Isprime c что это
Given a positive integer, check if the number is prime or not. A prime is a natural number greater than 1 that has no positive divisors other than 1 and itself.
Examples:
Input: n = 11
Output: trueInput: n = 15
Output: false
Input: n = 1
Output: false
Naive Approach:
Time Complexity: O(n)
Auxiliary Space: O(1)
Naive Approach (Optimised):
We need to check factors upto √n not till n. The resaon is suppose n has 2 factors and both are bigger than √n. Then n would be bigger than n , which is absurd! So n has at least one factor smaller than √n if it isn’t prime.
Below is the implementation of the above idea:
Time Complexity: O(sqrt(n))
Auxiliary Space: O(1)
Optimized School Method:
Time Complexity: O(sqrt(n))
Auxiliary Space: O(1)
Please refer complete article on Primality Test | Set 1 (Introduction and School Method) for more details!
Using Wilson’s theorem:
Given a number N, the task is to check if it is prime or not using Wilson Primality Test. Print ‘1’ if the number is prime, else print ‘0’.
Wilson’s theorem states that a natural number p > 1 is a prime number if and only if
Prime Number Program in C
Prime Number is an utmost important topic in number theory. They have a wide range of applications in real world like in cryptography, basically prime numbers are heart of cryptography.
In the Rabin-Karp string matching algorithm, prime numbers are used to hash the string into integers.
In this article, we will understand prime numbers in depth and will learn how to check whether a number is prime or not.
Scope
In this article, we will understand what prime numbers are and how to check if a number is prime or not.
We will look at multiple algorithms to check if a number is prime or not, from naive to the most efficient one.
What is Prime Number?
Prime Number is a positive integer greater than 1, which is divisible by 1 and itself. In other words, it has only 2 factors 1 and itself.
For example: 2 (it has only 2 factors, 1 and 2), 3 (it has only 2 factors, 1 and 3), 5 , 7 , 11 etc.
Note: 2 is the smallest prime number.
How to Check if a Number is Prime?
There are two methods to check if a number is prime or not :
1. Simple Algorithm
In this method, we will simply divide the given number n with every number from 1 to n (1 and n are included) and keep the count of numbers by which n is divisible. At last, we check if the count is exactly equals to 2, then n is prime otherwise not. For example :
n = 8 ,
1 divides 8
2 divides 8
4 divides 8
8 divides 8
so count is 4, therefore 8 is not prime.
n = 5 ,
1 divides 5
5 divides 5
so count is 2 , therefore 5 is prime .
This algorithm will take O(n) time to check whether a number is prime or not because we run a loop from 1 to n (1 and n inclusive).
Note: A little optimisation that we can do in the above approach is that we can loop from 2 to n/2 (2 and n/2 inclusive) and check if n is divisible by any number in the range [2,n/2] , if n is divisible then n is not prime otherwise prime.
2. Efficient Algorithm
This algorithm is based on the fact that if a number n is not prime, then n has at least one factor (greater than 1) which lies in the range 2 to sqrt(n) i.e [2, sqrt(n)] .
To be more clear, all divisors of a number n occur in pairs (a,b) such that b = n/a . eg. let's say n = 12 divisors of n = <1, 2, 3, 4, 6, 12>pairs =
So using the above observation, we can say that instead of looping from 1 to 12 , we can loop from to 2 to 3 to check if 12 is prime or not and this 3 is basically sqrt(12) = 3 .
So, in general to check if a number n is prime or not, run a loop from 2 to sqrt(n) and if n is divisible by any number in the range [2,sqrt(n)] then n is not prime otherwise prime.
For example: n = 36 -> sqrt(36) = 6
Now, if n has any factor in the range [2,6] then n is not prime. 2 lies in [2,6] and 2 divides 36 i.e 36 % 2 == 0 therefore 36 is not prime.
n = 5 -> sqrt(5) = 2 (taking floor value)
Now, if n has any factor in the range [2,2] then n is not prime. 2 lies in [2,2] but 2 does not divides 5 i.e 5 % 2 != 0 Therefore 5 is prime.
C Program to Check Prime Number
1. Prime number program in C Using Loops and Functions
In this method, we will follow the efficient algorithm as explained above. We are denoting this method as using functions because we are passing n (by value) to a function named isPrime(n) .
Steps:
If n is less than 2 then return 0.
Run loop from 2 to sqrt(n) and check if n is divisible by any number, if yes then return 0 showing n is not prime.
Otherwise, return 1 showing n is prime.
Output:
Since 5 has no divisors in the range [2,sqrt(5)] or [2,2] , therefore 5 is prime.
Time Complexity: O(sqrt n) because the loop runs from 2 to sqrt(n) .
Space Complexity: O(1) since we are not using any extra space.
2. Prime number program in C Using Pointers
In this method, we will follow the efficient algorithm as explained above.
We are denoting this method as using pointers because we are passing n(by reference) to a function named isPrime.
Steps:
- If n is less than 2 then return 0.
- Run loop from 2 to sqrt(n) and check if n is divisible by any number, if yes then return 0 showing n is not prime.
- Otherwise, return 1 showing n is prime.
Output:
Since 5 has no divisors in the range [2,sqrt(5)] or [2,2] , therefore 5 is prime.
Time Complexity: O(sqrt n) because the loop runs from 2 to sqrt(n).
Space Complexity: O(1) since we are not using any extra space.
3. Prime number program in C Using Recursion
In this method we will follow the efficient algorithm as explained above.
We are denoting this method as using recursion because we will use a recursive implementation to check if a number is prime.
Steps:
- If number n<=1 then it is not prime.
- Call the recursive function and pass n and sqrt(n) as arguments namely n and i .
- If i == 1 (base case) that means we explored all the numbers from [2, sqrt(n)] and we did not find any divisor of n therefore, we will return 1.
- Check if n is divisible by i , if yes then n is not prime, simply return 0 otherwise call the recursive function again with n and i — 1 as arguments.
Output :
Since 5 has no divisors in the range [2,sqrt(5)] or [2,2] , therefore 5 is prime.
Time Complexity: O(sqrt n) because recursive function will call itself at most sqrt(n) times.
Space Complexity: O(sqrt(n)) since recursion uses stack internally and there will be most sqrt(n) stack frames during recursion.
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Является ли число простым — Проверяем на языке Си
Напишем на языке Си программу, проверяющую является ли число простым. Для проверки будем использовать простейший алгоритм, основанный непосредственно на определении простого числа.
Простое число — определение
Простое число — это натуральное число (то есть целое и положительное), большее, чем единица, которое делится без остатка только на единицу и само на себя.
Список простых чисел (приведем до ста) начинается так: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97…
Функция на Си, проверяющая — является ли число простым
Напишем на языке Си функцию, которая будет проверять — простое ли число. И возвращать результат проверки в виде логической величины bool: true (да) или false (нет).
Алгоритм проверки числа n на простоту строится на определении термина простого числа.
Во-первых число n должно быть больше 1 (проверяем это в строке 5 с помощью условного оператора if), а во-вторых проверяемое число должно иметь только два делителя: 1 и n (проверяем это в строках 8-10 с помощью цикла for и оператора if).
Для работы данного метода требуется подключить заголовочный файл stdbool.h в начале файла с исходным кодом. В stdbool.h содержится определение логических констант true и false, поскольку в чистой версии языка Си они отсутствуют.
Check Prime Number in C#
In programming, writing algorithms to find positive integers greater than 1 which do not have any other factors except 1 or itself gives us the prime numbers. This tutorial will teach you three solutions to check prime numbers in C#.
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User Input to Check Prime Numbers in a Given Range in C#
In C#, the most efficient way to find prime numbers in a given range is by writing an algorithm using for loops and if conditions. The following C# program can help you understand algorithms to find prime numbers in a given range.
Use isPrime Boolean to Check Prime Number in C#
Use the isPrime Boolean to check whether the user input number is a prime number or not. In the case of a prime number, the value of isPrime will be true otherwise false .
It’s called trial division and consists of a for loop and if-else condition.
The for loop defines an integer n as a prime number if n > 1 and n is not divisible by i . The for loop output will either be a composite number or a prime number.
An if condition will further check if n is divisible or not by any other number. If it is divisible, the value of isPrime will be false ; otherwise, true .
After executing the for loop, the number will be prime if the value of isPrime is true .
Use Recursion to Check Prime Number in C#
It is a native solution to find a prime number in C#. A C# algorithm checks if a number between 2 to n — 1 divides n .
If it finds any number that divides, it will return false meaning n as a user-defined number is not a prime number.