Experimenting with Standard Young Tableaux

Using Symbolic Computation with Maple, we can discover lots of (rigorously-proved!) facts about Standard Young Tableaux, in particular the distribution of the entries in any specific cell, and the sorting probabilities.

One of the most iconic objects in mathematics, both concrete [K], and abstract [F], are Standard Young Tableaux [Wi].Recall that an integer partition, or partition, for short, aka shape, of a non-negative integer n, with k parts, is a non-increasing sequence of positive integers λ = (λ 1 , λ 2 , . . ., λ k ) , The Ferrers diagram (or Young diagram) of a partition λ is a left-justified array of dots (or empty boxes) where the top row has λ 1 dots (boxes), the second row has λ 2 boxes, . .., the k-th row has λ k dots.For example, the Ferrers diagram of (4, 4, 3, 1) is Given a shape λ = (λ 1 , . . ., λ k ) with n boxes, a standard Young tableau is a way of filling the boxes with the integers {1, . . ., n}, such that each of them shows up (necessarily once) and both rows and columns are increasing.More formally, it is an array such that T i,j < T i,j+1 and T i,j < T i+1,j whenever they exit.
To see the set of standard Young tableaux of shape L in our Maple package, type SYT(L); .For example to see the above five tableaux type SYT([2,2,1]); .
The total number of standard Young tableaux of shape λ, denoted by f λ , is famously given by the hook length formula, or equivalently (and more convenient for us) by the Young-Frobenius formula (see [K]).
Fix a shape λ and fix a cell [i, j], Who can be the occupant of that cell?
Calling that occupant r, we have: This gives a certain probability distribution.What is it?
For example, with the shape λ = (2, 2, 1) again, and the cell [2, 1], the set of possible occupants is {2, 3}, and the probability of it being 2 is 3 5 and of it being 3 is 2 5 .
Later on we will be interested not in specific shapes, but in general (mostly rectangular) shapes, with a fixed number of rows, but arbitrary (i.e.symbolic) shape.Fixing the number of rows to be k (where k is numeric), and regarding the shape (n, n, . . ., n), where n is repeated k times, we would be interested in deriving closed-form expressions (as rational functions of n), for the probability distribution of the possible occupants of a given first row cell [1, j], for any given numeric integer j.Note that the possible occupants of [1, j] are j, j + 1, j + 2, . . ., k(j − 1) + 1.
Once we found these expressions in n, we can ask about the limiting distribution, that Maple can find for us.Then we can also hope to see how it varies with i and look at the meta-limiting behavior as i gets larger.
Another kind of question, inspired by the beautiful work of Chan, Pak, and Panova [CPP1][CPP2], is to study the sorting probabilities.Given two cells c 1 and c 2 , draw a standard Young tableau uniformly at random.Who is bigger?
The occupant of c 1 or the occupant of c 2 ?
Of course if the two cells are related, i.e. one of them, say c 2 , is (weakly) below and (weakly) to the right of the other, c 1 , i.e. in the underlying poset c 1 < c 2 , then of course, always T c 1 < T c 2 , but what if they are not related i.e. writing The sorting probability is defined by: where T is a random standard Young tableau of shape λ.
In particular, following Chan-Pak-Panova, we are interested in the minimal (absolute value) of the sorting probabilities, over all pairs of cells, as the shapes get larger.

Simulation
One way to answer these questions, approximately, is via simulation.The amazing Greene-Nijenhuis-Wilf [GrNW] algorithm (that also lead to a beautiful probabilistic proof of the hook lenghth formula) inputs a shape, λ, and outputs, uniformly at random, one of the f λ standard Young tableaux of that shape.By sampling many of them, we can get approximations to the quantities of interest.
To get approximations for the sorting probability of cell c1 vs. cell c2, in the shape L, by sampling K tableaux: enter: SiPr(L,c1,c2,K); .
For example, would give something like 0.010400000.Of course, in this particular case the exact answer is obviously zero, by symmetry, so getting something close to 0 is a good sanity check.
Symbol Crunching in order to find The Probability Distribution of the Occupants of a Specific Cell in a Symbolic Shape For the sake of exposition, we will mostly be concerned with standard Young tableaux of rectangular shape.Fix the number of rows k, and consider the shape where n is repeated k times.More generally all our algorithms carry over to the general symbolic shape (n 1 , n 2 , . . ., n k ) , where n 1 ≥ n 2 ≥ n k ≥ 0, and they are all left symbolic.Using Young-Frobenius we get an explicit formula for their total number in the form of a certain rational function, of (n 1 , . . ., n k ) times the multinomial coefficient In the special case of a k × n rectangular shape: (i.e.(n, . . ., n), where n is repeated k times), it is a certain rational function of n (namely Now also fix a specific (numeric) cell, c = [i, j], and a specific (numeric) integer r.We want an explicit formula, as a rational function of n, for the probability that, when you draw (say using the GNW algorithm) uniformly at random, one of the (n k)!(k−1)!n!•••(n+k−1)!standard Young tableaux, of shape n k , that the occupant of the cell c = [i, j] is the integer r, in symbols: P r(T ij = r).
We will soon explain how to do it, but you are welcome to try it out first using our Maple package SYT.txt.Let's give a few examples.
• To get the explicit expression for the probability that the occupant of cell [1,3] in a random standard Young tableau of shape (n, n, n) happens to be 7, type: • For a more complicated example, to get the expression for the probability that the occupant of cell [3,3] in a random standard Young tableau of shape (n, n, n) happens to be 13, type: OcCs([n,n,n],13,[3,3]); ) .
• For yet another example, regarding the three-rowed shape (n 1 , n 2 , n 3 ), to get the rational function (in n 1 , n 2 , n 3 ) for the probability that cell [1,2] would be occupied by 3, type .
• If the cell is at the first row, c = [1, j], for some j > 1, then there are only finitely many possible occupants r, namely r = j, j + 1, . . ., k(j − 1) + 1, and to get the probability generating function, using the variable x, type OcGFs1(L,j,x); .
For example, entering: OcGFs1([n,n,n],2,x); gives you meaning that the cell [1,2] in a standard Young tableau of shape (n, n, n) is occupied by either 2, 3, or 4, with respective probabilities of To get the limiting distribution as n → ∞, as well as the expectation, variance, and the first few moments up to the K-th, try OcGFs1L(L,n,i,x,K); .
How does Maple perform these amazing calculations?In other words how does it compute P r(T i,j = r) for a random standard Young tableau of a symbolic shape?
Given a (symbolic, or numeric) shape λ, a cell c = [i, j], and an integer r, how can it happen that T ij = r?The cells occupied by {1, 2, . . ., r} form a certain standard Young tableau with r cells, that is a certain subshape, that must contain the cell c = [i, j], that must be a corner, of course.So let's ask our beloved computer to find all the shapes with r cells that contain the cell [i, j] as a corner, or equivalently the set of partitions, ν, of r with at least i rows such that ν i = j.Let's call this (finite) set S([i, j], r).
Let, as usual, f λ/ν denote the number of standard Young tableau of skew-shape λ/ν (recall that these are shapes where ν is a subshape of λ, and the cells of ν are removed).Then our quantity of interest is ν∈S([i,j];r) where ν ′ is the shape ν with the cell [i, j] removed.
The number f ν ′ is easily computed using the Young-Frobenius formula.How do we compute the (symbolic) expression f λ/ν ?
Recall that standard Young tableaux of shape λ = (λ 1 , . . ., λ k ) are in bijection with walks from the origin to the point λ in the k-dimensional discrete lattice N k , that always must stay in the region Similarly, standard Young tableaux of skew-shape λ/ν are in bijection with such 'sub-diagonal' walks from ν to λ.Now following the ideas of André [Z] (see also [GeZ]), put mirrors on the hyperplanes and look at the set of k! images of the point ν under the action of the group generated by these k − 1 reflections.As is well-known (and fairly easy to see), the underlying group is the symmetric group S k , and the sign is 1 or −1 according to whether the number of inversions is even or odd.
Calling the set of images IM AGE(ν), we have: where W (µ, λ) is the number of walks in the lattice from µ to λ given by the multinomial coefficient But since we are interested in probabilities, we can divide everything by f λ and stay in the realm of rational functions.
This is implemented in procedures Swee(L,M).
Computing the Sorting Probabilities for symbolic shape and any two cells where one of them is at the first row The numeric procedure Pr(L,c1,c2), for a random standard Young tableau of shape L, manually finds the sorting probability of cell c1 vs. cell c2, and the numeric procedure MinPr(L) finds the minimal sorting probabilities among all pair of cells, followed by the 'champions'.For example, if you type MinPr ([10,4,3]); , you would get meaning that the minimum (absolute value) of the sorting probabilities among all the 17 2 = 136 pairs of cells is 1 127 and it is achieved with the pair of cells [1,5] and [3,1].But we want to do things symbolically.Alas, things get complicated if neither cells are at the first row.
But we can, exactly, and symbolically, compute a closed-form expression, as a rational function of the symbols, of the sorting probabilities between any cell [1, j] on the first row and any cell below it (to the left, of course, or else the sorting probability is trivially −1).
This is implemented in procedure PrS(L,j,c2), where L is the symbolic shape and c2 is the cell below the first row that we compare it to [1,j].For example, to get the sorting probability of cell [1,3] How Does Maple find The Symbolic Sorting Probabilities?
How do we do it?Look at all the possible occupants of cell c 1 = [1, j] (there are finitely many of them).Suppose it happens to be r.How can it be larger than the occupant of cell c 2 = [m 1 , m 2 ]?.We find the (finite) set of shapes with r cells that include c 1 = [1, j], and in addition it is a corner.
In other words all the shapes ν with r cells such that ν 1 = j, ν has at least m 1 rows, and m 2 ≤ ν m 1 .
As before add-up f ν ′ times f λ/ν , and then add-them-up for all possible legal occupants of [1, j].
Getting a nice (or not so nice, but sill explicit) expression for P r(T 1,j > T m 1 ,m 2 ), and hence for the sorting probability 2 P r(T 1,j > T m 1 ,m 2 ) − 1 .
Of course, we always divide by f λ (but this is already built-in in all our macros).
A one-line proof that the Minimal sorting probabilities for the Catalan Poset is O( 1 n ) In a deep and beautiful work [CPP2], the authors proved that the minimal sorting probability of the Young lattice, as the shapes get larger, tends to 0. In the more specific paper [CPP1], they proved, by an ingenious and delicate asymptotic analysis, that for the two-rowed case, [n, n], (what they call the Catalan poset), it is O( ).But using our Maple package, we can get, without human effort, a (rigorous!)proof that it is at least O( 1 n ).
Indeed, entering in our Maple package SYT.txt, the command : we get, in one nano-second : 3 2n − 1 .
So we have the following computer-generated proposition (that admittedly could have been done by humans only using paper and pencil).
Proposition: The sorting probability of the cell [1, 3] and the cell [2, 1] in a random standard Young tableau of shape (n, n) is Hence the minimal sorting probability of the Catalan lattice is O( 1 n ).
Proposition: The sorting probability of the cell [1, 5] and the cell [2, 2] in a random standard Young tableau of shape (n, n) is Hence, again, the minimal sorting probability of the Catalan poset is O( 1 n ).
Procedure FindZero(L,n,K) searches for all pairs of cells c where c 1 is in the first row, and j, m 2 ≤ K, such that the sorting probability tends to 0 (and hence is, of course O(1/n)).Alas, except for the above two pairs (for the Catalan poset), none exists for K = 100.
Note that here we really lucked out, since the pairs {[1, 3], [2, 1]} and {[1, 5], [2, 2]} are numeric (and small), and give upper bound for the minimal sorting probability.In order to get to the true minimum, both c 1 and c 2 must be symbolic (that what was essentially done in [CPP1] and [CPP2] with great human effort).
The special case of the Catalan poset (2-rowed tableaux) For the Catalan case things can get much faster (as noticed in [CPP1]) and the procedures implementing this can be found by typing ezraD(); .Anij(n,i,j) is a faster version of PrS([n,n],i,[2,j]).It turns out that in this case we can get closed-form expressions, for the occupancy distribution of an arbitrary cell [1, i] at the first row of a standard Young tableau of shape (n, n) for symbolic i, that entail, in turn, closed-form expressions for the limiting distribution as n goes to ∞, and surprise!we can get explicit expressions for the average, variance, and higher moments for that limiting distribution for symbolic i, and even the meta-limiting behavior, as i goes to infinity.

We have
Proposition: The expectation of the occupant of cell [1, i] in a random standard Young tableau of shape (n, n), as n goes to infinity is confirming Richard Stanley's observation mentioned in [CPP1], Eq. (5.1)The asymptotics is The variance is The limiting (as i goes to infinity) skewness is 2(5π−16) √ 2 (3π−8) 3 2 = −0.4856928234 . . .