A combnatoral problem assocated wth nonograms Jessca Benton Ron Snow Nolan Wallach March 21, 2005 1 Introducton. Ths work was motvated by a queston posed by the second named author to the frst named author about a game that goes by many names but we wll refer to t here as the nonogram game. We frst descrbe a nonogram. The startng pont s an m n board wth all squares whte. One puts black squares n a selecton of postons on the board for example: We have a pcture n a 5 5 board. Now one looks at each row and puts together a sequence of postve ntegers that gves a lst of the numbers of contguous black squares and one gets a sequence of m lsts. One does the same for the columns gettng n lsts. The nonogram s ths par of lsts of lsts. Thus the nonogram assocated wth the above pcture s [[1, 1], [3], [1, 1], [2, 2], [3]], [[1], [5], [1, 1], [5], [1]]. The puzzle s to be gven a nonogram and to construct a pcture that yelds t. Thus the pcture above s a soluton to the correspondng nonogram. One can see easly that ths nonogram has a unque soluton (.e. pcture). We note that the two pctures below are solutons to [[1, 1], [1], [1, 1], [1, 1], [1, 1]], [[1, 1], [2], [1], [2], [1, 1]]. 1
There are also nonograms that are not related to pctures. For example, [[5], [1, 1, 1], [1], [1], [1]], [[1, 2], [5], [1], [1], [1]]. No good algorthms have been found to determne f a nonogram corresponds to a pcture and, f so, fnd a pcture. In fact, ths leads to an NP-complete problem. Havng descrbed the actual puzzle let us descrbe the queston. We smplfy the nonogram and replace the arrays of arrays wth row sums and column sums where we thnk of the orgnal pcture as an m n matrx wth entres consstng of 0 or 1. Thus the frst pcture above yelds (2, 3, 2, 4, 3) for the rows and (1, 5, 2, 5, 1) for the columns. The second par gve (2, 1, 2, 2, 2), (2, 2, 1, 2, 2). The queston s how many possble pars of row sums and column sums are there for m n matrces wth entres consstng only of 0 or 1? Is there a method of fndng all such possbltes? For example the mpossble nonogram above corresponds to (5, 3, 1, 1, 1), (3, 5, 1, 1, 1) whch s not even possble as a row and column sum of such a 5 5 matrx. The second named author found that for 1 1, 2 2, 3 3, 4 4 the number of such s respectvely, 2, 15, 328, 16145 In ths paper we gve a method of answerng both questons. It s very ntrgung that ths seemngly nnocent queston led us to look at farly deep aspects of the combnatorcs of the symmetrc group and a further property of Young s rasng operators (actually we do lowerng) and Schur functons. We also develop a q analogue of the queston and a conjecture about dvsblty by the q-analogue of n + 1, for the case of n n matrces (notce that the numbers above are respectvely dvsble by 2, 3, 4, 5). D. Goldsten and R. Stong [GS] have proved recurson formula for ths q analogue and n partcular gve a relatvely fast recurson to count the possble pars and a proof of the conjecture. Goldsten and Stong have ponted out to the authors that the man theorem n ths paper, Theorem 6, can be found n the standard lterature (cf. [BR] and [vlw]). Our method of proof s dfferent and t yelds an algorthm for constructng the pertnent matrces. 2 Row and column sum. We denote by B m,n the set of all m n matrces wth entres n the set {0, 1}. If M B m,n then we wrte M = [m j ]. We set x (M) = j m j for = 1,..., m and set y j (M) = m j for j = 1,..., n. We put x(m) = (x 1 (M), x 2 (M),..., x m (M)) and y(m) = (y 1 (M), y 2 (M),..., y n (M)). In what follows we the notaton wll not be consstent wth rght and left actons of groups. The lemma below should clarfy the nconsstences. If σ S m and f M = [m j ] s an m n matrx then we set σm = [m σ,j ] and f σ S n then we set Mσ = [m,σj ]. If v = (v 1,..., v n ) then we set vσ = (v σ1,..., v σn ) for σ S n. The followng result s proved by the obvous calculaton. 2
Lemma 1 Wth these notatons n place we have x(σm) = x(m)σ, y(σm) = y(m) and x(mσ) = x(m), y(mσ) = y(m)σ. We set RC(m, n) = {(x, y) x = x(m), y = y(m), M B m,n }. We are nterested n calculatng ths functon. We set RC(m, n) = {(x, y) x = x(m), y = y(m), M B m,n }. Clearly, RC(m, n) = RC(m, n). We wll now gve a prelmnary descrpton of RC(m, n). We say that an element x N n s domnant f x x +1 for = 1,..., n 1. If x N n then there exsts a unque domnant element of the form xσ wth σ S n. Set RC + (m, n) = {(x, y) RC(m, n) x, y domnant }. If x N n s domnant then we set orb n (x) = {xσ σ S n }. For such x we defne l 1,..., l p > 0 wth l 1 +... + l p = n and x 1 =... = x l1, x l1+1 =... = x l1+l 2, x l1+l 2+1 =... = x l1+l 2+l 3,... Then orb n (x) = n! l. We set λ 1! l p! n(x) = (l 1,..., l p ). If α N p then we set α! = α 1! α p!. Lemma 2 We have RC(m, n) = (x,y) RC +(m,n) orb m (x) orb n (y) = m!n! (x,y) RC +(m,n) 1 λ m (x)!λ n (y)!. Our problem s thus to determne the elements of RC + (m, n). If x = (x 1,..., x m ) s domnant and x 1 n then we can defne M B m,n by m 1 = 1 for = 1,..., x 1, m 2 = 1 for = 1,..., x 2, etc. and all other entres 0. Then x = x(m) and we set µ(x) = y(m). In the theory of parttons µ(x) s the dual partton of x possbly expanded to have n rows by ncludng 0 rows. We note that (x, µ(x)) RC + (m, n). If x s domnant and x 1 n then we set Y (x) = {y (x, y) RC + (m, n)}. Thus µ(x) Y (x). We now defne two orders on N n. (N ={0, 1, 2,...}). The frst s the lexcographc order that s x > y f x = y for < j and x j > y j. The other s the root order (or domnance order) whch s only a partal order that s x y f (x y ) 0 for all j = 1,..., n and at least one of these sums s postve. If j 0 r mn we defne ν r = ν r,m,n a domnant element of N n as follows: Let c 0 be defned by (c 1)n < r cn. Then snce r mn we see that 0 c m. Defne (ν r ) = c for = 1,..., r (c 1)n, and (ν r ) = c 1 for > r (c 1)n. Notce that f r = 0 then c = 0 and r (c 1)n = n so ν 0 = (0,..., 0). If 0 < r n then c = 1 and ν r = (1, 1,..., 1, 0,..., 0) wth r ones. If r > n then c > 1 we have ν r s domnant and (ν r ) = c(r (c 1)n) + (c 1)(n r + (c 1)n) = c(r (c 1)n) + (c 1)(cn r) = r. 3
In general f x N n then we set x = x 1 +... + x n. Lemma 3 Let 0 r mn and let P m,n (r) denote the set of all x domnant wth x m and x = r. Then ν r,m,n s the unque mnmal element n P m,n (r) relatve to both the lexcographc and the root order. Proof. Let x P m,n (r) and suppose that x 1 < c then x 1 c 1 thus x (c 1)n < r. Set k = r (c 1)n. Then f x = c for < j k and x j < c then the same argument shows that x < r. Thus we must have x = c for = 1,..., k. Now assume that x = r and x ν n the lexcographc order. Thus we have x = c for = 1,..., k and x k+1 c 1. If x = c 1 for = k + 1,..., k + l 1 but x k+l < c 1 then x < r. Thus the asserton about the lexcographc order follows. We wll now prove the asserton about the root order. We frst observe that f x, y N n and f x y and f x = y then x = y + <j a j (e e j ) wth e the usual vector wth a one n the -th poston and all the other entres 0 wth each a j a non-negatve nteger and some a j > 0. Assume that a j = 0 for < o and a = a oj > 0. Then x = y for < o and x o = y o + a > y o. j Thus f x y then x > y. Ths mples that ν r s a mnmal element relatve to the root order. We wll now show that t s the only one. We note that λ(ν r ) = (r (c 1)n, cn r) (f c = 0 then r = 0 and we should nterpret ths as only havng one entry, smlarly for r = un so c = u). If x P m,n (r) and f x x +1 2 then x e +e +1 P m,n (r). So x cannot be mnmal n P m,n (r). Thus f x s mnmal then x x +1 1. Suppose now that λ(x) = (l 1,..., l p ) wth p 3. Then x e l1 + e l1+l 2+1 P m,n (r). Thus f x s mnmal wth respect to the root order then p = 1 or 2. If p = 1 then x = (u,..., u) and so r = un and ν r = x. If p = 2 then f x were mnmal then x 1 = u and x l1+1 = u 1. Thus we have and ul 1 + (u 1)l 2 = r l 1 + l 2 = n. Hence l 2 = un r. Snce l 2 > 0 we see that u c. If u > c then l 1 = n l 2 = n un + r = r (u 1)n < 0. Thus u = c so x = ν r. The technque n the proof of the precedng lemma suggests some operatons on the elements of RC + (m, n) whch we wll make precse n the next secton. 3 Some operatons on domnant elements If x P m,n (r) recall that Y (x) = {y P m,n (r) (x, y) RC + (m, n)}. In ths secton we study two operatons on Y (x) that decrease elements n the root order. 4
Move 1. If y Y (x) and y y +1 > 1 then y e + e +1 Y (x). Indeed, let M B m,n be such that x = x(m) and y = y(m). Suppose that m k = 1 mples that m k+1 = 1 for all k = 1,..., n. Then y y +1. Snce we have assumed the contrary, there exsts k so that m k = 1 and m k+1 = 0. Thus f M = [m rs] wth m rs = m rs for (r, s) / {(k, ), (k, + 1)} and m k = 0, m k+1 = 1 then x(m ) = x and y(m ) = y e + e +1. Snce y y +1 + 2, y e + e +1 s domnant. Move 2. If y Y (x) and f y > y +1 y +2... y +k > y +k+1 then y e + e +k+1 Y (x). Indeed, let M B m,n be such that x = x(m) and y = y(m). Argung as n the justfcaton of Move 1 we see that there exsts 1 l n wth m l = 1 and m l,+k+1 = 0. Defne M as above to have all entres but the ones n the l, and the l, +k+1 postons the same as those of M but wth the two ndcated values nterchanged. Then as above x = x(m ) and y e + e +k+1 = y(m ) Y (x). Lemma 4 Let 0 r mn and let x P m (r). Then µ(x) s the maxmum element of Y (x) wth respect to the lexcographc order and t s the unque maxmal element of Y (x) wth respect to the root order. Also ν m,n,r s the mnmal element n Y (x) wth respect to the lexcographc order and the unque mnmal element n Y (x) wth respect to the root order. Proof. Let z = µ(x) and let y Y (x). Let M B n be such that x(m) = x and y(m) = y. Then we note that the number of j wth m j,1 = 1 s equal to y 1 and s less than the number of j such that x j 0. Thus y 1 z 1 and f y 1 = z 1 then m j1 = 1 precsely f x j 0. We show by nducton that f y = z for k 1 then y k z k and f y k = z k then m jk = 1 precsely when x j k. We have proved ths for k = 1. Assume for k l and we wll now prove t for k = l + 1. Suppose that y k > z k. Then then the number of j such that m jk = 1 s larger than the number of l such that x l k. Thus there exsts l wth x l k 1 and m l,k = 1. The nductve hypothess mples that m ls = 1 for s = 1,..., x l. But then f m l,k = 1 we would have x l > x l. Ths contradcton shows that y k z k and that m jk = 1 mples that x j k. We have observed that of α β then α > β. Ths shows that z = µ(x) s maxmal n Y (x) n the root order. Set Ỹ (x) = {y (x, y) RC(m, n)}. Suppose that y Ỹ (x) s maxmal n the root order. Then we assert that y Y (x) (we wll leave ths as an exercse to the reader). Thus the maxmal elements of Y (x) are exactly the same as those of Ỹ (x). Now let y Y (x) be maxmal n the root order. Let M B m,n be such that x(m) = x and y(m) = y. Suppose that y 1 < z 1. Then the number f ndces such that m j,1 = 1 must be less than the number of j such that x j > 0. Hence there s a j wth x j > 0 and m j,1 = 0. Hence there must be a k > 1 wth m j,k = 1. If we defne M to have the same entres as M except that m j,1 = 1and m j,k = 0 then x(m ) = x and y(m ) = y + e 1 e k. Thus y(m ) y n Ỹ (x). Ths s a contradcton. Now suppose that we have shown that y = z for k 1. But y k < z k. Then we can apply the argument n the 5
prevous part to see that x j k 1 f and only f m jl = 1 for j k 1. Now snce y k < z k there must be an ndex j such that x j k but m jk = 0. There must therefore be an ndex s > k wth m js = 1. We can therefore argue as n the case when k = 1 to see that y + e k e s Ỹ (x). The obvous nducton now shows that y s greater than z n the lexcographc order. The last asserton s mpled by Lemma 3. Theorem 5 Let 0 r mn and let x P m (r). P n (r) µ(x) y ν r }. Then Y (x) = {y We wll actually prove a much more general result. In the followng P n (r) can be replaced by the set of all domnant n-tuples wth non-negatve entres that sum to r. The condton that the entres need be at most m can be dropped. Theorem 6 Let z, y P n (r) wth z y then there exst elements z () P n (r), = 0,..., m such that z (0) = z and z (m) = y and z (+1) s obtaned from z () by a Move 1 or a Move 2. Theorem 5 follows from Theorem 6. Indeed, we have observed that these moves preserve Y (x). So applyng Theorem 6 to µ(x) we wll have proved Theorem 5. We note that the proof we gve of Theorem 6 actually gves an algorthm for the constructon of the connectng sequence. Here s a demonstraton. Consder z = (7, 5, 5, 3, 3, 3, 2) and y = (5, 5, 4, 4, 4, 4, 2). Then z y = (2, 0, 1, 1, 1, 1, 0). So z y. The method of the proof below would choose z (1) = (6, 6, 5, 3, 3, 3, 2) by Move 1, z (2) = (6, 5, 5, 4, 3, 3, 2) Move 2, z (3) = (5, 5, 5, 5, 3, 3, 2) Move 2, z (4) = (5, 5, 5, 4, 4, 3, 2) Move 1, z (5) = (5, 5, 4, 4, 4, 2) Move 2. We wll now prove Theorem 6. We wll prove the theorem by nducton on z n the order. If z s the mnmal element, ν r, of P n (r) the result s obvous snce then y = ν r and we take m = 0. So assume the result for all u P n (r) wth z u. We now prove the result for z. If z = y then there s nothng to prove. Thus there exsts o such that z = y for o and z o > y o. If z o > z o+1 + 1 then we may apply Move 1 to z and get z (1). We show that z (1) y. Indeed we have z (1) 1 = y 1, z (1) 2 = y 2,..., z (1) = y o 1 o 1 thus (z (1) y ) = 0 for j < o. (z (1) y ) = z o y o 1 0 and j o (z (1) y ) = (z y ) for k > o. Snce z z (1) the nductve hypothess k k mples the result n ths case. We may thus assume that z o z o+1 + 1. If z o = z o+1 + 1. Then snce we have z o > y o y o+1 we see that z o+1 y o+1. Now suppose that we have z j = z o+1 for all j o + 1. Then t s easly seen (argung as n the proof of the mnmalty of ν r ) that ths s 6
mpossble. Thus there exsts a frst j such that j o + 1 and z j > z j+1. We can apply Move 2 to z and get z (1) = z e o + e j+1. We wll now show that z (1) y whch wll complete the nducton n ths case. We note that we have z o+1 =... = z j > z j+1. Snce y o+1... y j y j+1 we have z k y k for o + 1 k j. Ths mples that (z (1) y ) 0 for k j. Now 0 and (z (1) k (z (1) j+1 k y ) = (z (1) y )+z j+1 y j+1 +1 = (z y ) j j+1 y ) = (z y ) 0 for k > j + 1. Thus z (1) y. k We are left wth the case when z o = z o+1. We note that snce z = x there must be a frst j such that z k = z o for k j and z j+1 < z j.. There are two cases. Frst, f z j > z j+1 +1. Then we can do Move 1 to get z (1) = z e j +e j+1. As n the other cases we have z k > y k for all k = o,..., j. So the argument for the frst part of the proof mples that z (1) y. We may thus assume that z j = z j+1 + 1. Now as above there must be another descent that s l > j such that z k = z j+1 for j + 1 k l and z l > z l+1. We now do Move 2 to get z (1) = z e j + e l+1. We note that as before z k y k for j + 1 k l and so the argument above mples that z (1) y. The proof s now complete. 4 A q-analogue In ths secton we wll study RC(q, m, n) = (x,y) RC(m.n) q x. The results of the prevous secton mply that RC(q, m, n) = m!n! r q r x P m(r) 1 λ m (x)! y P n (r) µ(x) y 1 λ n (y)!. We note that the polynomal RC(q, n, n) s of degree n 2 and that t s easly seen that the coeffcent of q j s the same as that of q n2 j for 0 j n 2. Conjecture 7 The polynomal RC(q, n, n) = RC(q, n) s evenly dvsble by 1 + q + q 2 +... + q n. Here are some examples. RC(q, 1) = 1 + q. RC(q, 2) = (1 + q + q 2 )(1 + 3q + q 2 ) 7
RC(q, 3) = (1 + q + q 2 + q 3 )(1 + 8q + 18q 2 + 28q 3 + 18q 4 + 8q 5 + q 6 ) Ths conjecture has been proved by D. Goldsten and R. Stong. They also prove that the polynomals P (m, n) = RC(q, m, n) satsfy the followng recurson. Set [m + 1] q = 1 + q +... + q m. Then 1. P (0, n) = 1 for n 0. 2. P (m, n) = P (n, m). 3. If m n then P (m, n) = m ( 1) +1( m =1 ) [m + 1] P (m, n ). Obvously, ths proves the conjecture. Note that ths mples that P (1, n) = (q + 1) n. so we could add ths to stop the recurson at m = 1. Ths recurson s easly mplemented n Mathematca or Maple. Here s a sample Mathematca code (you wll have to type out the exponents): P[m, n ] := Block[{}, If[m == 0, Return[1]]; If[n <m, Return[P[n, m]]]; If[m == 1, Return[(1 + q) n ]]; Return[ Expand[ Sum[(-1) +1 Bnomal[m, ](Factor[(1 - q m+1 )/(1 -q)] )* P[m, n - ], {, 1, m}]]]] References q [BR]R.H.Bruald and H.J.Ryser, Combnatoral Matrx Theory,Cambrdge Unversty Press, 1991. [GS] Danel Goldsten and Rchard Stong, On the number of possble row and column sums of 0,1-matrces, Preprnt. [vlw] J.H.van Lnt and R.M.Wlson. A course n combnatorcs. Cambrdge Unversty Press, 1992. 8