Counting Idempotent Relations

Transcription

1 Counting Idempotent Relations Beriht-Nr Florian Kammüller ISSN

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3 Abstrat This artile introdues and motivates idempotent relations. It summarizes haraterizations of idempotents and their relationship to transitive relations and quasi-orders. Finally it presents a ounting method for idempotent relations and lists the results for up to 6 points. 3

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5 Contents 1 Introdution 7 2 Bakground 9 3 Counting Idempotents 13 4 Outlook and Conlusions 19 5

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7 Chapter 1 Introdution An idempotent relation is one that is persistent under sequential relational omposition. Sequential omposition r; s of two relations r A B and s B C is defined as {(x, y) z. (x, z) r (z, y) r}. A relation r A A is idempotent iff r; r = r. In this paper we arrive at reduing the problem of ounting idempotents to ounting unlabelled and labelled partial orders and quasi-orders. The presented method works for ounting labelled idempotents. For the unlabelled ase it does not work. However, it arrives at ounting a sublass of unlabelled idempotents, those that have a partial order kernel. We reuse in this paper the following Sloane sequenes [9] and their numbers: A (labelled quasi-orders Q), A (unlabelled quasi-orders q), A (labelled partial orders PO), and A (unlabelled partial orders po). We also use Stirling numbers of the seond kind, A and A In the following we present a ounting method for a new sequene, that of labelled idempotent relations I. This paper begins in Setion 2 with notations, some fats that are needed about the automorphism group on partial orders, and a summary of previous results about idempotents that have been leading on to the ounting formula presented in this paper. Then in Setion 3 we present the theory that enables ounting idempotents. We give the ounting results for up to 6 points as far as we have been able to alulate by hand. The method we present is suitable for being implemented as a omputer program thereby inreasing the number of points. In an appendix we give details of the alulation for up to 5 points, for the alulation of up to 6 points see the author s web page [6]. 7

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9 Chapter 2 Bakground In the theory of software speifiation in partiular in refinement theory we onsider operations on a system speifiation as relations rather than funtions. If an element of the domain of a relation is related to more than one element in the range this is viewed as non-determinism and more pragmatially as a hoie or onsious loseness in speifiation. In order to build a proper alulus for speifiation languages, algebra for relations is needed. In an algebrai haraterization of operation omposition, idempotene enables to redue doubles, e.g. op 1 ; op 1 ; op 2 = op 1 ; op 2. In the paper [7] we presented a first approah to investigate the theory of idempotents using the Isabelle/HOL theorem prover. In [5] we used the example of idempotents as a basi appliation of interative theorem proving to develop simple algorithms rigorously. There we onstruted a generate-and-test method for idempotents. As generate-and-test is not effiient this method enables to ount idempotents only up to five elements in reasonable time. Already for six we have a run time of roughly ten weeks. Nevertheless, our generate-andtest algorithm is valuable as a ounting method as the implementation of the effiient test prediate is proved sound inside Isabelle/HOL. Therefore, we an use it here to verify our results. The three main steps of haraterizing finite idempotent relations presented in [7] and [5] are as follows. We use the notation r(x) to abbreviate the relational image of r at x, i.e., {y (x, y) r}. The set fix r annotates {x dom r (x, x) r}. Let r be reflexive and transitive. Then r is idempotent. (2.1) idempotent r idempotent r ( x. r(x) = y fix r r(x). r(y) transitive r n N. r(n) = x F r(n). r(x) r F transitive f F. y N r(f). r(y) r(f) 9 ) (2.2) (2.3)

10 where r F = {(x, y). (x, y) r x F }, F = fixp r, and N = dom r \ F in (2.3). These three properties are mehanially verified in Isabelle/HOL [5]; the proof sripts are available at the author s web page [6]. A relation is alled a quasi-order iff it is reflexive and transitive [8]. From (2.1) it follows immediately that a quasi-order is idempotent. We annotate labelled quasi-orders with Q(n), unlabelled with q(n), where n is the number of elements in the domain of the relation. To annotate the domain we sometimes use N n beause the initial segment of the naturals may well represent any set of (labelled) points. ( For onjuntions p q we sometimes use two-dimensional depition, i.e. p ) q. For disjointness of two sets A, B, i.e., A B =, we write A B. Counting partial orders Partial orders have been subjet to ounting methods many times. Most reent advanes are due to Pfeiffer and Brinkmann and MKay [1, 2, 3, 8]. In [8] ounting different kinds of relations R inluding transitive relations and quasi-orders has been ahieved using the automorphism group Aut(R). Using similar methods Brinkmann and MKay arrive at improving their earlier results [1] to up to 15 or 16 points [2]. In the following we introdue the relationship between unlabelled and labelled partial orders needed for ounting idempotents. We deviate from the way Pfeiffer introdued automorphisms and partial orders using ation groups as we need only some fats that an already be derived using equivalene relations and fatorizations. For a deeper and more omplete haraterization of partial orders and automorphisms see Pfeiffer s paper [8]. Proposition The relation = {(x, y) φ Aut(PO(n)). φ(x) = y} is an equivalene relation on labelled partial orders PO(n). It partitions PO into its orbits with respet to the automorphism group. The fatorization yields an isomorphism to the unlabelled partial orders. ι : PO(n)/ = po(n) (2.4) Proof. The relation is reflexive, as for any s PO(n) we have id s = s and id Aut(PO(n)). It is symmetri as with f Aut(PO(n)) also f 1 AutPO(n). Therefore, for any s, s PO(n) iff s s, i.e., f(s) = s for some f Aut(PO(n)), then also s s, as f 1 (s ) = s and f 1 PO(n). Finally, we use the same proof with the argument f 1, f 2 Aut(PO(n)) implies f 2 f 1 Aut(PO(n)) to prove transitivity of. Hene onstitutes an equivalene relation. For the seond part it follows immediately from the homomorphism theorems of universal algebra that the fatorization PO(n)/ with the indued relation 10

11 [x] ι [y] iff x y is well-defined. The fatorized relation is isomorphi to the unlabelled partial orders over n points po(n) as any relation with the same graph but different order of nodes is ontained in the same equivalene lass. We will usually write ζ for the natural epimorphism from PO PO/ and [s] R to annotate a representant of the lass of s PO. The proposition immediately implies that as onstitutes a partitioning we an ount PO using the size of the orbits and the representatives from PO. Let ι be the resulting isomorphism from PO(n)/ to po(n) introdued in the previous proposition. The labelled partial orders PO(n) are a disjoint union of the equivalene lasses of PO(n)/. We an build this union using ι as follows. PO(n) = s po(n) {r PO(n) ι[r] = s} (2.5) Hene the number of relations in PO(n) an be alulated from the ardinality of po(n) and the size of the lasses. #PO(n) = #ι 1 (s) (2.6) s po(n) The number of elements in an equivalene lass orresponds to the size of the orbit reated by the automorphism group. If the size of the lass [s] is one, i.e., any automorphisms maps s to itself, then the size of the orbit is 1. The maximal number of elements of an orbit is n! as this is the number of permutations of n elements, and hene also the maximal number of automorphisms for n elements. In the ase of suh a lass [s] with n! elements eah automorphism maps to a different relation the orbit does not repeat elements. The other ases are all fators of n! and may be alulated as follows. First we introdue the notion of indistinguishability of subrelations of a labelled partial order s PO. Indistinguishable subrelations are subsets of a relation s that are invariant under automorphisms. A family χ of sets χ j, j J with χ j s and χ j χ k for j k J is a set of indistinguishable subrelations iff φ Aut(PO(n)). j J.! k J. k j φ(χ j ) = χ k (x, y) s \ j J χ j. φ(x, y) = (φ(x), φ(y)) = (x, y) φ(s) = s We onsider only maximal indistinguishable sets of subrelations χ, i.e. if some subset u s is indistinguishable to some χ j then there is a k suh that u = χ k. Intuitively, indistinguishable subsets are subgraphs of a relation that may be exhanged without hanging the relation. Conerning the ardinality of the orbits, it diminishes proportional with the fatorial of the size of the sets of indistinguishable subsets. If there are indistinguishable subsets that ontain smaller indistinguishable subsets than the redution is proportional to the produt of the smaller subrelations. Hene, for the alulation of the orbits we onsider 11.

12 sets of minimal subrelations. Let s po(n) and χ j be a family of minimal indistinguishable subsets in s. Then the following inequality holds. #ι 1 s = n! j J (#χ j)! (2.7) Finally, we need as a provisional step the relationship between quasi-orders and partial orders also presented in Pfeiffer s paper [8]. For any quasi-order Q(n) the symmetri ore of is defined by ( ) x y x y iff. y x As pointed out by Pfeiffer [8] the indued relation i on the lasses of N n / is a partial order. To ount quasi-orders one an thus ount the number of partitions of a set with n elements into k nonempty subsets and multiply with the number of partial orders, i.e., #Q(n) = k i=1 { n i } #PO(k), where the number of partitions is a Stirling number of the seond kind (see the Appendix for the first few numbers). A Stirling distribution for a partial order s PO(n) we all aording to this the natural epimorphism S : Q(n) PO(n). 12

13 Chapter 3 Counting Idempotents Our ounting method for idempotents provides an effiient method for alulating the numbers of labelled idempotent relations by alulating partitions of partial orders and labelled quasi-orders. It is based on the basi onstrution idea that is already visible in properties (2.1) and (2.2) being made more preise in (2.3). Any quasi-order is an idempotent relation (2.1). Property (2.2) shows that an idempotent relation r over a base set A an be onstruted starting from a quasi-order over a subset of A the fixpoints of r by seleting a range extension and a domain extension, i.e., by hoosing the non-fixpoints that are elements of the ranges of the fixpoints (range extensions) and the non-fixpoints that share the ranges of fixpoints (domain extensions). If an element n shares the range of a fixpoint f, i.e., r(f) r(n) we say n hangs-onto f in r. The hangs-onto relation of an idempotent s is a quasi-order order, and vie versa any quasi-order may be seen as a hangs-onto relation for some idempotents s. This is the key to our ounting method for idempotents. As already noted in Setion 2, the quasi-orders an be partitioned to partial orders by fatorizing with the symmetri ore. We use this simplifiation as well to partition idempotents aording to the partial order representing the hangs-onto relationship of the fixpoints rather than the quasi-orders, as the symmetri ore does not have muh effet on idempotents. Taking as given the number of unlabelled partial orders representing the abstrated and fatorized hangs-onto relation of a set of idempotents, we define the alphabet of hoies of range extension, domain extension and range/domain extension represented by it the latter is a ombination aording to property (2.3). The alphabet represents for eah non-fixpoint all possibilities how it may be related to the partial order kernel of a relation. This alphabet enables the unique onstrution of idempotents. Its ardinality for eah partial order kernel thus gives rise to ounting idempotents. The onstrution proess starts from the quasi-order and the indued partial order that build the kernel of an idempotent. Definition The Q-kernel of an idempotent s I(n) is s (fix s 13

14 fix s). Similarly, the PO-kernel is ζ(s (fix s fix s)) where ζ is the natural epimorphism. For the unlabelled ases we assume the orresponding definitions. The natural epimorphism may be ontinued on idempotents. Definition Let s I(n), k Q(m), and k PO(m ), suh that k is the Q-kernel of s, and, k the PO-kernel of s, i.e. ζ(k) = k. The ontinuation ˆζ is the extension of ζ to I(n), i.e. ˆζ(s) = s/ζ. The order on s/ζ is the indued order ŝ given as ([x], [y]) ŝ iff (x, y) s whih is well-defined as non-fixpoints are not part of the symmetri ore. Otherwise, (n, n ) s and (n, n) s for non-fixpoints n, n would imply (n, n) s and (n, n ) s beause of transitivity and hene n, n would be fixpoints. We observe that idempotents are respeted by the transition from quasiorders to partial orders. Proposition Let s I(n), and t Q(m) the Q-kernel and t PO(m ) the PO-kernel of s. Then ˆζ(s) I(n ) for n = n (m m ). Proof. For the kernel this property is already given by the Stirling properties (see Setion 2). For the remaining ases of non-fixpoint z related to the kernel we use the above noted fat that non-fixpoints are not part of the symmetri ore. Hene the idempotents properties follow immediately beause whenever (z, x) s then also (z, y) s for all y x beause of transitivity. Also for the other diretion (x, z). Hene, in the fatorization, for all x, y, z, (x, y) and (y, z) iff (x, z). The size of the base set is n = n (m m ) again beause the non-fixpoints are not part of the symmetri ore, hene their lasses are singleton. Definition (domain/range extension) Given a base set N n, a set of fixpoints F N n, and a quasi-order or a partial order s on F, let n N n \ F be some arbitrary non-fixpoint. Then a range extension r is a subset of F suh that the extended relation s (r n) is idempotent, and a domain extension d is a subset of F suh that the extended relation s (n d) is idempotent. A down-set is defined for any partial order as a set d losed with respet to smaller elements, i.e., if x d and y x then y d. We write d or if the relation is lear fom ontext just d. Up-sets u are defined dually [4]. Lemma Let F N n be a set of fixpoints and be a partial order on F. Any d F is a domain extension iff d and any r is range extension r iff r. 14

15 Proof. Let d be a domain extension, x d, and x y. If for a non-fixpoint n we have that n x then also n y beause of transitivity, whene y d. Similarly, let r be a domain extension, x r and y x. If x n then again beause of transitivity, we have also y n whereby again y r. The other diretion follows for domain extension from (2.3) first onjunt and for range extension from transitivity. Lemma Let F N n be a set of fixpoints, s be a partial order on F, r, d F, and n N n \ F. Then s d, r s, r d, and x r, y d. (x, y) s iff s n d r n is idempotent. Proof. Assuming an idempotent of the form s n d r n we have by property (2.3) third onjunt that s(n) s(f) for any f r, i.e., for any x if (n, x) s then (f, x) s. Let f d. As (n, f ) s by assumption, we have (f, f ) s. Sine f r was arbitrary, we have f r, f d, (f, f ) s. For disjointness, assume r d. Then for any x r d we have by onstrution (n, x) s and (x, n) s whih ontradits transitivity if n is a non-fixpoint. For s d and s r the arguments from the proof of Lemma apply. Definition (Alphabet). Let s be a partial order in PO(n), and B its base set ran s = dom s. The alphabet (of possible hoies to extend to an idempotent) for s is the following set. A(s) = {(r, d) P(B) P(B) s r s d r d x r, y d.(x, y) s} The alphabet entails the following four ases for any (r, d) P(B) P(B): (r, d) (, ) ( ) ( ) ( ) r = r = d = s d s r d = s d s r r d x r, y d.. (x, y) s In order to effiiently ount we observe that labelled partial orders that belong to the same orbit with respet to the automorphism group have isomorphi alphabets. Proposition Let s, s PO(n) suh that s s. That is, there is f Aut(PO(n)) with f(s) = s. Now, let ˆf be the pointwise ontinuation of f on alphabets, i.e., f(r, d) = (f(r), f(d)). Then for any (r, d) A(s) we have that f(r, d) A(s ) = A(f(s)), i.e., the alphabets are isomorphi. Proof. Sine f is an automorphism we have that f(b) = B, hene f(r, d) P(B) P(B) = P(f(B)) P(f(B)). Now, we onsider the four ases impliit in the definition of alphabet. We show for eah ase that f(r, d) A(f(s)). First, if f(r, d) = (f(r), f(d)) = (, ) then it is in A(f(s)) as (f( ), f( )) = (, ). If (f(r), f(d)) = (, f(d)) and f(d), then r = and d. Hene, as (r, d) A(s), we have that d. As f is a homomorphism, we also have 15

16 f(d). Similarly, for f(d) = and r, we have d =, r, r, and f(r). Again, as f is homomorphism, we have f(r). Finally, for (f(r), f(d)) (, ), then also (r, d) (, ). Therefore, as (r, d) A(s), x r, y d. (x, y) s. Let x f(r) and y f(d) be arbitrary. Then as f is onto there are x r and y d with f(x) = x, f(y) = y, and (x, y) s. As f is homomorph (f(x), f(y)) in f(s) = s. Proposition simplifies the way we an ount alphabet sizes. It suffies to onstrut and ount the alphabet just for eah representative of a lass of partial orders ι 1 (s) R where s po(n) and ι the isomorphism introdued in (2.4). Corollary For any alphabet size j and points k we have the following equality. #{x PO(k) A(s) = j} = #ι 1 (s) s po(k) A(ι 1 (s) R )=j To alulate the size of the lasses we an use properties (2.6) and (2.7). A further observation that failitates the ounting of alphabets is that the number of down-sets (or pure range extensions) is equal to the number of up-sets (see the example alulation in the Appendix). Also, symmetri strutures have equally sized alphabets. An idempotent an be represented by its partial order kernel, a Stirling distribution and alphabet hoies for eah non-fixpoint. Theorem 1 (Representation). For any labelled idempotent i over N n with given fixpoints F and non-fixpoints N there are a unique Stirling distribution S : Q(m) PO(m ), s PO(m ), and α N A(s). Vie versa, any suh representation (S, s, α) orresponds to one idempotent relation for F and N. Proof. Let s q Q(m) be the Q-kernel of i and s PO(m ) be the orresponding PO-kernel ζ(s q ). We show that the non-fixpoints an be uniquely represented by α with respet to the ontinuation ˆζ(i). Proposition grants that we an abstrat from the symmetri ore, here represented by S. Let n N. Either n / dom i ran i then α(n) = (, ). If n ran i only, let r be the set of all fixpoints for whih n is a range extension. By Lemma we have r. Similarly, if n is only in the domains of elements in d we have d. Finally, if both ases are represented by (r, d) we have x r, y d. (x, y) s beause of the third onjunt in property (2.3). Eah of these ases orresponds to an element of A(s). Vie versa, we an onstrut an idempotent from the representation as S 1 (s) n N n S 1 (π 2 (α(n))) S 1 (π 1 (α(n))) n where π i are the projetions of a pair to its i-th omponent. Idempotene of the onstruted relation is a onsequene of Proposition together with Lemmata and

17 For n = 0 and n = 1 the number of idempotents is 1. For any larger number the following theorem enables to ount the number of labelled idempotents. Theorem 2 (Number of labelled idempotents). The number of idempotent relations I(n) an be alulated for n > 1 as n 1 ( ) 2 n i+1 1 #I(n) = 1 + ( p ij j n i ) + #Q(n) i i=1 j=3 where the p ij are given by the harateristi sequene p ij = i k=1 { } i #{s PO(k) #A(s) = j}. k Proof. The 1 stands for the empty relation whih is idempotent. Otherwise we partition the idempotents aording to their numbers of fixpoints. If there are n fixpoints, that is, the relation is reflexive, then aording to property (2.1), the number of labelled idempotents is given by the number of labelled quasi-orders Q(n). For any number of fixpoints i, 0 < i < n, there are ( n i) disjoint ases. For any seletion of i fixpoints, the representation given by Theorem 1 yields for eah of the n i non-fixpoints j seletions if there is an alphabet with j elements represented by some Q-kernel. So we need to know the number of labelled POkernels that have an alphabet with j elements. Therefore the harateristi sequene p ij ounts the number of labelled partial orders that have alphabets with j elements. For i fixpoints the maximal ardinality of an alphabet is 2 i+1 1 for i unrelated points. The minimum is 3 as the full relation (on any set of fixpoints F ) gives the alphabet {, (F, ), (, F )}. For any j, for whih exist some partial orders s with #A(s) = j, we have to onsider the distribution of the i fixpoints in 1 < k < i nonempty subsets where k orresponds to the number of points of the strutures s with #A(s) = j. The Stirling oeffiient represents the number of hoies for the symmetri ores. With Theorem 2 we an ount labelled idempotents given the alphabets for the partial order kernels for speifi sizes j. For the latter we an use Corollary in addition to onstrut the alphabet just one for eah lass of partial orders. For examples of the alphabets and alulations of the results up to 5, see the Appendix. In Table 3.1 we provide an overview over the results up to 6 together with the values for quasi-orders and partial orders and the numbers of all relations over n elements R(n) = (N n N n ) with #R(n) = 2 2n for omparison. In the remainder of this setion, we will briefly show in how far this method for ounting labelled idempotents an be used for ounting unlabelled ones. The alphabets onstruted for the labelled partial orders give rise to onstrut alphabets a(s) of unlabelled s po. From those we an in a similar way derive the number of unlabelled idempotents that have a partial order kernel, i.e. where the symmetri ore ontains only singleton elements. 17

18 # R PO po Q q I Table 3.1: Numbers of the examined relations up to 6 Theorem 3 (Number of unlabelled idempotents with po-kernel). The number of idempotents i po (n) whose q-kernel is already a partial order is given for n > 1 by n 1 #i po (n) = 1 + ( i=1 2 i+1 1 j=3 ( n i + j 1 #{s po(j) #a(s) = j} j 1 ) ) + #q(n). Proof. The formula derives in a similar way as that in Theorem 2. The hoies for the non-fixpoints, however, are alulated as multisets: for eah non-fixpoint it matters whih alphabet-element is hosen but the order does not matter. Hene, a seletion of (n i) non-fixpoints onsists of a multiset with n i elements out of j possible alphabet elements. For the hoies of i fixpoints 0 < i < n the order does not matter, hene we just sum up without binomial oeffiient. The Stirling seletion an unfortunately not be applied to derive fom this number the general number of unlabelled idempotents. The reason is that the reverse introdution of symmetri ores into the unlabelled partial order kernel may turn indistinguishable fixpoints into distinguishable sets of fixpoints. We believe it is not possible to generally alulate the numbers of unlabelled idempotents from the numbers of i po as the distinguishability it too dependent on the inividual struture of an s i po. 18

19 Chapter 4 Outlook and Conlusions We have presented a ounting method for labelled idempotent relations. This ounting method redues the problem to ounting labelled quasi-orders, unlabelled and labelled partial orders, and onstruting the alphabets for partial orders representing the kernel of an idempotent. For unlabelled idempotents we presented a partial result. We have applied the presented ounting method for labelled idempotents up to six points. The results are shown in Table 3.1 and the derivation steps are ontained in the Appendix. The results are additionally verified by the mehanially proved generate-and-test method [5]. The alulation of the number of idempotents over n points as desribed in this paper is an algorithm that takes some representation of unlabelled partial orders over m < n points, the number of labelled quasi-orders over n points, and outputs the number of labelled idempotents. The algorithm onstruts the alphabets for eah representant of a labelled partial orders and alulates the size of the orbit. Finally, it just alulates the number of idempotents using the formula of Theorem 2. We have not yet attempted to implement this algorithm. Suh an implementation will ertainly inrease the number of points for whih we an alulate the number of labelled idempotents, we believe up to more than 10. However, as we diretly reuse the number of Q we annot get further than 16 whih is the urrent state of the art for labelled quasi-orders [2]. 19

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21 Bibliography [1] G. Brinkmann and B. D. MKay. Posets on up to 16 points, Order 19 (2002), pp [2] G. Brinkmann and B. D. MKay. Counting unlabelled topologies and transitive relations, Journal of Integer Sequenes 8 (2005), Artile [3] Joseph C. Culberson and Gregory J. E. Rawlins. New results from an algorithm for ounting posets. Order 7 (1991), pp [4] R. A. Davey and H. A. Priestley. Introdution to Latties and Order, Seond Edition. Cambridge University Press, [5] F. Kammüller. Interative Theorem Proving in Software Engineering. Habilitationsshrift, Tehnishe Universität Berlin, [6] F. Kammüller. Web page at flokam, [7] F. Kammüller and J. W. Sanders. Idempotent relations in Isabelle/HOL. International Colloquium of Theoretial Aspets of Computing, ICTAC 04, LNCS 3407, Springer [8] Götz Pfeiffer. Counting transitive relations. Journal of Integer Sequenes 7, No. 3, (2004) Artile [9] N. J. A. Sloane The on-line enylopedia of integer sequenes. Published eletronially at njas/sequenes,

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23 Appendix In this appendix we illustrate our ounting method for labelled idempotents by giving the details of the alulation. The numbers of idempotents for n {0, 1} are #I(0) = #{ } = 1 and #I(1) = #{{(a, a)}} = 1. Labelled idempotents with two elements I(2) Following Theorem 2 we need as input #Q(2), whih is 4 aording to Sloane sequene A000798, see also Table 3.1. Now the formula of Theorem 2 instantiates for 2 to: #I(2) = 1 + where p 13 is given by: 1 k=1 1 i=1 ( ) 2 2 i+1 1 p ij j 2 i + Q(2) = 1 + 2p i j=3 { } 1 #{s PO(k) #A(s) = 3}] = #{s PO(1) #A(s) = 3}. k The set PO(1) is just the one point relation o = {(a, a)}. The alphabet over this relation is A(o) = {(, ), ({a}, ), (, {a})}. Sine #A(o) = 3 we have that p 13 is indeed 1, whereby #I(2) = = 11. Labelled idempotents with three elements I(3) The unlabelled partial orders with two elements are the following. The labelled.. versions are the following three elements s 0, s 1, and s 2. a. b. a b b 23 a

24 We depit the order from top to bottom, for example, the middle one, say s 1, is the relation {(a, a), (a, b), (b, b)}. A more elaborate representation of the three relations s 0, s 1, and s 2 is given below where we an omit labels a and b as they are represented by positions top and bottom. Although the seond representation is graphially more elegant we use the first one for ounting alphabets. The alphabet for the leftmost labelled partial order s 0 is: A(s 0 ) = {(, ), ({a}, ), ({b}, ), ({a, b}, ), (, {a}), (, {b}), (, {a, b})}. The number of elements of this alphabet is #A(s 0 ) = 7. Next, we build the alphabet for s 1 : A(s 1 ) = {(, ), ({a}, ), ({a, b}, }, (, {b}), (, {a, b}), ({a}, {b})}. The orders s 1 and s 2 are both in the same equivalene lass with respet to the natural isomorphism ζ (see Setion 2). Aording to Proposition it suffies to build one of the alphabets A(s 1 ) or A(s 2 ) as they are isomorphi and hene have the same number of elements. That is, #A(s 1 ) = #A(s 2 ) = 6. The numbers p 2j for 3 j 7 an now be alulated from these alphabet sizes. The harateristi number p 23 is again 1 as #{s PO(1) #A(s) = 3} is one (see previous setion) and { 2 1} is also one. There is no element of PO(2) that has an alphabet with three elements. The sequene elements p 24 and p 25 are zero as there are again no strutures in PO(1) or PO(2) with alpabets of those sizes. However, as we have have seen above there are strutures s 1, and s 2 with alphabet size six. Hene p 26 is two. For 7 the automorphism orbit ontains only s 0, therefore p 27 is 1. Summarizing, the number of idempotents with three elements an be alulated as follows. ) 2 i ( 3 #I(3) = 1 + p ij j 3 i + #Q(3) i i=1 j=3 ( ) ( ) 3 3 = 1 + p (3p p p 27 ) = ( ) + 29 = 123 Labelled idempotents with four points I(4) The labelled partial orders with three elements are represented by the following five representants s 0,..., s 4 PO(3) from left to right. a. b.. a. b a b 24 a b a b

25 The alphabets and sizes of the orbits for eah s i are given in the following table. For brevity we write for (, ), S r for (S, ), S d for (, S), and S 0, S 1 rd for (S 0, S 1 ). s i A(s i ) #A(s i ) #ι 1 (s i ) s 0 {, {a} d, {b} d, {} d, {a, b} d, {a, } d, {b, } d, {a, b, } d, 15 1 {a} r, {b} r, {} r, {a, b} r, {a, } r, {b, } r, {a, b, } r } s 1 {, {a} d, {} d, {b, } d, {a, } d, {a, b, } d, 12 6 {a} r, {b} r, {a, b} r, {b, } r, {a, b, } r, {b}, {} rd } s 2 {, {} d, {b, } d, {a, b, } d, {a} r, {a, b} r, {a, b, } r, 10 6 {a}, {} rd, {a}, {b, } rd, {a, b}, {} rd } s 3 {, {b} d, {} d, {b, } d, {a, b, } d, {a} r, {a, b} r, {a, } r, 12 3 {a, b, } r, {a}, {b} rd, {a}, {} rd, {a}, {b, } rd } s 4 {, {} d, {a, } d, {b, } d, {a, b, } d, {a} r, {b} r, {a, } r, 12 3 {a, b, } r, {a}, {} rd, {b}, {} rd, {a, b}, {} rd } As an illustration for sets of minimal indistinguishable subsets χ onsider s 4 : χ 1 = {a}, χ 2 = {b}, hene ι 1 (s 4 ) = 3! 2! = 3. Similarly for s 0 we have χ 1 = {a}, χ 2 = {b}, χ 3 = {} and ι 1 (s 4 ) = 3! 3! = 1. From this table we an diretly see that p 3,10 = 6, p 3,12 = 12, and p 3,15 = 1. The harateristi number p 33 is again 1 as { 3 1} = 1. For p36 we get a nonzero in the sum only for 2, i.e. { } 3 #{s PO(2) #A(s) = 6} = 6. 2 For p 37, similarly, { } 3 #{s PO(2) #A(s) = 7} = 3. 2 Now, the number of elements in I(4) an be alulated as follows. ) 2 i ( 4 #I(4) = 1 + p ij j 4 i + #Q(4) i i=1 j=3 ( ) ( ) 4 4 = 1 + p (3 2 p p p 27 ) ( ) 4 (3p p p p 3, p 3, p 3,15 ) = = 2360 Labelled idempotents with five points I(5) The labelled partial orders with four elements are represented by the following 16 representants s 0,..., s 15 PO(4). 25

26 a b a.. b. d. d a a. b. b d d a. a b d b d a b a b d d a b d a b a d. b a b d d a b b a d. d a b d b a d s 0 s 1 s 2 s 3 s 4 s 5 s 6 s 7 s 8 s 9 s 10 s 11 s 12 s 13 s 14 s 15 #A(s i ) #ι 1 (s i ) The new values for the harateristi sequene p 4j, 31 j 17 are simply given by the sum of the orbits of all strutures having j alphabet elements beause { 4 4} = 1. Other new values are p4j for 3 j 16 given by additional multipliation with the orresponding Stirling numbers: p 43 p 46 p 47 p 4,10 p 4,12 p 4, The ase p 4,15 shows why we build the harateristi sequene by a sum over k 1... i. We an have partial orders with a different number of points and equal alphabet sizes: the relation {(a, a), (b, b), (, )} PO(3) and the above s 4 PO(4) have alphabet size 15, hene ontribute to p 4,15. p 4,15 = { 4 3 = = 30 } #{s PO(3) #A(s) = 15} + The alulation of #I(5) is now as follows. #I(5) = i=1 ( ) 2 5 i+1 1 p ij j 5 i + #Q(5) i j=3 { } 4 #{s PO(4) #A(s) = 15} 4 = 1 + (5 ) p ( ) 5 (3 3 p p p 27 ) + 2 ( ) 5 (3 2 p p p p 3, p 3, p 3,15 ) + 3 ( ) 5 (3p p p p 4, p 4, p 4, p 4, p 4, p 4, p 4, p 4, p 4,31 ) = =

27 Used Stirling numbers The first few Stirling numbers { n k } representing the partition of n elements into k parts are given in the following table. k \ n