...... A note on a -enumeraton of antsymmetrc monotone trangles arxv:1410.811v3 [math.co] 18 Jul 015 Tr La Insttute for Mathematcs and ts Applcatons Unversty of Mnnesota Mnneapols, MN 55455 Emal: tmla@ma.umn.edu Webste: http://www.ma.umn.edu/~tmla/ Mathematcs Subject Classfcatons: 05A15, 05C70, 05E99 Abstract In ther unpublshed work, Jockusch and Propp showed that a -enumeraton of antsymmetrc monotone trangles s gven by a smple product formula. On the other hand, the author proved that the same formula counts the domno tlngs of the quartered Aztec rectangle. In ths paper, we explan ths phenomenon drectly by buldng a correspondence between the antsymmetrc monotone trangles and domno tlngs of the quartered Aztec rectangle. Keywords: Aztec damonds, Aztec rectangles, domno tlngs, monotone trangles. 1 Introducton A monotone trangle of order n s a trangular array of ntegers a 1,1 a,1 a, a 3,1 a 3, a 3,3....... a n,1 a n,... a n,n 1 a n,n Ths research was supported n part by the Insttute for Mathematcs and ts Applcatons wth funds provded by the Natonal Scence Foundaton (grant no. DMS-0931945). 1
whose entres are strctly ncreasng along the rows and weakly ncreasng along both rsng and descendng dagonals from left to rght. An antsymmetrc monotone trangle (AMT) s a monotone trangle, whch has a,k = a,+1 k, for any 1 n and 1 k (see examples of AMTs n Fgures.1 and.6). Let q be a number, then the q-weght of an AMT s q w f t has w postve entres, whch do not appear on the row above. Assume that n and 0 < a 1 < a <... < a n, the q-enumeraton A q n (a 1,a,...,a n ) of AMTs s defned as the sum of q-weghts of all AMTs of order n whose postve entres on the bottommost row are a 1,a,...,a n. Snce there s only one AMT of order 1 that conssts of a 0, we set A q 1( ) := 1, for any q. Jockusch and Propp proved a smple product formula for the -enumeraton of the AMTs. Theorem 1.1 (Jockusch and Propp [7]). Assume that k,a 1,a,...,a k are postve ntegers, such that a 1 < a <... < a k. The AMTs wth postve entres a 1,a,...,a k on the bottom are -enumerated by and A k (a 1,a,...,a k ) = A k+1(a 1,a,...,a k ) = k 0!!4!...(k )! k 1!3!5!...(k 1)! 1 <j k 1 <j k (a j a ) (a j a ) 1 <j k 1 j k (a +a j 1) (1.1) (a +a j 1), (1.) where the empty products (lke 1 <j k (a j a ) for k = 1) equal 1 by conventon. We denote by E(a 1,a,...,a k ) and O(a 1,a,...,a k ) the expressons on the rght-hand sdes of (1.1) and (1.), respectvely. The Jockusch-Propp s proof of Theorem 1.1 s algebrac, and no combnatoral proof has been known. Let R be a (fnte, connected) regon on the square lattce. A domno tlng of R s a coverng of R by domnoes so that there are no gaps or overlaps. We use the notaton T(R) for the number of domno tlngs of the regon R. The Aztec damond of order n s the unon of all unt squares nsde the contour x + y = n+1. The Aztec damond of order 9 s shown n Fgure 1.1. It has been proven that the number of domno tlngs of the Aztec damond of order n s n(n+1). See [4] for the four orgnal proofs; and e.g. [1], [], [5], [6], [8], [9], [13] for further proofs. Jockusch and Propp [7] ntroduced three types of quartered Aztec damonds (denoted by R(n), K a (n), and K na (n)) obtaned by dvdng the Aztec damond of order n nto four parts by two zgzag cuts passng the center as n Fgure 1.1 (for n = 9). They proved that the domno tlngs of the quartered Aztec damond are enumerated by E(a 1,a,...,a k ) or O(a 1,a,...,a k ), where a 1,a,...,a k are the frst k odd postve ntegers or the frst k even postve ntegers. The author gave a new proof for ths result by usng Cucu s Factorzaton Theorem (see [3, Theorem 1.]) n [10].
R(8) K na (8) R(8) R(8) K a (8) K a (8) R(8) K na (8) Fgure 1.1: Three knds of quartered Aztec damonds of order 9. AR 7,10 RE 7,10 (,3,6,9) 9 RO 7,10 (1,3,6) 6 6 3 1 3 (c) TR 7,10 TE 7,10 (3,5,8) 5 8 TO 7,10 (1,3,6,9) 6 9 3 3 (d) (e) (f) Fgure 1.: The Aztec rectangle, trmmed Aztec rectangle, and the four quartered Aztec rectangle of sze 7 10. 1 3
Besdes the Aztec damonds, we are also nterested n a smlar famly of regons called Aztec rectangles (see Fgure 1. for the Aztec rectangle of sze 7 10). Denote by AR m,n the Aztec rectangle regon havng m squares 1 along the southwest sde and n squares along the northwest sde. We also consder the trmmed Aztec rectangle TR m,n obtaned from the ordnary AR m,n by removng all squares runnng along ts northwest and northeast sdes (llustrated n Fgure 1.(d)). We generalze the Jockusch-Propp s quartered Aztec damonds as follows. Remove all squares at even postons (from the bottom to top) on the southwest sde of AR m,n, and remove arbtrarly n m+1 squares on the southeast sdes. Assume that we are removng all squares, except forthe a 1 -st, the a -nd,..., andthe a m+1 -thones, fromthe southeast sde, and denote by RE m,n (a 1,a,...,a m+1 ) the resultng regon (see Fgure 1.). We also have an odd-analog RO m,n (a 1,a,...,a m ) of the above regon when removng odd squares (nstead of the even ones) from the southwest sde, and removng all squares fromthe southeast sde, except for the squares at the postons a 1,a,...,a m (llustrated n Fgure 1.(c)). If we remove all even squares on the southwest sde of the trmmed Aztec rectangle TR m,n, and also remove the squares at the postons a 1,a,...,a m from ts southeast sde, we get the regon TE m,n (a 1,a,...,a m ) (shown n Fgure 1.(e)). Repeatng the process wth the odd squares on the southwest sde removed, we get the regon TO m,n (a 1,a,...,a m+1 ) (pctured n Fgure 1.(f)). We call each of the above four regons a quartered Aztec rectangle (QAR). We notce that the quartered Aztec damonds of order k are obtaned from the RE- and RO-QARs by specalzng m = n = k and a = or 1, dependng on the regon consdered; and the quartered Aztec damonds of order k + 1 are obtaned smlarly from the TE- and TO-QARs havng sze (k +1) (k +1). Surprsngly, the functons E(a 1,a,...,a k ) and O(a 1,a,...,a k ) n Theorem 1.1 also counts the domno tlngs of the QARs. Theorem 1. (La [11]). For any 1 k < n and 1 a 1 < a <... < a k n the domno tlngs of QARs are enumerated by T(RE k 1,n (a 1,a,...,a k )) = T(RE k,n (a 1,a,...,a k )) = E(a 1,a,...,a k ), (1.3) T(RO k,n (a 1,a,...,a k )) = T(RO k+1,n (a 1,a,...,a k )) = O(a 1,a,...,a k ), (1.4) T(TE k,n (a 1,a,...,a k )) = T(TE k+1,n (a 1,a,...,a k )) = k O(a 1,a,...,a k ), (1.5) T(TO k 1,n (a 1,a,...,a k )) = T(TO k,n (a 1,a,...,a k )) = k E(a 1,a,...,a k ). (1.6) Theorems 1.1 and 1. mply that Corollary 1.3. For any 1 k < n and 1 a 1 < a <... < a k n T(RE k 1,n (a 1,a,...,a k )) = T(RE k,n (a 1,a,...,a k )) = A k (a 1,a,...,a k ), (1.7) 1 From now on, we use the word square(s) to mean unt square(s). 4
T(RO k,n (a 1,a,...,a k )) = T(RO k+1,n (a 1,a,...,a k )) = A k+1(a 1,a,...,a k ), (1.8) T(TE k,n (a 1,a,...,a k )) = T(TE k+1,n (a 1,a,...,a k )) = k A k+1 (a 1,a,...,a k ), (1.9) T(TO k 1,n (a 1,a,...,a k )) = T(TO k,n (a 1,a,...,a k )) = k A k (a 1,a,...,a k ). (1.10) In ths paper, we gve a drect combnatoral proof of the denttes (1.7) (1.10) by gvng a correspondence between the AMTs and the domno tlngs of the QARs. The proof extends an dea of Jockusch and Propp for the case of quartered Aztec damonds. On the other hand, our combnatoral proof and Theorem 1. yeld a combnatoral proof for Jockusch-Propp s Theorem 1.1. Fnally, t s worth to notce that Cucu[3, Theore 4.1] showed a related correspondence between the domno tlngs of the Aztec rectangle wth holes (.e. Aztec rectangle wth certan squares removed) and a -enumeraton of the so-called alternatng sgn matrces, whch n turn are n bjecton wth the monotone trangles (see [1]). However, the - enumeraton of alternatng sgn matrces (or monotone trangles) does not mply the -enumeraton of AMTs n Theorem 1.1; and the enumeraton of tlngs of an Aztec rectangle wth holes does not mply drectly the tlng formula of the QARs ether. Drect proof of the denttes (1.7) (1.10) We prove only the equaltes (1.7) and (1.9), as (1.8) and (1.10) can be obtaned by a perfectly analogous manner. In ths proof, we always rotate the QARs by 45 clockwse to help the vsualzaton of our arguments. Frst, we can see that the domnoes on the top of RE k,n (a 1,a,...,a k ) are all forced; and by removng these domnoes, we get the regon RE k 1,n (a 1,a,...,a k ). Ths mples that the two QARs n (1.7) have the same number of domno tlngs. Thus, we only need to show that T(RE k 1,n (a 1,a,...,a k )) = A k(a 1,a,...,a k ). Denote by T(R) the set of all tlngs of a regon R, and A n (a 1,a,...,a n ) the set of all AMTs of order n havng postve entres a 1 < a <... < a n on the bottom. Next, we defne a map Φ : T (RE k 1,n (a 1,a,...,a k )) A k (a 1,a,...,a k ) as follows. Color the regon RE k 1,n (a 1,a,...,a k ) black and whte so that any two squares sharng an edge have dfferent colors, and that the bottommost squares are black. Gven a tlng T of the regon. Fgure.1 shows a sample domno tlng of the QAR (wth k = 4, n = 10, a 1 = 1, a = 3, a 3 = 6, and a 4 = 9). We say a black square s matched upward or matched downward, dependng on whether the whte square covered by the same domno s above or below t. We use the arrows to show n Fgure.1 the domnoes contanng matched-upward black squares. The QAR (rotated by 45 ) can be 5
0 4 0 4 4 4 4 3 0 3 4 9 3 1 1 3 9 9 3 0 3 9 9 6 3 1 1 3 6 9 Fgure.1: A domno tlng of RE 7,10 (,3,6,9) and the correspondng AMT. parttoned nto 4k 1 rows of squares; and we call these rows black or whte dependng on the color of ther squares. We now descrbe the AMT τ := Φ(T). We label all black squares on each row by 1,,...,n from left to rght (we also label all black squares removed on the bottommost row). Denote by B(,j) the black square at the poston j on the -th row (from the top). The postve entres n the -th row of τ are the labels (postons) of the matched-upward squares on the -th black row. By the antsymmetry, τ s completely determned by ts postve entres (see the llustraton n Fgure.1; the postve entres of the AMT are restrcted nsde the dotted rght trangle). We would lke to show that Φ s well defned,.e. we wll verfy that τ = Φ(T) A k (a 1,a,...,a k ). Consder the strp consstng of ( 1) top rows n the QAR, for k. There are ( 1)n black squares and ( 1)n+ whte squares n ths strp. Thus, there are whte squares n the strp, whch are matched wth some black squares outsde the strp. These must be whte squares on the ( 1)-th whte row, and those whte squares are matched wth some black squares on the -th black row. It means that there are matched-upward black squares on the -th black row; equvalently, the -th row of τ has exactly postve entres. Snce each row of τ records the postons of matched-upward squares n a black row, the entres n each row of τ are strctly ncreasng from left to rght. We now want to verfy that the entres n each dagonal are weakly ncreasng from left to rght. Snce τ s antsymmetrc, we only need to prove ths monotoncty for the rght half of τ, whch contans all postve entres. Assume that the -th row of τ has the postve entres 0 < t,1 < t, < < t,l(), where l() =. We only need to show that for any odd, and that for any even. t 1,1 t,1 t 1, t,... t 1,l( 1) t,l() (.1) t,1 t 1,1 t, t 1, t,3... t 1,l( 1) t,l() (.) 6
1 3 4 5 1 3 4 5 6 7 8 9 1 1 t,1 = 5 t, = 9 t,1 = 5 Fgure.: Illustratng the proof of the nequalty (.1). t 1,1 = 4 1 3 4 5 6 7 8 9 t 1,1 = 4 1 3 4 5 6 7 8 9 1 1 t,1 = 9 t,1 = 9 Fgure.3: Illustratng the proof of the nequalty (.1) (cont.). We consder frst the case when s odd. Consder the frst nequalty t 1,1 t,1. We assume otherwse that t 1,1 > t,1, then by defnton, all (black) squares B( 1,1),B( 1,),...,B( 1,t,1 ) are matched downward. However, the square B(,t,1 ) cannot be matched upward, snce all whte neghbor squares above t are already matched wth other black squares, a contradcton (see Fgure. for t,1 = 5; the upper black row s the ( 1)-th one; the matched-downward squares on the -th black row are not shown n the pcture; and the square havng black core ndcates the square that cannot be matched). We now show that Clam.1. t,1 t 1, t, f l(). Proof. There are two cases to dstngush. Case 1. t 1,1 = t,1. Snce t,1 = t 1,1 < t 1,, we only need to show that t 1, t,. Assume otherwse that t 1, > t,, then same stuaton as n the prevous paragraph happens wth the square B(,t, ): all of ts upper whte neghbors are matched wth other black squares, so t cannot be matched upward, a contradcton (see Fgure. for the case t,1 = 5 and t, = 9; the square havng black core ndcates the one cannot be matched). Case. t 1,1 < t,1. One can see that the domnoes contanng the black squares B( 1,t 1,1 +1),B( 1,t 1, +),...,B( 1,t,1 1) are all forced, and these black squares are all matched downward (see Fgures.3 and for the two possblty n the case when t 1,1 = 5 7
4 5 6 7 8 9 4 5 6 7 8 9 1 1 t,1 = 4 t, = 9 t,1 = 4 t, = 9 Fgure.4: Illustratng the proof of the nequalty (.1) (cont.). 0 - -4 0 4-4 - 4-4 -3 0 3 4-9 -3-1 1 3 9-9 -3-0 3 9-9 -6-3 -1 1 3 6 9 Fgure.5: The -blocks n the domno tlngs of RE 7,10 (,3,6,9) correspondng to the elements of S(τ) (the crcled entres on the rght). and t,1 = 9). Thus, the second match-upward square n the ( 1)-th black row must be on the rght of B( 1,t,1 1), ths means that t 1, t,1. Thus, we only need to show that t 1, t,. Agan, assume otherwse that we have the opposte nequalty t 1, > t,. In ths case, the square B(,t, ) cannot be matched upward, a contradcton (see the two possble cases n Fgure.4). By consderng smlarly the stuatons when t 1, = t, or t 1, < t,, we obtan t, t 1,3 t,3 f l() 3. Keep dong ths process, we get the remanng nequaltes n (.1). We now consder the case when s even. The leftmost nequalty n (.), t,1 t 1,1, follows easly from consderng forced domnoes. Smlar to the case of odd, by consderng two cases when t,1 = t 1,1 and when t,1 < t 1,1, we get now the nequalty t 1,1 t, t 1,. Then (.) s obtaned by applyng ths argument repeatedly. In summary, τ = Φ(T) s ndeed an AMT of order k. Moreover, snce all black squares on the bottom of the QAR must be matched upward, the postve entres n the bottommost row of τ are a 1,a,...,a k. Therefore, τ A k (a 1,a,...,a k ), and the map Φ s well defned. Next, we want to compute the cardnalty of the set Φ 1 (τ) for any gven AMT τ A k (a 1,a,...,a k ),.e. we want to fnd out how many dfferent domno tlngs of RE k 1,n (a 1,a,...,a k ) correspondng to τ. Let S(τ) be the set of postve entres of τ 8
whch do not appear n the row above (see the crcled entres n Fgure.5). We assgn to each black square of RE k 1,n (a 1,a,...,a k ) a label U or D, so that only the black squares B(,t,j ) have label U, and all other ones have label D. Atlng T corresponds to τ f and only f all U-squares are matched upward and all D-squares are matched downward n T. Vew the QAR RE k 1,n (a 1,a,...,a k ) as the unon of n + 1 columns of black or whte squares. Imagne that we are droppng domnoes to cover successvely the squares n the frst black column, so that the ones wth label U (resp., D) are matched upward (resp., downward). One can see that all the domnoes are forced, except for those cases when a U-square stays rght below a D-square (ths correspond to a 1 n one row of τ but not n the row above t). At each exceptonal place, we have two ways cover these two black squares by two parallel domnoes, whch create a block (see the - blocks restrcted by the bold contours n Fgure.5). If we now drop domnoes to cover the squares n the second black column, all domnoes are forced, except for some -block correspondng to a that appears n some row of τ but not n the precedng row. Contnung n ths way, one can see that the whole tlng s forced, except for certan -blocks correspondng to the elements of S(τ). Snce each block can be covered n two ways, the number of tlngs T correspondng to τ s S(τ). In partcular, Φ s surjectve; and we obtan T(RE k 1,n (a 1,a,...,a k )) = Φ 1 (τ) (.3) whch mples (1.7). = τ A k (a 1,...,a k ) τ A k (a 1,...,a k ) S(τ) (.4) = A k (a 1,...,a k ), (.5) Next, we prove the equalty (1.9). Smlar to (1.7), the frst equalty follows from the fact that TE k,n (a 1,a,...,a k ) s obtaned by removng forced domnoes on the top of TE k+1,n (a 1,a,...,a k ). Thus, we only need to show that We construct a map T(TE k,n (a 1,a,...,a k )) = k A k+1 (a 1,a,...,a k ). Ψ : T (TE k,n (a 1,a,...,a k )) A k+1 (a 1,a,...,a k ) smlar to Φ n the proof of (1.7), the only dfference s that the -th row of the AMT Ψ(T) records the postons of the squares on the -th black row, whch are not matched upward (ncludng the squares removed on the bottom of TE k,n (a 1,a,...,a k )). Fgure.6 llustrates the map Ψ for k = 3, n = 10, a 1 = 3, a = 5, and a 3 = 8. Smlar to the equalty (1.7), one can verfy that Ψ(T) s ndeed an AMT n A k+1 (a 1,a,...,a k ),.e. Ψ s well defned. 9
0-7 7-8 0 8-8 -5 5 8-8 -5 0 5 8-8 -5-1 1 5 8-8 -5-3 0 3 5 8 Fgure.6: A domno tlngs of TE 6,10 (3,5,8) and the correspondng AMT. Smlarly, gven an AMT τ A k+1 (a 1,a,...,a k ). By dong the same domnodroppng process, we get that there are now V(τ) tlngs of TE k,n (a 1,a,...,a k ) correspondng to τ, where V(τ) s the set of postve entres of τ, whch do not appear n the row below. It means that Ψ 1 (τ) = V(τ), and Ψ s surjectve. We now compare the cardnaltes of two sets S := S(τ) and V := V(τ). Denote by r the -th row of τ, and V := V(τ) r and S := S(τ) r. Partton τ nto two sets τ E, the set of all entres on the even rows, and τ O, the set of all entres on the odd rows. We wll compare the numbers of elements of V and S n each of these subsets. Removethepostveentres, whchappearnbothrowsr 1 andr. Bythedefnton, the remanng postve entres on r 1 are n V 1 and the remanng postve ones on r are n S. Snce r has one postve entry more than r 1 (snce each row r j has l(j) postve entres, and l() = = l( 1)+1), we have S = V 1 +1, for = 1,,...,k. Addng up all latter equaltes for = 1,,...,k, we have S τ E = V τ O +k. Smlarly, we have V = S +1, for = 1,,...,k, so S τ O = V τ E. Ths mples that By (.6), we have V = V τ E + V τ O = S τ O + S τ E k = S k. (.6) T(TE k,n (a 1,a,...,a k )) = Then (1.9) follows, and ths fnshes our proof. = τ A k+1 (a 1,...,a k ) τ A k+1 (a 1,...,a k ) Ψ 1 (τ) (.7) S(τ) k (.8) = k A k+1 (a 1,...,a k ). (.9) References [1] F. Boso and M. A. A. van Leeuwen, A bjecton provng the Aztec damond theorem by combng lattce paths, Electron. J. Combn. 0(4) (004), P4. 10
[] R. Bruald and S. Krkland, Aztec damonds and dgraphs, and Hankel determnants of Schröder numbers, J. Combn. Theory Ser. B 94() (005), 334 351. [3] M. Cucu, Enumeraton of perfect matchngs n graphs wth reflectve symmetry, J. Combn. Theory Ser. A 77 (1997), 67 97. [4] N. Elkes, G. Kuperberg, M.Larsen, and J. Propp, Alternatng-sgn matrces and domno tlngs, J. Algebrac Combn. 1 (199), 111 13, 19 34. [5] S.-P. Eu and T.-S. Fu, A smple proof of the Aztec damond theorem, Electron. J. Combn. 1 (005), R18. [6] M. Fendler and D. Greser, A new smple proof of the Aztec damond theorem (014). Preprnt arxv:1410.5590. [7] W. Jockusch and J. Propp, Antsymmetrc monotone trangles and domno tlngs of quartered Aztec damonds, Unpublshed work. [8] S. Kamoka, Laurent borthogonal polynomals, q-narayana polynomals and domno tlngs of the Aztec damonds, J. Combn. Theory Ser. A 13 (014), 14 9. [9] E. H. Kuo, Applcatons of graphcal condensaton for enumeratng matchngs and tlngs, Theor. Comput. Sc. 319 (004), 9 57. [10] T. La, A smple proof for the number of tlngs of quartered Aztec damonds, Electron. J. Combn. 1(1) (014), P1.6. [11] T. La, Enumeraton of tlngs of quartered Aztec rectangles, Electron. J. Combn. 1(4) (014), P4.46. [1] W. Mlls, D. Robbns, and H. Rumsey, Alternatng sgn matrces and descendng plane parttons, J. Combn. Theory Ser. A 34 (1983), 340 359. [13] J. Propp, Many faces of alternatng-sgn matrces, Dscrete Math. and Theoret. Comput. Sc., Specal ssue: Dscrete Models: Combnatorcs, Computaton, and Geometry, 001. [14] J. Propp, Generalzed domno-shufflng, Theor. Comput. Sc. 303 (003), 67 301. 11