unit viii searching and hashing

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Unit – VIII Searching and Hashing

Prepared By:Dabbal Singh Mahara

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• Introduction• Sequential search• Binary search• Comparison and efficiency of searching• Hashing

probing (Linear and Quadratic)

Contents

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Introduction• Searching is a process of finding an element within the list of elements

stored in any order or randomly. • Searching is divided into two categories Linear and Binary search.• Linear search

Small arrays Unsorted arrays

• Binary search Large arrays Sorted arrays

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• In linear search, access each element of an array one by one sequentially and see whether it is desired element or not. A search will be unsuccessful if all the elements are accessed and the desired element is not found.

• In brief, Simply search for the given element left to right and return the index of the element, if found. Otherwise return “Not Found”.

Algorithm:

LinearSearch(A, n,key){

for(i=0;i<n;i++){

if(A[i] == key) return i;}return -1; //-1 indicates unsuccessful search

}

Analysis: Time complexity = O(n)

Linear Search

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Binary Search • Binary search is an extremely efficient algorithm. • This search technique searches the given item in minimum possible comparisons.• To do this binary search, first we need to sort the a elements. The logic behind this

technique is given below:i. First find the middle element of the arrayii. Compare the middle element with an item.iii. There are three cases:

a. If it is a desired element then search is successfulb. If it is less than desired item then search only the first half of the

array.c. If it is greater than the desired element, search in the second half of

the array.• Repeat the same process until element is found or exhausts in the search area.• In this algorithm every time we are reducing the search area.

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Iterative Algorithm

BinarySearch(A, l, r, key){

while(l<=r){

m = (l + r) /2 ; //integer divisionif(key = = A[m]

print " Search successful"else if (key < A[m])

r = m - 1else

l = m+1}If(l>r)

print "unsuccessful search"}

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Algorithm: Recursive BinarySearch(A, l, r, key){

if(l= = r) //only one element{

if(key = = A[l]) print " successful Search"

else print "unsuccessful Search"

}else{

m = (l + r) /2 ; //integer divisionif(key = = A[m]

print "successful search"else if (key < A[m])

return BinarySearch(l, m-1, key) ;else

return BinarySearch(m+1, r, key) ;}

}

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Binary Search Example

• Ex. Binary search for 33.

641413 25 33 5143 53 8472 93 95 97966

low high

Example Tracing

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Running example: Take input array a[]

For Search key = 2l r mid remarks0 13 6 Key < a[6] i.e. 2 < 530 5 2 Key < a[2] i.e. 2 < 70 1 0 Key == a[0] i.e. 2 ==a[0]

Therefore, key found at index 0. Search Successful !!

2 5 7 9 18 45 53 59 67 72 88 95 101 1040 1 2 3 4 5 6 7 8 9 10 11 12 13

Exercise : Trace binary search algorithm for keys:i. 67ii. 50iii. 250

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Search for key = 67l r mid Remarks

0 13 6 Key < a[6] i.e. 67 > 537 13 10 Key < a[10] i.e. 67 < 887 9 8 Key == a[8] i.e. 67 ==a[8]Therefore, key found at index 8. Search Successful !!

2 5 7 9 18 45 53 59 67 72 88 95 101 1040 1 2 3 4 5 6 7 8 9 10 11 12 13

Input Array : a[ ]

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Search for key = 50l r mid Remarks

0 13 6 Key < a[6] i.e. 50 < 530 5 2 Key < a[2] i.e. 50 > 73 5 4 Key > a[4] i.e. 50 >185 5 5 Key > a[5] i.e. 50 > 456 5 l > r, terminateTherefore, key not found in the array. Search Unsuccessful !!

2 5 7 9 18 45 53 59 67 72 88 95 101 1040 1 2 3 4 5 6 7 8 9 10 11 12 13

Given Input Array a[]

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Efficiency:

From the above algorithm we can say that the running time of the algorithm is:

T(n) = T(n/2) + Ο(1) = Ο(log n)

In the best case output is obtained at one run i.e. Ο(1) time if the key is at middle.

In the worst case the output is at the end of the array, So running time is Ο(log n)

In the average case also running time is Ο(log n).

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Introduction to Hashing• Suppose that we want to store 10,000 students records (each with a 5-digit ID) in a

given container.

A linked list implementation would take O(n) time.

A height balanced tree would give O(log n) access time.

Using an array of size 100,000 would give O(1) access time but will lead

to a lot of space wastage.

• Is there some way that we could get O(1) access without wasting a lot of space?

• The answer is hashing.

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Introduction to Hashing

Hashing is a technique used for performing insertions, deletions and finds in constant average time O(1).

The techniques employed here is to compute location of desired record to retrieve it in a single access or comparison.

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.

Applications

• Database systems• Symbol table for compilers• Data Dictionaries• Browser caches

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The ideal hash table structure is 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.

Hash Table

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Example

HashFunction

mary 28200

dave 27500

phil 31250

john 25000Items

Hash Table

key

key

0123

456789

mary 28200dave 27500

phil 31250john 25000

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Hash Functions (cont’d)• A hash function, h, is a function which transforms a key from a set, K,

into an index in a table of size n:

h: K -> {0, 1, ..., n-2, n-1}

• A key can be a number, a string, a record etc.

• The size of the set of keys, |K|, to be relatively very large.

• It is possible for different keys to hash to the same array location. This situation is called collision and the colliding keys are called synonyms.

• A common hash function is

h(x)=x mod SIZE

• if key=27 and SIZE=10 then

hash address=27%10=7

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• A good hash function should:· Minimize collisions. · Be easy and quick to compute.· Distribute key values evenly in the hash table.· Use all the information provided in the key.

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Load Factor of a Hash Table

• Load factor of a hash table T:

= n/m– n = number of elements stored in the table

– m = number of slots in the table

• encodes the average number of elements

stored in a chain

• can be <, =, > 1

0

m - 1

T

chainchain

chain

chain

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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. i.e. For any two keys k1 and k2, H(k1) = H(k2) = β

• There are several methods for dealing with this:– Separate chaining– Open addressing

• Linear Probing• Quadratic Probing• Double Hashing

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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.

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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.

Exercise: Represent the keys {89, 18, 49, 58, 69, 78} in hash table using separate chaining.

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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.

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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.

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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

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Linear Probing• In linear probing, collisions are resolved by sequentially

scanning an array (with wraparound) until an empty cell is found.

• hi(x) = (hash(x) + f(i)) mod TableSize– i.e. f is a linear function of i, typically f(i)= i.

Example: Insert keys {89, 18, 49, 58, 69, 78} with the hash function: h(x)=x mod 10 using linear probing. Use table size 10. when x=89:

h(89)=89%10=9insert key 89 in hash-table in location 9

when x=18:h(18)=18%10=8insert key 18 in hash-table in location 8

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when x=49:h(49)=49%10=9 (Collision )so insert key 49 in hash-table in next

possible vacant location of 9 is 0

when x=58:h(58)=58%10=8 (Collision)insert key 58 in hash-table in next

possible vacant location of 8 is 1 (since 9, 0 already contains values).

when x=69:h(89)=69%10=9 (Collision )insert key 69 in hash-table in next

possible vacant location of 9 is 2 (since 0, 1 already contains values).

when x = 78h(78) = 78 % 10 = 8 ( Collision )search next vacant slot in the tablewhich is 3 (since 0,1,2 contain values)insert 78 at location 3.

0 49

1 58

2 69

3 78

4

5

6

7

8 18

9 89Fig. Hash table with keys

Using linear probing

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Disadvantage of linear probing is :Primary 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.

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Quadratic probing is a collision resolution method that eliminates the primary clustering problem take place in a linear probing.

Compute: Hash value = h(x) = x % table size

When collision occur then the quadratic probing works as follows:

(Hash value + 12)% table size,

if there is again collision occur then there exist rehashing.

(hash value + 22)%table size

if there is again collision occur then there exist rehashing.

(hash value = 32)% table size

In general in ith collision

hi(x)=(hash value +i2)%size

Quadratic Probing:

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solution:when x=89:

h(89)=89%10=9insert key 89 in hash-table in location 9

when x=18:h(18)=18%10=8insert key 18 in hash-table in location 8

when x=49:h(49)=49%10=9 (Collision )so use following hash function,h1(49)=(9 + 1)%10=0hence insert key 49 in hash-table in location 0

when x=58:h(58)=58%10=8 (Collision )so use following hash function,h1(58)=(8 + 1)%10=9again collision occur use again the following hash function ,h2(58)=(8+ 22)%10=2insert key 58 in hash-table in location 2

Example: Insert keys {89, 18, 49, 58, 69 78} with the hash-table size 10 using quadratic probing.

0 49

1

2 58

3 69

4

5

6

7 78

8 18

9 89Fig. Hash table with keys Using quadratic probing

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when x=69:h(89)=69%10=9 (Collision )so use following hash function,h1(69)=(9 + 1)%10=0again collision occurs use again the following hash function ,h2(69)=(9+ 22)%10=3insert key 69 in hash-table in location 3

when x=78:h(78)=78%10=8 (Collision )so use following hash function,h1(78)=(8 + 1)%10=9 ; again collision occursuse again the following hash function ,h2(78)=(8+ 22)%10=2 ; again collision occurs, compute following step h3(78)=(8+ 32)%10=7insert key 58 in hash-table in location 7

• 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.

• In above example: for keys 58 and 78 both follow the path 8, 9, 7 …• Techniques that eliminate secondary clustering are available,

the most popular is double hashing.

Quadratic Probing Problem

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Double Hashing To eliminate both types of clustering one way is double hashing. It involves two hash functions, h1(x) and h2(x),

where h1(x) is primary hash function, is first used to determine position of key and if it is occupied h2(x) is used.

Example: h1(x) = x % TABLESIZE h2(x) = R – (x % R), Where R is prime less than table size

hi(x) = h1 (x) + i.h2(x) ) % TABLESIZE Example: Insert keys {89, 18, 49, 58, 69 78} with the hash-

table size 10 using double hashing.

solution: when x=89:

h(89)=89%10=9insert key 89 in hash-table in location 9

when x=18:h(18)=18%10=8insert key 18 in hash-table in location 8

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when x=49:h(49)=49%10=9 (Collision )so use following hash function,h1(49)=(9 + 1(7- 49%7))%10

= (9 + (7-0) ) % 10 = 6hence insert key 49 in hash-table in location 6.

when x=58:h(58)=58%10=8 (Collision )so use following hash function,h1(58) = (8 + 1(7-(58%7))%10 =(8 + (7-2))% 10 =3INSERT 58 in the location 3.

Compute the location for keys: 69 78

0

1

2

3 58

4

5

6 49

7

8 18

9 89Fig. Hash table with keys

Using double hashing

Limitation: It takes extra time to compute hash function.

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Thank You !