Identify prime number - swift

Question

For some reason, it always gave me the wrong result. It's always isItPrime = true no matter what number was assigned to the "number" variable.
This is my code:
let number = 6
var i = 1
var isItPrime: Bool?
while i < number {
if number % i == 0 {
isItPrime = false
} else {
isItPrime = true
}
i += 1
}
print(isItPrime)
Can somebody explain to me what's wrong with my code and why the isItPrime bool outputs always true ?

Answer 1

Problem 1
The last iteration of your while loop
while i < number {
if number % i == 0 {
isItPrime = false
} else {
isItPrime = true
}
i += 1
}
does overwrite the result.
So you always end up with the following result
if number % (number-1) == 0 {
isItPrime = false
} else {
isItPrime = true
}
Problem 2
Finally every number can be divided by 1, so you should start i from 2.
So
let number = 6
var i = 2
var isItPrime = true
while i < number {
if number % i == 0 {
isItPrime = false
break
}
}
print(isItPrime)
Refactoring
You can write a similar logic using Functional Programming
let number = 5
let upperLimit = Int(Double(number).squareRoot())
let isPrime = !(2...upperLimit).contains { number % $0 == 0 }

Answer 2

Because isItPrime is overwritten in subsequent iterations, the last number which is checked, which is number - 1 will always set isItPrime to true, because number and number - 1 are coprime.
Instead of saving the value to a boolean, just end the loop when you found out that the number is not a prime:
let number = 6
var isItPrime: Bool = true
for i in 2 ..< number {
if number % i == 0 {
isItPrime = false
break // end the loop, as we know that the number is not a prime.
}
}
print(isItPrime)

Answer 3

When dealing with problems like this, don't be afraid to take out a piece of paper and manually see what is going on in your loop.
Your loop will go from i = 1 to number = 5 (because of the < operator.
With that in mind, we perform each iteration manually.
for i = 1, number = 6
6 mod 1 = 0, isItPrime = false
for i = 2, number = 6
6 mod 2 = 0, isItPrime = false
for i = 3, number = 6
6 mod 3 = 0, isItPrime = false
for i = 4, number = 6
6 mod 4 = 2, isItPrime = true
Last iteration of the loop, for i = 5, number = 6
6 mod 5 = 1, isItPrime = true
There we can see that the problem is that the last iteration will always have a module of 1, therefore resulting in in your else clause getting executed.

Answer 4

It is always returning true because your while loop isn't working the way you want. Currently, it loops until i is 1 less than number. During that final run through the loop, number % i == 0 is false, so your code sets isItPrime to true.
To fix this problem, try this code:
let number = 6
var i = 2
var isItPrime: Bool?
while (i < number || isItPrime == false) {
if number % i == 0 {
isItPrime = false
} else {
isItPrime = true
}
i += 1
}
print(isItPrime)
You may have noticed I set i to 2, because any number modulo (%) 1 is 0
I think it's worth pointing out, however, that:
You should probably make this a method
If you initially set isItPrime to true, you can dispense with the else part of your if-else statement
Hope this helps!

Answer 5

// MARK: - Function
func primeNo(_ num:Int,_ divisor :Int = 2){
if divisor == num{
print("Num is prime")
}else{
if num%divisor != 0{
primeNo(num, divisor + 1)
}else{
print("num is not prime")
}
}
}
// MARK: - Use
primeNo(6)
out Put
num is not prime