reasoning as above, we conclude that not only are there exactly two distinct internal values for "top" and "bottom", but there are also exactly two such values for "left" and "right", as well. The tiling is therefore a super-tiling of
If "B" is such a rectangle, and it does not form a separable tiling with "A" or with "C", then this is because the rectangle spanning "A" also spans "D" and the rectangle spanning "C" also spans "F". This indicates that "B" forms a separable tiling with the rectangle spanning "E".
If, on the other hand, both "A" and "C" are rectangles in the tiling, then "B" and "E" must be spanned by the same rectangle, so "A" and the rectangle spanning "D" form a separable tiling.
First, let us prove that no tilings exist for N<7.
The tiling with minimal N must be a non-decomposable tiling. If it had been a decomposable tiling, the M rectangles that form a separate tiling are already a solution. However, the non-decomposable solutions for N=2 and N=5 shown above are unique up to rotation and mirror reflection, and solving their parameters shows that neither has all-distinct rectangle perimeters.
For completion: here is a way to show that the solution for N=5 is unique up to reflection. Consider, again, that E=2, and therefore the tiling is a super-tiling of the 3x3 tiling shown in the diagram above.
Each rectangle must choose its top and its bottom from a set of size 4, for a total of 6 possibilities, minus 1 possibility because the global top and global bottom do not form a usable pair in a non-decomposable tiling. We therefore have exactly 1 representative of each possible choice.
Because each of the four corners must be covered by a different rectangle, this leaves the rectangle with its top and bottom both in internal edges as the fifth. For the same reason, it is also the rectangle with the internal left and right edges. So, "E" must be its own rectangle. However, now "B", "D", "F" and "H" cannot be their own rectangles. The rectangle covering each must also cover a corner. This leaves exactly the two choices equivalent to the non-decomposable tiling drawn for N=5 above.
A tiling with N=7 can be produced as is shown above. When tiling the unit square, the dimensions of the rectangles are as follows.
For N=8, extend the 7-tiling by adding a new rectangle of the same area that spans the entire top part of the rectangle.
For N=9, extend the 8-tiling by adding a new rectangle of the same area that spans the entire left part of the rectangle.
By repeating this procedure of adding to the top and adding to the left, one can extend this tiling to arbitrarily many rectangles.
In each such extension, the new rectangle is taller/wider than any previous rectangle, so its dimensions are not identical to any previous rectangle. The only possibility that two rectangles are of the same perimeter is therefore if the height of one equals the width of another. However, because one can scale the original rectangle to be tiled at will, this can always be avoided.
Credits for this month's riddle and its original solution are as follows.
On the 26th of March, over 3 weeks after this riddle was published, R. Nandakumar, who sent me the original e-mail query regarding this problem, discovered that the problem had already been studied in the past. Here's a link. The rest of the credits are for recent work, done independently from Descartes's. (This last link is a highly recommended one.)
R. Nandakumar's original question to me was whether there exists a tiling into rectangles of equal area but different perimeters. (To the best of his knowledge at the time, this question's first public appearance was in his blog.) This question was solved initially by myself. (And here I should add that I am indebted to the Discrete Math Research Group at Monash University for pointing me in the right direction for this solution, and especially to János Barát. I announced my solution, an explicit construction for N=7, shortly after consulting with the group.) Within twenty-four hours, János Barát also reached the same construction, followed by N. Ramana Rao.
Questions regarding the minimality of the N=7 construction, as well as its extendability to arbitrary N (and its relation to non-decomposable tilings) are my own results, as are results regarding tiling squares.
Here's a question posed by R. Nandakumar (also on the same blog post), which I
do not know the answer to:
Can a rectangle be tiled by different-perimeter same-area rectangles, when all rectangle sides must be rationals?
To this I add "Will this ever happen if we simply continue the spiral structure indefinitely?"
If any reader wants to continue exploring these questions, I'll be very happy to hear of progress.
Back to riddle
Back to main page