Introduction to Hashing in Data Structure using C++

CENG 213 Data Structures 1
Hashing

Hash Tables
• We’ll discuss the hash table ADT which supports only
a subset of the operations allowed by binary search
trees.
• The implementation of hash tables is called hashing.
• Hashing is a technique used for performing insertions,
deletions and finds in constant average time (i.e. O(1))
• This data structure, however, is not efficient in
operations that require any ordering information among
the elements, such as findMin, findMax and printing the
entire table in sorted order.

General Idea
• The ideal hash table structure is merely an array of some fixed
size, containing the items.
• A stored item needs to have a data member, called key, that will
be used in computing the index value for the item.
– Key could be an integer, a string, etc
– e.g. a name or Id that is a part of a large employee structure
• The size of the array is TableSize.
• The items that are stored in the hash table are indexed by values
from 0 to TableSize – 1.
• Each key is mapped into some number in the range 0 to
TableSize – 1.
• The mapping is called a hash function.

Example
Hash
Function
mary 28200
dave 27500
phil 31250
john 25000
Items
Hash
Table
key
key
0
1
2
3
4
5
6
7
8
9
mary 28200
dave 27500
phil 31250
john 25000

Hash Function
• The hash function:
– must be simple to compute.
– must distribute the keys evenly among the cells.
• If we know which keys will occur in
advance we can write perfect hash
functions, but we don’t.

Hash function
Problems:
• Keys may not be numeric.
• Number of possible keys is much larger than the
space available in table.
• Different keys may map into same location
– Hash function is not one-to-one => collision.
– If there are too many collisions, the performance of
the hash table will suffer dramatically.

Hash Functions
• If the input keys are integers then simply
Key mod TableSize is a general strategy.
– Unless key happens to have some undesirable
properties. (e.g. all keys end in 0 and we use
mod 10)
• If the keys are strings, hash function needs
more care.
– First convert it into a numeric value.

Some methods
• Truncation:
– e.g. 123456789 map to a table of 1000 addresses by
picking 3 digits of the key.
• Folding:
– e.g. 123|456|789: add them and take mod.
• Key mod N:
– N is the size of the table, better if it is prime.
• Squaring:
– Square the key and then truncate
• Radix conversion:
– e.g. 1 2 3 4 treat it to be base 11, truncate if necessary.

Hash Function 1
• Add up the ASCII values of all characters of the key.
int hash(const string &key, int tableSize)
{
int hasVal = 0;
for (int i = 0; i < key.length(); i++)
hashVal += key[i];
return hashVal % tableSize;
}
• Simple to implement and fast.
• However, if the table size is large, the function does not
distribute the keys well.
• e.g. Table size =10000, key length <= 8, the hash function can
assume values only between 0 and 1016

Hash Function 2
• Examine only the first 3 characters of the key.
int hash (const string &key, int tableSize)
{
return (key[0]+27 * key[1] + 729*key[2]) % tableSize;
}
• In theory, 26 * 26 * 26 = 17576 different words can be
generated. However, English is not random, only 2851
different combinations are possible.
• Thus, this function although easily computable, is also not
appropriate if the hash table is reasonably large.

Hash Function 3
int hash (const string &key, int tableSize)
{
int hashVal = 0;
for (int i = 0; i < key.length(); i++)
hashVal = 37 * hashVal + key[i];
hashVal %=tableSize;
if (hashVal < 0) /* in case overflows occurs */
hashVal += tableSize;
return hashVal;
};







1
0
37
]
1
[
)
(
KeySize
i
i
i
KeySize
Key
key
hash

Hash function for strings:
a l i
key
KeySize = 3;
98 108 105
hash(“ali”) = (105 * 1 + 108*37 + 98*372
) % 10,007 = 8172
0 1 2 i
key[i]
hash
function
ali
……
……
0
1
2
8172
10,006 (TableSize)
“ali”

Collision Resolution
• If, when an element is inserted, it hashes to the
same value as an already inserted element, then we
have a collision and need to resolve it.
• There are several methods for dealing with this:
– Separate chaining
– Open addressing
• Linear Probing
• Quadratic Probing
• Double Hashing

Separate Chaining
• The idea is to keep a list of all elements that hash to
the same value.
– The array elements are pointers to the first nodes of the
lists.
– A new item is inserted to the front of the list.
• Advantages:
– Better space utilization for large items.
– Simple collision handling: searching linked list.
– Overflow: we can store more items than the hash table
size.
– Deletion is quick and easy: deletion from the linked list.

Example
0
1
2
3
4
5
6
7
8
9
0
81 1
64 4
25
36 16
49 9
Keys: 0, 1, 4, 9, 16, 25, 36, 49, 64, 81
hash(key) = key % 10.

Operations
• Initialization: all entries are set to NULL
• Find:
– locate the cell using hash function.
– sequential search on the linked list in that cell.
• Insertion:
– Locate the cell using hash function.
– (If the item does not exist) insert it as the first item in the
list.
• Deletion:
– Locate the cell using hash function.
– Delete the item from the linked list.

Hash Table Class for separate chaining
template <class HashedObj>
class HashTable
{
public:
HashTable(const HashedObj & notFound, int size=101 );
HashTable( const HashTable & rhs )
:ITEM_NOT_FOUND( rhs.ITEM_NOT_FOUND ),
theLists( rhs.theLists ) { }
const HashedObj & find( const HashedObj & x ) const;
void makeEmpty( );
void insert( const HashedObj & x );
void remove( const HashedObj & x );
const HashTable & operator=( const HashTable & rhs );
private:
vector<List<HashedObj> > theLists; // The array of
Lists
const HashedObj ITEM_NOT_FOUND;
};
int hash( const string & key, int tableSize );
int hash( int key, int tableSize );

Insert routine
/**
* Insert item x into the hash table. If the item is
* already present, then do nothing.
*/
void HashTable<HashedObj>::insert(const HashedObj & x )
{
List<HashedObj> & whichList = theLists[ hash( x,
theLists.size( ) ) ];
ListItr<HashedObj> itr = whichList.find( x );
if( !itr.isValid() )
whichList.insert( x, whichList.zeroth( ) );
}

Remove routine
/**
* Remove item x from the hash table.
*/
void HashTable<HashedObj>::remove( const HashedObj & x )
{
theLists[hash(x, theLists.size())].remove( x );
}

Find routine
/**
* Find item x in the hash table.
* Return the matching item or ITEM_NOT_FOUND if not found
*/
const HashedObj & HashTable<HashedObj>::find( const
HashedObj & x ) const
{
ListItr<HashedObj> itr;
itr = theLists[ hash( x, theLists.size( ) ) ].find( x );
if(!itr.isValid())
return ITEM_NOT_FOUND;
else
return itr.retrieve( );
}

Analysis of Separate Chaining
• Collisions are very likely.
– How likely and what is the average length of
lists?
• Load factor  definition:
– Ratio of number of elements (N) in a hash table
to the hash TableSize.
• i.e.  = N/TableSize
– The average length of a list is also 
– For chaining  is not bound by 1; it can be > 1.

Cost of searching
• Cost = Constant time to evaluate the hash function
+ time to traverse the list.
• Unsuccessful search:
– We have to traverse the entire list, so we need to compare  nodes on
the average.
• Successful search:
– List contains the one node that stores the searched item + 0 or more
other nodes.
– Expected # of other nodes = x = (N-1)/M which is essentially
since M is presumed large.
– On the average, we need to check half of the other nodes while
searching for a certain element
– Thus average search cost = 1 +/2

Summary
• The analysis shows us that the table size is
not really important, but the load factor is.
• TableSize should be as large as the number
of expected elements in the hash table.
– To keep load factor around 1.
• TableSize should be prime for even
distribution of keys to hash table cells.

Hashing: Open Addressing

Collision Resolution with
Open Addressing
• Separate chaining has the disadvantage of
using linked lists.
– Requires the implementation of a second data
structure.
• In an open addressing hashing system, all
the data go inside the table.
– Thus, a bigger table is needed.
• Generally the load factor should be below 0.5.
– If a collision occurs, alternative cells are tried
until an empty cell is found.

Open Addressing
• More formally:
– Cells h0(x), h1(x), h2(x), …are tried in succession where
hi(x) = (hash(x) + f(i)) mod TableSize, with f(0) = 0.
– The function f is the collision resolution strategy.
• There are three common collision resolution
strategies:
– Linear Probing
– Quadratic probing
– Double hashing

Linear Probing
• In linear probing, collisions are resolved by
sequentially scanning an array (with
wraparound) until an empty cell is found.
– i.e. f is a linear function of i, typically f(i)= i.
• Example:
– Insert items with keys: 89, 18, 49, 58, 9 into an
empty hash table.
– Table size is 10.
– Hash function is hash(x) = x mod 10.
• f(i) = i;

Figure 20.4
Linear probing
hash table after
each insertion

Find and Delete
• The find algorithm follows the same probe
sequence as the insert algorithm.
– A find for 58 would involve 4 probes.
– A find for 19 would involve 5 probes.
• We must use lazy deletion (i.e. marking
items as deleted)
– Standard deletion (i.e. physically removing the
item) cannot be performed.
– e.g. remove 89 from hash table.

Clustering Problem
• As long as table is big enough, a free cell
can always be found, but the time to do so
can get quite large.
• Worse, even if the table is relatively empty,
blocks of occupied cells start forming.
• This effect is known as primary clustering.
• Any key that hashes into the cluster will
require several attempts to resolve the
collision, and then it will add to the cluster.

Analysis of insertion
• The average number of cells that are examined in
an insertion using linear probing is roughly
(1 + 1/(1 – λ)2
) / 2
• Proof is beyond the scope of text book.
• For a half full table we obtain 2.5 as the average
number of cells examined during an insertion.
• Primary clustering is a problem at high load
factors. For half empty tables the effect is not
disastrous.

Analysis of Find
• An unsuccessful search costs the same as
insertion.
• The cost of a successful search of X is equal to the
cost of inserting X at the time X was inserted.
• For λ = 0.5 the average cost of insertion is 2.5.
The average cost of finding the newly inserted
item will be 2.5 no matter how many insertions
follow.
• Thus the average cost of a successful search is an
average of the insertion costs over all smaller load
factors.

Average cost of find
• The average number of cells that are examined in
an unsuccessful search using linear probing is
roughly (1 + 1/(1 – λ)2
) / 2.
• The average number of cells that are examined in a
successful search is approximately
(1 + 1/(1 – λ)) / 2.
– Derived from:
dx
x













 0
x
2
)
1
(
1
1
2
1
1

Linear Probing – Analysis -- Example
• What is the average number of probes for a successful
search and an unsuccessful search for this hash table?
– Hash Function: h(x) = x mod 11
Successful Search:
– 20: 9 -- 30: 8 -- 2 : 2 -- 13: 2, 3 -- 25: 3,4
– 24: 2,3,4,5 -- 10: 10 -- 9: 9,10, 0
Avg. Probe for SS = (1+1+1+2+2+4+1+3)/8=15/8
Unsuccessful Search:
– We assume that the hash function uniformly
distributes the keys.
– 0: 0,1 -- 1: 1 -- 2: 2,3,4,5,6 -- 3: 3,4,5,6
– 4: 4,5,6 -- 5: 5,6 -- 6: 6 -- 7: 7 -- 8: 8,9,10,0,1
– 9: 9,10,0,1 -- 10: 10,0,1
Avg. Probe for US =
(2+1+5+4+3+2+1+1+5+4+3)/11=31/11
0 9
1
2 2
3 13
4 25
5 24
6
7
8 30
9 20
10 10

Quadratic Probing
• Quadratic Probing eliminates primary clustering
problem of linear probing.
• Collision function is quadratic.
– The popular choice is f(i) = i2
.
• If the hash function evaluates to h and a search in cell
h is inconclusive, we try cells h + 12
, h+22
, … h + i2
.
– i.e. It examines cells 1,4,9 and so on away from the
original probe.
• Remember that subsequent probe points are a
quadratic number of positions from the original
probe point.

Figure 20.6
A quadratic
probing hash
table after each
insertion (note
that the table size
was poorly
chosen because
it is not a prime
number).

Quadratic Probing
• Problem:
– We may not be sure that we will probe all locations in
the table (i.e. there is no guarantee to find an empty cell
if table is more than half full.)
– If the hash table size is not prime this problem will be
much severe.
• However, there is a theorem stating that:
– If the table size is prime and load factor is not larger
than 0.5, all probes will be to different locations and an
item can always be inserted.

Theorem
• If quadratic probing is used, and the table
size is prime, then a new element can
always be inserted if the table is at least half
empty.

Proof
• Let M be the size of the table and it is prime. We show that the first
M/2 alternative locations are distinct.
• Let two of these locations are h + i2
and h + j2
, where i, j are two
probes s.t. 0 i,j  M/2. Suppose for the sake of contradiction, that
these two locations are the same but
i  j. Then
h + i2
= h + j2
(mod M)
i2
= j2
(mod M)
i2
- j2
= 0 (mod M)
(i-j)(i+j) = 0 (mod M)
• Because M is prime, either (i-j) or (i+j) is divisible by M. Neither of
these possibilities can occur. Thus we obtain a contradiction.
• It follows that the first M/2 alternative are all distinct and since there
are at most M/2 items in the hash table it is guaranteed that an
insertion must succeed if the table is at least half full.

Some considerations
• How efficient is calculating the quadratic
probes?
– Linear probing is easily implemented.
Quadratic probing appears to require * and %
operations.
– However by the use of the following trick, this
is overcome:
• Hi = Hi-1+2i – 1 (mod M)

Some Considerations
• What happens if load factor gets too high?
– Dynamically expand the table as soon as the
load factor reaches 0.5, which is called
rehashing.
– Always double to a prime number.
– When expanding the hash table, reinsert the
new table by using the new hash function.

Analysis of Quadratic Probing
• Quadratic probing has not yet been
mathematically analyzed.
• Although quadratic probing eliminates primary
clustering, elements that hash to the same location
will probe the same alternative cells. This is know
as secondary clustering.
• Techniques that eliminate secondary clustering are
available.
– the most popular is double hashing.

Double Hashing
• A second hash function is used to drive the collision
resolution.
– f(i) = i * hash2(x)
• We apply a second hash function to x and probe at a
distance hash2(x), 2*hash2(x), … and so on.
• The function hash2(x) must never evaluate to zero.
– e.g. Let hash2(x) = x mod 9 and try to insert 99 in the
previous example.
• A function such as hash2(x) = R – ( x mod R) with R
a prime smaller than TableSize will work well.
– e.g. try R = 7 for the previous example.(7 - x mode 7)

The relative efficiency of
four collision-resolution methods

Hashing Applications
• Compilers use hash tables to implement the
symbol table (a data structure to keep track
of declared variables).
• Game programs use hash tables to keep
track of positions it has encountered
(transposition table)
• Online spelling checkers.

Summary
• Hash tables can be used to implement the insert
and find operations in constant average time.
– it depends on the load factor not on the number of items
in the table.
• It is important to have a prime TableSize and a
correct choice of load factor and hash function.
• For separate chaining the load factor should be
close to 1.
• For open addressing load factor should not exceed
0.5 unless this is completely unavoidable.
– Rehashing can be implemented to grow (or shrink) the
table.

Introduction to Hashing in Data Structure using C++

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