A new bijection on m -Dyck paths and application to random sampling - PowerPoint PPT Presentation

A new bijection on m -Dyck paths and application to random sampling Axel Bacher LIPN, Université Paris 13 June 3rd, 2016

Outline Introduction 1 The folding bijection 2 m -Łukasiewicz paths and m -Dyck prefixes Folding and unfolding Random sampling 3 Main algorithm Complexity and limit law Perspectives 4

m -Dyck paths and ( m +1) -ary trees m -Dyck path: path in N from 0 to 0 with steps in { +1 , − m } . � n 1 � The number of paths of length n = ( m +1) n ′ is mn ′ +1 n ′ (Fuß-Catalan number). Random sampling of general classes of trees in time O ( n log n ) , based on the cycle lemma. [Devroye ’12]

The folding bijection for Dyck paths One can sample a Dyck path in time O ( n ) by sampling a Dyck prefix and use the folding bijection to get a pointed Łukasiewicz path. [Barcucci, Pinzani, Sprugnoli ’92; B., Bodini, Jacquot ’15]

m -Łukasiewicz paths m -Łukasiewicz path: non-negative path except at its end. Paths of length n = ( m +1) n ′ + r have height h = r − ( m +1) . � n Their number is r � (Raney number). n n ′

m -Łukasiewicz paths m -Łukasiewicz path: non-negative path except at its end. Paths of length n = ( m +1) n ′ + r have height h = r − ( m +1) . � n Their number is r � (Raney number). n n ′ A pointed path has an associated factorization: � 0 ≤ h ( q i ) ≤ m − 1 , i < k w = pq 0 d · · · q k d , where 0 ≤ h ( q k ) ≤ r − 1 .

Decorated m -Dyck prefixes Paths of length n = ( m +1) n ′ + r have height h = ( m +1) h ′ + r .

Decorated m -Dyck prefixes a 2 = 2 a 1 = 3 a 0 = 1 Paths of length n = ( m +1) n ′ + r have height h = ( m +1) h ′ + r . A decoration of an m -Dyck prefix is defined as a sequence: � i < h ′ 1 ≤ a i ≤ m , ( a 0 , . . . , a h ′ ) , where 1 ≤ a k ≤ r . (a path has m h ′ r possible decorations).

Decorated m -Dyck prefixes a 2 = 2 a 1 = 3 a 0 = 1 Paths of length n = ( m +1) n ′ + r have height h = ( m +1) h ′ + r . A decoration of an m -Dyck prefix is defined as a sequence: � i < h ′ 1 ≤ a i ≤ m , ( a 0 , . . . , a h ′ ) , where 1 ≤ a k ≤ r . (a path has m h ′ r possible decorations). A decorated m -Dyck prefix has an associated factorization: w = p u q 0 · · · u q h ′ , where h ( q i ) = a i .

Folding and unfolding p u q 0 · · · u q k pq 0 d · · · q k d Theorem The folding operation is a bijection between decorated m -Dyck prefixes and pointed m -Łukasiewicz paths. Folding or unfolding only requires reading the part of the path after the point.

Random m -Dyck prefix m 1 We draw u and d steps with probabilities m +1 and m +1 .

Random m -Dyck prefix m 1 We draw u and d steps with probabilities m +1 and m +1 .

Random m -Dyck prefix m 1 We draw u and d steps with probabilities m +1 and m +1 .

Random m -Dyck prefix m 1 We draw u and d steps with probabilities m +1 and m +1 .

Random m -Dyck prefix m 1 We draw u and d steps with probabilities m +1 and m +1 .

Random m -Dyck prefix m 1 We draw u and d steps with probabilities m +1 and m +1 .

Random m -Dyck prefix m 1 We draw u and d steps with probabilities m +1 and m +1 . If we go in the negatives, we randomly point the path and unfold.

Random m -Dyck prefix m 1 We draw u and d steps with probabilities m +1 and m +1 . If we go in the negatives, we randomly point the path and unfold.

Random m -Dyck prefix m 1 We draw u and d steps with probabilities m +1 and m +1 . If we go in the negatives, we randomly point the path and unfold.

Random m -Dyck prefix m 1 We draw u and d steps with probabilities m +1 and m +1 . If we go in the negatives, we randomly point the path and unfold.

Random m -Dyck prefix m 1 We draw u and d steps with probabilities m +1 and m +1 . If we go in the negatives, we randomly point the path and unfold.

Random m -Dyck prefix m 1 We draw u and d steps with probabilities m +1 and m +1 . If we go in the negatives, we randomly point the path and unfold.

Random m -Dyck prefix m 1 We draw u and d steps with probabilities m +1 and m +1 . If we go in the negatives, we randomly point the path and unfold.

Random m -Dyck prefix m 1 We draw u and d steps with probabilities m +1 and m +1 . If we go in the negatives, we randomly point the path and unfold. At all times, paths of height ( m +1) h ′ + r appear with probability proportional to m h ′ .

Random m -Łukasiewicz path We draw a random m -Dyck prefix of length n = ( m +1) n ′ + r and height h = ( m +1) h ′ + r . We randomly decorate this prefix and we fold. The result is a uniform pointed m -Łukasiewicz path.

Complexity Random m -Łukasiewicz path w ← ε for i = 1 , . . . , n do m s ← u with probability m +1 , d otherwise w ← ws if h ( w ) < 0 then randomly point w unfold w (forget the decoration) end if end for randomly decorate w fold w (forget the point) return w

Complexity Random m -Łukasiewicz path w ← ε for i = 1 , . . . , n do m s ← u with probability m +1 , d otherwise w ← ws if h ( w ) < 0 then randomly point w unfold w (forget the decoration) end if end for randomly decorate w fold w (forget the point) return w We consider complexity in random bits and memory accesses.

Complexity Random m -Łukasiewicz path w ← ε for i = 1 , . . . , n do m s ← u with probability m +1 , d otherwise β w ← ws 1 if h ( w ) < 0 then randomly point w O (log i ) unfold w (forget the decoration) Unif { 1 , . . . , i } end if end for O ( √ n ) randomly decorate w fold w (forget the point) Unif { 1 , . . . , n } return w We consider complexity in random bits and memory accesses.

Complexity Random m -Łukasiewicz path w ← ε for i = 1 , . . . , n do m s ← u with probability m +1 , d otherwise β w ← ws 1 if h ( w ) < 0 then randomly point w O (log i ) unfold w (forget the decoration) Unif { 1 , . . . , i } end if end for O ( √ n ) randomly decorate w fold w (forget the point) Unif { 1 , . . . , n } return w We consider complexity in random bits and memory accesses. The if branches are independent with probability ∼ 1 2 i .

Complexity (cont.) Theorem The cost in random bits and memory accesses satisfies: B n M n d d − → β ; − → 1 + X + Unif[0 , 1] . n n 1 � � The number β is the cost in random bits of Bernoulli . m +1 According to [Knuth, Yao ’76] , we can take: 1 1 � � m � m � β ∼ − m +1 log 2 − m +1 log 2 . m +1 m +1

Complexity (cont.) Theorem The cost in random bits and memory accesses satisfies: B n M n d d − → β ; − → 1 + X + Unif[0 , 1] . n n 1 � � The number β is the cost in random bits of Bernoulli . m +1 According to [Knuth, Yao ’76] , we can take: 1 1 � � m � m � β ∼ − m +1 log 2 − m +1 log 2 . m +1 m +1 The law X is defined by: � X = Unif[0 , x ] , x ∈ S 1 where S is a Poisson point process of density λ ( x ) = 2 x on (0 , 1] .

Properties of the limit law Poisson x ∈ (0 , 1] ( 1 − x � The distribution X = 2 x ) Cumulant generating function K ( z ) = log E ( e zX ) : � z e y − 1 − y 1 K ( z ) = dy , κ n ( X ) = 2 n ( n + 1) . 2 y 2 0

Properties of the limit law Poisson x ∈ (0 , 1] ( 1 − x � The distribution X = 2 x ) Cumulant generating function K ( z ) = log E ( e zX ) : � z e y − 1 − y 1 K ( z ) = dy , κ n ( X ) = 2 n ( n + 1) . 2 y 2 0 Distribution function F ( x ) = P ( X ≤ x ) : F ( x ) + F ′ ( x ) + 2 xF ′′ ( x ) = F ( x − 1)

Properties of the limit law Poisson x ∈ (0 , 1] ( 1 − x � The distribution X = 2 x ) Cumulant generating function K ( z ) = log E ( e zX ) : � z e y − 1 − y 1 K ( z ) = dy , κ n ( X ) = 2 n ( n + 1) . 2 y 2 0 Distribution function F ( x ) = P ( X ≤ x ) : F ( x ) + F ′ ( x ) + 2 xF ′′ ( x ) = F ( x − 1) √ F ( x ) = sin 2 x , 0 ≤ x ≤ 1 .

Properties of the limit law Poisson x ∈ (0 , 1] ( 1 − x � The distribution X = 2 x ) Cumulant generating function K ( z ) = log E ( e zX ) : � z e y − 1 − y 1 K ( z ) = dy , κ n ( X ) = 2 n ( n + 1) . 2 y 2 0 Distribution function F ( x ) = P ( X ≤ x ) : F ( x ) + F ′ ( x ) + 2 xF ′′ ( x ) = F ( x − 1) � √ 2 e 1 − γ F ( x ) = sin 2 x , 0 ≤ x ≤ 1 . π

Properties of the limit law Poisson x ∈ (0 , 1] ( 1 − x � The distribution X = 2 x ) Cumulant generating function K ( z ) = log E ( e zX ) : � z e y − 1 − y 1 K ( z ) = dy , κ n ( X ) = 2 n ( n + 1) . 2 y 2 0 Distribution function F ( x ) = P ( X ≤ x ) : F ( x ) + F ′ ( x ) + 2 xF ′′ ( x ) = F ( x − 1) � √ 2 e 1 − γ F ( x ) = sin 2 x , 0 ≤ x ≤ 1 . π Tail distribution asymptotics: 1 − F ( x ) = x − x (log x ) − 2 x ( e/ 2) x + o ( x ) .

Graph of the distribution function F ( x ) 1 x 0 1 � 2 e 1 − γ √ 1 sin 2 x π x 0 1

Perspectives Can we use a similar method for other paths ( + a, − b )?