Circular and Necklace Permutations

Hiroshi KAJIMOTO and Mai OSABE

Department of Mathematical Science, Faculty of Education Nagasaki University, Nagasaki 852-8521, Japan e-mail: [email protected] (Received October 31, 2006)

Abstract We enumerate circular and necklace permutations which have beads of several colors, and count the related combinatorial numbers.

0 Introduction

Circular and necklace permutations are standard subjects of high school combinatorics. On the other hand they serve a good introduction to more ad- vanced subject of combinatorics, that is the Pólya theory of counting under group action. In fact these enumerations were historically a precursor of the theory [2, 4, p.558]. In this article we give the enumeration of circular and necklace permuations according to the given number of beads for each colors, and of the related combinatorial numbers. A circular permutation is defined to be an equivalence class of sequences under the cyclic group Cn, i.e., two sequences a1a2 . . . an and b1b2 . . . bn are equivalent if one is a cyclic shift of the other. A necklace (or necklace permutation) is defined to be an equivalence class of sequences under the dihedral group Dn, i.e., two sequences a1a2 . . . an and b1b2 . . . bn are equivalent if one is a cyclic shift of the other or if circular permutaions of them are mutually reflective with respect to some diameter. We make full use of the counting theory and look back the classical examples, and show how to use Pólya’s cycle index efficiently. We use combinatorial notation, especially relative to partitions, as in [4].

1 1 Cycle indices of cyclic and dihedral groups

We begin with a brief accout of the counting theory under group action. A good introduction to the theory is [3, chapter 6], that is a lecture note without detailed proof. To those who want the proof see [1, chapter 5] or [4, section 7.24]. Let G be a subgroup of the full permutation group SX of a finite set X. Let n = |X| be the number of elements of X. Each element g ∈ G is a permutation of X, so g is a product of several disjoint cycles of elements of X: g = (a1a2 . . . ak) ... (c1c2 . . . cm). The cycle type of g determines a monomial b1(g) b2(g) bn(g) p1 p2 . . . pn , that is, g has b1(g) cycles of length 1, b2(g) cycles of length 2, . . . in the cycle decomposition, so (1b1(g)2b2(g) ...) ⊢ n. Cycle index PG is a polynomial defined by 1 ∑ P = pb1(g)pb2(g) . . . pbn(g). G | | 1 2 n G g∈G Let R = {r, b, . . . , w} be a set of colors. Then Pólya’s main theorem is that all the inequivalent colorings f : X → R under G are expressible by a i i i generating function PG substituted by pi = r + b + ··· + w . For circular and necklace permutations let X = {1, 2, . . . n} and let the n cyclic group Cn = ⟨a|a = e⟩ act on X in an obvious way: a = (1, 2, . . . , n). The coloring f : X → R is a sequence of length n and then Cn act on X the set R of sequences as a cyclic shift by a(r1r2 . . . rn) = r2r3 . . . rnr1. An equivalence class of the sequences under Cn is exactly a circular permutation. n 2 −1 The dihedral group Dn = ⟨a, ε|a = ε = e, εaε = a ⟩ acts on X by putting ε being a reflection with respect to some symmetry axis. For instance put { − − ⌊ n ⌋ ⌈ n ⌉ (n)(1, n 1)(2, n 2) ... ( 2 , 2 ) when n odd ε = − n n (1, n)(2, n 1) ... ( 2 , 2 + 1) when n even.

An equivalence class of sequences under Dn is a necklace permutation. Lemma 1 Cycle indices of circular and necklace permutations are given by ∑ 1 n/d PCn = φ(d)pd , n d|n  (∑ )  1 n/d ⌊n/2⌋ d|n φ(d)pd + np1p2 when n odd P = 2n (∑ ) Dn  1 n/d n n/2 2 (n/2)−1 2n d|n φ(d)pd + 2 (p2 + p1p2 ) when n even, where φ(n) := #{d ∈ Z|gcd(n, d) = 1, 1 ≤ d ≤ n} is the Euler function.

2 ⊔ For proof, notice that Cn = d|n Cn(d) where Cn(d) is a subset of all the n/d ⊢ elements of order d and that all elements in Cn(d) have the cycle type⊔ (d ) n and |Cn(d)| = φ(d). Notice also a coset decomposition: Dn = Cn εCn and that all the elements of εCn are reflections. When n is odd all the reflections ⌊n/2⌋ of εCn have the cycle type (1)(2 ) ⊢ n. When n is even we have two kinds n/2 of symmetry axes, half the reflections of εCn have the cycle type (2 ) and another half the reflections have (12)(2n/2−1). The total number of circular and necklace permutations of length n with beads of k kinds of colors are given by substituting all pi = k in cycle indices. Corollary 2 The total number C(n, k) of circular permutations of length n with k colors of beads and N(n, k) of necklace permutations are ( ) 1 ∑ 1 ∑ n C(n, k) = φ(d)kn/d = φ kd, n d|n n d|n d  {  1 ∑ nk⌈n/2⌉ when n odd N(n, k) =  φ(d)kn/d +  . n (kn/2 + k(n/2)+1) when n even 2n d|n 2 We here take the simplest: Example (n = 3). Cycle indices: 1 1 P = (p3 + 2p ),P = (p3 + 2p + 3p p ). C3 3 1 3 D3 6 1 3 1 2 Total numbers: C(3, k) = (k3 + 2k)/3,N(3, k) = (k3 + 2k + 3k2)/6,

k 1 2 3 4 5 C(3, k) 1 4 11 24 45 . N(3, k) 1 4 10 20 35 k = 3: We have beads of three colors, red r, blue b and white w. Substituting i i i the ﬁgure inventory pi = r + b + w in the cycle index, we get 1 { } P = (r + b + w)3 + 2(r3 + b3 + w3) C3 3 = r3 + b3 + w3 + r2b + r2w + b2r + b2w + w2r + w2b + 2rbw by the multinomial expansion.

3 r b w r3 + b3 + w3 r r b b w w

b w r

r r b r2b + r2w + b2r + b2w + w2r + w2b r r b w r b

b b w w w w

r r

2rbw w b b w

The first 9 circles are self-reflective, the last 2 circles are pair-reflective each other, with respect to indicated symmetry axes. We can collect terms as ∑ ∑ ∑ r3 = r3+b3+w3, r2b = r2b+r2w+b2r+b2w+w2r+w2b, rbw = rbw. ∑ In these sums, for instance r2b indicates that terms with two beads of one 2 color and one bead of the second color are summed∑ up, r b is the typical term of the possible( ) color choices. The sum rbw has just one term when k ≥ k = 3, but has 3 terms when k 3 and has no terms hence = 0 when k < 3. Then we have

∑k ∑k ∑k 3 2 PC3 = r + r b + 2 rbw.

This expression is independent of the indicated number k of colors. In this example we easily have for necklaces,

∑k ∑k ∑k 3 2 PD3 = r + r b + rbw

4 ∑ λ by computation or by inspection. We will seek their coeﬃcient [ r ]PG, which is the number of essentially inequivalent color patterns under group action.

2 Circular permutations

To get all the circular permutations of length n with beads of k kinds of d d ··· d color, we have to expand PCn by substituting pd = r1 + r2 + + rk, the power sum symmetric polyomials. In this case the expansion is fairly easy. Remember the multinomial expansion and write that in the following way: ( ) ( ) ∑ n ∑ n ∑k (x + x + ··· + x )n = xα = xµ. 1 2 k α µ |α|=n,α∈Nk µ⊢n,l(µ)≤k ∑ Here l(µ) is the length of a partition µ and k xµ is a monomial symmetric polynomial mµ of k variables [4, section7.3] deﬁned by

∑k ∑ µ α x = mµ = x α where the last sum runs over all distinct permutations α = (α1, α2,...) of the ∑entries of the partition µ = (µ1, µ2,...). A monomial symmetric polynomial k xµ collects all the colorings whose choices of the numbers of diﬀerent k colors of beads are according to the partition µ ⊢ n. We have by the substitution  ( )  ∑ ∑ ∑ ∑k 1 n/d 1  n/d dµ PCn = φ(d)pd = φ(d) r n d|n n d|n µ⊢n/d µ where dµ = (dµ1, dµ2,...) ⊢ d|µ|. Putting λ = dµ ⊢ n, we get the generating function of circular permutations of length n with k colors of beads:

Theorem 3  ( )  ∑  1 ∑ n/d  ∑k P = φ(d) rλ. Cn n λ/d  λ⊢n d|gcd{λ1,λ2,...} where λ/d = (λ1/d, λ2/d, . . .) ⊢ n/d.

5 ∑ ∑ ( ) k λ n/d The coefficient [ r ]PCn = d|gcd{λ1,λ2,...} λ/d φ(d)/n is the number of essentially different circular permutations according to the number of each color indicated by the partition λ ⊢ n. For instance the number of circular permutations of∑ length n = 6 with two( ) red beads,( ) two blue beads and two white 2 2 2 { 6 3 } beads is [ r b w ]PC6 = φ(1) 23 +φ(2) 13 /6 = 16. A direct counting of this coefficient is possible by the Burnside lemma. The cycle index involves the Burnside lemma beforehand∑ in its polynomial form. The number of terms in k rλ is the number of color choices among k colors, according to the partition λ = (λ1, λ2,...) ⊢ n which indicates the numbers of each color. The length l(λ) is now not greater than k.

b1 b2 Lemma 4 Let the partition λ ⊢ n be λ = (1 2 ...) and let l(λ) = b1 + b2 + · · · ≤ ∑ k. Then the number of terms in a monomial symmetric polynomial k λ r is ( ) k (possibly = 0). b1, b2, . . . , k − l(λ)

3 Necklaces

⌊n/2⌋ 2 n/2−1 As to necklaces we have to expand the correction terms p1p2 and p1p2 in the cycle index PDn . When n is odd we have ( ) ( ) n−1 ∑ ∑ (n−1)/2 ∑ ∑ ∑ ⌊n/2⌋ 2 2 2µ p1p2 = r1 r1 = r1 r . µ⊢(n−1)/2 µ ∑ ∑ ∑ ∑ 2µ 2µ+ei For a product of monomial symmetric polynomials, r1 r = i r where ei is i-th unit vector. So ( ) k n−1 ⌊ ⌋ ∑ ∑ ∑ n/2 2 2µ+ei p1p2 = r . i=1 µ⊢(n−1)/2 µ ( ) ( ) ⊢ (n−1)/2 (n−1)/2 ⌊ ⌋ ⌊ ⌋ ⌊ ⌋ ⊢ Put λ = 2µ+ei n. Then µ = ⌊λ/2⌋ where λ/2 = ( λ1/2 , λ2/2 ,...) (n − 1)/2 = ⌊n/2⌋ and λ runs over all the partitions of n those have exactly one odd entry one by one, in the above summation. Hence ( ) ∑ ⌊ ⌋ ∑k ⌊n/2⌋ n/2 λ p1p2 = ⌊ ⌋ r λ λ/2

6 where the sum runs over all the partitions λ of n those have exactly one odd entry. When n is odd, the generating function of necklaces is   ∑ 1  n/d ⌊n/2⌋ PDn = φ(d)pd + np1p2 2n d|n   ( )  ( )  1 ∑ 1 ∑ n/d ∑k ∑ ⌊n/2⌋ ∑k  =  φ(d) rλ + rλ 2  n λ/d ⌊λ/2⌋  λ⊢n d|gcd{λ1,λ2,...} λ  ( ) ( )  ( )  ∑(1)  ∑ ⌊ ⌋  ∑k ∑(2) ∑ ∑k 1 1 n/d n/2 λ 1  n/d  λ =  φ(d) + ⌊ ⌋  r + φ(d) r λ 2 n d|gcd λ λ/d λ/2 λ 2n d|gcd λ λ/d ∑ (1) ⊢ ≤ where the ﬁrst sum runs over partitions∑ λ n, l(λ) k which have the exactly one odd entries, the second sum (2) runs over the other partitions λ ⊢ n, l(λ) ≤ k and gcd λ = gcd{λ1, λ2,...} is the great common divisor of nonzero entries of λ. 2 n/2−1 When n is even direct expansion of p1p2 is possible but clumsy. We quote the following proposition which is convenient for power sum expansions in the cycle index. Deﬁne the power sum symmetric polynomials as ∑ n ≥ pn = mn = xi , n 1 with p0 = 1 i ··· pµ = pµ1 pµ2 if µ = (µ1, µ2,...).

Proposition 5 ([4], 7.7.1) Let λ = (λ1, . . . , λl) ⊢ n, where l = l(λ), and set ∑ pλ = Rλµmµ. µ⊢n

Let k = l(µ). Then Rλµ is the number of ordered partitions π = (B1,...,Bk) of the set [l] such that ∑ µj = λi, 1 ≤ j ≤ k. i∈Bj

We then have ∑ 2 n/2−1 p1p2 = p(122n/2−1) = R(122n/2−1),λ mλ. λ

7 ̸ ̸ If Rµλ = 0 then µ is a reﬁnement of λ. So if R(122n/2−1),λ = 0 then either all (entries) of λ are even or exactly two entries of λ are odd. We( get R) (122n/2−1),λ = n/2 n/2−1 − λ/2 when all entries of λ are even and R(122n/2 1),λ = 2 ⌊λ/2⌋ when λ has exactly two odd entries, as an exercise of prop 5 (though the whole article is an exercise of the counting theory). We have ( ) ( ) ∑(3) ∑ ∑(4) ∑ − n/2 n/2 − 1 pn/2 + p2p(n/2) 1 = 2 rλ + 2 rλ 2 1 2 ⌊ ⌋ λ λ/2 λ λ/2 ∑ (3) ⊢ ≤ where∑ runs over partitions λ n, l(λ) k which have all even entries and (4) runs over partitions λ ⊢ n, l(λ) ≤ k which have exactly two odd entries. Substituting the power sums and gathering terms together in   ∑ 1  n/d n n/2 2 (n/2)−1  PDn = φ(d)pd + (p2 + p1p2 ) , 2n d|n 2 we get the generating function of necklaces of length n with beads of k kinds of color.

Theorem 6 When n is odd,  ( ) ( )  ( ) ∑(1) 1  1 ∑ n/d ⌊n/2⌋  ∑k ∑(2) 1 ∑ n/d ∑k P = φ(d) + rλ+  φ(d)  rλ. Dn  ⌊ ⌋  λ 2 n d|gcd λ λ/d λ/2 λ 2n d|gcd λ λ/d

When n is even,  ( ) ( ) ∑(3)  ∑  ∑k 1 1 n/d n/2 λ PDn =  φ(d) +  r + λ 2 n d|gcd λ λ/d λ/2  ( ) ( )  ( ) ∑(4)  ∑ n −  ∑k ∑(5) ∑ ∑k 1 1 n/d 2 1 λ  1 n/d  λ  φ(d) + ⌊ ⌋  r + φ(d) r . λ 2 n d|gcd λ λ/d λ/2 λ 2n d|gcd λ λ/d ⊢ ≤ Here all the∑ summations are for partitions λ n, l(λ) k with∑ the additional (1) (3) conditions: runs over λ those∑ have exactly one odd entry, runs over (4) λ those∑ have all even∑ entries, runs over λ those have exactly two odd entries, (2) and (5) run over the remaining partitions.

8 ∑ λ The coefficient [ r ]PDn is the number of essentially different necklaces according to λ ⊢ n which shows the number of beads of each colors. For instance the number∑ of necklaces∑ with 2 red beads, 2( blue) beads and 2 white beads is (3) 2 2 2 { 3 } in case and so [ r b w ]PD3 = 16 + 13 /2 = 11 (16 is the number of circular permutations already counted). We count this coefficient by the Burnside lemma in the last section. We find that the correction term of each generating function corresponds to the number of self-reflective circular permutations relative to the symmetry axes of each case. Try examples when n = 6 or 12, and enjoy the correspondence with necklaces and mathematical expressions, in the Pólya theory!

4 Burnside Lemma and Examples

We will give a direct counting by the Burnside lemma of the number of circular permutations and necklaces varying the number of beads for each colors. The Burnside(-Cauchy-Frobenius) lemma is the following, which en- ables us to replace counting equivalence classes by counting ﬁxed points of the permutation group.

Lemma 7 ([4], 7.24.5 or [1], Theorem 5.2) Let X be a ﬁnite set and G a subgroup of the full permutation group SX . For each g ∈ G let

Fix(g) = {x ∈ X|gx = x} be the ﬁxed point set of the permutation g. Let X/G be the set of orbits of G. Then ∑ | | 1 | | X/G = | | Fix(g) . G g∈G In other words, the average number of elements of X ﬁxed by an elements of G is equal to the number of equivalence classes.

Let R = {r1, r2,...} be a set of colors and let λ = (λ1, λ2,...) ⊢ n be a partition which shows the number of beads for each color. Let Rλ = {s = [n] s1s2 . . . sn ∈ R |s contains λ1 r1’s, λ2 r2’s,...} be the set of all the sequences of length n in which color r1 appears λ1-times, color r2 appears λ2-times and λ λ so on. Then G = Cn(or Dn respectively) acts on X = R , and X/G = R /G is the set of all the circular permutations of length n which have λ1-beads of color r1, λ2-beads of color r2 and so on when G = Cn (or the set of all

9 the necklaces which have the same numbers of beads for each color when G = Dn respectively). The number of elements of these sets are obtained as λ λ coeﬃcients [r ]PG = |R /G| of generating functions in section 2 and 3. We apply here the Burnside lemma 7.

Corollary 8 The number of circular permutations of length n which have λ1-beads of color r1, λ2-beads of color r2 and so on, is given by ( ) 1 ∑ n/d [rλ]P = φ(d). Cn n λ/d d|gcd{λ1,λ2,...} Proof. We know 1 ∑ 1 ∑ ∑ [rλ]P = |Rλ/C | = |Fix(g)| = |Fix(g)| Cn n n n g∈Cn d|n g∈Cn(d) by the Burnside lemma and ⊔ Cn = Cn(d) d|n where Cn(d) is a subset of all the elements of order d and all elements in n/d Cn(d) have the cycle type (d ) ⊢ n. Let g ∈ Cn(d) and s = s1s2 . . . sn ∈ λ Fix(g, R ). Then for cycles (i1, i2, . . . , id) of the permutation g, the corresponding elements in the sequence s must have the same color: si1 = s(i2 =) ··· | ∀ | λ | n/d = sid since g ﬁxes s. Hence we see d λi ( i), and get Fix(g, R ) = λ/d by considering selections of colors to the (n/d)-cycles in g. Since |Cn(d)| = φ(d) result follows.

λ Corollary 9 The number [r ]PDn of necklaces which have λ1-beads of color r1, λ2-beads of color r2 and so on, is given as follows. When n is odd, we have  ( ) ( ) 1  ∑ n/d (n − 1)/2  [rλ]P = φ(d) + n Dn  ⌊ ⌋  2n d|gcd λ λ/d λ/2 if λ has exactly one odd entry, and ( ) ∑ λ 1 n/d [r ]PDn = φ(d) 2n d|gcd λ λ/d

10 for other λ. When n is even, we have  ( ) ( )  ∑  λ 1 n/d n/2 [r ]PDn =  φ(d) + n  2n d|gcd λ λ/d λ/2 if λ has all even entries,  ( ) ( ) 1  ∑ n/d (n − 2)/2  [rλ]P = φ(d) + n Dn  ⌊ ⌋  2n d|gcd λ λ/d λ/2 if λ has exactly two odd entries, and ( ) ∑ λ 1 n/d [r ]PDn = φ(d) 2n d|gcd λ λ/d for other λ. Proof. We have to take reﬂections into consideration. We know   1 ∑ 1 ∑ ∑ [rλ]P = |Rλ/D | = |Fix(g)| =  |Fix(g)| + |Fix(g)| Dn n 2n 2n g∈Dn g∈Cn g∈εCn by the Burnside lemma and a coset decomposition ⊔ Dn = Cn εCn

n 2 −1 where Dn = ⟨a, ε|a = ε = e, εaε = a ⟩, Cn = ⟨a⟩ is a subgroup of all the rotations of Dn, and all the elements of εCn are reflections with respect to some axes. When n is odd all the reflections of εCn have the cycle type (1)(2(n−1)/2) ⊢ n. When n is even we have two kinds of symmetry axes, n/2 half the reflections of εCn have the cycle type (2 ) and another half the reflections have (12)(2(n−2)/2). 2 Take a case when n is even for instance. Then half the reflections εCn = 2 n−2 n/2 {ε, εa , . . . , εa } of εCn have the cycle type (2 ). The typical element is ( ) n n ε = (1, n)(2, n − 1) ... , + 1 . 2 2 ∈ λ Let s = s1s2 . . . sn Fix(ε, R ). Then s1 = sn, s2 = sn−1(, . . .) , sn/2 = s(n/2)+1. | ∀ ∈ 2 | λ | n/2 Hence 2 λi ( i). If g εCn, then we have Fix(g, R ) = λ/2 for λ which has

11 all even entries and Fix(g, Rλ) = ∅ for λ which has an odd entry. Another 2 { 3 n−1} 2 (n−2)/2 half εaCn = εa, εa , . . . , εa of εCn have the cycle type (1 )(2 ). The typical element is ( )( ) n n n εa = (n)(1, n − 1)(2, n − 2) ... − 1, + 1 . 2 2 2 ( ) The same reasoning shows that if g ∈ εC2, then |Fix(g, Rλ)| = n/2 for λ n( ) λ/2 | λ | (n−2)/2 which has all even entries, Fix(g, R ) = 2 ⌊λ/2⌋ for λ which has exactly two odd entries, and Fix(g, Rλ) = ∅ for other λ. Results for even n now follow. The simillar argument goes through the case when n is odd.

Examples. When n = 5 and( λ =) (2, 2, 1), the corresponding circular permu- 2 2 1 5 30 tations are, [r b w]PC5 = 5 2,2,1 = 5 = 6 as

r r r r

w r r w b b b b

b b b b w r r w pair-wise reﬂective

w w

r r b b self-reﬂective

b b r r

2 2 On{ the( other) ( hand)} the corresponding necklaces are of the number [r b (w])PD5 = 1 1 5 2 1 2 2 5 2,2,1 + 1 = 2 (6 + 2) = 4. We see that the correction term 1 = 2 is the number of self-reﬂective circular permutations. When n = 6, ﬁrst let λ =( (4, 1), 1). Then the corresponding circular 4 1 6 30 permutations are, [r bw]PC6 = 6 4,1,1 = 6 = 5 as

12 b b b b b w r r w r r r r r r

r r r r w r r w r r r r r r w pair-wise reﬂective self-reﬂective { ( ) ( )} The corresponding necklaces are of the number [r4bw]P = 1 1 6 + 2 = D6 2 {(6 4,1),1 ( 2 )} 1 ⊢ 2 2 2 1 6 3 2 (5+1) = 3. Next let λ = (2, 2, 2) 6. Then [r b w ]PC6 = 6 2,2,2 + 1,1,1 = 1 6 (90 + 6) = 16 as

r r r r r r b w w b b b b b w w w w w b b w w r r w b r r b r r w w b b r r r r b r r b b r r b pair-wise reﬂective b w w b w w w w w w b b

r b w b b r r r r

w w w w b b r b w r r r r b b self-reﬂective w b b w r r w w b b w w

13 [ {( ) ( )} ( )] And [r2b2w2]P = 1 1 6 + 3 + 3 = 1 (16 + 6) = 11. We D6 2 6 2,2,2 1(,1,)1 1,1,1( ) 2 2 3 see again that the correction terms 2 = 1 and 1,1,1 = 6 are the numbers of self-reﬂective permutations respectively. Incidentally here we can ask, is λ there eﬃcient way of writing out all the color patterns one by one in [r ]PG?

The observation of examples and the proof of corollary 9 show that the correction term counts the number of self-reflective, that is fixed by some reflexion, circular permutations.

Theorem 10 The number of self-reﬂective permutations among circular permutations of length n which have λ1-beads of color r1, λ2-beads of color r2 and so on, is given by λ − λ 2[r ]PDn [r ]PCn  ( ) ⌊ ⌋  n/2 when n is odd and λ has only one odd entry  ⌊λ/2⌋   ( )   n/2  λ/2 when n is even and λ has all even entries =  ( )   (n−2)/2  ⌊λ/2⌋ when n is even and λ has exactly two odd entries    0 otherwise.

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