Counting Quasigroup Words

Counting Quasigroup Words

Counting Words in Free Quasigroups Stefanie G. Wang Smith College Department of Mathematics Loops 2019 July 9, 2019 Overview I Background on quasigroup words and rooted trees I Quasigroup conjugates and nodal equivalence I s-peri-Catalan numbers s I The closed formula for Pn s I Asymptotics for Pn Quasigroup Conjugates In an equational quasigroup (Q; ·; =; n), we have the opposite operations: x ◦ y = y · x ; x==y = y=x ; xnny = ynx: (1) Basic and opposite operations yield the following combinatorial quasigroups known as conjugates or parastrophes: (Q; ·); (Q; =); (Q; n); (Q; ◦); (Q; ==); (Q; nn) (2) The identities (IR) in (Q; n) and (IL) in (Q,/) yield the respective identities (DL) x=(ynx) = y ; (DR) y = (x=y)nx Basic parsing trees In the free quasigroup on an alphabet fa1; a2;:::; as g,(basic) quasigroup words are repeated concatenations of the generators under the three basic quasigroup operations ·; =; n.A basic parsing tree Tw , defined recursively as follows: (a) For 1 ≤ i ≤ s, the tree Tai is a single vertex annotated by ai ; (b) For basic words u; v, the tree Tu·v has: (i) a root annotated by the multiplication, (ii) Tu as a left child, and Tv as a right child; (c) For basic words u; v, the tree Tu=v has: (i) a root annotated by the right division, (ii) Tu as a left child, and Tv as a right child; (d) For basic words u; v, the tree Tunv has: (i) a root annotated by the left division, (ii) Tu as a left child, and Tv as a right child. Basic parsing trees Let u = (a4=a2) · a1 and v = a3=a2. a4 a2 a4 a2 / a1 a3 a2 / a1 · / a3 a2 · / · Tu Tv Tu·v Action of S3 on quasigroup operations Writing σ and τ for the respective transpositions (12) and (23), the full set {·; n; ==;=; nn; ◦} of basic and opposite quasigroup operations is construed as the homogeneous space S3 g µ = fµ j g 2 S3g (3) for a regular right permutation action of the symmetric group S3. x · y = xy µ () xny = xy µτ ! x==y = xy µτσ l m x ◦ y = xy µσ () xnny = xy µστ ! x=y = xy µστσ Monoid of binary words Let M be the complete set of all derived binary operations on a quasigroup. A multiplication ∗ is defined on M by xy(α ∗ β) = x xyα β : (4) The right projection xy = y also furnishes a binary operation . Lemma The set M of all derived binary quasigroup operations forms a monoid (M; ∗; ) under the multiplication (4), with identity element . Proposition g For each element g of S3, the binary operation µ is a unit of the monoid M, with inverse µτg . Corollary The six quasigroup identities (SL); (IL); (SR); (IR); (DL); (DR) all take the form x xyµτg µg = y (5) for an element g of S3. Full parsing trees Full quasigroup words on the alphabet fa1; a2;:::; as g are repeated concatenations of the generators under the full set {·; n; ==;=; nn; ◦}. (a) For 1 ≤ i ≤ s, the full parsing tree Fai is a single vertex annotated by ai ; (b) For full parsing trees Fu; Fv and a basic or opposite operation g µ from the set (3), the tree Fuv µg has: (i) a base annotated by µg , along with (ii) Fu as a left child, and Fv as a right child. Nodal equivalence I The basic parsing tree represents a so-called nodal equivalence n−1 class Fu of 2 full parsing trees, sustaining a regular action n−1 of a permutation group (S2) known as the nodal group of the basic quasigroup word u. I At a given node of a full parsing tree with annotating operation µg , the non-trivial permutation of the nodal subgroup switches the two children of the node, and changes the node's annotation to µσg . It fixes the remainder of the tree. Nodal equivalence example Consider the basic quasigroup word (a · b)=c. It determines the nodal equivalence class fFab µ c µστσ ; Fba µσ c µστσ ; Fc ab µ µτσ ; Fc ba µσ µτσ g of full parsing trees, represented by the basic parsing tree T(a·b)=c = Fab µ c µστσ . a b b a · c ◦ c = = a b b a c · c ◦ == == s-peri-Catalan numbers Definition (a) A (basic or full) quasigroup word is reduced if it will not reduce further via the quasigroup identities. (b) A (basic or full) parsing tree representing a quasigroup word is reduced if its corresponding quasigroup word is reduced. Definition Let n and s be natural numbers. The n-th s-peri-Catalan number, s denoted Pn, gives the number of reduced basic quasigroup words of length n in the free quasigroup on an alphabet of s letters. Auxiliary bivariate function Definition Let s, a and b be positive integers. The auxiliary bivariate ms (a; b) denotes the number of (a + b)-leaf parsing trees representing reduced quasigroup words in s arguments, with an a-leaf basic parsing tree on the left child, a b-leaf basic parsing tree on the right child, and a given (basic or opposite) quasigroup operation at the root vertex. Auxiliary bivariates and nodal equivalence I The auxiliary bivariate ms (a; b) is invariant under any change of the choice of quasigroup operation at the root vertex of an (a + b)-leaf parsing tree of the type considered in Definition 6. I In particular, ms (a; b) = ms (b; a), since the left hand side counting certain trees with µg at the root corresponds to the right hand side counting certain trees with the opposite operation µσg at the root. I By convention, whenever one of the arguments s; a; b of an auxiliary bivariate is nonpositive, the output of the auxiliary bivariate is zero. To construct a length n quasigroup word: 1. Adjoin a reduced word u of length n − k to a reduced word v of length k, and 2. take one of the three basic quasigroup operations as the connective. Then n−1 n−1 s X s X s s Pn = 3 m (n − k; k) ≤ 3 Pn−k Pk (6) k=1 k=1 as an upper bound on the n-th s-peri-Catalan number. Number of reductions Proposition During the assembly of uv µg within the inductive process, cancellation occurs if and only if there is a (necessarily reduced) word v 0 of length n − 2k such that v = uv 0 µτg . Proof. The unique cancellations available are of the form u uv 0µτg µg = v 0 described in Corollary 3. Since the word v = uv 0 µτg is reduced, it follows that the subword v 0 is also reduced. Root vertex cancellation length k length n − k u length k v u length n − 2k g τg 0 µ µ v Figure: Root vertex cancellation Unpacking the auxiliary bivariate Proposition Consider 1 < n 2 N and 1 ≤ k < n. (a) The number of cancellations incurred during the inductive process when a word of length k is connected to a word of length n − k by a given operation µg is ms (n − 2k; k). In particular, no cancellation occurs when n = 2k. (b) The formula s s s s m (n − k; k) = Pn−k Pk − m (n − 2k; k) (7) holds. Euclidean Algorithm notation k k For 1 < n 2 N and 1 ≤ k ≤ n − 1, let r−1 = n and r0 = k. k k Consider the quotients ql and remainders rl for 1 ≤ l ≤ Lk+1 as given below, resulting from calls to the Division Algorithm in the computation of gcd(n; k) by the Euclidean Algorithm: k k k k k k k k r−1 = q1 r0 + r1 ;:::; rl−2 = ql rl−1 + rl ;:::; (8) r k = qk r k + r k : (9) Lk −1 Lk +1 Lk Lk +1 Here, r k = 0 and gcd(n; k) = r k . Lk +1 Lk k k k k 0 = 1 and l+1 = l + ql+1 : (10) Lemma k k k k k k k Using the notation r−1 = n, r0 = k, r−1 = q1 r0 + r1 , and 0 = 1, the formula k q1 −1 k X k k s q1 −1 s k k 0 +j0 s s m (n − k; k) = (−1) m (r0 ; r1 ) + (−1) P k k k P k r−1−j0 r0 r0 k j0 =1 (11) holds for 1 < n 2 N and 1 ≤ k ≤ n − 1. Proof of Lemma Proof: Through induction on i, we will show: i X k k s i s k k k 0 +j0 s s m (n−k; k) = (−1) m r0 ; r−1−(i+1)r0 + (−1) P k k k P k r−1−j0 r0 r0 k j0 =1 (12) k k for 0 ≤ i < q1 . Note that (12) for i = q1 − 1 yields (11). On the other hand, the base of the induction, namely (12) with i = 0, is given by ms (a; b) = ms (b; a). Proof cont. Now suppose that the induction hypothesis (12) holds for k 0 ≤ i < q1 − 1. Then i X k k ms (n − k; k) = (−1)i ms r k ; r k − (i + 1)r k + (−1)0 +j0 Ps Ps 0 −1 0 rk −jk rk rk −1 0 0 0 jk =1 0 h i = (−1)i Ps Ps − ms r k ; r k − (i + 2)r k rk −(i+1)rk rk 0 −1 0 −1 0 0 i X k k + (−1)0 +j0 Ps Ps rk −jk rk rk −1 0 0 0 jk =1 0 i+1 s k k k = (−1) m r0 ; r−1 − (i + 2)r0 i k X k k + (−1)0 +(i+1)Ps Ps + (−1)0 +j0 Ps Ps rk −(i+1)rk rk rk −jk rk rk −1 0 0 −1 0 0 0 jk =1 0 i+1 X k k = (−1)i+1ms r k ; r k − (i + 2)r k + (−1)0 +j0 Ps Ps 0 −1 0 rk −jk rk rk −1 0 0 0 jk =1 0 by (7) and ms (a; b) = ms (b; a), as required for the induction step.

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