Lucas Primality Test

// 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