introduction

This post is about a very powerful tool in combinatorics called Polya enumeration theorem. The paradigm question it answers is: How many distinct necklaces can be made out of N black and M white beads?

Suppose N=2 and M=2, it’s easy to see that there are only two possibilities, shown in Fig. 1. When the number of beads gets large, it will be quite difficult to enumerate them by brutal force.

Figure 1. Two distinct necklaces with 2 black and 2 white beads.

Before we proceed to the real content, I would like to point out a few random things. The mathematician George Polya who discovered this enumeration theorem has various important contributions to combinatorics, number theory, and probability theory. One interesting and popular example is the Polya urn model, commonly known as ‘the rich get richer’. He also wrote a popular book on problem solving called How to solve it. Also he lived to 97 years of age.

preliminary attempts

There are two complications in our necklace counting problem:

• symmetry caused by the ‘necklaceness’

The former is easy to deal with. In fact, if we ignore the necklace symmetry for now, elementary combinatorics would tell us that there are

$\frac{4!}{2!2!} = 6$

ways to arrange the 2 black and 2 white beads, which are listed in Fig. 2. Let’s call the necklaces in the first row $$a_i$$, and the ones in the second row $$b_i$$. From now on I will draw them as squares because for four beads the symmetry group is dihedral group of order 4, i.e., $$D_4$$.

There are 8 elements in $$D_4$$

$D_4 = \{R_0, R_{90}, R_{180}, R_{270}, H, V, D_1, D_2 \}$

where $$R_\theta$$ is rotation with $$\theta$$ degrees (say clockwise), $$H$$ and $$V$$ are horizontal and vertical flips, $$D_{1,2}$$ are diagonal flips. Specifically, let’s assume $$D_1$$ flips around the line connecting the two white balls in $$b_1$$ (actually I show the flip axis of $$D_2$$ in Fig. 3). Note also that $$R_0$$ is identity.

Figure 2. Six different ways to arrange 2 black and 2 white beads without considering $$D_4$$ symmetry. The necklaces in first ($$a_i$$) and second rows ($$b_i$$) belong to different orbits.

From Figure 2, we can see that the 4 necklaces in the first row are equivalent to each other under the symmetry of $$D_4$$ and the 2 necklaces in the second row are equivalent. Thus there are only two distinct necklaces with the given beads. These equivalent necklaces are examples of orbits in group theory.

As a summary, what we did in this section is to first explore the identical bead condition to limit the $$4!=24$$ possible necklaces to 6, and then check their equivalence by hand. This approach is not practical when the number of beads gets large.

the magic

To demonstrate the power of Polya enumerate theorem, let me first show you the procedure to get that answer of 2 for our necklace counting problem. Its explanations will be in the next section.

Each group is associated with a polynomial called cycle index, which is basically a fingerprint of the group in terms of how its individual element generates cycles (more details will be given in the next section). The cycle index of many famous groups are known. Even if it is unknown for a group, its calculation is very easy albeit somewhat tedious. For $$D_4$$ it is given by

$Z_{D_4}(z_1,z_2,z_3,z_4) = \frac{z_1^4 + 2z_1^2z_2 + 3z_2^2 + 2z_4}{8}$

Now we make the substitution

$z_1\rightarrow B+W, z_2\rightarrow B^2+W^2, z_4\rightarrow B^4 + W^4$

where $$B$$ and $$W$$ denote the color of the bead. After simplification, the polynomial becomes

$B^4 + B^3W + 2B^2W^2 + BW^3 + W^4$

and the coefficient of $$B^2W^2$$, i.e., 2, is the number of distinct necklaces with 2 black beads and 2 white beads. Similarly, the coefficients of other terms are the number of distinct necklaces with the corresponding bead situations. Also, if we set $$B=W=1$$, then the polynomial evaluates to $$6$$, which is the number of distinct necklaces with four beads of two colors. Furthermore, if we set $$z_1=z_2=z_4=m$$, then the polynomial evaluates to the number of distinct necklaces with four beads of $$m$$ colors.

I was quite shocked when I first learned this.

the details

In this section, I will explain why the previous magic works. Basic familarity with group theory is assumed. For example, you should know what coset is.

The following three concepts are essential for Polya enumeration theorem, given a set $$X$$, and a group $$G$$ that acts on it.

• orbit: the set of set elements connected by the group actions, i.e., $$orb_G(x)=\{g(x)|g\in G\}$$. For example, the two rows in Fig. 2 are two orbits.
• stabilizer: the set of group elements that stabilize a particular set element, i.e., $$stab_G(x)=\{g\in G|g(x)=x\}$$.
• fixed points: the set of set elements that are stabilized by a particular group element, i.e., $$fix_X(g)=\{x\in X|g(x)=x\}$$.

To be concrete, in our necklace example in Fig. 1, some of the stabilizers and fixed points are listed below

$stab_G(a_1)=\{R_0, H\}, stab_G(a_2)=\{R_0, V\}, stab_G(b_1)=\{R_0, R_{180},D_1, D_2\}\\ fix_X(R_{90})=\varnothing, fix_X(H)=\{a_1,a_3\}, fix_X(D_1) = \{b_1, b_2\}$

If you take the trouble to work out all the stabilizers for each necklace in Fig. 1, there is a pattern: all $$a_i$$ have 2 stabilizers, and all $$b_i$$ have 4 stabilizers. Also note $$D_3$$ has 8 elements. It turns out all these observations are no coincidence. It is summarized in the so-called orbit-stabilizer theorem:

$|G| = |orb_G(x)| |stab_G(x)|, \forall x \in X$

To prove it, notice that $$stab_G(x)$$ is a subgroup of $$G$$. Then all we need to do is to show the existence of a one-to-one correspondence between the orbit of $$x$$ and the left cosets of $$stab_G(x)$$. Such a mapping can be written as $$\phi: orb_G(x) \rightarrow G/stab_G(x)$$ and

$\phi(g(x)) = g * stab_G(x)$

where $$*$$ denotes the operation between group elements. I will omit it as long as it’s not confusing. The proof then has three parts, i.e., $$\phi$$ needs to be

• well-defined
• injective
• surjective

For $$\phi$$ to be well-defined, we need to show that for $$g, h\in G$$, if $$g(x)=h(x)$$, then $$\phi(g(x))=\phi(h(x))$$, i.e., same input should give same output. Note that the assumption immediately tells us that $$h^{-1}g\in stab_G(x)$$, thus $$h^{-1}g * stab_G(x) = stab_G(x)$$, which gives the desired result.

For $$\phi$$ to be injective, we need to show that if $$\phi(g(x))=\phi(h(x))$$, then $$g(x)=h(x)$$. This is essentially the same thing as the well-defineness proof.

Finally, the surjectiveness of $$\phi$$ is obvious.

I find the orbit-stabilizer therorem quite neat because set $$X$$ has little to do with $$G$$, but still any of its element can partition $$G$$. Furthermore, set elements in the same orbit have the same number of stabilizers. In other words, elements in the same set partition (recall that orbits partition the set) induce similar group partitions in terms of sizes.

With these terminologies, our necklace counting problem can be stated as: given a group G (symmetry, equivalence criterion) acting on a set X (beads), how many different orbits (necklaces) are there?

In fact, orbit-stabilizer theorem is powerful enough to answer this question. To get the number of orbits, we need to utilize an auxiliary quantity: the total number of (stabilizer, fixed point) pairs. There are two ways to count it, either from the stabilizers or from the fixed points. The results should agree, i.e.,

$\sum_{x\in X}|stab_G(x)| = \sum_{g\in G}|fix_X(g)|$

Since orbits partition the set, the first summation can be broken down as

$\sum_{o \in orbits}\sum_{x\in o}|stab_G(x)| = \sum_{o \in orbits}|G|$

Thus we have

$\text{# of orbits} = \frac{1}{|G|} \sum_{g\in G}|fix_X(g)|$

This is known as Burnside’s lemma. There are still a lot of work involved in the fixed points calculation. To simplify this calculation, we need to introduce the concepts of cycle index monomials and polynomials.

Figure 3. Labeling of the bead positions and the flip axis of $$D_2$$.

To proceed, let’s first give the four beads some labels, as shown in Fig. 3, so that we can refer to them more easily. Note that the group action on the set $$X$$ can be described as permutation of the labels, which can be further decomposes into cycles. For example,

$R_0: (1)(2)(3)(4) \rightarrow z_1^4\\ D_2: (13)(2)(4) \rightarrow z^2_1z_2\\ R_{90}: (1432) \rightarrow z_4 \\ R_{180}: (13)(24) \rightarrow z_2^2$

where the last term in each row is the cycle index monomial for the corresponding group element. It has the form

$\Pi_{i=1}^n z_i^{b_i}$

where $$n=|X|$$ is the size of set $$X$$, $$i$$ refers to the cycle length, and $$b_i$$ is the count of length-$$i$$ cycles. It’s easy to see that the following identity holds

$\sum_{i=1}^n i b_i = n$

Note also that in most cases, the 1-cycles are not written explicitly in cycle notations. Don’t forget to include them for cycle index monomials though.

The cycle index monomial of a group element $$g$$ is directly relevant to $$|fix_X(g)|$$. If the beads (labels) in the same cycle are of the same color, that necklace is fixed by $$g$$. Furthermore, simplying polynomials are essentially counting combinations because the same terms represent the same combinations. Thus we can make the replacement of

$z_i \rightarrow \sum_{k=1}^m c_k^i$

for the cycle index monomials to see all the possible combinations. Here $$m$$ is the number of colors and $$c_k$$ are color labels. For example, in the case of our 4-bead necklace with black and white colors

$D_4: z_1^2z_2=(B+W)^2(B^2+W^2) = B^4 +2B^3 W+2B^2 W^2+2B W^3 + W^4$

You can easily check graphically on Fig. 3 that all these coefficients are correct.

Let me further demonstrate the case of three colors, say black, white, and red with labels $$B, W, R$$. In this case, the replacement is

$z_1 \rightarrow B+W+R, z_2\rightarrow B^2+W^2+R^2, \ldots$

This is because for any 1-cycle, i.e., single bead, all 3 colors can be used. For 2-cycles, i.e., one bead becomes another bead under the action of the group element, both beads need to have the same color for the necklace to be a fixed point.

Finally, the cycle index polymonial of the group is the average over the whole group, i.e.,

$Z_G(z_1, z_2, \ldots) = \frac{1}{|G|}\sum_{g\in G}\Pi_i z_i^{g_i}$

where $$g_i$$ are the length-$$i$$ cycle counts of $$g$$. And Polya enumeration theorem states that all the possible colorings can be derived by the forementioned replacement of $$z_i$$, which is what we did in the previous section.