Hashing

ANJUMAN COLLEGE
OF ENGINEERING AND
TECHNOLOGY,
SADAR, NAGPUR.
DATA STRUCTURE AND PROGRAM
DESIGN
2017-18
Computer science and
Engineering
Seminar on:
Hashing
Guided by: Prof. Farhina Sheikh

COMPILED BY:
oAafaqueahmad Khan
oAsad Fazlani
oDanish Sheikh
oHimanshu Wasnik
oIrfan Sheikh
oJafar Sheikh
oRafiuddin
oRohit Raut
oSahil khan
oSourabh Phulpagar

CONTENTS
INTRODUCTION
HASHING
HASH FUNCTION
CHARACTERISTICS OF HASH
FUNCTION
HASH TABLE
STATIC HASHING
TYPES OF HASH FUNCTION
DIVISION METHOD

HASHING
Hashing is the process of indexing and
retrieving element (data) in a data structure to
provide faster way of finding the element
using the hash key.
In hashing time required to search an element
doesn’t depend on the number of element.
Using hashing data structure, an element is
searched with constant time complexity.
Hashing provides very fast access to records
on certain search conditions.

HASH FUNCTION
A hash function is a function which
takes a piece of data (i.e. key) as input
and outputs an integer (i.e. hash value)
which maps the data to a particular
index in the hash table.

CHARACTERISTICS OF
HASH FUNCTION
 It should be easy and quick to
compute.
 It should ideally be mathematically
one-to-one the set of relevant key
values.
 The number of collision should be less
while placing the records in the hash
table.
 It should achieve even distribution of
the key values that actually occur across

HASH TABLE
Hash Table also known as Hash Map is a
data structure which uses hash function
to generate key corresponding to the
associated value.
It is a tables from which an item can be
searched in o(1) time using a hash
function to form address from the key

STATIC HASHING
Primary Area: # primary pages fixed, allocated
sequentially, never deallocated; (say M buckets).
•A simple hash function: h(K) = f(K) mod M
Overflow area: disjoint from the primary area. It
keeps buckets which hold records whose key
maps to full bucket.
•Adding the address of an overflow bucket to a
primary area bucket is called chaining.
Collision does not cause a problem as long as
there is still room in the mapped bucket.
Overflow occurs during insertion when a record is
hashed to the bucket that is already full.

EXAMPLE
Assume f(k) = k. Let M = 5. So, h(k) = k mod
5
Bucket factor = 3 records
0 35 60
1 6 46
2 12 57 62
3 33
4 44
17

PROBLEMS OF STATIC
HASHING
 The main problem with static hashing: the number
of buckets is fixed:
• Long overflow chains can develop and degrade
performance.
• On the other hand, if a file shrinks greatly, a lot of
bucket space will be wasted.
 There are some other hashing techniques that
allow dynamically growing and shrinking hash
index. These include:
• Linear hashing
• Extendible hashing

TYPES OF HASH
FUNCTION:
A. Division method
B. MID square method
C. Folding Method

DIVISION METHOD
In this method, we choose a number m (i.e.,
memory locations) larger than the number of n
(i.e. number of records) of keys which uniquely
determine records). The number of usually
chosen to be a prime number or a number
without small divisors since this frequently
minimizes the number of The hash function H is
defined by:
H(K)=K(mod m)
Here K (mod m) denotes the remainder when K
is divided by m. The second formula is used

For example, a company has 68 employees and
they have been assigned 4-digit employee
number. Assume L (memory addresses in the
table) of 100 two digit addresses: 00, 01, 02...99.
Applying the above hash function, say, for
employee numbers:
3205, 7148, 2345
Here, we choose a prime number m close to 99,
such as 97 then
H (3205) = 4
H (7148) = 67
H(2345) = 17

That is, dividing 3205 by 97 gives a
remainder of 4, dividing 7148 by 97 gives a
remainder of 67 and dividing 2345 by 97
gives a remainder of 17. In the other case
where the memory addresses begin with 01
rather than 00, we choose the hash function:
H(3205) = 4+1=5
H(7148) = 67+1=68
H(2345) = 17+1=18

Ques:- Given following list of element
{63, 92, 84, 33, 90, 69, 97, 91}
Use Division method of hashing to sort the
hashing.
Solution:-
 Formula :-
• h(k) = k mode m
Given,
 k= keys element.
• The keys element is:
63 92 84 33 90 69 97 91
 m= Total no of Buckets.
• Total no of Buckets is 8 because we have 8

Calculation:-
h(k) = 63 % 8
h(k) = 7
h(k) = 92 % 8
h(k) = 6
h(k) = 84 % 8
h(k) = 4
h(k) = 32 % 8
h(k) = 0
h(k) = 90 % 8
h(k) = 2
h(k) = 69 % 8
h(k) = 5
h(k) = 97 % 8
h(k) = 1
h(k) = 91 % 8
h(k) = 3

Sr.
No.
Key Element,
(k)
Hash
Function
h(k) = k mod
m
Buckets
1. 63 63 % 8 0 32
2. 92 92 % 8 1 97
3. 84 84 % 8 2 90
4. 32 32 % 8 3 91
5. 90 90 % 8 4 84
6. 69 69 % 8 5 69
7. 97 97 % 8 6 92
Tabular
representation :-

Hashing

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