Forty years of tree enumeration

Transcription

1 October 24, 206

2 Canada John Moon (Alberta) Counting labelled trees Canadian mathematical monographs

3 What is our subject? Asymptotic enumeration or Probabilistic combinatorics?

4 What is our subject? Asymptotic enumeration AND Probabilistic combinatorics!

5 What is our subject? Exact enumeration!

6 Binary trees and exact enumeration Figure: Leaves labelled left to right

7 Theorem The expectation E n,j of the height of the leaf with label j in a binary tree of size n, is given by E n,j = 2(2j + )(2n 2j + ) n + 2 ( 2j j )( 2n 2j n j ) ( 2n n ) for 0 j n. ()

8 Binary trees and exact enumeration Figure: Nodes labelled via inorder traversal

9 Binary trees and exact enumeration Theorem The expectation E n,j of the number of descendants of the node with label j, where the nodes are labelled by inorder traversal, in a binary tree of size n, is given by E n,j = n + 4 ( 2j j )( 2(n+ j) n+ j ) ( 2n n ) for j n

10 Binary trees and exact enumeration Theorem The expectation E n,j of the number of descendants of the node with label j, where the nodes are labelled by inorder traversal, in a binary tree of size n, is given by E n,j = n + 4 ( 2j j )( 2(n+ j) n+ j ) ( 2n n ) for j n Many more results of this style are available (together with Alois Panholzer). Trivariate generating functions and Zeilberger s algorithm.

11 Preorder

12 Postorder

13 Lattice paths also belong to tree enumeration! Figure: A planar tree with 8 nodes (=7 edges) and the corresponding Dyck path of length 4 (=semi-length 7)

14 Dyck like paths Upsteps +, downsteps 2 (in general (t )) T T 2 T 3 Figure: The decomposition of generalized Dyck paths leading (recursively) to a ternary tree with subtrees T, T 2, T 3.

15 Returns and hills Figure: A ternary tree with 0 (internal) nodes. It has 6 returns and 3 hills.

16 explicit! Explicit forms of bivariate generating functions are available, and results about the limiting distribution (negative binomial distribution) Solving an open question.

17 Quickselect Permutation Pivot is

18 Quickselect

19 Quickselect

20 Quickselect

21 Quickselect If one only wants the find the jth ranked element (which is j, since we assume that the elements in question are {, 2,..., n}), one uses the same partitioning strategy as in Quicksort, but follows only the path which contains the sought element. This is the same as to say that one goes down recursively in only one subfile. This procedure is called Hoare s Find algorithm. Knuth has already computed the average number of comparisons C n,j. For this, it is assumed that every permutation of {, 2,..., n} is equally likely, and that the partitioning phase needs n comparisons. Then there is the recursion C n,j = n + n k<j C n k,j k + n j<k n C k,j. The solution is ) C n,j = 2 (n (n + )H n (j + 2)H j (n + 3 j)h n+ j.

22 Quickselect: Second factorial moment (comparisons) M n,j = 2(n + )(n + 6) H 2 n 8H n ( jh j + (n + j) H n+ j ) + 4 (3n2 + 8n + )j 2 + (n + )(3n 2 + 8n + )j + 4(n + ) H n j(n + j) + 2(j + 8)(j + )H 2 j + 2(n + 9 j)(n + 2 j) H 2 n+ j ( + 4 j 2 + (n + )j n 2 ) n + 4 H j H n+ j 2 ( 2(3n + 7)j 3 + (6n 2 + 4n + 3)j 2 ) + (n + )(6n + )j + 8(n + ) H j j(n + j) 2 ( 2(3n + 7)j 3 (2n n + 29)j 2 j(n + j) + 2(n + )(n + 6) H (2) n + (n + )(6n 2 ) + 26n + 5)j + 8(n + ) H n+ j 2(j 2 + 5j + 8) H (2) 2(j 2 (2n + 7)j + n 2 + 7n + 4) H (2) j n+ j + 0j4 20(n + )j 3 + (n 2 3n 66)j 2 + (9n n + 76)(n + )j j(n + j) ( + 4n (n + j) j H n k k= k n+ j + j k= ) H n k k

23 Digital search trees A : 00 B : 00 C : 0000 D : E : 000 F : 00 G : 0 H : 0 I : 00 C 0 B F 0 0 E A D I 0 G H

24 Digital search trees: path length Folklore: the average path length: n k=2 ( ) n ( ) k Q k 2 k with Q k = ( )( )... ( ) k (x; q) n = ( x)( xq)... ( xq n )

25 Digital search trees with l N = N N k=2 R N = Q N ( ) N ( ) k R k 2 k N Q k k=0 Asymptotic evaluation P.Flajolet and R.Sedgewick, Digital Search Trees Revisited, SIAM J. Computing,

26 Digital search trees: external internal nodes External internal nodes in digital search trees via Mellin transforms. H. Prodinger, SIAM Journal on Computing, 2:80 83, 992. Improved constant: nicer, fast convergent series: N Q log 2 + ( ) j 2 (j 2) / Q j (j ) 2 j j 2 Final result l N N ( α + R ( ) ) = N.

27 Digital search trees A big challenge: Dealing with n 2 ( ) n Q k 2 Q n k 2 k k=2 Digital search trees again revisited: The internal path length perspective. P. Kirschenhofer, H. Prodinger and W. Szpankowski, SIAM Journal on Computing, 23:598 66, 994. (paper written ) Analysis of the variance of the path length. Complicated expressions! Much improved a few years ago by Hwang, Fuchs, Zacharowitsch. Approximating much earlier, much nicer constant.

28 Digital search trees: protected nodes A : 00 B : 00 C : 0000 D : E : 000 F : 00 G : 0 H : 0 I : 00 C 0 B F 0 0 E A D I 0 G H Figure: A digital search tree with nine nodes, among which A and D are 2-protected.

29 Digital search trees: protected nodes Theorem The average number of 2-protected nodes in random DSTs of size N is exactly given by l N = N k=2 ( N k 2 )( ) k (n + )2 n n(n+) 4 Q k 2. k Q n= n Theorem The average number l N of 2-protected nodes in random DSTs of size N admits the asymptotic expansion l N = N N δ(log 2 N) + O(), The tiny periodic function δ(x) has a Fourier expansion that could be computed in principle.

30 Rice s method l N = Γ(N + )Γ( z) 2πi C Γ(N + z) ψ(z)dz, where C encircles the poles 2, 3,..., N and no others. The function ψ(z) is the extension of k 2 Q k 2 n= (n + )2 n n(n+) 4 Q n

31 q Combinatorics of geometrically distributed random variables.

32 q Combinatorics of geometrically distributed random variables. Flajolet: Prodinger s q-analogs

33 Batcher s odd-even exchange revisited Batcher s odd-even exchange revisited: a generating functions approach,, Theoretical Computer Science 636 (206), (0, 0) b 0 b 0 a 0 a 0 a a a 2 a 2 b b 2 b b 2 b 0 b b 2 b 0 b b 2... a 0 a 0 a a a 2 a 2 b 0 b 0... a 0 a 0 a a a 2 a 2 b b 2 b b 2 a 0 b 0 b a 0 b 0 a (n, n)

34 Batcher B n : the average number of exchanges Sedgewick: B n = ( 2n ) ( ) 2n (2F (k) + k), (2) n k n k where F (k) is the summatory function of f (j), which is the number of ones in the Gray code representation of j: F (k) := f (j). 0 j<k

35 Batcher (0, 0) b 0 b 0 a 0 a 0 a a a 2 a 2 b b 2 b b 2 b 0 b b 2 b 0 b b 2... a 0 a 0 a a a 2 a 2 b 0 b 0... a 0 a 0 a a a 2 a 2 b b 2 b b 2 a 0 b 0 b a 0 b 0 a (n, n) The weights in our problem are a k = f (k) and b k = f (k) +.

36 Batcher New: A generating function approach to derive Sedgewick s formula.

37 Batcher New: A generating function approach to derive Sedgewick s formula. Method: Write symbolic equations for families of lattice paths, then translate them into generating functions

38 Batcher A decorated path goes from (0, 0) to (n, n) and carries exactly one (vertical) label. W : the family of all paths (0, 0) to (n, n) D : (0, 0) to (n, n), staying on one (prescribed) side of the diagonal R p : the family of paths with vertical label a p, S p : the family of paths with vertical label b p. We treat a p as a fixed symbol (not depending on p). u Standard substition z = ( + u) 2.

39 Batcher W (z) = D(z) = 4z 2z 4z = n 0 = n 0 ( ) 2n z n = + u n u, n + R p = WA p W; ( ) 2n z n = + u. n A p = dda p Dh, p, A 0 = a p ddh.

40 Batcher A p = z( + u) 2 A p = ua p, p, A 0 = a p u + u. By iteration, A p = a p u p+ + u and therefore R p = a p u p+ ( + u) ( u) 2 by symmetry S p = b p u p+ ( + u) ( u) 2.

41 Batcher In the Batcher problem: b p = + a p, a p = f (p) f (k) : number of ones in the Gray code representation of k. B(z) := (R p + S p ) = p 0 u( + u) ( u) f (p) up+ ( + u) ( u) 2. p 0 B(z) = u( + u) + u) u 2k + 2u( ( u) 3 ( u) 3 + u. 2k+ k 0 Formula (with small mistake) given earlier by Knuth (without proof). Sedgewick s formula follows from this by standard extraction of coefficients.

42 Batcher Starting from B(z) = u( + u) + u) u 2k + 2u( ( u) 3 ( u) 3 + u. 2k+ k 0 one can also do asymptotics. Singularity of generating functions, Mellin transform.

43 Batcher Theorem The average number of exchanges in the odd-even merge of 2n elements satisfies B n 4 n log 2 n + nb(log 4 n), where B(x) is a continuous periodic function of period ; this function can be expanded as a Fourier series B(x) = k Z b ke 2kπix, with b 0 = 2 log 2 γ 4 log log 2 Γ ( 4) log2 π and for k 0, with the abbreviation χ k = 2πik log 2, b k = log 2 ζ( χ k, )Γ(χ k /2) 4. Furthermore, B(x) b χ k

44 Asymptotics in the early days De Bruijn s book Asymptotic methods in Analysis

45 Asymptotics in the early days De Bruijn s book Asymptotic methods in Analysis One chapter in Comtet s book Advanced Combinatorics (Darboux s lemma)

46 Asymptotics in the early days De Bruijn s book Asymptotic methods in Analysis One chapter in Comtet s book Advanced Combinatorics (Darboux s lemma) Edward A. Bender s survey article, SIAM Rev., 6 (974), (3 pages) Asymptotic Methods in Enumeration

47 Asymptotics in the early days De Bruijn s book Asymptotic methods in Analysis One chapter in Comtet s book Advanced Combinatorics (Darboux s lemma) Edward A. Bender s survey article, SIAM Rev., 6 (974), (3 pages) Asymptotic Methods in Enumeration much later Odlyzko s survey paper.

48 Asymptotics in the early days De Bruijn s book Asymptotic methods in Analysis One chapter in Comtet s book Advanced Combinatorics (Darboux s lemma) Edward A. Bender s survey article, SIAM Rev., 6 (974), (3 pages) Asymptotic Methods in Enumeration much later Odlyzko s survey paper. Now: The book Analytic Combinatorics and various survey papers by Philippe Flajolet (948 20)

49 Register function Work with Benjamin Hackl and Clemens Heuberger, in progress. Binary trees are either a leave or a root together with a left and a right subtree, which are binary trees. Symbolic equation: B = + B B

50 Reduction of binary trees; Register function A binary tree of size n has n internal nodes, and thus n + external nodes (leaves). The number b n of binary trees of size n is the nth Catalan number C n = n + ( ) 2n, n which follows the generating function B(z) = n 0 b n z n = +B 2 (z) = 4z 2z = n 0 ( ) 2n z n. n + n

51 Reduction of binary trees; Register function The equation B(z) = + z ( 2z B z 2 ) ( 2z) 2. (Touchard s identity) can also be seen as a recursive process to generate binary trees via B 0 (z) =, B r (z) = + z ( 2z B z 2 ) r ( 2z) 2, r. In this way we get B (z) = + z + 2z 2 + 4z 3 + 8z 4 + 6z z z z z z 0 +, B 2 (z) = + z + 2z 2 + 5z 3 + 4z z z z z z 9 +, B 3 (z) = + z + 2z 2 + 5z 3 + 4z z z z z z 9 +.

52 Reduction of binary trees; Register function The register function is recursively defined: Recursive description : reg( ) = 0, and if tree t has subtrees t and t 2, then { max{reg(t ), reg(t 2 )} if reg(t ) reg(t 2 ), reg(t) = + reg(t ) otherwise.

53 Reduction of binary trees; Register function Figure: A binary tree with 3 internal nodes. The numbers in the nodes are the register function of the subtree having this node as root. The register function of the tree is the value at the root, i. e., 3.

54 Reduction of binary trees; Register function Classical results (Flajolet et al.; Kemp, 979) The average value of the register function, assuming that all binary trees of size n (= n internal nodes), is asymptotically given as with a periodic function δ(x). log 4 n + δ(log 4 n)

55 Reduction of binary trees; Register function The register function is also known as Horton-Strahler numbers in the study of the complexity of river networks.

56 Reduction of binary trees; Register function Let R p denote the family of trees with register function = p, then one gets immediately from the recursive definition: R p =R p R p + R p + R j R j j<p j<p R p

57 Reduction of binary trees; Register function Let R p denote the family of trees with register function = p, then one gets immediately from the recursive definition: R p =R p R p + R p + R j R j j<p j<p R p R p = zr 2 p + 2R p j<p R j

58 Reduction of binary trees; Register function Easier to manipulate:

59 Reduction of binary trees; Register function Easier to manipulate: Let S p denote the family of trees with register function p, then one gets immediately from the recursive definition: S p =S p S p + S p B \ S p + B \ S p S p

60 Reduction of binary trees; Register function S p (z) = u2 u u 2p u 2p. u 2p R p (z) = u2. u u 2p+ u z = ( + u) 2

61 Reduction of binary trees; Register function Flajolet s approach: relatively elementary, using the dyadic valuation v 2 (n): If n = 2 i (2j + ), then i = v 2 (n). Can be linked to the sum of digits function S 2 (n). A result by H. Delange on the average value of the sum of digits function can be used.

62 Reduction of binary trees; Register function Kemp s approach: Mellin transform. At that period, people called it the Gamma function method.

63 Reduction of binary trees; Register function Today: Recursion at the bottom (at the leaves).

64 Reduction of binary trees; Register function Today: Recursion at the bottom (at the leaves). Compactification of binary trees, which we write as Φ(t):

65 Reduction of binary trees; Register function Today: Recursion at the bottom (at the leaves). Compactification of binary trees, which we write as Φ(t): The leaves (external nodes) will be erased.

66 Reduction of binary trees; Register function Today: Recursion at the bottom (at the leaves). Compactification of binary trees, which we write as Φ(t): The leaves (external nodes) will be erased. Then, if a node has only one off-spring, these two nodes will be merged; this operation will be repeated as long as there are such nodes. Finally, the endnodes are declared to be external nodes. This operation was introduced by two japanese physicists.

67 Reduction of binary trees; Register function

68 Reduction of binary trees; Register function

69 Reduction of binary trees; Register function

70 Reduction of binary trees; Register function

71 Reduction of binary trees; Register function Note that Φ( ) is undefined, so this is a partial function. Of course, many different trees are mapped to the same binary tree. However, they can all be obtained from a given reduced tree by the following operations:

72 Reduction of binary trees; Register function Note that Φ( ) is undefined, so this is a partial function. Of course, many different trees are mapped to the same binary tree. However, they can all be obtained from a given reduced tree by the following operations: Each leaf can be replaced by an internal node and an arbitrary chain of internal nodes on top, where the branches may be left or right ones. Thus, if the leaf is replaced by a chain of k internal nodes, this leads to 2 k choices. Similarly, an internal node is replaced by an internal and an arbitrary chain of internal nodes on top, where the branches may left or right ones. Eventually, the resulting tree is completed by external nodes in the usual way.

73 Reduction of binary trees; Register function If F (z) is a generating function counting some binary trees, then vf (zv) counts them with respect to size (variable z) and number of leaves (variable v).

74 Reduction of binary trees; Register function If F (z) is a generating function counting some binary trees, then vf (zv) counts them with respect to size (variable z) and number of leaves (variable v). The substitution process just described means that v z 2z and z z 2z.

75 Reduction of binary trees; Register function If F (z) is a generating function counting some binary trees, then vf (zv) counts them with respect to size (variable z) and number of leaves (variable v). The substitution process just described means that v z 2z and z z 2z z ( 2z F z 2 ) ( 2z) 2.. Altogether, this results in

76 Reduction of binary trees; Register function We have Φ r (t) B Reg(t) r

77 Reduction of binary trees; Register function Thus, if we set we get F (r) (z) = t: Reg(t) r z t, F (0) (z) = B(z), F (r) (z) = z 2z F (r )( z 2 ( 2z) 2 ), r. u The substitution z = is always a good idea when dealing (+u) 2 with the register function or Catalan numbers in general. Then, σ(z) := z2 = u2, so it just means u u 2. Furthermore, ( 2z) 2 (+u 2 ) 2 z 2z = u. Note also that F (0) (z) = B(z) = + u. +u 2 F (r) (z) = u u 2 + u 2 + u 4... u 2r F 0 (σ r (z)) + u 2r = u2 u u 2r ( + u 2r ) = u2 u 2r+ u u 2r u 2r.

78 Reduction of binary trees; Register function This formula for the generating function of the number of trees with register function r is of course well known. Likewise, the generating function B r (z) has the number of trees of size n and register function r as coefficients. B 0 (z) =, B r (z) = + z ( 2z B z 2 ) r ( 2z) 2, r.

79 Reduction of binary trees; Register function r-branches

80 Reduction of binary trees; Register function An r-branch is a maximal chain of nodes labelled r. This must be a chain, since the merging of two such chains would already result in the higher value r +. The nodes of the tree are partitioned into such chains, fromhelmut r = Prodinger 0,,....

81 Reduction of binary trees; Register function Parameter number of r-branches, in particular, the average number of them, assuming that all binary trees of size n are equally likely.

82 Reduction of binary trees; Register function Parameter number of r-branches, in particular, the average number of them, assuming that all binary trees of size n are equally likely. Explicit formula for the expectation (and, in principle, also for higher moments).

83 Reduction of binary trees; Register function Parameter number of r-branches, in particular, the average number of them, assuming that all binary trees of size n are equally likely. Explicit formula for the expectation (and, in principle, also for higher moments). Total number of r-branches, for any r, i.e., the sum over r 0.

84 Reduction of binary trees; Register function The r-branches are 0-branches after r iterations of Φ. The 0-branches are just the leaves; they are the only nodes labelled 0, and they form a branch for itself. So, we have again the generating function vb(zv). We start by computing average values. Then we have to compute v vb(zv) v= = Again we have the recursion G (0) (z) = = + u 4z u. 4z, G (r) (z) = z 2z G (r )( z 2 ( 2z) 2 ), r, this time for G (r) (z). Note that E n;r := C n [z n ]G (r) (z) is the average number of r-branches in a random tree of size n.

85 Reduction of binary trees; Register function Iteration leads now to G (r) (z) = u2 u u 2r + u u 2r+ u u u 2 r = u2 u u 2r ( u 2r ) 2. Expanding this function about u = means expanding it in terms of 4z. This can be done with a computer

86 Reduction of binary trees; Register function G (r) (z) 4 r 4z + 3 (4 r ) 4z + 5 (4 r r )( 4z) 3/2 +. Singularity analysis guarantees that one can read off coefficients in this expansion: [z n ]G (r) (z) ( ) 4 r 4n 2 ( ) n + ( ) n 3 (4 r )4 n 2 ( ) n n + ( 3 ) 5 (4 r r )4 n 2 ( ) n + n ( 4n π 4 r n + ( 7 ) 6 4 r+ n 3/2 ) +

87 Reduction of binary trees; Register function The asymptotics of C n are straight forward, especially for a computer, and eventually we find [z n ]G (r) (z) n C n 4 r + ( 5 ) 6 4 r + + (4 r ) 20n 4 r +. In principle, any number of terms would be available. Variance can also be computed.

88 Reduction of binary trees; Register function Theorem The number of r-branches in binary trees of size n has for expectation and variance the following asymptotic formulæ, which hold for fixed r and n : E n;r = n 4 r + 6 ( 5 4 r + ) + 20n (4 r ) ( ) 4 r + O n 2, V n;r = 4r 3 6 r n r r 45 + O ( n ).

89 Reduction of binary trees; Register function Theorem The expected number of r-branches in binary trees of size n is given by the explicit formula n + ( 2n ) [( ) ( ) ( )] 2n 2n 2n λ n + λ2 r 2 n λ2 r + n λ2 r. n λ

90 Reduction of binary trees; Register function Theorem The average value of the total number of branches in a random binary tree of size n admits the asymptotic expansion with 4n log 2 n 2ζ ( ) γ log 2 6 log δ(log 4 n) + o(), δ(x) = log 2 k 0 Γ( χ k 2 )ζ(χ k )(χ k )e 2πikx. The periodic function δ(x) is given by its Fourier series expansion; such functions are typical in a register context.

91 A Similar Recursive Scheme Involving Lattice Paths Simple two-dimensional lattice paths are sequences of the symbols {,,, }. It is easy to see that the generating function counting these paths (without the path of length 0) is L(z) = Theorem 4z 4z = 4z + 6z2 + 64z z z 5 +. The generating function L(z) = equation ( L(z) = 4L 4z 4z z 2 ) ( 2z) 2 Study of the compactification degree. satisfies the functional + 4z.

92 Reduction of planar trees Reduction of planar trees (ongoing research together with Benjamin Hackl and Sara Kropf) Leaves Paths Figure: Removal of leaves / paths.

93 Reduction of planar trees Proposition The generating function T (z, t) which enumerates rooted plane trees with respect to their internal nodes (marked by the variable z) and leaves (marked by t) is given explicitly by T (z, t) = (z t) 2(z + t) + (z t) 2. 2 Definition The Narayana numbers are defined as N n,k = ( )( ) n n n k k The Narayana polynomials are defined as N n (x) = n N n,k x k k=

94 Reduction of planar trees The operator that drives the reduction of leaves: ( ) z Φ(f (z, t)) := ( t)f ( t) 2, zt ( t) 2. Theorem The bivariate generating function G r (z, v) enumerating rooted plane trees whose leaves can be cut at least r-times, where z marks the tree size and v marks the size of the r-fold cut tree, is given by G r (z, v) = Φ r T (zv, tv) t=z and, equivalently, by ( ) u r+2 G r (z, v) = ( u r+ )( + u) T u( u r+ ) 2 ( u r+2 ) 2 v, ur+ ( u) 2 ( u r+2 ) 2 v, where z = u/( + u) 2.

95 Reduction of planar trees Many explicit results can be deduced from that. E.g. EXn,r d = [z n ] d C n v d G r (z, ) v= = [z n u d d! ] C n ( + u)( u r+ ) d ( u) d Ñd (u r )

96 Priority trees Average Case-Analysis of Priority trees: A structure for priority queue administration A. Panholzer and H. Prodinger, Algorithmica 22 (998),

97 Priority trees object of C with nodes k +,..., n k object of B with nodes,..., k Figure: Decomposition of the family A.

98 Priority trees + k object of A with nodes k +,..., n object of B with nodes,..., k Figure: Decomposition of the family B.

99 Priority trees object of C with nodes k +,..., n k object of A with nodes,..., k Figure: Decomposition of the family C.

100 Tree function The tree function y = ze y enumerates the labelled rooted trees: y(z) = n a variant of Lambert s W -function n zn n n!

101 Tree function Epidemics with two levels of mixing: The exact moments, H. Prodinger, SADIO 2 (999), 4. The probabilities ( ) n p(pk) k 2 ( pk) n k, ( k n) k where considered in the study of an epidemics model. E(z) = k 0 which is also given implicitly by A variant of the tree function. k zk (tk + ) k!, z = E t log E.

102 Tree function An identity conjectured by Lacasse via the tree function, H. Prodinger, Electronic Journal of Combinatorics 20 (3), 203, P7. β(n) = α(n) = n k=0 k +k 2 +k 3 =n ( ) n k k (n k) n k k n! k!k 3!k 3! kk kk 2 2 kk 3 3 β(n) α(n) = n n+ ( ) 2 ( ) 3 α(n) = n![z n ] β(n) = n![z n ] y y

103 Tree function Tight Bounds on Information Leakage from Repeated Independent Runs by Smith and Smith (206) Donald E. Knuth and Boris Pittel. A recurrence related to trees. Proceedings of the American Mathematical Society, 05(2):

104 Tree function On Ramanujan s Q(n)-function. P. Flajolet, P. Grabner, P. Kirschenhofer and H. Prodinger, Journal of Computational and Applied Mathematics, 58:03 6, 995. Q(n) = + n n + (n )(n 2) n 2 + Show that R(n) = + n n + + n 2 (n + )(n + 2) + R(n) Q(n) = (n + k) where k k(n) lies between 2 2 and 8 45

105 Tree function D(n) = R(n) Q(n) n= n zn D(n)n n! = log ( y)2 2( ez) = log ( y) 2 2( ye y ) y = ze y