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Sat, 22 Apr 2006
Counting squares
But there is a better way. It's called the Pólya-Burnside counting lemma. (It's named after George Pólya and William Burnside. The full Pólya counting theorem is more complex and more powerful. The limited version in this article is more often known just as the Burnside lemma. But Burnside doesn't deserve the credit; it was known much earlier to other mathematicians.) Let's take a slightly simpler example, and count the squares that have two colors, say blue and black only. We can easily pick them out from the list above:
Remember way back at the beginning where we decided that and and were the same because differences of a simple rotation didn't count? Well, the first thing you do is you make a list of all the kinds of motions that "don't count". In this case, there are four motions:
Now we temporarily forget about the complication that says that some squares are essentially the same as other squares. All squares are now different. and are now different because they are colored differently. This is a much simpler point of view. There are clearly 2^{4} such squares, shown below:
Which of these 16 squares is left unchanged by motion #3, a counterclockwise quarter-turn? All four wedges would have to be the same color. Of the 16 possible colorings, only the all-black and all-blue ones are left entirely unchanged by motion #3. Motion #1, the clockwise quarter-turn, works the same way; only the 2 solid-colored squares are left unchanged.
4 colorings are left unchanged by
a 180° rotation. The top wedge and the bottom wedges switch
places, so they must be the same color, and the left and right wedges
change places, so they must be the same color. But the top-and-bottom
wedges need not be the same color as the left-and-right wedges. We
have two independent choices of how to color a square so that it will
remain unchanged by a 180° rotation, and there are 2^{2} =
4 colorings that are left unchanged by a 180° rotation. These are
shown at right. So we have counted the number of squares left unchanged by each motion:
Next we take the counts for each motion, add them up, and average them. That's 2 + 4 + 2 + 16 = 24, and divide by 4 motions, the average is 6. So now what? Oh, now we're done. The average is the answer. 6, remember? There are 6 distinguishable squares. And our peculiar calculation gave us 6. Waaa! Surely that is a coincidence? No, it's not a coincidence; that is why we have the theorem. Let's try that again with three colors, which gave us so much trouble before. We hope it will say 24. There are now 3^{4} basic squares to consider. For motions #1 and #3, only completely solid colorings are left unchanged, and there are 3 solid colorings, one in each color. For motion 2, there are 3^{2} colorings that are left unchanged, because we can color the top-and-bottom wedges in any color and then the left-and-right wedges in any color, so that's 3·3 = 9. And of course all 3^{4} colorings are left unchanged by motion #4, because it does nothing.
The average is (3 + 9 + 3 + 81) / 4 = 96 / 4 = 24. Which is right. Hey, how about that? That was so easy, let's skip doing four colors and jump right to the general case of N colors:
Add them up and divide by 4, and you get (N^{4} + N^{2} + 2N)/4. So if we allow four colors, we should expect to have 70 different squares. I'm glad we didn't try to count them by hand! (Digression: Since the number of different colorings must be an integer, this furnishes a proof that N^{4} + N^{2} + 2N is always a multiple of 4. It's a pretty heavy proof if it were what we were really after, but as a freebie it's not too bad.) One important thing to notice is that each motion of the square divides the wedges into groups called orbits, which are groups of wedges that change places only with other wedges in the same orbit. For example, the 180° rotation divided the wedges into two orbits of two wedges each: the top and bottom wedges changed places with each other, so they were in one orbit; the left and right wedges changed places, so they were in another orbit. The "do nothing" motion induces four orbits; each wedge is in its own private orbit. Motions 1 and 3 put all the wedges into a single orbit; there are no smaller private cliques. For a motion to leave a square unchanged, all the wedges in each orbit must be the same color. For example, the 180° rotation leaves a square unchanged only when the two wedges in the top-bottom orbit are colored the same and the two wedges in the left-right orbit are colored the same. Wedges in different orbits can be different colors, but wedges in the same orbit must be the same color. Suppose a motion divides the wedges into k orbits. Since there are N^{k} ways to color the orbits (N colors for each of the k orbits), there are N^{k} colorings that are left unchanged by the motion. Let's try a slightly trickier problem. Let's go back to using 3 colors, and see what happens if we are allowed to flip over the squares, so that and are now considered the same. In addition to the four rotary motions we had before, there are now four new kinds of motions that don't count:
The diagonal reflections each have two orbits, and so leave 9 of the 81 squares unchanged. The horizontal and vertical reflections each have three orbits, and so leave 27 of the 81 squares unchanged. So the eight magic numbers are 3, 3, 9, and 81, from before, and now the numbers for the reflections, 9, 9, 27, and 27. The average of these eight numbers is 168/8 = 21. This is correct. It's almost the same as the 24 we got earlier, but instead of allowing both representatives of each pair like , we allow only one, since they are now considered "the same". There are three such pairs, so this reduces our count by exactly 3. Okay, enough squares. Lets do, um, cubes! How many different ways are there to color the faces of a cube with N colors? Well, this is a pain in the ass even with the Pólya-Burnside lemma, because there are 24 motions of the cube. (48 if you allow reflections, but we won't.) But it's less of a pain in the ass than if one tried to do it by hand. This is a pain for two reasons. First, you have to figure out what the 24 motions of the cube are. Once you know that, you then have to calculate the number of orbits of each one. If you are a combinatorics expert, you have already solved the first part and committed the solution to memory. The rest of the world might have to track down someone who has already done this—but that is not as hard as it sounds, since here I am, ready to assist. Fortunately the 24 motions of the cube are not all entirely different from each other. They are of only four or five types:
Unfortunately, the Pólya-Burnside technique does not tell you what the ten colorings actually are; for that you have to do some more work. But at least the P-B lemma tells you when you have finished doing the work! If you set about to enumerate ways of painting the faces of the cube, and you end up with 9, you know you must have missed one. And it tells you how much toil to expect if you do try to work out the colorings. 10 is not so many, so let's give it a shot:
Care to try it out? There are 4 ways to color the sides of a triangle with two colors, 10 ways if you use three colors, and N(N+1)(N+2)/6 if you use N colors. There are 140 different ways to color a the squares of a 3×3 square array, counting reflections as different. If reflected colorings are not counted separately, there are only 102 colorings. (This means that 38 of the colorings have some reflective symmetry.) If the two colors are considered interchangeable (so for example and are considered the same) there are 51 colorings. You might think it is obvious that allowing an exchange of the two colors cuts the number of colorings in half from 102 to 51, but it is not so for 2×2 squares. There are 6 ways to color a 2×2 array, whether or not you count reflections as different; if you consider the two colors interchangeable then there are 4 colorings, not 3. Why the difference?
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