positive integer, k n, by a k-restricted choice function we mean the restriction of some choice function to the collection of k?subsets of X. The symmetric group SX.
The Number of Classes of Choice Functions under Permutation Equivalence Spyros S. Magliveras and Wandi Weiy Department of Computer Science and Engineering University of Nebraska-Lincoln Lincoln, Nebraska 68588-0115 Abstract
A choice function f of an n?set X is a function whose domain is the power set P (X ), and whose range is X , such that f (A) 2 A for each A X . If k is a xed positive integer, k n, by a k-restricted choice function we mean the restriction of some choice function to the collection of k?subsets of X . The symmetric group SX acts, by natural extension on the respective collections C (X ) of choice functions, and Ck (X ) of k?restricted choice functions of X . In this paper we address the problem of nding the number of orbits in the two actions, and give closed form formulas for the respective numbers.
This work was supported in part by NSA grant MSFPF-95-G-091, and by CCIS { Univ. of Nebraska Lincoln y This work was supported in part by CCIS { Univ. of Nebraska - Lincoln
1
1 Introduction
A choice function f on an n?set X is de ned to be a function whose domain is the collection of all subsets of X , and whose range is X , such that for each subset A of X , f (A) is an element of A. If k is a xed positive integer, k n, by a k-restricted choice function we mean the restriction of some choice function to the collection of k?subsets of X . If f and g are two choice functions (or k?restricted choice functions) on X , we say that f and g are permutation-equivalent (or just equivalent) if there exists a permutation on the symbols of X which transforms f to g. Even though, as we shall see, it is easy to determine the number of classes under permutation equivalence in the collection of all choice functions, it is less trivial to determine the number of equivalence classes of k?restricted choice functions. It is perhaps surprising that solutions to these problems have not been given before, even though their statements have been known for awhile. In this paper we address both of these problems, and give closed form formulas for the respective number of equivalence classes.
2 Preliminaries For the most part we use standard notation. For example, if Y and Z are sets, Z Y denotes the collection of all functions f : Y ! Z . We denote the collection of all subsets of Z (i.e. Z the power set of Z ) by P (Z ), and the collection of all k?subsets of Z , by k , for 0 k n. Throughout, X will be an n?set, and SX , or just Sn will denote the symmetric group on X . Let C (X ) denote the collection of all choice functions on X . Thus, f 2 C (X ) if and only if X f 2 X P X and f (A) 2 A for allA2 P (X ). Let Ck (X ) denote the collection of all f 2 X ( k ) such that f (A) 2 A for all A 2 Xk . We call Ck (X ) the k-restricted choice functions of X . n By straightforward counting we easily see that j Ck (X ) j = k(k) , moreover the total number n of (unrestricted) choice functions is j C (X ) j = Qnk k(k). Let X be a set, and G a multiplicative group, with its identity element denoted by 1. Recall that a group action GjX is a function exp : X G ! X , (x; g) 7! xg , satisfying the two axioms (i) x = x, x 2 X , (ii) (xg )h = xgh; x 2 X; g; h 2 G. The G ? orbit of x 2 X is the set xG := fxg : g 2 Gg X , and the stabilizer of x 2 X the subgroup Gx = fg 2 G : xg = xg of G. It is elementary that jxG j = [G : Gx]. If GjX is a group action and g 2 G, the set of all elements of X xed by g, fx 2 X : xg = xg is called the x of g, and is denoted by Fix(g). The function : G ! N , (g) := jFix(g)j, is called the character of the group action. If GjX is a group action we denote by = (GjX ) the number of G-orbits on X . Any action GjX extends naturally to an action G on P (X ) de ned by (A; g) 7! Ag for A X , g 2 G. Here Ag = fag : a 2 Ag. It easily follows that jAg j = jAj. The G?orbit of A is the collection of subsets AG = fAg jg 2 Gg. If GjX is a group action, x 2 X , and 2 G, by the ? period of x, (x), we mean the length of the orbit of x under the cyclic group hi. If follows that (x) is the least positive integer j such that xj = x, and that (x) divides the order of . Similarly, if A X , we de ne the ? period of A, (A), to be the length of the orbit jAhi j. Note that ( ) = (X ) = 1, for any 2 Sn. (
)
=1
1
2
For a general group action GjX , the following facts are easy to see: For x 2 X andPg; h 2 G, (i) Gxg = fGxgg , (ii) Fix(gh) = fFix(g)gh, (iii) (gh) = (g), (iv) = jGj g2G (g). The last equality providing us with the number of G?orbits in X is known as the CauchyFrobenius lemma. For the remainder of this paper we let X = f1; 2; : : : ; ng. 1
3 Main Results
To simplify our discussion, let C denote either of the sets C (X ) or Ck (X ). If f 2 C it will be useful to write f = Aaii , where ai = f (Ai) 2 Ai, thus: !
f = Aa i = Aa Aa :::::: Aa i ::::::Aar i r i 1
2
1
2
!
(3.1)
where r = 2n or nk accordingly. If 2 G, x 2 X , and A X , we denote by x , and A the images under of x and action A respectively. We now extend the GjX canonically to A A i i an action GjC . If f = ai 2 C , and 2 G we de ne f by ai = Aaii . Thus,
f
!
A : : : A : : : A A A i = a = a a : : : ai : : : ar i i r 1
2
1
2
!
(3.2)
? It is now clear that f is the composition of functions f . Thus, a permutation 2 G A xes f = aii 2 C if and only if for all A in the domain of f , the following diagram commutes: 1
A
f
e
A
- eAf
?e
f
- ?e
Af = Af
Let : Sn ! N denote the character of the action Sn j C (X ), and k : Sn ! N the character of Sn j Ck (X ). The following lemma is evident: Lemma 3.1 Let 2 Sn, and f 2 C . Then, the following statements are equivalent: (i) f is xed by (ii) For each A X , f (A) = a if and only if f (A ) = a We immediately have, Lemma 3.2 If is a non-identity element of Sn, then xes none of the choice functions in C (X ), i.e. () = 0. 3
Proof: Suppose f 2 C (X ), and let 0 = (a ; : : : ; at ), t 2, be a non-trivial cycle in the decomposition of as the product of disjoint cycles. If A = fa ; : : : ; at g, then, A = A, while ai 6= ai for 1 i t. If f (A) = ai, then f (A ) = f (A) = ai 6= ai , so by ??, f is not xed by . 2 Theorem 3.1 The number of equivalence classes in C (X ) under permutation equivalence is 1
1
(n) =
n Y
k=1
k(k)? n
1
Proof: Note that a non-identity permutation of Sn xes () = 0 functions of C (X ), while the identity permutation xes all choice functions. Hence, by the Cauchy-Frobenius lemma, the number of Sn?orbits in C (X ) is n n n n Y Y X 1 1 ) ( k (n) = n! () = n! k = k(k)? 2Sn k k 1
=1
=1
2
If a 2 A X , and 2 Sn , let (a) and (A) be the orbit lengths under hi of the point-orbit ahi and set-orbit Ahi respectively. It can be easily seen that, in general, both cases (i) (a) (A), (ii) (a) > (A) are possible. Suppose however, that f 2 Ck (X ) is xed by , and f (A) = a. Then, it is clear that (ii) can not occur, otherwise the statement (ii) of lemma ?? would be violated. Moreover, again under the assumptions that f is xed by and f (A) = a, it is easily seen that (a) must divide (A), otherwise condition (ii) of lemma ?? would again be violated. Consequently, we have the following Lemma 3.3 Let 2 Sn , and suppose that f 2 Ck (X ) is xed by . If (a) is the orbit length of a = f (A) under the cyclic group hi, and (A) the orbit length of A under hi, then, (a) divides (A). Let k be a xed positive integer, k n, an arbitrary element of Sn, and suppose that = m is the factorization of as the product of disjoint cycles, where i = (ai; ; : : : ; ai;vi). Let Vi = fai; ; : : : ai;vi g be the set of elements of X occuring in i . Now, for any A 2 Xk , A = A [ [ Am, where Ai = A \ Vi. Let ki = jAij. Even though some of the Ai may be empty, with corresponding ki = 0, we still call the collection fA ; : : : ; Am g a partition of A, and fk ; : : : ; kmg a partition of the natural number k. Since the Ai are disjoint, and A = A [ [ Am , the following lemma is straightforward: Lemma 3.4 Under the notation just established, we have: (i) i ( ) = 1 (ii) (Ai ) = i (Ai) (iii) (A) = lcm [ (A ); : : : ; (Am)]. Let : N ! N be a mobius-like function de ned by Y Y (3.3) ( pki i ) = (?1)ki 1
2
1
1
1
1
1
1
1
i
i
where the pi are primes. Then, we have the following: 4
Proposition 3.1 Suppose that = (1; 2; : : : ; v), and t; k, and v are positive integers such that t; k v, and v = rt. Let N (v; k; t) denote the number of subsets A of V = f1; 2; : : : ; vg such that jAj = k, and (A) = t. Then, ! N (v; k; t) =
v=d (d=r) k=d
X
(3.4)
d r j d j (k; v)
Proof: Since (A) = t, the stabilizer of A in hi is of order jhij=t = v=t = r. But hi has a unique subgroup of order r, namely the cyclic group generated by y = t . Now, y is a permutation of type rt, i.e. it consists of t cycles of length r. Since A is xed by y, A is the union of some of the t r?cycles of y . In particular, jAj = sr = k, so that rjk, t and s = k=r. Now, there are s = v=r k=r ways of selecting these s r?cycles to form A. However, some of these k?sets are also the of k=d d?cycles in the cyclic subgroup of union v=d hi order d, where r j d j k. There are k=d ways of forming the latter k?sets, and they fall into orbits of length v=d. Hence, by the inclusion-exclusion principal the result follows. 2
Proposition 3.2 Let be a permutation in Sn whose factorization as the product of disjoint cycles is = m , ji j = vi . Then, the number of k?restricted choice functions in Ck (X ) xed by is k () =
1
Y
Y
X
f
ki g f
Qm
k1 + + km = k (t1 ; : : : ; tm ) vi j lcm [t1 ; : : : ; tm ] 0 k i vi ti j vi 1im ki = 0 ) ti = 1 1im
i=1 N (vi ;ki ;ti ) =
t ;:::;tm ] g
LCM [ 1
Proof: For 0 i m, let i = (ai; ; ai; ; : : : ; ai;vi ), and Vi = fai; ; ai; ; : : : ; ai;vi g. For a xed partition of k, k + + km = k with 0 ki vi, and a xed (t ; : : : ; tm ) where the ti are positive integers, ti j vi, there are N (v ; k ; t ) N (vm ; km; tm) ways of selecting an mtuple of fragments (A ; A ; : : : ; Am ) such that Ai Vi, [mi Ai = A, jAj = k, and i (Ai) = ti . But, for this xed (t ; : : : ; tm ), the collection F of all m?tuples of fragments (A ; : : : ; Am), and the collection F 0 of corresponding k-sets A, are partitioned into hi?orbits all of length lcm [t ; : : : ; tm ]. Now, to form a choice function f 2 Ck (X ) xed by it is necessary and sucient to assign a value to a single orbit representative, from each hi-orbit of k?sets. Moreover, by lemma ??, the number of choices is P vi j A ki . The result follows. 2 1
2
1
1
2
1
1
1
1
1
2
1
1
1
( )
4 Acknowledgement The authors would like to thank Professor David A. Klarner for inspired discussions of this and related problems, some as early as 1981. 5
References [1] Marshall Hall, Jr., Combinatorial Theory, Willey-Interscience in Discrete Math., John Willey & Sons, 1986 [2] H. Wielandt, Finite Permutation Groups, Academic Press, New York, London, 1964.
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