The enumeration results in this section assumes an infinite checkerboard-coloured plane but also applies to sufficiently large finite regions. This holds as long as each polyomino can be placed in all necessary positions without crossing the boundary, ensuring all distinct chrominoes are generated. Under these conditions, the enumeration formulas remain valid.
Proof. For each symmetry class in
Table 3 we must determine the number of distinct chrominoes formed by placing a polyomino
P on the checkerboard-coloured plane
. This count depends on the number of orientations of
P and whether different orientations yield equivalent chrominoes. We also determine when different translations of
P produce equivalent chrominoes. Equivalent chrominoes indicate duplicates, which must be excluded from the total count.
Consider placing a polyomino P on . The resulting chromino has colour indices determined by the indexed modulo Formula (5). Without loss of generality, let this chromino be , corresponding to in Equation (5). Keeping the orientation of P fixed, translating it one square right (or downward) shifts the colour indices forward by one step, yielding a new chromino . Each step forward produces a new chromino until, after C steps, the original chromino reappears. Further translations repeat this cycle, while reversing the translation direction reverses the colour index cycle. Thus, for each orientation of P, the chrominoes cycle through at most C distinct variants: . With orientations, this gives at most free chrominoes.
The remainder of the proof identifies duplicate chrominoes based on the symmetries of the polyomino and the embedding plane. We treat and separately, as has colour-reversing symmetries not present in .
This case is simpler than
to analyse, as every isometry of a polyomino placed in
—whether a translation or a transformation in
—either preserves or reverses the colouring of its squares. As a result, each polyomino generates at most two chrominoes,
and
, which may or may not be distinct. If a chromino possesses a colour-reversing symmetry, then
, meaning there is only one distinct chromino. If only colour-preserving symmetries exist, then
, i.e., there are two distinct chrominoes. The two-colour symmetries of
are defined in
Table 4, and
Table 5 summarises the analysis, classifying chrominoes into 15 two-colour symmetry classes.
For example, a class
polyomino in
Table 3 is asymmetric, so
and
are always distinct. Chrominoes are also distinct for class
and class
polyominoes, since
and
are always colour-preserving. All other classes include chrominoes with both colour-preserving and colour-reversing symmetries.
If
, then the chromino is ‘balanced’ [
8], meaning it contains an equal number of black and grey squares. However, the converse does not hold; there exist balanced chrominoes with
.
Unlike the case, where colour-reversing symmetries create equivalences, the cyclic nature of three or more colour indices and the fact that the only symmetry of for is ensure that the fixed chrominoes are distinct. In other words, a fixed polyomino yields exactly C distinct chrominoes.
However, two chrominoes with different orientations may still be equivalent under the diagonal reflection
. An example of this was illustrated in
Figure 7. To determine which orientations of a polyomino yield equivalent chrominoes, we analyse each symmetry class in
Table 3. Using the dihedral group actions listed in
Table 2, we first identify the distinct orientations of a polyomino
P by checking whether any of its eight possible orientations in (
2) are equivalent. This process is demonstrated in Example 1. Since it is straightforward, we omit the details and directly list the unique orientations for each symmetry class. We then apply
to each orientation, again using the dihedral group actions, to determine which ones yield the same chrominoes.
For an asymmetric polyomino
P of class #1 in
Table 3, the eight possible orientations are shown in
Figure 8, illustrated with a representative example.
Using
Table 2 and applying the symmetry
yields the following correspondences:
i.e., the pairs (f) and (a), (e) and (b), (h) and (c), and (g) and (d) are equivalent for chromino generation. Therefore, class #1 polyominoes always yield chrominoes.
For a polyomino
P of class #2 in
Table 3 with symmetry
, the four possible orientations are given in
Figure 9, illustrated with a representative example.
Using
Table 2 we apply the symmetry
and obtain the following correspondences:
Thus, orientations (a) and (d), and (b) and (c), are equivalent for chromino generation. So class #2 polyominoes yield chrominoes.
For a polyomino
P of class #3 in
Table 3 with symmetry
, the four possible orientations are shown in
Figure 10, illustrated with a representative example.
Using
Table 2, we apply the symmetry
and obtain the following correspondences:
Thus, orientations (a) and (b), and (c) and (d), are equivalent for chromino generation. Therefore, class #3 polyominoes with symmetry yield chrominoes. The argument for polyominoes of class #3 with symmetry is similar and gives the same count.
For a given polyomino
P of class #4 in
Table 3 with the symmetry
, the four possible orientations are given in
Figure 11 and illustrated for a representative example.
Using
Table 2 and applying the symmetry
yields the following correspondences:
Thus, the pairs (a) and (c), are equivalent orientations as far as generating chrominoes is concerned; however, orientations (b) and (d) are different. Thus, polyominoes P with the symmetry belonging to class #4 always yield a total of chrominoes when placed in the checkerboard region . The argument for a polyomino belonging to class #4 with the symmetry is similar and yields the same total counts for .
For a polyomino
P of class #5 in
Table 3 with symmetry
, the two possible orientations are shown in
Figure 12, illustrated with a representative example.
Using
Table 2 and applying the symmetry
, we obtain the following correspondence:
Thus, the two orientations of a class #5 polyomino always yield a total of chrominoes.
For a polyomino
P of class #6 in
Table 3 with symmetries
and
, the two possible orientations are shown in
Figure 13, illustrated with a representative example.
Using
Table 2 and applying the symmetry
, we obtain the following correspondence:
Thus, the two orientations of a class #6 polyomino always yield a total of chrominoes.
For a polyomino
P of class #7 in
Table 3 with symmetries
and
, the two possible orientations are shown in
Figure 14, illustrated with a representative example.
Using
Table 2 and applying the symmetry
, we obtain:
Thus, the two orientations of a class #7 polyomino are different and always yield a total of chrominoes.
For a polyomino
P of class #8 in
Table 3 with all the symmetries of
, the single orientation is shown in
Figure 15, illustrated with a representative example.
Different placements of a polyomino of class #8 in the checkerboard region yield a total of chrominoes. □