Non-crossing tree realizations of ordered degree sequences

We investigate the enumeration of non-crossing tree realizations of integer sequences, and we consider a special case in four parameters, that can be seen as a four-dimensional tetrahe-dron that generalizes Pascal’s triangle and the Catalan numbers.


Introduction
Enumeration of non-crossing tree realizations of integer compositions A non-crossing tree t is a labeled tree on a sequence of vertices v 0 , v 1 , . . ., v n drawn in counterclockwise order on a circle, and whose edges are straight line segments that do not cross.For any index 0 ≤ i ≤ n, let d i stand for the number of edges incident with v i (that is the degree of v i ).Then as any other tree on n+ 1 vertices, t satisfies n i=0 d i = 2n.Thus, the sequence d 0 , d 1 , . . ., d n defines a composition of 2n into n + 1 positive summands (two sequences of integers that differ only in the order of their elements define distinct compositions of the same integer).Stated otherwise, t is a non-crossing tree realization of the composition d 0 , d 1 , . . ., d n .
For any composition c = d 0 , d 1 , . . ., d n of 2n into n + 1 positive summands, let nct(c) stand for the number of non-crossing tree realizations of c, that is the number of non-crossing trees on n + 1 vertices v 0 , v 1 , . . ., v n such that vertex v i has degree d i for any index 0 ≤ i ≤ n (there always exists at least one, see Proposition 2.3).We aim at computing nct.Note that here the input is more specific than the degree partition, as for example in [5].
From proof nets to non-crossing trees Our interest for these non-crossing tree realizations comes from linguistics and proof theory.The starting point for this work was the following linguistic problem: How many different readings can an ambiguous sentence at most have?Particularly, which sentence of a given length has the most different readings?When using categorial grammars based on the Lambek calculus [9] or related systems, a parse tree is a formal proof in a deductive system.Thus, our questions become: How many different formal proofs can a formula have?Particularly, which formula of a given length has the most different formal proofs?In category theoretical terms these questions come down to the cardinality of the Hom-sets in a free non-commutative star-autonomous category [2].The corresponding logic is a variant of non-commutative intuitionistic linear logic [16,8] for which formal proofs can be represented as planar proof nets.It would go too far beyond the scope of this paper to go into the details of this correspondence.However, to give the reader an idea, we have shown in Figure 1 the transformation of a parse tree into a proof net.The first step transforms the parse tree into a formal proof according to Lambek's work [9].In the second step, this proof is embedded into a one-sided multiple conclusion system using the From parse trees via proof trees to proof nets binary connectives and [16].In order not to lose the information on positive and negative positions in the formulas we use polarities (see, e.g., [8] for details).The final three steps show how this one-sided sequent proof is translated into a proof net by simply drawing the flow graph on the atoms appearing in the proof (for more details, see [8,3,14]).It is a well-known fact of linear logic that such a graph G does indeed correspond to a sequent proof if and only if every switching (that is, every graph obtained from G by removing for each -node one of the two edges that it to its children) is a connected and acyclic graph [4].If G does not contain any -nodes, as in our example, then G itself has to be connected and acyclic.Furthermore, we have that G is planar if and only if the sequent proof does not contain the exchange rule, as it is the case for the Lambek calculus [16].
Thus, our questions become: How many different planar proof nets can at most be defined over a given sequent?Particularly, over which sequent of a given length can the most different planar proof nets be defined?
In that respect, we can ignore the names of the atoms, and only -free sequents are of interest: on the one hand, occurrences of lying above an occurrence of can moved down by the transformations which preserve correctness without affecting linkings (see [6,7]); on the other hand, root occurrences of are irrelevant and can be removed.Hence, for every sequent Γ there is a -free sequent Γ ′ , such that for Γ ′ exist at least as many different planar proof nets as for Γ.
Finally, up to the associativity of , planar -free proof nets are in bijection with non-crossing trees as shown in Figure 2.
We were not able to find a closed formula for nct d 0 , d 1 , . . ., d n depending only on the input composition d 0 , d 1 , . . ., d n .However experiments show that at least up to n = 25, nct is which we write as 12 p (13) q 12 r (13) s and are the first step of our study (compositions maximizing nct for higher values of n may be of a different shape or involve summands higher then 3).The input is now reduced to four parameters p, q, r and s such that n + 1 = 1 + p + 2q + 1 + r + 2s.We write nct 12 p (13) q 12 r (13) s as N p,q,r,s , and we are interested in computing N p,q,r,s .
A four-dimensional Generalization of Pascal's and Catalan's triangles Recall that Pascal's triangle P p,r = p+r p,r can be generated recursively by: The first few values are shown below (see [13, A007318]): r ?? ???The Catalan numbers C q = 1 q+1 2q q,q are generated recursively by C 0 = 1 and C q+1 = q j=0 C j • C q−j .A combination of Pascal's triangle and the Catalan numbers is known as Catalan's triangle Q p,q = p+1 p+q+1 p+2q p+q,q which can be generated recursively by: The first few values are shown below (see [13, A009766]): q ?????It is also possible to generalize the recursive formula of the Catalan numbers into a triangle R q,s generated by (assuming that C −1 = 0): The  (7) As an example, We shall establish that N p,q,r,s is a four-dimensional "tetrahedron" that generalizes the three triangles P , Q and R above, insofar as: N 0,q,0,s = N 0,s,0,q = R q,s .
Outline The organization of this paper is as follows: First, in Section 2, we study the general case of enumerating non-crossing tree realizations of integer compositions.Then, in Sections 3-7, we concentrate on the four-parameter case.In particular, we will prove identities ( 8)- (10) in Sections 4 and 5. Finally, we will provide the generating function for N p,q,r,s in Section 7.

General case
Any labeled tree on a sequence of vertices can be drawn in such a way that its vertices lie in counterclockwise order on a circle and its edges are straight line segments lying inside that circle.
In that case, of course, some of its edges may cross each other.Let us call such a labeled tree a crossing tree.The order of summands in a composition does not matter regarding the number of its labeled tree realizations (there are six for any composition of 2 • 4 into 4 + 1 summands in the multiset {1, 1, 2, 2, 2}).But it does as soon as we distinguish between non-crossing and crossing realizations.As an example, there are one non-crossing and five crossing realizations of 1, 1, 2, 2, 2 , while there are three non-crossing and three crossing realizations of 1, 2, 1, 2, 2 (these are shown on Figure 3).
Remark 2.1.A proof of Cayley's formula (see e.g., [1]), which asserts that the number of labeled trees on n + 1 vertices is (n + 1) n−1 (see [13, A000272]) relies on: where the sum ranges over the  The image under rotation of a non-crossing tree t on vertices v 0 , v 1 , . . ., v n is a non-crossing tree on vertices v σ(0) , v σ(1) , . . ., v σ(n) for some cyclic permutation σ in the symmetric group S n+1 .Moreover, t realizes a composition d 0 , d 1 , . . ., d n iff its image under rotation realizes the composition d σ(0) , d σ(1) , . . ., d σ(n) .Thus, for any composition d 0 , d 1 , . . ., d n and any cyclic permutation σ ∈ S n+1 , We shall refer to this property as stability under rotation. 1he same way, the mirror image of a non-crossing tree t on vertices v 0 , v 1 , . . ., v n is a noncrossing tree on vertices v n , v n−1 , . . ., v 0 , and t realizes a composition d 0 , d 1 , . . ., d n iff its mirror image realizes the composition d n , d n−1 , . . ., d 0 .Thus, for any composition d 0 , d 1 , . . ., d n , We shall refer to this property as stability under mirror image. 2e will now establish that for any positive integer n and any composition c of 2n into n + 1 positive summands, there exists a non-crossing tree realization of c (Proposition 2.3).Lemma 2.2.For any positive integer n and any sequence 1, d 1 , . . ., d n , d n+1 of n + 2 positive integers such that 1+ n+1 i=1 d i < 2(n+1), there is an index 1 ≤ k < n+1 such that 1+ k i=1 d i = 2k.
Proof.For any index 1 ≤ l ≤ n + 1, let S l stand for 1 + l i=1 d i .We prove the following implication by induction on l: if there is no index 1 ≤ k < l such that S k = 2k, then S l ≥ 2l.Since by hypothesis S n+1 < 2(n + 1), there must exist an index 1 ≤ k < n + 1 such that S k = 2k.Base.Since d 1 ≥ 1, S 1 = 1 + d 1 ≥ 2 • 1 and the stated implication holds trivially.Induction.Assume that the stated implication holds for l (IH), and that there exists no index 1 ≤ k < l + 1 such that S k = 2k.We reformulate the latter hypothesis as: (i) there exists no index 1 ≤ k < l such that S k = 2k, and (ii) S l = 2l.By (IH) we get from (i), that S l ≥ 2l, and from (ii), that S l > 2l, i.e., S l ≥ 2l + 1.Since d l+1 ≥ 1, S l+1 = S l + d l+1 ≥ 2(l + 1).Proposition 2.3.For any positive integer n and any composition c of 2n into n + 1 positive summands, there exists a non-crossing tree realization of c.
Proof.We proceed by induction on n.Base.The unique composition 1, 1 of 2 • 1 into 1 + 1 positive summands is realized by the unique (trivially non-crossing) tree on 1 + 1 vertices.Induction.Assume that the stated property holds for any positive integer up to n (IH), and let d 0 , d 1 , . . ., d n+1 be a composition of 2(n + 1) into n + 2 positive summands.Since n is a positive integer, 2(n + 1) > n + 2 and there must exist at least one summand d k > 1.By stability under • there exists a non-crossing tree on vertices t 0 , t 1 , . . ., t k realizing the composition 1, d 1 , . . ., d k of 2k into k + 1 positive summands, • there exists a non-crossing tree on vertices u 0 , u 1 , . . ., u n−k+1 realizing the composition Let T and U stand for the respective edge sets of these non-crossing trees (where edges are defined as couples of vertices).We "merge" t 0 and u 0 into a single vertex v 0 to get a tree on vertices v 0 , v 1 , . . ., v n+1 which edge set is defined as (see Figure 4).This tree is non-crossing and it realizes the composition d 0 , d 1 , . . ., d n , d n+1 of 2(n + 1) into n + 2 positive summands.
The previous proof suggests a recursive definition of nct: • The unique composition 1, 1 of 2 • 1 into 1 + 1 positive summands is realized by the unique non-crossing tree on 1 + 1 vertices.Thus, • Let n be strictly greater than 1 and d 0 , d 1 , . . ., d n be a composition of 2n into n + 1 positive summands.Let k be the smallest index such that d k > 1 (there exists at least one).By stability under rotation, Thus we can assume that d 0 > 1.In that case, where the sum ranges over the set of indices 1 ≤ k < n such that 1 + k i=1 d i = 2k (there exists at least one).
where the sum ranges over the set of sequences k 0 , . . ., k d0 of d 0 + 1 indices such that 0 = k 0 < • • • < k d0 = n and such that for all 0 < j ≤ d 0 , 3 The four parameters case We focus now on the special case where compositions are of the shape 1, 2, 2, . . ., which we write as 12 p (13) q 12 r (13) s .Recall that N p,q,r,s stands for nct 12 p (13) q 12 r (13) s .
Lemma 3.1.For any p, q, r and s, N p,q,r,s = N r,q,p,s = N p,s,r,q .
Proof.This follows from stability under rotation and mirror image.We give the formal calculations here in full, because we use similar arguments later on without showing them explicit.
The same way, N p,q,r,s = N p,s,r,q .
where C s stands for the s-th Catalan number.

Pascal's & Catalan's triangles
In this section we are going to establish identities ( 8) and ( 9 Proof.First we prove (30) by induction on p, then we prove (33) by induction on s.From this (31) and (32) follow as a special case with r = 0.By Lemma 3.1 and (26) we have for any r that N 0,0,r,0 = N r,0,0,0 = 1.Now assume that N p,0,r,0 = p+r p,r for any r (IH1).Then

A triangular Catalan recurrence
In this section we establish the identity (10) from the introduction.We need the following lemma: Lemma 5.1.For any t and u, Proof.By induction on u (the base case is the usual recurrence for Catalan numbers).

Generating functions
We can use the identities (24) and (58) for calculating the generating function for N p,q,r,s .Recall that we use the following abbreviations: P p,r = N p,0,r,0 C q = N 0,q,0,0 Q p,q = N p,q,0,0 T p,q,r = N p,q,r,0 R q,s = N 0,q,0,s V p,q,s = N p,q,0,s Theorem 7.1.We have Q p,q x p y q = C(y) R(y, w) = q,s R q,s y q w s = C(y) • C(w) T (x, y, z) = p,q,r T p,q,r x p y q z r = (1 V (x, z, w) = p,q,s V p,q,s x p y q w s = C(y) Proof.The formulas in (i) and (ii) are well-known.For the others, the calculation follows below.
(iii) By letting r = s = 0 in (24) and using (69) we get If we plug (70) into the general formula for Q(x, y), we get Q(x, y) = q Q 0,q y q + x p,q ( q i=0 C i • Q p,q−i ) x p y q = C(y) + x • C(y) • Q(x, y) From this, (iii) follows immediately.
(iv) We let R ′ (y, w) = q,s R q+1,s y q w s and R ′′ (y, w) = q,s R q,s+1 y q w s Then, we get immediately from (58) and ( 69)

Figure 4 :
Figure 4: Merging two non-crossing trees into a single one