Copyright c by the paper's authors. Copying permitted for private and academic purposes. In: L. Ferrari, M. Vamvakari (eds.): Proceedings of the GASCom 2018 Workshop, Athens, Greece, 18{20 June 2018, published at http://ceur-ws.org
In this article, we consider a generalization of Young tableaux in which we allow some consecutive pairs of cells with decreasing labels. We show that this leads to a rich variety of combinatorial formulas, which suggest that these new objects could be related to deeper structures, similarly to the ubiquitous Young tableaux. Our methods rely on variants of hook-length type formulas, and also on a new e cient generic method (which we call the density method) which allows not only to generate constrained combinatorial objects, but also to enumerate them. We also investigate some repercussions of this method on the D- niteness of the generating functions of combinatorial objects encoded by linear extension diagrams, and give a limit law result for the average number of local decreases.
As predicted by Anatoly Vershik in [Ver01], the 21st century should see a lot of challenges and advances on the links of probability theory with (algebraic) combinatorics. A key role is played here by Young tableaux, because of their ubiquity in representation theory [Mac15] and in algebraic combinatorics, as well as their relevance in many other di erent elds (see e.g. [Sta11]).
Young tableaux are tableaux with n cells labelled from 1 to n, with the additional constraint that these labels increase among each row and each column (starting from the lower left cell). Here we consider the following variant: What happens if we allow exceptionally some consecutive cells with decreasing labels? Does this variant lead to nice formulas if these local decreases are regularly placed? Is it related to other mathematical objects or theorems? How to generate them? This article gives some answers to these questions.
As illustrated in Figure 1, we put a bold red edge between the cells which are allowed to be decreasing. Therefore these two adjacent cells can have decreasing labels (like 19 and 12 in the top row of Figure 1, or 11 and 10 in the untrustable Fifth column), or as usual increasing labels (like 13 and 15 in the bottom row of Figure 1). We call these bold red edges \walls".
7 18 19 12 21 20 17 2 6 8 9 10 14 16 1 3 4 5 11 13 15
For Young tableaux of shape1 n 2 several cases lead directly to nice enumerative formulas for the total number of speci c tableaux with 2n cells: 1. Walls everywhere: (2n)! 2. Horizontal walls everywhere: (2n)!
2n 3. Horizontal walls everywhere in left (or right) column: (2n 1)!! = (22nnn)!! 4. Vertical walls everywhere: 2n n
= ((2nn!))2! 1 2n 5. No walls: n+1 n
(2n)! = (n+1)(n!)2
In this article we are interested in the enumeration and the generation of Young tableaux (of di erent rectangular shapes) with such local decreases, and we investigate to which other mathematical notions they are related. Section 2 focuses on the case of horizontal walls: We give a link with the Chung{Feller Theorem, binomial numbers and a Gaussian limit law. Section 3 focuses on the case of vertical walls: We give a link with hook-length type formulas. Section 4 presents a generic method, which allows us to enumerate many variants of Young tableaux (or more generally, linear extensions of posets), and which also o ers an e cient uniform random generation algorithm, and links with D- niteness.
1We will refer to \n m Young tableaux", or \Young tableaux of shape n and m columns. They are trivially in bijection with m n Young tableaux.
m", for rectangular Young tableaux with n rows
Vertical walls, Chung{Feller and binomial numbers In this section we consider a family of Young tableaux having some local decreases at places indicated by vertical walls, see Figure 2.
Theorem 2.1. The number of n 2 Young tableaux with k vertical walls is equal to
1 vn;k = n + 1
n k k 2n n Proof. We apply a bijection between two-column Young tableaux of size 2n with k walls and Dyck paths without the positivity constraint of length 2n and k coloured down steps.
These paths start at the origin, end on the x-axis and are composed out of up steps (1; 1), and coloured down steps (1; 1) which are either red or blue.
Given an arbitrary two-column Young tableau, the m-th step of the associated path is an up step if the entry m appears in the left column, while the m-th step is a down step, if the m-th entry appears in the right column. Furthermore, we associate colours to the down steps: If the m-th down step is in a row with a wall we colour it red, and blue otherwise.
Thus, vn;k counts the number of paths with exactly k red down steps. Note that the down steps of a path below the x-axis are always red because a wall has to be involved, yet above the x-axis down steps can have any colour. We decompose paths with k coloured down steps with respect to the number of steps which are below the x-axis. By the Chung{Feller Theorem [CF49] (see also [Che08] for a bijective proof) the number of Dyck paths of length 2n with i down steps below the x-axis is independent of i and equal to the Catalan number Catn = n+11 2nn . When i steps are below the x-axis we have to colour k i of the remaining n i steps above the x-axis red. This gives vn;k = Xk nk i=0 i i
Catn = n + 1 k
and the claim follows.
Corollary 2.2. The average number of linear extensions of a random n the location of these walls is chosen uniformly at random, is Proof. In a two-column Young tableau of size 2n we have nk possibilities to add k walls.
Theorem 2.3. Let Xn be the random variable for the number of walls in a random n 2 Young tableau chosen uniformly at random. The rescaled random variable Xpnnn=4=2 converges to the standard normal distribution N (0; 1).
Proof. We see that the total number of two-column Young tableaux of size n with walls is equal to
n X vn;k = Catn 2n+1 k=0
1 : P (Xn = k) = n + 1 k
Horizontal walls and the hook-length formula The hook-length formula is a well-known formula to enumerate standard Young tableaux of a given shape (see e.g. [Mac15, Sta11]). What happens if we add walls in these tableaux? Let us rst consider the case of a Young tableau of size n such that its walls cut the corresponding tableau into m disconnected parts without walls of size k1; : : : ; km (e.g., some walls form a full horizontal or vertical line). Then, the number of llings of such a tableau is trivially: n! m
Y HookLengthFormula(subtableau of size ki): k1! : : : km! i=1 So in the rest of article, we focus on walls which are not trivially splitting the problem into subproblems: They are the only cases for which the enumeration (or the random generation) is indeed challenging.
We continue our study with families of Young tableaux of shape m n having some local decreases at places indicated by horizontal walls in the left column. We will need the following lemma counting special llings of Young tableaux.
Lemma 3.1. The number of n 2 \Young tableaux" with 2 cells lled with the numbers 1; 2; : : : ; 2n for n such that (the number 2n is used and) all consecutive numbers between the minimum of the second column and 2n are used is equal to 2n 2n 1 Proof. The constraint on the maximum implies that all not used numbers are smaller than the number in the bottom right cell. Therefore it is legitimate to add these numbers to the tableaux. In particular, we create a standard Young tableau of shape ( ; 2n ) (i.e., the rst column has cells and the second one 2n ) which is in bijection with the previous tableau.
Next we build a bijection between standard Young tableaux of shape ( ; 2n ) and Dyck paths with up steps (1; 1) and down steps (1; 1) starting at (0; 2(n )), always staying above the x-axis and ending on the x-axis after 2n steps. In particular, if the number i appears in the left column, the i-th step is an up step, and if it appears in the right column, the i-th step is a down step.
Finally, note that these paths can be counted using the re ection principle [And87]. In particular, there are 2n possible paths from (0; 2(n )) to (2n; 0). Yet, 2n1 \bad" paths cross the x-axis at some point. This can be seen, by cutting such a path at the rst time it reaches altitude 1. The remaining path is re ecting along the horizontal line y = 1 giving a path ending at (2n; 2). It is easy to see that this is a bijection between bad paths from (0; 2(n )) to (2n; 0) and all paths from (0; 2(n )) to (2n; 2). The latter is obviously counted by 2n1 , as 1 of the 2n steps have to be up steps.
Theorem 3.2. The number of n 2 Young tableaux of size 2n with k walls in the rst column at heights 0 < hi < n, i = 1; : : : ; k with hi < hi+1 is equal to with h0 := 0 and hk+1 := n.
Remark 3.3. Denoting consecutive relative distances of the walls by i := hi previous result can also be stated as hi 1 for i = 1; : : : ; k + 1 the 1 2n + 1 1 2n + 1 k+1 Y i=1 2( 1 + : : : + i) + 1 i
: Proof. We will show this result by induction on the number of walls k. For k = 0 the result is clear as we are counting two-column standard Young tableaux which are counted by Catalan numbers (for a proof see also Lemma 3.1 with = n).
Next, assume the formula has been shown for k 1 walls and arbitrary n. Choose a proper lling with k walls and cut the tableau at the last wall at height hk into two parts. The top part is a Young tableau with 2(n hk) elements and no walls, yet labels between 1 and 2n. Furthermore, it has the constraint that all numbers larger than the element in the bottom right cell have to be present. This is due to the fact that all elements in lower cells must be smaller. In other words, these are the objects of Lemma 3.1 and counted by (1).
The bottom part is a Young tableau with k 1 walls and 2hk elements (after proper relabelling). By our induction hypothesis this number is equal to
and set := n
hk. Multiplying the last two expressions then shows the claim.
Remark 3.4. Note that so far we have not found a direct combinatorial interpretation of this formula. However, note that in general 2n+1 does not have to be divisible by 2n + 1.
Let us now also give the general formula for n m Young tableaux with walls of lengths m 1 from columns 1 to m 1, i.e., a hole in column m and nowhere else in a row with walls. Before we state the result, let us de ne for integers n; k the falling factorial (n)k := n(n 1) (n k + 1) and for integers n; m1; : : : ; mk such that n m1 + + mk the (shortened) multinomial coe cient2 m1;m2n;:::;mk := m1!m2! mk!(nn! m1 ::: mk)! . Theorem 3.5. The number of n m Young tableaux of size mn with k walls from column 1 to m 0 < hi < n, i = 1; : : : ; k with hi < hi+1 is equal to 1 at heights (m (mn + m 1)! 0k+1 m 2
Y Y i=1 j=1 i + j j 11 A hi 1 and the i's in the multinomial coe cients appear m
Proof (Sketch). First derive an extension of Lemma 3.1 proved by the hook-length formula and then compute the product. Note that this gives a telescoping factor, giving the rst factor.
Just as one more example, here is a more explicit example of what it gives.
i = 1; : : : ; k with hi < hi+1 is equal to
4 Young tableaux with k walls from column 1 to 3 at heights 0 < hi < n, 6 (4n + 3)(4n + 2)(4n + 1)
Let us consider some other special cases. For example, consider tableaux with walls between every row and a hole in the last column. For this case we set i = 1 for all i. This gives the general formula n(!m(mn!))!n , for n m tableaux, see OEIS A001147 for m = 2 and OEIS A025035 to OEIS A025042 for the special cases m = 3; : : : ; 10.
Now that we gave several examples of closed-form formulas enumerating some families of Young tableaux with local decreases, we go to harder families which do not necessarily lead to a closed-form result. However, we shall see that we have a generic method to get useful alternative formulas (based on recurrences), also leading to an e cient uniform random generation algorithm.
2In the literature, one more often nds the notation m1;m2;:::;mk;nn m1 ::: mk := m1!m2! mk!(nn! m1 ::: mk)! . But we opted in this article for a more suitable notation to the eyes of our readers!
The density method, D- niteness, random generation In this section, we present a generic approach which allows us to enumerate and generate any shape involving some walls located at periodic positions. To keep it readable, we illustrate it with a speci c example (without loss of generality).
So, we now illustrate the method on the case of a 2n 3 tableau where we put walls on the right and on the left column at height 2k (for 1 k n 1), see the leftmost tableau in Figure 3. In order to have an easier description of the algorithm (and more compact formulas), we generate/enumerate rst similar tableaux with an additional cell at the bottom of the middle column, see the middle tableau in Figure 3: It is a polyomino Polyon with 6n + 1 cells. There are trivially (6n + 1)! llings of this polyomino with the numbers 1 to 6n + 1. Some of these llings are additionally satisfying the classical constraints of Young tableaux (i.e., the labels are increasing in each row and each column), with some local decreases allowed between cells separated by a wall (as shown with bold red edges in Figure 3). Let fn be the number of such constrained llings.
To compute fn we use a generic method which we call the density method, which we introduced and used in [Mar18, Mar16, BMW18]. It relies on a geometric point of view of the problem: Consider the hypercube [0; 1]6n+1 and associate to each coordinate a cell of Polyon. To almost every element of [0; 1]6n+1 (more precisely, every element with all coordinates distinct) we can associate a lling of Polyon: Put 1 into the cell of Polyon corresponding to the smallest coordinate of , 2 into the cell of Polyon corresponding to the second smallest coordinate of and so on. The reverse operation associates to every lling of Polyon a region of [0; 1]6n+1 (which is actually a polytope). We call P the set of all polytopes corresponding to correct llings of Polyon (i.e., respecting the order constraints). This P is also known as the \order polytope" in poset theory.
Let us explain how the density method works. It requires two more ingredients. The rst one is illustrated in Figure 3: It is a generic building block with 7 cells with names X,Y,Z,R,S,V,W. We put into each of these cells a number from [0; 1], which we call x; y; z; r; s; v; w, respectively. The second ingredient is the sequence of polynomials pn(x), de ned by the following recurrence (which in fact encodes the full structure of the problem, building block after building block): Let us now give a more algorithmic presentation of our method:
1 Initialization: We order the building blocks from k = n 1 (the top one) to k = 0 (the bottom one). We start with the value k := n 1, i.e. the building block from the top. Put into its cell Z a random number z with density pn(z)= R01 pn(t) dt. We repeat the following process until k = 0: 2 Filling: Now that Z is known, put into the cells X; Y; R; S; V; W random numbers x; y; r; s; v; w with conditional density where 1P z is the indicator function of the k-th building block (with value z in cell Z): 1 gk;z(x; y; r; s; v; w) := pk+1(z) pk(x)1P z; 1P z := 1f0 x y z;0 r y;r s z;z w 1;y v wg: 3 Iteration: Consider X as a the Z of the next building block. Set k := k
Theorem 4.1. The density method algorithm is a uniform random generation algorithm with quadratic time complexity and linear space complexity.
Proof. Let us indeed prove that the algorithm gives a random element of our set of polytopes P with the uniform measure. Our algorithm yields a (6n + 1)-tuple x := (xj ; yi; ri; si; vi; wi; 0 j n; 0 i n 1) whose density is the product of the conditional densities:
n pn(xn)
Y gn i;xn i+1 (xn i; yn i; rn i; sn i; vn i; wn i)
R01 pn(t)dt i=1 The crucial point is that this product is telescopic and equal to pn(xn) n 1
Y pk(xk)1P xk = R01 pn(t)dt k=0 pk+1(xk+1) p0(x0)1P R01 pn(t) dt =
1P R01 pn(t) dt (as p0 = 1); where 1P xk is as in the algorithm above the indicator function of the k-th block (where the local variables x; y; r; s; v; w; z of the algorithm are now xk; yk; rk; sk; vk; wk; zk) and where the product 1P of these indicator functions is the indicator function of the full polytope (with n blocks): 1P = Qkn=01 1P xk .
Therefore, this density is constant on our set P of polytopes and zero elsewhere, which is exactly what we wanted. The fact that it is a density implies that its integral is 1, whence
Z [0;1]6n+1 1P dx =
Z 1 0
pn(t) dt: Z
[0;1]6n+1 1P dx: Now if we choose a random uniform element in [0; 1]6n+1, the probability that it belongs to our set P of polytopes is But due to the reasoning above, this is also the probability that a random uniform lling of our building block is correct (i.e., respects the order constraints). Thus this probability is given by R01 pn(t)dt=(6n + 1)!.
This implies that fn = (6n + 1)! R01 pn(t)dt.
Finally, as each step relies on the computation and the evaluation of the associated polynomial pn(z) (of degree proportional to n), this gives a quadratic time complexity, and takes linear space. (4) (5) (6) (7) Remark 4.2. If one wants to generate many diagrams and not just one, then it is valuable to make a precomputation phase computing and storing all the polynomials pn. The rest of the algorithm is the same. For each new object generated, this is saving O(n2) time, to the price of O(n2) memory. The algorithm is globally still of quadratic time complexity (because of the evaluation at each step of pk(x), while pk+1(z) was already evaluated). Remark 4.3. If one directly wants to generate 2n 3 Young tableaux with decreases instead of our strange polyomino shapes Polyon, then one still uses the same relation between pn+1 and pn but p0 is not de ned and p1 has a more complicated form. Another way is to generate Polyon, and to reject all the ones not having a 1 in the bottom cell, then to remove this bottom cell and to relabel the remaining cells from 1 to 6n (see Figure 3). This still gives a fast algorithm of O(n2) time complexity (the only di erence being the cost of the initial algorithm which is the multiplicative constant included in the big-O).
Using dynamic programming or clever backtracking algorithms allows hardly to compute the sequence fn (the number of llings of the diagram) for n 3. In the same amount of time, the density method allows us to compute thousands of coe cients via the relation fn = (6n + 1)! R01 pn(z), where the polynomial pn(z) is computed via the recurrence
Z z 1 (z 1)(x z)(3x3 7x2z xz2 z3 2x2 + 4xz + 4z2)pn(x) dx: (8)
This gives the sequence ffngn 0: f1, 12, 8550, 39235950, 629738299350, 26095645151941500, 2323497950101372223250, 392833430654718548673344250, 115375222087417545717234273063750, 55038140590519890608190921051205837500, 40460077456664688766902540022810130044068750, 4393840235884118464495128448703896167747914784375, . . . g. As far as we know, there is no further simple expression for this sequence. This concludes our analysis of the model given by Figure 3.
We can additionally mention that the generating function associated to the sequence of polynomials pn(x) has a striking property: Theorem 4.4. The generating function G(t; z) := Pn 0 pn(z)tn is D- nite3 in z.
Proof. The general scheme (whenever one has one hole between the walls) is where Q is a polynomial in x and z, given by Q(x; z) := RPz 1. The fact that there is just one hole between the walls guarantees that all the other variables encoding the faces of the polytope Pz will disappear in this integration. Let d be the degree of Q in z, applying @@zdd++11 to both sides of Formula 9 leads to a relation between the (d + 1)-st derivative of pn+1 and the rst (d + 1) derivatives of pn. Multiplying this new relation by tn+1 and summing over n 0 leads to the D- nite equation for G(t; z).
Note that G(t; z) is D- nite in z, but is (in general) not D- nite in t. When it is D- nite in t, our algorithm has a better complexity (namely, a O(n3=2) time complexity), because it is then possible to evaluate pn(z) in time O(pn ln n) instead of O(n). See [BCGLLSS17, Chapter 15] for more details on these complexity issues. 5
We presented a new way to enumerate and generate Young tableaux with local decreases (and, more generally, linear extensions of posets). Our approach is di erent from the classical way to generate Young tableaux (e.g. via the Greene{Nijenhuis{Wilf algorithm, see [GNW84]), which relies on the existence of an enumeration by a simple product formula (given by the hook-length formula). As there is no such simple product formula for the more general cases we considered here, such an approach cannot work anymore. Obviously, in order to generate these objects, there is the alternative to use some naive \brute-force-like" methods (like e.g. dynamic programming with backtracking). However this leads to an exponential time algorithm. The density method which we presented here is the only method we are aware of which leads to a quadratic cost uniform random generation algorithm.
3A function F (z) is D- nite if it satis es a linear di erential equation, with polynomial coe cients in z. See e.g. [FS09] for their role in enumeration and asymptotics of combinatorial structures.
This work was initiated during the postdoctoral position of Michael Wallner at the University of Paris Nord, in September-December 2017, thanks a MathStic funding. The subsequent part of this collaboration was funded by the Erwin Schrodinger Fellowship of the Austrian Science Fund (FWF): J 4162-N35. [And87] [Che08] K. L. Chung and W. Feller. On uctuations in coin-tossing. Proceedings of the National Academy of Sciences of the United States of America, 35:605{608, 1949.
P. Flajolet and R. Sedgewick. Analytic Combinatorics. Cambridge University Press, 2009.
I. G. Macdonald. Symmetric functions and Hall polynomials. Oxford Classic Texts in the Physical Sciences. The Clarendon Press, Oxford University Press, second edition, 2015.
P. Marchal. Rectangular Young tableaux and the Jacobi ensemble. Discrete Mathematics and Theoretical Computer Science Proceedings, BC:839{850, 2016.
P. Marchal. Permutations with a prescribed descent set. This Proceedings Volume, 2018.
R. P. Stanley. Enumerative combinatorics. Volume 1. Cambridge University Press, 2nd edition, 2011. A. M. Vershik. Randomization of Algebra and Algebraization of Probability. In: B. Engquist, W. Schmid (Eds.): Mathematics unlimited|2001 and beyond., pp. 1157{1166. Springer, 2001.