Binary trees, super-Catalan numbers and 3-connected Planar Graphs - PowerPoint PPT Presentation

Binary trees, super-Catalan numbers and 3-connected Planar Graphs Gilles Schaeffer LIX, CNRS/´ Ecole Polytechnique Based in part on a joint work with E. Fusy and D. Poulalhon

Mon premier souvenir d’un cours de combinatoire... Mots de Dyck � 2 n � 1 Arbres binaires n +1 n Arbres planaires Mots de � Lukasievicz Mini-jardin de Catalan (D’apr` es photocopies de transparents de Viennot, 1993)

Today’s subject: Super Catalan numbers ( Catalan, Gessel ) 1 (2 n )!(2 m )! 2 ( n + m )! n ! m ! These numbers are integers for all positive m, n . ⇒ They deserve a combinatorial interpretation! (2 n )! � 2 n 1 � – For m = 1 , Catalan numbers: n !( n +1)! = . n +1 n 1 , 2 , 5 , 14 , 42 , 132 , 429 , 1430 . . . 6(2 n )! � 2 n � 6 1 – For m = 2 , the numbers are: ( n +2)! n ! = . n +2 n +1 n 2 , 3 , 6 , 14 , 36 , 99 , 286 , 858 , . . . We shall discuss some interpretations for m = 2 .

”More precisely”, we aim at the following diagram: � 2 n � 6 1 n +2 n +1 n

A one-page preliminary...

In the Catalan garden, I pluck the... Binary trees with n nodes and Dyck paths with length 2 n . � 2 n 1 � They are counted by Catalan numbers: n +1 n Here is a bijection: Turn around the tree, ( ( ) write up or down ) ( when entering or exiting ) ( ( a left subtree. ) ) ( )

First interpretations: unrooted binary trees

Colors make pictures more fun... Edge-3-colored binary tree = a binary tree with colors on the edge and nodes such that there are two type of nodes: Choose colors Take a binary for the root tree edge and the root vertex: � 2 n 1 � ⇒ # { colored tree with n nodes } = 6 · . n +1 n

Agriculture hors sol unrooted 3-colored tree = like a 3-colored binary tree, but without the root... These trees have no symmetries: indeed symmetries of planar trees must leave the center invariant. Here the center can be: or: each tree has n + 2 distinct rootings. ⇒ # { unrooted 3-c trees with n nodes } � 2 n 1 � � 2 n 6 1 = 6 · � = n +2 · n +1 n n +1 n

Here is our first super-Cat-structure: � 2 n � 6 1 n +2 n +1 n

An elegant restatment: Trees on the hexagonal lattice (Pippenger & Schleich’03) Up to translation and rotations, there is a unique way to embed an unrooted colored tree on the colored hexagonal lattice (possibly with overlaps). � 2 n � 6 1 = n +2 n +1 n

An elegant restatment: Trees on the hexagonal lattice (Pippenger & Schleich’03) Up to translation and rotations, there is a unique way to embed an unrooted colored tree on the colored hexagonal lattice (possibly with overlaps). � 2 n � 6 1 = n +2 n +1 n

An elegant restatment: Trees on the hexagonal lattice (Pippenger & Schleich’03) Up to translation and rotations, there is a unique way to embed an unrooted colored tree on the colored hexagonal lattice (possibly with overlaps). � 2 n � 6 1 = n +2 n +1 n = # { hexagonal trees with n nodes } .

We got an arrow ! � 2 n � 6 1 n +2 n +1 n

Hexagonal trees are Mireille’s embedded trees... Turn counterclockwise around the tree, and label +1 each side of edges: • at each corner ℓ = ℓ +1 • at each leaf: ℓ = ℓ − 3. -3

Hexagonal trees are Mireille’s embedded trees... Turn counterclockwise around the tree, and label +1 each side of edges: • at each corner ℓ = ℓ +1 • at each leaf: ℓ = ℓ − 3. -3 Exemple: 0

Hexagonal trees are Mireille’s embedded trees... Turn counterclockwise around the tree, and label +1 each side of edges: • at each corner ℓ = ℓ +1 • at each leaf: ℓ = ℓ − 3. -3 Exemple: 0 1

Hexagonal trees are Mireille’s embedded trees... Turn counterclockwise around the tree, and label +1 each side of edges: • at each corner ℓ = ℓ +1 • at each leaf: ℓ = ℓ − 3. -3 Exemple: 0 1-2

Hexagonal trees are Mireille’s embedded trees... Turn counterclockwise around the tree, and label +1 each side of edges: • at each corner ℓ = ℓ +1 • at each leaf: ℓ = ℓ − 3. -3 Exemple: 0 1-2 -1

Hexagonal trees are Mireille’s embedded trees... Turn counterclockwise around the tree, and label +1 each side of edges: • at each corner ℓ = ℓ +1 • at each leaf: ℓ = ℓ − 3. -3 Exemple: 0 -4 1-2 -1

Hexagonal trees are Mireille’s embedded trees... Turn counterclockwise i around the tree, and label +1 each side of edges: • at each corner ℓ = ℓ +1 i + 1 i - 2 • at each leaf: ℓ = ℓ − 3. -3 -3 Exemple: 0 -4 1-2 -1 -3 0 More generally: -4 1-2 -1

Hexagonal trees are Mireille’s embedded trees... Turn counterclockwise i around the tree, and label +1 each side of edges: • at each corner ℓ = ℓ +1 i + 1 i - 2 • at each leaf: ℓ = ℓ − 3. -3 -3 Exemple: 0 Read Mireille’s labels on the left of inner edges. -4 1-2 -1 These labels should be viewed as -3 0 More generally: angles (multiples of π/ 3 ). -4 1-2 After a full turn around the tree, the -1 angle variation is − 2 π = − 6 · ( π/ 3) .

A bigger example. -3 0 -4 1 -2 -1 2 -1 -5 0 -3 0 -2 -3 -6 1 -3 -1 -4 -2 -4 -1 -5 0 -2 -3

A bigger example. Rerooting changes the actual -3 0 label, but not the variations! -4 1 -2 -1 2 -1 -5 0 -3 0 -2 -3 -6 1 -3 -1 -4 -2 -4 -1 -5 0 -2 -3

A bigger example. Rerooting changes the actual 3 0 label, but not the variations! 2 1 -2 -1 2 -1 1 0 -3 0 -2 -3 0 1 -3 -1 -4 -2 -4 -1 -5 0 -2 -3

A bigger example. Rerooting changes the actual 3 0 label, but not the variations! 2 1 -2 By rerooting, the sequence of -1 2 variations are cyclically permuted. -1 1 0 -3 0 -2 -3 0 1 -3 -1 -4 -2 -4 -1 -5 0 ⇒ apply the cycle lemma. -2 -3 � 2 n � 6 1 n +2 n +1 n

This should give Mireille’s formula for positive binary trees: recall Theorem (part of her) Let B n be the number of rooted binary trees with n nodes with label ≥ 0 . Then 6(2 n )! + B ≥ 0 B ≥ 0 n +1 = n !( n + 2)! . n In other terms: 6(2 n )! B ≥ 0 + B ≥− 1 = n !( n + 2)! . n n

We got an arrow ! � 2 n � 6 1 n +2 n +1 n

Second interpretation: Dyck paths

Let a Gessel-Xin pair be a pair of Dyck paths such that the height of the two paths differ at most by one. Example: 2 n = 1 3 n = 2 + symmetric 6 n = 3

Let a Gessel-Xin pair be a pair of Dyck paths such that the height of the two paths differ at most by one. Example: 2 n = 1 3 n = 2 + symmetric 6 n = 3 Theorem (Gessel & Xin). The number of Gessel-Xin pairs with total length 2 n is: 6(2 n )! 4 C n − C n +1 = ( n +2)! n ! .

Let a Gessel-Xin pair be a pair of Dyck paths such that the height of the two paths differ at most by one. Example: 2 n = 1 3 n = 2 + symmetric 6 n = 3 Theorem (Gessel & Xin). The number of Gessel-Xin pairs with total length 2 n is: 6(2 n )! 4 C n − C n +1 = ( n +2)! n ! . Can we relate this to the previous binary trees ?

Decomposition at the center of the tree peeling

Decomposition at the center of the tree peeling k T k ( z ) 2 � ⇒ Two binary trees with equal height:

Decomposition at the center of the tree peeling k T k ( z ) 2 � ⇒ Two binary trees with equal height: The center can also be a node:

Decomposition at the center of the tree peeling k T k ( z ) 2 � ⇒ Two binary trees with equal height: The center can also be a node: � k T k ( z ) T k − 1 ( z ) ⇒ Two binary trees with almost the same height:

Decomposition at the center of the tree peeling k T k ( z ) 2 � ⇒ Two binary trees with equal height: The center can also be a node: � k T k ( z ) T k − 1 ( z ) ⇒ Two binary trees with almost the same height: But this approach does not yield the relation to Dyck paths: • Colors are not taken into account correctly... • Not the right notion of height!

A notion of center inherited from Dyck paths. 2 Recall the bijection... 1 ( ( 2 ( ( ) ) 0 ) ) ( ( ) ) 0 ( ( 1 1 ( ( � 2 n 1 � ) ) ) ) n +1 n ( ( 0 0 0 ) ) 1 k 0 0 hence the rule for computing the height: i k = max( i + 1 , j ) j j i # { } = ( ) | � 2 n 6 1 | i − j | ≤ 1 � . n +2 n +1 n

Depending on the position of the root, each edge can get two labels: there is a height labelling of an unrooted tree! 2 0 0 1 1 Exemple: 0 2 1 1 2 2 0 3 1 0 0 0 0 Theorem. Exactly one of the following two cases occur: • there is one edge with the 2 labels that are equal, • or there is one vertex with the 3 incident labels that are equal.

Decomposition at the center of the tree i i The center is an edge: k D k ( z ) 2 3 � ⇒ Two binary trees with equal height: i The center is a node: i i The center can also be a node: k zD k ( z ) 3 ⇒ Three binary trees with the same height: 2 � k 3 D k ( z ) 2 + 2 zD k ( z ) 3 = � 6(2 n )! n !( n +2)! z n . This is correct: � But what we want are pairs of Dyck paths with almost the same height.

i i i +1 i i i i +1 i +1 i i i +1 i +1 i i i i i i +1

i i i +1 i i i i +1 i +1 i i i +1 i +1 i i i i i i +1

i i i +1 i i +1 = i i i i +1 i +1 i i i +1 i +1 i i i i i i +1