Searching algorithms

Searching
Objective : To find out the location of any given element in array. If element found then
successful, otherwise unsuccessful.
Two approaches –
1. Linear Search (Sequential Search): if array elements in random order
2. Binary Search: if array elements in sorted order
Ashish Gupta, Data Structures @ JUET, Guna

Linear Search
Algorithm: Let a is given array and no. of elements is n. Pos is representing the position of
desired element k in array a, if element found otherwise, set Pos to -1. Variable i is used as
simple variable in loop.
1. Initialize Pos = -1
2. For i = 0 to n-1
3. If a[i] == k
4. Set Pos = i+1
5. If Pos < 0
6. Print “Element is not found”
7. Else Print Pos
C1
C2
C3
C4
C5
C6
C7

Analysis of Linear Search
IDENTICAL ELEMENTS:
On successful search or unsuccessful search
Time Complexity = C1 + (n+1)*C2 + n * C3 + n* C4 + C5 + C6 +C7 = 3n + 5
= O(n)
DISTINCT ELEMENTS:
Time Complexity = C1 + (n+1)*C2 + n * C3 + C4 + C5 + C6 +C7 = 2n + 6
= O(n)
That means in worst case, at most n number of comparisons are required to find
out the desired element location.

Binary Search
 Best searching approach
Constraint:
 Given array should be in sorted order
OR
 First sort the given array and then perform search, but then sorting time will be
added

1. Initialize low =0, high = n-1
2. While low <= high
3. mid = (low+high)/2
4. If a[mid] == Item
5. Set Pos = mid
6. Break and Jump to step 10
7. Else if Item < a[mid]
8. high = mid -1
9. Else low =mid +1
10. If Pos < 0
12. Else Print Pos
Binary Search (Iterative)
Algorithm: Let a is given array which is sorted in ascending order and no. of elements is n.
Pos is representing the position of desired element Item in array a, if element found
otherwise, set Pos to -1. Variable mid is used to keep the mid index of array.

Binary Search Illustration
mid = (low+high)/2
= (0 + 9) /2 = 4
item to be searched, Item = 94
If a[mid]<Item
low = mid + 1
7 8 15 27 31 57 68 94 103 104
low = 0 high = 9mid = 4

Binary Search Illustration
mid = (low+high)/2
= (5 + 9) /2 = 7
item to be searched, Item = 94
If a[mid]==Item
Pos = mid
7 8 15 27 31 57 68 94 103 104
low = 5 high = 9mid = 7

1. Initialize low =0, high = n-1
2. While low <= high
5. Set Pos = mid
6. Break and Jump to step 10
8. high = mid -1
9. Else low =mid +1
10. If Pos < 0
12. Else Print Pos
Binary Search (Iterative)
C1
C2
C3
C4
C5
C6
C7
C8
C9
C10
C11
C12

Analysis of Iterative Binary Search
DISTINCT & SORTED ELEMENTS:
Time Complexity =
= C1 + ( log2n +1)*C2 + log2n * (C3 + C4 +C7 +C8 +C9) + C5 + C6 +C10 + C11 +C12
= 6 log2n + 7
= O (log2n)
That means in worst case, at most log2n number of comparisons are required to
find out the desired element location which is better than the linear search.

1. If low<=high
4. Set Pos = mid
5. Return Pos
7. Return Bin_Search(a,Item,low,mid-1)
8. Else Return Bin_Search(a,Item,mid+1,high)
9. Return Pos
Binary Search (Recursive)
Algorithm Bin_Search(a, Item,low, high): Let a is given array which is sorted in ascending
order and no. of elements is n. Pos is representing the position of desired element Item in
array a, if element found otherwise, set Pos to -1. Variable mid is used to keep the mid index
of array. Here, initially low and high are assigned the lowest and highest index of array a.

Recurrence Relation
Algorithm
Iterative Recursive
T(n) = Total cost of steps T(n) = Recurrence relation
Order of Complexity

Recurrence Relation (Cont…)
Let T(n) is a recurrence relation. Then,
T – a recurrence function defining a function n – parameter to function T
Therefore,
T(n) – Sequence term generated by function T for parameter n
Examples :
1.
2.
3.
T(n) =
0 for n=0
T(n-1) + n n>0
T(n) =
1 for n=0
T(n/2) + 1 n>0
T(n) =
0 for n=0
1 n=1
2T(n-1) +1 n>1

Illustration 1: Factorial Recursive
Algorithm Fact_Rec(n): Let n is number for which factorial is to be calculated.
1. If n==1
2. Return 1
3. Return Fact_Rec(n-1) * n
C1
C2
C3
Therefore,
Now,
How to solve this recurrence relation ?
T(n) =
1 for n=1
T(n-1) + 1 n>1

Solving Recurrence Relation
Substitution method – try to find the pattern
Let’s take a instance –
n = 10 i.e.
T(10) = T(9) + 1
= T(8) + 1 + 1
= T(7) + 1 + 1 +1
………..
………..
= T(1) + 1 + 8
= 10
= n
Hence, Time complexity = O(n)

Illustration 2: Fibonacci Recursive
Algorithm Fibo_Rec(n): Let n is number means nth is to be calculated.
1. If n==1
2. Return 0
3. If n==2
4. Return 1
5. Return Fibo_Rec(n-1) + Fibo_Rec(n-2)
C1
C2
C3
Therefore,
T(n) =
0 for n=1
1 n=2
T(n-1) + T(n-2) n>2

Solving Recurrence Relation
Let’s take a instance –
n = 5 i.e.
T(5) = T(4) + T(3)
= T(3) + T(2) + T(2) + T(1)
= T(2) + T(1) + 1 + 1 + 0
= 1 + 0 + 1 +1
= 3
 It grows like exponential tree.
Time complexity = O(2n)

Analysis of Binary Search Recursive
Recurrence relation –
T(n) = T(n/2) + 1
= T(n/2*2) + 1 + 1
= T(n/2*2*2) + 3
= T(n/24) + 4
…………….
= T(n/2k) + k
Let n/2k = 1, n = 2k , log2n = k * log22
i.e. k = log2n
Now,
T(n) = T(1) + log2n = 1 + log2n
= O(log2n)
T(n) =
1 for n=1
T(|_n/2_|) + 1 n>1

Summary
 Linear search with illustration
 Analysis of linear search
 Binary search (iterative) and its analysis
 Binary search (recursive) and its analysis using recurrence relation
 Discussion on recurrence relation

Searching algorithms

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