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Efficient algorithms and data structures for VLSI CAD

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Efficient algorithms and data structures for VLSI CAD
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xii, 141 leaves : ill. ; 29 cm.

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Thesis (Ph. D.)--University of Florida, 1996.
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Includes bibliographical references (leaves 138-140).
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EFFICIENT ALGORITHMS AND DATA STRUCTURES FOR VLSI CAD











By

SEONGHUN CHO










A DISSERTATION PRESENTED TO THE GRADUATE SCHOOL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF
DOCTOR OF PHILOSOPHY UNIVERSITY OF FLORIDA 1996














ACKNOWLEDGMENTS


My heartfelt appreciation goes to my advisor Professor Sartaj Sahni for giving me continued guidance in my thesis work. I thank him for the help, patience and support he provided throughout my stay in the University of Florida. Weekly meetings and discussions with him have spawned many ideas, for which I am thankful.

I would like to thank other members in my supervisory committee, Dr Li-Mmn Fu, Dr Theodore Johnson, Dr Sanguthevar Rajasekaran, and Dr Paul W. Chun for their interest and comments.

Thanks go to Venkat Thanvantri for his willingness to discuss the general subject of algorithms.

Thanks go to my wife Joungyim for her love, encouragement and patience. Finally, I would like to thank my parents for the love and support, without which I could not have pursued my doctoral studies. To them I dedicate this work.













TABLE OF CONTENTS




ACKNOWLEDGMENTS ............................. ii

LIST OF TABLES ................................. vii

LIST OF FIGURES ................................ x

ABSTRACT .... ....... .. ........... ............ xi

CHAPTERS

I INTRODUCTION ............................. 1
1.1 Background ..... .. ....... .. .. .......... 1
1.2 Dissertation Outline ......................... 2

2 MINIMUM AREA JOINING OF COMPACTED CELLS ........ 4
2.1 Introduction ..... ......... .. .. ...... .... 4
2.2 1-layer River Routing ......................... 9
2.3 Constraint Graph Representation .................. 16
2.4 Heuristics to Minimize Area ..................... 20
2.4.1 Heuristic I .. . . 20
2.4.2 Heuristic 2 .. .. ..... .. .. .. ... .... 21
2.4.3 Heuristic 3 ....... .. .. .. ... .... 23
2.5 Experimental Results . . . 25
2.6 Conclusion . . . 30

3 A NEW WEIGHT BALANCED BINARY SEARCH TREE ...... 32
3.1 Introduction ..... ......... .. .. .. ..... .... 32
3.2 Balanced Trees and Rotations .................... 34
3.3 fl-BBSTs ................................ 39
3.4 Search, Insert, and Delete in a O-BBST . 44
3.4.1 Search . . . 44
3.4.2 Insertion . . . 44
3.4.3 Deletion . . . 57
3.4.4 Enhancements . . . 63
3.4.5 Top Down Algorithms . . 64








3.5 Simple O-BBSTs . . . 66
3.6 BBSTs without Deletion . . 70
3.7 Experimental Results . . . 72
3.8 Conclusion . . . 98

4 WEIGHT BIASED LEFTIST TREES AND MODIFIED SKIP LISTS 105
4.1 Introduction . . . 105
4.2 Weight Biased Leftist Trees . . 106
4.3 Modified Skip Lists . . . 109
4.4 MSLs As Priority Queues . . 118
4.5 Experimental Results For Priority Queues . 122
4 6 Conclusion . . . 129

5 CONCLUSIONS . . . 135

A ABBREVIATIONS . . . 136


REFERENCES . . . . 138

BIOGRAPHICAL SKETCH . . . 141




























i %,
















LIST OF TABLES




2.1 Error rate (%) over optimal, I = 1 . . 27

2.2 Improvement (%) over Fang, 1 = 1 . . 27

2.3 Time taken, I = I . . . 28

2.4 Error rate (%) over optimal, I = 2 . . 29

2.5 Improvement over I = I cases . . 29

2.6 Improvement over Fang, I = 2 . . 30

2.7 Time taken, 1 = 2 . . . 31

3.1 The number of key comparisons on random inputs (version 1 code) 76

3.2 The number of key comparisons on ordered inputs (version 1 code) 77

3.3 Height of the trees on random inputs (version 1 code) . 77

3.4 Height of the trees on ordered inputs (version 1 code) . 77

3.5 The number of rotations on random inputs (version I code) 79

3.6 The number of rotations on ordered inputs (version I code) 80

3.7 Run time on random inputs using integer keys (version I code) 81

3.8 Run time on ordered inputs using integer keys (version 1 code) 82

3.9 Run time on random real inputs (version 1 code) . 83



V








3.10 Run time on ordered real inputs (version 1 code) . 84

3.11 The number of key comparisons on random inputs (version 1 code) 87

3.12 The number of key comparisons on ordered inputs (version 1 code) 88

3.13 Height of the trees on random inputs (version 1 code) . 89 3.14 Height of the trees on ordered inputs (version I code) . 89 3.15 The number of rotations on random inputs (version 1 code) 90 3.16 The number of rotations on ordered inputs (version I code) 91 3.17 Run time on random inputs using integer keys (version I code) 93 3.18 Run time on ordered inputs using integer keys (version I code) 94 3.19 Run time on random real inputs (version 1 code) . 95

3.20 Run time on ordered real inputs (version 1 code) . 96

3.21 The number of key comparisons on random inputs (version 2 code) 99

3.22 The number of key comparisons on ordered inputs (version 2 code) 100

3.23 Run time on random real inputs (version 2 code) . 101

3.24 Run time on ordered real inputs (version 2 code) . 103

4.1 The number of key comparisons . . 117

4.2 Num ber of levels . . . 118

4.3 Run tim e . . . . 119

4.4 The number of key comparisons . . 124

4.5 Height/level of the structures . . 126

4.6 Run time using integer keys . . 127

4.7 The number of key comparisons . . 130



VI








4.8 Height/level of the structures . . 131

4.9 Run time using integer keys . . 132

A.1 Abbreviations used in tables . . 137














































VII

















LIST OF FIGURES




2.1 Cell joining...................................... 5

2.2 I-layer river routing. .. .. .. .. ... .... ... ... ... .....8

2.3 Round robin and greedy layer assignments. .. .. .. ... ... ....11

2.4 Minimizing the number of tracks or layers. .. .. ... ... ... ..14

2.5 Constraint graph representation. .. .. .. ... ... ... ... ....18

2.6 Merge in constraint graph. .. .. .. .... ... ... ... ... ..19

2.7 Heuristic 1. .. .. .. .. ... ... .... ... ... ... ... ....20

2.8 Heuristic 2. .. .. .. .. ... ... ... .... ... ... ... ....22

2.9 Heuristic 3. .. .. .. .. ... ... ... .... ... ... ... ....24

3.1 LL and RL rotations. .. .. .. ... ... ... .... ... ... ..35

3.2 A tree in WVB(1/4) that is not L-balanced. .. .. ... ... .. ....42

3.3 !-balanced tree that is not a COST .. .. .. .. .... ... ... ..43

3.4 Lb rotation for insertion .. .. .. .. .... ... ... ... ... ..45

3.5 Substep (i) of insertion LR rotation .. .. .. .. .... ... ... ..46

3.6 Case Lb, for LR(ii) rotation .. .. .. ... .... ... ... ... ..49

3.7 Case LR for LR(ii) rotation .. .. .. ... ... .... ... ... ..51








3.8 LL rotation for deletion. .. .. .. .. ... ... ... .... ... ..58

3.9 LR rotation for deletion. .. .. ... ... ... ... ... ... ...60

3.10 Restructuring procedure. .. .. .. ... .... ... ... ... ....65

3.11 Simple restructuring procedure for insertion .. .. .. ... ... ....68

3.12 Simple restructuring procedure for deletion .. .. .. .. ... ... ..68

3.13 Simple restructuring procedure without a #i value .. .. .. ... ....70

3.14 Run time on real inputs (version 1 code) .. .. .. ... ... ... ..85

3.15 Run time on random real inputs (version 1 code) .. .. .. ... ...92

3.16 Run time on ordered real inputs (version 1 code) .. .. .. .... ..97

3.17 Run time on random real inputs (version 2 code). .. .. .. ... ..102

3.18 Run time on ordered real inputs (version 2 code). .. .. ... .. ..102

4.1 Example min-WBLTs .. .. .. .. ... ... ... ... ... ... ..108

4.2 min-WBLT Insert. .. .. .. .. .... ... ... ... ... ... ..109

4.3 min-WILT Delete-mmn.. .. .. .... ... ... ... ... ... ..110

4.4 Skip Lists. .. .. .. ... .... ... ... ... ... ... ... ...11

4.5 Modified Skip Lists .. .. .. .. ... .... ... ... ... ... ..113

4.6 MSL Search. .. .. .. ... ... ... ... ... ... .... ....113

4.7 MSL Insert .. .. .. ... ... ... .... ... ... ... ... ..114

4.8 MSL Delete. .. .. .. ... ... ... .... ... ... ... ....114

4.9 Run time. .. .. ... ... ... ... ... ... ... .... .....120

4.10 TMSL Insert. .. .. ... ... ... ... ... ... .... ... ..121

4.11 TMSL Delete-min.. .. .. ... ... .... ... ... ... ... ..121




ix








4.12 TMSL Delete-max ....... ............................ 122

4.13 Run time on randomly ....... .......................... 128

4.14 Run time on random2 ...... .......................... 129

4.15 Run time on randomly ...... .......................... 131

4.16 Run time on random2 ...... .......................... 133











































x
















Abstract of Dissertation Presented to the Graduate School
of the University of Florida in Partial Fulfillment of the Requirements for the Doctor of Philosophy EFFICIENT ALGORITHMS AND DATA STRUCTURES FOR VLSI CAD By


Seonghun Cho


May 1996



Chairman: Dr. Sartaj Sahni
Major Department: Computer and Information Science and Engineering


In this dissertation, we develop efficient algorithms and data structures for problems that arise in electronic computer aided design (ECAD).

We consider the problem of joining a row of compacted cells so as to minimize the area occupied by the cells and the interconnects. The cell joining process includes cell stretching and river routing. We propose several heuristics to join a row of cells in such a way that area is minimized. The proposed heuristics are compared experimentally with the previously proposed heuristic.

We develop a new class of weight balanced binary search trees called f-balanced binary search trees (0i-BBSTs). 3-BBSTs are designed to have reduced internal path


xi








length. As a result, they are expected to exhibit good search time characteristics. Individual search, insert, and delete operations in an n node 13-BBST take O(log n) time for 0 < j3 < V2- 1. Experimental results comparing the performance of #I3BSTs, W13(a) trees, AVL-trees, red/black trees, treaps, deterministic skip lists and skip lists are presented. Two simplified versions of /3-BBSTs are also developed.

We propose the weight biased leftist tree as an alternative to traditional leftist trees for the representation of mergeable priority queues. A modified version of skip lists that uses fixed size nodes is also proposed. Experimental results show our modified skip list structure is faster than the original skip list structure for the representation of dictionaries. Experimental results comparing weight biased leftist trees and competing priority queue structures as well as experimentat results for double ended priority queues are presented.



























X11















CHAPTER 1
INTRODUCTION


1.1 Background


In VLSI layout, we are concerned with transforming a circuit from its logical design to a physical implementation. The layout problem for VLSI circuits is generally decomposed into smaller problems such as partitioning, floorplanning, placement, routing and compaction.

The partitioning process decomposes a large circuit/module into a collection of smaller sub-ci rcui ts/ modules. In floorplanning, logical components of a circuit are assigned relative positions on a chip. The physical realization of each component (i.e., its area and aspect ratio) is also selected. The objectives of floorplanning include overall area minimization, minimization of power consumption. etc. The precise locations for the components of a design are then determined during the placement process to optimize the area and the timing. After the components are placed, the pins are connected during the routing process. During the process of compaction, the components and interconnections are moved so as to further optimize the layout in terms of area and delay.

The routing process is usually divided into three smaller sub-problems of global routing, detailed routing, and specialized routing. Global routing decomposes the




1




2


complex routing problem into small and manageable sub-problems. It assigns each net to a set of routing regions such as channel, switcbbox, and planar routing to minimize a combination of criteria such as area, circuit delay, etc. Steiner trees and spanning trees are the commonly used approaches for net connection in global routing. Specialized routing is used to connect power-ground nets or clock nets. Detailed routing has two types of routing, general and restricted and there are three types of detailed restricted routing, channel, switchbox, and planar.

In channel routing, all terminals of nets are located in two parallel rows across a routing region called channel. In switchbox routing, terminals of nets are located on the four sides of the routing region. Planar routing is a problem in which the interconnection topology of the nets is planar. That is, all connections can be realized on a single layer. Vias allow wires to change layers but the presence of vias reduces reliability and performance of a circuit. Single layer routing is not always possible.

River routing is a special case of planar routing in which all nets have exactly two terminals, one on each side of the channel, and the net sequence on each side of the channel is the same. River routing is used in PCB routing, particularly in dataflow architectures with multi-bit buses connecting a series of logic blocks, and symbolic IC design systems.

1.2 Dissertation Outline

This dissertation is divided into four chapters. In Chapter 2, we consider the problem of joining a row of compacted cells so as to minimize the area occupied by the cells and the interconnects. The cell joining process includes cell stretching and




3


river routing. We propose several heuristics to join a row of cells in such a way that area is minimized. The proposed heuristics are compared experimentally with the previously proposed one.

VLSI Physical Design Automation is essentially the study of algorithms and data structures related to the physical design process. Specific data structures can be used to improve the performance of algorithms. For example, maze routing algorithms and line-probe algorithms in global routing [30] use search structures, retiring algorithms [6, 291 use priority queue structures, and layout compaction algorithms [13] use radix priority search trees.

Chapters 3 and 4 are about search and priority queue structures. As specific data structures could be used to produce better performance of algorithms in VLSI design automation, new search and priority queue structures are proposed and thoroughly compared with other data structures.

Finally, in the last chapter, we present conclusions of this work.














CHAPTER
MINIMUM AREA JOINING OF COMPACTED CELLS


2.1 Introduction

When designing circuits with compacted symbolic sticks basic cells, the circuit is realized by a collection of compacted cells that tile a two-dimensional area. The intercell interconnects are such that each interconnect connects two terminals that are on adjacent boundaries of neighboring cells. So, for example, if cells A and B (Figure 2.1(a)) are neighboring cells of the circuit, then the right boundary of A is adjacent to the left boundary of B. The number of terminals on each of these boundaries will be the same and the Z'th terminal (from the bottom) on the right boundary of A is to be connected to the i'th terminal (from the bottom) on the left boundary of B.

Since the cells are available in compacted form, it is not possible to reduce the distance between any pair of terminals on any side of a cell. However, this distance can be increased by stretching the cell. In the example of Figure 2. 1 (a), we can stretch either cell vertically by defining a horizontal cut line at any position and pulling the two cell pieces apart by any desired amount (the cell can also be stretched horizontally by using a vertical cut line).






4




5





5 5 E- 5 35
4 15
A- 4 3 B 10
3 2
2


(a) Horizontal adjacent cells (b) Joining by stretching





12




(c) Joining by river routing (d) Combination cell joining

Figure 2. 1. Cell joining


The required interconnects between cells A and B of Figure 2.1(a) can be accornplished by stretching cells A and B so that the terminals of A and B line up as in Figure 2.1(b). The broken lines in Figure 2.1(a) indicate the cut lines used for stretching. The stretching enables us to join cells A and B using no routing tracks (by "join" we mean make the interconnects between cells A and B). This method of joining cells is also called pitch matching.

Another way to join cells A and B is to river route the interconnects as in Figure 2.1(c). This uses routing tracks in a channel between cells A and B but does riot increase cell height. The pitch matching and river routing approaches to cell joining have been studied in Boyer [51 and Weste [33]. Algorithms for single-layer




6



river routing can be found in several works [15, 19, 23. 24] and those for multilayer river routing can be found in Baratz [3]. Single-layer gridless river routing is studied in Tompa [32]. Two applications of river routing are hybrid circuit design and structured design (DSP).

Cell stretching (or pitch matching) increases the height of the layout while river routing increases its width. Both affect the layout area. The layout of Figure 2.1(b) has area 150. To compute the area of the layout of Figure 2.1(c), we assume tracks have unit separation. So, the layout width is 14 and height is 11. The layout has area 154. Cheng and Despain [S) have proposed using a combination of cell stretching and river routing so as to obtain layouts with smaller area than possible when only one of these joining methods is used. Figure 2.1(d) shows the result of joining cells A and B using both stretching and river routing. The area of this layout is 144. This is minimum for the instance of Figure 2.1(a).

Cheng and Despain [8] have proposed a heuristic for single layer joining of compacted cells. At each step of their heuristic either a row or column of compacted cells is joined. Following this, the row or column of joined cells is replaced by a composite cell that represents the result of joining. Notice that when a row (column) of cells is joined, cells may be stretched vertically (horizontally) and river routing is done in a vertical (horizontal) channel. To join a row of cells, Cheng and Despain [81 bound the maximum height to which a cell may be stretched. This bound is hmax + hb1v9/(4 h...)




7


where h,,,, is the height of the tallest compacted cell being joined and h,,,-, is the average height of the cells being joined.

Using this bound, cells are joined one-at-a-time using a penalty/reward scheme to determine if a pair of terminals is to be joined by stretching or by river routing.

Lim, Cheng, and Sahni [17] have considered the case when only two cells are to be joined. They develop fast polynomial time algorithms to obtain the minimum area join of two cells. In addition, they are able to obtain, in low order polynomial time, minimum area joins that rninirf ze the length of the longest wire or the total wire length. Lim [16] has proposed an 0(n(n/c)-') algorithm to find the minimum area join of c cells having a total of n terminals. This algorithm does an exhaustive search over all possible numbers of tracks in the c I routing channels between adjacent cells. A constraint graph is used to determine the minimum height layout for each assignment of number of tracks to routing channels. The time required per track assignment is O(n) and the worst case number of track assignments is 0((n/c)-1). The algorithm of Lim [16] is flawed as it handles channels with zero routing tracks by joining the adjacent cells using minimum height cell stretching and then considers the joined cells as one. This problem is easily fixed, however, by combining, in the constraint graph, pairs of vertices that represent corresponding terminals of the two cells (i.e., ZAli terminals of each cell) with zero routing tracks in between.

In this chapter, we consider the case when I > 1 routing layers are available to river route the inter cell connections. Note that while multiple layers do not affect layout area when cell stretching alone is used, a reduction in area is possible when




8




















(a) 1-layer (b) 2-layer

Figure 2.2. I-layer river routing


cell stretching is combined with river routing or when river routing alone is used. We assume that in each layer of each routing channel, the interconnects are to be accomplished using river routing. An alternative is to use HV routing when I = 2, HVH or VHV routing when I 3, and extensions of HVH and VHV routing for

1 > 3. However, for river routing instances, using routing layers in this way has no advantage over river routing in each layer (see Theorem 5, Section 2). When the number of layers available for river routing is increased, one may see a dramatic reduction in the number of routing tracks needed per layer. Figure 2.2 shows an instance that needs n tracks when routed in one layer but only one track/layer when routed in two layers.

We begin, in section 2, by stating the necessary and sufficient conditions for a river routing instance to be routable in I layers using at most t tracks per layer arid stating how to perform I-layer river routing when such a routing is possible. In




9



this section, we also show that IIV style routing has no advantage over river routing in each layer. In Section 2.3, we describe the constraint graph used to determine minimum height stretching of c cells. Heuristics for the minimum area joining of c cells are proposed in Section 2.4 and the results of experiments with these are provided in Section 2.5. Our conclusions appear in Section 2.6.



2.2 1-layer River Routing


Let (Ai, B,), 1 < I < mn be a set of terminal pairs such that the A,'s are on one side (say left or top) of a routing channel and the Bi's are on the other (right or bottom) side. Terminal Ai is to be connected to terminal B2, 1 < i < m. For this channel routing instance to be an instance of river routing, it must be the case that a, < a2 < ... < a,~ and b, < b2 < ... < bm where a, and bi, respectively, give the positions of terminals Ai and Bi, 1 K I < m. We may assume an underlying grid with each terminal being at a grid position. In the case of a horizontal (vertical) channel the ais and bis are grid column (row) numbers. Leiserson and Pinter [151 have obtained the following necessary and sufficient condition for a river routing instance to be routable in a single layer using at most t > 0 tracks. Theorem 1 [15] The river routing instance defined above is routable in a single layer using at most t > 0 tracks if and only if

(a) ai~t 6, > t (b) b~ ai > t

for every i < 771 t.




10


For the general case of I > 1 layers, we obtain the necessary and sufficient condition of Theorem 2.


Theorem 2 The river routing instance defined above is routable in I > 1 layers (each layer routing whole nets) using at most t > 0 tracks per layer if and only if

(a) ai+it bi > t (b) bj+lt a > t

for every i < m- lt.


Proof: First, we establish that (a) and (b) are necessary for I layer routing. Since the proofs for (a) and (b) are similar, we provide that for (a) only. Suppose that ai+lt bi < t for some i. Consider the It + 1 terminal pairs (Aj, Bi), i < j < i + It. When routing these on I layers, at least one layer has to be assigned > t + 1 terminal pairs. So, suppose that terminal pairs (A',B ), (A'2, B), ..., (A+ 1,B'+), ... are assigned to the same layer for river routin g. We may assume that a' < a'2 < ... < at+1 andb < b2 < < bt+1. Since a' < a1+< t and b' > b -b' ai+it-bi 1. 2 a+1l -- ait+t

From Theorem 1, it follows that the terminal pairs (A, B), 1 < j < t + 1, cannot be river routed on a single layer. Hence, (Aj, Bj), i < j < i + It, cannot be river routed on 1 layers. So, (Aj, Bj), 1 < J < m, cannot be river routed on I layers. As a result,

(a) is a necessary condition.

To show that (a) and (b) are sufficient conditions for routability, we present two algorithms (RoundRobin and Greedy) that assign the nets to layers in such a way that each layer is river routable when both (a) and (b) are satisfied. The correctness








procedure RoundRobin
{ Assign the m nets to I layers. }
begin
fori 1 to m do
assign net (Ai, B1) to layer (i mod 1) + 1;
end ;

procedure Greedy;
{ Assign the m nets to I layers. }
begin
fori:= 1 tomdo
assign net (Ai, B1) to layer q such that q is the smallest integer < I
for which the conditions of Theorem 1 are not violated on layer q
(if there is no such q, then fail)
end ;

Figure 2.3. Round robin and greedy layer assignments

of these algorithms is established in Theorems 3 and 4, respectively. 0



We later discovered that Baratz [3] has not only obtained the same condition but also proposed the same two algorithms for I-layer river routing. One assigns nets to layers in a round robin fashion and the other uses a greedy strategy. The corresponding procedures are given in Figure 2.3.


Theorem 3 The layer assignment produced by the RoundRobin procedure is river routable if

(a) ai+lt bi > t (b) bi+t ai > t for all i < m-It.




12



Proof: Let (A', B'), (A,, B'), ... = (Aj,Bj), (Aj+t, Bj+,), (Aj+2, Bj+21), ... be the nets assigned to layer (j mod 1) + 1, j < 1. So, a' = aj+(i-) and b' = bj+(i-I. Hence,
I
ai+t b = aj+(i+t_)l bj+(i-)l

aj+(i-1)I+ti bj+(i-1)l

> t (from (a))


Similarly, b, a > t. So, the layer assignment satisfies the conditions of Theorem 1 and is river routable using t tracks. 0


Theorem 4 If (a) ai+t bi t and (b) bi+tt ai t for all i < m lt, then procedure Greedy assigns nets to layers such that the assignment to each layer is routable using t tracks.


Proof: If procedure Greedy is able to assign each of the mn nets to a layer, then the layer assignments satisfy the conditions of Theorem 1 and so are routable using t tracks. Suppose the algorithm fails while trying to assign net (A,, B,) to a layer. At this time nets (Ai, Bi), 1 < i < r, have been assigned to layers so as to satisfy the conditions of Theorem 1 and the assignment of net (A,, Br) to each of these layers violates these conditions. Consider first those layers, L,, on which condition (a) is violated. For a layer s E La, suppose that the assigned nets are ..., (A-t, B-t),
..., (A1,Bj). Let (A, BJ) = (A,B,.). Since s E L,, we have a, bj_, < t. Now, if b,.- > bj'_t, then a. br-it < t which violates condition (a) of this theorem. So, b-t < b'>_. Since, b-., < b'. < ... < b6'2 < b'-_, I of the It 1 nets




13


(Ar -t+ I, Br..it+ ). ...., (Ar1, B,_1) have been assigned to layers E L,,. Consequently. the layers in L,, account for tiLal of these it 1 nets.

In a similar way, we can show that the remaining 1 ILail layers account for another W~ 1L.1) of these nets. This gives us a total of ti nets, whereas we had only ti 1. This contradiction implies that procedure Greedy cannot fail unless conditions

(a) and (b) are not satisfied. 11



Procedure RoundRobin is easily seen to have complexity of 0(m). A straightforward im-plementation of procedure Greedy will have complexity of 0(ml). However, by using priority search trees [18] the complexity can be reduced to O(m log 1). In practice, since I is quite small, it is unlikely that the priority search tree implementation will run faster than the straightforward implementation in which the 1 layers are checked in sequence. The actual routing for all I layers can be done in 0(mt) time using the computed layer assignment and the single layer routing algorithm of Leiserson and Pinter [15].

Using Theorem 2, we can develop a linear time algorithm to determine the minimum number of tracks needed to route an instance in I layers as well as to find the minimum number of layers needed for a t track routing. The algorithm for the former is given in Figure 2.4. This figure also shows the changes needed in case t is given and we wish to determine the minimum number of layers. The correctness of the algorithm follows from that of Theorem 2 and the fact that t (or 1) is increased only if the current t (1) is found to be infeasible. The complexity is 0(m) as neither




14


procedure MinimizeTracks; {or MinimizeLayers}
{ Determine the minimum number of tracks per layer (or minimum number
of layers) needed for multilayer river routing }
begin
t:0O; {or I := 1}

while (i < m It) do
if (ai+lt b, < t) or (bi+t ai < t)
thent:=t+ 1 {orl:=l+1}
else i :=i + 1;
end ;

Figure 2.4. Minimizing the number of tracks or layers

i nor t (1) can exceed m. So, neither clause of the if statement can be executed more than m 1 times.

Using the multilayer river routing results of Baratz [3], one can trivially extend all the results of Lim, Cheng and Sahni [17] to the case of multilayer joining of compacted cells. So, the multilayer minimum area join of two compacted cells with m nets can be obtained in O(m2) time. If we wish to minimize the maximum wire length while keeping area minimum, the asymptotic time complexity is still O(m2). The total wire length can be minimized while keeping area minimum in O(m2 log m) time.

In HV style routing, each routing layer is assigned a routing direction (either H or V). In an H (V) layer only horizontal (vertical) wire segments can be laid out. Horizontal segments on one layer connect to vertical segments, of the same net, on another layer by means of vias. In the case of river routing instances, one can see that there is no advantage to having more than two V-layers (i.e., two V-layers are sufficient to route all river routing instances).




15


Let RR(I, t) be the set of all river routing instances that can be routed in 1 layers, using t tracks per layer and using river routing in each layer. Let HV(I, t) be all river routing instances that can be routed using HV style routing, 1 layers, and t tracks per layer. Note that HV(I,t) includes instances routable with 0, 1, and 2 V-layers. Let HVV(1, t) be all river routing instances using HV style routing, 1 2 H-layers, and 2 V-layers. Theorems 5 and 6 below hold for both the knock-knee [25] and directional HV models. Theorem 7 holds only for the directional model. Theorem 5 HV(I,t) C RR(I,t) for every I > 1 and every t > 1. Proof: HV(l, t) RR(/, t) follows from a more general result obtained by Baratz [3]. Baratz [3] has shown that, for river routing instances, there is no advantage to using any routing scheme that wires a net on more than one layer. Since it is easy to construct river routing instances X such that X E RR(1, t) and X HV(l,t), it follows that HV(I, t) C RR(I, t).

We provide a simpler proof of HV(I, t) g RR(l, t). This proof will also establish our next result. We shall show that if X is a river routing instance such that X RR(/,t), then X 0 HV(/,t). Hence, HV(/,t) C RR(/,t).

Suppose that X RR(l, t). From Theorem 2, it follows that ai+t bi < t or b,+tj aj < t for some i. Suppose that a,+Ii bi < t (the proof is similar when bi+tt ai < t). So, ai+lt < bi t. Since X is a river routing instance, at least nets i + t,. ,i + It intersect a vertical cut line drawn at ai+lt. Hence, the density of X at aj+it is > i + It (i + t) + 1 = (I 1)t + 1. When HV style routing is used with




16


1, 1 > 1, layers, at most I 1 layers are available for horizontal routes. With t tracks per layer, densities of at most (I 1)t can be accommodated. So, X HV(/, t). 0 Theorem 6 HVV(l,t) C RR(l- 1,t) for everyI > 2 and everyt > 1. Proof: As in Theorem 5, suppose that X 0 RR(I- 1, t). Let i be such that ai+(-1)t bi < t. The net density at a1+(1-1)t is > (1 2)t + 1. In HVV routing, two layers are V-layers. So only I 2 layers are available for horizontal segments. This is not enough as the horizontal segment density is > (1 2)t + 1 at ai+(-1)t. Hence, X HVV(!, t).

One may easily construct river routing instances that are in RR(l 1, t) but not in HVV(l,t). 0


Theorem 7 RR(2, t) r HV(l, t) HVV(1, t) for every I > 1 and every t > 1. Proof: Consider the RR instance (a,,bl) = (1,2) and (a2,b2) = (2,3). This is in RR(2,t) for every t > 1 but is not in HV(/, t) HVV(/, t) for any 1. El



As remarked earlier, Theorem 7 holds only for the directional model. For the knock-knee model, one can show that RR(/, t) C HV(/ + 1, t) for every I > 1 and every t > 1.


2.3 Constraint Graph Representation

Lim [16] has proposed the use of a constraint graph to determine the terminal positions in a row of compacted cells. This is for the case when the number of tracks




17


in each routing channel is given and we wish to minimize the layout height. In the constraint graph, each cell is represented by a directed chain of vertices. Each cell terminal is represented by a vertex. The exception is when a compacted cell has terminals at the same y-position on both sides of the cell. In this case, the two terminals at the same y-position are represented by a single vertex. The vertex chain is linked in the direction of increasing y-position. The chain edges are labeled by the minimum allowable terminal separation. In addition, the constraint graph contains a source vertex that represents the bottom of the layout and a sink vertex that represents the layout top. The source vertex connects to the bottom of each chain and the top of each chain is connected to the sink vertex.

Figure 2.5(b) shows the chains (solid edges) for the four cell row of Figure 2.5(a). To complete the constraint graph, directed edges are added to introduce the channel routing constraints of Theorem 2. These are represented by the broken edges of Figure 2.5(b). Figure 2.5(b) is for the two layer case.

Lim [16] has shown that the constraint graph is acyclic provided the number of tracks in each routing channel is > 0. He has proposed handling channels with zero tracks by finding first the minimum area joining of the adjacent cells (only cell stretching is permitted now) and then combining these two cells into one. Le., the two cells are replaced by their minimum area join. This strategy can be shown to result in non-optimality of the algorithm proposed in Lim [16). To preserve optimality, it is necessary to merge the vertices that represent terminals that are the endpoints of




18



sink


2

C 9 M



C M f 2
b g b
L f

a d h k 3 2

a d h k

(a) Cell joining with 2 layers



source

(b) Constraint graph representation Figure 2.5. Constraint graph representation


nets that are to be routed using no tracks as in Figure 2.6. The resultant constraint graph is also acyclic.

It is easy to see that the number of vertices and edges in the constraint graph is O(n) where n is the total number of terminals. Furthermore, the graph can be constructed in O(n) time given the number of routing layers and the number of tracks in each channel. The constraint graph described by us is identical to that of Lim [16] except in the way channels with zero tracks are handled and in that our graph is defined for 1 > 1 routing layers while that of Lim [161 is only for I = 1.




19



sink


2

C 9 M


f 2

b


3 di 2
i V
a h k





source

Figure 2.6. Merge in constraint graph


The length of the longest path from the source vertex of the constraint graph to each of the remaining vertices can be computed in O(n) time by doing this in topological order [14, Section 6.5]. It is easy to see that if each terminal is placed at a vertical position given by the longest path length from the source, then all nets can be routed in the given number of tracks (as the conditions of Theorem 2 are satisfied in each routing channel). Furthermore, Lim [16] has shown that such a positioning of terminals results in a stretched layout of minimum height for the given channel widths. As a result, when channel widths are known, cells can be stretched to minimize area in O(n) time. The channel widths that result in minimum area can be determined in 0(n(n/c)c-1) time where c is the number of cells by trying out all




20


procedure Ileuristici
begin
for i: 1 to c -1 do
begin
determine the minimum area join of each pair of adjacent cells;
select the pair that has minimum area and replace it with its
minimum area join;
end;
end ;

Figure 2.7. Heuristic 1

possible channel widths [16]. Since this is feasible only for small c, we propose several heuristics in the next section.



2.4 Heuristics to Minimize Area


We formulate three greedy heuristics to obtain the minimum area join of a row of c compacted cells that have a total of n terminals.


2.4.1 Heuristic 1

The heuristic is described in Figure 2.7.

In each iteration of the for loop we examine every pair of adjacent cells. For each pair, the minimum area join is found using the algorithm of Lim, Cheng and Sahni [17] extended to the multilayer case as discussed in Section 2.2. The pair which has the minimum area join is replaced by a single cell that represents this join. So following each iteration of the for loop the number of cells decreases by one. When the for loop terminates, we are left with a Single cell that represents the join of all c cells. The time needed to determine the minimum area join of a pair of cells with ni nets between them is O(n?). The time to do this for all pairs of adjacent




21


cells is O(>ji n') 0 (n 2). So, the for ioop iteration with i = 1 takes 0(n2) time. On subsequent iterations, only the two pairs that include the cell introduced in the previous iteration need to have their minimum area join computed. Since each cell pair being considered includes at least one composite cell, the minimum area join is computed by considering the portion of the constraint graph that represents all the basic cells in the cell pair. Channel widths for channels within a composite cell are not changed while obtaining the minimum area join of the cell pair. However, as different channel widths for the channel between the two (composite) cells being joined are tried, the constraint graph is used to determine the minimum height of the combined cell. So, the time to combine two (composite) cells with ni terminals in the channel between them is O(nn1). Hence the time for the remaining c 2 iterations is O(n Xj7i n,) = 0(n'). The overall complexity of Heuristic 1 is therefore 0(n2). In case the terminals are uniformly distributed over the cells, ni = O(n/c) for all i. The time for the first iteration of the for loop is now O(n2/c) and that for each of the remaining iterations is 0(n2/c). The overall time is 0(n2).

2.4.2 Heuristic 2

In this heuristic, we begin by assigning each channel the number of tracks needed to route the channel with no cell stretching. This number can be determined in 0(n1) time for a channel with ni nets as described in Section 2.2. The time taken to do this for all c 1 channels is 0(n). The configuration obtained in this way is the maximum width layout. Starting from this configuration, we reduce the total number of tracks available across all c 1 channels by one on each iteration. For this, the




22


procedure Heuristic2
begin
for each channel determine the number of tracks, ti,
needed to route with no stretching, 1 < i < C;
t =E-1t
set up the constraint graph using ti tracks in channel i, 1 < i < c
compute layout area, A ;
for tracks :=t downto 1 do {reduce by 11
begin
for 1 i=1to c-i1 do
begin
reduce the number of tracks in channel i by 1
modify the constraint graph to reflect this ;
determine the length of the longest path in the graph
and from this the layout area, a1
end;
select 1 such that a1 = min{ a,
reduce the number of tracks in channel j' by I
A =min{f A, a,
end
end ;

Figure 2.8. Heuristic 2

effect of a one track reduction is computed for each channel. The minimum layout

height is determined by computing the length of the longest path in the constraint

graph of Section 2.3. The track reduction is done in the channel that results in the

smallest layout height (hence the minimum area for the given number of tracks). The

algorithm is stated more formally in Figure 2.8.

When the algorithm terminates, A is the area of the minimum area join found

by the heuristic. To reconstruct the layout, it is necessary to store the tracks per

channel each time A is updated in the statement A = min{f A, a, }.

For the time complexity, we see that the steps that precede the outer for loop

take O(n) time. Each iteration of the outer loop takes 0(nc) time. Hence this ioop




23


contributes a total of 0(net) to the time. Since t = O(n), the overall time complexity of Heuristic 2 is O(n 2 C).

2.4.3 Heuristic 3

Unlike Heuristic 2 which attempts to minimize the layout height for each value of t, the totaJ number of tracks, Heuristic 3 attempts to minimize the width (i.e., total number of tracks) for each choice of layout height. The heuristic begins with a layout height, ht, equal to the height of the tallest compacted cell. At each iteration, the next layout height to use is computed as described later. During each iteration, cells are combined in groups of at most k (k > 1 is a parameter to the heuristic). Each group of combined cells is replaced by its minimum area join subject to the constraint that the height of the join does not exceed ht. This joining of < k cells at a time continues until only one cell remains. Its area is computed and recorded. The minimum area obtained over all heights tried is then reported as the best. Heuristic

3 is given in Figure 2.9.

In our implementation of Heuristic 3, the minimum area join of k cells is found by considering the portion of the constraint graph for all the basic cells included in these k cells. So, for this purpose composite cells are not handled as single cells. Rather, as in Heuristic 1, the basic cells they are composed of are considered and channel widths previously assigned to the associated channels are not changed. Track assignment is done only for the k 1 channels between the k composite cells. We found this to give better results than when composite cells were regarded as atomic. For the case k = 2, the minimum area is determined by a binary search over the




24


procedure Heuristic3
begin
ht :=height of the tallest cell;
repeat I minimize width subject to height < ht}
repeat I do this by combining k cells at a time}
select k adjacent cells such that the minimum height cell is selected
and the height of the tallest selected cell is minimum
(if there are fewer than k cells, then select all of them)
obtain the minimum area layout for the selected cells under
the constraint that the layout height does not exceed ht
during the preceding step record the next value of ht
that is possible for a layout
until one cell remains;
compute the area of the remaining cell and record it
if it is less than the minimum area found so far;
if there is no next height then terminate;
hti: next height;
until false
end ;

Figure 2.9. Heuristic 3

number of tracks in the single channel. This takes O(n log n,) time where ni is the

number of nets in channel i. Thus the time needed for the inner repeat loop when

k = 2 is O(cn log n) (for uniform terminal distribution it is O(cn log(n/c)). During

the binary search, the heights corresponding to channel widths that require height

> ht are recorded. The minimum of these heights yields the next value of hi.

When k > 2, all track combinations for the k 1 channels are tried as in

Section 2.3. Again, each composite cell is broken up into its basic cells. As different

track combinations are tried, we record the minimum height > ht that results from

any track combination. This gives the next value of ht. The time for the inner

repeat loop is O((c/(k 1))n(n/k)k-1) (or O((c/(k 1))n(n/c)k1') when terminals

are uniformly distributed).




25


In all our experiments, the outer repeat loop was iterated fewer than (k 1)n times. To ensure that the number of iterations is O(kn), one may adopt the following scheme. When the number of iterations first reaches (k 1)n, compute a set of at most n new heights by beginning with the current constraint graph. This uses the current assignment of number of tracks in each channel. Heuristic 2 is next used to reduce the total number of available tracks by one and determine the height needed to complete the routing with the reduced number of tracks. This process gives us at most n new heights h, < h2 < ... < hp,. Heuristic 3 is now resumed with h, as the next height. Only two iterations are performed. Then Heuristic 3 is resumed with max{ h2, ht I as the next height. Again two iterations of the outer repeat loop are done. Next the heuristic is resumed with max I h3, ht } as the next height. This continues until we have gone through p resumptions of the heuristic. With this scheme to limit the number of iterations, the complexity of Heuristic 3 becomes O(cn 2 log n) when k = 2 and O((c/(k 1))kn 2(n/k)k1') = O(cn2(n/k)k1') when k > 2. For the case when the n terminals are uniformly distributed over the c cells, the complexity is O(cn'log(n/c)) when k = 2 and O((c/(k 1))kn(n/c)-1) = O(cn(n/c)k1') when k > 2. One may verify that since Heuristic 3 tries the maximum useful height (i.e., the height needed when no routing tracks are available), it generates optimal solutions when k = c.



2.5 Experimental Results

We programmed our three heuristics as well as the heuristic Fang [8) in C and ran tests on a single KSR processor. Optimal solutions for instances with up to nine




26


cells were obtained using the corrected version of the exhaustive search algorithm of Lim [161. Our test set consisted of instances that had a number of cells, c, equal to one of the numbers in the set 13, ..., 9, 10, 20, 50, 1001. For each value of c, there were twenty instances and the results were averaged over these instances. An instance with c cells had c I routing channels. The number, t, of terminals on either side of each routing channel was equal to c for 3 < c < 9 and was 10 for the other values of c. In addition, when c = 100, we also had instances with 20 terminals on either side. In our experiments, we considered only single layer and two layer routing.

Table 2.1 gives the average percentage by which the area of the single layer solutions generated by each of the heuristics exceeded the area of the single layer optimal solution. As is evident, each of the heuristics proposed in this chapter gave noticeably better solutions than did Fang. This table is only for the cases 3 < c < 9 as for c > 9 the optimal algorithm of Lim [16] required too much time to complete. When k > c, Heuristic 3 is guaranteed to generate an optimal solution. So, we did not run these cases.

In table 2.2, we have used the single layer solution produced by Fang as the benchmark against which the solutions obtained by our three heuristics are compared. This table gives the average percentage by which the area of the solutions produced by our heuristics is less than that of the solutions produced by Fang. Our solutions have area 9 to 18% less.

Table 2.3 compares the computing time requirements of the various algorithms for the case of one layer. The optimal algorithm is useful only for small values of c




27








Table 2.1. Error rate (%) over optimal, 1 = 1

cells___t Fang_ Hersi Heuisic He 31k =41
[~ ~ cel 21 Fag Huitc Huitc4__Huitc__3 59 0.5 0.2 0 *
4 4 10.0 0.9 0.1 0 0 *
5 5 11.0 3.7 0.3 0.3 0.1 0.1
6 6 12.7 2.4 0.3 0.2 0.2 0.0
7 7 16.1 3.5 0.3 0.1 0.1 0.1
8 8 17.9 2.7 0.4 0.3 0.3 0.1
9 9 18.8 3.5 0. 3 0.4 10.3 0.

t = number of terminals on each side of each routing channel
:k> c












Table 2.2. Improvement (%) over Fang, I = 1

cell I t 21 Heriti Heriti __rs
[~ ~~ cel 2f~ Hersi J Hersi 3__eur=ti 4
10 10 9.4 14.0 14.0 14.2 14.3
20 10 10.5 15.4 15.4 15.6 15.6
50 10 10.6 16.1 16.1 16.2
100 10 9.1 16.5 16.5 16.6
100 20 if 9.3 18.3 18.0

excessive run time




28



Table 2.3. Time taken, I

cells I t Fang Heuristic Heuristic Heuristic3 Optimal
1 2 k 2 k 3 k 4
3 3 0.0 0.01 0.02 0.02 0.01
4 4 0.0 0.02 0.02 0.04 0.09 0.05
5 5 0.0 0.03 0.04 0.09 0.36 1.2 0.77
6 6 0.0 0.03 0.08 0.21 0.66 3.7 13
7 7 0.0 0.04 0.14 0.50 3.4 27 278
8 8 0.0 0.07 0.24 0.84 4.4 34 1.8t
9 9 0.0 0.09 0.42 1.8 16 72.3 47t
10 10 0.0 0.14 0.78 3.1 19 334
20 10 0.01 0.32 5.8 16 117 1196
50 10 0.01 1.4 93 127 896
100 10 0.03 4.4. 782 590 4873
100 20 0.06 11 3022 4087

Times are in seconds.
t : Times are in hours.


(say up to 7). While Fang is significantly faster than the heuristics proposed here, the quality of the solutions generated by our heuristics is superior.

Table 2.4 is the analog of table 2.1 for the case of two layers. Again, our heuristics performed considerably better than did Fang. Table 2.5 gives the improvement in area due to increasing the number of routing layers from one to two. This is influenced somewhat by the width of cells which in our case ranged from 5 to 30 times the track separation. With narrower cells, the impact of the second layer would have been greater and with wider cells, it would have been less. Also, the impact of the second layer is more when more routing tracks are needed. For the smaller instances of table 2.4, for example, the optimal solutions with 1 = 2 required, on average, only

1.8% less area than when I = 1.




29









Table 2.4. Error rate (%) over optimal, 1 = 2

cells I Fang Heuristic Heuristic Heuristic3

3 3 6.3 0.5 0 0
4 4 9.8 1.1 0.1 0 0
5 5 11.0 2.7 0.1 0.2 0.1 0.0
6 6 13.0 1.4 0.1 0.0 0.0 0.1
7 7 15.9 2.3 0.1 0 0 0
8 8 16.8 1.7 0.2 0 0.1 0.0
9 9 18.4 2.2 0.1 0.1 0.1 0.0














Table 2.5. Improvement (%) over I = 1 cases

cells Fang Heuristic Heuristic Heuristic3
1 2 V=-k=[=
~~I I_________ k = 2k=_3 1 k = 4
10 10 4.1 4.7 3.3 3.3 3.0 3.0
20 10 4.1 5.8 3.4 3.4 3.3 3.2
50 10 4.6 5.2 3.4 3.2 3.2
100 10 4.7 6.3 3.5 3.4 3.4
100 20 7.1 10.4 4.8 5.2 -




30



Table 2.6. Improvement (%) over Fang, I = 2


[cells t Heuristic I Heuristic Heurisfic3
1 2 k = 2 1 k = 3 k =- 4
10 10 10.0 13.3 13.2 13.2 13.3
20 10 12.2 14.8 14.8 14.9 14.8
50 10 11.2 15.1 14.9 15.0
100 10 10.7 15.4 15.4 15.4
100 20 12.5 16.3 16.3 :__J



Table 2.6 is the analog of table 2.2 for the case of two layers. The results are similar to those in table 2.2. 'Fable 2.7 gives the average computing times for the two layer instances. These are less than for the one layer case as the constraint graph has fewer edges.

For large c, we recommend the use of heuristic 2 or 3 (with k = 2) and for small c we recommend using heuristic 3 (with k = 3 or 4).


2.6 Conclusion

vVe have considered the problem of joining a row of compacted cells and developed heuristics to stretch cells and river-route the nets so that the layout area is minimized. Our proposed heuristic was compared, experimentally, with Fang [81 and found to produce layouts with less area. However, Fang is faster. We recommend the use of our Heuristic 3 with k = 3 or 4 in practice.




31

















Table 2.7. Time taken, 1 = 2

~cellsj IJt Fanj Heuristic ersi1k=leuristic3 =_ Optimal

3 3 0.0 0.01 0.02 0.02 *0.0
4 4 0.0 0.02 0.02 0.03 0.06 0.04
5 5 0.0 0.02 0.02 0.07 0.25 0.75 0.60
6 6 0.0 0.02 0.04 0.13 0.38 2.0 11
7 7 0.0 0.04 0.06 0.29 1.9 14 223
8 8 0.0 0.06 0.12 0.53 2.5 19 15
9 9 0.0 0.06 0.20 0.96 7.5 34 39
10 10 0.01 0.09 0.32 1.7 9.9 15320 10 0.01 0.24 2.7 9.9 6.5 65150 10 0.01 1.1 42 85 586
100 10 0.02 3.7 357 472 3069 -1 00 20 0,05 7.9 1377 3166 -Times are in seconds.
t:Times are in hours.















CHAPTER 3
A NEW WEIGHT BALANCED BINARY SEARCH TREE


3.1 Introduction


A dictionary is a set of elements on which the operations of search, insert, and delete are performed. Many data structures have been proposed for the efficient representation of a dictionary [141. These include direct addressing schemes such as hash tables and comparison schemes such as binary search trees, AVL-trees, red/black trees [121, trees of bounded balance [21], treaps [1], deterministic skip lists [20], and skip lists [26]. Of these schemes, AVL-trees, red/black trees, and trees of bounded balance (WB(a)) are balanced binary search trees. When representing a dictionary with n elements, using one of these schemes, the corresponding binary search tree has height O(log n) and individual search, insert, and delete operations take O(log n) time. When (unbalanced) binary search trees, treaps, or skip lists are used, each operation has an expected complexity of O(log n) but the worst case complexity is O(n). When hash tables are used, the expected complexity is 0(1) per operation. However, the worst case complexity is 0(n). So, in applications where a worst case complexity guarantee is critical, one of the balanced binary search tree schemes is to be performed.






:32




33


In this chapter, we develop a new balanced binary search tree called fl-BBST (/3balanced binary search tree). Like WB(a) trees, this achieves balancing by controlling the relative number of nodes in each subtree. However, unlike WB(a) trees, during insert and delete operations, rotations are performed along the search path whenever they reduce the internal path length of the tree (rather than only when a subtree is out of balance). As a result, the constructed trees are expected to have a smaller internal path length than the corresponding WB(a) tree. Since the average search time is closely related to the internal path length, the time need to search in a /3-BBST is expected to be less than that in a WB(a) tree.

In Section 3.2, we define the total search cost of a binary search tree and show that the rebalancing rotations performed in AVL and red/black trees might increase this metric. We also show that while similar rotations in WB(a) trees do not increase this metric, insert and delete operations in WB(a) trees do not avail of all opportunities to reduce the metric. In Section 3.3, we define O-BBSTs and show their relationship to WB(a) trees. Search, insert, and delete algorithms for /3-BBSTs are developed in Section 3.4. A simplified version of 83-BBSTs is developed in Section 3.5. Search, insert and delete operations for this version also take O(log n) time each. An even simpler version of f3-BBSTs is developed in Section 3.6. For this version, we show that the average cost of an insert and search operation is O(log n) provided no deletes are performed.

An experimental evaluation of f3-B BSTs and competing schemes for dictionaries (AVL, red/black, skip lists, etc.) was done and the results of this are presented in




34



Section 3.7. This section also compares the relative performance of P-BBSTs and the two simplified versions of Sections 3.5 and 3.6.

3.2 Balanced Trees and Rotations

Following an insert or delete operation in a balanced binary search tree (e.g., AVL, red/black, WB(a), etc.), it may be necessary to perform rotations to restore balance. The rotations are classified as LL, RR, LR, and Rb [14]. Lb and RR rotations as well as bR and Rb rotations are symmetric. While the conditions under which the rotations are performed vary with the class of balanced tree considered, the node movement patterns are the same. Figure 3.1 shows the transformation performed by an bb and an bR rotation. In this figure, nodes whose subtrees have changed as a result of the rotation are designated by a prime. So, p' is the original node p however its subtrees are different.

bet h(x) be the height of the subtree with root x. bet s(x) be the number of nodes in this subtree. When searching for an element x, x is compared with one element at each of I(x) levels, where I(x) is the level at which x is present (the root is at level 1). So, one measure of the "goodness" of the binary search tree, T, for search operations (assuming each element is searched for with equal probability) is its total search cost defined as:



C(T) =Z1(x).
xET




35




9P
P

P d
qI 9P I

q
c a A d
b c

a b

(a) LL rotation


9P
qI

p d
I
P 9P
a q A. X
a b c d

b c

(b) LR rotation Figure 3.1. LL and RL rotations


Notice that C(T) = I(T) + n where I(T) is the internal path length of T and n is the number of elements/nodes in T. The cost of unsuccessful searches is equal to the external path length E(T). Since E(T) = I(T) + 2n, minimizing C(T) also minimizes E(T).

Total search cost is important as this is the dominant operation in a dictionary (note that insert can be modeled as an unsuccessful search followed by the insertion of a node at the point where the search terminated and deletion can be modeled by




36


a successful search followed by a physical deletion; both operations are then followed by a rebalancing/restructuring step).

Observe that in an actual implementation of the search operation in programming languages such as C++, C, and Pascal, the search for an x at level I(x) will involve upto two comparisons at levels 1, 2,..., 1(x). If the code first checks x = ej where ej is the element at level i to be compared and then x < e to decide whether to move to the left or right subtree, then the number of element comparisons is exactly 21(x) 1. In this case, the total number of element comparisons is



NC(T) = 2 E 1(x) n = 2C(T) n
XET


and minimizing C(T) also minimizes NC(T). If the code first checks x < ej and then x = ei (or > ei), the number of element comparisons done to find x is 1(x) + r(x) + 1 where r(x) is the number of right branches on the path from the root to x. The total number of comparisons is bounded by 2C(T). For simplicity, we use C(T) to motivate our data structure.

In an AVL tree, when an LL rotation is performed, h(q) = h(c)+1 = h(d)+1 (see Figure 3.1(a)). At this time, the balance factor at gp is h(p) h(d) = 2. The rotation restores height balance which is necessary to guarantee O(log n) search, insert, delete operations in an n node AVL tree. The rotation may, however, increase the total search cost. To see this, notice that an LL rotation affects the level numbers of only those nodes that are in the subtree with root gp prior to the rotation. We see that l(q') = l(q) 1, (p') = l(p) 1, 1(gp') = l(gp) + 1, the total search cost of the subtree




37


with root a is decreased by s(a) as a result of the rotation, etc. Hence, the increase in C(T) due to the rotation is:



1(p') 1(p) + l(q') I(q) + l(gp') l(gp) s(a) s(b) + s(d)



= -1 1 + 1 s(q) + 1 + s(d) = s(d) s(q). A similar analysis shows that an LR rotation increases C(T) by s(d) s(q).

If the LL rotation was triggered by an insertion, s(q) is at least one more than the minimum number of nodes in an AVL tree of height t = h(q)- 1. So, s(q) > 0t+2/v/5 where 4 = (1 + V/5)/2. The maximum value for s(d) is 2' 1. So, an LL rotation has the potential of increasing total search cost by as much as



2t 1 Ot+2/,/-5 -- 2t 1 1.62t+2/2.24.



This is negative for t < 2 and positive for t > 2. When t = 10, for example, an LL rotation may increase total search cost by as much as 877. As t gets larger, the potential increase in search cost gets much greater. This analysis is easily extended to the remaining rotations and also to red/black trees. Definition (WB(a) [21]) The balance, B(p), of a node p in a binary tree is the ratio (s(l) + 1)/(s(p) + 1) where I is the left child of p. For a E [0, 1/2], a binary tree T is in WB(a) iff a < B(p) < 1 a for every node p in T. By definition, the empty tree is in WB(a) for all a.




38


Lemma 1 (1) The maximum height, hmaz(n), of an n node tree in WB(a) is ~ log.s (n + 1) [21]

(2) Inserts and deletes can be performed in an n node tree in WB(a) in O(log n) time for 2/11 < a < 1 V2-/2 [4].

(3) Each search operation in an n node tree in WB(a) takes O(log n) time [21].


In the case of weight balanced trees WB(a), an LL rotation is performed when B(gp) = 1 a and B(p) > a/(1 a) (see Figure 3.1(a)) [21]. So,


s(p)+1 s(p) + 1
1 -as
s(gp) + 1 s(p) + s(d) + 2


or
a 2a- 1
s(d) s(p) a
1 a 1 a and

a s(q) + 1
-- < B(p) = o1-a B _s(p)+1

or
a 20 1
s(q) > s(p)-2- + 2a1-a 1-a

So, LL rotations (and also RR) do not increase the search cost. For LR rotations [211, B(gp) ,~ 1 a and B(p) < a/(1 a). So, s(d) ~ s(p)_ + and with respect to Figure 3.1(b),


a s(p) s(q)
> B(p)1 -a s(p) +1




39


or

s(q) >s(p) 1 2a a


For a < 1/3, s(q) > s(d) and LR (RL) rotations do not increase search cost. Thus, in the case of WB(a) trees, the rebalancing rotations do not increase search cost. This statement remains true if the conditions for LL and LR rotation are changed to those in Blum and Mehlhorn [4].

While rotations do not increase the search cost of WB(a) trees, these trees miss performing some rotations that would reduce search cost. For example, it is possible to have a < B(gp) < 1 a, B(p) > -a, and s(q) > s(d). Since B(gp) isn't high enough, an LL rotation isn't performed. Yet, performing such a rotation would reduce search cost.

3.3 U-BBSTs

Definition A cost optimized search tree (COST) is a binary search tree whose search cost cannot be reduced by performing a single LL, RR, LR, or RL rotation. Theorem 8 If T is a COST with n nodes, its height is at most log,(v/5(n + 1)) 2. Proof Let Nh be the minimum number of nodes in a COST of height h. Clearly, No = 0 and N1 = 1. Consider a COST Q of height h > 2 having the minimum number of nodes Nh. Q has one subtree R whose height is h 1 and another, S, whose height is < h 1. R must be a minimal COST of height h 1 and so has Nh-1 nodes. R, in return, must have one subtree, U, of height h 2 and another, V, of height < h 2. Both U and V are COSTs as R is a COST. Since R is a minimal




40


COST, U is a minimal COST of height h 2 and so has Nh-2 nodes. Since Q is a COST, IS max{IjUI, IVI}. We may assume that Nh is a nondecreasing function of h. So, ISI Nh-2. Since Q is a minimal COST of height h, ISI = Nh-2. So, Nh = Nh-1 + Nh-2 + 1, h > 2



N0 =0, N = 1.


This recurrence is the same as that for the minimum number of nodes in an AVL tree of height h. So, Nh = Fh+2 -1 where F is the i'th Fibbonacci number. Consequently, Nh a qh+2/v/5 1 and h < log,(v5(n + 1)) 2. O




Corollary 1 The maximum height of a COST with n nodes is the same as that of an AVL tree with this many nodes.


Definition Let a and b be the root of two binary trees. a and b are 3-balanced,

0

(a) fl(s(a) 1) < s(b)

(b) fl(s(b) 1) < s(a)


A binary tree T is fl-balanced iff the children of every node in T are 6-balanced.



A full binary tree is 1-balanced and a binary tree whose height equals its size (i.e., number of nodes) is 0-balanced.




41


Lemma 2 If the binary tree T is /-balanced, then it is 1-balanced for 0 < 7 < /.


Proof Follows from the definition of balance. 0




Lemma 3 If the binary tree T is #-balanced, 0 < /# 5 1/2, then it is in WB(a) for a = #/(1 + ). Proof Consider any node p in T. Let 1 and r be node p's left and right children.


s(l) + 1 1
B(p) = ---------- = 1
B(p) = s(1) + s(r) + 2 1 + s(L)+1


Since T is #-balanced, s(l) 1 < s(r)/4 or s(l) + 1 < s(r)/ + 2. So, s()+1 < 1/3 + 21-1
s(r) + 1 (s(r) + 1) 1/


or
s(r) + 1 >
s(l) + 1

So, B(p) < 1/(1 + /). Further, s(r) 1 < s(1)/3. So, s(r) + 1
< 1/#t.
s(1) + 1


And, B(p) 1/(1 + 1//) = #/(1 + f). Hence 8/(1 + #) < B(p) < 1/(1 + /) for every p in T. So, T is in WB(a) for a = 8/(1 + /3). 0




42




00
00
0/O O
0 0


Figure 3.2. A tree in WB(1/4) that is not --balanced


Remark While every #-balanced tree, 0 < < 1/2, is in WB(a) for a = /(1+ /), there are trees in WB(a) that are not 3-balanced. Figure 3.2 shows an example of a tree in WB(1/4) that is not -balanced.


Lemma 4 If T is a COST then T is -balanced. Proof If T is a COST, then every subtree of T is a COST. Consider any subtree with root p, left child 1, and right child r. If neither 1 nor r exist, then s(1) = s(r) = 0 and p is -balanced. If s(l) = 0 and s(r) > 1, then r has a nonempty subtree with root t and s(t) > s(1). So p is not a COST. Hence, s(r) < 1 and p is -balanced. The same is true when s(r) = 0. So, assume s(l) > 0 and s(r) > 0.

If s(1) = 1, then s(r) < 3 as otherwise, one of the subtrees of r has m > 2 nodes and m > s(l) implies p is not a COST. Since s(r) 5 3, 1(s(r) 1) < s(1) and (s(l)- 1) s(r). So, p is -balanced. The same proof applies when s(r) = 1. When s(1) > 1 and s(r) > 1, let a and b be the roots of the left and right subtrees of 1. Since p is a COST, s(a) < s(r) and s(b) 5 s(r). So, s(l) = s(a) + s(b) + 1 < 2s(r) + 1 and (s(l) 1) s(r). Similarly, (s(r) 1) s(l). So, !-(l,r). Since this proof applies to every nodes in T, the children of every p are -balanced and T is -balanced. O




43




~0
0 0 0
0 --0 0__Figure 3.3. '-balanced tree that is not a COST


Remark There are !-balanced trees that are not COSTs (see Figure 3.3).



While a COST is in WB(1/3) and WB(a) trees can be maintained efficiently only for 2/11 < a < 1 1/V2- z 0.293, a COST is better balanced than WB(a) trees with a in the usable range. Unfortunately, we are unable to develop O(log n) insert/delete algorithms for a COST.

In the next section, we develop insert and delete algorithms for #-balanced binary search trees (P-BBST) for 0 < 0 V2 1. Note that every (v'2- 1)-BBST is in WB(a) for a = 1 1/v/r- which is the largest permissible a. Since our insert and delete algorithms perform rotations along the search path whenever these result in improved search cost, BBSTs are expected to have better search performance than WB(a) trees (for a = //(1 + fl)).

Each node of a /-BBST has the fields LeftChild, Size, Data, and RightChild. Since every /-BBST, P3 > 0, is in WB(a), for a > 0, fl-BBSTs have height that is logarithmic in n, the number of nodes (provided # > 0).




44



3.4 Search, Insert, and Delete in a j8-BBST

To reduce notational clutter, in the rest of the chapter, we abbreviate s(a) by a (i.e., the node name denotes subtree size).

3.4.1 Search

This is done exactly as in any binary search tree. Its complexity is 0(h) where h is the height of the tree. Notice that since each node has a size field, it is easy to perform a search based on index (i.e., find the 1O'th smallest key). Similarly, our insert and delete algorithms can be adapted to indexed insert and delete.

3.4.2 Insertion


To insert a new element x into a /3-BBST, we first search for x in the /3-BBST. This search is unsuccessful (as x is not in the tree) and terminates by falling off the tree. A new node y containing x is inserted at the point where the search falls off the tree. bet p' be the parent (if any) of the newly inserted node. We now retrace the path from p' to the root performing rebalancing rotations.

There are four kinds of rotations LL, LR, Rb, and RR. Lb and RR rotations are symmetric and so also are bR and Rb rotations. The typical configuration before an Lb rotation is performed is given in Figure 3.4(a). p' denotes the root of a subtree in which the insertion was made. bet p be the (size of the) subtree before the insertion. Then, since the tree was a /3-BBST prior to the insertion, 03-(p, d). Also, for the Lb rotation to be performed, we require that (q > c) and (q > d). Note that q > d implies q > 1. 'Ae shall see that #~-(q, c) follows from the fact that the insertion is




45




9P p

LL
d q gp


q c c d

(a) before (b) after

Figure 3.4. LL rotation for insertion


made into a /3-BBST and from properties of the rotation. Following an LL rotation, p is updated to be the node p".


Lemma 5 [LL insertion lemma] If [f6-(p, d) A /3-(q, c) A (q > c) A (q > d) for 0 < < 1/2 before the rotation, then 0-(q, gp') and 0-(c,d) after the rotation. Proof Assume the before condition.

(a) f(q- 1) < c (as f-(q,c)) < gp'. Also, f(gp'- 1) =/(c+d) < 2/)3q (as />0, q > c and q > d) < q (as 3 < 1/2). So, 0-(q, gp').

(b) d < q :, d- 1 < q- 1 =, 0(d- 1) <5 (q- 1) < c (as #-(q,c)). Also, ,8(c-1)


In an LR. rotation, the before configuration is as in Figure 3.4(a). However, this time q < c. Figure 3.4(a) is redrawn in Figure 3.5(a). In this, the node labeled c in Figure 3.4(a) has been labeled q and that labeled q in Figure 3.4(a) has been labeled




46




gp
q'

P d LR(i)


a q
a b c d

b c

(a) before (b) after substep (i)

Figure 3.5. Substep (i) of insertion LR rotation


a. With respect to the labelings of Figure 3.5(a), rotation LR is applied when



[(q > a) A (q > d)].



The other conditions that apply when an LR rotation is performed are



[/-(p, d) A /3-(a,q) A /3-(b, c)].



Here p denotes the (size of the) left subtree of gp prior to the insertion. An LR rotation is accomplished in two substeps (or two subrotations). The first of these is shown in Figure 3.5(b). Following an LR rotation, p' is updated to be node q'. Lemma 6 [LR substep(i) insertion lemma] If [f-(p,d) A 3-(a,q) A /3-(b,c) A (q > a)A(q > d)]forO < P < 1/2 before the subrotation, then [3-(p",gp')A{(3-(a,b)A
(cd V ( -(ab) A -(cd0 after the subrotation. (c, d)) V (--(a, b) A 03-(c, d))}] after the subrotation.




47



Proof Assume the before condition. First, we show that 0-(p",gp') after the rotation. Note that O(p"- 1) = #(a + b) = #(a + b+ c + 1)- 3(c + 1) = p(p'- 1) /3(c+1) = 8(p-1)-8c < d-Oc < d< gp'. Also, /3(gp'- 1) = 8(c+d) b+#+#d (as #-(b,c)) 5 b + Sq (as q > d) 5 b + a + P (as #-(a,q)) < p" (as # 5 1/2 and p" = a + b + 1). So, #-(p", gp').

Next, we prove two properties that will be used to complete the proof. Pl: #(b- 1) < a.

To see this, note that fl(b 1) /3(q 1) < a (as 0-(a, q)). P2: (c 1) < d.

For this, observe that p' 1 = a + q 8(q 1) + q (as /-(a, q)) = ( + 1)(q 1) + 1. So, q- 1 < 2 = ~+. Similarly, q- 1 = b + c > fl(c- 1) + c (as #-(b,c))
--P--OP-1 <) d (---aq )< o
= (/3 + 1)(c- 1) + 1. So, 9(c- 1) < 'L(q 2) (q 1) 5 < 1 (as

/-(p, d)) < d.

To complete the proof of the lemma, we need to show


{(-(ab)A -(cd)) V( (a,b)A #-(c,d))}.
1+/ 1+/


We do this by considering the two cases b > c and b < c. Case b > c: Since a < q = b + c + (a -1) 0(b + c) < 2b < b. This and P1 imply 3-(a, b). Also, d < q = b + c + 1. So, +O(d 1) 5 + (b + c -) = c + -(b- 1) < -Lc + (as #-(b,c)) = c. This, together with P2 implies

1-(c, d). So, 0-(a, b) A -(c, d).

Case b < c: Since a < q = b + c + 1,a- 1 < b + c. So, a- 1 < b+c- 1 or




48


A- 1+0 1 13 + 1 + 0 1s-+b

Also, d- 1< q-2= b+c-1. So, fl(d- 1) : 3(b+c-1) <3(2c- 1) c. This, together with P2 implies #3-(c, d). So, ---- (a, b)A/3-(c, d). 0



Since an LR(i) rotation can cause the tree to lose its #-balance property, it is necessary to follow this with another rotation that restores the /-balance property. It suffices to consider the two cases of Figures 3.6 and 3.7 for this follow up rotation. The remaining cases are symmetric to these. In Figures 3.6 and 3.7, p and d denote the nodes that do not satisfy #-(p, d). Note, however, that these nodes do satisfy

1 1 16(p, d).

Since the follow up rotation to LR(i) is done only when


d~, ) A d)),
1+


either #(p- 1) > d or 3(d-1) > p. When fl(p- 1) > d, the second substep rotation is one of the two given in Figures 3.6 and 3.7. When 93(d 1) > p, rotations symmetric to these are performed. In the following, we assume /(p 1) > d. Further, we may assume d > 0, as d = 0 and 1+--(p, d) imply p < 1. Hence, /3(p, d). Also, d > 0 and fl(p- 1)> dimplyp> 1.

The LR(ii) LL rotation is done when the condition



A=(q>d)A(c<(1+3)q+(1-3))AB where




49




gp p
LR(ii)
d
d q gp'
LL q


q c c d

(a) before (b) after
Figure 3.6. Case LL for LR(ii) rotation


B = (p,d) A (-#-(p,d)) A -(q,c)A(fl(p 1) > d > 0).
1+0

Lemma 7 [Case LR(ii) LL rotation] If A holds before the rotation of Figure 3.6, then 0-(q, gp') and #-(c, d) after the rotation provided 0 < < V2 1. Proof (a) fl-(q,gp'):

O(q-1) c (as 0-(q,c)) < gp'. Also, 0(gp'-1) = 0(c+d) < #((1+P)q+(1-0)+d) 5 #(1 + O)q + 3(1 0) + #(q 1) (as q> d) = 0(2+ 8)q /2 < q (as 0(2+ 0) < 1 for 0 < p < v'2 1). So, fl-(q, gp').

(b) f-(c, d):

fl(d 1) < #(q 1) 5 c (as P-(q, c)). And, fl(c 1) = (c 1) + 0+(c
q + (c 1) = (q + c 1) = (P- 2) < (p 1) < d (as '3+-(p, d)).


So, fl-(c, d). O




Lemma 8 If (c < (1 + fl)q + (1 0)) A (fl(p 1) > d) in Figure 3.6, then d < q provided 0 < 3 < V2 1.




50


Proof Sinced < 0(p-1) = O(q+c) < O(q+(1+O)q+1-0) = (3+2)q+0(1-3) < q + 1 (as (# + 2) 1 and 0(1- ) < 1 for 0 < /_ v/2-1). So, d q. O



So, the only time an LR(ii) LL rotation is not done is when C = (Cl V C2) A B holds where

C1, = (q= d) A (c < (1 + #)q+1 ) C2=c> (1 +/3)q+(1- #).


At this time, the LR rotation of Figure 3.7 is done. In terms of the notation of Figure 3.7, the condition C becomes D = (D1 V D2) A E where



D = (a d) A (q < (1 + ~3)a + 1 /0)


D2=q > (1 + #)a + 1-/0


E =1 (p,d) A -0-(p,d) A fl-(a,q) A fl-(b,c) A (i(p- 1) > d > 0).
1+#


Lemma 9 When an LR(ii) LR rotation is performed and/3 < V2 1, q > d and so search cost is reduced.


Proof If D1, thensince d d/l3 d > d as S5 V 1. If D2, then d < (p- 1) = (a + q) < (q- + q) = q <
q q 1+(a s-)3 1 ). O
@A< : (as 0< Vf2--1). []
1+0 -




51




gp


pd LR(ii)

LR
a qxx
a b c d

b c

(a) before (b) after

Figure 3.7. Case LR for LR(ii) rotation


Lemma 10 When (d = a) A #3-(b, c) A (fl(p 1) > d) A (3 V2- 1) (see Figure S. 7), /3(a 1) < b and 03(d 1):5 c. Proof Since f(p -1) >dand d =a, O(p -1) > aorl#(a +q) > aor a( -8)


1(a -1) < 2(+b#,1-fl /3

1-0l 1-0l
#- + Il)b + #(#2 +# Ii+ p)
1-0l 1-0l
#_ +l I)b + 0(#2 2 -1)
1-f 1-fl





52


Since O(c- 1) < b,c < + 1. So,


#" f1" b #(3( + 1)b 302O(a-1)< 12(b+c+1)< # (b+ -2)_ 1- + 1- .
1-01- $1-13 1-13


So,
S+ 1 3# -1
a-1< b+ .
1-0 1- 0

However, since #12 + 2# 1 < 0 for f< vf 1, (1 + #)/(1 #) < and

(30-1)/(1- ) <10. So,a-l < b/y+#. Ifa>c+1, thenc



Lemma 11 [Case LR(ii) LR rotation] If D holds before the rotation of Figure 3.7, then #-(p', gp'), #-(a, b), and #-(c, d) following the rotation provided 0 < 1 5 1. Proof (a) -(p',gp'):

1(gp' 1) = #(c + d) 5 b + 3 + 8d (as 0-(b,c)) < b + 0 + 3q (from Lemmas 9 and 10, q >d) < b+ + a+ 13= a+ b+ 20 < a+b+ 1 = p'. Also, since +--(p,d) and q d,,3(p 1) 5 (1 + 1)d or #(a + q) 5 (,8 + 1)d or a + q < (1 + )d or a<(1+ -)d-q5(1+ )d-d=d/f. So, #(p'-1)=/3(a+b)
(as 0-(b,c)) < d+ c + 1 = gp'.

(b) 0-(a, b):

Since b < q and 13-(a,q),(b- 1) B (q- 1) < a.




53


When DI, 13(a 1) b was proved in Lemma 10. So, f3- (a, b). When D2, q a( +3)+ 1 So,







So,



#(aI 1fib +1c+/ O b+3 1+-P13 b



So, #-(a, b).

(c) 0-(c, d):

Note that 13(c 1) < 8(q 1) < (q- 1< '3(p -1) <5d. When DI, fl(d 1) c was proved in Lemma 10. So, 03-(c, d). When D2, if d < b +1, then d< b and /3(d -1) 138(b- 1) < c. So, assume d> b +1. Now, b < d 1 < O(p 1) 1. So,



b < 3(a+b+c )-i

q1 + # +b+c+1)-1


1P (b~c13(1 3)(b+c+ 1)) -1
1+3

1 + 1 c+ +0+ ( +3 + c+ 1)) 1


c+13(1+fl)c+213-1

= (2+13)c+3,3-1 <(2+L)c+3(asI3 V2--1)




54


f< c+ (as < V2- 1).


Also, from d < f/(p 1) and the above derivation, we get


d < (b+c+#+(1+#)(b+c+1))

< 1 + + C+, +T(1+0)( + 0 +c+ l))
1+ #1

j ,/311 2/32 1+13
(2+/ + + 22c) + ( + 1)
= (2+/)c+ 1+c+ (1 ) +1) 1++ 0
2/32 1 2+3+ + 2
= (2 1 f+)c + 1+c+

S(2 + )c+ + 42 + 1

(2+03)c+ 1 (aS#334/32/3< 1 +# for# v'2- 1).


So, f(d- 1) # (2 + P)c < c (as P V 1). So, -(c,d). 0



Theorem 9 If T is fl-balanced, 0 < #3 < V/ 1, prior to insertion, it is so following the insertion.

Proof First note that since all binary search trees are balanced for 1 = 0, the rotations (while unnecessary) preserve 0-balance. So, assume /3> 0. Consider the tree T' just after the new element has been inserted but before the backward restructuring pass begins.

If the newly inserted node, z, has no parent in T', then T was empty and T' is /-balanced. If z has a parent but no grandparent, then T has at most one nonempty




55


subtree X. Since T is fl-balanced, fl([XI 1) < 0. So, IXf < 1. Following the insertion, T' has one subtree with < 1 nodes and one with exactly one. So, T' is #l-balanced. We may therefore assume that z has a grandparent in T'.

From the downward insertion path, it follows that all nodes u in T' that have children 1 and r for which -#fl-(l, r) must lie on the path from the root to z. During the backward restructuring pass, each node on this path (other than z and its parent) play the role of gp in Figures 3.4 and 3.5. The #-property cannot be violated at z as z has no children. It cannot be violated at the parent, s, of z as s satisfied the fl-property prior to insertion. As a result its other subtree has < 1 element. So, following the insertion, s satisfies the fl-property. As a result, each node in T' that might possibly violate the fl-property becomes the gp node during the restructuring pass. Consider one such gp node. It has children in T' denoted by p' and d. Its children in T are p and d. Figures 3.4 and 3.5 show the case when d is the right subtree of gp in both T and T'. The cases RR and RL arise when d is the left subtree.

During the restructuring pass, gp begins at the grandparent of z and moves up to the root of T'. If z is at level r in T', (the root being at level 1), then gp takes on r 2 values during the restructuring pass. We shall show that at each of these r 2 positions either

(a) no rotation is performed and all descendants of gp satisfy the fl-property or

(b) a rotation is performed and following this, all descendants of node p" (Figure 3.4) or of node q' (Figure 3.5) satisfy the fl-property.




56


As a result, following the rotation (if any) performed when gp becomes the root of T', the restructured tree is /-balanced. The proof is by induction on r. When r = 3 (recall, we assume z has a grandparent), gp begins at the root of T' and its descendants satisfy the fl-property.

Without loss of generality, assume that the insertion took place in the left subtree of gp. With respect to Figure 3.4, we have three cases: (i) q c and q > d, (ii) q < c and c > d, and (iii) q < d and c < d. In case (i), all conditions for an LL rotation hold and such a rotation is performed. In case (ii), an LR rotation is performed. Following either rotation, T' is /-balanced. In case (iii), /3(p' 1) = /3(q + c) < 2/3d < d (as ,6 _< v/2- -1). Also, fl(d 1) < p < p + 1 = p'. So, fl(d 1) < p'. Hence, #-(p', d) and T' is /-balanced.

For the induction hypothesis, assume (a) and (b) whenever r < k. In the induction step, we show (a) and (b) for trees T with r = k + 1. The subtree in which the insertion is done has r = k. So, (a) and (b) hold for all gp locations in the subtree. We need to show (a) and (b) only when gp is at the root of T'. This follows from Lemmas 5, 6, 7, and 11.

The theorem now follows. 0




Lemma 12 The time needed to do an insertion in an n node /3-BBST is O(log n) provided 0



57



Proof Follows from the fact that insertion takes 0(h) time where h is the tree height and h = O(log n) when /3> 0 (Lemmas 1 and 3). 0




3.4.3 Deletion

To delete element x from a /3-BBST, we first use the unbalanced binary search tree deletion algorithm of Horowitz and Sahni [14] to delete x and then perform a series of rebalancing rotations. The steps are: Step 1 [Locate x] Search the f3-BBST for the node y that contains x. If there is no

such node, terminate.


Step 2 [Delete x] If y is a leaf, set d to nil, gp, to the parent of y, and delete node

y. If y has exactly one child, set d' to be this child; change the pointer from the parent (if any) of y to point to the child of y; delete node y; set gp to be the parent of d'. If y has two children, find the node z in the left subtree of y that has largest value; move this value into node y; set y = z; go to the start

of Step 2. { note that the new y has either 0 or 1 child }


Step 3 [Rebalance] Retrace the path from d' to the root performing rebalancing

rotations.


There are four rebalancing rotations LL, LR, RR, and Rb. Since Lb and RR as well as LR and Rb are symmetric rotations, we describe LL and LR only. The discussion is very similar to the case of insertion. The differences in proofs are due to the fact that a deletion reduces the size of encountered subtrees by 1 while an




58







g p p 1
LL
pd' qgp


q c c 9

(a) before (b) after

Figure 3.8. LL rotation for deletion insertion increases it by 1. In an LL rotation, the configuration just before and after the rotation is shown in Figure 3.8. This rotation is performed when q > c and q > d'. Following the rotation, d' is updated to the node p'.

Let d denote the size of the right subtree of gp before the deletion. So, d = d + 1. Since prior to the deletion the I3-BBST was #-balanced, it follows that f-(p, d) and 0-(q, c).


Lemma 13 [LL deletion lemma] If [f#-(p,d)A 0-(q,c)A(q c)A(q > d)A(1/3 < 3 1/2)] before the rotation, then [#-(q, gp') A 8-(c, d')] after the rotation. Proof (a) fl-(q, gp'):

/3(q -1) !5 c (as 8)-(q,c)) < gp'. Also, B3(gp' -1) = fl(c+ d') < 2,8q (as c < q and d' < q) < q (as )3 < 1/2). So, 0-(q, gp').

(b) #-(c,d'):

d' < q = d'-1 < q-1 =: (d'-1) < fl(q-1) < c. Also, when c < 1, 0(c-1) < 0 < d' (as d' > 0). When c > 1,q > c =z q > 2 and p = q+c+ 1 > c+3. So, 0(c- 1) < /(p- 1)- 3/3< d- 3/3 (as 0-(p,d)) < d- 1 (as fl > 1/3) = d'. Hence,




59


/3-(c, d'). 0



In an LR rotation, the before configuration is as in Figure 3.8(a). However, this time q < c. Figure 3.8(a) is redrawn in Figure 3.9(a). In this, the node labeled c in Figure 3.8(a) has been relabeled q and that labeled q in Figure 3.8(a) has been relabeled a. With respect to the labelings of Figure 3.9(a), rotation LR is applied when

[(q > a) A (q > d')].


The other conditions that apply when an LR rotation is performed are



[fl-(p, d) A f8-(a, q) A fl-(b, c)J.



Here d denotes the (size of) right subtree of gp prior to the deletion. As in the case of insertion, an LR rotation is accomplished in two substeps (or two subrotations). The first of these is shown in Figure 3.9. Following an LR rotation, d' is updated to node q'.


Lemma 14 [LR substep(i) deletion lemma] If [f3-(p, d) A #3-(a, q) A #-(b, c) A (q > a) A (q > d')] before the subrotation LR(i), then [3-(p', gp') A{ (/3-(a, b) A '3--(c, d))V( (a, b) A /-(c, d'))}] after the subrotation provided 1/3 < 3 < 1/2 Proof Assume the before condition.

(a) If b =c= 0, then q b+c+ 1 = 1. Furthermore, (q > a) and (q > d') imply a d'= 0. So, gp' = p'= 1. Hence, [!-(p',gp') A !-(a,b) A !-(c,d')]




60




gp
q'

P d' LR(i)


a q
a b c d'

b c

(a) before (b) after substep (i)

Figure 3.9. LR rotation for deletion


(b)Ifb=landc=0,thenq=2,a<1, andd'<1. So, 1
(c) Ifb=O0andc= 1, thenq=2, a<1, andd' < 1. So, 1 1 and c > 1. So, q >3, a> 1 (as -(a,q) i 6(q- 1) < a or a> 20 > 0), p = a + q + 1 > 5, d >2 (as #-(p, d) = (p 1) < d and f > 1/3), and d' = d 1 > 1.

First, we show that #-(p', gp'). For this, note that a + b + c + 1 = p 1. From #-(p,d), it follows that #(a + b+ c + 1) = /(p- 1) < d. So, 8(a + b) 5 d- c c- f. From Figure 3.9(b), we see that (p'- 1) = 3(a + b). Hence, I(p'- 1) 5 d- flc- = d'-,c+1-0 5 d'+1-2,8


(gp'- 1) =,(c + d') 5< b + + +d' (as ,-(b,c)) < b+Oq+O(asq>d')




61


< b + a + 2 (as -(a,q))

< p.



So, 0-(p', gp').

Next, we prove two properties that will be used to complete the proof. PI: /(b- 1) < a.

To see this, note that fl(b 1) < /(q 1) !5 a (as #-(a, q)). P2: l(c- 1) < d'.

For this, observe that fl(c 1) < P(q 2) (as c < q 1) < #(p 4) (as q = p a 1 and a > 1) = #/(p- 1)- 3/3 1/3) = d'.

To complete the proof of the lemma, we need to show



1043-(a, b) A # d)) V -(a, b) A #-(c, ))}.



For this, consider the two cases b > c and b < c (as in Lemma 6). Case b > c: Since a < q = b + c + 1,/#(a 1) < /(b + c) < 203b < b. This, together with P1 implies 3-(a, b). Also, d' < q =b+c+ 1. So, -6+(d' -1)< --(b +c- 1) =
--g-c~+ + +0(-+1+c
+ --- 1) c + c = c. This, together with P2 implies a --(c, d'). So,


/-(a,b) A 1 -(c, d').

Case b < c: Since a < q = b + c + 1,a 1 < b + c. So, a 1 < b + c 1 or or1+ + 1+ 1+ /3+1/3
d- 1 < q 2 = b + c 1. So, fl(d 1) < P(b + c 1) < 3l(2c 1) < c. This and




62


P2 imply 3-(c, d'). Hence, +-(a, b) A 3-(c,d'). 0



The substep(ii) rotations are the same as for insertion.


Theorem 10 If T is #-balanced, then following a deletion the resulting tree T' is also 3-balanced provided 1/3 < fl < -' 1.


Proof Similar to that of Theorem 9. 0



When 0 < # < 1/3, we need to augment the LL rotation by a transformation for the cased' = 0. Whend'= 0,/(p- 1) < d = d' + 1 = 1. So, p < 1/fl + 1 and gp = p + d' + 1 < 1/ + 2. To fl-balance at gp, the at most 1/ + 2 nodes in gp are rearranged into any fl-BBST in constant time (as 1/# + 2 is a constant). When d' > 0, the proof of Lemma 13 part (b) can be changed to show 0(c 1) < d' for 0 < < v2 1. The new proof is: since c < q,c < (p 1)/2 and fl(c 1) < (p- 1)/2 !5 d/2- # = d- d/2 0 < d- 1 0 < d'. The LR rotation needs to be augmented by a transformation for the case d' = d- 1 < 1. At this time, fl(p- 1) 5 d < (2) So, gp = p + d< 2( )+ 1 + To #-balance at

gp, we rearrange the fewer than + 1 + nodes in the subtree, in constant

time, into any #-balanced tree. When d' > 1 1, the proof for 3(c 1) 5 d' in 0(2+0)
Lemma 14 needs to be changed to show that the LR substep(i) lemma holds. The new proof is:



d > 01(p 1) = (a + b + c + 1) >(1(q-1) + b + c + 1bc+1)




63


= 1(j3(b+c)+b+c+1)

N + ((1+/)(c- 1) + (1 + O)c+ 1)

= #((1 +/3)2(c- 1)+2+/).


So, 1) d-2-2 So ~~~(1+,6)2-d-1(ad>

Also, note that when = 0, all trees are 8-balanced so the rotations (while not needed) preserve balance.


Theorem 11 With the special handling of the case d' = 0, the tree T' resulting from a deletion in a fl-BBST is also #-balanced for 0 < fl < V2- 1. Lemma 15 The time needed to delete an element from an n node fl-BBST is O6og n) provided 0 < /3 < V 1.


3.4.4 Enhancements

Since our objective is to create search trees with minimum search cost, the rebalancing rotations may be performed at each positioning of gp during the backward restructuring pass so long as the conditions for the rotation apply rather than only at gp positions where the tree is unbalanced.

Consider Figure 3.4(a). If p' < d, then the conditions of Lemmas 5 and 6 cannot apply as q < p' < d. However, it is possible that e > p' where e is the size of either the left or right subtree of d. In this case, an RR or RL rotation would reduce the total search cost. The proofs of Lemmas 5 and 6 are easily extended to show that these rotations would preserve balance even though no insertion was done in the subtree




64



d. The same observation applies to deletion. Hence the backward restructuring pass for the insert and delete operations can determine the need for a rotation at each gp location as below (I and r are, respectively, the left and right children of gp).

if s(l) > s(r) then check conditions for an Lb and LR rotation

else check conditions for an RR and RL rotation.

The enhanced restructuring procedure used for insertion and deletion is given in Figure 3.10. In the RR and RL cases, we have used the relation '>' rather than'> as this results in better observed run time.

Since it can be shown that the rotations preserve balance even when there has been no insert or delete, we may check the rotation conditions during a search operation and perform rotations when these improve total search cost.

Finally, we note that it is possible to use other definitions of 18-balance. For example, we could require 83(s(a) 2) < s(b) and 86(s(b) 2) < s(a) for 83-(a, b). One can show that the development of this chapter applies to these modifications also. Furthermore, when this new definition is used, the number of comparisons in the second substep of the LR and Rb rotations is reduced by one.

3.4.5 Top Down Algorithms


As in the case of red/black and WB(a) trees, it is possible to perform, in O(log n) time, inserts and deletes using a single top to bottom pass. The algorithms are similar to those already presented.




65



procedure Restructuring ; begin
while (gp) do
begin
if (s(gp.left) > s(gp.right)) then
begin {check conditions for an LL and LR rotation}
p = gp.left ;
if (s(p.left) > s(p.right)) then
begin if (s(p.left) > s(gp.right)) then do LL rotation; end
else
begin
if (s(p.right) > s(gp.right)) then {LR}
begin
do LR rotation ;
{ now notations a, b,c, and d follow from figure 3.1(b) }
if ((s(a) 1) > s(b)) then
if ((s(a.right) < (1 + )s(a.left) + 1 8) and (s(b) < s(a.left))) then do LL rotation else do LR rotation else if (P(s(d) 1) > s(c)) then
if ((s(d.left) < (1 + )s(d.right) + 1 4) and (s(c) < s(d.right))) then do RR rotation else do RL rotation; end
end
end
else {check conditions for an RR and RL rotation}
begin
p = gp.right;
if (s(p.left) > s(p.right)) then
begin
if (s(p.left) > s(gp.left)) then {RL}
do symmetric to the above LR case;
end
else
begin if (s(p.right) > s(gp.left)) then do RR rotation; end
end ;
gp = gp.parent;
end;
end;

Figure 3.10. Restructuring procedure




66


3.5 Simple 9-BBSTs

The development of Section 3.4 was motivated by our desire to construct trees with minimal search cost. If instead, we desire only logarithmic performance per operation, we may simplify the restructuring pass so that rotations are performed only at nodes where the fl-balance property is violated. In this case, we may dispense with the LL/RR rotations and the first substep of an LR/RL rotation. Only LR/RL substep (ii) rotations are needed. To see this, observe that Lemmas 7 and 11 show that the second substep rotations rebalance at gp (see Figures 3.6 and 3.7) provided '- l+13

(p, d) (The remaining conditions are ensured by the bottom-up nature of restructuring and the fact the tree was fl-balanced prior to the insert or delete).

If the operation that resulted in loss of balance at gp was an insert, then fl(p 2) < d (as p > d, the insert took place in subtree p and gp was 8l-balanced prior to the insert) and fl(p 1) > d (gp is not f-balanced following the insert). For the substep (ii) rotation to restore balance, we need fl(p 1) < (1 + f3)d. This is assured if d+/ : (fl+ 1)d (asfl(p-2) < d). So, we need d> 1. Ifd < 1, then d= 0. Now 0(p 2) < d and f(p 1) > d imply p = 2. One may verify that when p = 2, the LR(ii) rotations restore balance.

If the loss of f-balance at gp is the result of a deletion (say from its right subtree), then 0(p 1) < d + 1 (as gp was f-balanced prior to the delete). For the substep

(ii) rotation to accomplish the rebalancing, we need fl(p 1) < (f8 I- 1)d. This is guaranteed if d + 1 < (f8 + 1)d or d > 1/fl. When d < 1/fl and 8 > 1/3, d < 2. Since 0(p-1) < d+1 and 9> 1/3, when d=2, p < 10; whend= 1, p< 7; and when




67


d =0, p < 4. We may verify that for all these cases, the LR(ii) rotations restore balance. Hence, the only problematic case is when P3 < 1/3 and d < 1/f8.

When /3 < 1/3, an Lb rotation fails to restore balance only when d = 0 (see discussion following Theorem 10). So we need to rearrange the at most 1/0l + 2 nodes in gp into any fl-balanced tree when d = 0. An LR rotation fails only when d < -1. To see this, note that in the terminology of Lemma 14, d is d.
0(2+03)

The proof of P2 is extended to the case #3 1/3 when d'> 1 _1. Also, since d <1/fl, for the case b > c, we get fl(d'- 1) < 1-f < C (as c > 1). For the case b < c, we need to show fl(a 1) < b. Since an LR rotation is done only when condition D1 V D2 holds, from Lemmas 10 and 11, it follows that 8l(a 1) < b. So, an LR rotation rebalances when P3 < 1/3 provided d > 1 1. For smaller d, the at most 12(2+,) 0 1(2+0) nodess in the subtree gp may be directly rearranged into a fl-balanced tree.

The restructuring algorithm for simple fl-BBSTs is given in Figures 3.11 and 3.12. The algorithm of Figure 3.11 is used following an insert and that of Figure 3.12 after a delete.

Simple fl-BBSTs are expected to have higher search cost than the fl-BBSTs of Section 3.4. However, they are a good alternative to traditional WB(a) trees as they are expected to be "better balanced". To see this, note that from the proof of Lemma 3, the balance, B(p), at any node p in a fl-balanced tree satisfies


1 1+s(r)+ 1
B(p) ls(1) 1




68



procedure Restructuring2; begin
while (gp) do
begin
if (fl(s(gp.left) 1) > s(gp.right)) then {do an LL or LR rotation}
begin
p = gp.left;
if ((s(p.right) < (1 + )s(p.left) + 1 /3) and
(s(gp.right) < s(p.left))) then
do LL rotation
else do LR rotation ;
end
else
do symmetric to the above L case;
gp = gp.parent ;
end;
end;

Figure 3.11. Simple restructuring procedure for insertion


procedure Restructuring3 begin
while (gp) do
begin
if (fl(s(gp.left) 1) > s(gp.right)) then
if (/3 < 1/3) and (s(gp.right) < 1//3(2 + /3) 1) then
rearrange the subtree rooted at gp into any /3-balanced tree
else {do an LL or LR rotation}
begin
p = gp.left ;
if ((s(p.right) < (1 + fl)s(p.left) + 1 /3) and
(s(gp.right) < s(p.left))) then
do LL rotation
else do LR rotation ;
end
end
else
do symmetric to the above L case;
gp = gp.parent;
end;
end ;

Figure 3.12. Simple restructuring procedure for deletion




69


1
S1+ 1/3 + p(s(r)+1)
1 + 2)3-+
0 03((r)+x) 120-1
i +
13 1(3)1)


So,
1
B(p) < 1i + ((r)+1)

Also, sinces(r)- 1 < s(1)/fl, s(r) + 1 < s(l)/p + 2. Hence, 1 + < 1 + (( +
8Q+ 03((1)+'
2 So s(1)+1' So


1
B(p) 1 + +

1
-(1) T1 (s(l)+1)


Consequently,


1 1
I1 2-1 < B(p) < 1 + + 2-1
1 + (s(I)+1) -1 ( (r)+1)


When 3 = 0- 1,


1 1
< B(p) < 1
2 + v/ + 2 + v/ + 1 -+v


If s(p) 5 10, 0.296 < B(p) < 1 0.296. So, every #-balanced subtree with 10 or

fewer nodes is in WB(a) for a 0.296. Similarly, every subtree with 100 or fewer

nodes is in WB(a) for a 0.293. In fact, for every fixed k, subtrees of size k or less




70


procedure Restructuring4
begin
while (gp) do
begin
if (s(gp.left) > s(gp.right)) then
begin {check conditions for an LL and LR rotation}
p = gp.left ;
if (s(p.left) > s(p.right)) and (s(p.left) > s(gp.right)) then
do LL rotation
else if (s(p.left) 5 s(p.right)) and (s(p.right) > s(gp.right)) then
do LR rotation ;
end
else {check conditions for an RR and RL rotation}
do symmetric to the above L case;
gp = gp.parent;
end;
end;

Figure 3.13. Simple restructuring procedure without a f8 value

are in WB(a) for a slightly higher than 1 0.2929 which is the largest value

of a for which WB(a) trees can be maintained.

3.6 BBSTs without Deletion

In some applications of a dictionary, we need to support only the insert and

search operations. In these applications, we can construct binary search trees with

total cost

C(T) nlogo(V(n + 1))


by using the simpler restructuring algorithm of Figure 3.13.


Theorem 12 When the only operations are search and insert and restructuring is done

as in Figure 3.13, C(T) < nlog,(v'5(n + 1)).




71


Proof Suppose T currently has m 1 elements and a new element is inserted. Let u be the level at which the new element is inserted. Suppose that the restructuring pass performs rotations at q < u of the nodes on the path from the root to the newly inserted node. Then C(T) increases by at most v = u q as a result of the insertion. The number of nodes on the path from the root to the newly inserted node at which no rotation is performed is also v. Let these nodes be numbered 1 through v bottom to top. Let Si denote the number of elements in the subtree with root i prior to the restructuring pass. We see that S, > 1 and S2 > 2. For node i, 2 < i < v, one of its subtrees contains node i 1. Without loss of generality, let this be the left subtree of i. Let the root of the right subtree of i be d. So,


Si >! Si_ + s(d) + 1.



If i 1 is not the left child of i, then since no rotation is done at i, s(d) > Si-1. If i 1 is the left child of i, then consider node i 2. This is in one of the subtrees of i. Since no rotation is performed at i 1, s(d) > Si-2. Since Si-I > Si_2, we get



Si > Si-1 + S,-2 2 1.



Hence, S,, N where N,, is the minimum number of elements in a COST of height v. So, v < log,(v"5(m + 1)). So, when an element is inserted into a tree that has m 1 elements, its cost C(T) increases by at most logo(v'5(m + 1)). Starting with an empty tree and inserting n elements results in a tree whose cost is at most




72


n logj(v' (n + 1)). 0




Corollary 2 The expected cost of a search or insert in a BBST constructed as above is O(log n).

Proof Since C(T) < nlogj(v'5(n + 1)), the expected search cost is C(T)/n < log,(V5(n + 1)). The cost of an insert is the same order as that of a search as each insert follows the corresponding search path twice (top down and bottom up). 0



3.7 Experimental Results

For comparison purposes, we wrote C programs for BBSTs, SBBSTs (simple BBSTs), BBSTDs (BBSTs in which procedure Restructuring4 (Figure 3.13) is used to restructure following inserts as well as deletes), unbalanced binary search trees (BST), AVL-trees, top-down red-black trees (RB-T), bottom-up red-black trees (RBB) [31], weight balanced trees (WB), deterministic skip lists (DSL), treaps (TRP), and skip lists (SKIP). For the BBST and SBBST structures, we used P = 207/500 while for the WB structure, we used a = 207/707. While these are not the highest permissible values of # and a, this choice permitted us to use integer arithmetic rather than the substantially more expensive real arithmetic. For instance, ,l-(a, b) for 3 = 207/500 can be checked using the comparisons 207(s(a) 1) > 500s(b) and 207(s(b) 1) > 500s(a). The randomized structures TRP and SKIP used the same random number generator with the same seed. SKIP was programmed with probability value p = 1/4 as in Pugh [26].




7 3



To minimize the impact of system call overheads on run time measurements, we programmed all structures using simulated pointers (i.e., an array of nodes with integer pointers [271. Skip lists use variable size nodes. This requires more complex storage management than required by the remaining structures which use nodes of the same size. For our experiments, we implemented skip lists using fixed size nodes, each node being of the maximum size. As a result, our run times for skip lists are smaller than if a space efficient implementation had been used. In all our tree structure implementations, null pointers were replaced by a pointer to a tail node whose data field could be set to the search /insert/ delete key and thus avoid checking for falling off the tree. Similar tail pointers are part of the defined structure of skip and deterministic skip lists. Each tree also had a head node. WB(a) trees were implemented with a bottom-up restructuring pass. Our codes for SKIP and DSL are based on the codes of Pugh [261 and Papadakis [22], respectively. Our AVL and RB-T codes are based on those of Papadakis [22] and Sedgewick [28]. The treap structure was implemented using joins and splits rather than rotations. This results in better performance. Furthermore, AVL, RB-B, WB, and BBST were implemented with parent pointers in addition to left and right child pointers. For BBSTs, the enhancements described in Section 3.4.4 for insert and delete (see Figure 3.10) were employed. No rotations were performed during a search when using any of the structures.

For our experiments, we tried two versions of the code. These varied in the order in which the 'equality' and 'less than' or 'greater than' check between x and e (where x is the key being searched/inserted/deleted and e is the key in the current




74


node) is done. In version 1, we conducted an initial experiment to determine if the total comparison count is less using the order L:

if x < e then move to left child

else if x e then move to right child

else found

or the order R:

if x > e then move to right child

else if x je then move to left child

else found.

Our experiment indicated that doing the 'left child' check first (i.e. order L) worked better for AVL, BBST, BBSTD, and DSL structures while R worked better for the RB-T, RB-B, WB, SBBST, and TRP structures. No significant difference between L and R was observed for BSTs. For skip lists, we do not have the flexibility to change the comparison order. The version 1 codes performed the comparisons in the order determined to be better. For BSTs, the order R was used.

In the version 2 codes the comparisons in each node took the standard form

if x =e then found

else if x < e then move to left child

else move to right child.

The version 2 restructuring code for BBSTs differed from that of Figure 3.10 in that the '>' test in the second, third, and forth if statements was changed to ''




75


No change was made in the corresponding if statements for RR and RL rotations. While this increased the number of comparisons, it reduced the run time.

We experimented with n = 10,000, 50,000, 100,000, and 200,000. For each n, the following experiments were conducted:

(a) start with an empty structure and perform n inserts;

(b) search for each item in the resulting structure once; items are searched for in the order they were inserted

(c) perform an alternating sequence of n inserts and n deletes; in this, the n elements inserted in (a) are deleted in the order they were inserted and n new elements are inserted

(d) search for each of the remaining n elements in the order they were inserted

(e) delete the n elements in the order they were inserted.

For each n, the above five part experiment was repeated ten times using different random permutations of distinct elements. For each permutation, we measured the total number of element comparisons performed and then averaged these over the ten permutations.

First, we report on the relative performance of SBBSTs, BBSTDs, and 1313STs. For this comparison, we used only version I of the code. Table 3.1 gives the average number of key comparisons performed for each of the five parts of the experiment. The three versions of our proposed data structure are very competitive on this measure. BBSTI)s and BBSTs generally performed fewer comparisons than did SBBSTs. All three structures had a comparison count within 2% of one another.




76



Table 3.1. The number of key comparisons on random inputs (version 1 code)

n operation 11SBBST IBBSTD IBBST]
insert 212495 212223 212111
search 194661 191599 191578
10,000 ins/del 416125 416967 416862 search 194957 191666 191676
delete 168033 166441 166487
insert 1241080 1236682 1236114 search 1152137 1135131 1134969 50,000 ins/del 2437918 2438083 2437639 search 1153821 1134277 1134062
_____ delete 1018675 1007766 1007688 insert 2635913 2624829 2623792 search 2458079 2423988 2423613 100,000 ins/del 5183619 5180383 5179653 search 2461221 2420282 2419990
____ delete 2190798 2168049 2168110 insert 5580139 5555190 5553256 search 5223989 5148220 5147698 200,000 ins/del 10981441 10969578 10968053 search 5229172 5144808 5144148 delete 4692447 4641349 4641389



However, when we used ordered data rather than random data (Table 3.2), SBBSTs

performed noticeably inferior to BBSTDs and BBSTs; the later two remained very

competitive.

Tables 3.3 and 3.4 give the average heights of the trees using random data and

using ordered data, respectively. The first number gives the height following part (a)

of the experiment and the second following part (c). The numbers are identical for

BBSTDs and BBSTs and slightly higher (lower) for SBBSTs using random (ordered)

data.




77



Table 3.2. The number of key comparisons on ordered inputs (version 1 code)

n operation~ SBBST IBBSTD I BBST
insert 170182 150554 150554
search 188722 185530 185530
10,000 ins/del 425305 315177 314998
search 191681 184155 184155
____ delete 215214 135311 135131
insert 991526 872967 872967
search 1117174 1101481 1101481 50,000 ins/del 2472808 1806346 1805439 search 1116390 1098065 1098065 delete 1277756 792717 791815 insert 2103808 1850548 1850548 search 2384327 2354757 2354757 100,000 ins/del 5249194 3823415 3821594 search 2382759 2346118 2346128
_____ delete 2738294 1686397 1684584
insert 4449143 3903083 3903083 search 5068632 4946753 4946753 200,000 ins/del 11105525 8051695 8048058 search 5065496 5001967 5001967
_____ delete 5842168 3580856 13577223



Table 3.3. Height of the trees on random inputs (version 1 code)

n SBBST 1 BBSTD I BBST 10,000 fi17,17 ~f16,16 16,16
50,000 j~20,20 19,19 19,19
100,000 21,21 20,20 20,20
200,000 22,23 21,21 21,21



Table 3.4. Height of the trees on ordered inputs (version 1 code)

II 33SBBST I BBSTD [ BBST 10,000 3]16,15 17,17 17,17
50,000 20,20 20,20 20,20 100,000 21,21 21,21 21,21
200,000 22,22 23,22 23,22




78


The average number of rotations performed by each of the three structures is given in Tables 3.5 and 3.6. A single rotation (i.e., LL or RR) is denoted 'S' and a double rotation (i.e., LR or Rb) denoted 'D'. In the case of BBSTs, double rotations have been divided into three categories: D = LR and Rb rotations that do not perform a second substep rotation; DS = LR and Rb rotations with a second substep rotation of type Lb and RR; DD = LR and Rb rotations with a second substep rotation of type bR and Rb. BBSTIs and BBSTs performed a comparable number of rotations on both data sets. However, on random data SBBSTs performed about half as many rotations as did BBSTIs and BBSTs. On ordered data, SBBSTs performed 15 to 20% fewer rotations on part (a), 34% fewer on part (c), and 51% fewer on part (e).

The run-time performance of the structures is significantly influenced by compiler and architectural features as well as the complexity of a key comparison. The results we report are from a SUN SPARC-5 using the UNIX C compiler cc with optimization option. Because of instruction pipelining features, cache replacement policies, etc., the measured run times are not always consistent with the compiler and architecture independent metrics reported in Tables 3.1 through 3.6 and later in Tables 3.11 through 3.16. For example, since the search codes for all tree based methods are essentially identical, we would expect methods with a smaller comparison count to have a smaller run time for parts (b) and (d) of the experiment. This was not always the case.

Tables 3.7 and 3.8 give the run times of the three BBST structures using integer keys and Tables 3.9 and 3.10 do this for the case of real (i.e., floating point) keys. The





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Table 3.6. The number of rotations on ordered inputs (version 1 code)



n operation S1B1U S B TD ~ S D B DS IDDIinsert 9984 0 9985 2387 9985 2387 0 0
10,000 ins/del 14997 0 16567 6130 16644 5797 25 154
delete 4989 0 6570 3726 6647 3392 26 154
insert 49980 0 49983 11956 49983 11956 0 0
50,000 ins/del 74996 0 82862 30659 83247 28982 137 770
delete 24987 0 32859 18686 33242 17018 136 766
insert 99979 0 99983 23917 99983 23917 0 0 100,000 ins/del 149996 0 165738 61327 166504 57969 280 1540
_____ delete 49986 0 65733 37392 66505 34040 278 1536
insert 199978 0 199982 47839 199982 47839 0 0 200,000 ins/del 299996 0 331473 122653 333012 115938 559 3078
_____ delete 999 85 0 131478 74795 133016 68086 1557 13076J



sum of the run time for parts (a) (e) of the experiment is graphed in Figure 3.14. For random data, SBBSTs significantly and consistently outperformed BBSTDs and BBSTs. On ordered data, however, BBSTDs were slightly faster than BBSTs and both were significantly faster than SBBSTs.

Since BBSTs generated trees with the least search cost, we expect BBSTs to outperform SBBSTs and BBSTDs in applications where the comparison cost is very high relative to that of other operations and searches are done with a much higher frequency than inserts and deletes. However, with the mix of operations used in our tests, SBBSTs are the clear choice for random inputs and BBSTDs for ordered inputs.

In comparing with the other structures, our tables repeat the data for BBSTs. The reader may make the comparison with SBBSTs and BBSTDs.




81














Table 3.7. Run time on random inputs using integer keys (version 1 code)

n~ operation SBBST BBSTD [BBSTI insert 0.27 0.30 0.34
search 0.06 0.06 0.07
10,000 ins/del 0.57 0.62 0.70
search 0.06 0.06 0.06
delete 0.22 0.25 0.26
insert 1.48 1.61 1.75
search 0.35 0.36 0.37
50,000 ins/del 2.90 3.47 3.84
search 0.36 0.38 0.39
delete 1.13 1.47 1.62
insert 3.00 3.5 380
search 0.78 0.83 0.84
100,000 ins/del 6.28 7.78 8.41
search 0.83 0.87 0.88
delete 2.54 3.31 3.58
insert 6.56 7.74 8.37
search 1.80 1.89 1.89
200,000 ins/del 13.89 17.32 18.57 search 1.86 1.98 1.98
delete 5.64 7.41 8.02

Time Unit :sec




89















Table 3.8. Run time on ordered inputs using integer keys (version 1 code)

n operation SBBS BBSTD BBST
insert 0.32 0 -20 0.27
search 0.05 0.03 0.05
10,000 ins/del 0.58 0.43 0.57
search 0.07 0.03 0.03
delete 0.20 0.17 0.23
insert 1.8 1.20 1.10
search 0.25 0.20 0.20
50,000 ins/del 2.63 2.18 2.40 search 0.25 0.20 0.20
delete 0.95 0.92 1.05
insert 3- .43 2.23 2.53
search 0.72 0.45 0.42
100,000 ins/del 5.97 4.70 5.13
search 0.55 0.47 0.42
delete 2.10 1.98 2.15
insert 6.65 4.95 5.25
search 1.20 0.92 0.90
200,000 ins/del 13.13 10.23 10.88 search 1.17 0.90 0.90
delete 4.63 4.25 4.58

Time Unit: see




83













Table 3.9. Run time on random real inputs (version 1 code)

n operation II SBBSTT-[ BBSTD BBST
insert 0.23 0.34 0.36
search 0.07 0.10 0.10
10,000 ins/del 0.44 0.75 0.79
search 0.08 0.10 0.10
delete 0.17 0.29 0.30
insert 1.43 1.76 1.93
search 0.47 0.53 0.52
50,000 ins/del 2.76 3.89 4.22
search 0.50 0.54 0.55
delete 1.13 1.62 1.76
insert 2.96 3.94 4.36
search 1.08 1.17 1.16
100,000 ins/del 6.11 8.58 9.30
search 1.12 1.20 1.22
delete 2.50 3.66 3.95
insert 6.85 8.92 9.33
search 2.41 2.58 2.57
200,000 ins/del 13.86 19.49 20.46
search 2.49 2.69 2.66
delete 5.61 8.25 8.80

Time Unit: sec




84














Table 3.10. Run time on ordered real inputs (version 1 code)

n Operation SBBST BBSTD) BBST]
insert 0.27 0.23 0.20
search 0.08 0.07 0.07
10,000 ins/del 0.53 0.50 0.43
search 0.08 0.07 0.05
delete 0.18 0.23 0.20
insert 1.43 1.25 1.12
search 0.40 0.30 0.30
50,000 ins/del 2.80 2.17 2.37
search 0.40 0.30 0.30
_____ delte 1.07 0.90 0.97
insert 3.28 2.58 2.77
search 0.90 0.62 0.63
100,000 ins/del 6.15 4.70 5.13
search 0.87 0.62 0.63
delete 2.35 1.93 2.10
insert 7.37 4.55 4.92
search 1.85 1.32 1.32
200,000 ins/del 13.35 10.03 10.93
search 1.87 1.33 1.33
_____ delete 5.08 4.17 4.43

Time Unit: sec




85




Time is sum of time for parts (a)-(e) of the experiment 45

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15 10

50000 100000 150000 200000
n

Figure 3.14. Run time on real inputs (version 1 code)



The average number of comparisons for each of the five parts of the experiment are given in Table 3.11 for the version 1 implementation. On the comparison measure, AVL, RB-B, WB, and BBSTs are the front runners and are quite competitive with one another. On parts (a) (insert n elements) and (c) (insert n and delete n elements), AVL trees performed best while on the two search tests ((b) and (d)) and the deletion test (e), BBSTs performed best.

Table 3.12 gives the number of comparisons performed when ordered data (i.e., the elements in part (a) are 1, 2,... ,n and are inserted in this order) and those in part (c) are n + 1,.. ,2n (in this order) is used instead of random permutations of distinct elements. This experiment attempts to model realistic situations in which the inserted elements are in "nearly sorted order". BSTs were not included in this test as they perform very poorly with ordered data taking 0(n') time to insert n




86


times. The computer time needed to perform this test on BSTs was determined to be excessive. This test exhibited greater variance in performance. Among the deterministic structures, BBSTs outperformed the others in parts (a) (d) while AVL trees were ahead in part (e). For part (a), BBSTs performed approximately 45% fewer comparisons than did "L trees and approximately 12% fewer than WB trees. The randomized structure TRP was the best of the eight structures reported in Table 3.12 for part (a). It performed approximately 10% fewer comparisons than did BBST trees. However, the BBST remained best overall on parts (b), (c), and

(d).

The heights of the trees (number of levels in the case of DSL and SKIP) for the experiments with random and ordered data are given in Tables 3.13 and 3.14 respectively. The first number in each table entry is the tree height after part (a) of the experiment and the second, the height after part (c). In all cases, the number of levels using skip lists is fewest. However, among the tree structures, AVL and BBST trees have least height on random data and AVL has least with ordered data.

Tables 3.15 and 3.16, respectively, give the number of rotations performed by each of the deterministic tree schemes for experiment parts (a), (c), and (e). Note that none of the schemes performs rotations during a search.

On ordered data, BBSTs perform about 25% more rotations than do the remaining structures. These remaining structures perform about the same number of rotations. On random data, "L trees, bottom-up red-black trees and WB trees perform a comparable number of rotations. Top-down red-black trees and BBST trees






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