Determine if a number is a prime without using loops and mutations? - c

Question

How can I determine is a number is a prime without using a loop? I'm a beginner student and so far I've only been taught functional C. All I am allowed to use is the most basic code like + -, if, ==,!,||....Also, no variable mutations are allowed, all variables must be constant.
Here is my code with mutation:
bool prime(const int i, int a) {
if (a > i/2){
return false;
}
if (i%a == 0){
return true;
}
else {
return prime(i, a+1);
}
}
# int a is a counter its given initial value is always 2
#prime (5,2) --> false
#prime (9,2) --> true
However, as the question requires, no mutation is allowed, and only one variable is consumed so it should look like
bool prime(const int i) {
...
}
I'm thinking of a recursive approach but I just can't quite figure it out.
Also efficiency is not a factor to be considered in this problem.
Any help will be appreciated, thanks!

Answer 1

With an auxiliary function it should be more or less easy, although terrible innefficient:
bool prime(const int i)
{
prime_aux(i, i-1);
}
bool prime_aux(const int i, const int d)
{
if (d == 1) return true;
if (i % d == 0) return false;
return prime_aux(i, d - 1);
}
There are some easy optimizations out there, but I omitted them because the code is prettier this way. Also some checks to the input parameters may be needed before proceeding.

Answer 2

Only way I can think of is:
bool prime(const int i)
{
assert(i >= 0);
return char_array_odd_primes_const[i / 8] & (1U << (i % 8)) ? true : false;
}
Where char_array_primes_const contains 231 bits which are set to 1 for prime numbers and 0 for non-prime.
More efficient would be:
bool prime(const int i)
{
assert(i >= 0);
if(2 == i) return true;
if(i % 2) return false;
return char_array_primes_const[i / 16] & (1U << ((i/2) % 8)) ? true : false;
}

Answer 3

check a given number prime or not without loop,Recursion,Auxiliary Function
import java.math.RoundingMode;
import java.util.Scanner;
public class PrimeNumberwithoutloop {
static void PrimeNumber(double num)
{
double n=num/6;
if(n<Math.round(n))
{ n=6*(Math.round(n))-1;
if(n==num)
System.out.println("Prime");
else
System.out.println("Not Prime");
}
else
{ n=6*(Math.round(n))+1;
if(n==num)
System.out.println("Prime");
else
System.out.println("Not Prime");
}
}
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
System.out.println("Enter number to check prime or not");
double num=sc.nextDouble();
PrimeNumber(num);
}
}

Answer 4

Randomized Algorithms are known to exist for determining whether a no. is prime or not (and most probably they use only a single input(i do not have idead regarding all such algorithms), i.e., the number itself that is to be tested for its primality). These algorithms are proven to be accurate in maximum no. of situations, i.e., the probability with which they can give the correct result is very high(although not 100% like the usual deterministic algorithms that are written). If you want, you can use them.
Refer to the following link for more details:
http://en.wikipedia.org/wiki/Primality_test#Probabilistic_tests

Answer 5

This is the best possible way without using loop.
const isPrimeNumber = (num) => {
if(num % 2 === 0 || num % 3 === 0 || num % 5 === 0 || num % 7 === 0 || (Math.sqrt(num) - Math.floor(Math.sqrt(num))) === 0 ) {
console.log(`${num} = is not a prime number`)
}
else {
console.log(`${num} = is a prime number`)
}
}
isPrimeNumber(161);
isPrimeNumber(25689);
isPrimeNumber(11);