Un número p mayor que uno es primo si y solo si los únicos divisores de p son 1 y p . Los primeros números primos son 2, 3, 5, 7, 11, 13, …
La prueba de Lucas es una prueba de primalidad para un número natural n, puede probar la primalidad de cualquier tipo de número.
Se sigue del pequeño teorema de Fermat: si p es primo y a es un número entero, entonces a^p es congruente con a (mod p )
Test de Lucas : Un número positivo n
es primo si existe un entero a (1 < a < n) tal que:
Y para todo factor primo q de (n-1),
Ejemplos :
Input : n = 7 Output : 7 is Prime Explanation : let's take a = 3, then 3^6 % 7 = 729 % 7 = 1 (1st condition satisfied). Prime factors of 6 are 2 and 3, 3^(6/2) % 7 = 3^3 % 7 = 27 % 7 = 6 3^(6/3) % 7 = 3^2 % 7 = 9 % 7 = 2 Hence, 7 is Prime Input : n = 9 Output : 9 is composite Explanation : Let's take a = 2, then 2^8 % 9 = 256 % 9 = 4 Hence 9 is composite
lucasTest(n): If n is even return composite Else Find all prime factors of n-1 for i=2 to n-1 pick 'a' randomly in range [2, n-1] if a^(n-1) % n not equal 1: return composite else // for all q, prime factors of (n-1) if a^(n-1)/q % n not equal 1 return prime Return probably prime
Los problemas asociados con la prueba de Lucas son :
- Conociendo todos los factores primos de n-1
- Encontrar una opción apropiada para un
C++
// C++ Program for Lucas Primality Test #include <bits/stdc++.h> using namespace std; // function to generate prime factors of n void primeFactors(int n, vector<int>& factors) { // if 2 is a factor if (n % 2 == 0) factors.push_back(2); while (n % 2 == 0) n = n / 2; // if prime > 2 is factor for (int i = 3; i <= sqrt(n); i += 2) { if (n % i == 0) factors.push_back(i); while (n % i == 0) n = n / i; } if (n > 2) factors.push_back(n); } // this function produces power modulo // some number. It can be optimized to // using int power(int n, int r, int q) { int total = n; for (int i = 1; i < r; i++) total = (total * n) % q; return total; } string lucasTest(int n) { // Base cases if (n == 1) return "neither prime nor composite"; if (n == 2) return "prime"; if (n % 2 == 0) return "composite1"; // Generating and storing factors // of n-1 vector<int> factors; primeFactors(n - 1, factors); // Array for random generator. This array // is to ensure one number is generated // only once int random[n - 3]; for (int i = 0; i < n - 2; i++) random[i] = i + 2; // shuffle random array to produce randomness shuffle(random, random + n - 3, default_random_engine(time(0))); // Now one by one perform Lucas Primality // Test on random numbers generated. for (int i = 0; i < n - 2; i++) { int a = random[i]; if (power(a, n - 1, n) != 1) return "composite"; // this is to check if every factor // of n-1 satisfy the condition bool flag = true; for (int k = 0; k < factors.size(); k++) { // if a^((n-1)/q) equal 1 if (power(a, (n - 1) / factors[k], n) == 1) { flag = false; break; } } // if all condition satisfy if (flag) return "prime"; } return "probably composite"; } // Driver code int main() { cout << 7 << " is " << lucasTest(7) << endl; cout << 9 << " is " << lucasTest(9) << endl; cout << 37 << " is " << lucasTest(37) << endl; return 0; }
Java
// Java Program for Lucas Primality Test import java.util.*; class GFG { static ArrayList<Integer> factors = new ArrayList<Integer>(); // function to generate prime factors of n static ArrayList<Integer> primeFactors(int n) { // if 2 is a factor if (n % 2 == 0) factors.add(2); while (n % 2 == 0) n = n / 2; // if prime > 2 is factor for (int i = 3; i <= Math.sqrt(n); i += 2) { if (n % i == 0) factors.add(i); while (n % i == 0) n = n / i; } if (n > 2) factors.add(n); return factors; } // this function produces power modulo // some number. It can be optimized to // using static int power(int n, int r, int q) { int total = n; for (int i = 1; i < r; i++) total = (total * n) % q; return total; } static String lucasTest(int n) { // Base cases if (n == 1) return "neither prime nor composite"; if (n == 2) return "prime"; if (n % 2 == 0) return "composite1"; // Generating and storing factors // of n-1 primeFactors(n - 1); // Array for random generator. This array // is to ensure one number is generated // only once int[] random = new int[n - 2]; for (int i = 0; i < n - 2; i++) random[i] = i + 2; // shuffle random array to produce randomness Collections.shuffle(Arrays.asList(random)); // Now one by one perform Lucas Primality // Test on random numbers generated. for (int i = 0; i < n - 2; i++) { int a = random[i]; if (power(a, n - 1, n) != 1) return "composite"; // this is to check if every factor // of n-1 satisfy the condition boolean flag = true; for (i = 0; i < factors.size(); i++) { // if a^((n-1)/q) equal 1 if (power(a, (n - 1) / factors.get(i), n) == 1) { flag = false; break; } } // if all condition satisfy if (flag) return "prime"; } return "probably composite"; } // Driver code public static void main(String[] args) { System.out.println(7 + " is " + lucasTest(7)); System.out.println(9 + " is " + lucasTest(9)); System.out.println(37 + " is " + lucasTest(37)); } } // This code is contributed by phasing17
Python3
# Python3 program for Lucas Primality Test import random import math # Function to generate prime factors of n def primeFactors(n, factors): # If 2 is a factor if (n % 2 == 0): factors.append(2) while (n % 2 == 0): n = n // 2 # If prime > 2 is factor for i in range(3, int(math.sqrt(n)) + 1, 2): if (n % i == 0): factors.append(i) while (n % i == 0): n = n // i if (n > 2): factors.append(n) return factors # This function produces power modulo # some number. It can be optimized to # using def power(n, r, q): total = n for i in range(1, r): total = (total * n) % q return total def lucasTest(n): # Base cases if (n == 1): return "neither prime nor composite" if (n == 2): return "prime" if (n % 2 == 0): return "composite1" # Generating and storing factors # of n-1 factors = [] factors = primeFactors(n - 1, factors) # Array for random generator. This array # is to ensure one number is generated # only once rand = [i + 2 for i in range(n - 3)] # Shuffle random array to produce randomness random.shuffle(rand) # Now one by one perform Lucas Primality # Test on random numbers generated. for i in range(n - 2): a = rand[i] if (power(a, n - 1, n) != 1): return "composite" # This is to check if every factor # of n-1 satisfy the condition flag = True for k in range(len(factors)): # If a^((n-1)/q) equal 1 if (power(a, (n - 1) // factors[k], n) == 1): flag = False break # If all condition satisfy if (flag): return "prime" return "probably composite" # Driver code if __name__=="__main__": print(str(7) + " is " + lucasTest(7)) print(str(9) + " is " + lucasTest(9)) print(str(37) + " is " + lucasTest(37)) # This code is contributed by rutvik_56
C#
// C# Program for Lucas Primality Test using System; using System.Linq; using System.Collections.Generic; class GFG { static List<int> factors = new List<int>(); // function to generate prime factors of n static List<int> primeFactors(int n) { // if 2 is a factor if (n % 2 == 0) factors.Add(2); while (n % 2 == 0) n = n / 2; // if prime > 2 is factor for (int i = 3; i <= Math.Sqrt(n); i += 2) { if (n % i == 0) factors.Add(i); while (n % i == 0) n = n / i; } if (n > 2) factors.Add(n); return factors; } // this function produces power modulo // some number. It can be optimized to // using static int power(int n, int r, int q) { int total = n; for (int i = 1; i < r; i++) total = (total * n) % q; return total; } static string lucasTest(int n) { // Base cases if (n == 1) return "neither prime nor composite"; if (n == 2) return "prime"; if (n % 2 == 0) return "composite1"; // Generating and storing factors // of n-1 primeFactors(n - 1); // Array for random generator. This array // is to ensure one number is generated // only once int[] random = new int[n - 2]; for (int i = 0; i < n - 2; i++) random[i] = i + 2; // shuffle random array to produce randomness Random rand = new Random(); random = random.OrderBy(x => rand.Next()).ToArray(); // Now one by one perform Lucas Primality // Test on random numbers generated. for (int i = 0; i < n - 2; i++) { int a = random[i]; if (power(a, n - 1, n) != 1) return "composite"; // this is to check if every factor // of n-1 satisfy the condition bool flag = true; foreach (var factor in factors) { // if a^((n-1)/q) equal 1 if (power(a, (n - 1) / factor, n) == 1) { flag = false; break; } } // if all condition satisfy if (flag) return "prime"; } return "probably composite"; } // Driver code public static void Main(string[] args) { Console.WriteLine(7 + " is " + lucasTest(7)); Console.WriteLine(9 + " is " + lucasTest(9)); Console.WriteLine(37 + " is " + lucasTest(37)); } } // This code is contributed by phasing17
Javascript
// JavaScript Program for Lucas Primality Test // A function to shuffle the array. function shuffle(arr){ for(let i = arr.length-1; i>0;i--){ // have a random index from [0, arr.length-1] let j = Math.floor(Math.random() * (i+1)); // swap the original and random index element let temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } return arr; } // function to generate prime factors of n function primeFactors(n, factors) { // if 2 is a factor if (n % 2 == 0){ factors.push(2); } while (n % 2 == 0){ n = n / 2; } // if prime > 2 is factor for (let i = 3; i <= Math.sqrt(n); i += 2) { if (n % i == 0){ factors.push(i); } while (n % i == 0){ n = n / i; } } if (n > 2){ factors.push(n); } } // this function produces power modulo // some number. It can be optimized to // using function power(n, r, q) { let total = n; for (let i = 1; i < r; i++){ total = (total * n) % q; } return total; } function lucasTest(n) { // Base cases if (n == 1){ return "neither prime nor composite"; } if (n == 2){ return "prime"; } if (n % 2 == 0){ return "composite1"; } // Generating and storing factors // of n-1 const factors = []; primeFactors(n - 1, factors); // Array for random generator. This array // is to ensure one number is generated // only once const random = []; for (let i = 0; i < n - 2; i++){ // random[i] = i + 2; random.push(i+2); } // shuffle random array to produce randomness shuffle(random); // Now one by one perform Lucas Primality // Test on random numbers generated. for (let i = 0; i < n - 2; i++) { let a = random[i]; if (power(a, n - 1, n) != 1){ return "composite"; } // this is to check if every factor // of n-1 satisfy the condition let flag = true; for (let k = 0; k < factors.length; k++) { // if a^((n-1)/q) equal 1 if (power(a, (n - 1) / factors[k], n) == 1) { flag = false; break; } } // if all condition satisfy if (flag){ return "prime"; } } return "probably composite"; } // Driver code { console.log( 7 + " is " + lucasTest(7)); console.log( 9 + " is " + lucasTest(9)); console.log( 37 + " is " + lucasTest(37)); return 0; } // The code is contributed by Gautam goel (gautamgoel962) Javascript
Producción:
7 is prime 9 is composite 37 is prime
Complejidad de tiempo: O (nlogn)
Espacio Auxiliar: O(n)
Este método es bastante complicado e ineficiente en comparación con otras pruebas de primalidad. Y los principales problemas son los factores de ‘n-1’ y la elección de ‘a’ apropiado.
Otras pruebas de Primalidad:
- Prueba de primalidad | Conjunto 1 (Introducción y Método Escolar)
- Prueba de primalidad | Juego 2 (Método Fermat)
- Prueba de primalidad | Conjunto 3 (Miller-Rabin)
- Prueba de primalidad | Conjunto 4 (Solovay-Strassen)
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA