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A polyomino is a collection of equal-sized squares joined edge-to-edge in the plane (think Tetris pieces, but with an arbitrary number of squares instead of just four). A heptomino is a polyomino composed of seven squares. There are $108$ total heptominos, of which $104$ can perfectly tile the plane (no gaps and no overlaps). Of the four that can't, the impossiblity of three of them are nontrivial to prove, but the impossibility of the fourth is immediately obvious to anyone who sees it. Without looking up the shapes or- heavens forbid- trying to draw them all out, tell me which one is obvious and why.

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The one that

has a hole:

\begin{matrix} X &X &. \\ X &. &X \\ X &X &X \end{matrix}

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  • $\begingroup$ 25 upvotes for this? This seems overrated... +1 regardless. $\endgroup$
    – mathlander
    Feb 23 at 18:04

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