Coding elements related to Catalan numbers by Dyck words

Transcription

1 Coding elements related to Catalan numbers by Dyck words Zoltán Kása January 9, 009 Catalan numbers The Catalan numbers, named after the mathematician E. C. Catalan, defined as C n = ( ) n, n + n are as known as the Fibonacci numbers. The Catalan numbers are the solution of the following recurrence equation: C n+ = C 0 C n + C C n C n C 0 for n 0, with C 0 =. () Another recurrence equation for the Catalan numbers is: (n + )C n+ = (n + )C n for n 0, with C 0 =. () Catalan numbers can be expressed also as ([]): C n+ = + ( ) n k ( ) k C n k. () k + k 0 C n being equal to the number of binary trees with n vertices, can be expressed [] as C n = b k n = ( )( ) k n n k, n k k k 0 k where b k n is the number of binary trees with n vertices and k leaves. From this a new recurrence equation will result: C n = ( ) n (k + ) n k C k. () n k k Sapientia Hungarian University of Transylvania, Cluj/Kolozsvár, Department of Mathematics and Informatics, Târgu Mureş/Marosvásárhely kasa@ms.sapientia.ro

2 The generating function of Catalan numbers is n 0 C n z n = z. z Catalan numbers arise in a lot of combinatorial problems as the number of some elements. The Catalan number C n describe, among other things [], the number of binary trees with n vertices, the number of ways in which parantheses can be placed in a sequence of n + numbers to be multiplied two at a time, the number of well-formed reverse Polish expressions with n operands and n + operators, the number of paths in a grid from (0, 0) to (n, n), increasing just one coordinate by one at each step, without crossing the main diagonal, the number of n-bit sequences that the number of s never exceeds the number of 0s in each position from left to right, the number of ways you can draw non-crossing segments between n points on a circle in the plane, the number of sequences (x, x,..., x n ), with x i {, } for all i between and n and having the following properties for all partial sums: x 0, x +x 0,..., x +x +...+x n 0, x +x +...+x n = 0, the number of ways a polygon with n + sides can be cut into n triangles, using n diagonals, the number of frieze pattern with n + rows, the number of mountain ranges you can draw using n upstrokes and n downstrokes, the number of ways n votes can come in for each of two candidates in an election, with the first never behind the second. Dyck words Let B = {0, } be a binary alphabet and x x... x n B n. Let h : B {, } be a valuation function with h(0) =, h() =, and h(x x... x n ) = n h(x i ). i=

3 A word x x... x n B n is called a Dyck word [] if it satisfy the following conditions: n is the semilength of the word. h(x x... x i ) 0, for i n h(x x... x n ) = 0. Encoding and decoding In [] a method for coding elements whose number is the Catalan number C n is presented. If O is a set of C n elements, Dyck words can be used for encoding the elements of O. In the following encoding and decoding algorithms are presented for different elements.. Binary trees Algorithm for encoding a binary tree The encoding of a binary tree is the following: when a vertex has only one descendant, we put the sequence 0 for a single left edge, 0 for a single right edge, and 00 for the left edge resp. for the right edge when there are two descendants. We complete the resulting sequence with 0 at the beginning and at the end. The encoding is made using a preorder traversal of the binary tree Encoding of binary trees for n =.

4 Let B L be the left and B R the right subtree of the binary tree B. w0 means the concatenation of word w with 0, and w is considered a global variable. EncodingBT(B) if B L and B R = then w w0 EncodingBT(B L ) if B L = and B R then w w0 EncodingBT(B R ) 7 if B L and B R 8 then w w00 9 EncodingBT(B L ) 8 w w 9 EncodingBT (B R ) 0 return Call: w 0 EncodingBT(B) w w Algorithm for decoding a Dyck word into a binary tree At the beginning the root of the generated binary tree is the current vertex. When an edge is drawn, its endvertex becomes the current vertex. DecodingBT(w) Let ab be the first two letters of w. Delete ab from w. if ab = 0 then draw a left edge from the current vertex DecodingBT(w) if ab = 0 7 then draw a right edge from the current vertex 8 DecodingBT(w) 9 if ab = 00 0 then put in the stack the position of the current vertex draw a left edge from the current vertex DecodingBT(w) if ab = then get from the stack the position of the new current vertex draw a right edge from the current vertex DecodingBT(w) 7 return

5 Call: delete 0 from the beginning and from the end of the input word w draw a vertex (the root of the tree) as current vertex DecodingBT(w). Paths in grid A Dyck path is a path in the grid from (0,0) to (n, n) using only horizontal and vertical segments, not crossing the diagonal x = y. For encoding we shall put 0 for a horizontal unit of the path and for a vertical one. The resulting word is a Dyck word because the path never cross the main diagonal of the grid. The decoding is immediate Encoding of paths in grid. Expressions with multiplications To encode expressions we first attach to each expression for multiplication a binary tree by a very simple method. If we multiple a by b, this yields a binary tree with a root and two descendant nodes a and b. A multiplication of two expressions yields a binary tree with two subtrees which are the binary trees corresponding to the two expressions. In the resulting binary tree each internal nodes has exactly two descendants. Such trees are called extended binary trees. To encode an extended binary tree we shall omit all leaves (with of course the corresponding edges) in the tree corresponding to the multiplication expression and use the encoding method presented before for the resulting binary tree. For n = we shall have the expressions and the corresponding extended binary trees in Fig.. If we omit all leaves with the adjacent edges in these extended trees the binary trees and the corresponding encoding result.

6 ((( ) ) ) (( ( )) ) ( (( ) )) ( ( ( ))) (( ) ( )) Encoding of multiplications For decoding we first draw the corresponding binary tree, complete it to having two descendants for each node. The resulting extended binary tree give us the order of multiplications.. Sequences We encode sequences (x, x,..., x n ), with x i {, } for all i between and n and having the following properties for all partial sums: x 0, x + x 0,..., x +x +...+x n 0, x +x +...+x n = 0. We shall code in the sequence by and by 0. It is easy to see that in any positions the number of s never exceeds the number of 0s, and they are equals in the sequence (because the sum of all n elements is equal to 0), so the resulting word is a Dyck one. For example:,,,, coded by: 000,,,, coded by: 000,,,, coded by: 000,,,, coded by: 000,,,, coded by: 000. Segments If we have n points on a circle in the plane and n non-crossing segments between them, the encoding is the following: Let us mark the points clockwise on the circle with numbers from to n. For a segment between i and j (i < j) put 0 in the i th position and in the j th position in the code sequence. For n = see Fig.. It is easy to see that the resulting word is a Dyck one.

7 Encoding of segments For decoding we search for the first subword 0, trace the corresponding segment, omit it from the word and continue with the remaining word (keeping the original positions).. Reverse Polish expressions We shall code each operand by 0 and each operator by, and then delete one 0 from ther beginning. For example, if we have the reverse Polish expression aaa a which corresponds to the expression (a ((a a) a)) the resulting code is 000. For decoding let us add an 0 at the begining of the Dyck word, and put an operand for an 0, and operator for an. 7

8 .7 Polygons The polygon is divided into triangles. We consider one node in each triangle, and one outside of each side of the polygon. Join by an edge two nodes if the corresponding triangles (or a triangle and the outside of the polygon) have a side in common. We shall get a tree, on which the encoding will be made. If we mark one side of the polygon and the corresponding edge of the tree, and eliminate all edges from the tree that have an endpoint as a leave, we shall get a binary tree (the root will be the node which is adjacent with the marked edge). The exemplification will be made for n = (pentagon). The marked side is AB. a b a b a b a b a b Encoding of polygons References [] Bege, A., Kása, Z., Coding objects related to Catalan numbers, Studia Universitatis Babes-Bolyai, Informatica,, (00), pp. 0. [] Duchon, P., On the enumeration and generation of generalized Dyck words, Discrete Mathematics, (000). [] Kása, Z., Ţâmbulea L., Binary trees and the number of states in buddy systems, Annales Universitatis Scientiarium Budapestinensis de Rolando Eötvös Nominatae Sectio Computatorica, 7 (987) 0. 8

9 [] Kása, Z., Generating and ranking of Dyck words, Acta Universitatis Sapientiae, Informatica,, (009) [] Stanley, R. P., Enumerative combinatorics II., Cambridge University Press, 00. 9