So far, we have looked only at pillows with concave and convex edges. Today, we begin also to allow straight edges. To keep it simple, let’s look at the three different pillows that have two straight edges, one concave, and one convex edge. Here they are. I call them the *arrow pillows*.

Because they have straight edges, we can finally tile rectangles that have straight edges, too, like so:

There are a few immediate questions: Is this always possible? Can we say something about the number of arrows of each type we need? The key to the answers is indicated in the right image. The convex edge of one arrow pillow (the *predecessor*) fits snugly into the concave edge of a second arrow pillow (the *successor*), thus providing us with a recipe to move from one pillow to a neighbor. If we have a tiling of a rectangle just by arrow pillows, this sequence of consecutive successors must form a closed cycle. Therefore, the entire rectangle will be covered by possibly several such closed cycles, so we have what is called a *Hamiltonian circuit*. Readers of my blog have seen these before.

Vice versa, given any Hamiltonian circuit and a direction for each component, we can lay out the arrow pillows along each path to obtain a tiling. Below are two more examples with two components each that use only right and straight arrow pillows.

Can you tile a 5×5 square with arrow pillows? If you checkerboard color the rectangle black and white, any path alternates between black and white squares, so a closed path will cover the same number of white and black squares. Thus in particular Hamiltonian circuits must have an even length on rectangles.

Let’s look at a single closed cycle, and let’s assume we follow it clockwise. Then there must be four more right turns than left turns. We have seen examples with no left turn arrow pillow, and with two left turns. Below are examples with just one and just three left turns.

These little insights not only help to show that some tiling is impossible, they also give hints to design tilings. For instance, suppose you want to tile a square using the same number of straight, right, and left arrow pillows. Then the smallest square for which this could work is the 6×6 square. We also see that we need an even number of cycles in our Hamiltonian circuit in order to balance the left and right arrow pillows. The simple solution below uses two mirror symmetric tilings of 3×6 rectangles.

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