What is the best algorithm for checking if a numbe

Question

Just an example of what I am looking for: I could represent every odd number with a bit e.g. for the given range of numbers (1, 10], starts at 3:

1110

The following dictionary can be squeezed more right? I could eleminate multiples of five with some work, but numbers that end with 1, 3, 7 or 9 must be there in the array of bits. Hope this would clarify what I want.

I am looking for the best algorithm, to check if a number is prime i.e. a boolean function:

bool isprime(number);

I would like to know the best algorithm to implement this functionality. Naturally, there would be a data structure I could query. I define the best algorithm, to be the algorithm that produces a data structure with lowest memory consumption for the range (1, N], where N is a constant.

Answer 1

In Python:

def is_prime(n):
    return not any(n % p == 0 for p in range(2, int(math.sqrt(n)) + 1))

A more direct conversion from mathematical formalism to Python would use all(n % p != 0... ), but that requires strict evaluation of all values of p. The not any version can terminate early if a True value is found.

Answer 2

best algorithm for Primes number javascript

 function isPrime(num) {
      if (num <= 1) return false;
      else if (num <= 3) return true;
      else if (num % 2 == 0 || num % 3 == 0) return false;
      var i = 5;
      while (i * i <= num) {
        if (num % i == 0 || num % (i + 2) == 0) return false;
        i += 6;
      }
      return true
    }

Answer 3

myInp=int(input("Enter a number: "))
if myInp==1:
    print("The number {} is neither a prime not composite no".format(myInp))
elif myInp>1:
    for i in range(2,myInp//2+1):
        if myInp%i==0:
            print("The Number {} is not a prime no".format(myInp))
            print("Because",i,"times",myInp//i,"is",myInp)
            break
    else:
        print("The Number {} is a prime no".format(myInp))
else:
    print("Alas the no {} is a not a prime no".format(myInp))

Answer 4

We can use java streams to implement this in O(sqrt(n)); Consider that noneMatch is a shortCircuiting method that stops the operation when finds it unnecessary for determining the result:

Scanner in = new Scanner(System.in);
int n = in.nextInt();
System.out.println(n == 2 ? "Prime" : IntStream.rangeClosed(2, ((int)(Math.sqrt(n)) + 1)).noneMatch(a -> n % a == 0) ? "Prime" : "Not Prime");

Answer 5

You could try something like this.

def main():
    try:
        user_in = int(input("Enter a number to determine whether the number is prime or not: "))
    except ValueError:
        print()
        print("You must enter a number!")
        print()
        return
    list_range = list(range(2,user_in+1))
    divisor_list = []
    for number in list_range:
        if user_in%number==0:
            divisor_list.append(number)
    if len(divisor_list) < 2:
        print(user_in, "is a prime number!")
        return
    else:
        print(user_in, "is not a prime number!")
        return
main()

Answer 6

public static boolean isPrime(int number) {
 if(number < 2)
   return false;
 else if(number == 2 || number == 3)
        return true;
      else {
        for(int i=2;i<=number/2;i++)
           if(number%i == 0)
             return false;
           else if(i==number/2)
                return true;
      }
    return false;
}