卢卡斯原始性测试

当且仅当p的唯一除数是1和p时，大于1的p是质数。前几个质数是2、3、5、7、11、13…
Lucas检验是自然数n的素数检验，它可以检验任何数字的素数。
遵循费马小定理：如果p为素数， a为整数，则a ^ p与a(mod p)一致。

卢卡斯测验：正数n
如果存在整数a (1

$a^{{n-1}}\ \equiv \ 1{\pmod n}$

对于(n-1)的每个素数q，

$a^{{({n-1})/q}}\ \not \equiv \ 1{\pmod n}$

例子：

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

与卢卡斯考试有关的问题是：

• 知道n-1的所有素数
• 找到一个合适的选择

C++

// C++ Program for Lucas Primality Test
#include 
using namespace std;
 
// function to generate prime factors of n
void primeFactors(int n, vector& 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 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;
}

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

输出：

7 is prime
9 is composite
37 is prime

与其他原始性测试相比，此方法非常复杂且效率低下。而主要问题是“ n-1”因子和选择合适的“ a”因子。

其他Primality测试：

• 原始性测试|第一组(介绍和学校方法)
• 原始性测试|套装2(Fermat方法)
• 原始性测试|第三组(米勒-拉宾)
• 原始性测试|第4组(索洛维·斯特拉森)