In this article, the trial division method to check whether a number is a prime or not is discussed. Given a number N, the task is to check whether the number is prime or not.

**Examples:**

Input:N = 433

Output:Prime

Explanation:

The only factors of 433 are 1 and 433. Therefore, it is a prime.

Input:N = 1263

Output:Composite

Explanation:

The factors of 1263 are 1, 3, 421, 1263. Therefore, it is a composite number.

**Naive Approach:** By definition, a prime number is a whole number greater than 1, which is only divisible by 1 and itself. Therefore, we initialize a loop from 2 to N – 1 to and check the divisibility. The following is the pseudo-code for the approach:

N <- input initialise: i <- 2 while(i ≤ N - 1): if(N % i == 0): return "Composite" return "Prime"

**Time Complexity Analysis:**

- For any given number
**N**, the while loop runs for**N – 2**times. Therefore, the time complexity for the while loop is**O(N)**. - The divisibility check is done in constant time. Therefore, the time complexity for the if condition in the while loop is
**O(1)**. - Therefore, the overall time complexity of the above approach is
**O(N)**.

**Trial Division Method:** The primality check can be performed more efficiently by the concept of trial division method. The Trial Division method is one of the crucial but one of the easiest factorization techniques when dealing with integer factorization.

**Observation:** The above method works with the observation that the maximum factor for any number N is always less than or equal to the square root(N). This conclusion can be derived in the following way:

- From the school arithmetics, it is a known fact that any composite number is built out of two or more prime numbers.
- Let the factors of
**N**be n1, n2 and so on. The factors are largest only when there exist two factors**n1**and**n2**for the number**N**. - Therefore, lets assume
**n1**and**n2**are the two largest factors for the number**N**. These numbers**n1**and**n2**can be largest only when both**n1 and n2 are equal**. - Let
**n1 = n2 = n**. Therefore,**N = n * n**. Hence, the largest possible factor for N is**square root(N)**.

**Approach:** From the above observation, the approach for this algorithm is straight forward. The idea is instead of checking till N – 1 for a factor, we only check until **square root(N)**.

Below is the implementation of the above approach:

## C++

`// CPP implementation of ` `// Trial Division Algorithm ` `#include <bits/stdc++.h> ` ` ` `using` `namespace` `std; ` ` ` `// Function to check if a number is ` `// a prime number or not ` `int` `TrialDivision(` `int` `N){ ` ` ` ` ` `// Initializing with the value 2 ` ` ` `// from where the number is checked ` ` ` `int` `i = 2; ` ` ` ` ` `// Computing the square root of ` ` ` `// the number N ` ` ` `int` `k = ` `ceil` `(` `sqrt` `(N)); ` ` ` ` ` `// While loop till the ` ` ` `// square root of N ` ` ` `while` `(i<= k){ ` ` ` ` ` `// If any of the numbers between ` ` ` `// [2, sqrt(N)] is a factor of N ` ` ` `// Then the number is composite ` ` ` `if` `(N % i == 0) ` ` ` `return` `0; ` ` ` `i += 1; ` ` ` `} ` ` ` ` ` `// If none of the numbers is a factor, ` ` ` `// then it is a prime number ` ` ` `return` `1; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `N = 49; ` ` ` `int` `p = TrialDivision(N); ` ` ` ` ` `// To check if a number is a prime or not ` ` ` `if` `(p) ` ` ` `cout << (` `"Prime"` `); ` ` ` `else` ` ` `cout << (` `"Composite"` `); ` ` ` ` ` `return` `0; ` `} ` ` ` `// This code is contributed by mohit kumar 29 ` |

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## Java

`// Java implementation of ` `// Trial Division Algorithm ` `import` `java.util.*; ` ` ` `class` `GFG{ ` ` ` `// Function to check if a number is ` `// a prime number or not ` `static` `int` `TrialDivision(` `int` `N){ ` ` ` ` ` `// Initializing with the value 2 ` ` ` `// from where the number is checked ` ` ` `int` `i = ` `2` `; ` ` ` ` ` `// Computing the square root of ` ` ` `// the number N ` ` ` `int` `k =(` `int` `) Math.ceil(Math.sqrt(N)); ` ` ` ` ` `// While loop till the ` ` ` `// square root of N ` ` ` `while` `(i<= k){ ` ` ` ` ` `// If any of the numbers between ` ` ` `// [2, sqrt(N)] is a factor of N ` ` ` `// Then the number is composite ` ` ` `if` `(N % i == ` `0` `) ` ` ` `return` `0` `; ` ` ` `i += ` `1` `; ` ` ` `} ` ` ` ` ` `// If none of the numbers is a factor, ` ` ` `// then it is a prime number ` ` ` `return` `1` `; ` `} ` ` ` `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ` ` ` `int` `N = ` `49` `; ` ` ` `int` `p = TrialDivision(N); ` ` ` ` ` `// To check if a number is a prime or not ` ` ` `if` `(p != ` `0` `) ` ` ` `System.out.print(` `"Prime"` `); ` ` ` `else` ` ` `System.out.print(` `"Composite"` `); ` ` ` `} ` `} ` ` ` `// This code is contributed by shivanisinghss2110 ` |

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## Python3

`# Python3 implementation of ` `# Trial Division Algorithm ` ` ` `# Function to check if a number is ` `# a prime number or not ` `def` `TrialDivision(N): ` ` ` ` ` `# Initializing with the value 2 ` ` ` `# from where the number is checked ` ` ` `i ` `=` `2` ` ` ` ` `# Computing the square root of ` ` ` `# the number N ` ` ` `k ` `=` `int` `(N ` `*` `*` `0.5` `) ` ` ` ` ` `# While loop till the ` ` ` `# square root of N ` ` ` `while` `(i<` `=` `k): ` ` ` ` ` `# If any of the numbers between ` ` ` `# [2, sqrt(N)] is a factor of N ` ` ` `# Then the number is composite ` ` ` `if` `(N ` `%` `i ` `=` `=` `0` `): ` ` ` `return` `0` ` ` `i ` `+` `=` `1` ` ` ` ` `# If none of the numbers is a factor, ` ` ` `# then it is a prime number ` ` ` `return` `1` ` ` `# Driver code ` `if` `__name__ ` `=` `=` `"__main__"` `: ` ` ` `N ` `=` `49` ` ` `p ` `=` `TrialDivision(N) ` ` ` `# To check if a number is a prime or not ` ` ` `if` `(p): ` ` ` `print` `(` `"Prime"` `) ` ` ` `else` `: ` ` ` `print` `(` `"Composite"` `) ` ` ` |

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## C#

`// C# implementation of ` `// Trial Division Algorithm ` `using` `System; ` ` ` `class` `GFG{ ` ` ` `// Function to check if a number is ` `// a prime number or not ` `static` `int` `TrialDivision(` `int` `N){ ` ` ` ` ` `// Initializing with the value 2 ` ` ` `// from where the number is checked ` ` ` `int` `i = 2; ` ` ` ` ` `// Computing the square root of ` ` ` `// the number N ` ` ` `int` `k =(` `int` `) Math.Ceiling(Math.Sqrt(N)); ` ` ` ` ` `// While loop till the ` ` ` `// square root of N ` ` ` `while` `(i<= k){ ` ` ` ` ` `// If any of the numbers between ` ` ` `// [2, sqrt(N)] is a factor of N ` ` ` `// Then the number is composite ` ` ` `if` `(N % i == 0) ` ` ` `return` `0; ` ` ` `i += 1; ` ` ` `} ` ` ` ` ` `// If none of the numbers is a factor, ` ` ` `// then it is a prime number ` ` ` `return` `1; ` `} ` ` ` `// Driver Code ` `public` `static` `void` `Main() ` `{ ` ` ` ` ` `int` `N = 49; ` ` ` `int` `p = TrialDivision(N); ` ` ` ` ` `// To check if a number is a prime or not ` ` ` `if` `(p != 0) ` ` ` `Console.Write(` `"Prime"` `); ` ` ` `else` ` ` `Console.Write(` `"Composite"` `); ` ` ` `} ` `} ` ` ` `// This codee is contributed by AbhiThakur ` |

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**Output:**

Composite

**Time Complexity Analysis:**

- The while loop is executed for a maximum of square root(N) times. Therefore, the time complexity of the while loop is
**O(sqrt(N))**. - The running time of all the if conditions are constant. Therefore, the time complexity of the if statements are
**O(1)**. - Therefore, overall time complexity is
**O(sqrt(N))**.

**Optimised Trial Division Method:** The above trial division method can be further optimized by eliminating all even numbers in the range **[2, K]** where K = square root(N) as 2 is the only even prime number. The overall complexity still remains the same but the number of executions gets reduced by half.

**Note:** The optimization made in the Trial Division method might seem very small as this method is almost similar to Naive Approach except the number of iterations. However, this drastically reduces the number of computations for higher values of N. This is explained by the following graph plotted against the corresponding running times of the algorithms:

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