Styjun

I have been assigned with a task to print prime numbers from a range 2...100. I've managed to get most of the prime numbers but can't figure out how to get rid of 9 and 15, basically multiples of 3 and 5. Please give me your suggestion on how can I fix this.

for n in 2...20 
 if n % 2 == 0 && n < 3
 print(n)
 else if n % 2 == 1 
 print(n)
 else if n % 3 == 0 && n > 6 
 

This what it prints so far:

2
3
5
7
9
11
13
15
17
19

Your "algorithm" is wrong. You should store somewhere your prime numbers and then just check if n satisfies the condition n % prime != 0 for each prime. If it does, just append this n to an array of primes and continue to next n.

Since Swift 4.2 you can use this very simple algorithm using allSatisfy

var primes = [Int]()

for n in 2...100 
 if primes.allSatisfy( n % $0 != 0 ) 
 primes.append(n)
 


print(primes) // [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]

Or, you can implement more effective Sieve of Eratosthenes algorithm.

Below is my simple implementation (there are ofc more effective solutions)

var range = [Int](2...100)
var p = 0

while true 
 guard let newPrimeIndex = range.firstIndex(where: $0 > p ) else break 
 p = range[newPrimeIndex]
 range.removeAll(where: $0 % p == 0 && $0 != p )


print(range) // [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]

Made for Swift 5

Unlike other answers, this is easily scaleable. I did not include a pre-made list of prime numbers, which would not work for very large numbers which have not been included.

Here is the algorithm I came up with. I made it as efficient as I can, with the factors only going up to the floor of the square root of the number. This reduces wasted time comparing factors which will never work.

import Foundation


// Function to determine whether a number is prime
func isPrime(_ num: Int) -> Bool 
 let max = Int(floor(sqrt(Double(num))))
 guard max >= 2 else return true 

 for factor in 2...max 
 if num.isMultiple(of: factor) // Write num % factor == 0 for older versions of Swift
 return false
 
 

 return true



// Find if a number is prime for numbers 2 to 200
for i in 2...100 
 if isPrime(i) 
 print(i)
 

I finally figured it out lol, It might be not pretty but it works haha, Thanks for everyone's answer. I'll post what I came up with if maybe it will help anyone else.

for n in 2...100 
 if n % 2 == 0 && n < 3
 print(n)
 else if n % 3 == 0 && n > 6 
 else if n % 5 == 0 && n > 5 
 else if n % 7 == 0 && n > 7
 else if n % 2 == 1 
 print(n)