Burnside lemma and Polay Count learning little Memory

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In combinatorial mathematics there is such a problem, for example, with red and blue color pairs of 2*2 lattice staining, rotation after the same as a. How many different staining schemes are there? We list, then, a total of 16. But we found that the 3,4,5,6 is the same, 7,8,9,10 is in one, 11,12 is the same, 13,14,15,16 is the same species, that is, only 6 kinds of essentially different staining. Small scale We can enumerate all the options and then choose, on a large scale it is difficult to enumerate all the scenarios. Below, we explain the use of Burnside lemma and Polay count to solve this kind of problem.

The permutation Group G: refers to all permutations. In the example above, there are only 4 permutations, that is, rotation 0, 90, 180, and 270 degrees. Where the size of G is recorded as | G|=4.

Second, (1) for each element K, here the K satisfies the 1<=k<=16,g in which the K remains unchanged the permutation composed of the whole, called. For example, four permutations can make 1 remain the same, and only the first and third permutations make 11 remain the same, so we have (the elements in G are represented by G):

(2) for each element K, under the action of four kinds of permutation the element is called K under G trajectory, expressed as, represents an equivalence class. For example, 1 in four kinds of permutations are obtained 1,3 under four kinds of substitution in turn to get 3,4,5,6. So we have:

We find that the equivalence classes of the elements in the same equivalence class are the same, for example, the equivalence class of 3,4,5,6 is the same as 3. Here we give a formula that is clearly established under its own calculation:

Next we calculate the sum of the number of times each element is invariant under each permutation. As shown in.

can be obtained (note is to use a to represent the permutation, we are all using the G), such as 1 in the first substitution has not changed, then, and in the first substitution of 16 species have not changed, the second type of replacement only 1, 2 did not change, the third substitution only 1,2,11,12 unchanged, the fourth type of replacement only 1, 2 unchanged, Then there are:

In fact, we have the following formula to set up:

Below, we use N to represent the total number of elements, above n=16, we use | G| to indicate the number of permutations, above | G|=4. We have an L equivalence class (which we said above, such as 3,4,5,6 's equivalence class is the same, there are actually only 5 equivalence classes, {1},{2},{3,4,5,6},{11,12},{13,14,15,16}), we have:

So

This is the famous Burnside lemma. Note that the L here is actually the number of different staining schemes we require. Because l represents a different equivalence class, the same equivalence class is like a stain. We get L= (16+2+4+2)/4=6, and the answer we get from the beginning is the same. With this lemma calculation, the first step is to find all permutations, and the second step calculates the number of invariant elements under each permutation. However, we found that sometimes the second step of this calculation is not so easy. Next we say Polay theorem.

First of all, we explain what the circular festival is.

For example, for a n=5, under a certain permutation (1,2,3,4,5) becomes (3,5,1,4,2), we will remember it as (13) (25) (4), that is, the permutation of the Loop section is 3. That is, there is no intersection between the two replacement nodes. For the above question, we have four squares labeled 1,2,3,4.

Construction permutation:

We found that the number of elements in the same permutation section in the same color (which we now assume to be stained with M color, just m=2) is the number of elements that are unchanged at the top of the permutation:

From this we get:

This is the world-famous Polay theorem. With this theorem, we just need to find the loop section of each permutation.

Burnside lemma and Polay Count learning notes