Ranked Schrder Trees Olivier Bodini , Antoine Genitrini and Mehdi - PowerPoint PPT Presentation

Ranked Schröder Trees Olivier Bodini ∗ , Antoine Genitrini † and Mehdi Naima ∗ March 18th 2019 ∗ University of Paris Nord † Sorbonne University Laboratoire d’Informatique de Paris Nord. Laboratoire d’Informatique de Paris 6.

Outline 1. Introduction Schröder Trees 2. Strongly Increasing Schröder Trees 3. Weakly Increasing Schröder Trees 4. Relaxed Weakly Increasing Schröder Trees 2

Schröder trees: The simplest combinatorial model 1870: Original model by Schröder S = Z + Seq ≥ 2 S . First elements of the enumeration ( s n ) n = ( 0 , 1 , 1 , 3 , 11 , 45 , 197 , 903 , 4279 , 20793 , . . . ) 3

Schröder trees: The simplest combinatorial model Limitations 1. almost a simply generated tree: long trunk 2. many nodes of arity larger than 3 3. no temporal steps 2

Strongly Increasing Schröder trees

First extension: tree with chronology Strongly increasing Schröder trees: • tree-structure: a Schröder tree 1 • internal nodes increasingly labeled • uniqueness of each label 2 • size: number of leaves 3 6 4 5 8 7 9 3

First extension: tree with chronology Strongly increasing Schröder trees: 1 • tree-structure: a Schröder tree • internal nodes increasingly labeled • uniqueness of each label 2 • size: number of leaves In Analytic Combinatorics [BFS92, FS09] 3 6 4 S = Z + U � ⋆ Seq ≥ 2 S . 5 8 Through the symbolic method, we get an integral equation in 2 variables. (ordinary in z and exponential in u ) 7 9 We are interested in [ z n ]( L u S ) u = 1 . 2

First extension: tree with chronology Strongly increasing Schröder trees: 1 • tree-structure: a Schröder tree • internal nodes increasingly labeled • uniqueness of each label 2 • size: number of leaves In Analytic Combinatorics [BFS92, FS09] 3 6 4 S = Z + U � ⋆ Seq ≥ 2 S . 5 8 Through the symbolic method, we get an integral equation in 2 variables. (ordinary in z and exponential in u ) 7 9 We are interested in [ z n ]( L u S ) u = 1 . Another combinatorial specification allows 2 to introduce an evolution process.

An evolution process Strongly increasing Schröder trees: 1 • tree-structure: a Schröder tree • internal nodes increasingly labeled 2 • uniqueness of each label • size: number of leaves 3 6 4 How does a tree grow ? • Start with a single (unlabeled) leaf; 5 8 • Iterate the following process: at step ℓ (for ℓ ≥ 1 ), select a leaf and replace it by an internal node with 7 9 label ℓ attached to a sequence of at least two leaves. 3

An evolution process Strongly increasing Schröder trees: 1 • tree-structure: a Schröder tree • internal nodes increasingly labeled 2 • uniqueness of each label • size: number of leaves 3 6 4 How does a tree grow ? • Start with a single (unlabeled) leaf; 5 10 8 • Iterate the following process: at step ℓ (for ℓ ≥ 1 ), select a leaf and 7 9 replace it by an internal node with label ℓ attached to a sequence of at least two leaves. 4

Strongly increasing Schröder trees as a growth process 1 20 2 40 5 3 6 12 4 9 7 18 17 13 28 8 16 24 35 10 22 43 54 27 29 63 11 23 62 44 58 14 51 71 34 50 32 55 30 73 48 15 25 46 69 68 72 64 60 86 74 78 70 37 42 81 59 19 31 26 80 61 85 67 91 90 75 39 77 82 21 53 33 65 57 92 84 89 79 36 87 41 93 38 66 52 47 Combinatorial specification: 94 45 49 56 � � S = Z + Θ S × Seq ≥ 1 ( Z ) . 76 83 88 95 z 2 1 − z S ′ ( z ) . S ( z ) is an ordinary formal series: S ( z ) = z + 5

Strongly increasing Schröder trees as a growth process 1 20 2 40 5 3 6 12 4 9 7 18 17 13 28 8 16 24 35 10 22 43 54 27 29 63 11 23 62 44 58 14 51 71 34 50 32 55 30 73 48 15 25 46 69 68 72 64 60 86 74 78 70 37 42 81 59 19 31 26 80 61 85 67 91 90 75 39 77 82 21 53 33 65 57 92 84 89 79 36 87 41 93 38 66 52 47 Combinatorial specification: 94 45 49 56 � � S = Z + Θ S × Seq ≥ 1 ( Z ) . 76 83 88 95 z 2 1 − z S ′ ( z ) . S ( z ) is an ordinary formal series: S ( z ) = z + First elements of the enumeration ( s n ) n = ( 0 , 1 , 1 , 3 , 12 , 60 , 360 , 2520 , 20160 , 181440 , 1814400 , . . . ) Size- n strongly increasing Schröder trees are in bijection with 5 a half of the permutations of { 1 , 2 , . . . , n } . (for n ≥ 2)

Uniform sampling with an incremental process From the specification we have, � t n = n · t n − 1 for n > 2 t 1 = 1 , t 2 = 1 . RandomTree(n): If n=1: Return single leaf Else If n=2: Return a cherry Else T:=cherry For i in 3,. . . ,n-1: k := rand_int(1,i) If k<i: Create a new binary node at position k in T Else: \\ k = i 6 Add a new leaf to the last added internal node in T

Example, RandomTree(8) RandomTree(n): If n=1: Return single leaf Else If n=2: Return a cherry Else T:=cherry For i in 3,. . . ,n-1: k := rand_int(1,i) If k<i: Create a new binary node at position k in T Else: \\ k = i Add a new leaf to the last added internal node in T 7

Example, RandomTree(8) RandomTree(n): If n=1: Return single leaf Else If n=2: Return a cherry Else T:=cherry For i in 3,. . . ,n-1: k := rand_int(1,i) If k<i: Create a new binary node at position k in T Else: \\ k = i Add a new leaf to the last added internal node in T 7

Example, RandomTree(8) RandomTree(n): If n=1: Return single leaf Else If n=2: Return a cherry Else T:=cherry For i in 3,. . . ,n-1: k := rand_int(1,i) If k<i: Create a new binary node at position k in T Else: \\ k = i Add a new leaf to the last added internal node in T 7

Example, RandomTree(8) RandomTree(n): If n=1: Return single leaf Else If n=2: Return a cherry Else T:=cherry For i in 3,. . . ,n-1: k := rand_int(1,i) If k<i: Create a new binary node at position k in T Else: \\ k = i Add a new leaf to the last added internal node in T 7

Example, RandomTree(8) RandomTree(n): If n=1: Return single leaf Else If n=2: Return a cherry Else T:=cherry For i in 3,. . . ,n-1: k := rand_int(1,i) If k<i: Create a new binary node at position k in T Else: \\ k = i Add a new leaf to the last added internal node in T 7

Example, RandomTree(8) RandomTree(n): If n=1: Return single leaf Else If n=2: Return a cherry Else T:=cherry For i in 3,. . . ,n-1: k := rand_int(1,i) If k<i: Create a new binary node at position k in T Else: \\ k = i Add a new leaf to the last added internal node in T 7

Example, RandomTree(8) RandomTree(n): If n=1: Return single leaf Else If n=2: Return a cherry Else T:=cherry For i in 3,. . . ,n-1: k := rand_int(1,i) If k<i: Create a new binary node at position k in T Else: \\ k = i Add a new leaf to the last added internal node in T 7

Strict Increasing Schröder trees Mean Variance Limit law Internal nodes n − ln n ln n Normal distribution Binary nodes n − 2 ln n 4 ln n Normal distribution Degree of root 2 e − 3 0.499.. Difference of 2 poissons 2 e 2 e Leaves attached root n n Table 1: limiting parameters on Strict Increasing Schröder trees 7

Weakly Increasing Schröder trees

Weakly increasing Schröder trees as a growth process 1 4 2 11 8 14 17 3 14 12 19 42 65 6 5 27 22 30 15 26 23 71 58 7 9 6 6 35 39 29 32 33 20 18 38 40 64 8 44 13 48 7 16 49 56 41 44 48 37 47 67 65 67 54 25 51 21 49 53 10 50 27 62 61 70 49 66 63 34 59 45 24 58 57 11 20 52 54 26 40 31 28 56 36 60 43 51 55 69 70 46 68 68 70 S ( z ) is an ordinary formal series: 57 62 � � z 2 S ( z ) = z + S z + − S ( z ) . 1 − z First elements of the enumeration ( s n ) n = ( 0 , 1 , 1 , 3 , 13 , 75 , 541 , 4683 , 47293 , 545835 , 7087261 , . . . ) s n corresponds to the ( n − 1 ) -th ordered Bell numbers. Size- n weakly increasing Schröder trees are in bijection with ordered partitions of { 1 , 2 , . . . , n − 1 } . 8

Ordered Bell numbers Stirling partition numbers Number of ways to partition a set of n objects into k non-empty subsets Example � 3 � = | {{ 1 , 2 }{ 3 }} , {{ 1 , 3 }{ 2 }} , {{ 1 }{ 2 , 3 }} | = 3. 2 Ordered Bell numbers n � n � 1 n ! � B n = k ! B ( z ) = B n ∼ , 2 − e z , 2 (ln 2 ) n + 1 . k k = 0 9

Bijection s n , k is the number of trees with n leaves and k iteration steps Theorem � n − 1 � s n , k = k ! k Result � 1 � n s n = B n − 1 = ( n − 1 )! 2 ln 2 10

Example of the bijection [ { 3 , 4 } , { 1 , 6 , 7 } , { 2 , 8 } , { 5 } ] The mapping M 2 • p ′ = [ p 1 ] p ′ = [ { 3 , 4 } ] • the degree of the root | p 1 | + 1 • For i ∈ [ 2 , . . . , ℓ ] :  p ′ =add p i to p ′ � M 2 p ′ = normalize ( p ′ ) For each e ∈ runs ( p ′ i ) : Create node labeled i with | e | + 1 leaves Add the node at position e 1 . 11

Example of the bijection [ { 3 , 4 } , { 1 , 6 , 7 } , { 2 , 8 } , { 5 } ] The mapping M 2 • p ′ = [ p 1 ] p ′ = [ { 3 , 4 } , { 1 , 6 , 7 } ] • the degree of the root | p 1 | + 1 • For i ∈ [ 2 , . . . , ℓ ] : ? p ′ =add p i to p ′ p ′ = normalize ( p ′ ) For each e ∈ runs ( p ′ i ) : Create node labeled i with | e | + 1 leaves Add the node at position e 1 . 12