P lya? But I Hardly Know Ya!faculty.etsu.edu/gardnerr/4127/algebra-club/polya-talk.pdfIn 1921, Major...

Pólya? But I Hardly Know Ya!

Robert A. Beeler, Ph.D.

East Tennessee State University

February 8, 2019

Robert A. Beeler, Ph.D. (East Tennessee State University )Pólya? But I Hardly Know Ya! February 8, 2019 1 / 43

A Counting Problem

Suppose that we want to make necklaces with six beads. Each beadmay either be black or white.

How many different necklaces can we make?


What do we mean by “different”?

Note that we can rotate the strand of beads to obtain an equivalentone. Likewise, we can flip the strand to obtain an equivalent one.


Brute Force Enumeration

Assuming that flips and rotations do not result in different necklaces,we end up with fewer than the 26 = 64 necklaces that we wouldexpect through elementary counting. Brute force enumeration revealsthat there are in fact thirteen different arrangements.


Brute Force Enumeration (Part 2)

Generally, we don’t want to do brute force enumeration. We wantsomething that can be generalized to n beads and m colors. In whichcase, things can get bad pretty quickly.

Example: 26 beads and 2 colors yields 1296858 different necklaces.Example: 6 beads and 16 colors yields 1415896 different necklaces.


Table for n beads and m colors (small values)

m \ n 3 4 5 6 7 8 92 4 6 8 13 18 30 463 10 21 39 92 198 498 12194 20 55 136 430 1300 4435 150845 35 120 377 1505 5895 25395 1100856 56 231 888 4291 20646 107331 5637867 84 406 1855 10528 60028 365260 22503118 120 666 3536 23052 151848 1058058 74729849 165 1035 6273 46185 344925 2707245 2155296910 220 1540 10504 86185 719290 6278140 55605670


Burnside’s Lemma

Burnside’s Lemma, and its generalization Pólya’s EnumerationTheorem, allow for “common sense” counting in cases such as above.

Let X be a set and let G be a group permuting (acting on) theelements of X . Two elements of X , say a and b, are equivalent underG if there exists π ∈ G such that π(a) = b. Note that like allequivalence relations, this partitions X into classes.


Burnside’s Lemma

For each element π ∈ G , the set of invariants of π is the set ofelements of X that are fixed under π. In other words,

Inv (π) = {x ∈ X : π(x) = x}.

Burnside’s Lemma - Let G be a group acting on a set X . Thenumber of equivalence classes in X induced G is given by

1

|G |

∑

π∈G

|Inv (π)|.


An Example

Consider the following graph:

4 5 6 7

2 3

1

We want to know the number of equivalence classes induced by itsautomorphism group.


An Example (Part 2)

4 5 6 7

2 3

1

π Inv (π) |Inv (π)|(1)(2)(3)(4)(5)(6)(7) {1, 2, 3, 4, 5, 6, 7} 7(1)(2)(3)(4)(5)(6, 7) {1, 2, 3, 4, 5} 5(1)(2)(3)(4, 5)(6)(7) {1, 2, 3, 6, 7} 5(1)(2)(3)(4, 5)(6, 7) {1, 2, 3} 3(1)(2, 3)(4, 6)(5, 7) {1} 1(1)(2, 3)(4, 6, 5, 7) {1} 1(1)(2, 3)(4, 7, 5, 6) {1} 1(1)(2, 3)(4, 7)(5, 6) {1} 1


An Example (Part 3)

We can use the table, along with Burnside to find the number ofequivalence classes:

1

|G |

∑

π∈G

|Inv (G )| =

1

8(7 + 2(5) + 3 + 4(1)) = 3.

Note that the three classes are {1}, {2, 3}, and {4, 5, 6, 7}.


A Bit of History

Note that the name “Burnside’s Lemma” is a bit misleading. Thelemma was in fact known by Cauchy as early as 1845. However,Cauchy’s version made several unnecessary assumptions and waspractically unusable.


A Bit History (Part 2)

In 1887, Frobenius published a paper that includes the lemma,without Cauchy’s unnecessary assumptions. It is also considerablymore efficient than Cauchy’s formulation. However, Frobenius givesCauchy credit for it. Sometimes, the lemma is referred to as the“Cauchy-Frobenius Counting Formula”.


A Bit of History (Part 3)

Most people seemed to learn the formula from Burnside’s bookTheory of Groups of Finite Order. Published in 1911, this is one ofthe first books on group theory. It should be noted that Burnsideattributes the lemma to Frobenius. It does not seem to be referred toas “Burnside’s Lemma” until 1961. Because of its convoluted history,the lemma is sometimes referred to as “The Lemma that is notBurnside’s”.


Coloring Problems

When applying Burnside’s Lemma to coloring problems (such as ournecklace problem earlier), we need a bit of group theory.

Recall that every permutation can be written as a product of disjointcycles. The number of disjoint cycles (including fixed points) is calledthe cycle index of the permutation. The cycle index of π is denotedcyc(π).


Coloring Problems (Part 2)

Suppose that we are coloring n objects using m possible colors. Notethat our set X is the set of all mn possible “naive” colorings.

So how many colorings are invariant under a permutation π? Toanswer this, we will make use of the cycle index.


Coloring Problems (Part 3)

As an example, take the permutation

π = (1, 2)(3, 4, 5)(6, 7, 8, 9).

For a coloring to be invariant under π, 1 and 2 must receive the samecolor. Likewise, 3,4, and 5 must receive the same color (though notnecessarily the same color as 1 and 2). Finally, 6, 7, 8, and 9 must allreceive the same color (again, not necessarily the same as {1, 2} or{3, 4, 5}).

Thus,|Inv (π)| = mcyc(π) = m3.


Burnside’s Lemma - Cycle Index Version

The above observation allows us to construct a version of Burnside’sLemma using the cycle index.

Burnside’s Lemma (Cycle Index Version) - Suppose that we arecoloring n objects using an m-set of colors. Let X be the set of allmn colorings of the objects. If G is a group acting on X , then thenumber of distinct colorings is given by

1

|G |

∑

π∈G

mcyc(π).


An example

How many distinct ways can we color the vertices of the followinggraph using three colors?

4 5 6 7

2 3

1


An example (Part 2)

The automorphism group is as follows:

π cyc(π) π cyc(π)(1)(2)(3)(4)(5)(6)(7) 7 (1)(2)(3)(4)(5)(6, 7) 6(1)(2)(3)(4, 5)(6)(7) 6 (1)(2)(3)(4, 5)(6, 7) 5(1)(2, 3)(4, 6)(5, 7) 4 (1)(2, 3)(4, 6, 5, 7) 3(1)(2, 3)(4, 7, 5, 6) 3 (1)(2, 3)(4, 7)(5, 6) 4

Using Burnside and the above calculations gives the number ofdistinct colorings as

1

|G |

∑

π∈G

mcyc(π)

=1

8

(

m7 + 2m6 +m5 + 2m4 + 2m3)

.

For m = 3, this gives 513 distinct colorings.Robert A. Beeler, Ph.D. (East Tennessee State University )Pólya? But I Hardly Know Ya! February 8, 2019 20 / 43

Returning to our necklace example

Our original problem was to determine the number of distinctnecklaces on six beads, where each bead could be either white orblack.

The strand of beads can be either flipped or rotated. Thus the groupacting on the beads is the sixth dihedral group, D6.


Returning to our necklace example (Part 2)

The elements of D6:

π cyc(π) π cyc(π)(1)(2)(3)(4)(5)(6) 6 (1)(2, 6)(3, 5)(4) 4(1, 2)(3, 6)(4, 5) 3 (1, 2, 3, 4, 5, 6) 1(1, 3)(2)(4, 6)(5) 4 (1, 3, 5)(2, 4, 6) 2(1, 4)(2, 3)(5, 6) 3 (1, 4)(2, 5)(3, 6) 3(1, 5)(2, 4)(3)(6) 4 (1, 5, 3)(2, 6, 4) 2(1, 6, 5, 4, 3, 2) 1 (1, 6)(2, 5)(3, 4) 3

This gives the number of distinct m colorings as

1

|G |

∑

π∈G

mcyc(π) =1

12

(

m6 + 3m4 + 4m3 + 2m2 + 2m)

.

For m = 2, we have 13 colorings.Robert A. Beeler, Ph.D. (East Tennessee State University )Pólya? But I Hardly Know Ya! February 8, 2019 22 / 43

A application to puzzles

In 1921, Major Percy MacMahon invented a puzzle. The puzzlesconsists of a set of one-sided square tiles that have been divided intoquadrants along the diagonals. Each of the resulting quadrants isassigned one of three colors. The set of tiles consists of everypossible 3-coloring of the squares, up to rotation. How many tiles dowe need?


An application to puzzles (Part 2)

Note that the group acting on the quadrants of the square is thecyclic group C4. Its permutations are:

π cyc(π) π cyc(π)(1)(2)(3)(4) 4 (1, 2, 3, 4) 1(1, 3)(2, 4) 2 (1, 4, 3, 2) 1

Using Burnside and the above calculations give the number ofdistinct m-colorings as

1

|G |

∑

π∈G

mcyc(π) =1

4

(

m4 +m2 + 2m)

.

In particular, when m = 3 this yields 24 distinct colorings.



And here they are:

The goal is to arrange the pieces in such a way that if two tilestouch, the color on their borders match. Further, the outside borderis a solid color. Other puzzles include duplication (make two shapesof the same size according to above restrictions) and triplication(make three shapes of same size according to above restrictions).



Here are some solutions:


An interesting example

Consider the group acting on the parallelogram below. The groupconsists of four elements:

(i) The identity, e

(ii) The reflection across the x-axis, α

(iii) The reflection across the y -axis, β

(iv) The reflection across both axes, αβ

Note that this group is isomorphic to Klein-4.

c

b

a

d3 2

4 1


An interesting example (Part 2)

We are going to look at how the this group induces permutationgroups on three sets:

(i) G1 acting on the set of vertices {a, b, c, d}

(ii) G2 acting on the set of edges {1, 2, 3, 4}

(iii) G3 acting on the set vertices and edges {a, b, c, d , 1, 2, 3, 4}

Note that all of these are isomorphic. However, G1 and G2 areembedded in S4, while G3 is embedded in S8.

c

b

a

d3 2

4 1



First the identity:

c

b

a

d3 2

4 1

7→e c

b

a

d3 2

4 1

e cyc(e)G1 (a)(b)(c)(d) 4G2 (1)(2)(3)(4) 4G3 (a)(b)(c)(d)(1)(2)(3)(4) 8



Now α:

c

b

a

d3 2

4 1

7→α a

b

c

d4 1

3 2

α cyc(α)G1 (a,c)(b)(d) 3G2 (1,2)(3,4) 2G3 (a,c)(b)(d)(1,2)(3,4) 5



Now β

c

b

a

d3 2

4 1

7→β c

d

a

b2 3

1 4

β cyc(β)G1 (a)(b,d)(c) 3G2 (1,4)(2,3) 2G3 (a)(b,d)(c)(1,4)(2,3) 5



Finally, αβ

c

b

a

d3 2

4 1

7→αβ a

d

c

b1 4

2 3

αβ cyc(αβ)G1 (a,c)(b,d) 2G2 (1,3)(2,4) 2G3 (a,c)(b,d)(1,3)(2,4) 4


An Interesting Example (Part 7)

From the above tables, we can use Burnside to give the number of mcolorings of each of the respective sets:

G1,14(m4 + 2m3 +m2)

G2,14(m4 + 3m2)

G3,14(m8 + 2m5 +m4)

For m = 2, this yields 9, 7, and 84 distinct colorings of the abovesets, respectively.


An interesting implication of the above example

In the above example, G1, G2, and G3 were all isomorphic to Klein-4.However, they yielded a different number of distinct coloringsbecause of how they acted on their respective sets.


An application to graph theory

Suppose that we want to know the number of non-isomorphic graphsof order n.

Whether an edge is or isn’t in the graph can be thought of as a twocoloring of the “potential” edges in the complete graph Kn.

The group acting on the vertices of the complete graph is thesymmetric group Sn. However, this induces a group on the potentialedges, the pair group, S

(2)n .


Constructing the Pair Group

Type No. Ex. π ∈ S4 gπ ∈ S(2)4 cyc(gπ)

[1, 1, 1, 1] 1 (1)(2)(3)(4) (12)(13)(14)(23)(24)(34) 6[2, 1, 1] 6 (1, 2)(3)(4) (12)(13, 23)(14, 24)(34) 4[2, 2] 3 (1, 2)(3, 4) (12)(13, 24)(14, 23)(34) 4[3, 1] 8 (1, 2, 3)(4) (12, 23, 13)(14, 34, 24) 2[4] 6 (1, 2, 3, 4) (12, 23, 34, 14)(13, 24) 2

This gives that the number of non-isomorphic graphs of order four is

1

24

(

m6 + 6m4 + 3m4 + 8m2 + 6m2)

=1

24

(

m6 + 9m4 + 14m2)

.

For m = 2, this gives eleven non-isomorphic graphs.


And here they are!


Pólya’s Enumeration Theorem

Pólya’s Enumeration Theorem is a generalized version of Burnside’sLemma. Basically, Pólya’s Enumeration Theorem allows you to findthe number of colorings that satisfy a given condition.

For example, how many distinct necklaces can you make with 17orange beads, 23 purple beads, 15 blue beads, and 5 gold beads?

Pólya’s Enumeration Theorem is one of the contenders for “TheFundamental Theorem of Combinatorics,” but let’s save it for(maybe) another talk. (BTW - There are48056241625130292004826575200 such necklaces.)


A little more history...

Given the controversy regarding the name of Burnside’s Lemma, Ithought you might be interested that something similar happenedwith Pólya’s Enumeration Theorem. Originally, this was discovered in1927 by Redfield. Pólya independently rediscovered it in 1937. Pólyawas the bigger name and had better examples (e.g., chemicalcompounds) so it was popularized due to him.


Anticipating the Questions of My Colleagues

For the “standard groups” that come up in combinatorics (cyclic,dihedral, symmetric, pair group). YES!!! (For a certain value of“generalize”). Hint: The general formula for the cyclic and thedihedral involves the Euler φ-function.


Anticipating the Questions of My Colleagues

Some of the related stuff that might make for possible studentresearch: history, some of the weird Pólya applications (i.e.,chemistry), and working through the book by Harary and Palmer.


Additional Readings

(i) Beeler. How to Count.

(ii) Harary and Palmer. Graphical Enumeration.

(iii) Neumann. “A Lemma that is not Burnside’s”. In TheMathematical Scientist.

(iv) Williamson. Pólya Counting Theory.


Questions?


P lya? But I Hardly Know Ya!faculty.etsu.edu/gardnerr/4127/algebra-club/polya-talk.pdfIn 1921, Major...

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