The Oldest Child Tree

Transcription

1 The Oldest Child Tree Dennis E. Davenport Mathematics Department Howard University, Washington, DC Louis W. Shapiro Mathematics Department Howard University, Washington, DC Leon C. Woodson Department of Mathematics Morgan State University, Baltimore, MD October, 0 Abstract An ordered tree, also known as a plane tree or a planar tree, is defined recursively, as having a root and an ordered set of subtrees. An oldest child tree is an ordered tree where the rightmost edge above any vertex can either red or green. If we are considering family trees, then oldest child can be spoiled or not. In this paper we show that the number of oldest child trees gives the Large Schröder numbers. We also investigate other properties of oldest child trees, such as the average root degree and the average number of oldest children. Introduction Consider the following ordered tree. Each vertex is labeled to show the order.

2 h i e f g b c d a Figure For vertex a, {a,d} is the righmost edge, for vertex d, {d,g} is the rightmost edge, for vertex b, {b,e} is the rightmost edge, and for vertex e, {e,i} is the rightmost edge. Note that leaves have no rightmost edges, since there are no edges with leaves as vertices that are above them. An oldest child tree is an ordered tree where the rightmost edge of each vertex can be either green or red. This corresponds to a family situation where the oldest child can be either spoiled or not. The twenty two oldest child trees with 3 edges are illustrated below. (8) (4) () (4) (4) Figure Note that there are Oldest Child Trees when n = 3, since there are two choices for each rightmost edge above each vertex. Generating Function for the number of Oldest Child Trees If an edge (child) is rightmost we allow two possibilities, the child can be spoiled or not. Thus if O(z), or more briefly O, is the generating function

3 counting Oldest Child Trees, then O(z) = + zo(z) + z O (z) + z 3 O 3 (z) +... O(z) = + zo(z)( + zo(z) + z O (z) + z 3 O 3 (z) +... = + zo(z) zo(z) = + zo(z) zo(z) The quadratic formula gives us O(z) = z 6z + z z = + z + 6z + z z z which is the generating function for the Large Schröder numbers. We next describe a bijection between the Oldest Child Trees and Schröder paths. Given any Oldest Child Tree, T we proceed as follows. Let R be a point on the left of T (R is on the opposite side of the oldest children). For each edge e, pick a point P(e) on e. If e is an oldest child, we use o instead. Using the points P(e) and R we create a new tree with R as the root. In the new tree, T, RP(e) is an edge iff the line segment RP(e) intersects T only at the point P(e). Given P(e ) and P(e ), then P(e )P(e ) is an edge, if P(e )P(e ) intersects T at no other point and P(e ) is on a branch that is either the at the same level or higher than the branch that contains P(e ). Finally all paths terminate at an oldest child. We now use the well-known preorder traversal algorithm, (or the worm crawling around the tree, see []), to generate a Schröder path. Note that each P(o) point corresponds to a peak in the Schröder path. If the oldest child is spoiled, we use the peak. And if the child is not spoiled (level headed) we use a level step. The following example illustrates our process. Example: Consider the following Oldest Child Tree T. Let R be a point to the left of T. (R) Figure 3 3

4 We next pick a point on each edge. Note that for each generation, the rightmost edge is the oldest child. (R) Figure 4 We now draw edges from R to the new points on T that it can see and the remaining edges to get T. The bold lines are the edges of T (R) Figure 5 From here we get the following ordered tree T with R as the root. Note that each leaf corresponds to an oldest child from T. (R) Figure 6 T, is an ordered tree with bicolored leaves. These trees are mentioned in [4] Exercise 6.39.c. and []. 4

5 Using the preorder traversal algorithm, with U denoting up and D down, we get the following UUDDUDUUUUDDUDDD. This corresponds to the Schröder path given below. Figure 7 Each peak of the path relates to an oldest child. There are two choices for oldest children, spoiled or not. If the child is not spolied, then we have a level step; if the oldest child is spoiled, then we have a peak. For example, suppose that only the rightmost oldest child is not spoiled. Then we get the following Schröder path. Figure 8 We now give the generating function for the number of oldest children. The A-numbers refer to Sloane s OEIS,[3]. Let V be the generating function counting ordered trees with a marked vertex and L the generating function counting such trees with a marked leaf. Then the generating function for the number of oldest children is V L. We note that V = (zo) = ( ΣO n z n+) = Σ(n + )On z n since any tree with n edges has n + vertices. Also note that 5

6 O = + z(o + O ) = z zo and Recall that Starting with O z V (z) = d(zo(z)) dz = O + O = O + z do dz = O + zo O = ( z zo) We have Thus O = ( z zo) ( + O + zo ) = O ( + O + zo ) O = O ( + O) = O O + O zo zo = O O z V = O + zo L = O + O O zo = V/O zo = + O zo = + z + 8z + 38x 3 + 9z z z , A364 Hence, for the generating function for the total number of oldest children we get ( V L = LO L = L(O ) = (O ) + O ) zo ( ) O zo = (O ) zo = z + 0z + 50z z z z , A070 Referring to the 5 trees in figure going from left to right we have 8 3, 4,, 4, and 4 oldest children, for a total of 50. 6

7 3 Average Root Degree Recall that O(z) = + zo(z) + z O (z) + z 3 O 3 (z) +... Let u be the sequence,,,,... Then u(z) = + z + z + z 3 + z = + z z. Let T denote the total root degree. Then T = zou (zo) = zo ( zo) = z + 8z + 34z 3 + 5z z z Again referring to the 5 trees in figure going from left to right we have 8, 4, 3, 4, and 4 root degrees, for a total of 34. Let N be the generating function for O trees where the edge at the root is not spoiled. Then Recall that So that O z N = O +. O = + z ( O + O ). = O + O = O (O + ). When considering vertices of height one for O trees, there are two possible cases. The vertex at height is on the left most branch. In this case we get zon. The vertex at height is not on the left most branch. In this case we get zon(o ). Hence the generating function counting such vertices is zon + zon(o ) = zo O + (O + ) = z ( ) O (O + ) z = O zo z z 7

8 So, for n Thus [z n ] O zo z z = O n+ O n (O n+ O n ) ( 3 + ) = + O n We next consider the leaf-vertex ratio. Let V = Σv n, L = Σl n, and D = = Σd 6z+z nz n. The generating function D counts Delannoy paths from (0, 0) to (n, n). These are paths using steps E = (, 0), N = (0, ), and D = (, ). Then Thus Note that Hence l n = (d n + d n ) and v n = 3 d n d n l n v n = (d n + d n ) 3 d n d = n d n d n 3 + l n 3 + v n 3(3 + ) = d n d n 3 dn d n 0.93 We can now show quickly that V L counts Delannoy paths that do not start with a (, ) step. This follows from v n l n = References ( 3 d n ) d n (d n + d n ) = d n d n. [] M. Gardner, Time Travel and Other Mathematical Bewilderments, W.H. Freeman and Co.,

9 [] L.W. Shapiro, R.A. Sulanke, Bijections for the Schröder numbers, Math. Mag. 73n(000) [3] Sloane s Online Encyclopedia of Integer Sequences, [4] R.P. Stanley, Enumerative combinatorics, Vol., Cambridge University Press, Cambridge, 999 9