On the shape of binary trees

Transcription

1 On the shape of binary trees Mireille Bousquet-Mélou, CNRS, LaBRI, Bordeaux ArXiv math.co/ ArXiv math.pr/ (with Svante Janson) bousquet

2 A complete binary tree n internal vertices, called nodes (size n) n + external vertices, called leaves

3 n nodes (size n) An (incomplete) binary tree

4 A plane tree Each node has a (possibly empty) ordered sequence of children n edges, n + nodes (size n)

5 What is the shape of a (large) random binary tree?

6 The (horizontal) profile of a binary tree height A binary tree of height 4 (or 5...)

7 The horizontal profile of a plane tree A plane tree of height 3.

8 NEW! The vertical profile of a binary tree LABEL = ABSCISSA A binary tree of right width 3, and vertical profile [2,2;4,2,,].

9 Why study the vertical profile of trees?

10 Why study the vertical profile of trees? Wait and see!

11 Limit results on the shape of trees: general approach Enumerative combinatorics (decompositions of trees, recurrence relations, generating functions...) Singularity analysis [Flajolet-Odlyzko 90]: extract from the generating functions the asymptotic behaviour of their coefficients

12 Decomposition and enumeration of binary trees Let a n be the number of binary trees with n nodes (internal vertices): a 0 = a n = n m=0 a m a n m Let A(t) := n 0 a n t n be the associated generating function: A(t) = + ta(t) 2. m n m = + A(t) = + ta(t) 2

13 Decomposition and enumeration of plane trees Let a n be the number of plane trees with n edges (n + vertices): a 0 = a n = n m=0 a m a n m Let A (t) := n 0 a nt n be the associated generating function: A (t) = + ta (t) 2. m n m = + A (t) = + ta (t) 2

14 The Catalan numbers There are as many binary trees with n nodes as plane trees with n edges: The associated generating function is a n = a n A(t) = A (t) = n a n t n It satisfies Hence A(t) = + ta(t) 2 A(t) = 4t 2t is the nth Catalan number. a n = a n = ( 2n n + n ) = n 0 ( 2n) t n. n + n

15 Part. Height and width of binary trees??????

16 Counting trees of bounded height Let H j (t) H j be the generating function of binary trees of height at most j: H j = h n, j t n n 0 H 0 = H j = + th 2 j = + Let H j (t) H j be the generating function of plane trees of height at most j, where t counts edges. H 0 = H j = + th j H j = +

17 Counting trees of bounded (right) width Let W j (t) W j be the generating function of binary trees of right width at most j. j + = + j j For j 0, j W = W j = + tw j+ W j

18 Trees of bounded height or bounded width: functional equations Family of trees Bounded height Bounded (right) width H 0 = W = H j = + th j 2 W j = + tw j+ W j (j 0) H 0 = H j = + th j H j

19 Generating functions for trees of bounded height/width Proposition: The generating function of plane trees of height j is: H j = A Qj+ Q j+2 where A counts plane trees and Q = A [de Bruijn, Knuth, Rice 72].

20 Generating functions for trees of bounded height/width Proposition: The generating function of plane trees of height j is: H j = A Qj+ Q j+2 where A counts plane trees and Q = A [de Bruijn, Knuth, Rice 72]. Proposition [mbm 06]: The generating function of binary trees of right width j is: W j = A ( Zj+2 )( Z j+7 ) ( Z j+4 )( Z j+5 ), where A counts binary trees and Z Z(t) is the unique series in t such that ( + Z 2 ) 2 Z = t Z + Z 2 and Z(0) = 0. But... w h y?

21 Limit results on the shape of trees: general approach Enumerative combinatorics: Using a recursive decomposition of trees, write equations for the relevant generating functions and solve them... Use the results (or the technique) of singularity analysis [Flajolet-Odlyzko 90] to extract from these series the asymptotic behaviour of their coefficients Results: If B(t) = n b n t n behaves like this in the neighborhood of its dominant singularity t s, then its coefficients b n behave asymptotically like that. Technique: Cauchy s formula b n = 2iπ for a carefully chosen contour C n. C n B(t) dt t n+

22 The average height of binary trees: experimental approach Consider the average height of plane trees of size n: E(H n) := h(τ) = j (h n, j a n a h n, j ). n τ =n j H n

23 Convergence in law of the height: experimental approach The number of trees of size n and height at most j is h n, j. For x 0, consider ( ) H F n (x) := P n x = h n a n, x n, n the proportion of trees of size n having height at most x n. It is the probability that a random tree of size n has height at most x n. Graph of F n (x) : n = 5 n = 0 n = 20 n = 30 5 n 00.

24 Limit results for the height of trees Convergence in law: For all x 0, ( H P n x n ) F(x), where F(x) = i π H coth(x z) ze z dz i H = + k= e k2 x 2 ( 2k 2 x 2 ) 0 Convergence of the moments: Let H n be the (random) height of a plane tree with n edges. As n, E(H n) n π, E ) k k(k )Γ(k/2)ζ(k) for k 2. n ( H n [Flajolet-Odlyzko 82], [Brown-Schubert 84] + similar results for binary trees

25 Limit results for the right width of binary trees [mbm 06] Convergence of the moments: Let W n be the (random) right width of a binary tree with n nodes. Then, as n, E(W n ) n /4 3 ( ) 2 π Wn, E 2Γ(3/4) n /4 6 π, E ( ) k Wn n /4 = 24 πk!ζ(k ) 2 k Γ((k 2)/4) for k 3. Convergence in law: P ( ) Wn n /4 x G(x) G(x) = 3 i π H ze z i sinh 2 (x( z /4 / 2)) dz 0 H

26 Height and width of binary trees β n /4 α n

27 Part 2. What about the profiles?

28 Gallery of (horizontal and vertical) profiles y 60 y 60 y 60 y 60 y x x x x x Horizontal profiles of random binary trees with 000 nodes y 00 y 00 y 00 y 00 y x x x x x Vertical profiles of random binary trees with 000 nodes.

29 The average (horizontal and vertical) profiles Average horizontal profile of plane trees of size 0, 20, 30, 40, 50: Average vertical profile of binary trees of size 5, 0, 20:

30 One new ingredient: bivariate generating functions Before: Given j, how many plane trees of size n have height at most j? series H j (t) = n h n, j t n Now: Given j, how many plane trees of size n have exactly k nodes at height j? series Pj (t, u) = p n,k,j t n u k n,k

31 The number of nodes at height j (plane trees) Let Pj P j (t, u) be the generating function of plane trees, counted by edges (variable t) and by the number of nodes at height j (variable u): P 0 = ua(t) P j = + tp j P j = + Proposition [???]: P j = A MQj MQ j+ where A counts plane trees, Q = A and M = A u tua 2 u + A( u) + tua 2 ( A).

32 The number of nodes at abscissa j (binary trees) Let V j V j (t, u) be the generating function of binary trees, counted by nodes (t) and by the number of nodes at abscissa j (variable u). V 0 = + utv 2 For j, V j = + tv j+ V j j = + j Proposition [mbm 06]: the series V j are algebraic. Moreover, V j = A ( + MZj )( + MZ j+5 ) ( + MZ j+2 )( + MZ j+3 ) where A counts binary trees, ( + Z 2 ) 2 Z = t Z + Z 2, and M M(t, u) is the unique power series in t such that Z( + MZ) 2 ( + MZ 2 )( + MZ 6 ) M = (u ) ( + Z) 2 ( + Z + Z 2 )( Z) 3 ( M 2 Z 5 ).

33 The horizontal profile A tree of size n has, on average, height O( n). How many nodes are there at a given (horizontal) level????

34 The horizontal profile A tree of size n has, on average, height O( n). How many nodes are there at a given (horizontal) level? About n.

35 The horizontal profile A tree of size n has, on average, height O( n). How many nodes are there at a given (horizontal) level? About n. Let X n (j) be the number of nodes located at height j in a random plane tree with n edges. We study the quantity X n( λ n ) n.

36 The vertical profile A binary tree of size n has, on average, width O(n /4 ). How many nodes are there on a given (vertical) layer? About n 3/4. Let Y n (j) be the number of nodes located at abscissa j in a random binary tree with n nodes. We study the quantity Y n ( λn /4 ) n 3/4.

37 The horizontal profile of plane trees The sequence X n ( λ n ) n converges in law for each λ (and as a process [Drmota-Gittenberger 97] ) The first moment: E ( X n ( λ n ) n ) 2λe λ On average, there are about 2λe λ2 n nodes at height λ n in a plane tree having n edges In other words, the height of a random node in a random tree, once normalized by n, follows a law of density 2λe λ2.

38 The vertical profile of binary trees Let Y n (j) be the (random) number of nodes at abscissa j in a binary tree having n nodes. The sequence Y n( λn /4 ) n 3/4 converges in law to a random variable Y(λ) described explicitly by its Laplace transform for every λ (and as a process [mbm-janson 06]) Moreover, E ( Yn ( λn /4 ) n 3/4 ) 2π m 0 ( 2 λ ) m m! cos ( (m + )π m + 3 Γ 4 4 This gives the average number of nodes at abscissa λn /4 in a random binary tree having n nodes. ) x [mbm 06]

39 An example: nodes at abscissa 0 The random variable Y n (0) n 3/4, which gives the (normalized) number of nodes at abscissa 0, converges in law to a variable Y(0) such that E ( Y (0) k) = ( 2 3 ) k Γ( + 3k/4) Γ( + k/2). Hence (...) Y (0) = parameter 2/ T 2/3, where T 2/3 follows a unilateral stable law of E(e at 2/3 ) = e a2/3 for a 0. Merci Alain Rouault!

40 An example: nodes at abscissa 0 The random variable Y n (0) n 3/4, which gives the (normalized) number of nodes at abscissa 0, converges in law to a variable Y(0) such that E ( Y (0) k) = ( 2 3 ) k Γ( + 3k/4) Γ( + k/2). Hence (...) Y (0) = parameter 2/ T 2/3, where T 2/3 follows a unilateral stable law of E(e at 2/3 ) = e a2/3 for a 0. Merci Alain Rouault! The number of nodes lying at a positive abscissa, normalized by n, converges to U(0,) [Aldous].

41 Part 3. Why study the vertical profile? Binary trees (drawn in a canonical way) form a family of embedded trees. There are many other families! Random increments 0, ± along edges The same questions can be asked for any class of embedded trees: - what is the largest label? (width) - How many nodes have label j? (profile)

42 Why study embedded trees? They have interesting combinatorial properties (mysterious algebraic series)

43 Why study embedded trees? They have interesting combinatorial properties (mysterious algebraic series) The GF of embedded plane trees with increments ± having largest label at most j is where T = + 2tT 2 and T j = T ( Zj+ )( Z j+5 ) ( Z j+2 )( Z j+4 ), Z = t ( + Z)4 + Z 2.

44 Why study embedded trees? They have interesting combinatorial properties (mysterious algebraic series) Embedded trees with nonnegative labels are related bijectively to planar maps [Cori-Vauquelin 8], [Chassaing-Schaeffer 04], [Del Lungo, Del Ristoro, Penaud 00], [Bouttier-Di Francesco-Guitter 03]

45 Why study embedded trees? They have interesting combinatorial properties (mysterious algebraic series) Embedded trees with nonnegative labels are related bijectively to planar maps [Cori-Vauquelin 8], [Chassaing-Schaeffer 04], [Del Lungo, Del Ristoro, Penaud 00], [Bouttier-Di Francesco-Guitter 03] They are related to ISE (the Integrated SuperBrownian Excursion) [Aldous 93], [Borgs et al. 99]

46 The Integrated SuperBrownian Excursion The ISE is a (random) probability distribution on R d that occurs (almost) everywhere. At least, as soon as a branching structure (tree) is combined with an embedding of the nodes in the space. The ISE describes how the space is occupied by the nodes [Aldous 93], [Marckert-Mokkadem 03]

47 Embedded trees and ISE Let T n denote a random embedded tree with n nodes. Let µ n be the occupation measure of T n : µ n = v T n δ(l(v) ), where l(v) denotes the label (position) of the node v, and δ(x) is the Dirac measure at x Histogram of µ n

48 Embedded trees and ISE Let T n denote a random embedded tree with n nodes. Let µ n be the occupation measure of T n : µ n = n v T n δ(l(v) ), where l(v) denotes the label (position) of the node v, and δ(x) is the Dirac measure at x. Then µ n is a probability distribution (total weight ) Histogram of µ n

49 Embedded trees and ISE Let T n denote a random embedded tree with n nodes. Let µ n be the occupation measure of T n : µ n = n v T n δ(l(v)n /4 ), where l(v) denotes the label (position) of the node v, and δ(x) is the Dirac measure at x. Then µ n is a probability distribution (total weight ). It is random since it depends on the random tree T n Histogram of µ n

50 Embedded trees and ISE Let T n denote a random embedded tree with n nodes. Let µ n be the occupation measure of T n : µ n = n v T n δ(l(v)n /4 ), where l(v) denotes the label (position) of the node v, and δ(x) is the Dirac measure at x. Thm. As n grows, µ n µ ise, where µ ise is the ISE. [Aldous 93, Borgs et al. 99, Janson-Marckert 04] This holds for many families of random embedded trees! Histogram of µ n

51 Another occurrence of ISE: Properly embedded trees In dimension d 8, the occupation measure of properly embedded trees converges also to the d-dimensional ISE. [Derbez-Slade 98]

52 Main objective: study ISE (in D) via embedded trees y 00 y 00 y 00 y 00 y x x x x x Largest label law of the maximum of the support of the ISE Number of nodes at abscissa j = λn /4 law of the density of the ISE at a fixed point λ Number of nodes at abscissa j = λn /4 law of the distribution function of the ISE at a fixed point λ

53 A beautiful slide for Wjcch. Let M i be the ith moment of the ISE (M i is a random variable!) M i = lim n n v T n ( l(v)n /4 ) i

54 A beautiful slide for Wjcch. Let M i be the ith moment of the ISE (M i is a random variable!) Then E ( M 2k for k > 0, M i = lim n n v T n ( l(v)n /4 ) i ) ( ) = 0 and E M 2k a = k Γ(/2) 2 k/2 Γ((5k )/2), where a 0 = 2 and 4 a k = k i= ( 2k) ai a k i + k(2k )(5k 4)(5k 6)a k 2i

55 A beautiful slide for Wjcch. Let M i be the ith moment of the ISE (M i is a random variable!) Then E ( M 2k for k > 0, Also, 4 a k = M i = lim n n ) = 0 and E ( M 2k k i= v T n ( l(v)n /4 ) i ) = a k Γ(/2) 2 k/2 Γ((5k )/2), where a 0 = 2 and ( 2k) ai a k i + k(2k )(5k 4)(5k 6)a k 2i E(M 2k Ml 2 ) = a k,l Γ(/2) 2 (k+l)/2 Γ((5k + 3l )/2), where a 0,0 = 2 and the a k,l are determined by induction on k + l: a k,l = 4 (0,0)<(i,j)<(k,l) ( 2k 2i )( l j ) ai,j a k i,l j + 2l(l )a k+,l k(2k )(5k +3l 4)(5k +3l 6)a k,l + 2 (4k +)l(5k +3l 4)a k,l.

56 That s it!

57 The Laplace transform of Y (λ) Let λ 0. The sequence Y n ( λn /4 ) converges in distribution to a nonnegative random variable Y (λ) whose Laplace transform is given, for a < 4/ 3, by where L(λ, a) = + 48 i π A(x) A is the unique solution of E ( e ay (λ)) = L(λ, a) Γ A(a/v 3 )e 2λv ( + A(a/v 3 )e 2λv ) 2v5 e v4 dv, A = x ( + A) 3 24 A satisfying A(0) = 0, and the integral is taken over Γ = { te iπ/4, t (,0]} { + te iπ/4, t [0, )}.

58 Did you say ISE?