A Combinatorial Proof of the Recurrence for Rook Paths

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1 A Combinatorial Proof of the Recurrence for Rook Paths Emma Yu Jin and Markus E. Nebel Department of Computer Science niversity of Kaiserslautern, Germany Submitted: Feb 23, 2012; Accepted: Mar 5, 2012; Published: Mar 19, 2012 Mathematics Subject Classifications: 05A19, 68R05 Abstract Let a n count the number of 2-dimensional rook paths R n,n from (0, 0) to (2n, 0). Rook paths R m,n are the lattice paths from (0, 0) to (m+n, m n) with allowed steps (x, x) and (y, y) where x, y N +. In answer to the open question proposed by M. Erickson et al. (2010), we shall present a combinatorial proof for the recurrence of a n, i.e., (n + 1)a n+1 + 9(n 1)a n 1 = 2(5n + 2)a n with initial conditions a 0 = 1 and a 1 = 2. Furthermore, our proof can be extended to show the recurrence for the number of multiple Dyck paths d n, i.e., (n + 2)d n+1 + 9(n 1)d n 1 = 5(2n + 1)d n with d 0 = 1 and d 1 = 1, where d n = N n (4) and N n (x) is Narayana polynomial. 1 Introduction The set of rook paths R m,n is given by all 2-dimensional lattice paths from (0, 0) to (m + n, m n) with allowed steps (x, x) and (y, y) where x, y N +. They owe their name to the following view: consider a rook placed on the lower left corner (0, 0) of a chess board of size m n. Then R m,n counts the number of ways the rook can reach the opposite corner if in each single step it can move an arbitrary number of squares in the horizontal or vertical direction. Generalizations to d-dimensional chess boards are available [3]. There the number of corresponding paths can also be interpreted as the number of ways a debitor can repay all of its d creditors if the extension of the chess board in dimension i equals the specific amount owing creditor i. The work of this author has been supported by the Alexander von Humboldt Foundation by a postdoctoral research fellowship. Author to whom correspondence should be addressed. the electronic journal of combinatorics 19 (2012), #P59 1

2 We set R m,n = a m,n and R n = R n,n for simplicity. M. Erickson et al. [1] gave the following generating function f(s, t) for the number of rook paths: f(s, t) = m 0,n 0 a m,n s m t n = (1 s)(1 t) 1 2(s + t) + 3st. (1) For the special case a n = a n,n the generating function f(x) = n 0 a nx n can be derived from eq. (1) by first substituting t = x and then finding the coefficient of s s0 using partial fractions and Laurent series [1, 6]. Explicitly, the double generating function f(s, t) is rational and it converges for sufficiently small s and t. As a result, f(x) = [s 0 ]F (s, x) s converges for x small. We shall fix a small x; then the series f(s, t), as a function of s, converges in some circle s = r > 0. According to the Cauchy integral theorem and the Residue theorem, we have f(x) = [s 0 ]f ( s, x ) = 1 s 2πi s =r f ( s, x ) ds s s = S 0 [ ( Res f s, x ) 1 ] s s where S 0 is the set of all the poles of 1f(s, x ) inside the circle s = r and for any pole s s s S 0, lim x 0 s = 0. Therefore we arrive at f(x) = S 0 Res = 1 2πi s =r [ ( f s, x ) 1 s s ] (1 s)(1 x) s ds 1 2(s + x ) + 3x s. s The only pole of 1f(s, x) approaching 0 as x 0 is at s s s 0 = 3x x2 10x + 1. Consequently, [ (1 s)(1 x f(x) = Res ) s s=s0 1 2(s + x) + 3x 1 ] ( + 1 s 2 = 1 ) 1 x 1 + (2) 2 1 9x s by taking the initial condition f(0) = 1 into account. The recurrence relation for a n = [x n ]f(x) can be derived from eq. (2)[1], i.e., a 0 = 1, a 1 = 2, a n = n + 1 2(5n + 2) a 9(n 1) n+1 + 2(5n + 2) a n 1. (3) The set of multiple Dyck paths MD n is given by the 2-dimensional lattice paths from (0, 0) to (2n, 0) with allowed steps (x, x) and (y, y) where x, y N + that never go below x-axis. MD n = d n. Similarly the generating function of d n is given by g(x) = d n x n = 1 ( x 1 ) (1 x)(1 9x), (4) x n 0 the electronic journal of combinatorics 19 (2012), #P59 2

3 and the recurrence for d n is d 0 = 1, d 1 = 1, d n = n + 2 5(2n + 1) d 9(n 1) n+1 + 5(2n + 1) d n 1. (5) In the next section we shall present the combinatorial proof of eq. (3). Eq. (5) follows analogously. Part of the proof employs the bijective proof of recurrence for Dyck paths [2, 5, 7]. Furthermore, we refer the readers to the related combinatorial proof of some famous numbers, like the recurrence for the Delannoy numbers was proved bijectively by P. Peart et al.[4], the recurrence for the large Schröder numbers was shown by Foata et al.[2], and R.A. Sulanke [7] independently. 2 The Main Proof We can represent each rook path by the combination of (i.e., up-step) and D (i.e., down-step) with multiplicities. Accordingly, let i = (resp. D }{{} i = D D) be a }{{} i-times i-times single up-step (resp. down-step) of length i (the underline is used to represent the step s atomicity). For our proof we will count the occurrence of, D, D and DD ignoring the underlines. Let S 1 S 2 (S 1 S 2 ) denote the steps S 1 S 2 above x-axis (resp. below x-axis) where S i {, D} for i = 1, 2 subject to the condition that the y-the coordinates of both, the start point and the end point of S 1, are non-negative (resp. non-positive). Let m 11 count the occurrence of and DD, m 22 count the occurrence of DD and, m 12 count the occurrence of D and D, and m 21 count the occurrence of D and D. For example, m 11, m 12, m 21, m 22 of the rook path DDDDDD is 4, 3, 0, 4. i,j=1,2 m ij = 2n 1 holds for each path M R n. Here is the sketch of the idea of the combinatorial proof. (n+1)a n+1 counts the family of rook paths R n+1 having one unit -step labeled with z, denoted by ( z, R n+1 ). In Lemma 1 we focus on the map from ( z, R n+1 ) to (s x, R n, {a, b}) with multiplicity on each patten x, x and D x D, D x D, and x D, D x. In Lemma 2, 3(n 1)a n 1 counts the family of ( z, R n 1, [i, j]) where [i, j] = [1, 2], [2, 1], [2, 2] and having one unit -step labeled with z, accordingly we present map from ( z, R n 1, [i, j]) where [i, j] = [1, 2], [2, 1], [2, 2] to ( x, R n, {s, t}) with multiplicity on each pattern s x and s x D for s = or D. Lemma 1. For each M R n, let m 11 count the occurrence of and DD, m 22 count the occurrence of DD and, m 12 count the occurrence of D and D, and m 21 count the occurrence of D and D. Then for a n+1 = R n+1 we have (n + 1)a n+1 = M R n 5(m 11 + m 21 ) + 8m m (6) the electronic journal of combinatorics 19 (2012), #P59 3

4 Proof. Let F n be the family of free Dyck paths, i.e., the lattice paths from (0, 0) to (2n, 0) with allowed steps (1, 1) and (1, 1), and pair ( z, F n+1 ) be the set of paths F n+1 having one of the unit up-steps labeled with z. Tuple (s x, F n, {a, b}) represents the set of paths F n having one of the unit steps s ( or D) labeled with x, or having nothing labeled. We set f i (F n ) = f(s x, F n, {i}) for i = a, b. In the sequel we will construct a bijection f : (s x, F n, {a, b}) ( z, F n+1 ). Case 1: (s x, F n, {a}): For the paths having labeling as P 1 = s x F n, we set f a (P 1 ) = s z D F n+1 if s is above x-axis, otherwise f a (P 1 ) = sd z F n+1. For the paths without labeling as P 2 we have f a (P 2 ) = z DP 2. x-axis, a: x D x x-axis, a: z f a D z D D f a f a D x D x f a D z D z For step s, we define the level of the step s, denoted by l 1 (s) as the y-th coordinates of its final point. Case 2: (s x, F n, {b}) and l 1 (s x ) 0: For an arbitrary path P 1 = s x (s x, F n ) and l 1 (s x ) > 0, let s be the first -step preceding s for which l 1 ( s ) = l 1 (s) and let D s be the first D-step that follows s such that l 1 (D s ) = l 1 (s) 1. Note that s = s if s is a -step. As a result, we can write P 1 = s s x RD s where R is the Dyck path between s x and D s. Then f b (P 1 ) = z s srdd s F n+1. x f b z D D D x D f b D z D D Similarly, for the paths P 2 = s x F n and l 1 (s) < 0, let s be the first -step after s for which l 1 ( s ) = l 1 (s) + 1 and let D s be the first D-step preceding s such the electronic journal of combinatorics 19 (2012), #P59 4

5 that l 1 (D s ) = l 1 (s). Note that D s = s if s is a D-step. As a result, we can write P 2 = D s s x R s where R is the Dyck path between s x and s. Then f b (P 2 ) = D s sdr z s F n+1. x-axis: D x f b D x f b D D z D z Case 3: (, F n, {b}) or (s x, F n, {b}) and l 1 (s) = 0: In the former case, let P 1 F n, f b (P 1 ) = D z P 1. In the later case, if P 2 = D x then f b (P 2 ) = DD z holds, if P 3 = x then f b (P 3 ) = z D. We omit the proof that the map f is bijective since that proof follows similarly to that for Dyck paths [2, 5, 7]. The next step we shall extend f 1 to the map g : ( z, R n+1 ) (s x, R n, {a, b}) by allowing the atomic up step and down step to be of arbitrary length, according to which g is surjective but not injective. We start from studying the number of labeled paths in ( z, R n+1 ) induced by f(s x, R n, {a, b}) via integrating the newly added 2-steps to the paths in R n. Case 1: (, R n, {a, b}): W.l.o.g. we assume path p 1 = m p 0 R n where the first step of p 1 is -step of length m (m 1). Then f a (p 1 ) = z D m p 0 and f b (p 1 ) = D z m p 0. D z m p 0 ( z, R n+1 ) is obtained by gluing z and m (m 1), therefore we set g( z, z D m p 0 ) = (, p 1, {a}), g( z, D z m p 0 ) = g( z, D z m p 0 ) = (, p 1, {b}). For each image (, p 1 ) (, R n ) by map g, there are three preimages in ( z, R n+1 ) such that the first unit -step is labeled with z, which has cardinality 3a n. Case 2: (s x, R n, {a, b}) and s is the last unit step: W.l.o.g. we assume path p 2 = D x. Then f a (p 2 ) = D z D and f b (p 2 ) = DD z. DD z ( z, R n+1 ) is obtained by gluing two D-steps, therefore we set g( z, D z D) = (D x, p 2, {a}), g( z, DD z ) = g( z, DD z ) = (D x, p 2, {b}). For each image (s x, p 2 ) (s x, R n ) by map g, there are three preimages in ( z, R n+1 ) such that the last unit -step is labeled with z, which has cardinality 3a n. In what follows we consider all the possible scenarios for s x Y P where Y is the step immediately following s x for an arbitrary but fixed path p R n. the electronic journal of combinatorics 19 (2012), #P59 5

6 Let represent xor, i.e., the two exclusive cases of either an atomic up-step of length two or of two up-steps which for sure belong to two different atomic up-steps. For a multiple Dyck path A that has a t-step of length m 1 as its first atomic step and s-step of length m 2 as its last atomic step, i.e., A = t m 1 A s m 2 where t =, s = D or t = D, s = and m 1, m 2 1, we set tas = t m 1+1 A s m 2 s tas = t m 1+1 A s m 2+1 tas = tt m 1 A s m 2+1 tas = tt m 1 A s m 2 s ta s = tas xor tas t As = tas xor tas Case 3: x : Let p 3 = x, then f a (p 3 ) = z D and recall that D x is the step that after x for which l 1 ( x ) = l 1 (D x) + 1. Let A x 1 (resp. A 1 ) be the sub multiple Dyck paths from x to D x with (resp. without) labeling x, then we can assume p 3 = A x 1, and therefore f b (p 3 ) = z A 1 D. By setting g( z, z D ) = ( x, p 3, {a}) xor g( z, z D ) = ( x, p 3, {a}), g( z, z A 1 D ) = g( z, z A 1 D ) = ( x, p 3, {b}), g( z, z A 1 D ) = g( z, z A 1 D ) = ( x, p 3, {b}), all the five preimages in ( z, R n+1 ) are mapped by g to ( x, R n, {a, b}). We name the active point to be the point that connects two S-steps (S = or D) as SS or SS, Furthermore, the hollow circled dot represents the point that connects one of the patterns from {SS, SS} exclusively. The figure below shows the images ( x, p 3, {a, b}) representing the active points by big black dots. x a g z b g D x D x A x z A 1 1 Case 4: x D : Let p 4 = x D, then f a (p 4 ) = z DD and f b (p 4 ) = the electronic journal of combinatorics 19 (2012), #P59 6

7 z DD holds. Accordingly we set g( z, z DD ) = g( z, z DD ) = ( x D, p 4, {a}), g( z, z DD ) = g( z, z DD ) = ( x D, p 4, {a}), g( z, z DD ) = g( z, z DD ) = ( x D, p 4, {b}), g( z, z DD ) = g( z, z DD ) = ( x D, p 4, {b}). Every eight preimages in ( z, R n+1 ) are mapped by g to one image ( x D, p 4, {a, b}). z a x g b x g z Case 5: D x : First we discuss the subcase in which D x touches the x-axis, i.e., l 1 (D x ) = 0, and is followed by -step. Let p 5 = D x, then f a (p 5 ) = D z D and f b (p 5 ) = DD z hold. Thus, we set g( z, D z D ) = (D x, p 5, {a}), g( z, DD z ) = g( z, DD z ) = (D x, p 5, {b}), g( z, DD z ) = g( z, DD z ) = (D x, p 5, {b}). Next we study the subcase in which D x is above x-axis. Let p 5 = D x, then f a (p 5 ) = D z D holds. Let D D x be the first step after D x such that l 1 (D x ) = l 1 (D D x) + 1. Let D x be the first step preceding D x such that l 1 ( D x) = l 1 (D x ). Then we assume A x 2 (resp. A z 2) is the path from D x to D x with labeling D x (resp. D z = z ) x and A 3 is the path from the end of D x to D D x. Therefore we can interpret p 5 as p 5 = A x 2A 3, and f b (p 5 ) = A z 2A 3 D. Thus, we set g( z, D z D ) = (D x, p 5, {a}), g( z, A z 2A 3 D ) = g( z, A z 2A 3 D ) = (D x, p 5, {b}), g( z, A z 2A 3 D ) = g( z, A z 2A 3 D ) = (D x, p 5, {b}). In both cases, every five preimages in ( z, R n+1 ) are mapped by g to one image in (D x, R n, {a, b}). D x a D x b A x 2 A 3 g g D z D A 3 D z D D x A z 2 the electronic journal of combinatorics 19 (2012), #P59 7

8 Case 6: D x D : First we discuss the subcase in which D x touches the x-axis, i.e., l 1 (D x ) = 0, and is followed by a D-step. Let p 5 = D x D, then f a (p 5 ) = D z DD and f b (p 5 ) = DD z D. We set g( z, D z DD ) = (D x D, p 5, {a}) g( z, D z DD ) = (D x D, p 5, {a}), g( z, DD z D ) = (D x D, p 5, {b}) g( z, DD z D ) = (D x D, p 5, {b}). xor xor Next we study the subcase that D x is above x-axis. For p 6 = D x D, f a (p 6 ) = D z D D holds. Let D x be the first step preceding D x such that l 1 ( D x) = l 1 (D x ). Then we follow the assumption of A x 2 and A z 2 in Case 5, and represent p 6 as p 6 = A x 2 D, therefore f b (p 6 ) = A z 2D D. Consequently, by setting g( z, D z DD ) = (D x D, p 6, {a}) g( z, D z DD ) = (D x D, p 6, {a}), g( z, A z 2DD ) = (D x D, p 6, {b}) g( z, A z 2DD ) = (D x D, p 6, {b}), xor xor every two preimages in ( z, R n+1 ) are mapped by g to one image in (D x D, R n, {a, b}). D x a g z D x b g z For those labeled step s x Y, the case discussion follows similarly. To sum up, for each path p R n, the number of paths in ( z, R n+1 ) resulted from p is 5(m 11 + m 21 ) + 8m m Hence eq. (6) follows and the proof is complete. Lemma 2. nder the assumption of Lemma 1 we have Proof. We shall construct the map M R n [(m 11 + m 21 ) + 2m 12 ] + 3(n 1)a n 1 = 2na n (7) µ : ( x D x, R n ) ( x D, R n, {a, b}) (D x D x D, R n ) (D x, R n, {a, b}) ( x, R n 1, {[1, 2], [2, 1], [2, 2]}) ( z, R n, {s, t}). Note that (D x D x D, R n ) and (D x, R n, {a, b}) are the mirror images of ( x D x, R n ) and ( x D, R n, {a, b}) under reflection K : y = 0, i.e., for a ( x D x, R n ), then K(a) (D x D x D, R n ) and g(k(a)) = K(g(a)), for which we omit the case discussion for (D x D x D, R n ) and (D x, R n, {a, b}). the electronic journal of combinatorics 19 (2012), #P59 8

9 Case 1: For w 1 ( x, R n ) or w 1 ( x D, R n, {a, b}), µ simply replaces the labeling x with z, i.e., µ( x, w 1 ) = ( z, w 1, s) and µ( x D, w 1, a) = ( z D, w 1, s), µ( x D, w 1, b) = ( z D, w 1, t). Case 2: w 2 (D x, R n ): Let x be the first -step preceding D x such that l 1 ( x ) = l 1 (D x ) + 1. Similarly, let D be the first D-step that follows and for which l 1 () = l 1 (D ) + 1 holds. We assume A x 4 (resp. A 4 ) is the path from x to D x with (resp. without) labeling x, and A 5 is the path from the end of to D. Then we consider path w 2 = A x 4 A 5, µ(w 2 ) = ( z, z A 4 A 5, t). We observe that in the case A 5 D, µ is a one-to-one correspondence. In the case A 5 = D, three patterns are missing from ( z, R n, t), i.e., ( z A 4 A 5, t), ( z A 4 A 5, t) and ( z A 4 A 5, t). Case 3 will make up for this. A µ D B A D x B z D Case 3: w 3 ( x, R n 1, {[1, 2], [2, 1], [2, 2]}): If x R n 1 (thus w 3 = x A D in which A is a multiple Dyck path (can be )), then we set µ(w 3, [2, 1]) = ( z, z A DD, t) µ(w 3, [1, 2]) = ( z, z A DD, t) µ(w 3, [2, 2]) = ( z, z A DD, t), which make up three other possible patterns in Case 2. It remains to prove the map µ is bijective. Note the union of the image set for these 3 cases is exactly the set ( z, R n, {s, t}) and µ is injective by construction. Consequently, eq. (7) holds. Theorem 3. Let a n denote the number of 2-dimensional rook paths from (0, 0) to (2n, 0). Then the recurrence of a n is given by a 0 = 1, a 1 = 2, a n = n + 1 2(5n + 2) a 9(n 1) n+1 + 2(5n + 2) a n 1. (8) Proof. In combination of Lemma 1 and Lemma 2, we obtain (n + 1)a n+1 + 9(n 1)a n 1 = 2(2n 1) = 2(5n + 2)a n, with initial condition a 0 = 1 and a 1 = 2. M R n [(m 11 + m 21 ) + 2m 12 ] + 9(n 1)a n 1 the electronic journal of combinatorics 19 (2012), #P59 9

10 The combinatorial proof for the recurrence of rook paths can also be extended to the combinatorial proof for the recurrence of multiple Dyck paths. Eq. (5) follows according to the lemmas below. Lemma 4. For each M D n, let n 11, n 12, n 21, n 22 count the occurrence of, D, D and DD. Then for d n+1 = D n+1 we have (n + 2)d n+1 = M MD n 5(n 11 + n 21 ) + 8n n 22. (9) Lemma 5. nder the assumptions of Lemma 4 we have References M MD n 2n (n 1)d n 1 = (n + 1)d n. (10) [1] M. Erickson, S. Fernando and K. Tran. Enumerating rook and Queen paths. Bull. Inst. Combin. Appl., 60:37-48, [2] D. Foata and D. Zeilberger. A classic proof of a recurrence for a very classical sequence. J. Comb. Theor., Ser. A., 80: , [3] M. Kauers and D. Zeilberger. The computational challenge of enumerating highdimensional rook walks, arxiv: v1 [math.co] 21 Nov [4] P. Peart and W. J. Woan. A bijective proof of the delannoy recurrence. Proceeding of the Thirty-third Southeastern International Conference on Combinatorics, Graph Theory and Computing (Boca Raton, FL), 158: Congressus Numerantium, [5] J-L. Rémy. n procédé itératif de dénombrement d arbres binaires et son application à leur génération aléatpoire. RAIRO Inform. Théor., 19: , [6] R.P. Stanley. Enumerative Combinatorics (Vol.2). Cambridge niversity Press, New York, [7] R.A. Sulanke. Bijective Recurrences concerning Schröder paths. Electron. J. Combin., 5:#R47, the electronic journal of combinatorics 19 (2012), #P59 10