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DefinitionsInteger Partitions

Set Partitions

Partitions and Permutations

Gordon Royle

Semester 1, 2004

Gordon Royle Partitions and Permutations


Set Partitions

PartitionsThe word partition is shared by (at least) two different concepts,although both refer to the process of dividing an object intosmaller sub-objects.

I Integer PartitionsA partition of an integer n is a way to write it as a sum ofsmaller integers, such as

5, 4+1, 3+2, 3+1+1, 2+2+1, 2+1+1+1, 1+1+1+1+1.

I Set PartitionsA partition of a set is a way to divide it into a number ofsubsets, such as

{1, 2, 3}, {1, 2}{3}, {1, 3}{2}, {1}{2, 3}, {1}{2}{3}



Set Partitions

GeneratingConjugacyCounting

Lexicographic order

We will represent a partition of an integer n by a sequencea1a2 . . ., where a1 a2 > 0. Thus the partitions of 7 wouldbe represented as follows.

1111111 211111 22112221 3111 3211322 331 4111421 43 51152 61 7

In this table the partitions are given in lexicographic order.



Set Partitions


Reverse Lexicographic Order

The simplest algorithm to generate partitions actually generatesthem in reverse lexicographic order, starting with the partition nand ending with

I Find the largest index i such that ai 6= 1; suppose thatai = a + 1 so that

= (a + 1)11 . . . 1.

I Replace the suffix (a + 1)11 . . . 1 by aa . . . ar where r < a,thus obtaining

succ() = aa . . . ar.



Set Partitions


Example 1

What is the successor of = 22211?

I The largest i such that ai 6= 1 is i = 3

a1 a2 a3 a4 a52 2 2 1 1

I Replace the suffix 211 with 1111

a1 a2 a3 a4 a5 a62 2 2 1 12 2 1 1 1 1

Therefore we getsucc(22211) = 221111.



Set Partitions


Example 2

What is the successor of = 3311?

I The largest i such that ai 6= 1 is i = 2

a1 a2 a3 a43 3 1 1

I Replace the suffix 311 with as many 2s as possible, and aremainder if necessary.

a1 a2 a3 a43 3 1 13 2 2 1

Therefore we getsucc(3311) = 3221.



Set Partitions


Fixed number of parts

Suppose we are interested in generating the number of partitionsof n into exactly k parts, for example if n = 11 and k = 4 we get

3332 4322 4331 44215222 5321 5411 62216311 7211 8111



Set Partitions


Successors

The easiest algorithm for generating partitions of fixed size k startswith the partition

(n k + 1)1111

and then computes the successor of = a1a2 . . . ak as follows:

I Let i be the smallest index such that ai < a1 1.I Assign ai + 1 to each of a2, a3, . . ., ai and then set a1 as

needed to maintain a partition of n.

The algorithm terminates if there are no values i such thatai < a1 1; in this case each part of the partition is bnk c or d

nk e.



Set Partitions


Example

Suppose that n = 15 and k = 5 and we want the successor of = 66111.

I The smallest index i such that ai < 5 is i = 3a1 a2 a3 a4 a56 6 1 1 1

I Set a2 and a3 to the value a3 + 1 which is 2a1 a2 a3 a4 a5 2 2 1 1

I Set a1 to the required value to maintain a partition of n = 15a1 a2 a3 a4 a59 2 2 1 1



Set Partitions


Colex ordering

This ordering of partitions with a fixed number of parts is notreverse lexicographic, but rather colexicographic.

Lexicographic Reverse lex Colexicographic3322 7111 71113331 6211 62114222 5311 53114321 5221 44114411 4411 52215221 4321 43215311 4222 33316211 3331 42227111 3322 3322



Set Partitions


Conjugate permutations

Suppose that f and g are two permutations in the symmetricgroup Sym(n). Then they are said to be conjugate if

f = h1gh

for some h Sym(n).

ExampleThe permutations f = (1, 2, 3)(4, 5) and g = (1, 3, 4)(2, 5) areconjugate because

(1, 2, 3)(4, 5) = (2, 3, 4) (1, 3, 4)(2, 5) (2, 4, 3)

and therefore we may take h = (2, 4, 3) in the above definition.



Set Partitions


Conjugacy is an equivalence relation

I Every permutation is conjugate to itself

f = e1fe

I If f is conjugate to g, then g is conjugate to h

f = h1gh = g = hfh1

I If a is conjugate to b and b conjugate to c, then a is conjugateto c

a = h1bh and b = g1cg = a = h1g1cgh = (gh)1c(gh).



Set Partitions


Conjugacy Classes

This means that the permutations fall into disjoint classes calledconjugacy classes such that two permutations are conjugate if andonly if they lie in the same class.

e (3, 4) (1, 2)(3, 4) (2, 3, 4) (1, 2, 3, 4)(2, 3) (1, 3)(2, 4) (2, 4, 3) (1, 2, 4, 3)(2, 4) (1, 4)(2, 3) (1, 2, 3) (1, 3, 4, 2)(1, 2) (1, 2, 4) (1, 3, 2, 4)(1, 3) (1, 3, 2) (1, 4, 3, 2)(1, 4) (1, 3, 4) (1, 4, 2, 3)

(1, 4, 2)(1, 4, 3)



Set Partitions


Cycle Structure

The cycle structure of a permutation is the number of cycles ofeach length in its cycle decomposition.

TheoremTwo permutations are conjugate in Sym(n) if and only if theyhave the same cycle structure.

We need to prove two things

I If two permutations are conjugate, then they have the samecycle structure, and

I If two permutations have the same cycle structure, then theyare conjugate.



Set Partitions


Conjugate permutations have the same cycle structure

If the permutation g maps

i ig

then h1gh mapsih igh.

Therefore for every cycle

(a, b, . . . , c)

in the cycle decomposition of g, there is a corresponding cycle

(ah, bh, . . . , ch)

in the cycle decomposition of f = h1gh.



Set Partitions


Permutations with the same cycle structure are conjugate

If two permutations f and g have the same cycle structure, thenwe can find a conjugating permutation h such that f = h1gh.

For each cycle(f1, f2, . . . fk)

of f there is a corresponding cycle

(g1, g2, . . . , gk)

of g.

Then define h by the rule

h : gi fi.



Set Partitions


Example

Letf = (1, 4, 5)(2, 7)(3, 6) g = (2, 4, 6)(1, 5)(3, 7)

Then define h (in image notation) as follows:

h =(

1 2 3 4 5 6 72 1 3 4 7 5 6

)Then h = (1, 2)(5, 7, 6) and it is easy to check that

f = h1gh

and so h is a conjugating permutation.



Set Partitions


In GAP

gap> f := (1,4,5)(2,7)(3,6);(1,4,5)(2,7)(3,6)gap> g := (2,4,6)(1,5)(3,7);(1,5)(2,4,6)(3,7)gap> h := PermList([2,1,3,4,7,5,6]);(1,2)(5,7,6)gap> h^-1*g*h;(1,4,5)(2,7)(3,6)gap> f = last;truegap>



Set Partitions


Cycle structure

The cycle structure of a permutation can be represented in anumber of different ways. One way would be to simply list thelengths of the cycles. For example, if f Sym(10) has cycledecomposition

f = (1, 2, 4)(5, 6)(8, 9, 10)

then its cycle structure could be written as

33211

In this representation, the cycle structure of a permutation issimply a partition of n, the degree of the permutation!



Set Partitions


Partitions and Conjugacy Classes

We have therefore essentially proved the following:

TheoremThe number of conjugacy classes of Sym(n) is equal to thenumber of partitions of n.

For Sym(4) the correspondence is

Partition Representative

1111 e211 (1, 2)22 (1, 2)(3, 4)31 (1, 2, 3)4 (1, 2, 3, 4)



Set Partitions


Counting partitions

Let p(n) be the number of partitions of n, and let pk(n) be thenumber of partitions of n that have k parts.Then

p(n) =n

k=1

pk(n).

We can write these numbers out in a triangle reminiscent ofPascals triangle:

p1(1)p1(2) p2(2)

p1(3) p2(3) p3(3)p1(4) p2(4) p3(4) p4(4)



Set Partitions


Top of the triangle

By direct calculation we see that the triangle starts

11 1

1 1 11 2 1 1

1 2 2 1 11 3 3 2 1 1

The total number of partitions p(n) of each number n is obtainedby summing the entries in each row of this triangle.



Set Partitions


A recurrence

We can express the values of pk(n) in terms of smaller suchnumbers by splitting up the partitions of n according to thenumber of parts that are equal to 1. So if P is the set of allpartitions of n with k parts, then define P0, P1, . . . , Pk by

Pi = { P | has exactly i 1s}

Then clearly

|P | =k

i=0

|Pi|.

Therefore we need to count the number of partitions in each set Pi.



Set Partitions


Partitions with i 1sGiven a partition Pi, consider the partition obtained bysubtracting one from each part of . This is a partition of n kwith k i parts, and conversely any partition of n k with k iparts yields a partition in Pi if we add 1 to each part, and adjoin iparts equal to 1.Therefore

|Pi| = pki(n k).

ExampleThe partitions of 10 into 4 parts with two 1s are

4411 5311 6211

and the partitions of 6 into 2 parts are

33 42 51.



Set Partitions


Putting it all together

Therefore

pk(n) =k

i=0

pki(n k)

and replacing i by k i we get

pk(n) =k

i=0

pi(n k).

This sum involves terms of the form p0(x); this value must betaken to be 0 unless x = 0 in which case p0(0) is defined to be 1.



Set Partitions


Extending the triangleWe will use this to extend the triangle to n = 7

11 1

1 1 11 2 1 1

1 2 2 1 11 3 3 2 1 1

1 3 4 3 2 1 1

The blue value p3(7) is obtained by adding the first 3 values on theline for n = 7 3 = 4, thus

4 = 1 + 2 + 1.



Set Partitions


Size of conjugacy classes

Given a partition of n, how many permutations have that cyclestructure?

For example, how big is the conjugacy class of Sym(7) with cyclestructure 322? A permutation with this cycle structure has theform

(a, b, c)(d, e)(f, g).

There are 7! ways of assigning the numbers 1, 2, . . ., 7 to the 7positions, but many of them yield the same permutation.

For example

(1, 2, 3)(4, 5)(6, 7) and (2, 3, 1)(6, 7)(4, 5)

are the same.



Set Partitions


Overcounting

For the conjugacy class of Sym(7) with cycle structure 322, thereare

I 3 ways to write the cycle (a, b, c)I 2 ways to write the cycle (d, e)I 2 ways to write the cycle (f, g)I 2 ways to order the two 2-cycles

Therefore every permutation arises in

3 2 2 2 = 24

ways, and so the total number of distinct permutations in thisconjugacy class is

7!/24 = 210.



Set Partitions


In general

In general, suppose that a partition has ci parts equal to i, so that

c1 + 2c2 + 3c3 + + ncn = n.

Then the number of permutations in the conjugacy class with thiscycle structure is

n!(

i ci!ici) .

Proof.Each cycle of length i can be written in i different ways, and the cicycles of length i can be written in ci! different orders. Hence thedenominator of this expression counts the number of different waysof writing the permutation.



Set Partitions


Example

Check that this all works for n = 5 where there are 7 partitions:

Partition c1 c2 c3 c4 c5 Overcount Number

11111 5 0 0 0 0 120 12111 3 1 0 0 0 12 10221 1 2 0 0 0 8 15311 2 0 1 0 0 6 2032 0 1 1 0 0 6 2041 1 0 0 1 0 4 305 0 0 0 0 1 5 24

120



Set Partitions

Stirling Numbers of the Second KindStirling Numbers of the First Kind

Stirling Numbers

The Stirling numbers are the two series of numbers s(n, k) andS(n, k) defined as follows:

I Stirling numbers of the 1st kind:(1)nks(n, k) is the number of permutations of degree nwith k cycles.

I Stirling numbers of the 2nd kind:S(n, k) is the number of partitions of an n-set into knon-empty parts.



Set Partitions


First, the second kindIt is easy to list all the set partitions for small n and k:

I n = 1I k = 1: 1

I n = 2I k = 1: 12I k = 2: 1|2

I n = 3I k = 1: 123I k = 2: 12|3, 13|2, 1|23I k = 3: 1|2|3

I n = 4I k = 1: 1234I k = 2: 1|234, 134|2, 124|3, 123|4, 12|34, 13|24, 14|23I k = 3: 12|3|4, 13|2|4, 14|2|3, 1|23|4, 1|24|3, 1|2|34I k = 4: 1|2|3|4



Set Partitions


The triangleWe can write the numbers out in a triangle (similar to Pascalstriangle) as follows:

S(1, 1)S(2, 1) S(2, 2)

S(3, 1) S(3, 2) S(3, 3)S(4, 1) S(4, 2) S(4, 3) S(4, 4)

From the previous slide we get

11 1

1 3 11 7 6 1



Set Partitions


A recurrence

We can express the Stirling number S(n, k) in terms of smallerStirling numbers by noting that a set partition of {1, . . . , n} with kparts is obtained either by

I Adding n as a singleton to a partition of {1, . . . , n 1} withk 1 parts, or

I Putting n into one of the cells of a partition of {1, . . . , n 1}with k parts

Counting the number of set partitions of each type yields

S(n, k) = S(n 1, k 1) + kS(n 1, k).



Set Partitions


Illustration

For (n, k) = (4, 3), this formula is

S(4, 3) = S(3, 2) + 3 S(3, 3)

11 1

1 3 311 7 6 1

and the corresponding set partitions in the two groups are

I 12|3|4, 13|2|4, 1|23|4I 14|2|3, 1|24|3, 1|2|34



Set Partitions


An explicit formula

We can find an explicit formula for S(n, k) by using the principleof inclusion and exclusion:

Let A denote the set of all functions

f : {1, 2, . . . , n} {1, 2, . . . , k}

Now, let Ai be the set

Ai = {f A | f1(i) = }.

In other words, Ai is the set of functions whose range does notinclude i.



Set Partitions


Principle of Inclusion/Exclusion

If A1, A2, . . ., Ak are all subsets of a set A then for any index setI {1, 2, . . . , k}, define

AI =iI

Ai

(where we take A = X).

Principle of Inclusion/ExclusionThe number of elements of X that do not belong to any of thesets A1, A2, . . ., Ak is given by

I{1,2,...,k}

(1)|I||AI |.



Set Partitions


Set partitions with k parts

By PIE, the number of functions that map {1, 2, . . . , n} onto{1, 2, . . . , k} is given by:

I{1,2,...,k}

(1)|I||AI |.

For each possible size 0 j k there are(kj

)possible subsets I of

size j, and each of them makes the same contribution to this sum.More precisely, if |I| = j then

|AI | = (k j)n

because this just counts the number of functions that avoid aparticular set of j elements.



Set Partitions


The formulaPutting this together, we see that the number of functions from{1, . . . , n} onto {1, . . . , k} is given by

j=kj=0

(1)j(

k

j

)(k j)n

which (on replacing j by k j) is equal toj=kj=1

(1)kj(

k

j

)jn.

Now each such function determines a partition of {1, 2, . . . , n} intok parts, but we have counted each partition k! times and so

S(n, k) =1k!

j=kj=1

(1)kj(

k

j

)jn.



Set Partitions


Bell numbers

The total number of partitions of a set of size n into any numberof non-empty parts is called the Bell number B(n).Thus

B(n) =k=nk=1

S(n, k).

The first few Bell numbers are

1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975

GAP has built-in functions Stirling1(n,k), Stirling2(n,k)and Bell(n) giving these numbers.



Set Partitions


Now, the first kindThe triangle of Stirling numbers of the first kind

s(1, 1)s(2, 1) s(2, 2)

s(3, 1) s(3, 2) s(3, 3)s(4, 1) s(4, 2) s(4, 3) s(4, 4)

starts as follows:

11 1

2 3 16 11 6 1



Set Partitions


A recurrence

We can find a recurrence expressing s(n, k) in terms of smallerStirling numbers, by noting that a permutation of degree n with kcycles can be obtained either by

I Adjoining element n as a fixed point to a permutation ofdegree n 1 with k 1 cycles, or

I Inserting n into one of the cycles of a permutation of degreen 1 with k cycles.

and then counting the number of permutations in each category.



Set Partitions


Finding s(4, 2)

The permutations of degree 4 with 2 cycles consist of

I The permutations of degree 3 with 1 cycle, with 4 adjoined asa fixed point

(1, 2, 3)(4) (1, 3, 2)(4)

I The permutations of degree 3 with 2 cycles, with 4 insertedinto one of the cycles

(1, 2)(3) (1, 3)(2) (1)(2, 3)(1, 4, 2)(3) (1, 4, 3)(2) (1, 4)(2, 3)(1, 2, 4)(3) (1, 3, 4)(2) (1)(2, 4, 3)(1, 2)(3, 4) (1, 3)(2, 4) (1)(2, 3, 4)

There are 2 in the first group and 9 = 3 3 in the second group,thus giving us 11 altogether.



Set Partitions


The formula

This arguments shows us that

|s(n, k)| = |s(n 1, k 1)|+ (n 1)|s(n 1, k)|

but it does not take into account the sign of s(n, k) (that is,whether it is positive or negative).

Now (1)nk is equal to (1)(n1)(k1) and so taking the signsinto consideration we get

s(n, k) = s(n 1, k 1) (n 1)s(n 1, k).



Set Partitions


An astonishing connection

The reason for associating signs with the Stirling numbers isrelated to the following astonishing connection between these twosets of numbers, illustrated here for n = 4.

1 0 0 01 1 0 01 3 1 01 7 6 1

1 0 0 01 1 0 02 3 1 06 11 6 1

= I4


DefinitionsInteger PartitionsGeneratingConjugacyCounting

Set PartitionsStirling Numbers of the Second KindStirling Numbers of the First Kind

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