A Practical Introduction to Data

Structures and Algorithm Analysis -

JAVA Edition

slides derived from material

by Clifford A. Shaffer

1
The Need for Data Structures
[A primary concern of this course is efficiency.]

Data structures organize data

⇒ more efficient programs. [You might
believe that faster computers make it unnecessary to be
concerned with efficiency. However...]

• More powerful computers ⇒ more complex

applications.
• YET More complex applications demand
more calculations.
• Complex computing tasks are unlike our
everyday experience. [So we need special
training]

Any organization for a collection of records can

be searched, processed in any order, or
modified. [If you are willing to pay enough in time delay.
Ex: Simple unordered array of records.]

• The choice of data structure and algorithm

can make the difference between a program
running in a few seconds or many days.

2
Efficiency

A solution is said to be efficient if it solves the

problem within its resource constraints. [Alt:
Better than known alternatives (“relatively” efficient)]

• space [These are typical contraints for programs]

• time
[This does not mean always strive for the most efficient
program. If the program operates well within resource
constraints, there is no benefit to making it faster or smaller.]

The cost of a solution is the amount of

resources that the solution consumes.

3
Selecting a Data Structure

Select a data structure as follows:

1. Analyze the problem to determine the
resource constraints a solution must meet.
2. Determine the basic operations that must
be supported. Quantify the resource
constraints for each operation.
3. Select the data structure that best meets
these requirements.
[Typically want the “simplest” data struture that will meet
requirements.]

Some questions to ask: [These questions often help

to narrow the possibilities]

• Are all data inserted into the data structure

at the beginning, or are insertions
interspersed with other operations?
• Can data be deleted? [If so, a more complex
representation is typically required]

• Are all data processed in some well-defined

order, or is random access allowed?

4
Data Structure Philosophy

Each data structure has costs and benefits.

Rarely is one data structure better than

another in all situations.

A data structure requires:

• space for each data item it stores, [Data +
Overhead]

• time to perform each basic operation,

• programming effort. [Some data
structures/algorithms more complicated than others]

Each problem has constraints on available

space and time.

Only after a careful analysis of problem

characteristics can we know the best data
structure for the task.

Bank example:
• Start account: a few minutes
• Transactions: a few seconds
• Close account: overnight

5
Goals of this Course

1. Reinforce the concept that there are costs

and benefits for every data structure. [A
worldview to adopt]

2. Learn the commonly used data structures.

These form a programmer’s basic data
structure “toolkit.” [The “nuts and bolts” of the
course]

3. Understand how to measure the

effectiveness of a data structure or
program.
• These techniques also allow you to judge
the merits of new data structures that
you or others might invent. [To prepare
you for the future]

6
Definitions

A type is a set of values.

[Ex: Integer, Boolean, Float]

A data type is a type and a collection of

operations that manipulate the type.
[Ex: Addition]

A data item or element is a piece of

information or a record.
[Physical instantiation]

A data item is said to be a member of a data

type.
[]

A simple data item contains no subparts.

[Ex: Integer]

An aggregate data item may contain several

pieces of information.
[Ex: Payroll record, city database record]

7
Abstract Data Types

Abstract Data Type (ADT): a definition for a

data type solely in terms of a set of values and
a set of operations on that data type.

Each ADT operation is defined by its inputs

and outputs.

Encapsulation: hide implementation details

A data structure is the physical

implementation of an ADT.
• Each operation associated with the ADT is
implemented by one or more subroutines in
the implementation.

Data structure usually refers to an

organization for data in main memory.

File structure: an organization for data on

peripheral storage, such as a disk drive or tape.

An ADT manages complexity through

abstraction: metaphor. [Hierarchies of labels]
[Ex: transistors → gates → CPU. In a program, implement an
ADT, then think only about the ADT, not its implementation]

8
Logical vs. Physical Form

Data items have both a logical and a physical

form.

Logical form: definition of the data item within

an ADT. [Ex: Integers in mathematical sense: +, −]

Physical form: implementation of the data item

within a data structure. [16/32 bit integers: overflow]

Data Type
ADT: Data Items:
 Type
 Operations
Logical Form

Data Structure: Data Items:

{ Storage Space Physical Form
{ Subroutines
[In this class, we frequently move above and below “the line”
separating logical and physical forms.]

9
Problems

Problem: a task to be performed.

• Best thought of as inputs and matching
outputs.
• Problem definition should include
constraints on the resources that may be
consumed by any acceptable solution. [But
NO constraints on HOW the problem is solved]

Problems ⇔ mathematical functions

• A function is a matching between inputs
(the domain) and outputs (the range).
• An input to a function may be single
number, or a collection of information.
• The values making up an input are called
the parameters of the function.
• A particular input must always result in the
same output every time the function is
computed.

10
Algorithms and Programs

Algorithm: a method or a process followed to

solve a problem. [A recipe]

An algorithm takes the input to a problem

(function) and transforms it to the output. [A
mapping of input to output]

A problem can have many algorithms.

An algorithm possesses the following properties:

1. It must be correct. [Computes proper function]

2. It must be composed of a series of

concrete steps. [Executable by that machine]
3. There can be no ambiguity as to which
step will be performed next.
4. It must be composed of a finite number of
steps.
5. It must terminate.

A computer program is an instance, or

concrete representation, for an algorithm in
some programming language.
[We frequently interchange use of “algorithm” and “program”
though they are actually different concepts]

11
Mathematical Background
[Look over Chapter 2, read as needed depending on your
familiarity with this material.]

Set concepts and notation [Set has no duplicates,

sequence may]

Recursion

Induction proofs

Logarithms [Almost always use log to base 2. That is our

default base.]

Summations

12
Algorithm Efficiency

There are often many approaches (algorithms)

to solve a problem. How do we choose between
them?

At the heart of computer program design are

two (sometimes conflicting) goals:
1. To design an algorithm that is easy to
understand, code and debug.
2. To design an algorithm that makes efficient
use of the computer’s resources.

Goal (1) is the concern of Software

Engineering.

Goal (2) is the concern of data structures and

algorithm analysis.

When goal (2) is important, how do we

measure an algorithm’s cost?

13
How to Measure Efficiency?

1. Empirical comparison (run programs).

[Difficult to do “fairly.” Time consuming.]

2. Asymptotic Algorithm Analysis.

Critical resources:
• Time
• Space (disk, RAM)
• Programmer’s effort
• Ease of use (user’s effort).

Factors affecting running time:

• Machine load
• OS
• Compiler
• Problem size or Specific input values for
given problem size

For most algorithms, running time depends on

“size” of the input.

Running time is expressed as T(n) for some

function T on input size n.

14
Examples of Growth Rate

Example 1: [As n grows, how does T(n) grow?]

static int largest(int[] array) { // Find largest val

// all values >=0
int currLargest = 0; // Store largest val
for (int i=0; i<array.length; i++) // For each elem
if (array[i] > currLargest) // if largest
currLargest = array[i]; // remember it
return currLargest; // Return largest val
}

[Cost: T(n) = c1 n + c2 steps]

Example 2: Assignment statement [Constant cost]

Example 3:
sum = 0;
for (i=1; i<=n; i++)
for (j=1; j<=n; j++)
sum++;

[Cost: T(n) = c1 n2 + c2 Roughly n2 steps, with sum being n2 at

the end. Ignore various overhead such as loop counter
increments.]

15
Growth Rate Graph
[2n is an exponential algorithm. 10n and 20n differ only by a
constant.]
2n 2n2 5n log n
1400

1200
20n
1000

800

600 10n
400

200

0
0 10 20 30 40 50

2n 2n2
400

20n
300

5n log n
200

10n
100

0
0 5 10 15
Input size n

16
Important facts to remember

• for any integer constants a, b > 1 na grows

faster than logb n
[any polynomial is worse than any power of any
logarithm]

• for any integer constants a, b > 1 na grows

faster than log nb
[any polynomial is worse than any logarithm of any
power]

• for any integer constants a, b > 1 an grows

faster than nb
[any exponential is worse than any polynomial]

17
Best, Worst and Average Cases

Not all inputs of a given size take the same

time.

Sequential search for K in an array of n

integers:
• Begin at first element in array and look at
each element in turn until K is found.

Best Case: [Find at first position: 1 compare]

Worst Case: [Find at last position: n compares]

Average Case: [(n + 1)/2 compares]

While average time seems to be the fairest

measure, it may be difficult to determine.
[Depends on distribution. Assumption for above analysis:
Equally likely at any position.]

When is worst case time important?

[algorithms for time-critical systems]

18
Faster Computer or Algorithm?

What happens when we buy a computer 10

times faster? [How much speedup? 10 times. More
important: How much increase in problem size for same time?
Depends on growth rate.]

T(n) n n′ Change n′/n

10n 1, 000 10, 000 n′ = 10n 10
20n 500 5, 000 n ′ = 10n 10
√
5n log n 250 1, 842 10n√< n′ < 10n 7.37
2n2 70 223 n′ = 10n 3.16
2n 13 16 n′ = n + 3 −−
[For n2 , if n = 1000, then n′ would be 1003]

n: Size of input that can be processed in one

hour (10,000 steps).

n′: Size of input that can be processed in one

hour on the new machine (100,000 steps).
[Compare T(n) = n2 to T(n) = n log n. For n > 58, it is faster to
have the Θ(n log n) algorithm than to have a computer that is
10 times faster.]

19
Asymptotic Analysis: Big-oh

Definition: For T(n) a non-negatively valued

function, T(n) is in the set O(f (n)) if there
exist two positive constants c and n0 such that
T(n) ≤ cf (n) for all n > n0.

Usage: The algorithm is in O(n2) in [best,

average, worst] case.
Meaning: For all data sets big enough (i.e.,
n > n0), the algorithm always executes in less
than cf (n) steps [in best, average or worst
case].
[Must pick one of these to complete the statement. Big-oh
notation applies to some set of inputs.]

Upper Bound.

Example: if T(n) = 3n2 then T(n) is in O(n2).

Wish tightest upper bound:

While T(n) = 3n2 is in O(n3), we prefer O(n2).
[It provides more information to say O(n2 ) than O(n3 )]

20
Big-oh Example

Example 1. Finding value X in an array. [Average

case]

T(n) = csn/2. [cs is a constant. Actual value is irrelevant]

For all values of n > 1, csn/2 ≤ csn.
Therefore, by the definition, T(n) is in O(n) for
n0 = 1 and c = cs.

Example 2. T(n) = c1n2 + c2n in average case

c1n2 + c2n ≤ c1n2 + c2n2 ≤ (c1 + c2)n2 for all

n > 1.

T(n) ≤ cn2 for c = c1 + c2 and n0 = 1.

Therefore, T(n) is in O(n2) by the definition.

Example 3: T(n) = c. We say this is in O(1).

[Rather than O(c)]

21
Big-Omega

Definition: For T(n) a non-negatively valued

function, T(n) is in the set Ω(g(n)) if there
exist two positive constants c and n0 such that
T(n) ≥ cg(n) for all n > n0.

Meaning: For all data sets big enough (i.e.,

n > n0), the algorithm always executes in more
than cg(n) steps.

Lower Bound.

Example: T(n) = c1n2 + c2n.

c1n2 + c2n ≥ c1n2 for all n > 1.

T(n) ≥ cn2 for c = c1 and n0 = 1.

Therefore, T(n) is in Ω(n2) by the definition.

Want greatest lower bound.

22
Theta Notation

When big-Oh and Ω meet, we indicate this by

using Θ (big-Theta) notation.

Definition: An algorithm is said to be Θ(h(n))

if it is in O(h(n)) and it is in Ω(h(n)).
[For polynomial equations on T(n), we always have Θ. There
is no uncertainty, a “complete” analysis.]

Simplifying Rules:

1. If f (n) is in O(g(n)) and g(n) is in O(h(n)),

then f (n) is in O(h(n)).
2. If f (n) is in O(kg(n)) for any constant
k > 0, then f (n) is in O(g(n)). [No constant]
3. If f1(n) is in O(g1(n)) and f2(n) is in
O(g2(n)), then (f1 + f2)(n) is in
O(max(g1(n), g2(n))). [Drop low order terms]
4. If f1(n) is in O(g1(n)) and f2(n) is in
O(g2(n)) then f1(n)f2(n) is in
O(g1(n)g2(n)). [Loops]

23
Running Time of a Program
[Asymptotic analysis is defined for equations. Need to convert
program to an equation.]

Example 1: a = b;

This assignment takes constant time, so it is

Θ(1). [Not Θ(c) – notation by tradition]

Example 2:
sum = 0;
for (i=1; i<=n; i++)
sum += n;
[Θ(n) (even though sum is n2 )]

Example 3:
sum = 0;
for (j=1; j<=n; j++) // First for loop
for (i=1; i<=j; i++) // is a double loop
sum++;
for (k=0; k<n; k++) // Second for loop
A[k] = k;
P
[First statement is Θ(1). Double for loop is i = Θ(n2 ). Final
for loop is Θ(n). Result: Θ(n2 ).]

24
More Examples

Example 4.
sum1 = 0;
for (i=1; i<=n; i++) // First double loop
for (j=1; j<=n; j++) // do n times
sum1++;

sum2 = 0;
for (i=1; i<=n; i++) // Second double loop
for (j=1; j<=i; j++) // do i times
sum2++;
[First loop, sum is n2 . Second loop, sum is (n + 1)(n)/2. Both
are Θ(n2 ).]

Example 5.
sum1 = 0;
for (k=1; k<=n; k*=2)
for (j=1; j<=n; j++)
sum1++;

sum2 = 0;
for (k=1; k<=n; k*=2)
for (j=1; j<=k; j++)
sum2++;
Plog n Plog n−1
[First is k=1
n = Θ(n log n). Second is k=0
2k = Θ(n).]

25
Binary Search

Position 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Key 11 13 21 26 29 36 40 41 45 51 54 56 65 72 77 83

static int binary(int K, int[] array,

int left, int right) {
// Return position in array (if any) with value K
int l = left-1;
int r = right+1; // l and r are beyond array bounds
// to consider all array elements
while (l+1 != r) { // Stop when l and r meet
int i = (l+r)/2; // Look at middle of subarray
if (K < array[i]) r = i; // In left half
if (K == array[i]) return i; // Found it
if (K > array[i]) l = i; // In right half
}
return UNSUCCESSFUL; // Search value not in array
}

invocation of binary
int pos = binary(43, ar, 0, 15);

Analysis: How many elements can be examined

in the worst case? [Θ(log n)]

26
Other Control Statements

while loop: analyze like a for loop.

if statement: Take greater complexity of

then/else clauses.
[If probabilities are independent of n.]

switch statement: Take complexity of most

expensive case.
[If probabilities are independent of n.]

Subroutine call: Complexity of the subroutine.

27
Analyzing Problems

Use same techniques to analyze problems, i.e.

any possible algorithm for a given problem
(e.g., sorting)

Upper bound: Upper bound of best known

algorithm.

Lower bound: Lower bound for every possible

algorithm.
[The examples so far have been easy in that exact equations
always yield Θ. Thus, it was hard to distinguish Ω and O.
Following example should help to explain the difference –
bounds are used to describe our level of uncertainty about an
algorithm.]

Example: Sorting
1. Cost of I/O: Ω(n)
2. Bubble or insertion sort: O(n2)
3. A better sort (Quicksort, Mergesort,
Heapsort, etc.): O(n log n)
4. We prove later that sorting is Ω(n log n)

28
Multiple Parameters
[Ex: 256 colors (8 bits), 1000 × 1000 pixels]

Compute the rank ordering for all C (256) pixel

values in a picture of P pixels.

for (i=0; i<C; i++) // Initialize count

count[i] = 0;
for (i=0; i<P; i++) // Look at all of the pixels
count[value(i)]++; // Increment proper value count
sort(count); // Sort pixel value counts

If we use P as the measure, then time is

Θ(P log P ).

But this is wrong because we sort colors

More accurate is Θ(P + C log C).
If C << P , P could overcome C log C

29
Space Bounds

Space bounds can also be analyzed with

asymptotic complexity analysis.

Time: Algorithm
Space: Data Structure

Space/Time Tradeoff Principle:

One can often achieve a reduction in time is
one is willing to sacrifice space, or vice versa.
• Encoding or packing information
Boolean flags
• Table lookup
Factorials

Disk Based Space/Time Tradeoff Principle:

The smaller you can make your disk storage
requirements, the faster your program will run.
(because access to disk is typically more costly
than ”any” computation)

30
Algorithm Design methods:
Divide et impera

Decompose a problem of size n into (one or

more) problems of size m < n

Solve subproblems, if reduced size is not

”trivial”, in the same manner, possibly
combining solutions of the subproblems to
obtain the solution of the original one ...

... until size becomes ”small enough” (typically

1 or 2) to solve the problem directly (without
decomposition)

Complexity can be typically analyzed by means

of recurrence equations

31
Recurrence Equations(1)

we have already seen the following

T(n) = aT(n/b)+cnk , for n > 1

T(1) = d,

Solution of the recurrence depends on the ratio

r = bk /a

T(n) = Θ(nlogb a), if a > bk

T(n) = Θ(nk log n), if a = bk
T(n) = Θ(nk ), if a < bk

Complexity depends on
• relation between a and b, i.e., whether all
subproblems need to be solved or only some
do
• value of k, i.e., amount of additional work
to be done to partition into subproblems
and combine solutions

32
Recurrence Equations(2)

Examples
• a = 1, b = 2 (two halves, solve only one),
k = 0 (constant partition+combination
overhead): e.g., Binary search: T(n) =
Θ(log n) (extremely efficient!)
• a = b = 2 (two halves) and (k=1)
(partitioning+combination Θ(n)) T(n) =
Θ(n log n); e.g., Mergesort;
• a = b (partition data and solve for all
partitions) and k = 0 (constant
partition+combining) T(n) = Θ(nlogb a) =
Θ(n), same as linear/sequential processing
(E.g., finding the max/min element in an
array)

Now we’ll see

1. max/min search as an example of linear
complexity
2. other kinds of recurrence equations
• T(n)=T(n − 1)+n leads to quadratic
complexity: example bubblesort;
• T(n)=aT(n − 1)+k leads to exponential
complexity: example Towers of Hanoi

33
MaxMin search(1)

”Obvious” method: sequential search

public class MinMaxPair {
public int min;
public int max;
}

public static MinMaxPair minMax (float [] a) {

//guess a[0] as min and max
MaxMinPair p = new MaxMinPair(); p.min = p.max = 0;
// search in the remaining part of the array
for (int i = 1; i<a.length; i++) {
if (a[i]<a[p.min]) p.min = i;
if (a[i]>a[p.max]) p.max = i;
}
return p;
}

Complexity is T(n)=2(n − 1)=Θ(n)

Divide et impera approach: split array in two,

find MinMax of each, choose overall min among
the two mins and max among the two maxs

34
MaxMin search(2)

public static MinMaxPair minMax (float [] a,

int l, int r) {
MaxMinPair p = new MinMaxPair();
if(l==r) {p.min = p.max = r; return p;}
if (l==r-1) {
if (a[l]<a[r]) {
p.min=l; p.max=r;
}
else {
p.min=r; p.max=l;
}
return p;
}
int m = (l+r)/2;
MinMaxPair p1 = minMax(a, l, m);
MinMaxPair p2 = minMax(a, m+1, r);
if (a[p1.min]<a[p2.min]) p.min=p1.min else p.min=p2.min;
if (a[p1.max]>a[p2.max]) p.max=p1.max else p.max=p2.max;
return p;
}

Asymptotic complexity analyzable by means of

recurrence
T(n) = aT(n/b)+cnk , for n > 1
T(1) = d,

We have a = b and k = 0 hence T(n) = Θ(n),

apparently no improvement: we need a more
precise analysis

35
MaxMin search(3)
16

# of elements
of the array slice 8 8 depth of the
recursion

4 4 4 4

2 2 2 2 2 2 2 2

Assume for simplicity n is a power of 2. Here is

the tree of recursive calls for n = 16. There are
• n/2 leaf nodes, each of which takes 1
comparison
• n/2 − 1 internal nodes each of which takes
2 comparison
• hence #comparisons =
2(n/2 − 1) + n/2 = (3/2)n − 2, a 25%
improvement wrt linear search

36
bubblesort as a
divide et impera algorithm

To sort an array of n element, put the smallest

element in first position, then sort the
remaining part of the array.

Putting the smallest element to first position

requires an array traversal (Θ(n) complexity)
static void bubsort(Elem[] array) { // Bubble Sort
for (int i=0; i<array.length-1; i++) // Bubble up
//take i-th smallest to i-th place
for (int j=array.length-1; j>i; j--)
if (array[j].key() < array[j-1].key())
DSutil.swap(array, j, j-1);
}

i=0 1 2 3 4 5 6
42 13 13 13 13 13 13 13
20 42 14 14 14 14 14 14
17 20 42 20 15 15 15 15
13 17 20 42 20 17 17 17
28 14 17 15 42 20 20 20
14 28 15 17 17 42 23 23
23 15 28 23 23 23 42 28
15 23 23 28 28 28 28 42

37
Towers of Hanoi

Move stack of rings form one pole to another,

with following constraints
• move one ring at a time
• never place a ring on top of a smaller one
Divide et impera approach: move stack of n − 1
smaller rings on third pole as a support, then
move largest ring, then move stack of n − 1
smaller rings from support pole to destination
pole using start pole as a support
static void TOH(int n,
Pole start, Pole goal, Pole temp) {
if (n==1) System.out.println("move ring from pole " +
+ start + " to pole " + goal);
else {
TOH(n-1, start, temp, goal);
System.out.println("move ring from pole " +
+ start + " to pole " + goal);
TOH(n-1, temp, goal, start);
}
}

Time complexity as a function of the size n of

the ring stack: T(n)=2n-1

38
Exponential complexity
of Towers of Hanoi

Recurrence equation is T(n)=2T(n − 1)+1 for

n > 1, and T(1)=1.

A special case of the more general recurrence

T(n)=aT(n − 1)+k, for n > 1, and T(1)=k.

It is easy to show that the solution is

T(n)=k n−1 i
i=0 a hence T(n)=Θ(a )
n
P

Why? A simple proof by induction.

P0 i
Base: T(1)=k= k i=0 a

Induction:

T(n + 1)=aT(n)+k=
Pn−1 i Pn Pn
=ak i=0 a + k = k i=1 a + k = k i=0 ai=
i

P(n+1)−1 i
=k i=0 a

In the case of Towers of Hanoi a = 2, k = 1,

Pn−1 i
hence T(n)= i=0 2 = 2n-1

39
Lists
[Students should already be familiar with lists. Objectives: use
alg analysis in familiar context, compare implementations.]

A list is a finite, ordered sequence of data

items called elements.
[The positions are ordered, NOT the values.]

Each list element has a data type.

The empty list contains no elements.

The length of the list is the number of

elements currently stored.

The beginning of the list is called the head,

the end of the list is called the tail.

Sorted lists have their elements positioned in

ascending order of value, while unsorted lists
have no necessary relationship between element
values and positions.

Notation: ( a0, a1, ..., an−1 )

What operations should we implement?

[Add/delete elem anywhere, find, next, prev, test for empty.]

40
List ADT

interface List { // List ADT

public void clear(); // Remove all Objects
public void insert(Object item); // Insert at curr pos
public void append(Object item); // Insert at tail
public Object remove(); // Remove/return curr
public void setFirst(); // Set to first pos
public void next(); // Move to next pos
public void prev(); // Move to prev pos
public int length(); // Return curr length
public void setPos(int pos); // Set curr position
public void setValue(Object val); // Set current value
public Object currValue(); // Return curr value
public boolean isEmpty(); // True if empty list
public boolean isInList(); // True if curr in list
public void print(); // Print all elements
} // interface List
[This is an example of a Java interface. Any Java class using
this interface must implement all of these functions. Note
that the generic type “Object” is being used for the element
type.]

41
List ADT Examples

List: ( 12, 32, 15 )

MyLst.insert(element);

[The above is an example use of the insert function.

“element” is an object of the list element data type.]

Assume MyLst has 32 as current element:

MyLst.insert(99);
[Put 99 before current element, yielding (12, 99, 32, 15).]

Process an entire list:

for (MyLst.setFirst(); MyLst.isInList(); MyLst.next())

DoSomething(MyLst.currValue());

42
Array-Based List Insert

[Push items up/down. Cost: Θ(n).]

Insert 23:

13 12 20 8 3 13 12 20 8 3
0 1 2 3 4 5 0 1 2 3 4 5
(a) (b)

23 13 12 20 8 3
0 1 2 3 4 5
(c)

43
Array-Based List Class

class AList implements List { // Array-based list

private static final int defaultSize = 10;

private int msize; // Maximum size of list

private int numInList; // Actual list size
private int curr; // Position of curr
private Object[] listArray; // Array holding list

AList() { setup(defaultSize); } // Constructor

AList(int sz) { setup(sz); } // Constructor

private void setup(int sz) { // Do initializations

msize = sz;
numInList = curr = 0;
listArray = new Object[sz]; // Create listArray
}

public void clear() // Remove all Objects from list

{numInList = curr = 0; } // Simply reinitialize values

public void insert(Object it) { // Insert at curr pos

Assert.notFalse(numInList < msize, "List is full");
Assert.notFalse((curr >=0) && (curr <= numInList),
"Bad value for curr");
for (int i=numInList; i>curr; i--) // Shift up
listArray[i] = listArray[i-1];
listArray[curr] = it;
numInList++; // Increment list size
}

44
Array-Based List Class (cont)

public void append(Object it) { // Insert at tail

Assert.notFalse(numInList < msize, "List is full");
listArray[numInList++] = it; // Increment list size
}

public Object remove() { // Remove and return Object

Assert.notFalse(!isEmpty(), "No delete: list empty");
Assert.notFalse(isInList(), "No current element");
Object it = listArray[curr]; // Hold removed Object
for(int i=curr; i<numInList-1; i++) // Shift down
listArray[i] = listArray[i+1];
numInList--; // Decrement list size
return it;
}

public void setFirst() { curr = 0; } // Set to first

public void prev() { curr--; } // Move curr to prev
public void next() { curr++; } // Move curr to next
public int length() { return numInList; }
public void setPos(int pos) { curr = pos; }
public boolean isEmpty() { return numInList == 0; }

public void setValue(Object it) { // Set current value

Assert.notFalse(isInList(), "No current element");
listArray[curr] = it;
}

public boolean isInList() // True if curr within list

{ return (curr >= 0) && (curr < numInList); }
} // Array-based list implementation

45
Link Class

Dynamic allocation of new list elements.

class Link { // A singly linked list node

private Object element; // Object for this node
private Link next; // Pointer to next node
Link(Object it, Link nextval) // Constructor
{ element = it; next = nextval; }
Link(Link nextval) { next = nextval; } // Constructor
Link next() { return next; }
Link setNext(Link nextval) { return next = nextval; }
Object element() { return element; }
Object setElement(Object it) { return element = it; }
}

46
Linked List Position

head curr tail

20 23 12 15

(a)

head curr tail

20 23 10 12 15

[Naive approach: Point to current

(b) node. Current is 12. Want

to insert node with 10. No access available to node with 23.

How can we do the insert?]

head curr tail

20 23 12 15

(a)

head curr tail

20 23 10 12 15

(b)

[Alt implementation: Point to node preceding actual current

node. Now we can do the insert. Also note use of header
node.]

47
Linked List Implementation
public class LList implements List { // Linked list
private Link head; // Pointer to list header
private Link tail; // Pointer to last Object in list
protected Link curr; // Position of current Object

LList(int sz) { setup(); } // Constructor

LList() { setup(); } // Constructor
private void setup() // allocates leaf node
{ tail = head = curr = new Link(null); }

public void setFirst() { curr = head; }

public void next()
{ if (curr != null) curr = curr.next(); }

public void prev() { // Move to previous position

Link temp = head;
if ((curr == null) || (curr == head)) // No prev
{ curr = null; return; } // so return
while ((temp != null) && (temp.next() != curr))
temp = temp.next();
curr = temp;
}

public Object currValue() { // Return current Object

if (!isInList() || this.isEmpty() ) return null;
return curr.next().element();
}

public boolean isEmpty() // True if list is empty

{ return head.next() == null; }
} // Linked list class

48
Linked List Insertion

// Insert Object at current position

public void insert(Object it) {
Assert.notNull(curr, "No current element");
curr.setNext(new Link(it, curr.next()));
if (tail == curr) // Appended new Object
tail = curr.next();
}

curr

... 23 12 ...

Insert 10: 10

(a)

curr

... 23 12 ...

3
10
1 2
(b)

49
Linked List Remove

public Object remove() { // Remove/return curr Object

if (!isInList() || this.isEmpty() ) return null;
Object it = curr.next().element(); // Remember value
if (tail == curr.next()) tail = curr; // Set tail
curr.setNext(curr.next().next()); // Cut from list
return it; // Return value
}

curr

... 23 10 15 ...

(a)

curr 2

... 23 10 15 ...

it 1
(b)

50
Freelists

System new and garbage collection are slow.

class Link { // Singly linked list node with freelist
private Object element; // Object for this Link
private Link next; // Pointer to next Link
Link(Object it, Link nextval)
{ element = it; next = nextval; }
Link(Link nextval) { next = nextval; }
Link next() { return next; }
Link setNext(Link nextval) { return next = nextval; }
Object element() { return element; }
Object setElement(Object it) { return element = it; }

// Extensions to support freelists

static Link freelist = null; // Freelist for class

static Link get(Object it, Link nextval) {

if (freelist == null)//free list empty: allocate
return new Link(it, nextval);
Link temp = freelist; //take from the freelist
freelist = freelist.next();
temp.setElement(it);
temp.setNext(nextval);
return temp;
}
void release() {
// add current node to freelist
element = null; next = freelist; freelist = this;
}
}

51
Comparison of List Implementations

Array-Based Lists: [Average and worst cases]

• Insertion and deletion are Θ(n).

• Array must be allocated in advance.
• No overhead if all array positions are full.

Linked Lists:
• Insertion and deletion Θ(1);
prev and direct access are Θ(n).
• Space grows with number of elements.
• Every element requires overhead.

Space “break-even” point:

DE
DE = n(P + E); n=
P +E
n: elements currently in list
E: Space for data value
P: Space for pointer
D: Number of elements in array (fixed in the
implementation)
[arrays more efficient when full, linked lists more efficient with
few elements]

52
Doubly Linked Lists

Simplify insertion and deletion: Add a prev

pointer.

class DLink { // A doubly-linked list node

private Object element; // Object for this node
private DLink next; // Pointer to next node
private DLink prev; // Pointer to previous node
DLink(Object it, DLink n, DLink p)
{ element = it; next = n; prev = p; }
DLink(DLink n, DLink p) { next = n; prev = p; }
DLink next() { return next; }
DLink setNext(DLink nextval) { return next=nextval; }
DLink prev() { return prev; }
DLink setPrev(DLink prevval) { return prev=prevval; }
Object element() { return element; }
Object setElement(Object it) { return element = it; }
}
head curr tail

20 23 12 15

53
Doubly Linked List Operations
curr

... 20 23 12 ...

Insert 10: 10

(a)

curr
4 5
... 20 23 12 ...
10
3 1 2
(b)

// Insert Object at current position

public void insert(Object it) {
Assert.notNull(curr, "No current element");
curr.setNext(new DLink(it, curr.next(), curr));
if (curr.next().next() != null)
curr.next().next().setPrev(curr.next());
if (tail == curr) // Appended new Object
tail = curr.next();
}

public Object remove() { // Remove/return curr Object

Assert.notFalse(isInList(), "No current element");
Object it = curr.next().element(); // Remember Object
if (curr.next().next() != null)
curr.next().next().setPrev(curr);
else tail = curr; // Removed last Object: set tail
curr.setNext(curr.next().next()); // Remove from list
return it; // Return value removed
}

54
Circularly Linked Lists

• Convenient if there is no last nor first

element (there is no total order among
elements)
• The ”last” element points to the ”first”,
and the first to the last
• tail pointer non longer needed
• Potential danger: infinite loops in list
processing
• but head pointer can be used as a marker

55
Stacks

LIFO: Last In, First Out

Restricted form of list: Insert and remove only

at front of list.

Notation:
• Insert: PUSH
• Remove: POP
• The accessible element is called TOP.

56
Array-Based Stack

Define top as first free position.

class AStack implements Stack{ // Array based stack class
private static final int defaultSize = 10;
private int size; // Maximum size of stack
private int top; // Index for top Object
private Object [] listarray; // Array holding stack
AStack() { setup(defaultSize); }
AStack(int sz) { setup(sz); }

public void setup(int sz)

{ size = sz; top = 0; listarray = new Object[sz]; }

public void clear() { top = 0; } // Clear all Objects

public void push(Object it) // Push onto stack

{ Assert.notFalse(top < size, "Stack overflow");
listarray[top++] = it; }

public Object pop() // Pop Object from top

{ Assert.notFalse(!isEmpty(), "Empty stack");
return listarray[--top]; }

public Object topValue() // Return top Object

{ Assert.notFalse(!isEmpty(), "Empty stack");
return listarray[top-1]; }

public boolean isEmpty() { return top == 0; }

};

57
Linked Stack

public class LStack implements Stack {

// Linked stack class
private Link top; // Pointer to list header

public LStack() { setup(); } // Constructor

public LStack(int sz) { setup(); } // Constructor

private void setup() // Initialize stack

{ top = null; } // Create header node

public void clear() { top = null; } // Clear stack

public void push(Object it) // Push Object onto stack

{ top = new Link(it, top); }

public Object pop() { // Pop Object from top

Assert.notFalse(!isEmpty(), "Empty stack");
Object it = top.element();
top = top.next();
return it;
}

public Object topValue() // Get value of top Object

{ Assert.notFalse(!isEmpty(), "No top value");
return top.element(); }

public boolean isEmpty() // True if stack is empty

{ return top == null; }
} // Linked stack class

58
Array-based vs linked stacks

• Time: all operations take constant time for

both
• Space: linked has overhead but is flexible;
array has no overhead but wastes space
when not full

Implementation of multiple stacks

• two stacks at opposite ends of an array
growing in opposite directions
• works well if their space requirements are
inversely correlated

top1 top2

59
Queues

FIFO: First In, First Out

Restricted form of list:

Insert at one end, remove from other.

Notation:
• Insert: Enqueue
• Delete: Dequeue
• First element: FRONT
• Last element: REAR

60
Array Queue Implementations

Constraint: all elements

1. in consecutive positions
2. in the initial (final) portion of the array
If both (1) and (2) hold: rear element in pos 0,
dequeue costs Θ(1), enqueue costs Θ(n)
Similarly if in final portion of the array and/or
in reverse order
If only (1) holds (2 is released)
• both front and rear move to the ”right”
(i.e., increase)
• both enqueue and dequeue cost Θ(1)
front rear

20 5 12 17

(a)

front rear

12 17 3 30 4

”Drifting queue” problem:

(b)
run out of space
when at the highest posistions
Solution: pretend the array is circular,
implemented by the modulus operator, e.g.,
front = (front + 1) % size

61
Array Q Impl (cont)

A more serious problem: empty queue

indistinguishable from full queue
[Application of Pigeonhole Principle: Given a fixed (arbitrary)
position for front, there are n + 1 states (0 through n elements
in queue) and only n positions for rear. One must distinguish
between two of the states.]
front
front
20 5
12 12
17 17
rear
3
30
4

rear
(a) (b)

2 solutions to this problem

1. store # elements separately from the queue
2. use a n + 1 elements array for holding a
queue with n elements an most
Both solutions require one additional item of
information
Linked Queue: modified linked list.
[Operations are Θ(1)]

62
Binary Trees

A binary tree is made up of a finite set of

nodes that is either empty (then it is an empty
tree) or consists of a node called the root
connected to two binary trees, called the left
and right subtrees, which are disjoint from
each other and from the root.

B C

D E F

G H I

[A has depth 0. B and C form level 1. The tree has height 4.

Height = max depth + 1.]

63
Notation

(left/right) child of a node: root node of the

(left/right) subtree

if there is no left (right) subtree we say that

left/(right) subtree is empty

edge: connection between a node and its child

(drawn as a line)

parent of a node n: the node of which n is a

child

path from n1 to nk : a sequence n1 n2 ... nk ,

k >= 1, such that, for all 1 <= i < k, ni is
parent of ni+1

length of a path n1 n2 ... nk is k − 1 (⇒ length

of path n1 is 0)

if there is a path from node a to node d then

• a is ancestor of d
• d is descendant of a

64
Notation (Cont.)

hence
- all nodes of a tree (except the root) are
descendant of the root
- the root is ancestor of all the other nodes
of the tree (except itself)

depth of a node: length of a path from the

root (⇒ the root has depth 0)

height of a tree: 1 + depth of the deepest

node (which is a leaf)

level d of a tree: the set of all nodes of depth

d (⇒ root is the only node of level 0)

leaf node: has two empty children

internal node (non-leaf): has at least one

non-empty child

65
Examples

B C

D E F

G H I

- A: root
- B, C: A’s children
- B, D: A’s subtree
- D, E, F: level 2
- B has only right child (subtree)
- path of length 3 from A to G
- A, B, C, E, F internal nodes
- D, G, H, I leaves
- depth of G is 3, height of tree is 4

66
Full and Complete Binary Trees

Full binary tree: each node either is a leaf or is

an internal node with exactly two non-empty
children.

Complete binary tree: If the height of the tree

is d, then all levels except possibly level d − 1
are completely full. The bottom level has
nodes filled in from the left side.

(a) full but not complete

(b) complete but not full
(c) full and complete

(a) (b)

(c)

[NB these terms can be hard to distinguish

Question: how many nodes in a complete binary tree?

A complete binary tree is ”balanced”, i.e., has minimal height

given number of nodes

A complete binary tree is full or almost full or ”almost full”

(at most one node with one son) ]

67
Making missing children explicit

A A

B B

A A

B EMPTY EMPTY B

for a (non-)empty subtree we say the node has

a (non-)NULL pointer

68
Full Binary Tree Theorem

Theorem: The number of leaves in a

non-empty full binary tree is one more than the
number of internal nodes.

[Relevant since it helps us calculate space requirements.]

Proof (by Mathematical Induction):

• Base Case: A full binary tree with 0
internal node has 1 leaf node.
• Induction Hypothesis: Assume any full
binary tree T containing n − 1 internal
nodes has n leaves.
• Induction Step: Given a full tree T with
n − 1 internal nodes (⇒ n leaves), add two
leaf nodes as children of one of its leaves ⇒
obtain a tree T’ having n internal nodes
and n + 1 leaves.

69
Full Binary Tree Theorem Corollary

Theorem: The number of empty subtrees in a

non-empty binary tree is one more than the
number of nodes in the tree.

Proof: Replace all empty subtrees with a leaf

node. This is a full binary tree, having #leaves
= #empty subtrees of original tree.

alternative Proof:
- by definition, every node has 2 children,
whether empty or not
- hence a tree with n nodes has 2n children
- every node (except the root) has 1 parent
⇒ there are n − 1 parent nodes (some
coincide)
⇒ there are n − 1 non-empty children
- hence #(empty children) = #(total
children) - #(non-empty children) =
2n − (n − 1) = n + 1.

70
Binary Tree Node ADT

interface BinNode { // ADT for binary tree nodes

// Return and set the element value
public Object element();
public Object setElement(Object v);

// Return and set the left child

public BinNode left();
public BinNode setLeft(BinNode p);

// Return and set the right child

public BinNode right();
public BinNode setRight(BinNode p);

// Return true if this is a leaf node

public boolean isLeaf();
} // interface BinNode

71
Traversals

Any process for visiting the nodes in some

order is called a traversal.

Any traversal that lists every node in the tree

exactly once is called an enumeration of the
tree’s nodes.

Preorder traversal: Visit each node before

visiting its children.

Postorder traversal: Visit each node after

visiting its children.

Inorder traversal: Visit the left subtree, then

the node, then the right subtree.

NB: an empty node (tree) represented by

Java’s null (object) value
void preorder(BinNode rt) // rt is root of subtree
{
if (rt == null) return; // Empty subtree
visit(rt);
preorder(rt.left());
preorder(rt.right());
}

72
Traversals (cont.)

This is a lef t − to − right preorder: first visit lef t

subtree, then the right one.

Get a right − to − lef t preorder by switching last

two lines

To get inorder or postorder, just rearrange the

last three lines.

73
Binary Tree Implementation

B C

D E F

G H I
[Leaves are the same as internal nodes. Lots of wasted
space.]

 c

 +

4 x  a

2 x

[Example of expression tree: (4x ∗ (2x + a)) − c. Leaves are

different from internal nodes.]

74
Two implementations of BinNode
class LeafNode implements BinNode { // Leaf node
private String var; // Operand value

public LeafNode(String val) { var = val; }

public Object element() { return var; }
public Object setElement(Object v)
{ return var = (String)v; }
public BinNode left() { return null; }
public BinNode setLeft(BinNode p) { return null; }
public BinNode right() { return null; }
public BinNode setRight(BinNode p) { return null; }
public boolean isLeaf() { return true; }
} // class LeafNode

class IntlNode implements BinNode { // Internal node

private BinNode left; // Left child
private BinNode right; // Right child
private Character opx; // Operator value

public IntlNode(Character op, BinNode l, BinNode r)

{ opx = op; left = l; right = r; } // Constructor
public Object element() { return opx; }
public Object setElement(Object v)
{ return opx = (Character)v; }
public BinNode left() { return left; }
public BinNode setLeft(BinNode p) {return left = p;}
public BinNode right() { return right; }
public BinNode setRight(BinNode p)
{ return right = p; }
public boolean isLeaf() { return false; }
} // class IntlNode

75
Two implementations (cont)

static void traverse(BinNode rt) { // Preorder

if (rt == null) return; // Nothing to visit
if (rt.isLeaf()) // Do leaf node
System.out.println("Leaf: " + rt.element());
else { // Do internal node
System.out.println("Internal: " + rt.element());
traverse(rt.left());
traverse(rt.right());
}
}

76
A note on polymorphism and
dynamic binding

The member function isLeaf() allows one to

distinguish the “type” of a node
- leaf
- internal
without need of knowing its subclass
This is determined dynamically by the JRE
(Java Runtime Environment)

77
Space Overhead

From Full Binary Tree Theorem:

Half of pointers are NULL.

If leaves only store information, then overhead

depends on whether tree is full.

All nodes the same, with two pointers to

children:
Total space required is (2p + d)n.
Overhead: 2pn.

If p = d, this means 2p/(2p + d) = 2/3 overhead.

[The following is for full binary trees:]

Eliminate pointers from leaf nodes:

n (2p)
2 p
n (2p) + dn =
2 p+d
[Half the nodes have 2 pointers, which is overhead.]

This is 1/2 if p = d.
2p/(2p + d) if data only at leaves ⇒ 2/3
overhead.

Some method is needed to distinguish leaves

from internal nodes. [This adds overhead.]

78
Array Implementation
[This is a good example of logical representation vs. physical
implementation.]

For complete binary trees.

1 2

3 4 5 6

7 8 9 10 11

(a)

Node 0 1 2 3 4 5 6 7 8 9 10 11

• Parent(r) = [(r − 1)/2 if r 6= 0 and r < n.]

• Leftchild(r) = [2r + 1 if 2r + 1 < n.]

• Rightchild(r) = [2r + 2 if 2r + 2 < n.]

• Leftsibling(r) = [r − 1 if r is even, r > 0 and r < n.]

• Rightsibling(r) = [r + 1 if r is odd, r + 1 < n.]

[Since the complete binary tree is so limited in its shape,

(only one shape for tree of n nodes), it is reasonable to
expect that space efficiency can be achieved.

NB: left sons’ indices are always odd, right ones’ even, a
node with index i is leaf iff i > n.of.nodes/2 (Full Binary
Tree Theorem)]

79
Binary Search Trees

Binary Search Tree (BST) Property

All elements stored in the left subtree of a node

whose value is K have values less than K. All
elements stored in the right subtree of a node
whose value is K have values greater than or
equal to K.
[Problem with lists: either insert/delete or search must be
Θ(n) time. How can we make both update and search
efficient? Answer: Use a new data structure.] 120

37 42

24 42 7 42

7 32 40 42 2 32

2 120 24 37

40

(a) (b)

80
BinNode Class

interface BinNode { // ADT for binary tree nodes

// Return and set the element value
public Object element();
public Object setElement(Object v);

// Return and set the left child

public BinNode left();
public BinNode setLeft(BinNode p);

// Return and set the right child

public BinNode right();
public BinNode setRight(BinNode p);

// Return true if this is a leaf node

public boolean isLeaf();
} // interface BinNode

We assume that the datum in the nodes

implements interface Elem with a method key
used for comparisons (in searching and sorting
algorithms)

interface Elem {
public abstract int key();
} // interface Elem

81
BST Search

public class BST { // Binary Search Tree implementation

private BinNode root; // The root of the tree

public BST() { root = null; } // Initialize root

public void clear() { root = null; }
public void insert(Elem val)
{ root = inserthelp(root, val); }
public void remove(int key)
{ root = removehelp(root, key); }
public Elem find(int key)
{ return findhelp(root, key); }
public boolean isEmpty() { return root == null; }

public void print() {

if (root == null)
System.out.println("The BST is empty.");
else {
printhelp(root, 0);
System.out.println();
}
}

private Elem findhelp(BinNode rt, int key) {

if (rt == null) return null;
Elem it = (Elem)rt.element();
if (it.key() > key) return findhelp(rt.left(), key);
else if (it.key() == key) return it;
else return findhelp(rt.right(), key);
}

82
BST Insert

private BinNode inserthelp(BinNode rt, Elem val) {

if (rt == null) return new BinNode(val);
Elem it = (Elem) rt.element();
if (it.key() > val.key())
rt.setLeft(inserthelp(rt.left(), val));
else
rt.setRight(inserthelp(rt.right(), val));
return rt;
}

37

24 42

7 32 40 42

2 35 120

83
Remove Minimum Value

private BinNode deletemin(BinNode rt) {

if (rt.left() == null)
return rt.right();
else {
rt.setLeft(deletemin(rt.left()));
return rt;
}
}

private Elem getmin(BinNode rt) {

if (rt.left() == null)
return (Elem)rt.element();
else return getmin(rt.left());
}

10
rt

5 20

84
BST Remove

private BinNode removehelp(BinNode rt, int key) {

if (rt == null) return null;
Elem it = (Elem) rt.element();
if (key < it.key())
rt.setLeft(removehelp(rt.left(), key));
else if (key > it.key())
rt.setRight(removehelp(rt.right(), key));
else {
if (rt.left() == null)
rt = rt.right();
else if (rt.right() == null)
rt = rt.left();
else {
Elem temp = getmin(rt.right());
rt.setElement(temp);
rt.setRight(deletemin(rt.right()));
}
}
return rt;
}
37 40

24 42

7 32 40 42

2 120

85
Cost of BST Operations

Find: the depth of the node being found

Insert: the depth of the node being inserted

Remove: the depth of the node being removed,

if it has < 2 children, otherwise depth of node
with smallest value in its right subtree

Best case: balanced (complete tree): Θ(log n)

Worst case (linear tree): Θ(n)

That’s why it is important to have a balanced

(complete) BST

Cost of constructing a BST by means of a

series of insertions
- if elements inserted in in order of increasing
value n 2
i=1 i = Θ(n )
P

- if inserted in ”random” order almost good

enough for balancing the tree, insertion cost
is in average Θ(log n), for a total Θ(n log n)

86
Heaps

Heap: Complete binary tree with the

Heap Property:

• Min-heap: all values less than child values.

• Max-heap: all values greater than child
values.

The values in a heap are partially ordered.

Heap representation: normally the array based

complete binary tree representation.

87
Building the Heap
[Max Heap

NB: for a given set of values, the heap is not unique]

1 7

2 3 4 6

4 5 6 7 1 2 3 5

(a)

1 7

2 3 5 6

4 5 6 7 4 2 1 3

(b)

(a) requires exchanges (4-2), (4-1), (2-1),

(5-2), (5-4), (6-3), (6-5), (7-5), (7-6).

(b) requires exchanges (5-2), (7-3), (7-1),

(6-1).
[How to get a good number of exchanges? By induction.
Heapify the root’s subtrees, then push the root to the correct
level.]

88
The siftdown procedure

To place a generic node in its correct position

Assume subtrees are Heaps

If root is not greater than both children, swap

with greater child

Reapply on modified subtree

1 7 7

5 7 5 1 5 6

4 2 6 3 4 2 6 3 4 2 1 3

Shift it down by exchanging it with the greater

of the two sons, until it becomes a leaf or it is
greater than both sons.

89
Max Heap Implementation

public class MaxHeap {

private Elem[] Heap; // Pointer to the heap array
private int size; // Maximum size of the heap
private int n; // Number of elements now in heap

public MaxHeap(Elem[] h, int num, int max)

{ Heap = h; n = num; size = max; buildheap(); }

public int heapsize() // Return current size of heap

{ return n; }

public boolean isLeaf(int pos) // TRUE if pos is leaf

{ return (pos >= n/2) && (pos < n); }

// Return position for left child of pos

public int leftchild(int pos) {
Assert.notFalse(pos < n/2, "No left child");
return 2*pos + 1;
}

// Return position for right child of pos

public int rightchild(int pos) {
Assert.notFalse(pos < (n-1)/2, "No right child");
return 2*pos + 2;
}

public int parent(int pos) { // Return pos for parent

Assert.notFalse(pos > 0, "Position has no parent");
return (pos-1)/2;
}

90
Siftdown

For fast heap construction:

• Work from high end of array to low end.
• Call siftdown for each item.
• Don’t need to call siftdown on leaf nodes.
public void buildheap() // Heapify contents of Heap
{ for (int i=n/2-1; i>=0; i--) siftdown(i); }

private void siftdown(int pos) { // Put in place

Assert.notFalse((pos >= 0) && (pos < n),
"Illegal heap position");
while (!isLeaf(pos)) {
int j = leftchild(pos);
if ((j<(n-1)) && (Heap[j].key() < Heap[j+1].key()))
j++; // j now index of child with greater value
if (Heap[pos].key() >= Heap[j].key()) return;
DSutil.swap(Heap, pos, j);
pos = j; // Move down
}
}

91
Cost for heap construction

log
Xn n
(i − 1) i = Θ(n).
i=1 2

[(i − 1) is number of steps down, n/2i is number of nodes at

that level. ]

cfr. eq(2.7) p.28:

Pn i = 2 − n+2
i=1 2i 2n

notice that
Plog n n Pn i
i=1 (i − 1) 2i ≤ n i=1 2i

Cost of removing root is Θ(log n)

Remove element too (root is a special case

thereof)

92
Priority Queues

A priority queue stores objects, and on request

releases the object with greatest value.

Example: Scheduling jobs in a multi-tasking

operating system.

The priority of a job may change, requiring

some reordering of the jobs.

Implementation: use a heap to store the

priority queue.

To support priority reordering, delete and

re-insert. Need to know index for the object.
// Remove value at specified position
public Elem remove(int pos) {
Assert.notFalse((pos >= 0) && (pos < n),
"Illegal heap position");
DSutil.swap(Heap, pos, --n); // Swap with last value
while (Heap[pos].key() > Heap[parent(pos)].key())
DSutil.swap(Heap, pos, parent(pos)); // push up
if (n != 0) siftdown(pos); // push down
return Heap[n];
}

93
General Trees

A tree T is a finite set of nodes such that it is

empty or there is one designated node r called
the root of T , and the remaining nodes in
(T − {r}) are partitioned into n ≥ 0 disjoint
subsets T1, T2, ..., Tk , each of which is a tree.
[Note: disjoint because a node cannot have two parents.]
Root R
Ancestors of V
Parent of V P

V
S1 S2
C1 C2 Siblings of V

Subtree rooted at V

Children of V

95
General Tree ADT
[There is no concept of “left” or “right” child. But, we can
impose a concept of “first” (leftmost) and “next” (right).]

public interface GTNode {

public Object value();
public boolean isLeaf();
public GTNode parent();
public GTNode leftmost_child();
public GTNode right_sibling();
public void setValue(Object value);
public void setParent(GTNode par);
public void insert_first(GTNode n);
public void insert_next(GTNode n);
public void remove_first(); // remove first child
public void remove_next(); // remove right sibling
}

public interface GenTree {

public void clear();
public GTNode root();
public void newroot(Object value, GTNode first,
GTNode sib);
}

96
General Tree Traversal

[preorder traversal]

static void print(GTNode rt) { // Preorder traversal

if (rt.isLeaf()) System.out.print("Leaf: ");
else System.out.print("Internal: ");
System.out.println(rt.value());
GTNode temp = rt.leftmost_child();
while (temp != null) {
print(temp);
temp = temp.right_sibling();
}
}
R

A B

C D E F

[RACDEBF]

97
General Tree Implementations

Lists of Children
Index Val Par
0 R 1 3
1 A 0 2 4 6
2 C 1
3 B 0 5
4 D 1
5 F 3
6 E 1
7

[Hard to find right sibling.]

98
Leftmost Child/Right Sibling
Left Val Par Right
1 R
R 0
3 A 0 2
6 B 0
R X C 1 4
D 1 5
A B E 1
F 2
C D E F 8 R 0

X 7
[Note: Two trees share same array.]

Left Val Par Right

1 R 7 8
R0
3 A 0 2
6 B 0
R X C 1 4
D 1 5
A B E 1
F 2
C D E F 0 R 0
-1
X 7

99
Linked Implementations
Val Size
R 2
R

A 3 B 1
A B

C D E F C 0 D 0 E 0 F 0
(a) (b)

[Allocate child pointer space when node is created.]

R A B

A B

C D E F C D E F
(a) (b)

100
Sequential Implementations

List node values in the order they would be

visited by a preorder traversal.

Saves space, but allows only sequential access.

Need to retain tree structure for reconstruction.

For binary trees: Use symbol to mark NULL links.

A

B C

D E F

G H I

AB/D//CEG///F H//I//

101
Sequential Implementations (cont.)

Full binary trees: Mark leaf or internal.

A

B C

D E F

G H I

[Need NULL mark since this tree is not full.]

A′B ′/DC ′E ′G/F ′HI

General trees: Mark end of each subtree.

R

A B

C D E F

RAC)D)E))BF )))

102
Convert to Binary Tree

Left Child/Right Sibling representation

essentially stores a binary tree.

Use this process to convert any general tree to

a binary tree.

A forest is a collection of one or more general

trees.

root

(a) (b)

[Dynamic implementation of “Left child/right sibling.”]

103
K-ary Trees

Every node has a fixed maximum number of

children

fixed # children ⇒ easy to implement, also in

array

K high ⇒ potentially many empty subtrees ⇒

different implementation for leaves becomes
convenient

Full and complete K-ary trees similar to binary

trees

full, not complete complete, not full

full and complete

Theorems on # empty subtrees and on relation

between # internal nodes and # leaves similar
to binary trees

104
Graphs

graph G = (V, E): a set of vertices V, and a

set of edges E; each edge in E is a connection
between a pair of vertices in V, which are called
adjacent vartices.

# vertices written |V|; # edges written |E|.

0 ≤ |E| ≤ |V|2.

A graph is
- sparse if it has ”few” edges
- dense if it has ”many” edges
- complete all possible edges
- undirected as in figure (a)
- directed as in figure (b)
- labeled (figure (c))if it has labels on
vertices
- weighted (figure (c))if it has (numeric)
labels on edges

0 2
4 1

3 4 7
1
2
1 3
(a) (b) (c)

105
Graph Definitions (Cont)

A sequence of vertices v1, v2, ..., vn forms a path

of length n − 1 (⇒ length = # edges) if there
exist edges from vi to vi+1 for 1 ≤ i < n.

A path is simple if all vertices on the path are

distinct.

In a directed graph
• a path v1, v2, ..., vn forms a cycle if n > 1
and v1 = vn. The cycle is simple if, in
addition, v2, ..., vn are distinct
• a cycle v, v is a self-loop
• a directed graph with no self-loops is simple

In an undirected graph
• a path v1, v2, ..., vn forms a (simple) cycle if
n > 3 and v1 = vn (and, in addition, v2, ..., vn
are distinct)
– hence the path ABA is not a cycle, while
ABCA is a cycle

106
Graph Definitions (Cont)

Subgraph S = (VS, ES) of a graph G = (V, E):

VS ⊂ V and ES ⊂ E and both vertices of any
edge in ES are in VS

An undirected graph is connected if there is at

least one path from any vertex to any other.

The maximal connected subgraphs of an

undirected graph are called
connected components.

A graph without cycles is acyclic.

A directed graph without cycles is a

directed acyclic graph or DAG.

A free tree is a connected, undirected graph

with no cycles. Equivalently, a free tree is
connected and has |V − 1| edges.

107
Connected Components

A graph with (composed of) 3 connected

components

0 2 6 7

1 3 5

108
Graph Representations

Adjacency Matrix: space required Θ(|V|2).

Adjacency List: space required Θ(|V| + |E|).
0 1 2 3 4
0 2 0 1 1
1 1
4 2 1
3 1
1 3 4 1

(a) (b)

0 1 4
1 3
2 4
3 2
4 1

(c)
0 1 2 3 4
0 2 0 1 1
1 1 1 1
4 2 1 1
3 1 1
1 3 4 1 1 1

(a) (b)

0 1 4
1 0 3 4
2 3 4
3 1 2
4 0 1 2
(c)

[Instead of bits, the graph could store edge, weights.]

109
Graph Representatiosn (cont)

Adjacency list efficient for sparse graphs (only

existing edges coded)

Matrix efficient for dense graphs (no pointer

overload)

Algorithms visiting each neighbor of each

vertex more efficient on adjacency lists,
especially for sparse graphs

110
Graph Interface

interface Graph { // Graph class ADT

public int n(); // Number of vertices
public int e(); // Number of edges
// Get first edge having v as vertex v1
public Edge first(int v);
// Get next edge having w.v1 as the first edge
public Edge next(Edge w);
public boolean isEdge(Edge w); // True if edge
public boolean isEdge(int i, int j); // True if edge
public int v1(Edge w); // Where from
public int v2(Edge w); // Where to
public void setEdge(int i, int j, int weight);
public void setEdge(Edge w, int weight);
public void delEdge(Edge w); // Delete edge w
public void delEdge(int i, int j); // Delete (i, j)
public int weight(int i, int j); // Return weight
public int weight(Edge w); // Return weight

// Set Mark of vertex v

public void setMark(int v, int val);

// Get Mark of vertex v

public int getMark(int v);
} // interface Graph

Edges have a double nature:

seen as pairs of vertices or as aggregate
objects.

Vertices identified by an integer i, 0 ≤ i ≤ |V |

111
Implementation: Edge Class

interface Edge { // Interface for graph edges

public int v1(); // Return the vertex it comes from
public int v2(); // Return the vertex it goes to
} // interface Edge

// Edge class for Adjacency Matrix graph representation

class Edgem implements Edge {
private int vert1, vert2; // The vertex indices

public Edgem(int vt1, int vt2) //the constructor

{ vert1 = vt1; vert2 = vt2; }
public int v1() { return vert1; }
public int v2() { return vert2; }
} // class Edgem

112
Implementation: Adjacency Matrix

class Graphm implements Graph { // Adjacency matrix

private int[][] matrix; // The edge matrix
private int numEdge; // Number of edges
public int[] Mark; // The mark array, initially all 0

public Graphm(int n) { // Constructor

Mark = new int[n];
matrix = new int[n][n];
numEdge = 0;
}

public int n() { return Mark.length; }

public int e() { return numEdge; }

public Edge first(int v) { // Get first edge

for (int i=0; i<Mark.length; i++)
if (matrix[v][i] != 0)
return new Edgem(v, i);
return null; // No edge for this vertex
}

public Edge next(Edge w) { // Get next edge

if (w == null) return null;
for (int i=w.v2()+1; i<Mark.length; i++)
if (matrix[w.v1()][i] != 0)
return new Edgem(w.v1(), i);
return null; // No next edge;
}

Class Graphm implements interface Graph

Class Edgem implements interface Edge

113
Adjacency Matrix (cont)

public boolean isEdge(Edge w) { // True if an edge

if (w == null) return false;
else return matrix[w.v1()][w.v2()] != 0;
}

public boolean isEdge(int i, int j) // True if edge

{ return matrix[i][j] != 0; }

public int v1(Edge w) {return w.v1();} // Where from

public int v2(Edge w) {return w.v2();} // Where to

public void setEdge(int i, int j, int wt) {

Assert.notFalse(wt!=0, "Cannot set weight to 0");
if (matrix[i][j] == 0) numEdge++;
matrix[i][j] = wt;
}

public void setEdge(Edge w, int weight) // Set weight

{ if (w != null) setEdge(w.v1(), w.v2(), weight); }

public void delEdge(Edge w) { // Delete edge w

if (w != null)
if (matrix[w.v1()][w.v2()] != 0)
{ matrix[w.v1()][w.v2()] = 0; numEdge--; }
}

public void delEdge(int i, int j) { // Delete (i, j)

if (matrix[i][j] != 0)
{ matrix[i][j] = 0; numEdge--; }
}

NB: matrix[i][j]==0 iff there is no edge (i,j)

If there is no edge (i,j) then
weight(i,j)=Integer.MAX VALUE (INFINITY)
114
Adjacency Matrix (cont 2)

public int weight(int i, int j) { // Return weight

if (matrix[i][j] == 0) return Integer.MAX_VALUE;
else return matrix[i][j];
}

public int weight(Edge w) { // Return edge weight

Assert.notNull(w,"Can’t take weight of null edge");
if (matrix[w.v1()][w.v2()] == 0)
return Integer.MAX_VALUE;
else return matrix[w.v1()][w.v2()];
}

public void setMark(int v, int val)

{ Mark[v] = val; }

public int getMark(int v) { return Mark[v]; }

} // class Graphm

115
Graph Traversals

Some applications require visiting every vertex

in the graph exactly once.

Application may require that vertices be visited

in some special order based on graph topology.

Example: Artificial Intelligence

• Problem domain consists of many “states.”
• Need to get from Start State to Goal State.
• Start and Goal are typically not directly
connected.

To insure visiting all vertices:

void graphTraverse(Graph G) {
for (v=0; v<G.n(); v++)
G.setMark(v, UNVISITED); // Initialize mark bits
//next for needed to cover all the graph in case
//of graph composed of several connected components
for (v=0; v<G.n(); v++)
if (G.getMark(v) == UNVISITED)
doTraverse(G, v);
}
[Two traversals we will talk about: DFS, BFS.]

116
Depth First Search

static void DFS(Graph G, int v) { // Depth first search

PreVisit(G, v); // Take appropriate action
G.setMark(v, VISITED);
for (Edge w = G.first(v); G.isEdge(w); w = G.next(w))
if (G.getMark(G.v2(w)) == UNVISITED)
DFS(G, G.v2(w));
PostVisit(G, v); // Take appropriate action
}

Cost: Θ(|V| + |E|).

A B A B

C C

D D
F F
E E
(a) (b)

[The directions are imposed by the traversal. This is the

Depth First Search Tree.]

If PreVisit simply prints and PostVisit does

nothing then DFS prints
A C B F D E

117
Breadth First Search

Like DFS, but replace stack with a queue.

Visit the vertex’s neighbors before continuing
deeper in the tree.
static void BFS(Graph G, int start) {
Queue Q = new AQueue(G.n()); // Use a Queue
Q.enqueue(new Integer(start));
G.setMark(start, VISITED);
while (!Q.isEmpty()) { // Process each vertex on Q
int v = ((Integer)Q.dequeue()).intValue();
PreVisit(G, v); // Take appropriate action
for (Edge w=G.first(v); G.isEdge(w); w=G.next(w))
if (G.getMark(G.v2(w)) == UNVISITED) {
G.setMark(G.v2(w), VISITED);
Q.enqueue(new Integer(G.v2(w)));
}
PostVisit(G, v); // Take appropriate action
}
}

If PreVisit simply prints and PostVisit does

nothing then BFS prints A C E B D F
A B A B

C C

D D
F F
E E
(a) (b)

118
Topological Sort

Problem: Given a set of jobs, courses, etc.

with prerequisite constraints, output the jobs in
an order that does not violate any of the
prerequisites. (NB: the graph must be a DAG)

J6
J1 J2 J5 J7

J3 J4

static void topsort(Graph G) { // Topo sort: recursive

for (int i=0; i<G.n(); i++) // Initialize Mark array
G.setMark(i, UNVISITED);
for (int i=0; i<G.n(); i++) // Process all vertices
if (G.getMark(i) == UNVISITED)
tophelp(G, i); // Call helper function
}

static void tophelp(Graph G, int v) { // Topsort helper

G.setMark(v, VISITED);
for (Edge w = G.first(v); G.isEdge(w); w = G.next(w))
if (G.getMark(G.v2(w)) == UNVISITED)
tophelp(G, G.v2(w));
printout(v); // PostVisit for Vertex v
}
[Prints in reverse order: J7, J5, J4, J6, J2, J3, J1

It is a DFS with a PreVisit that does nothing]

119
Queue-based Topological Sort

static void topsort(Graph G) { // Topo sort: Queue

Queue Q = new AQueue(G.n());
int[] Count = new int[G.n()];
int v;
for (v=0; v<G.n(); v++) Count[v] = 0; // Initialize
for (v=0; v<G.n(); v++) // Process every edge
for (Edge w=G.first(v); G.isEdge(w); w=G.next(w))
Count[G.v2(w)]++; // Add to v2’s count
for (v=0; v<G.n(); v++) // Initialize Queue
if (Count[v] == 0) // Vertex has no prereqs
Q.enqueue(new Integer(v));
while (!Q.isEmpty()) { // Process the vertices
v = ((Integer)Q.dequeue()).intValue();
printout(v); // PreVisit for Vertex V
for (Edge w=G.first(v); G.isEdge(w); w=G.next(w)) {
Count[G.v2(w)]--; // One less prerequisite
if (Count[G.v2(w)] == 0) // This vertex now free
Q.enqueue(new Integer(G.v2(w)));
}
}
}

120
Sorting

Each record is stored in an array and contains a

field called the key.
Linear (i.e., total) order: comparison.
[a < b and b < c ⇒ a < c.]

The Sorting Problem

Given a sequence of records R1, R2, ..., Rn with

key values k1, k2, ..., kn, respectively, arrange the
records into any order s such that records
Rs1 , Rs2 , ..., Rsn have keys obeying the property
ks1 ≤ ks2 ≤ ... ≤ ksn .
[Put keys in ascending order. ]

NB: there can be records with the same key

A sorting algorithm is stable if after sorting

records with the same key have the same
relative position as before
Measures of cost:
• Comparisons
• Swaps (when records are large)

121
Sorting (cont)

Assumptions: for every record type there are

functions
- R.key returns the key value for record R
- DSutil.swap(array, i, j) swaps records in
positions i and j of the array

Measure of the ”degree of disorder” of an array

in the number of INVERSIONS
∀el = a[i],
#inversions = #elements > el which are in a
position j < i

#inversions for the entire array =

= #inversions of each array element
P

For a sorted array #inversions = 0

For an array with elements in decreasing order

#inversions = Θ(n2)

122
Insertion Sort
static void inssort(Elem[] array) { // Insertion Sort
for (int i=1; i<array.length; i++) // Insertrecord
for (int j=i; (j>0) &&
(array[j].key()<array[j-1].key()); j--)
DSutil.swap(array, j, j-1);
}

i=1 2 3 4 5 6 7
42 20 17 13 13 13 13 13
20 42 20 17 17 14 14 14
17 17 42 20 20 17 17 15
13 13 13 42 28 20 20 17
28 28 28 28 42 28 23 20
14 14 14 14 14 42 28 23
23 23 23 23 23 23 42 28
15 15 15 15 15 15 15 42

Best Case: [0 swaps, n − 1 comparisons]

Worst Case: [n2/2 swaps and compares]
Average Case: [n2/4 swaps and compares: # inner loop
iterations for an element in position n = #inversione = n/2 in
the average]

[At each iteration takes one element to its place and does only
that; it works only on the sorted portion of the array ]

[Nearly best performance when input ”nearly sorted” ⇒ used

in conjunction with mergesort and quicksort small array
segments]

123
Bubble Sort

static void bubsort(Elem[] array) { // Bubble Sort

for (int i=0; i<array.length-1; i++) // Bubble up
for (int j=array.length-1; j>i; j--)
if (array[j].key() < array[j-1].key())
DSutil.swap(array, j, j-1);
}

[Using test “j > i” saves a factor of 2 over “j > 0”.]

i=0 1 2 3 4 5 6
42 13 13 13 13 13 13 13
20 42 14 14 14 14 14 14
17 20 42 20 15 15 15 15
13 17 20 42 20 17 17 17
28 14 17 15 42 20 20 20
14 28 15 17 17 42 23 23
23 15 28 23 23 23 42 28
15 23 23 28 28 28 28 42

Best Case: [n2 /2 compares, 0 swaps]

Worst Case: [n2 /2 compares, n2 /2 swaps]

Average Case: [n2 /2 compares, n2 /4 swaps]

[At each iteration takes the smallest to its place, but it moves
also other ones;

NB: it works also on the unsorted part of the array;

No redeeming features to this sort.]

124
Selection Sort

static void selsort(Elem[] array) { // Selection Sort

for (int i=0; i<array.length-1; i++) { // Select i’th
int lowindex = i; // Remember its index
for (int j=array.length-1; j>i; j--) // Find least
if (array[j].key() < array[lowindex].key())
lowindex = j; // Put it in place
DSutil.swap(array, i, lowindex);
}
}
[Select the value to go in the ith position.]
i=0 1 2 3 4 5 6
42 13 13 13 13 13 13 13
20 20 14 14 14 14 14 14
17 17 17 15 15 15 15 15
13 42 42 42 17 17 17 17
28 28 28 28 28 20 20 20
14 14 20 20 20 28 23 23
23 23 23 23 23 23 28 28
15 15 15 17 42 42 42 42
Best Case: [0 swaps (n − 1 as written), n2 /2 compares.]

Worst Case: [n − 1 swaps, n2 /2 compares]

Average Case: [O(n) swaps, n2 /2 compares]

[It minimizes # swaps]

125
Pointer Swapping

Key = 42 Key = 42
Key = 5 Key = 5

(a) (b)

[For large records.]

This is what done in Java, when records are

objects

126
Exchange Sorting

Summary

Insertion Bubble Selection

Comparisons:
Best Case Θ(n) Θ(n2) Θ(n2)
Average Case Θ(n2) Θ(n2) Θ(n2)
Worst Case Θ(n2) Θ(n2) Θ(n2)
Swaps:
Best Case 0 0 Θ(n)
Average Case Θ(n2) Θ(n2) Θ(n)
Worst Case Θ(n2) Θ(n2) Θ(n)

127
Mergesort

List mergesort(List inlist) {

if (inlist.length() <= 1) return inlist;;
List l1 = half of the items from inlist;
List l2 = other half of the items from inlist;
return merge(mergesort(l1), mergesort(l2));
}

Analyze first the algorithm for merging sorted

sublists
• examine first element of each sublist
• pick the smaller element (it is the smallest
overall)
• remove it from its sublist and put it in the
output list
• when one sublist is exhausted, pick from
the other
Complexity of merging two sorted sublist: Θ(n)

36 20 17 13 28 14 23 15

20 36 13 17 14 28 15 23

13 17 20 36 14 15 23 28

13 14 15 17 20 23 28 36

128
Mergesort Implementation

Mergesort is tricky to implement.

Main question: how to represent lists?

Linked lists
• merging does not require direct access,
but...
• splitting requires a list traversal (Θ(n)),
whether list size is known (take as two
sublists the first and second halves) or
unknown (assign elements alternating
between the two lists)

Lists represented by arrays

• splitting very easy (Θ(1)) if array bounds
are known
• merging easy (Θ(n)) only if sub-arrays
merged into a second array (hence double
the space requirement!)
• avoid the need for a distinct additional array
for each recursive call by f irst copying
sub-arrays into auxiliary array and then
merging them back to the original array
(hence can use only one array for the
overall process)

129
Mergesort Implementation (2)

static void mergesort(Elem[] array, Elem[] temp,

int l, int r) {
if (l == r) return; // One element list
int mid = (l+r)/2; // Select midpoint
mergesort(array, temp, l, mid); // Ssort first half
mergesort(array, temp, mid+1, r); // Sort second half
merge(array, temp, l, mid, mid+1, r);
}

static void merge(Elem[] array, Elem[] temp,

int l1, int r1, int l2, int r2) {
for (int i=l1; i<=r2; i++) // Copy subarrays
temp[i] = array[i];
// Do the merge operation back to array
int i1 = l1; int i2 = l2;
for (int curr=l1; curr<=r2; curr++) {
if (i1 > r1) // Left sublist exhausted
array[curr] = temp[i2++];
else if (i2 > r2) // Right sublist exhausted
array[curr] = temp[i1++];
// else choose least of the two front elements
else if (temp[i1].key() < temp[i2].key())
array[curr] = temp[i1++]; // Get smaller val
else array[curr] = temp[i2++];
}
}

130
Complexity of Mergesort

Ad hoc analysis
• depth of recursion is log n
• at each recursion depth i
– 2i recursive calls
– each recursive call has array length n/2i,
hence...
– total length of merged arrays is n at
every depth
• therefore total cost is T(n) = Θ(n log n)

Alternative analysis: use recurrence equation

T(n) = aT(n/b)+cnk = 2T(n/2)+cn, T(1)=d

We have a = b = 2, k = 1 and therefore a = bk ,

hence

T(n) = Θ(n log n)

131
Heapsort

Heapsort uses a max-heap.

static void heapsort(Elem[] array) { // Heapsort
MaxHeap H = new MaxHeap(array, array.length,
array.length);
for (int i=0; i<array.length; i++) // Now sort
H.removemax(); // Put max value at end of heap
}

Cost of Heapsort: [Θ(n log n)]

Cost of finding k largest elements: [Θ(k log n + n).

Time to build heap: Θ(n).

Time to remove least element: Θ(log n).]

[Compare to sorting with BST: this is expensive in space

(overhead), potential bad balance, BST does not take
advantage of having all records available in advance.]

[Heap is space efficient, balanced, and building initial heap is

efficient.]

132
Heapsort Example
Original Numbers 73
73 6 57 88 60 42 83 72 48 85 6 57
88 60 42 83
72 48 85

Build Heap 88
88 85 83 72 73 42 57 6 48 60 85 83
72 73 42 57
6 48 60

Remove 88 85
85 73 83 72 60 42 57 6 48 88 73 83
72 60 42 57
6 48

Remove 85 83
83 73 57 72 60 42 48 6 85 88 73 57
72 60 42 48
6

Remove 83 73
73 72 57 6 60 42 48 83 85 88 72 57
6 60 42 48

133
Empirical Comparison
[MS Windows – CISC]

Algorithm 10 100 1000 10,000

Insert. Sort .10 9.5 957.9 98,086
Bubble Sort .13 14.3 1470.3 157,230
Select. Sort .11 9.9 1018.9 104,897
Shellsort .09 2.5 45.6 829
Quicksort .15 1.8 23.6 291
Quicksort/O .10 1.6 20.9 274
Mergesort .12 2.4 36.8 505
Mergesort/O .08 1.8 28.0 390
Heapsort – 50.0 60.0 880
Radix Sort/1 .87 8.6 89.5 939
Radix Sort/4 .23 2.3 22.5 236
Radix Sort/8 .19 1.2 11.5 115
[UNIX – RISC]

Algorithm 10 100 1000 10,000

Insert. Sort .66 65.9 6423 661,711
Bubble Sort .90 85.5 8447 1,068,268
Select. Sort .73 67.4 6678 668,056
Shellsort .62 18.5 321 5,593
Quicksort .92 12.7 169 1,836
Quicksort/O .65 10.7 141 1,781
Mergesort .76 16.8 234 3,231
Mergesort/O .53 11.8 189 2,649
Heapsort – 41.0 565 7,973
Radix Sort/1 7.40 67.4 679 6,895
Radix Sort/4 2.10 18.7 160 1,678
Radix Sort/8 4.10 11.5 97 808

[Clearly, n log n superior to n2 . Note relative differences on

different machines.]

134
Upperbound and Lowerbound for a
Problem

Upperbound: asymptotic cost of the fastest

known algorithm

lowerbound best possible efficiency of any

possible (known or unknown) algorithm

open problem: upperbound different from

(greater than) lowerbound

closed problem: upperbound equal to

lowerbound

135
Sorting Lower Bound

Want to prove a lower bound sorting problem

based on key comparison.

Sorting I/O takes Ω(n) time. (no algorithm

can take less than I/O time)

Sorting is O(n log n).

Will now prove Ω(n log n) lower bound.

Form of proof:
- Comparison based sorting can be modeled
by a binary tree.
- The tree must have Ω(n!) leaves (because
there are n! permutations of n elements).
- The tree cannot be less than Ω(n log n)
levels deep (a tree with k nodes has at least
log k levels).

this comes from the fact that

log n! = Θ(n log n)

which is due to Stirling’s approximation of n!:

√  n
n! ≈ 2πn ne

from which log n! ≈ n log n

136
Decision Trees
XYZ
XYZ YZX
XZY ZXY
YXZ ZYX
Yes A[1]<A[0]? No
(Y<X?)
YXZ XYZ
YXZ XYZ
YZX XZY
ZYX ZXY
Yes No Yes No
A[2]<A[1]? A[2]<A[1]?
YZX (Z<X?) YXZ XZY (Z<Y?) XYZ
YZX XZY
ZYX ZXY
Yes No Yes No
A[1]<A[0]? A[1]<A[0]?
ZYX (Z<Y?) YZX ZXY (Z<X?) XZY
[Illustration of Insertion sort. Lower part of table shows
possible output (sorted version of input array) after each
check]

There are n! permutations, and at least 1 node

for each permutation.

Where is the worst case in the decision tree?

137
Primary vs. Secondary Storage

review following sections of textbook

9.1 on Primary vs secondary storage

9.2 on Disk & Tape drives

9.3 on Buffers and Buffer Pools

138
Buffer Pools

A series of buffers used by an application to

cache disk data is called a buffer pool.

Virtual memory uses a buffer pool to imitate

greater RAM memory by actually storing
information on disk and “swapping” between
disk and RAM.

Caching Same technique to imitate greater

CACHE memory by storing info on RAM and
swapping between RAM and CACHE.

Organization for buffer pools: which one to use

next?
• First-in, First-out: Use the first one on the
queue.
• Least Frequently Used (LFU): Count buffer
accesses, pick the least used.
• Least Recently Used (LRU):
Keep buffers on linked list.
When a buffer is accessed, bring to front.
Reuse the one at the end.

139
Programmer’s View of Files

Logical view of files:

• An array of bytes.
• A file pointer marks the current position.

Three fundamental operations:

• Read bytes from current position (move file
pointer).
• Write bytes to current position (move file
pointer).
• Set file pointer to specified byte position.

140
Java File Functions

RandomAccessFile(String name, String mode)

close()

read(byte[] b)

write(byte[] b)

seek(long pos)

141
External Sorting

Problem: Sorting data sets too large to fit in

main memory.
• Assume data stored on disk drive.

To sort, portions of the data must be brought

into main memory, processed, and returned to
disk.

An external sort should minimize disk accesses.

142
Model of External Computation

Secondary memory is divided into equal-sized

blocks (512, 2048, 4096 or 8192 bytes are
typical sizes).

The basic I/O operation transfers the contents

of one disk block to/from main memory.

Under certain circumstances, reading blocks of

a file in sequential order is more efficient.
(When?) [1) Adjacent logical blocks of file are physically
adjacent on disk. 2) No competition for I/O head.]

Typically, the time to perform a single block

I/O operation is sufficient to Quicksort the
contents of the block.

Thus, our primary goal is to minimize the

number of block I/O operations.

Most workstations today must do all sorting on

a single disk drive.

143
Key Sorting

Often records are large while keys are small.

• Ex: Payroll entries keyed on ID number.

Approach 1: Read in entire records, sort them,

then write them out again.

Approach 2: Read only the key values, store

with each key the location on disk of its
associated record.

If necessary, after the keys are sorted the

records can be read and re-written in sorted
order.
[But, this is not usually done. (1) It is expensive (random
access to all records). (2) If there are multiple keys, there is
no “correct” order.]

144
External Sort: Simple Mergesort

Quicksort requires random access to the entire

set of records.
Better: Modified Mergesort algorithm
• Process n elements in Θ(log n) passes.

1. Split the file into two files.

2. Read in a block from each file.
3. Take first record from each block, output
them in sorted order.
4. Take next record from each block, output
them to a second file in sorted order.
5. Repeat until finished, alternating between
output files. Read new input blocks as
needed.
6. Repeat steps 2-5, except this time the
input files have groups of two sorted
records that are merged together.
7. Each pass through the files provides larger
and larger groups of sorted records.

A group of sorted records is called a run.

145
Problems with Simple Mergesort

36 17 28 23 20 36 14 28 13 17 20 36

20 13 14 15 13 17 15 23 14 15 23 28

Runs of length 1 Runs of length 2 Runs of length 4

Is each pass through input and output files

sequential? [yes]

What happens if all work is done on a single

disk drive? [Competition for I/O head eliminates
advantage of sequential processing.]

How can we reduce the number of Mergesort

passes? [Read in a block (or several blocks) and do an
in-memory sort to generate large initial runs.]

In general, external sorting consists of two

phases:
1. Break the file into initial runs.
2. Merge the runs together into a single sorted
run.

146
Breaking a file into runs

General approach:
• Read as much of the file into memory as
possible.
• Perform and in-memory sort.
• Output this group of records as a single run.

147
General Principals of External
Sorting

In summary, a good external sorting algorithm

will seek to do the following:
• Make the initial runs as long as possible.
• At all stages, overlap input, processing and
output as much as possible.
• Use as much working memory as possible.
Applying more memory usually speeds
processing.
• If possible, use additional disk drives for
more overlapping of processing with I/O,
and allow for more sequential file
processing.

148
Search

Given: Distinct keys k1, k2, ... kn and

collection T of n records of the form
(k1, I1), (k2, I2), ..., (kn, In)
where Ij is information associated with key kj
for 1 ≤ j ≤ n.

Search Problem: For key value K, locate the

record (kj , Ij ) in T such that kj = K.

Exact match query: search records with a

specified key value.

Range query: search records with key in a

specified range.

Searching is a systematic method for locating

the record (or records) with key value kj = K.

A successful search is one in which a record

with key kj = K is found.

An unsuccessful search is one in which no

record with kj = K is found (and presumably
no such record exists).

149
Approaches to Search

1. Sequential and list methods (lists, tables,

arrays).
2. Direct access by key value (hashing).
3. Tree indexing methods.

[recall: sequences: duplicate key values allowed; sets no key

duplication]

150
Searching Ordered Arrays

Sequential Search

Binary Search
static int binary(int K, int[] array,
int left, int right) {
// Return position of element (if any) with value K
int l = left-1;
int r = right+1; // l and r are beyond array bounds
while (l+1 != r) { // Stop when l and r meet
int i = (l+r)/2; // Look at middle of subarray
if (K < array[i]) r = i; // In left half
if (K == array[i]) return i; // Found it
if (K > array[i]) l = i; // In right half
}
return UNSUCCESSFUL; // Search value not in array
}

Position 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Key 11 13 21 26 29 36 40 41 45 51 54 56 65 72 77 83

Improvement: Dictionary Search, expected

record position computed from key value; value
of key found there used as in binary search

151
Lists Ordered by Frequency

Order lists by (expected) frequency of

occurrence ⇒ Perform sequential search.

Cost to access first record: 1; second record: 2

Expected (i.e., average) search cost:

C n = 1p1 + 2p2 + ... + npn

[pi is probability of ith record being accessed.]

Example: all records have equal frequency

n
X
Cn = i/n = (n + 1)/2.
i=1

Example: Exponential frequency

1/2i
(
if 1 ≤ i ≤ n − 1
pi =
1/2n−1 if i = n
[Second line is to make proabilities sum to 1.]

n
(i/2i) ≈ 2.
X
Cn ≈
i=1

[very good performance, because assumption (exp. freq.) is

strong]

152
Zipf Distributions

Applications:
• Distribution for frequency of word usage in
natural languages.
• Distribution for populations of cities, etc.

Definition: Zipf frequency for item i in the

distribution for n records as 1/iHn.
Pn 1
[Hn = i=1 i
≈ loge n.]

n
X
Cn = i/iHn = n/Hn ≈ n/ loge n
i=1

80/20 rule: 80% of the accesses are to 20% of

the records.

For distributions following the 80/20 rule,

C n ≈ 0.122n.

153
Self-Organizing Lists

Self-organizing lists modify the order of records

within the list based on the actual pattern of
record access.

Based on assumption that past searches

provide good indication of future ones

This is a heuristic similar to those for

managing buffer pools.
• Order by actual historical frequency of
access. (Similar to LFU buffer pool
replacement strategy.) [COUNT method: slow
reaction to change]

• Move-to-Front: When a record is found,

move it to the front of the list. [Not worse
than twice “best arrangement”; easy to implement with
linked lists, not arrays]

• Transpose: When a record is found, swap

it with the record ahead of it. [A pathological,
though unusual case: keep swapping last two elements.]

154
Advantages of self-organizing lists

• do not require sorting

• cost of insertion and deletion low
• no additional space
• simple (hence easy to implement)

155
Example of Self-Organizing Tables

Application: Text compression.

Keep a table of words already seen, organized

via Move-to-Front Heuristic.

If a word not yet seen, send the word.

Otherwise, send the (current) index in the
table.
[NB: sender and receiver maintain identical lists, so they agree
on indices]

The car on the left hit the car I left.

The car on 3 left hit 3 5 I 5.

156
Hashing

Hashing: The process of mapping a key value

to a position in a table.

A hash function maps key values to positions

It is denoted by h.

A hash table is an array that holds the

records. It is denoted by T .
[NB: records not necessarily ordered by key value or frequency]

The hash table has M slots, indexed from 0 to

M − 1.

For any value K in the key range and some

hash function h,
h(K) = i, 0 ≤ i < M , such that T [i].key() = K.

157
Hashing (continued)

Hashing is appropriate only for sets (no

duplicates).

Good for both in-memory and disk based

applications.
[Very good for organizing large databases on disk]

Answers the question “What record, if any, has

key value K?”
[Not good for range queries.]

Trivial Example: Store the n records with keys

in range 0 to n − 1.
• Store the record with key i in slot i.
• Use hash function h(K) = K.

Typically, there are however many more values

in the key range than slots in the hash table

158
Collisions

More reasonable example:

• Store about 1000 records with keys in
range 0 to 16,383.
• Impractical to keep a hash table with
16,384 slots.
• We must devise a hash function to map the
key range to a smaller table.

Given: hash function h and keys k1 and k2.

β is a slot in the hash table.
If h(k1) = β = h(k2), then k1 and k2 have a
collision at β under h.

Perfect Hashing: hash function devised so

that there are no collisions

Often impractical, sometimes expensive but

worthwhile

It works when the set is very stable (e.g., a

database on a CD)

159
Collisions (cont)

Search for the record with key K:

1. Compute the table location h(K).
2. Starting with slot h(K), locate the record
containing key K using (if necessary) a
collision resolution policy.

Collisions are inevitable in most applications.

• Example: 23 people are likely to share a
birthday (p = 1
2 ).

Example: store 200 students, in a table T with

365 records, using hash function h: birthday

160
Hash Functions

A hash function MUST return a value within

the hash table range.

To be practical, a hash function SHOULD

evenly distribute the records stored among the
hash table slots.

Ideally, the hash function should distribute

records with equal probability to all hash table
slots. In practice, success depends on the
distribution of the actual records stored.

If we know nothing about the incoming key

distribution, evenly distribute the key range
over the hash table slots.

If we have knowlege of the incoming

distribution, use a distribution-dependant hash
function.

161
Hash Functions (cont.)

Reasons why data values are poorly distributed

- Natural distributions are exponential (e.g.,
populations of cities)
- collected (e.g., measured) values are often
somehows skewed (e.g., rounding when
measuring)
- coding and alphabets introduce uneven
distributions (e.g., words in natural
language have first letter poorly distributed)

162
Example Hash Functions

static int h(int x) {

return(x % 16);
}

This function is entirely dependant on the lower

4 bits of the key, likely to be poorly distributed.

Mid-square method: square the key value,

take the middle r bits from the result for a
hash table of 2r slots.
[Works well because all bits contribute to the result.]

Sum the ASCII values of the letters and take

results modulo M .
static int h(String x, int M) {
int i, sum;
for (sum=0, i=0; i<x.length(); i++)
sum += (int)x.charAt(i);
return(sum % M);
}

[Only good if the sum is large compared to the size of the

table M .]

[This is an example of a folding method]

[NB: order of characters in the string is immaterial]

163
Open Hashing

What to do when collisions occur?

Open hashing treats each hash table slot as a
bin.

Open: collisions result in storing values outside

the table

Each slot is the head of a linked list

0 1000 9530
1
2
3 3013
4
5
6
7 9877 2007 1057
8
9 9879

164
Open Hashig Performance

Factors influencing performance

- how records are ordered in a slot’s list (e.g.,
by key value or frequency of access)
- ration N/M (records/slots)
- distribution of record key values

NB: Open Hash table must be kept in main

memory (storing on disk would defeat the
purpose of hashing)

165
Bucket Hashing

Divide the hash table slots into buckets.

• Example: 8 slots/bucket.

Include an overflow bucket.

Records hash to the first slot of the bucket,

and fill bucket. Go to overflow if necessary.

When searching, first check the proper bucket.

Then check the overflow.

166
Closed Hashing

Closed hashing stores all records directly in the

hash table.

Each record i has a home position h(ki).

If i is to be inserted and another record already

occupies i’s home position, then another slot
must be found to store i.

The new slot is found by a

collision resolution policy.

Search must follow the same policy to find

records not in their home slots.

167
Collision Resolution

During insertion, the goal of collision resolution

is to find a free slot in the table.

Probe Sequence: the series of slots visited

during insert/search by following a collision
resolution policy.

Let β0 = h(K). Let (β0, β1, ...) be the series of

slots making up the probe sequence.
void hashInsert(Elem R) { // Insert R into hash table T
int home; // Home position for R
int pos = home = h(R.key());// Initial pos on sequence
for (int i=1; T[pos] != null; i++) {
// Find next slot: p() is the probe function
pos = (home + p(R.key()), i)) % M;
Assert.notFalse(T[pos].key() != R.key(),
"Duplicates not allowed");
}
T[pos] = R; // Insert R
}

168
Collision Resolution (cont.)

// p(K, i) probe function returns offset from home position

// for ith slot of probe sequence of K
ELEM hashSearch(int K) { // Search for record w/ key K
int home; // Home position for K
int pos = home = h(K); // Initial pos on sequence
for (int i = 1; (T[pos] != null) &&
T[pos].key() != K); i++)
pos = (home + p(K, i)) % M; // Next pos on sequence
if (T[pos] == null) return null; // K not in hash table
else return T[pos]; // Found it
}

169
Linear Probing

Use the probe function

int p(int K, int i) { return i; }

This is called linear probing.

Linear probing simply goes to the next slot in

the table.

If the bottom is reached, wrap around to the

top.

To avoid an infinite loop, one slot in the table

must always be empty.

170
Linear Probing Example

Assuming hash function h(x) = x mod 11

0 1001 0 1001
1 9537 1 9537
2 3016 2 3016
3 3
4 4
5 5
6 6
7 9874 7 9874
8 2009 8 2009
9 9875 9 9875
10 10 1052

(a) (b)

Primary Clustering: Records tend to cluster

in the table under linear probing since the
probabilities for which slot to use next are not
the same.
[notation: prob(x) is the probability that next element goes to
position x]

[For (a): prob(3) = 4/11, prob(4) = 1/11, prob(5) = 1/11,

prob(6) = 1/11, prob(10) = 4/11.]

[For (b): prob(3) = 8/11, prob(4,5,6) = 1/11 each.]

[small clusters tend to merge ⇒ long probe sequences]

171
Improved Linear Probing

Instead of going to the next slot, skip by some

constant c.

Warning: Pick M and c carefully.

The probe sequence SHOULD cycle through all

slots of the table.
[If M = 10 with c = 2, then we effectively have created 2 hash
tables (evens vs. odds).]

Pick c to be relatively prime to M .

There is still some clustering.

• Example: c = 2. h(k1) = 3. h(k2) = 5.
• The probe sequences for k1 and k2 are
linked together.

172
Pseudo Random Probing

The ideal probe function would select the next

slot on the probe sequence at random.

An actual probe function cannot operate

randomly. (Why?)
[Execution of random procedure cannot be duplicated when
searching]

Pseudo random probing:

• Select a (random) permutation of the
numbers from 1 to M − 1:
r1, r2, ..., rM −1
• All insertions and searches use the same
permutation.

Example: Hash table of size M = 101

• r1 = 2, r2 = 5, r3 = 32.
• h(k1) = 30, h(k2) = 28.
• Probe sequence for k1 is: [30, 32, 35, 62]

• Probe sequence for k2 is: [28, 30, 33, 60]

[The two probe sequences diverge immediately]

173
Quadratic Probing

Set the i’th value in the probe sequence as

(h(K) + i2) mod M.

Example: M = 101.
• h(k1) = 30, h(k2) = 29.
• Probe sequence for k1 is:
[30, 31, 34, 39] =
[30, 30+12 , 30 + 22 , 30 + 32 ]

• Probe sequence for k2 is:

[29, 30, 33, 38] =
[29, 29+12 , 29 + 22 , 29 + 32 ] =

Problem: not all slots in the hash table are

necessarily in the probe serquence

174
Double Hashing

Pseudo random probing eliminates primary

clustering.

If two keys hash to same slot, they follow the

same probe sequence. This is called
secondary clustering.

To avoid secondary clustering, need a probe

sequence to be a function of the original key
value, not just the home position.

Double hashing:

p(K, i) = i ∗ h2(K) for 0 ≤ i ≤ M − 1. ]

Be sure that all probe sequence constants are

relatively prime to M [just like in improved linear
probing] .

Example: Hash table of size M = 101

• h(k1) = 30, h(k2) = 28, h(k3) = 30.
• h2(k1) = 2, h2(k2) = 5, h2(k3) = 5.
• Probe sequence for k1 is: [30, 32, 34, 36]

• Probe sequence for k2 is: [28, 33, 38, 43]

• Probe sequence for k3 is: [30, 35, 40, 45]

175
Analysis of Closed Hashing

The expected cost of hashing is a function of

how full the table is

The load factor is α = N/M where N is the

number of records currently in the table.

Expected # accesses (NB: accesses are due to

collisions) vs α
- solid lines: random probing
- dashed lines: linear probing

Insert Delete
4

1
0 .2 .4 .6 .8 1.0

176
Deletion

1. Deleting a record must not hinder later

searches.
2. We do not want to make positions in the
hash table unusable because of deletion.

Both of these problems can be resolved by

placing a special mark in place of the deleted
record, called a tombstone.

A tombstone will not stop a search, but that

slot can be used for future insertions.

Unfortunately, tombstones do add to the

average path length.

Solutions:
1. Local reorganizations to try to shorten the
average path length.
2. Periodically rehash the table (by order of
most frequently accessed record).

177
Indexing

Goals:
• Store large files.
• Support multiple search keys.
• Support efficient insert, delete and range
queries.
Entry sequenced file: Order records by time
of insertion. [Not practical as a database organization.]

Use sequential search.

Index file: Organized, stores pointers to actual

records. [Could be a tree or other data structure.]

Primary key: A unique identifier for records.

May be inconvenient for search.

Secondary key: an alternate search key, often

not unique for each record. Often used for
search key.

178
Linear Indexing

Linear Index: an index file organized as a

simple sequence of key/record pointer pairs
where the key values are in sorted order.

Features:
• If the index is too large to fit in main
memory, a second level index may be used.
• Linear indexing is good for searching
variable length records.
• Linear indexing is poor for insert/delete.

179
Tree Indexing

Linear index is poor for insertion/deletion.

Tree index can efficiently support all desired

operations (typical of a database):
• Insert/delete
• Multiple search keys [Multiple tree indices.]

• Key range search

Storing a (BST) tree index on disk causes

additional problems:
1. Tree must be balanced. [Minimize disk accesses.]

2. Each path from root to a leaf should cover

few disk pages.

Use buffer pool to store recently accessed

pages; exploit locality of reference

But only mitigates the problem

180
Tree indexing (cont.)

Rebalance a BST after insertion/deletion can

require much rearranging

Example of insert(1)

5 4

3 7 2 6

2 4 6 1 3 5 7

(a) (b)

181
2-3 Tree

A 2-3 Tree has the following shape properties:

1. A node contains one or two keys.
2. Every internal node has either two children
(if it contains one key) or three children (if
it contains two keys).
3. All leaves are at the same level in the tree,
so the tree is always height balanced.

The 2-3 Tree also has search tree properties

analogous to BST
1. values in left subtree < first node value
2. values in center subtree ≥ first node value
3. values in center subtree < second node
value (if existing)
4. (if both existing) values in right subtree ≥
first node value

182
2-3 Tree(cont.)

The advantage of the 2-3 Treeover the BST is

that it can be updated at low cost.
- always insert at leaf node
- search position for key to be inserted
- if there is room (1 free slot) then finished
- otherwise must add a node (split operation)
- from 1 node with 2 keys get 2 nodes with 1
key and promote middle valued key
- recursively, insert promoted key into parent
node
- if splitting repeated until root of the tree
then its depth increases (but tree remains
balanced)

18 33

12 23 30 48

10 15 20 21 24 31 45 47 50 52

183
2-3 Tree Insertion

18 33

12 23 30 48

10 15 15 20 21 24 31 45 47 50 52
14 [Insert 14]

18 33

12 23 30 48 52

10 15 20 21 24 31 45 47 50 55

[Insert 55. Always insert at leaf node.]

184
2-3 Tree Splitting

[Insert 19 into node 20-21 ⇒ split and promote 20 into node

23-30 ⇒ split and promote 23, this becomes new root, tree is
1 level deeper]

[NB: All operations are local to original search path.]

23
20 23 30 20 30

19 21 24 31 19 21 24 31

(a) (b)

23

18 33

12 20 30 48

10 15 19 21 24 31 45 47 50 52

(c)

185
B-Trees

The B-Tree is a generalization of the 2-3 Tree.

The B-Tree is now the standard file

organization for applications requiring insertion,
deletion and key range searches.

1. B-Trees are always balanced.

2. B-Trees keep related records on a disk
page, which takes advantage of locality of
reference.
3. B-Trees guarantee that every node in the
tree will be full at least to a certain
minimum percentage. This improves space
efficiency while reducing the typical number
of disk fetches necessary during a search or
update operation.

186
B-Trees (Continued)

A B-Tree of order m has the following

properties.
• The root is either a leaf or has at least two
children.
• Each node, except for the root and the
leaves, has between ⌈m/2⌉ and m children.
• All leaves are at the same level in the tree,
so the tree is always height balanced.

NB: A 2-3 Tree is a B-Tree of order 3

A B-Tree node is usually selected to match the

size of a disk block.

A B-Tree node could have hundreds of children

⇒ depth is ≈ log100 n.

A block implemented in a disk block

A pointer implemented by a disk block

reference

187
B-Tree Example

Search in a B-Tree is a generalization of search

in a 2-3 Tree.
1. Perform a binary search on the keys in the
current node. If the search key is found,
then return the record. If the current node
is a leaf node and the key is not found,
then report an unsuccessful search.
2. Otherwise, follow the proper branch and
repeat the process.

A B-Tree of order 4

Example: search for record with key 47

24

15 20 33 45 48

10 12 18 21 23 30 31 38 47 50 52 60

188
B-Tree Insertion

Obvious extension of 2-3 Tree insertion

NB: split and promote process ensures all nodes

are half full

Example: node with 4 keys + add one key

SPLIT ⇒ promote middle key + 2 nodes with

2 keys each

189
B+-Trees

The most commonly implemented form of the

B-Tree is the B+-Tree.

Internal nodes of the B+-Tree do not store

records – only key values to guide the search.

Leaf nodes store records or pointers to records.

A leaf node may store more or less records than

an internal node stores keys.

Requirement: leaf nodes always half full.

Leaf nodes doubly linked in a list ⇒ can

traverse it in any order ⇒ very good for range
queries.

Search: similar to B-Tree search: must always

go to the leaf (internal nodes do not store
records)

190
B+-Tree Example: search

[Assume leaves can store 5 values, internal notes 3 (4

children).]

[Example: search key 33]

33

18 23 48

10 12 15 18 19 20 21 22 23 30 31 33 45 47 48 50 52

191
B+-Tree Insertion

Insertion similar to B-Tree insertion:

- find leaf that should contain inserted key
- if not full, insert and finish
- else split and promote a copy of least
valued key of the newly formed right node

192
B+-Tree Example: Insertion
[Note special rule for root: May have only two children.]

33

10 12 23 33 48 10 12 23 33 48 50
[(b) Add (a50.]
) [Add 45, 52, 47 (split),(b)18, 15, 31 (split), 21,
20.]
18 33 48

10 12 15 18 20 21 23 31 33 45 47 48 50 52
(c)
[Add 30 (split).]

33

18 23 48

10 12 15 18 20 21 23 30 31 33 45 47 48 50 52
(d)

193
B+-Tree Deletion

- locate level N containing key to be deleted

- if more than half full, remove and finish
- else (underflow) must restructure the tree
- if possible get spare values from adjacent
siblings (⇒ possibly keys in parent node
must be updated)
- if siblings cannot give values (they are
only half full)
- N goves its values to them and is
removed (possible because its siblings
are only half full and N is
underflowing)
- this can cause underflow in the parent
node (⇒ propagate upwards, possibly
eventually causing two chindren or root
to merge and tree to lose one level)

194
B+-Tree Example: Deletion
[Simple delete – delete 18 from original example.] [NB: do
not need to delete 18 from internal node: it is a placeholder,
can still be used to guide the search]

33

18 23 48

10 12 15 19 20 21 22 23 30 31 33 45 47 48 50 52

[Delete of 12 form original example: Borrow from sibling.]

33

19 23 48

10 15 18 19 20 21 22 23 30 31 33 45 47 48 50 52

195
B-Tree Space Analysis

B+-Tree nodes are always at least half full.

The B∗-Tree splits two pages for three, and

combines three pages into two. In this way,
nodes are always 2/3 full.

Improves performance, makes implementation

very complex

Tradeoff between space utilization and

efficiency and complexity of impolementation

Asymptotic cost of search, insertion and

deletion of records from B-Trees, B+-Trees and
B∗-Trees is Θ(log n). (The base of the log is
the (average) branching factor of the tree.)

Ways to reduce the number of disk fetches:

• Keep the upper levels in main memory.
• Manage B+-Tree pages with a buffer pool.

196
B-Tree Space Analysis: Examples

Example: Consider a B+-Tree of order 100

with leaf nodes containing 100 records.
1 level B+-Tree: [Max: 100]

2 level B+-Tree: [Min: 2 leaves of 50 for 100 records.

Max: 100 leaves with 100 for 10,000 records.]

3 level B+-Tree: [Min: 2 × 50 nodes of leaves for 5000

records. Max: 1003 = 1, 000, 000 records.]

4 level B+-Tree: [Min: 250,000 records (2 * 50 * 50 *

50). Max: 100 million records (100 * 100 * 100 * 100).]

197