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The Dictionary ADT Definition A dictionary is an ordered or unordered list of key-element pairs, where keys are used to locate elements in the list.

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the dictionary adt
The Dictionary ADT

Definition A dictionary is an ordered or unordered list of key-element pairs,

where keys are used to locate elements in the list.

Example: consider a data structure that stores bank accounts; it can be viewed as a dictionary, where account numbers serve as keys for identification of account objects.

Operations (methods) on dictionaries:

size () Returns the size of the dictionary

empty () Returns true is the dictionary is empty

findItem (key) Locates the item with the specified key. If

no such key exists, sentinel value NO_SUCH_KEY is returned. If more

than one item with the specified key exists, an arbitrary item is returned.

findAllItems (key) Locates all items with the specified key. If

no such key exists, sentinel value NO_SUCH_KEY is returned.

removeItem (key) Removes the item with the specified key

removeAllItems (key) Removes all items with the specified key

insertItem (key, element) Inserts a new key-element pair

additional methods for ordered dictionaries
Additional methods for ordered dictionaries

closestKeyBefore (key) Returns the key of the item with largest key

less than or equal to key

closestElemBefore (key) Returns the element for the item with largest

key less than or equal to key

closestKeyAfter (key) Returns the key of the item with smallest

key greater than or equal to key

closestElemAfter (key) Returns the element for the item with smallest

key greater than or equal to key

Sentinel value NO_SUCH_KEY is always returned if no item in the dictionary

satisfies the query.

Note Java has a built-in abstract class java.util.Dictionary In this class,

however, having two items with the same key is not allowed. If an application

assumes more than one item with the same key, an extended version of the

Dictionary class is required.

example of unordered dictionary
Example of unordered dictionary

Consider an empty unordered dictionary and the following set of operations:

Operation Dictionary Output

insertItem(5,A) {(5,A)}

insertItem(7,B) {(5,A), (7,B)}

insertItem(2,C) {(5,A), (7,B), (2,C)}

insertItem(8,D) {(5,A), (7,B), (2,C), (8,D)}

insertItem(2,E) {(5,A), (7,B), (2,C), (8,D), (2,E)}

findItem(7) {(5,A), (7,B), (2,C), (8,D), (2,E)} B

findItem(4) {(5,A), (7,B), (2,C), (8,D), (2,E)} NO_SUCH_KEY

findItem(2) {(5,A), (7,B), (2,C), (8,D), (2,E)} C

findAllItems(2) {(5,A), (7,B), (2,C), (8,D), (2,E)} C, E

size() {(5,A), (7,B), (2,C), (8,D), (2,E)} 5

removeItem(5) {(7,B), (2,C), (8,D), (2,E)} A

removeAllItems(2) {(7,B), (8,D)} C, E

findItem(4) {(7,B), (8,D)} NO_SUCH_KEY

example of ordered dictionary
Example of ordered dictionary

Consider an empty ordered dictionary and the following set of operations:

Operation Dictionary Output

insertItem(5,A) {(5,A)}

insertItem(7,B) {(5,A), (7,B)}

insertItem(2,C) {(2,C), (5,A), (7,B)}

insertItem(8,D) {(2,C), (5,A), (7,B), (8,D)}

insertItem(2,E) {(2,C), (2,E), (5,A), (7,B), (8,D)}

findItem(7) {(2,C), (2,E), (5,A), (7,B), (8,D)} B

findItem(4) {(2,C), (2,E), (5,A), (7,B), (8,D)} NO_SUCH_KEY

findItem(2) {(2,C), (2,E), (5,A), (7,B), (8,D)} C

findAllItems(2) {(2,C), (2,E), (5,A), (7,B), (8,D)} C, E

size() {(2,C), (2,E), (5,A), (7,B), (8,D)} 5

removeItem(5) {(2,C), (2,E), (7,B), (8,D)} A

removeAllItems(2) {(7,B), (8,D)} C, E

findItem(4) {(7,B), (8,D)} NO_SUCH_KEY

implementations of the dictionary adt
Implementations of the Dictionary ADT

Dictionaries are ordered or unordered lists. The easiest way to implement a list

is by means of an ordered or unordered sequence.

Unordered sequence implementationItems are added to the initially empty

dictionary as they arrive. insertItem(key, element) method is O(1) no matter whether the

new item is added at the beginning or at the end of the dictionary. findItem(key),

findAllItems(key), removeItem(key) and removeAllItems(key) methods, however, have

O(n) efficiency. Therefore, this implementation is appropriate in applications where the

number of insertions is very large in comparison to the number of searches and removals.

Ordered sequence implementationItems are added to the initially empty

dictionary in nondecreasing order of their keys. insertItem(key, element) method is O(n),

because a search for the proper place of the item is required. If the sequence is implemented

as an ordered array, removeItem(key) and removeAllItems(key) take O(n) time, because

all items following the item removed must be shifted to fill in the gap. If the sequence is

implemented as a doubly linked list , all methods involving search also take O(n) time.

Therefore, this implementation is inferior compared to unordered sequence implementation.

However, the efficiency of the search operation can be considerably improved, in which case

an ordered sequence implementation will become a better choice.

implementations of the dictionary adt contd
Implementations of the Dictionary ADT (contd.)

Array-based ranked sequence implementation A search for an item in a

sequence by its rank takes O(1) time. We can improve search efficiency in an

ordered dictionary by using binary search; thus improving the run time efficiency

of insertItem(key, element), removeItem(key) and removeAllItems(key) to

O(log n).

More efficient implementations of an ordered dictionary are binary search trees

and AVL trees which are binary search trees of a special type. The best way to

implement an unordered dictionary is by means of a hash table. We discuss AVL

trees and hash tables next.

avl trees
AVL trees

Definition An AVL tree is a binary tree with an ordering property where the

heights of the children of every internal node differ by at most 1.

Example

44 (4)

17 (2) 78 (3)

32 (1) 50 (2) 88 (1)

48 (1) 62 (1)

Note: 1. Every subtree of an AVL tree is also an AVL tree.

2. The height of an AVL tree storing n keys is O(log n).

insertion of new nodes in avl trees
Insertion of new nodes in AVL trees

Assume you want to insert 54 in our example tree.

Step 1: Search for 54 (as if it were a binary search tree), and find where the

search terminates unsuccessfully

44 (5)

17 (2) 78 (4)

These two children

32 (1) 50 (3) 88 (1) are unbalanced

48 (1) 62 (2)

54 (1)

Step 2: Restore the balance of the tree.

rotation of avl tree nodes
Rotation of AVL tree nodes

To restore the balance of the tree, we perform the following restructuring. Let z be the first

“unbalanced” node on the path from the newly inserted node to the root, y be the child of z

with higher height, and x be the child of y (x may be the newly inserted node). Since z became

unbalanced because of the insertion in the subtree rooted at its child y, the height of y is 2

greater than its sibling.

Let us rename nodes x, y, and z as a, b, and c, such that a precedes b and b precedes c in

inorder traversal of the currently unbalanced tree. There are 4 ways to map x, y, and z to

a, b, and c, as follows:

z = a

y = b y = b

T0

x = c z = a x = c

T1

T2 T3 T0 T1 T2 T3

rotation of avl tree nodes contd
Rotation of AVL tree nodes (contd.)

z = c

y = b y = b

x = a T3 x = a z = c

T2

T0 T1 T0 T1 T2 T3

z = a

y = c x = b

T0 x = b z = a y = c

T3

T1 T2 T0 T1 T2 T3

rotation of avl tree nodes contd11
Rotation of AVL tree nodes (contd.)

z = c

y = a x = b

x = b T3 y = a z = c

T0

T1 T2 T0 T1 T2 T3

the restructure algorithm
The restructure algorithm

Algorithm restructure(x):

Input: A node x that has a parent node y, and a grandparent node z.

Output: Tree involving nodes x, y and z restructured.

1. Let (a,b,c) be inorder listing of nodes x, y and z, and let (T0, T1, T2, T3) be

inorder listing of the four children subtrees of x,y, and z.

2. Replace the subtree rooted at z with a new subtree rooted at b.

3. Let a be the left child of b and let T0 and T1 be the left and right subtrees of

a, respectively.

4. Let c be the right child of b and let T2 and T3 be the left and right subtrees of

c, respectively.

If y = b, we have a single rotation, where y is rotated over z. If x = b, we have a

double rotation, where x is first rotated over y, and then over z.

deletion of avl tree nodes
Deletion of AVL tree nodes

Consider our example tree and assume that we want to delete 32.

44 (4)

These children are

17 (1) 78 (3) unbalanced

50 (2) 88 (1)

48 (1) 62 (1)

Note: Search for the node to delete is performed as in the binary search tree.

To restore the balance of the tree, we may have to perform more than one rotation

when we move towards the root (one rotation may not be sufficient here).

deletion of avl tree nodes contd
Deletion of AVL tree nodes (contd.)

After the restructuring of the tree rooted in node 44:

44 (4) z=a 50

17 (1) 78 (3) y=c 44 78

x=b 50 (2) 88 (1) 17 48 62 88

48 (1) 62 (1)

implementation of unordered dictionaries hash tables
Implementation of unordered dictionaries: hash tables

Hashing is a method for directly referencing an element in a table by performing

arithmetic transformations on keys into table addresses. This is carried out in two

steps:

Step 1: Computing the so-called hash function H: K -> A.

Step 2: Collision resolution, which handles cases where two or more different keys

hash to the same table address.

K1

K2

K3

...

Kn

A1

A2

...

An

implementation of hash tables
Implementation of hash tables

Hash tables consist of two components: a bucket array and a hash function.

Consider a dictionary, where keys are integers in the range [0, N-1]. Then, an

array of size N can be used to represent the dictionary. Each entry in this array is

thought of as a “bucket” (which is why we call it a “bucket array”). An element e

with key k is inserted in A[k]. Bucket entries associated with keys not present in

the dictionary contain a special NO_SUCH_KEY object. If the dictionary contains

elements with the same key, then two or more different elements may be mapped

to the same bucket of A. In this case, we say that a collision between these

elements has occurred. One easy way to deal with collisions is to allow a sequence

of elements with the same key, k, to be stored in A[k]. Assuming that an arbitrary

element with key k satisfies queries findItem(k) and removeItem(k), these

operations are now performed in O(1) time, while insertItem(k, e) needs only to

find where on the existing list A[k] to insert the new item, e. The drawback of this is

that the size of the bucket array is the size of the set from which key are drawn,

which may be huge.

hash functions
Hash functions

We can limit the size of the bucket array to almost any size; however, we must

provide a way to map key values into array index values. This is done by an

appropriately selected hash function, h(k). The simplest hash function is

h(k) = k mod N

where k can be very large, while N can be as small as we want it to be. That is,

the hush function converts a large number (the key) into a smaller number

serving as an index in the bucket array.

Example. Consider the following list of keys: 10, 20, 30, 40,..., 220.

Let us consider two different sizes of the bucket array:

(1) a bucket array of size 10, and

(2) a bucket array of size 11.

example contd
Example (contd.)

Case 1: Case 2:

Position Key Position Key

0 10, 20, 30,..., 220 0 110, 220

1 1 100, 210

2 2 90, 200

3 3 80, 190

4 4 70, 180

5 5 60, 170

6 6 50, 160

7 7 40, 150

8 8 30, 140

9 9 20, 130

10 10, 120

example 2
Example 2

Consider a dictionary of strings of characters from a to z. Assume that each

character is encoded by means of 5 bits, i.e.

character code

a 00001

b 00010

c 00011

d 00100

e 00101

......

k 01011

......

y 11001

Then, the string akey has the following code

(00001 01011 00101 11001)2 = (44217)10

Assume that our hash table has 101 buckets. Then,

h(44217) = 44217 mod 101 = 80

That is, the key of the string akey hashes to position 80. If you do the same with

the string barh, you will see that it hashes to the same position, 80.

hash functions contd
Hash functions (contd.)

These examples suggest that if N is a prime number, the hash function helps

spread out the distribution of hashed values. If dictionary elements are spread

fairly evenly in the hash table, the expected running times of operations

findItem, insertItem and removeItem are O(n/N), where n is the number of

elements in the dictionary, and N is the size of the bucket array. These efficiencies

are ever better, O(1), if no collision occurs (in which case only a call to the hash

function and a single array reference are needed to insert or find an item).

collision resolution
Collision resolution

There are 2 main ways to perform collision resolution:

  • Open addressing.
  • Chaining.

In our examples, we have assumed that collision resolution is performed by

chaining, i.e. traversing the linked list holding items with the same key in order to

find the one we are searching for, or insert a new item with that key.

In open addressing we deal with collision by finding another, unoccupied location

elsewhere in the array. The easiest way to find such a location is called linear

probing. The idea is the following. If a collision occurs when we are inserting a

new item into a table, we simply probe forward in the array, one step at a time,

until we find an empty slot where to store the new item. When we remove an item,

we start by calculating the hash function and test the identified index location. If

the item is not there, we examine each array entry from the index location until:

(1) the item is found; (2) an empty location is encountered, or (3) the array end is

reached.