A *polystick* is a connected finite subgraph of the grid graph, and a *tetrastick* is a polystick with four edges. There are 25 of them, counting mirror copies.

In other words, these are squiggles you can make with four strokes. They’d make a nice alphabet for people who are addicted to abstraction.

Today we are focussing on six of them, fattened and colored above. They are denoted by the letters H, O, F, +, and the mirrors of H and F. For reasons to become clear later we consider O and + also as mirrors of each other in a certain sense. The goal is to tile rectangles with them, like in the 3×7 and 2×12 rectangles below.

There are many constraints on what tiles one can use, and how many. For instance, an a x b rectangular grid has a(b+1) vertical and (a+1)b horizontal edges, for a total of 2ab+a+b edges. This number is divisible by 4 only if a-b is divisible by 4, so squares are good for tiling, as are the two rectangles above. They both consist of 52 segments and thus require 13 tetrasticks. Below is a different example.

Note that all our 6 letter except for H and its mirror use two horizontal and two vertical segments. As the 3×7 rectangle has 4 more horizontal edges than vertical ones, we need at least four H-tetrasticks (or its mirrors) to tile this rectangle. We can use more, but then only an even number of them. Likewise, we need at least two H-tetrasticks to tile the 2×12 rectangle.

This brings us to today’s puzzle: Tile the 3×7 rectangle with your choice of 13 tetrasticks from our selection of six, and then use the same set to tile the 2×12 rectangle. The examples on this page are attempts that require to flip an H or an F into its mirror (or an O into +). Can you find a perfect solution that doesn’t require flipping a tetrastick over?

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13 solutions for 7×3 :

F, F, F, F, F, F, o,

o, h(turn -90°), h(+90°), o, F(180°), F(180°)

(+12 variations)

2 solutions for 10×2 (you wrote “12” and draw 10)

o, h-90°, o, F-90°, F-90°, F-90°,

F+90°, o, h+90°, o, h-90°, h+90°, o

(and viceversa)

LG Markus 🙂

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