Ballot Paths Avoiding Depth Zero Patterns

Transcription

1 Ballot Paths Avodng Depth Zero Patterns Henrch Nederhausen and Shaun Sullvan Florda Atlantc Unversty, Boca Raton, Florda 1 Introducton In a paper by Sapounaks, Tasoulas, and Tskouras [8], the authors count the number of occurrences of patterns of length four n Dyck paths In ths paper we specfy n one drecton and generalze n another We only count ballot paths that avod a gven pattern, where a ballot path stays weakly above the dagonal y = x, starts at the orgn, and takes steps from the set {, } = {u, r} A pattern s a fnte strng made from the same step set; t s also a path Notce that a ballot path endng at a pont along the dagonal s a Dyck path Consder the followng enumeraton of ballot paths avodng the pattern rur m n The number of ballot paths to (n, m avodng rur s n (m, the entry at the pont (n, m, s the number of ballot paths from the orgn to that pont avodng the pattern rur We can generate ths 1

2 table by usng the recurrence formula s n (m = s n (m 1 + s n 1 (m s n 1 (m 1 + s n 1 (m 2 (1 where s 0 (m = 1, and s n (n 1 = 0 represents the weak boundary y = x The above table s calculated usng ths recurson and suggests the columns are the values of a polynomal sequence It wll be shown that s n (m s a polynomal n m We shall present a theorem guaranteeng the exstence of a polynomal sequence for a gven recurrence relaton and boundary condtons If the number of ballot paths reachng (n, x s the values of a polynomal sequence s n (x, the recurrence relaton obtaned can be transformed nto an operator equaton Usng the tools from Fnte Operator Calculus n Secton 2, we can fnd an explct formula for the polynomal sequence s n (x In ths paper, we wll only consder patterns p such that ts reverse pattern p s a ballot path For example, the reverse pattern of p = uururrr s p = uuururr We call such patterns depth-zero patterns Ths s a requrement n Theorem 4, although for some patterns that are not depth-zero, the soluton may be a polynomal sequence, but not for the entre enumeraton For patterns of nonzero depth see [5] To develop the recursons, we need to nvestgate the propertes of the pattern we wsh to avod Defnton 1 A nonempty strng o {u, r} s a bfx of a pattern p f t can be wrtten n the form p = op and p = p o for nonempty strngs p, p {u, r} If a pattern has no bfxes, then we call t bfx-free Defnton 2 If a s the number of r s n p and c s the number of u s, then we say p has dmensons a c, and d(p = a c If o s a bfx for the pattern p, then we wll see n Secton 4 that we need that d(p 0 The followng lemma shows that ths restrcton s not necessary when p s depth zero Lemma 3 If p s a depth zero pattern and o s a bfx of p, then d(p 0 Proof Frst we note the addtve property of d, that s d(ab = d(a+d(b By defnton p = op = p o, whch mples d(p = d(p Snce p s depth zero, any suffx of p s also depth zero, n partcular p s depth zero Therefore, d(p 0, whch mples d(p 0

3 2 Man Tools In ths secton we wll present the man tools from Fnte Operator Calculus that wll be used to solve these enumeraton problems As we noted before, not every pattern we choose to avod wll have a polynomal sequence soluton The followng theorem wll llustrate why we want a depth-zero pattern Theorem 4 Let x 0, x 1, be a gven sequence of ntegers where F n (m = 0 for m < x n For m > x n defne F n (m recursvely for all n 0 by F n (m = F n (m 1 + n a,j F n (m b,j, =1 where a 1,j 0 and wth ntal values F n (x n If F 0 (m = c 0 for j all m x 0, and f for each j holds b,j x n x n + 1 for n 1 and = 1,, n, then there exsts a polynomal sequence f n (x where deg f n = n such that F n (m = f n (m for all m x n and n 0 (wthn the recursve doman Before we prove ths theorem, we look back at the example of ballot paths avodng the pattern rur j m n F n (m = F n (m 1 + F n 1 (m F n 1 (m 1 + F n 1 (m 2 Here, x n = n 1 and F n (x n = δ n,0 The recurson satsfes the condtons b,j x n x n + 1 = + 1 and j a 1,j = 1 0 In the table, below the ntal values are the polynomal extensons, whch are not all 0 Clearly, f the recurson requred us to use values below these ntal

4 values, the values generated would not agree wth the values generated by the polynomal extensons, thus we need the condton b,j x n x n + 1, n general We now prove Theorem 4 Proof We frst let f 0 (x = F 0 (x 0, snce F 0 (m = F 0 (m 1 Now suppose F n (m has been extended to a polynomal of degree n The recurson tells us that F n+1 (m F n+1 (m 1 can be extended to a polynomal of degree n snce a 1,j 0 But f the backwards dfference of a functon s a j polynomal, then the functon tself s a polynomal of one degree hgher (see [3] The objects of Fnte Operator Calculus are polynomal sequences called Sheffer sequences A Sheffer sequence s defned by ts generatng functon, whch can be wrtten n the followng way p(x, t = n 0 s n (xt n = σ(te xβ(t where σ(t s a power seres wth a multplcatve nverse and β(t s a delta seres, that s, a power seres wth a compostonal nverse We say (s n s a Sheffer sequence for the basc sequence (b n where b(x, t = n 0 b n (xt n = e xβ(t Notce that b n (0 = δ n,0 and b 0 (x = 1 In order to fnd the soluton to the recurrence, we frst transform the recurson nto an equaton wth fnte operators For every Sheffer sequence, the lnear operator B = β 1 (D, where D = d dx, maps s n to s n 1 The operator B also maps ts basc sequence b n to b n 1 (Rota, Kahaner, and Odlyzko [7] Also f B s a delta operator, that s B = n 1 a n D n where a 1 0, then t corresponds to a class of Sheffer sequences assocated to a unque basc sequence We do not have to worry about convergence snce these operators are only appled to polynomals, thus only a fnte number of the terms wll be used for a gven polynomal, and thus the name Fnte Operator Calculus An example of a delta operator that wll be frequently used n ths paper s = 1 E 1, where E a s n (x = s n (x + a s a shft operator By Taylor s Theorem, we can wrte

5 f(x + a = f (n (x an n! n 0 or E a = n 0 D n a n n! = e ad n operators Clearly s a delta operator If we consder s n (x for x n 1 n (1, we obtan the operator equaton = B BE 1 + BE 2 We have wrtten one delta operator n terms of an unknown delta operator The followng theorem [6, Theorem 1] lets us fnd the basc sequence for the unknown operator B n terms of the basc sequence of the known operator Theorem 5 (Tranfer Formula Let A be a delta operator wth basc sequence a n (x Suppose A = τ(b = j 1 T j B j, where the T j are lnear operators that commute wth shft operators, and T 1 s nvertble Then B s a delta operator wth basc sequence b n (x = x n [ τ ] 1 n x a (x =1 where [ τ ] n s the coeffcent of tn n τ (t Notce that ths Theorem mples that the unknown operator B s a delta operator, and so the underlyng polynomal sequence s a Sheffer sequence The fnal step s to transform ths basc sequence nto the correct Sheffer sequence usng the ntal values For ballot paths we have that s n (n 1 = δ n,0, thus the followng lemma gves us the soluton based on these ntal values Lemma 6 If (b n s a basc sequence for B, then t n (x = (x an c b n(x c x c s the Sheffer sequence for B wth ntal values t n (an + c = δ n,0

6 In our case, we would obtan the soluton from the basc sequence b n gven n the transfer formula as s n (x = (x n + 1 b n(x + 1 x + 1 called Abelzaton [6, (25] We want to remark that gven the generatng functon e xβ(t = n 0 b n (x t n, we fnd s n (x t n = e (x+1β(t t d x + 1 dt e(x+1β(t = (1 tβ (t e (x+1β(t n 0 All the examples connected to pattern avodance are of the form 1 E 1 = = k 1 a k E c k B k (2 where a k, c k Z If ths equaton can be solved for E, E = 1 + τ (B, where τ s a delta seres, then b n (x t n = (1 + τ (t x n 0 We frst present the results n [5] whch covered bfx-free patterns and patterns wth exactly one bfx We then present results for patterns wth any number of bfxes, hence for any depth-zero pattern We also gve some examples and specal cases 3 Bfx-free patterns Let s n (x; p be the number of ballot paths avodng the pattern p; we occasonally wll drop the p n the notaton when convenent If the pattern s bfx-free and depth-zero, we need only to subtract paths that would end n the pattern, thus we have the recurrence s n (m; p = s n 1 (m; p + s n (m 1; p s n a (m c; p (3 where p has dmensons a c For example uurrurrur has dmensons 5 4, and depth zero If the pattern has depth zero, and a c 1, a 2 (f a = 1 then p = ur, a pattern we do not want to avod, then t s easy

7 to check that the condtons of the Theorem 4 are satsfed; the soluton s polynomal In operator notaton, = B(1 B a 1 E c Snce the delta operator can be wrtten as a delta seres n B, the operator B s also a delta operator The basc sequence can be expressed va the Transfer Formula n a 1 b n (x = x ( 1 x c Usng (2 we obtan n a 1 s n (x = (x n + 1 ( ( n (a 1 x + n (a + c 1 1 ( 1 x c + 1 n (a 1 ( ( n (a 1 x + n (a + c 1 n (a 1 Therefore the number of ballot paths avodng p and returnng to the dagonal (Dyck paths equals s n (n = n a 1 ( 1 n c + 1 ( n (a 1 ( 2n (a + c 1 n (a 1 4 Patterns wth exactly one bfx If the pattern p has exactly one bfx, then there exsts a unque nonempty pattern o such that p = op o If p has depth 0 and dmensons a c, and p has dmensons b d, we have a recurrence of the form s n (x = s n 1 (x + s n (x 1 0( 1 s n a b (x c d (4 For example let p = urruurr Ths pattern s depth-zero, wth dmensons 4 3 and bfx o = urr, so b = 2 and d = 2 From the paths reachng (n a, x c, those endng n p = urru cannot be ncluded n the recurrence and must be subtracted, and from those agan we cannot nclude paths endng n urru, and so on The p pece of the pattern that s responsble for ths excluson-ncluson process may not go below the dagonal; hence t must have d(p 0 whch lemma 3 guarantees If d(p < 0, then at some pont we would be usng numbers below the y = x boundary, whch are only the polynomal extensons and do not count paths

8 In operators, we have or 1 = B + E 1 0 ( 1 B a+b E c d = B + E 1 Ba E c 1 + B b E d = B + B b+1 E d + B b E d 1 B a E c B b E d (5 = B B a E c + B b+1 E d B b E d Usng the Transfer Formula, we obtan b n (x = x ( n (a 1j bk (b 1l ( 1 j+l j, k, l n (a 1j bk (b 1l j,k,l 0 ( n (a + c 1j (b + d(k + l 1 + x n (a 1j b(k + l 1 Because s n (n 1 = δ 0,n we get s n (x = ( n (a 1j bk (b 1l ( 1 j+l (x n + 1 j, k, l n (a 1j bk (b 1l j,k,l 0 ( n (a + c 1j (b + d(k + l + x n (a 1j b(k + l 1 So for the pattern urruurr, we obtan s n (x = ( n 3j 2k l ( 1 j+l ( (x n + 1 n 6j 4(k + l + x j, k, l n 3j 2k l n 3j 2(k + l 1 j,k,l 0 m n The number of ballot paths avodng urruurr

9 5 A Specal Case The above operator equaton (5 smplfes when a = b + 1 and c = d Ths corresponds to a pattern of the form rp r For ths case we get = B(1 + B b E d 1, and b n (x = x 0 s n (x = (x n ( 1 ( n (b 1 1 x d ( 1 ( n (b 1 1 x d + 1 ( x + n (d + b 1 n b, ( x + n (d + b n b The number of Dyck paths equals s n (n = ( 1 ( ( n (b 1 1 2n (d + b n d + 1 n b 0 The frst pattern we consdered, rur, s of ths form where b = d = 1 The soluton n ths case s s n (x = (x n + 1 ( 1 ( ( n 1 x + n 2 x + 1 n 0 and the number of Dyck paths equals s n (n = ( 1 ( ( n 1 2n 2 = ( n 1 ( 1 C n n + 1 n 0 0 where C n s the nth Catalan number 6 Patterns wth any number of bfxes Now we suppose that the pattern p has m dstnct bfxes o 1,, o m Let b d be the dmensons of p, the pattern p wthout the suffx o, and a c be the dmensons of p The number of ways to reach the lattce pont (n a j b j, x c j d j wth k 0 combnatons of the p j wth

10 ( m j=0 k j = k s wth sgn ( 1 k Then the recurrence for the 1,, m ballot paths avodng p s s n (x = s n 1 (x + s n (x 1 k 0( 1 k (6 In operators, = B k m=k ( 1 k = B B a E c k 0 ( k s n a Σ jb 1,, j (x c j d j m 1+ + m=k ( k B a+ jb j E c jd j 1,, m ( m k ( 1 k B b E d B a E c = B 1 + m (7 B b E d In general, we can fnd the basc sequence for any depth zero pattern va the transfer formula, although t would nvolve many summatons We now consder an example and some specal cases Example 7 The pattern rururrur has two bfxes, rur and r In ths case, a = 5, c = 3, b 1 = 3, d 1 = 2, b 2 = 4 and d 2 = 3 Usng the above operator equaton, we obtan = B 5 E 3 B 1 + B 3 E 2 + B 4 E 3 = B + B 4 E B 3 E 2 + B 4 E 3 Usng the transfer formula and (2, b n (x = n x ( ( ( k 1 j k n 3j 3k x n + + j + k j 0 k 0 ( x n j + k 1

11 and n ( ( ( k s n (x = (x n + 1 j k n 3j 3k j 0 k 0 ( 1 x n j + k x n + + j + k Patterns of the form r a We had prevously determned n [4] by other methods that the recurrence formula for ths pattern s gvng the operator equaton s n (x = s n 1 (x + s n (x 1 s n a (x 1 a 1 = B B a E 1 or E = B Usng the new formula (7, we vew r a as a pattern wth a 1 bfxes Here c = d = 0 and b = for all 0 a 1, thus the operator equaton becomes as expected B a = B = a 1 B =1 a 1 1 or E = B a 1 B 62 Patterns of the form r(ur k If the pattern s of the form r(ur k, then a = k + 1, c = k, and for each 0 < k there s the bfx r(ur So, b j = d j = j for 1 j k Let S = BE 1 We obtan the followng operator equaton for ths pattern, = B = B BS k 1 + S + S S k ( 1 Usng the Transfer Formula, n b n (x = x ( ( ( 1 j j j n kj j 0 S k 1 + S + S S k k+1 ( 1 x n x n +

12 ( x where n s n (x = (x n + 1 k+1 := [t n ](1 + t + + t k x [4] The number of Ballot paths s n ( ( ( 1 j j j n kj j 0 k+1 ( 1 x n + 2 x n We obtan a very nce formula for the Dyck paths n ths case, n ( ( s n (n = C ( 1 j j j n kj j 0 where C n s the nth Catalan number k+1 References [1] Deutsch, E, 1999, Dyck path enumeraton Dscrete Math 204, [2] Euler, L, 1801, De evolutone potestats polynomals cuuscunque (1 + x + x 2 + x 3 + x 4 + etc n Nova Acta Academae Scentarum Imperals Petropoltnae 12, [3] Jordan, C, 1939, Calculus of Fnte Dfferences Chelsea Publ Co, New York, 3rd edton 1979 [4] Nederhausen, H, Sullvan, S, 2007, Euler Coeffcents and Restrcted Dyck Paths Congr Numer 188, (arxv: [5] Nederhausen, H and Sullvan, S, 2007, Pattern Avodng Ballot Paths and Fnte Operator Calculus To appear n J Statstcal Inference and Computng (arxv: [6] Nederhausen, H, 2003, Rota s umbral calculus and recursons Algebra Unvers 49, [7] Rota, G-C, Kahaner, D, Odlyzko, A, 1973, On the Foundatons of Combnatoral Theory VIII: Fnte operator calculus J Math Anal Appl 42, [8] Sapounaks, A, Tasoulas, I, Tskouras, P, 2007, Countng strngs n Dyck paths Dscrete Math 307, [9] Gubas, L J, Odlyzko, A M, 1981, Strng overlaps, pattern matchng, and nontranstve games J Comb Theory 30,