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Searching (and into Trees)
Trees
1
Searching techniques

Searching algorithms in unordered “lists”



Searching algorithms in ordered “lists”



“Lists” may be arrays or linked lists
The algorithms are adaptable to either
Primarily arrays with values sorted into order
We can exploit the order to search more efficiently
Let’s start with arrays

The data held in the array might be
 Simple (e.g. numbers, strings) or more complex objects: the
search will probably be based on a chosen key field in the
objects (e.g., search student records by registration
number)

We will only consider simple data – the searching
techniques are the same
Trees
2
Sequential search in an
unordered list

In these algorithms we will assume:







The basic technique to be used is “sequential search”


The data is integers
Held in an array variable numbers,
Is in random order
The number of data values is indicated by a variable size
The data is in elements indexed 0 to (size-1)
We are seeking the integer held in variable val
Compare val with the value in numbers[0], then with that in
numbers[1], etc
We will look at two versions of an algorithm encoding
this

Other adaptations are possible
Trees
3
Algorithm 1: Standard sequential
search




Here is a basic search algorithm. It leaves its result in a variable
called position:
int position = 0;
while (position < size)
{
if (numbers[position] == val)
break;
// Exit loop if found
position++;
}
If val is not present:
 The entire array will be scanned - taking size steps

position will have a final value of size
But if val is present:

break; -> the while loop terminates immediately
 The average number of scanning steps expected is size/2
Easy to adapt to return a boolean, or throw an exception
Trees
4

If we are careful, we can combine the loop test
and the array element check:



int position = 0;
while (numbers[position] != val &&
position < size)
position++;
Is this correct?
int position = 0;
while (numbers[position++] != val &&
position < size);
Trees
5
Corrected Version

If we are careful, we can combine the loop test
and the array element check:
int position = 0;
while (position < size && numbers[position] != val)
position++;
1.
2.
3.
4.
The && test checks position < size first,
and if it is false does not check numbers[position]
!= val
otherwise would get
ArrayIndexOutOfBoundsException if val is not
present!
This is called "conditional" or "short-circuit" behaviour:
it applies to && and ||
Trees
6
Algorithm 2: Sequential search with
a “sentinel”

We can improve the basic search algorithm if the array numbers has one
extra element, numbers[size], that is never used for actual data
 Instead we place a copy of the sought value there, so the search always
succeeds. This means that the loop does not need to carry out the “end
of array” test - less work, so quicker.

int[] numbers = new int[size+1];
...
int position = 0;
numbers[size] = -1; // Insert "sentinel"
while (numbers[position] != val)
position++;
return position;
As before, position has the final value size if val is not present
Trees
7


We may be interested in an algorithm’s best case,
worst case or average execution time:
For the sequential search algorithm (with or
without sentinel):



Best case is 1 step: O(1)
Worst case is N steps: O(N)
The actual average number of steps depends on ratio of
successful/unsuccessful searches:
The average of successful searches is N/2 steps, and so
is O(N)
All unsuccessful searches take N steps, which is O(N),
So overall the average complexity is O(N)
Trees
8
Searching an Ordered List

Again we will assume:






The data is integers, held in an array sequence,
So the data is in elements indexed 0 to (length-1)
But this time we assume that the values are held in
ascending numerical order
We are seeking the integer held in val
We could use the sequential search algorithm, but this does
not take advantage of the knowledge that the data is
ordered. (The complexity remains O(N). )
Instead, we will take advantage of the ordering to improve
search efficiency (i.e., to reduce the complexity)
Trees
9
Binary Search



If the data is already ordered, we can do much better than a
linear time algorithm. Here is the scheme:
 Pick the middle element in the array
 If it is equal to val, stop the search
 If it is greater than val, search the lower half of the
remaining array
 If it is less than val, search the remaining upper half
At each iteration:
 We are searching in a remaining partition of the array
 We cut the remaining partition in half, rather than just
removing one element
Example: Searching for 11 in
1, 3, 5, 7, 9, 11, 13
 First compare with 7, so search in 9, 11, 13
 Now compare with 11 - found it - in two steps
Trees
10

Concretely:


Let variable low indicate the lowest element of the partition
(index 0 initially)
high (h) indicate the highest element (size-1 initially)
middle (m) indicate the next element being tested
The search for 11 proceeds like this:
low=0 h=6  m=3
0
1
2
3
4
1
3
5
7
9
low
Not found, and 11>7, so
low=(m+1)=4 h=6
 m=5
1
5
11 13
h
m
3
5
7
6
9
11 13
low m
h
Found it, at index 5
Trees
11
An unsuccessful search: search for 10
0
low=0
h=6
 m=3
Not found, and 10>7, so
low=(m+1)=4 h=6
 m=5
1
2
3
4
1 3
low
5
7
m
9
11 13
h
5
7
9
11 13
1
low m
Not found, and 10<11, so
1
low=4 h=(m-1)=4
 m=4
Not found, and 10>9, so
low=(m+1)=5 h=4
3
5
1
3
5
7
9
6
h
11 13
low
h
m
3
5
7
9
11 13
h low
Now low>h, and the partition has
Trees
“vanished”: the search has failed
12

Algorithm binarySearch:
INPUT: val – value of interest, sequence – sorted data
OUTPUT: object or value of interest if exists, null otherwise
int low = 0, middle = 0, high = seq.length;
while (high >= low) {
middle = (high + low) / 2;
if (sequence[middle] == val)
return sequence[middle];
// Found it
else if (sequence[middle] < val)
low = middle + 1;
// Search upper half
else
high = middle - 1;
// Search lower half
}
return -1; // or null if an object-type

The outcomes:
 Ordinary loop exit when the indexes “cross”  not found (i.e. high < low)
 Loop exit on return  found
(detect this by testing high >= low)
Trees
13
The Complexity of Binary Search


Best case: val is exactly sequence[middle] at the first step
 The search stops after first step, so complexity O(1)
Worst case:
 This will be when we continue dividing until the “partition”
contains only one value: then it is either equal to val or not
 For 250 elements this turns out to be about 8 iterations
 For 500 it is about 9
 For 1000 it is about 10
 Double the amount of data  Add one step!

In general: the size is approximately 2steps
So the number of steps is approximately log2 size
 Complexity is O(logN)
 For emphasis: double the amount of data  Add one step!
Average case:
 Don’t need to consider this: the worst case is very good!


Trees
N
log2N
1
0
2
1
4
2
8
3
16
4
32
5
64
6
128
7
256
8
512
9
1024
10
2048
11
4096
12
8192
13
16384 14
32768 15
65536 16
131072 17
262144 18
524288 19
1048576 20
14
Trees
Make Money Fast!
Stock
Fraud
Ponzi
Scheme
Trees
Bank
Robbery
15
What is a Tree



In computer science, a
tree is an abstract model
of a hierarchical
structure
A tree consists of nodes
with a parent-child
relation
US
Applications:



Computers”R”Us
Sales
Manufacturing
International
Organization charts
File systems
Europe
Programming
environments
Trees
Asia
Laptops
R&D
Desktops
Canada
16
Tree Terminology







Root: node without parent (A)
 Subtree: tree consisting
of a node and its
Internal node: node with at least
descendants
one child (A, B, C, F)
External node (a.k.a. leaf ): node
A
without children (E, I, J, K, G, H, D)
Ancestors of a node: parent,
grandparent, grand-grandparent,
B
C
D
etc.
Depth of a node: number of
ancestors
E
F
G
H
Height of a tree: maximum depth
of any node (3)
Descendant of a node: child,
I
J
K
subtree
grandchild, grand-grandchild, etc.
Trees
17
Tree ADT


We use positions to abstract
nodes
Generic methods:





integer size()
boolean isEmpty()
Iterator iterator()
Iterable positions()





boolean isInternal(p)
boolean isExternal(p)
boolean isRoot(p)
Update method:

element replace (p, o)
Additional update methods
may be defined by data
structures implementing the
Tree ADT
Accessor methods:

Query methods:
position root()
position parent(p)
Iterable children(p)
Trees
18
Preorder Traversal



A traversal visits the nodes of a
tree in a systematic manner
In a preorder traversal, a node
is visited before its descendants
Application: print a structured
document
1
Algorithm preOrder(v)
visit(v)
for each child w of v
preorder (w)
Make Money Fast!
2
5
1. Motivations
9
2. Methods
3
4
1.1 Greed
1.2 Avidity
6
7
2.1 Stock
Fraud
Trees
2.2 Ponzi
Scheme
References
8
2.3 Bank
Robbery
19
Postorder Traversal


In a postorder traversal, a
node is visited after its
descendants
Application: compute space
used by files in a directory and
its subdirectories
9
Algorithm postOrder(v)
for each child w of v
postOrder (w)
visit(v)
cs16/
3
8
7
homeworks/
todo.txt
1K
programs/
1
2
h1c.doc
3K
h1nc.doc
2K
4
5
DDR.java
10K
Trees
Stocks.java
25K
6
Robot.java
20K
20
Ordered Binary Trees

A binary tree is a tree with the
following properties:






Each internal node has at most
two children (exactly two for
proper binary trees)
The children of a node are an
ordered pair



a tree consisting of a single node,
or
a tree whose root has an ordered
pair of children, each of which is a
binary tree
Trees
arithmetic expressions
decision processes
searching
A
We call the children of an internal
node left child and right child
Alternative recursive definition: a
binary tree is either

Applications:
B
C
D
E
H
F
G
I
21
Arithmetic Expression Tree

Binary tree associated with an arithmetic expression



internal nodes: operators
external nodes: operands
Example: arithmetic expression tree for the
expression (2  (a - 1) + (3  b))
+


-
2
a
3
b
1
Trees
22
Decision Tree

Binary tree associated with a decision process



internal nodes: questions with yes/no answer
external nodes: decisions
Example: dining decision
Want a fast meal?
No
Yes
How about coffee?
On expense account?
Yes
No
Yes
No
Starbucks
Spike’s
Al Forno
Café Paragon
Trees
23
BinaryTree ADT


The BinaryTree ADT
extends the Tree
ADT, i.e., it inherits
all the methods of
the Tree ADT
Additional
methods:




position left(p)
position right(p)
boolean hasLeft(p)
boolean hasRight(p)Trees

Update methods
may be defined by
data structures
implementing the
BinaryTree ADT
25
Inorder Traversal


In an inorder traversal a
node is visited after its left
subtree and before its right
subtree
Application: draw a binary
tree


Algorithm inOrder(v)
if hasLeft (v)
inOrder (left (v))
visit(v)
if hasRight (v)
inOrder (right (v))
x(v) = inorder rank of v
y(v) = depth of v
6
2
8
1
4
3
7
9
5
Trees
26
Print Arithmetic Expressions

Specialization of an inorder
traversal



print operand or operator
when visiting node
print “(“ before traversing
left subtree
print “)“ after traversing
right subtree
+


-
2
a
3
Algorithm printExpression(v)
if hasLeft (v)
print(“(’’)
inOrder (left(v))
print(v.element ())
if hasRight (v)
inOrder (right(v))
print (“)’’)
b
((2  (a - 1)) + (3  b))
1
Trees
27
Evaluate Arithmetic Expressions

Specialization of a postorder
traversal


recursive method returning
the value of a subtree
when visiting an internal
node, combine the values
of the subtrees
+

Algorithm evalExpr(v)
if isExternal (v)
return v.element ()
else
x  evalExpr(leftChild (v))
y  evalExpr(rightChild (v))
  operator stored at v
return x  y

-
2
5
3
2
1
Trees
28
Linked Structure for Trees

A node is represented by
an object storing
1.
2.
3.
Element
Parent node
Sequence of children
nodes

B

A
B
D
A
C

D
F
F

E
C
Trees

E
30
Linked Structure for Binary Trees

A node is represented
by an object storing
1.
2.
3.
4.

Element
Parent node
Left child node
Right child node
B

B
A

A
D
C
D

E

C
Trees


E
31
Array-Based Representation of
Binary Trees

Nodes are stored in an array A
1
A
0

A
B
D
1
2
3
…
G
H
10
11
…
2
Node v is stored at A[rank(v)]
4
 rank(root) = 1
E
 if node is the left child of parent(node),
rank(node) = 2  rank(parent(node))
 if node is the right child of parent(node),
10
rank(node) = 2 rank(parent(node)) + 1
Trees
3
B
D
5
6
7
C
F
J
11
G
H
32