# Tower of tiles revisited

An S-tileset is a collection of n oriented tiles, where no two tiles have the same size, each tile is one unit thick, and its non-zero-integer length and width add up to n+1. (So, an S-tileset has n tiles, of sizes 1×n, 2×n−1, ...n×1.).

Conjecture: No two disjoint subsets of any S-tileset can be tiled to form two areas of identical size and shape.

Can you either prove the conjecture or find a counter-example.

• Another conjecture: A subset (of an S-tileset) having more than one tile in it cannot tile a rectangle Aug 17, 2014 at 23:18
• I know this is a math puzzle, but the way it's written feels like it should go on Math.SE under the "puzzle" tag rather than actually here.
– user88
Aug 17, 2014 at 23:49
• Do you allow rotation? If so, the answer is trivial - simply use the 1xn tile normally, and the nx1 tile rotated 90 degrees. They are disjoint, yet form two areas of identical size and shape. Oct 3, 2014 at 18:41
• @Trenin No, the tiles are oriented, so rotation is specifically disallowed. Oct 5, 2014 at 20:30
• @jwpat7 - a counterexample to your conjecture: imgur.com/k11J8zm. This is actually parametric in the n=14 set, there's a number of rectangles with the same layout. I suspect this might help find a counterexample to the posted conjecture. Oct 8, 2014 at 1:19

Here's a counter-example to the conjecture, for N=860

(you may need to open the image in a new tab to read the sizes in the image) The set of rectangles used:

Shape 1

• 602x259
• 210x651
• 28x833
• 133x728
• 469x392
• 679x182
• 707x154
• 840x21

Shape 2

• 196x665
• 70x791
• 161x700
• 413x448
• 609x252
• 770x91
• That is an excellent result. Did you solve it using Diophantine equations? Oct 20, 2014 at 22:41
• Sort of. I set up systems of linear equations, and used sympy to solve or simplify them as much as possible. This ruled out a lot of candidates on its own (two sides would have the same algebraic length, for example). The system describing these two shapes ends up having only 1 free variable, so it's just a matter of picking the first one that gives integers for all sides. Oct 21, 2014 at 0:29