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Commit 7ea7b3f

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Merge pull request #251 from Priyangshuyogi/myone
Algorithms on Primality Test
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Given a number n, check if it is prime or not. This method is a probabilistic method and is based on below Fermat’s Little Theorem.
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Fermat's Little Theorem:
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If n is a prime number, then for every a, 1 <= a < n,
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a^n-1 ~ 1 mod (n)
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OR
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a^n-1 % n ~ 1
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Example:
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Since 5 is prime, 2^4 ≡ 1 (mod 5) [or 2^4%5 = 1],
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3^4 ≡ 1 (mod 5) and 4^4 ≡ 1 (mod 5).
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Since 7 is prime, 2^6 ≡ 1 (mod 7),
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3^6 ≡ 1 (mod 7), 4^6 ≡ 1 (mod 7)
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5^6 ≡ 1 (mod 7) and 6^6 ≡ 1 (mod 7).
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We will take help of this Fermat's Little Theorem as a function to calculate a no is prime or not.
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#include <bits/stdc++.h>
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using namespace std;
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/* Iterative Function to calculate (a^n)%p in O(logy) */
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int power(int a, unsigned int n, int p)
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{
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int res = 1; // Initialize result
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a = a % p; // Update 'a' if 'a' >= p
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while (n > 0)
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{
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// If n is odd, multiply 'a' with result
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if (n & 1)
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res = (res*a) % p;
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// n must be even now
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n = n>>1; // n = n/2
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a = (a*a) % p;
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}
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return res;
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}
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// If n is prime, then always returns true, If n is
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// composite than returns false with high probability
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// Higher value of k increases probability of correct result.
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bool isPrime(unsigned int n, int k)
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{
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// Corner cases
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if (n <= 1 || n == 4) return false;
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if (n <= 3) return true;
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// Try k times
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while (k>0)
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{
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// Pick a random number in [2..n-2]
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// Above corner cases make sure that n > 4
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int a = 2 + rand()%(n-4);
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// Fermat's little theorem
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if (power(a, n-1, n) != 1)
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return false;
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k--;
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}
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return true;
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}
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// Driver Program
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int main()
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{
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int k = 3;
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isPrime(11, k)? cout << " true\n": cout << " false\n";
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isPrime(15, k)? cout << " true\n": cout << " false\n";
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return 0;
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}
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Output:
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true
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false
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Given a positive integer, check if the number is prime or not. A prime is a natural number greater than 1 that has no
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positive divisors other than 1 and itself.
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Examples of first few prime numbers are {2, 3, 5, 7,...}
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Examples:
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**Input:** n = 11
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**Output:** true
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**Input:** n = 15
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**Output:** false
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**Input:** n = 1
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**Output:** false
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A simple solution is to iterate through all numbers from 2 to n-1 and for every number check if it divides n. If we find
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any number that divides, we return false. Instead of checking till n, we can check till √n because a larger factor of n
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must be a multiple of smaller factor that has been already checked. The algorithm can be improved further by observing
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that all primes are of the form 6k ± 1, with the exception of 2 and 3. This is because all integers can be expressed
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as (6k + i) for some integer k and for i = ?1, 0, 1, 2, 3, or 4; 2 divides (6k + 0), (6k + 2), (6k + 4); and 3 divides
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(6k + 3). So a more efficient method is to test if n is divisible by 2 or 3, then to check through all the numbers of
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form 6k ± 1.
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Time complexity of this solution is O(root(n)).
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// A optimized school method based C++ program to check if a number is prime or not.
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#include <bits/stdc++.h>
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using namespace std;
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bool isPrime(int n)
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{
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// Corner cases
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if (n <= 1) return false;
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if (n <= 3) return true;
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// This is checked so that we can skip
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// middle five numbers in below loop
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if (n%2 == 0 || n%3 == 0) return false;
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for (int i=5; i*i<=n; i=i+6)
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if (n%i == 0 || n%(i+2) == 0)
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return false;
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return true;
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}
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// Driver Program
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int main()
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{
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isPrime(23)? cout << " true\n": cout << " false\n";//use of ternary operator
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isPrime(35)? cout << " true\n": cout << " false\n";
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return 0;
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}
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Output:
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true
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false

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