Testing *My* Intuition (34): Tiling High Dimension with an Arbitrary Low-Dimensional Tile.

Test your intuition 34 asked the following:

A tile T is a finite subset of \mathbb Z^d. We can ask if \mathbb Z^d can or cannot be partitioned into copies of T. If  \mathbb Z^d can be partitioned into copies of T we say that T tiles \mathbb Z^d.

Here is a simpe example. Let T consists of 24 points of the 5 by 5 planar grid minus the center point. T cannot tile \mathbb Z^2.

Test your intuition: Does T tiles \mathbb Z^d for some d>2?

We had a poll and 58% of voters said YES. The answer is


As a matter of fact Adam Chalcraft have made the beautiful conjecture that every tile T in \mathbb Z^d tiles \mathbb Z^n for some large n.  This conjecture was proved by Vytautas Gruslys, Imre Leader,  and Ta Sheng Tan in their remarkable paper Tiling with arbitrary tiles.

Theorem (Vytautas Gruslys, Imre Leader,  and Ta Sheng Tan): Let T \subset  \mathbb Z^d be a tile. Then T tiles \mathbb Z^n for some n \ge d.

But wait, what about our tile T? After seeing the abstract of Imre Leader’s lecture, looking briefly at the paper which contained the 5 by 5 minus the middle example, I posted the question on my blog.  But then driving to Jerusalem I suddenly was sure that there is no way in the world the hole in T can be filled up by another tile of the same shape.  T is simply too fat, I thought. I must have missed something –  some extra condition or subtelty that I overlooked. It turned out that my intuition was wrong already after I saw the right answer. (This does happen from time to time.)

When I had a chance I looked again at the paper, and saw a beautiful picture explaining how the hole can be filled in four dimension. (BTW, I don’t know what is the minimum dimension that T can tile.)

This entry was posted in Combinatorics, Test your intuition and tagged , , , . Bookmark the permalink.

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