Even Order Polyominoes

## Techniques For Establishing An Even Order

Evidence suggests every nontrivial polyominoe has an even order. Although this remains an open conjecture, we can prove it for several specific shapes, and several classes of shapes. Below are some combinatorial tricks used to assert an even order in certain situations.

#### Notation

Throughout this writeup, and in subsequent pages, a concise hexidecimal notation is used to represent a polyomino. The hex digits are grouped into pairs as usual, each pair describing the bits of a byte. Each byte describes one row of the polyomino, starting at the bottom. Each bit in this byte indicates an occupied square, where the high order bit denotes a square in the leftmost position. Thus ff, f0f0, and c0c0c0c0 are all rectangles consisting of 8 squares. The shape f8e0e0 consists of a row of 5 squares, with two rows of three squares above it.

#### The Checkerboard Argument

Several polyominoes, such as f840, f0e040, e0f080, and fe70, yield to the checkerboard argument. Given a tiled rectangle, overlay a traditional black-and-white checkerboard. Because the base polyomino consists of an even number of squares, the rectangle has at least one even dimension, and the number of black squares in the rectangle equals the number of whites. Yet each piece, no matter how it is embeded, introduces either an excess or a deficiency of 2 white squares. Since excesses and deficiencies must cancel, the number of pieces in the rectangle is even.

#### Orthogonal Stripes

The shape fcf030 evades the checkerboard argument, since it tiles as many white squares as black. Suppose its order is odd while both dimensions of its rectangle are even. This is possible because the shape contains 12 = 2×2×3 squares. Overlay vertical black and white stripes upon the rectangle. Shapes that are oriented horizontally consume 6 white squares and 6 black squares, but each vertical piece introduces an excess or deficiency of 4. These must sum to 0, hence there are an even number of vertical pieces. By overlaying horizontal stripes, there are an even number of horizontal pieces as well. Therefore, if the order is odd, one of the dimensions is odd, and the other is 4 mod 8. The same argument holds for other shapes such as fffc3c30.

#### Sparse Checkerboard

Continuing our analysis of fcf030, assume the width of its rectangle is a multiple of 4, while the height is odd. Overlay a checkerboard upon this rectangle, but this time the white squares are replaced with alternating red and blue squares, and the black squares play no role. Thus each row is colored black red black blue ... in sequence. Because each row is divisible by 4, red and blue are represented equally. Now each piece, no matter its orientation, introduces an excess or deficiency of two red squares, hence there are an even number of pieces. The same proof works for fffc3c30. As of this writing, the order of the latter piece has not been established, but we know that the order is even.

#### Larger Checkerboard Squares

Because f0e020 has 8 squares, we can immediately posit a rectangle whose width is a multiple of 4. Overlay a black and white checkerboard, but this time the squares are two units on a side. Such a checkerboard can be placed in one of 4 positions, as determined by the coordinates of the lower left corner of the lower left 2×2 square (black or white) that is wholly contained within the rectangle. No matter the phase of the checkerboard, black and white areas are distributed evenly across the rectangle. First align the checkerboard at 0,0. Each embeded piece introduces an excess or deficiency of 2 iff it is centered on a point whose x and y coordinates have opposite parity. Hence there are an even number of these pieces. Shift the checkerboard up to force an even number of pieces centered on points whose x and y coordinates have the same parity. We have partitioned the rectangle into two even subsets, hence the order is even.

#### Travel Along The Edge

Consider the shape fcf0f0, which resembles Oklahoma. Drop down to the floor of the rectangle and note that Oklahoma cannot stand on its pan handle. Nor can it lie on its face. It must either rest on its back or stand up on its blunt end. These orientations consume 6 and 3 squares respectively. Thus both dimensions are divisible by 3, and the order of this polyomino is a multiple of 9. One can actually prove that pairs of pieces on the floor face or point towards each other, consuming 6 and 12 squares respectively, hence each dimension is divisible by 6. The reasoning is rather detailed, and is beyond the scope of this discussion.

The polyomino f8f880 provides another example. If a piece stands on the floor, with its tab at the lower right, this tab is necessarily adjacent to the tab of another vertical piece, or the narrow end of a horizontal piece. A horizontal piece, lying face-up on the floor, has, at its narrow end, a copy of itself reflected, or the tab of a vertical piece. Thus the pieces along the floor can be grouped together in pairs, each pair consuming 6, 8, or 10 squares. Each dimension is even, and the shape has order divisible by 4.

#### Three-Colored Stripes

Since the shape fefe80 contains 15 squares, we may assume the width of its housing rectangle is a multiple of three, and overlay a repeating pattern of red, blue, and yellow vertical stripes. Each horizontal piece introduces an excess of 3 red, blue, or yellow squares. Let hr, hb, and hy denote the quantity of pieces in these three categories. Meantime, each vertical piece introduces a deficiency of 6 squares. Let vr, vb, and vy denote the quantity of these pieces. The excess of red must equal the excess of blue, or 3hr - 6vr = 3hb - 6vb. Divide by 3 and reduce mod 3 to obtain hr + vr = hb + vb. By symmetry, each side of this equation is also equal to hy + vy. The entire rectangle has been partitioned into three subsets, each having the same cardinality mod 3. Thus the order is a multiple of 3.

The shape f8f8f8e0 admits the same sets hr hb hy, and vr vb vy. This time the vertical pieces have an excess of 3 (of red blue or yellow), while the horizontals show a deficiency of 3. Excess red = excess blue, or 3vr-3hr = 3vb-3hb. Divide by 3 and invoke symmetry to get vr-hr = vb-hb = vy-hy. This seems like it should tell us something, but I guess it doesn't. 