RECURSIVE TREES. Michael Drmota. Institue of Discrete Mathematics and Geometry Vienna University of Technology, A 1040 Wien, Austria


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1 RECURSIVE TREES Michael Drmota Institue of Discrete Mathematics and Geometry Vienna University of Technology, A 040 Wien, Austria Enrage Topical School on GROWTH AND SHAPES, Paris, IHP, June 2 6, 2008
2 Contents Combinatorics on Recursive Trees The Shape or Random Recursive Trees Cutting down Recursive Trees Plane Oriented Recursive Trees Methodology Mixture of combinatorial, analytic and probabilistic methods Heavy use of generating function
3 Recursive Trees
4 Recursive Trees....
5 Recursive Trees.. 2..
6 Recursive Trees
7 Recursive Trees
8 Recursive Trees
9 Recursive Trees
10 Recursive Trees
11 Recursive Trees Combinatorial Description labelled rooted tree labels are strictly increasing no lefttoright order (nonplanar)
12 Recursive Trees Motivations spread of epidemics pyramid schemes familiy trees of preserved copies of ancient texts convex hull algorithms...
13 Enumeration of Recursive Trees All recursive trees of size 4:
14 Enumeration of Recursive Trees Number of recursive trees y n = number of recursive trees of size n = (n )! The node with label j has exactly j possibilities to be inserted = y n = 2 (n ).
15 Enumeration of Recursive Trees Bijection to permutations root degree = number of cycles subtree sizes = cycle lengths
16 Enumeration of Recursive Trees Generating Functions: y(x) = n y n x n n! = n x n n = log x y (x) = + y(x) + y(x)2 2! + y(x)3 3! + = e y(x) R = + R + R R + R R R +... A recursive tree can be interpreted as a root followed by an unordered sequence of recursive trees. (y (x) = y n+ x n /n!) n 0
17 Random Recursive Trees Probability Model: Process of growing trees: The process starts with the root that is labeled with. At step j a new node (with label j) is attached to any previous node with probability /(j ). After n steps every tree (of size n) has equal probability /(n )!.
18 Random Recursive Trees....
19 Random Recursive Trees.. p =..
20 Random Recursive Trees.. 2 p =..
21 Random Recursive Trees.. p = /2 2 p = /2..
22 Random Recursive Trees.. 2 p = /2 p = /
23 Random Recursive Trees Remark: lefttoright order is irrelevant =
24 Profile of Recursive Trees First sample of shape parameters: insertion depth of the nth node: D n path length: I n (sum of alle distances to the root) height H n (maximal distance to the root) degree distribution profile X n,k (number of nodes at level k)
25 Profile of Recursive Trees First sample of shape parameters: insertion depth of the nth node: D n path length: I n (sum of alle distances to the root) height H n (maximal distance to the root) degree distribution PROFILE X n,k (number of nodes at level k)
26 Profile of Recursive Trees Relevance of the profile X n,k : P{D n = k} = n E X n,k I n = k 0 kx n,k H n = max{k 0 : X n,k > 0} The profile descibes the shape of the tree.
27 Profile of Recursive Trees Average profile: E X n,k = ( ( )) n n)2 (k log e 2 log n + O. 2π log n log n 0,5 00,50 2 x 3 4
28 Profile of Recursive Trees Central limit theorem for the insertion depth [Devroye, Mahmoud] D n log n log n N(0, ) E D n = log n + O(), V D n = log n + O().
29 Profile of Recursive Trees Lemma [Dondajewski+Szymánski] E X n,k = [u k ] ( n + u ) s n,k = n (n )! s n,k... Stirling numbers of the first kind s n,k... number of permutations of {,... n} with k cycles: n k=0 n k=0 s n,k u k = u(u ) (u n + ) s n,k u k = u(u + ) (u + n )
30 Stirling Numbers s n+,k = s n,k ns n,k, s n+,k = s n,k + n s n,k s n,k k = 0 k = k = 2 k = 3 k = 4 k = 5 k = 6 k = 7 k = 8 n = 0 n = 0 n = n = n = n = n = n = n =
31 Profile of Recursive Trees Remark ( n + α ) = ( ) n ( α) n α = n n Γ(α) ( + O ( )) n Corollary s n,k = (n )!(log n)k k k!γ( log n + ) n! n)2 (k log e 2 log n 2π log n ( + O ( )) n Proof: ( n + u ) n u n Γ(u + ) = [u k ] ( n + u n ) (log n) k k k!γ( log n + ).
32 Profile of Recursive Trees Corollary P{D n = k} = E X n,k n n)2 (k log e 2 log n 2π log n This implies the central limit theorem for D n.
33 Cycles in Permutations s n,k = number of permutations of {,... n} with k cycles Corollary C n... random number of cycles in permutations C n log n log n N(0, ) Corollary R n... root degree of random recursive trees R n log n log n N(0, )
34 Profile of Recursive Trees Profile polynomial W n (u) = X n,k u k k 0 Lemma. The normalized profile polynomial M n (u) = W n(u) E W n (u) is a martingale (with respect to the natural filtration related to the tree evolution process).
35 Profile of Recursive Trees Theorem [Chauvin+Drmota+Jabbour for binary search trees] ( ) Wn (u) E W n (u), u B (M(u), u B) for a suitable domain B C. Remarks (M(u), u B) stochastic process of random analytic functions. Fixed point equation: M(u) uu u M () (u) + ( U) u M (2) (u), where M () (u) and M (2) (u) are independent copies of M(u), U is uniform in [0, ] and (U, M () (u), M (2) (u)) are independent.
36 Profile of Recursive Trees Theorem [Chauvin+Drmota+Jabbour for binary search trees], α I (M(α), α I). E X n, α log n X n, α log n Idea X n,k = [u k ]W n (u) = [u k ]M n (u) E W n (u) [u k ]M(u) E W n (u) M(α)[u k ]E W n (u) = M(α)E X n,k. α = k/ log n... saddle point the function n u u k.
37 Path Length Remark M n() = I n E I n n Corollary I n E I n n M () The random variable M () is not normal. Note also that E I n n log n.
38 Leaves in Recursive Trees Theorem [Najock+Heyde] L n... number of leaves in a random recursive tree of size n n P{L n = k} =. (n )! k n k... Eulerian numbers Remark n k = number or recursive trees of size n with k leaves.
39 Eulerian Numbers n k = k n k + (n k) n k n k k = 0 k = k = 2 k = 3 k = 4 k = 5 k = 6 k = 7 k = 8 n = 0 n = n = 2 4 n = 3 n = n = n = n = n =
40 Leaves in Recursive Trees Corollary L n n n N(0, ) E L n = n 2, E (L n) 2 = 2 (3(n )2 + 3(n ) + 4),
41 Leaves in Recursive Trees Generating functions l n,k... number or recursive trees of size n with k leaves. y(x, u) = n,k l n,k u k xn n! = P{L n = k} u k xn n n,k y(x, u) x = u + e y(x,u) y(x, u) = (x )(u ) + log ( u e (x )(u ) )
42 Degree Distribution l (d) n,k... number or r.t. s of size n with k nodes of outdegree d. L (d) n... number of nodes of outdegree d in a random r.t. of size n: P{L (d) n = k} = l(d) n,k (n )! D n... degree of a random node in a random r.t. of size n P{D n = d} = n E L(d) n Theorem P{D n = d} = ( 2 d+ + O (2 log n) d ) n 2 d!
43 Degree Distribution Generating functions y(x, u) = n,k l (d) n,k uk xn n! = P{L n (d) n,k = k} u k xn n y(x, u) x = e y(x,u) y(x, u)d + (u ) d! Y (x) = y(x, u) u = u= n 0 E L (d) n xn n = P{D n = d} x n n 0 Y (x) = x Y (x) + d! ( log ) d x
44 Degree Distribution Y (x) = x Y (x) + d! ( log ) d x = Y (x) = 2 d+ x + (x ) d j=0 j!2 d+ j ( log ) j x = P{D n = d} = ( 2 d+ + O (2 log n) d ) n 2 d! uniformly for d (2 ε) log n. Corollary P{D n > d} = ( 2 d+ + O (2 log n) d ) n 2 d!
45 Maximum Degree Theorem [Szymánski, Pittel] n... maximum node degree in random r.t. s. E n log 2 n Remark. This degree is much larger than the expected root degree with is about log n.
46 First Moment Method X... discrete random variable on nonnegative integers. = P{X > 0} min{, E X}. Proof E X = k P{X = k} P{X = k} = P{X > 0}. k 0 k
47 Maximum Degree Upper bound: first moment method X d... number of nodes of degree > d: E X d = np{d n > d} n > d X d > 0 = E n = P{ n > d} d 0 = P{X d > 0} d 0 min{, E X d } d 0 d log 2 n + n = log 2 n + O(). d>log 2 n P{D n > d}
48 Second Moment Method Theorem X... nonnegative random variable with bounded second moment = P{X > 0} (E X)2 E (X 2 ). Proof E X = E ( ) X [X>0] E (X 2 ) E ( 2[X>0] ) = E (X 2 ) P{X > 0} Remark In order to apply the second moment method to obtain a lower bound for E n one needs estimates for E (X d ) 2 which can be derived in a similar fashion as above.
49 Maximum Degree Theorem [Goh+Schmutz] P{ n d} = exp ( 2 (d log 2 n+) ) + o() Remark. The limiting behaviour of n is related the the exteme value (= Gumbel) distribution (F (t) = e e t ). The distribution of n is extremely concentrated around d log 2 n. The proof is an analytic tour de force.
50 Height of Recursive Trees Height H n Theorem [Devroye 987, Pittel 994] H n log n e (a.s.)
51 Height Distribution F (z) solution of y F (y/e /e ) = y 0 F (z/e/e )F (y z) dz Recursive sequence of generating functions: y k+ (x) = eyk(x), y 0 (x) = 0, y k (0) = 0. Theorem [Drmota] E H n = e log n + O ( log n (log log n) ). P{H n k} = F (n/y k ()) + o() P{ H n E H n η} e cη (c > 0)
52 Height Distribution y n,k... number of r.t. s of size n and height k: P{H n k} = y n,k /(n )! y k (x) = n 0 P{H n k} xn n = x n y n,k n 0 n! y k+ (x) = ey k(x) Y k (x) = y k (x) = n 0 P{H n+ k}x n (Y k+ (0) = ) Y k+ (x) = Y k+(x)y k (x)
53 Height Distribution y F (y/e /e ) = y 0 F (z/e/e )F (y z) dz Ψ(u) = 0 F (y)e yu dy Y k (x) = e k/e Ψ ( e k/e ( x) )
54 Height Distribution Y k (0) Ck ( 2 e ) k, Y k () = e k/e. Y k+(x) = Y k+ (x)y k (x) For every positive integer l and for every real number k > 0 the difference Y l (x) Y k (x) has exactly one zero ( Intersection Property ).
55 Height Distribution Y k (x) = n 0 Y n,k x n is an entire function with coefficients Y n,k = F ( ve k/e ) v n e v dv 0 and asymptotically we have Y n,k = F ( ne k/e) + o()
56 Height Distribution Remark: The functions satisfy the recurrence y k (x) = x 0 Y k(t) dt = log Y k+ (x) y k+ (x) = e y k(x)
57 Height Distribution Proof idea Y k (x) is approximated by the auxiliary function Y ek (x): Y k () = Y ek () e k = e log Y k (ρ) k. Y k (x) Y ek (x) in a neighbourhood of x = = P{H n k} Y n,ek = F ( n/y k () ) + o()
58 Plane Oriented Trees....
59 Plane Oriented Trees.. p =..
60 Plane Oriented Trees.. 2 p =..
61 Plane Oriented Trees.. p = /3 p = /3 2 p = /3..
62 Plane Oriented Trees p = /3 p = /3 p = /3 3..
63 Plane Oriented Trees Remark: lefttoright order is relevant =
64 Plane Oriented Trees Number of Plane Oriented Trees: y n = number of plane oriented trees of size n = 3 5 (2n 3) = (2n 3)!! = (2n 2)! 2 n (n )! The node with label j has exactly 2j 3 possibilities to be inserted = y n = 3 (2n 3).
65 Plane Oriented Trees Generating Functions: y(x) = n y n x n n! = n 2 n ( 2(n ) n ) x n n = 2x y (x) = + y(x) + y(x) 2 + y(x) 3 + = y(x) R = + R + R R + R R R +... A plane oriented tree can be interpreted as a root followed by an ordered sequence of plane oriented trees. (y (x) = y n+ x n /n!) n 0
66 Plane Oriented Trees Probability Model: Process of growing trees The process starts with the root that is labeled with. At step j a new node (with label j) is attached to any previous node of outdegree d with probability (d + )/(2j 3). After n steps every tree (of size n) has equal probability /(2n 3)!!.
67 Plane Oriented Trees Depth D n of the nth node E D n = H 2n 2 H n = 2 log n + O() V D n = H 2n 2 H n H (2) 2n + 4 H(2) n = 2 log n + O() Central limit theorem: D n 2 log n 2 log n N(0, )
68 Plane Oriented Trees Number L n of leaves E L n = 2n 3 V L n = n 9 8 6(2n ) Central limit theorem: L n 2 3 n n9 N(0, )
69 Plane Oriented Trees Distribution of outdegrees D n... degree of a random node in a random p.o.r.t. of size n P{D n = d} = 4 (d + )(d + 2)(d + 3) + o() Remark. 4 (d + )(d + 2)(d + 3) 4 d 3 as d.
70 Plane Oriented Trees Root degree R n P{R n = k} = (2n 3 k)! 2 n k (n k)! 2 πn e k2 /(4n) E R n = πn + O()
71 Plane Oriented Trees Height H n [Pittel 994] H n log n 2s = (a.s.) where s = is the positive solution of se s+ =. Precise results (as above) are also available ([Drmota]).
72 Dary Recursive Trees
73 Dary Recursive Trees
74 Dary Recursive Trees 2
75 Dary Recursive Trees 2 3
76 Dary Recursive Trees 2 4 3
77 Dary Recursive Trees
78 Dary Recursive Trees
79 Dary Recursive Trees
80 Dary Recursive Trees
81 Cutting down Recursive Trees
82 Cutting down Recursive Trees
83 Cutting down Recursive Trees
84 Cutting down Recursive Trees
85 Cutting down Recursive Trees
86 Cutting down Recursive Trees
87 Cutting down Recursive Trees
88 Cutting down Recursive Trees
89 Cutting down Recursive Trees....
90 Cutting down Recursive Trees X n... number of random cuts to cut down a random r.t. of size n. X 0 = X = 0, X n X In + (n 2), where I n is a discrete random variable with P{I n = k} = (n k)(n k + ) that is independent of (X 0, X,..., X n ). n n (0 k < n)
91 Cutting down Recursive Trees Lemma The probability to that the remaining tree has size = k if we cut a random edge in a random recursive tree of size n equals (n k)(n k + ) n n Proof (k )!(n k )! (n )(n )! k j= ( n j) (k )!(n k )! = ( n n k (n )(n )! n k + n = (n k)(n k + ) n )
92 Cutting down Recursive Trees Theorem [Drmota+Iksander+Möhle+Rösler] X n log n n log log n n (log n) 2 n (log n) 2 Y, where Y is a stable random variable with characteristic function E e iλy = e iλ log λ π 2 λ. E X n = n ( ) ( ) n log n + O log 2, V X n = n2 n 2 n 2 log 3 n + O log 4 n
93 Cutting down Recursive Trees Stable distributions The distribution of random variable X is stable, if for all real a, b and independent copies X, X 2 of X there exists c, d with ax + bx 2 cx + d Examples: normal distribution, Cauchy distribution, Levy distribution All stable distributions can be characterized in term of the characteristic function E e iλx.
94 Coalescent Process
95 Coalescent Process
96 Coalescent Process
97 Coalescent Process
98 Coalescent Process
99 Coalescent Process
100 Coalescent Process
101 Coalescent Process
102 Coalescent Process Stochastic model Let Λ be a measure on [0, ]. Continuous Markov process of partitions of {, 2..., n}, Initial partition: {{}, {2},..., {n}}. If ξ and η are two partitions with a resp. b equivalence classes, where b a + classes of ξ are merged to obtain η. Then the rate q ξ,η that ξ merges to η is q ξ,η = { [0,]( ( x)b bx( x) b )x 2 dλ(x) if ξ = η, [0,] xb a ( x) a dλ(x) if ξ η.
103 Coalescent Process Kingmancoalescent Λ = δ 0 BolthausenSznitmancoalescent Λ = univ[0, ]
104 Coalescent Process Remark The process of number of classes is also a Markov process with rates ( a < b n) ) g ba = ( b a [0,] xb a ( x) a dλ(x)
105 Coalescent Process BolthausenSznitmancoalescent X n... number of collisions until there is a single block: X 0 = X = 0, X n X In + (n 2), where I n is a discrete random variable with P{I n = k} = (n k)(n k + ) that is independent of (X 0, X,..., X n ). n n (0 k < n)
106 Cutting down Recursive Trees Lemma f(s, t) = n E s Xn t n satisfies the partial differential equation f(s, t) t ( t + with initial condition f(s, 0) =. t log( t) ( s ) ) = f(s, t)
107 Cutting down Recursive Trees Expected Value g(t) := f(s, t) = E X s n t n s= = g(t) = = g (t) g(t) t = t ( t) 3 log t ( t) 2 log t log log t t + O ( t) 2 log 2 t = E X n = n ( ) n log n + O log 2 n.
108 Thank You!
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