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Analysis of Algorithms | Big – Θ (Big Theta) Notation

  • Last Updated : 07 Aug, 2021
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In the analysis of algorithms, asymptotic notations are used to evaluate the performance of an algorithm, in its best cases and worst cases. This article will discuss Big – Theta notations represented by a Greek letter (Θ).

Definition: Let g and f be the function from the set of natural numbers to itself. The function f is said to be Θ(g), if there are constants c1, c2 > 0 and a natural number n0 such that c1* g(n) ≤ f(n) ≤ c2 * g(n) for all n ≥ n0

Mathematical Representation:

Θ (g(n)) = {f(n): there exist positive constants c1, c2 and n0 such that 0 ≤ c1 * g(n) ≤ f(n) ≤ c2 * g(n) for all n ≥ n0}
Note: Θ(g) is a set

The above definition means, if f(n) is theta of g(n), then the value f(n) is always between c1 * g(n) and c2 * g(n) for large values of n (n ≥ n0). The definition of theta also requires that f(n) must be non-negative for values of n greater than n0



Graphical Representation:

Graphical Representation

In simple language, Big – Theta(Θ) notation specifies asymptotic bounds (both upper and lower) for a function f(n) and provides the average time complexity of an algorithm. 

Follow the steps below to find the average time complexity of any program:

  1. Break the program into smaller segments.
  2. Find all types and number of inputs and calculate the number of operations they take to be executed. Make sure that the input cases are equally distributed.
  3. Find the sum of all the calculated values and divide the sum by the total number of inputs let say the function of n obtained is g(n) after removing all the constants, then in Θ notation its represented as Θ(g(n))

Example: Consider an example to find whether a key exists in an array or not using linear search. The idea is to traverse the array and check every element if it is equal to the key or not.

The pseudo-code is as follows:

bool linearSearch(int a[], int n, int key)
{
    for (int i = 0; i < n; i++) {
        if (a[i] == key)
            return true;
    }

    return false;
}

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find whether a key exists in an
// array or not using linear search
bool linearSearch(int a[], int n, int key)
{
    // Traverse the given array, a[]
    for (int i = 0; i < n; i++) {
 
        // Check if a[i] is equal to key
        if (a[i] == key)
            return true;
    }
 
    return false;
}
 
// Driver Code
int main()
{
    // Given Input
    int arr[] = { 2, 3, 4, 10, 40 };
    int x = 10;
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
   
    if (linearSearch(arr, n, x))
        cout << "Element is present in array";
    else
        cout << "Element is not present in array";
 
    return 0;
}

Java




// Java program for the above approach
import java.lang.*;
import java.util.*;
 
class GFG{
 
// Function to find whether a key exists in an
// array or not using linear search
static boolean linearSearch(int a[], int n,
                            int key)
{
     
    // Traverse the given array, a[]
    for(int i = 0; i < n; i++)
    {
         
        // Check if a[i] is equal to key
        if (a[i] == key)
            return true;
    }
    return false;
}
 
// Driver code
public static void main(String[] args)
{
     
    // Given Input
    int arr[] = { 2, 3, 4, 10, 40 };
    int x = 10;
    int n = arr.length;
 
    // Function Call
    if (linearSearch(arr, n, x))
        System.out.println("Element is present in array");
    else
        System.out.println("Element is not present in array");
}
}
 
// This code is contributed by avijitmondal1998

Python3




# Python3 program for the above approach
 
# Function to find whether a key exists in an
# array or not using linear search
def linearSearch(a, n, key):
 
    # Traverse the given array, a[]
    for i in range(0, n):
 
        # Check if a[i] is equal to key
        if (a[i] == key):
            return True
     
    return False
 
# Driver Code
 
# Given Input
arr =  2, 3, 4, 10, 40
x = 10
n = len(arr)
 
# Function Call
if (linearSearch(arr, n, x)):
    print("Element is present in array")
else:
    print("Element is not present in array")
     
# This code is contributed by shivanisinghss2110

C#




// C# program for above approach
using System;
 
class GFG{
 
// Function to find whether a key exists in an
// array or not using linear search
static bool linearSearch(int[] a, int n,
                            int key)
{
     
    // Traverse the given array, a[]
    for(int i = 0; i < n; i++)
    {
         
        // Check if a[i] is equal to key
        if (a[i] == key)
            return true;
    }
    return false;
}
 
 
// Driver Code
static void Main()
{
    // Given Input
    int[] arr = { 2, 3, 4, 10, 40 };
    int x = 10;
    int n = arr.Length;
 
    // Function Call
    if (linearSearch(arr, n, x))
        Console.Write("Element is present in array");
    else
        Console.Write("Element is not present in array");
}
}
 
// This code is contributed by sanjoy_62.

Javascript




<script>
// JavaScript program for the above approach
// Function to find whether a key exists in an
// array or not using linear search
function linearSearch(a, n, key)
{
     
    // Traverse the given array, a[]
    for(var i = 0; i < n; i++)
    {
         
        // Check if a[i] is equal to key
        if (a[i] == key)
            return true;
    }
    return false;
}
 
// Driver code
    // Given Input
    var arr = [ 2, 3, 4, 10, 40 ];
    var x = 10;
    var n = arr.length;
 
    // Function Call
    if (linearSearch(arr, n, x))
        document.write("Element is present in array");
    else
        document.write("Element is not present in array");
 
 
// This code is contributed by shivanisinghss2110
</script>
Output
Element is present in array

In a linear search problem, let’s assume that all the cases are uniformly distributed (including the case when the key is absent in the array). So, sum all the cases (when the key is present at position 1, 2, 3, ……, n and not present, and divide the sum by n + 1. 



Average case time complexity = \frac{\sum_{i=1}^{n+1}\theta(i)}{n + 1}

⇒ \frac{\theta((n+1)*(n+2)/2)}{n+1}

⇒ \theta(1 + n/2)

⇒ \theta(n)           (constants are removed)

When to use Big – Θ notation: Big – Θ analyzes an algorithm with most precise accuracy since while calculating Big – Θ, a uniform distribution of different type and length of inputs are considered, it provides the average time complexity of an algorithm, which is most precise while analyzing, however in practice sometimes it becomes difficult to find the uniformly distributed set of inputs for an algorithm, in that case, Big-O notation is used which represent the asymptotic upper bound of a function f.

 

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