UNIVERSITY OF 
 
 ILLINOIS LIBRARY 
 
 AT URBANA-CHAMPAIGN 
 
Digitized by the Internet Archive 
 in 2013 
 
 http://archive.org/details/generatingkarytr901zaks 
 
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 UILU-ENG 77 1758 
 
 GENERATING K-ARY TREES LEXICOGRAPHICALLY 
 
 by 
 
 Shmuel Zaks 
 
 November 1977 
 
t 
 
 GENERATING K-ARY TREES LEXICOGRAPHICALLY 
 
 by 
 Shmuel Zaks 
 
 Department of Computer Science 
 
 University of Illinois at Urbana-Champaign 
 
 Urbana, Illinois 61801 
 
 November, 1977 
 
 This work was supported in part by the National Science Foundation under 
 grant number NSF MCS-73-03408. 
 
TABLE OF CONTENTS 
 
 1. Introduction 
 
 2. Definitions and Notations 
 
 3. Existing Algorithms and Results; The Generating Algorithm 
 
 4. The Ranking Function and the Unranking Procedure 
 
 5. Summary 
 
I. INTRODUCTION 
 
 We show a 1-1 correspondence between all regular k-ary trees with 
 
 kn 
 n internal nodes, all 0,1-sequences x = {x. K with n Ts and (k-l)n O's in 
 
 which k-1 times the number of l's in any prefix Cx.}, (1 < i < kn) is at 
 
 least as the number of O's in that prefix, and all integer sequences z = {z^}-. 
 
 in which < z, < z 2 < ... < z n and z.. <_ ki - (k-1) for i = 1 , 2, . . . , n. 
 
 Working with the reverses of these sequences, we deal with the 
 
 following three steps: 
 
 1. Generating: the sequences are generated one-by-one in lexico- 
 graphic order. 
 
 2. Ranking: we show how to compute the function that, given a 
 sequence, determines its position in that lexicographic order. 
 
 3. Unranking: a procedure is discussed that, given a position 
 in that lexicographic order, constructs the sequence which 
 occupies this position. 
 
 We also discuss relations to existing related algorithms. 
 
II. DEFINITIONS AND NOTATIONS 
 
 2.1 We state here definitions and notations that we shall follow through- 
 out this paper. We follow [KN] and [LI] for our basic terms. 
 
 A tree means an ordered tree . The terms root , son , internal node 
 and leaf are used as defined in the literature. A k-ary tree is a tree all 
 of which internal nodes have at most k sons each. A regular k-ary tree is 
 a tree all of which internal nodes have exactly k sons each. |T| denotes 
 the number of vertices in a tree T, and T. is the subtree rooted at the i 
 son of the root of T. T(k,n) denotes the set of all the regular k-ary trees 
 with n internal nodes. The number of elements in T(k,n) is known to be 
 
 WTWT ("n") < see [ RN: P- 584 ]» 
 
 2.2 In order to generate trees "lexicographically", we first establish a 
 1-1 correspondence between them and certain integer sequences, and then we 
 generate these sequences lexicographically. Given a tree T e T(k,n) we 
 label each internal node with 1 and each leaf with 0. We then read these 
 labels in preorder (root-left-right), and thus obtain a sequence of n l's 
 and (k-l)n + 1 0's. The last visited node is a leaf, and we omit the 
 corresponding for simplifying matters. Denote the resulting sequence by 
 f(T) = x = Uj}^", and let F(T) = X be the reverse of f(T). 
 
 Instead of working with the sequences x and X of length kn, one 
 can prefer using the sequences g(T) = z = {z.j}, in which z^ = position of 
 the i th 1 in f(T) = x,and its reverse G(T) = Z. Note that both z and 
 Z are of length n. See figure 1 for an example. The 0,1-sequence x = f(T) 
 is also discussed in [BM], [GA] and [KL]. 
 
 * 
 
 The sons are read from left to right, 
 
f(T) = x = 11000001000100100000 
 
 F(T) = X = 00000100100010000011 
 
 g(T) = z = 1, 2, 8, 12, 15 
 
 G(T) = Z = 15, 12, 8, 2, 1 
 
 
 Sequences Corresponding to a k-ary Tree 
 Figure 1 
 
 We note that the reconstruction of a tree T given X = F(T) or 
 x = f(T), and the transformations x +* z and X ■*-»• Z, are all quite easy, and 
 the details are left to the reader. 
 
 A sequence x is called feasible if there is a tree T e T(k,n) s.t. 
 x = f(T). The same definition holds for X, z and Z. x(k,n), X(k,n), z(k,n) 
 and Z(k,n) denote the classes of the feasible x, X, z and Z sequences, 
 
 respectively, corresponding to the trees in T(k,n). 
 
 kn 
 Let a = {a..-}-] be a sequence consisting of n l's and (k-l)n 0's. 
 
 We say that a has the k-dominating property if in each prefix {a,-}-i» 1 £ A < 
 
 kn, the number of l's is at least 
 
 ; in other words, the number of 0's is 
 
 Ttl denotes the smallest integer not smaller than t. UJ denotes the 
 largest integer not larger than t. 
 
at most (k-1) times the number of l's in any such prefix. This property 
 is closely related to the g eneralized box-office problem* (see [DM], [ZA]). 
 The case k = 2 is discussed in various references (see, for example, [YY: 
 Problem 83], or [GN: p. 37]). 
 
 2.3 In this paper we work with the reverse sequences, for which we obtain 
 the following: 
 
 1. Generating: we generate Z(k,n) lexicographically (corresponds 
 to a lexicographic generation of X(k,n); see Theorem 2). 
 
 2. Ranking: we show how to compute the function index(X) that 
 assigns to a sequence X e X(k,n) its position in the lexi- 
 cographic ordering of X(k,n). 
 
 3. Unranking: given an integer t, we construct the sequence 
 X e X(k,n) s.t. index (X) = t. 
 
 In a previous work [ZA], we proved that generating x(k,n) lexico- 
 graphically corresponds to generating the trees in T(k,n) according to the 
 following ordering (see also Theorem 2): 
 Definition 1 : Given T, T' e T(k,n), we say that T < T' if 
 
 1 . T is empty, or 
 
 2. T is not empty, and for some i, 1 <_ i <_ k, we have 
 
 a) T. = T'. for j = 1, 2, ... , i-1, and 
 
 b) T, <T|. 
 
 Consequently, generating X(k,n) (= the reverses of the feasible x 
 
 * 
 
 The generalized box-office problem: kn people are waiting in a line at 
 a box-office, n of them have k-1 1-dollar bills, and the other (k-l)n 
 have one k-dollar bill. A ticket costs k-1 dollars, and each customer 
 buys one ticket. When the box-office opens, there is no money in the 
 till. In how many ways can they stand so that none of them will have 
 to wait for change? 
 
sequences) corresponds to generating T(k,n) according to the "reverse" order- 
 ing, defined as follows: 
 Definition 2 : Given T, T' e T(k,n), we say that T < T' if 
 
 1. T is not empty, and for some i, 1 <_ i <_ k, we have: 
 
 a) T. = T'. for j = k, k-1, ... , i+1 , and 
 
 b) T. < Tl, or 
 
 2. T is empty. 
 
 As defined, the x = f(T) sequence corresponds to a preorder travers- 
 ing of the 0,1 labels of the vertices of T; therefore, the X = F(T) corresponds 
 to a postorder (right-left-root) traversing of these labels. This preorder/ 
 postorder nature of these sequences is \jery briefly reflected in these two 
 definitions. 
 
III. EXISTING ALGORITHMS AND RESULTS; THE GENERATING ALGORITHM 
 
 We mention here related algorithms, and discuss results that will 
 be used in the sequel. Theorems 1, 2 and 3, and the generating algorithm 
 following them are not proved here. They are all modifications of results 
 we proved in [ZA]. 
 
 3. 1 In [RH] a binary tree is represented by the sequence of the levels 
 
 of its leaves from left to right, and the generating, ranking and unranking 
 steps are discussed. In [KN: 2.2.1, ex. 2,4,5; 2.3.1, ex. 6], [TR1 , 2] and 
 [KNO] a tree with n vertices is represented by an appropriate permutation 
 of {1, 2, ... , n). The generating, ranking and unranking steps are dis- 
 cussed in [TR1 , 2]. Both discussions use definition 1 (as proved in [ZA: 
 Theorems 4.8 and 7.8]). 
 
 Ranking and unranking for k = 2 are discussed in [KNO], using a 
 different ordering which is also used - for an arbitrary k - in [TR1]. 
 
 In [ZR] definition 1 is used to generate, rank and unrank k-ary 
 trees - and larger classes of trees - using a pure combinatorial approach. 
 
 3.2 In [ZA] we proved that there is a 1-1 correspondence between T(k,n), 
 
 kn 
 all 0,1-sequences x = ix.}, with n l's and (k-l)n O's having the k-dominat- 
 
 ing property, and all integer sequences z = (z-}, s.t. < z. < z ? < ... 
 
 < z and z. < ki - (k-1) for i = 1, 2, ... , n. Modification of this re- 
 n l — 
 
 suit to the reverse sequences gives a characterization of X(k,n) and Z(k,n), 
 
 as follows: 
 
 Theorem 1 : The following sets are in a 1-1 correspondence with one another: 
 
 (1) T(k,n) 
 
 (2) All 0,1-sequences X = {X,} 1 ? 1 , with n l's and (k-l)n O's, the 
 reverses of which have the k-dominating property. 
 
 (3) All integer sequences Z = {Z i }" s.t. Z-j > 1^ > ... > Z n > 
 
 and Z .,, < ki - (k-1) for i = 1, 2, ... , n. 
 n-i+1 - v 
 
 Following the results proved in [ZA] for the x and z sequences, and 
 
using the definitions of the X and Z sequences, we get: 
 
 Theorem 2 : Let T, T' e T(k,n), and let x, Z, z, Z and x\ X', z', V be the 
 
 corresponding sequences associated with them. Then 
 
 1. T < T' according to definition l«>x<x'«>z>z'. 
 
 2. T < T 1 according to definition 2 <» X < V <=> 1 < V . 
 
 kn 
 
 3.3 In [ZA], we proved that a 0,1-sequence x = {x-}-i is feasible iff 
 
 k-1 
 erasing any 10 pattern*, as long as possible, results in the empty sequence, 
 
 A graphical interpretation of this reduction is discussed there. Modify- 
 ing this to the X sequences yields the following: 
 
 kn k-1 
 
 Theorem 3 : A 0,1-sequence X = (X.)-, is feasible iff erasing any 1 
 
 pattern, as long as possible, results in the empty sequence. 
 
 It is Theorem 3 that enables us to compute the ranking function 
 
 for X(k,n) (see Theorem 5 in the next section), while the mentioned result 
 
 enabled us to compute it for x(k,n) only for k = 2. 
 
 3.4 In [ZA], we generated the feasible z sequences lexicographically. A 
 modification of this algorithm will generate the Z sequences, as characterized 
 by Theorem 1 : 
 
 Algorithm GENERATE (Generating the sequences Z e Z(k,n) lexicographically): 
 
 1. Begin with Z = {I. } = {n, n-1 , . . . , 1 } 
 
 2. Find the largest I s.t. Z £ < k(n-£+l ) - (k-1). 
 If i = 1 then goto 3. 
 
 If i > 1 then: if Z < Z , - 1 then goto 3 else go on looking for such 
 
 an £. 
 
 If no such i exists - goto 5. 
 
 3. The sequence V = {!'.} next to Z is built as follows: 
 
 a means a string consisting of m consecutive a's. 
 
z \ - z i 
 
 Zl <- n - i + 1 
 Let V be called Z. 
 Goto 2. 
 End. 
 
 for i < i 
 
 for i > £ 
 
 Example : In figure 2 we list the sequences x, X, z and Z corresponding to 
 the ternary trees with 3 internal nodes (T(3,3)). The table is arranged 
 according to lexicographic ordering of the x sequences ( = anti lexicographic 
 ordering of the z sequences) on the left, and lexicographic ordering of the 
 X sequences (= lexicographic ordering of the Z sequences) on the right. 
 
 z 
 
 X 
 
 index 
 
 X 
 
 z 
 
 1,4,7 
 
 10 10 10 
 
 1 
 
 111 
 
 3,2,1 
 
 1,4,6 
 
 10 10 10 
 
 2 
 
 10 11 
 
 4,2,1 
 
 1,4,5 
 
 10 1 10 
 
 3 
 
 110 1 
 
 4,3,1 
 
 1,3,7 
 
 10 10 10 
 
 4 
 
 10 11 
 
 5,2,1 
 
 1,3,6 
 
 10 10 10 
 
 5 
 
 10 10 1 
 
 5,3,1 
 
 1,3,5 
 
 10 10 10 
 
 6 
 
 110 1 
 
 5,4,1 
 
 1,3,4 
 
 10 110 
 
 7 
 
 10 11 
 
 6,2,1 
 
 1,2,7 
 
 110 10 
 
 8 
 
 10 10 1 
 
 6,3,1 
 
 1,2,6 
 
 110 10 
 
 9 
 
 10 10 1 
 
 6,4,1 
 
 1,2,5 
 
 110 10 
 
 10 
 
 10 11 
 
 7,2,1 
 
 1,2,4 
 
 1 10 10 
 
 11 
 
 10 10 1 
 
 7,3,1 
 
 1,2,3 
 
 1110 
 
 12 
 
 10 10 1 
 
 7,4,1 
 
 The Sequences Corresponding to T(3,3) 
 Figure 2 
 
IV. THE RANKING FUNCTION AND THE UNRANKING PROCEDURE 
 
 4.1 In this section, we find the position Index(T) of a given tree T e 
 
 T(k,n). For this purpose, we take X = F(T), and find the position index(X) 
 
 of X in the lexicographic ordering of X(k,n). The modification to Z(k,n) 
 
 is left to the reader. First we observe the following: 
 
 Theorem 4 : There is a 1-1 correspondence between all the sequences in 
 
 X(k,n) which begin with £+k_1 l, and those in X(k,n-1) which begin with . 
 
 £+k-l 
 Proof: Suppose there are u sequences in X(k,n) beginning with 1. Call them 
 
 A Q £+k-l 1Y 5 A 2 =0 £+k-1 lY 2 , ... ,A u =0 £+k " 1 lY u . By omitting the first occurence of 
 k_1 l from each of the A^s we get B ] = £ Y ] , B 2 = £ Y 2> ... , B y = £ Y u , 
 each of which is a sequence in X(k,n-1) by Theorem 3. It is clear that 
 B- =f B. for i j j , and that all the sequences in X(k,n-1) beginning with 
 are matched in this way (by Theorem 3); the theorem has thus been proved. D 
 Note : It is clear that the correspondence mentioned above preserves lexi- 
 cographic order, i.e. A. < A. iff B. < B.. This fact will be used in the 
 next theorem - which is used recursively to compute the ranking function - 
 as follows: 
 Theorem 5 : Let X = £ 1Y belong to X(k,n)*. Then 
 
 index(X)** = 
 
 if X = a l for some a and b 
 a((k-l)n - U+l),n,k) + index(0 £_k+1 Y) otherwise 
 
 where a((k-l)n - (£+l),n,k) is the number of sequences in X(k,n) which begin 
 
 with £+1 . 
 
 Proof : Immediate, by the nature of lexicographic order and theorem 4. D 
 
 * 
 
 By Theorem 4, we have £ > k - 1 
 
 ** 
 
 We use here the same name for the functions index(X) corresponding to a given 
 n and k. index(X) gives the position of X in X(k,n), while 
 
 index(0 £ ~ k+1 Y) gives the position of the feasible sequence o £-k+1 Y in X(k,n-1) 
 
10 
 
 4.2 The following discussion concerns the evaluation of the numbers a(i,j,k) 
 
 as defined in Theorem 5. We make use of the following geometric interpreta- 
 
 kn 
 tion (see [YY: Problem 83]): a sequence a = {a.}, , with n l's and (k-l)n 
 
 0's, corresponds to a path from B((k-l)n,n) to A(0,0) in the rectangular 
 
 lattice defined by B and A (see figure 3), where 1 means "one step down" 
 
 and means "one step to the left." This sequence is feasible iff this path 
 
 never goes below the line AB(see Theorem 1), which is i = (k-l)j. 
 
 J 
 
 5 15 34 65 108 163 228 283 283 283 B((k-l)n,n) 
 
 -+— —9— —9— — # ♦ ♦ • - 
 
 3- 
 
 ■lO-o — 19-<» — 3V-*^ -43hk 55-o — 55- 
 
 A(0,0) 
 
 -9- 
 
 -3- 
 
 -h*^~ -CH» 
 
 
 
 12- 
 
 
 
 ->*^ — 0-^> 
 
 
 
 Of 
 
 <H> G-*> (H» 0-<' 
 
 <H' 0-<> (H» (H» 
 
 -O-i » 0- 1 1- — 0-i r O-i i- 
 
 
 
 
 
 
 
 
 a(i,j,k) for k = 3 and j < 5 
 Figure 3 
 The following labeling of the lattice points gives the number of 
 ways to go from a point (i,j) to A, without going below the line AB: 
 
 > i 
 
 a(i,j,k) = < 
 
 a(i,j-l ,k) + a(i-l ,j,k) 
 see example in figure 3. 
 
 i = 
 
 1 > (k-l)j 
 
 otherwise 
 
11 
 
 r 
 
 The solution to this recurrence relation is as follows: 
 Theorem 6: The solution to the recurrence relation 
 
 i = 
 i > (k-l)j 
 
 .a(i,j-l,k) +a(i-l,j,k) otherwise 
 
 a(i,j,k) = \ 
 
 where i, j ^ 0, is given by 
 
 i-1 
 
 *■'.« - or) - L S T^) 
 
 i+j-l-kt\ 1 /kt 
 j-t I (k-l)t+l I t 
 
 2 , where s < 1, is taken to be 0) 
 v t=l > 
 
 Proof : We prove by induction on i and j. For i = and any j, we have a(0,j,k)=l 
 We assume the formula holds for i-1, and prove it for i, as follows: 
 Let i = (k-l)x + y, 1 < y ^k - 1. For j <_ x, we get i = (k-1 )x + y >_ (k-1 )j 
 + y > (k-l)j and in this case the formula must give us the boundary condition 
 
 0, and it really does: as i = (k-l)x + y, and 1 <y < k - 1, thus 
 
 So we get 
 
 i-1 
 
 i-1 
 
 k-1 
 
 = x. 
 
 v J /i+j-l-kt\ 1 /kt\ . 
 
 t z } \ j-t I (k-i)t+i [tj - 
 
 y /l+kt\/[i-(k-l)j-l] + k(j-t)\ _±_ 
 
 t Z^\ t )[ j-t I l+kt 
 
 i+j 
 j 
 
 i+j-1 
 j 
 
 r+s-tn 
 n 
 
 In [KN: p. 58] we have 
 
 2 /r-tk\ /s-t(n-k)\ _r_ 
 k>^0\ k j \ n-k / r-tk 
 
 for integer n and r =)= tk, <_ k <_ n. Setting k -<- t, r ■*■ 1 , t *- -k, 
 
 s -*- i - (k-l)j - 1 and n «- j we get our summation ( . J j minus the term 
 
 corresponding to t = 0. 
 
12 
 
 So the formula gives 
 
 a(i,j,k) = ( i+ J'- ] ) ♦ ( 1t j" 1 ) - ( 1 }J) - 0, as desired. 
 
 Assuming it holds for j - 1 , we continue as follows: if j > x , we have 
 
 i = (k-l)x + y < (k-l),i + y, or i <_ (k-l)j, and we show that a(i,j,k) satisfies 
 
 a(i,j,k) = a(i,j-l,k) + a(i-l,j,k). By the induction hypothesis we have 
 
 1.-1 
 
 •^•«-ni-.^(w^TFiW(?) 
 
 and 
 
 a(i-l,j,k) = f 1 "^" 2 
 
 i-2 
 
 ■ k ; 1 -l /i+j-2-kt\ 1 /kt 
 
 £1 \ j-t ; (k-ut+i 1 1 
 
 Therefore, 
 
 i-1 
 
 k-1 
 a(i,j-l,k) + a(i-l.j.k) = ( i+ H - 2 
 
 v ] ' t=l 
 
 i-2 
 
 k-1 
 
 (...) + "2' 
 t=l 
 
 (...) 
 
 It remains to show that 
 
 i-1 
 
 i-2 
 
 - k ; 1 J/i+j-2-kt\ 1 /kt\ . L k ;U/i+j-2-kt\ 1 /kt\ 
 
 ^ \ j-i-t J (k-i)t+i U^ ^ I j-t 1 (k-ut+i U; 
 
 i-1 
 
 k-1 
 
 = Lyj/ 1+ j-l-kt\ 1 
 
 ^ I j-t j (k-l)t+l \t J . 
 
 kt 
 
 (*) 
 
 If y > 1 then 
 
 i-1 
 _k-l_ 
 
 = 
 
 i-2 
 _k-l_ 
 
 = x, and all the summations are 2 (...), hence 
 
 t=l 
 
 (*) is correct. If y = 1, we have 
 
 i-1 
 
 x-1 
 
 = x, 
 
 i-2 
 
 k-1 
 
 ■ x - 1, 
 
 but then a term corresponding to t = x in the second summation on the left 
 
 side of (*) is ( J **~ ); but j > x, so this is 0. The proof is thus completed. □ 
 
13 
 
 4.3 As for the unranking procedure, we find it in a "reverse" interpretation 
 of Theorem 5. We are given an integer t, and look for a tree T e T(k,n) s.t. 
 Index(T) = t. For this we find a sequence X e X(k,n) s.t. index(X) = t. 
 The modification for the Z sequences is left to the reader. 
 Algorithm UNRANK (Finding the sequence X e X(k,n) s.t. index(X) = t for a 
 given t): 
 
 1. A «- t, j *■ n 
 
 2. Find* a. s.t. a(£. j,k) < A <_ aU.+l ,j,k) 
 
 3. A + t - a(£, ,j,k) 
 
 j -*- J - 1 
 
 If A > 1 then goto 2. 
 
 n 
 Comment: we have now t - 1 + 2 a(£.,j,k) 
 
 C^o j o 
 
 4. X «- ° 1 ° 
 
 m - j + 1 
 
 s 4- (k-l)m - i 
 m 
 
 5. Change X as follows: 
 
 X. unchanged fori ^s - k 
 X s+1 X s+2 ••• X km "" X s-k+l X s-k+2 '■• X k(m-1) 
 X s-k + l X s-l + 2 ■•• X s^ ^ ] 
 
 6. m *- m + 1 
 
 If m <^ n then set s «- (k-l)m - i m and goto 5 
 
 7. End. 
 
 * 
 
 This step is performed by using a table look-up for the pre-computed values 
 of the a(i,j,k)'s, or by computing them on-line. In both cases, we use the 
 formula derived in Theorem 6. Note that I. is unique because a(i,j,k) is 
 increasing in i. 
 
14 
 
 ■f"h 
 
 Example : For the sequence X = 000100101 (the 8 sequence in figure 2), we 
 have - by our ranking function - the following: 
 index(X) = index(000100101 ) = 
 
 = a(2,3,3) + index(000101) = 
 = a(2,3,3) + a(0,2,3) + index(OOl) = 
 = a(2,3,3) + a(0,2,3) +1=6+1+1=8. 
 As for the unranking procedure: we look for a tree T e T(3,3) s.t. 
 Index(T) = 8. For this purpose we will find X s.t. index(X) = 8. Applying 
 our unranking algorithm, we have after its 4 step: 
 
 8 = a(2,3,3) + a(0,2,3) + 1. 
 By step 5 we first set X to 001. Step 6 changes X to 000101 - by insert- 
 ing 001 after the first - and then to 000100101. 
 
15 
 
 V. SUMMARY 
 
 5.1 After discussing the feasibility criteria for the integer sequences 
 associated with the k-ary trees with n internal nodes, we introduced the 
 algorithm GENERATE (in 3.4) which generates the sequences Z e Z(k,n) 
 lexicographically. The recursive ranking function index(X) (computed by 
 Theorem 5) finds the position of X in the lexicographic ordering of X(k,n), 
 and is based on the reduction criterion (Theorem 4) and on the solution to 
 the recurrence relation for a(i,j,k) (Theorem 6). The algorithm UNRANK 
 
 (in 4.3) finds the sequence X e X(k,n) s.t. index(X) = t, and - as is 
 always the case - is done by using the ranking procedure backwards. 
 
 5.2 There is a simple 1-1 correspondence between all regular k-ary trees 
 with n internal vertices and all k-ary trees with n vertices (see [KN: p. 559] 
 or [TR1 : p. 3]), so our algorithms can be applied to these classes as well. 
 
 ACKNOWLEDGEMENT 
 
 I would like to thank Professor C. L. Liu for making valuable 
 suggestions. 
 
16 
 
 REFERENCES 
 
 [BM] N. G. Debruijn and B. J. M. Morselt, "A Note on Plane Trees," 
 Journal of Combinatorial Theory 2, (1967), 27-34. 
 
 [DM] A. Dvoretzky and T. Motzkin, "A Problem of Arrangements," Duke Math. 
 Journal 14 (1947), 305-313. 
 
 [GA] M. Gardner, "Mathematical Games": Catalan Numbers, Scientific 
 American , June 1976, 120-122. 
 
 [GN] B. N. Gnedenko, The Theory of Probability , Chelsea Publishing Company, 
 N.Y., 1962. 
 
 [KL] D. A. Klarner, "Correspondences Between Plane Trees and Binary Sequences," 
 Journal of Combinatorial Theory 9, (1970) 401-411. 
 
 [KNO] Gary D. Knott, "A Numbering System for Binary Trees," Communications 
 of the ACM , 20, No. 2, February 1977. 
 
 [KN] Donald E. Knuth, The Art of Computer Programming Vol. 1: Fundamental 
 Algorithms, Addi son-Wesley, Reading, MA 1968. 
 
 [LI] C. L. Liu, Elements of Discrete Mathematics , McGraw-Hill, 1977. 
 
 [RH] F. Ruskey and T. C. Hu, "Generating Binary Trees Lexicographically" 
 SI AM Journal on Computing , to appear. 
 
 [TR1] Anthony E. Trojanowski , "On the Ordering, Enumeration and Ranking of 
 k-ary Trees," Tech. Report UIUCDCS-R-77-850, Department of Computer 
 Science, University of Illinois at Urbana-Champaign, February, 1977. 
 
 [TR2] Anthony E. Trojanowski, "Ranking and Listing Algorithms for k-ary 
 Trees," SIAM Journal on Computing , to appear. 
 
 [WH] W. A. Whitmorth, "Arrangements of m Things of One Sort and m Things 
 of Another Sort Under Certain Conditions of Priority," Messenger 
 of Math. 8 (1878), 105-114. 
 
 [YY] A. M. Yaglom and I. M. Yaglom, Challenging Mathematical Problems with 
 Elementary Solutions Vol. 1: Combinatorial Analysis and Probability 
 Theory, Hoi den-Day, 1964. 
 
 [ZA] S. Zaks, "Generating Binary Trees Lexicographically," Tech. Report 
 
 UIUCDCS-R-77-888, Department of Computer Science, University of Illinois 
 at Urbana-Champaign, August, 1977. 
 
 [ZR] S. Zaks and D. Richards, "Generating Trees and Other Combinatorial 
 
 Objects Lexicographically," Tech. Report UIUCDCS-R-77-903, Department 
 of Computer Science, University of Illinois at Urbana-Champaign, 
 November, 1977. 
 
IBLIOGRAPHIC DATA 
 HEET 
 
 1. Report No. |2. 
 
 UIUCDCS-R-77-901 
 
 3. Recipient's Accession No. 
 
 Title anJ Subtitle- 
 
 5- Report Date 
 
 
 November 1977 
 
 Generating k-ary Trees Lexicographically 
 
 6. 
 
 Auchor(s) 
 
 Shmuel Zaks 
 
 8. Performing Organization Rept. 
 
 No. 
 
 IVrlorming Organization Name and Address 
 
 10. Project/Task/Work Unit No. 
 
 Department of Computer Science 
 
 11. Contract /Grant No. 
 
 University of Illinois at Urbana-Champaign 
 Urbana, IL 61801 
 
 MCS-73-03408 
 
 1 Sponsoring Organization Name and Address 
 
 13. Type of Report & Period 
 Covered 
 
 National Science Foundation 
 
 
 Washington, D.C. 
 
 14. 
 
 >. Supplementary Notes 
 
 i. Abstracts 
 
 We show a one-one correspondence between all the regular k-ary 
 
 trees with n internal nodes and certain integer sequences. We then generate 
 
 these sequences lexicographically, and discuss the ranking function and the 
 
 unranking procedure. Relations to existing algorithms are discussed. 
 
 '. Key Words and Document Analysis. 17a. Descriptors 
 
 k-ary tree, ordered tree, lexicographic order, ranking, unranking. 
 
 b. Identifiers/Open-Ended Terms 
 
 c COSATI Field/Group 
 
 ,'Availability Statement 
 
 ! 
 1 
 
 19. Security Class (This 
 Report) 
 
 UNCLASSIFIED 
 
 21- No. of Pages 
 
 20. Security Class (This 
 Page 
 
 UNCLASSIFIED 
 
 22. Price 
 
 )*M NTIS-35 (10-70) 
 
 
 
 
 USCOMM'DC 40329-P7 1 
 
FCD 
 
 o lrt*7n 
 
AUG l 5 198G