I am trying to create an app that will have the user input a number and the app will determine if it is prime or not. Here is my code:
var arr: [Int] = [2, 3, 4, 5, 6, 7, 8, 9]
let text = Int(textField.text!)!
for x in arr {
if text == 1 {
iLabel.text = "NOT PRIME!"
}
else if x == 2 || x == 3 || x == 5 || x == 7 {
iLabel.text = "IT'S PRIME!"
}
else if text % x == 0 {
iLabel.text = "NOT PRIME!"
break
}
else if text % x != 0 {
iLabel.text = "IT'S PRIME!"
break
}
}
For the most part, this works. However, when I set variable "text" to equal something like 82, the result is "IT'S PRIME!" despite it being evenly divisible by 2...can anyone explain the flaw in my code?
if text == 1 else if x == 2 else if text % x == 0
^ ^ ^
it will trigger x=2 and set "IT'S PRIME` before it gets to the modulo/remainder check, which it will then skip. It will trigger the same thing at loop 3, 5, 7.
82 is not divisible by 8 or 9 so it won't happen to trigger the remainder check at the end of the loop and accidentally output the correct value, leaving "IT'S PRIME" in the output box, without actually checking whether it's prime.
Related
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
One of effective algorithms to find prime numbers is Sieve of Eratosthenes. It is based on idea that you have sorted array of all numbers in given range and you go from the beginning and you remove all numbers after current number divisible by this number which is prime number. You repeat this until you check last element in the array.
There is my algorithm which should do what I described above:
func primes(upTo rangeEndNumber: Int) -> [Int] {
let firstPrime = 2
guard rangeEndNumber >= firstPrime else {
fatalError("End of range has to be greater than or equal to \(firstPrime)!")
}
var numbers = Array(firstPrime...rangeEndNumber)
// Index of current prime in numbers array, at the beginning it is 0 so number is 2
var currentPrimeIndex = 0
// Check if there is any number left which could be prime
while currentPrimeIndex < numbers.count {
// Number at currentPrimeIndex is next prime
let currentPrime = numbers[currentPrimeIndex]
// Create array with numbers after current prime and remove all that are divisible by this prime
var numbersAfterPrime = numbers.suffix(from: currentPrimeIndex + 1)
numbersAfterPrime.removeAll(where: { $0 % currentPrime == 0 })
// Set numbers as current numbers up to current prime + numbers after prime without numbers divisible by current prime
numbers = numbers.prefix(currentPrimeIndex + 1) + Array(numbersAfterPrime)
// Increase index for current prime
currentPrimeIndex += 1
}
return numbers
}
print(primes(upTo: 100)) // [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]
print(primes(upTo: 2)) // [2]
print(primes(upTo: 1)) // Fatal error: End of range has to be greater than or equal to 2!
what is the Prime num : Prime numbers are the positive integers having only two factors, 1 and the integer itself,
//Funtion Call
findPrimeNumberlist(fromNumber: 1, toNumber: 100)
//You can print any range Prime number using this fucntion.
func findPrimeNumberlist(fromNumber:Int, toNumber: Int)
{
for i in fromNumber...toNumber
{
var isPrime = true
if i <= 1 { // number must be positive integer
isPrime = false
}
else if i <= 3 {
isPrime = true
}
else {
for j in 2...i/2 // here i am using loop from 2 to i/2 because it will reduces the iteration.
{
if i%j == 0 { // number must have only 1 factor except 1. so use break: no need to check further
isPrime = false
break
}
}
}
if isPrime {
print(i)
}
}
}
func getPrimeNumbers(rangeOfNum: Int) -> [Int]{
var numArr = [Int]()
var primeNumArr = [Int]()
var currentNum = 0
for i in 0...rangeOfNum{
currentNum = i
var counter = 0
if currentNum > 1{
numArr.append(currentNum)
for j in numArr{
if currentNum % j == 0{
counter += 1
}
}
if counter == 1{
primeNumArr.append(currentNum)
}
}
}
print(primeNumArr)
print(primeNumArr.count)
return primeNumArr
}
Then just call the function with the max limit using this
getPrimeNumbers(rangeOfNum: 100)
What is happening in above code:
The numArr is created to keep track of what numbers have been used
Any number that is prime number is added/appended to primeNumArr
Current number shows the number that is being used at the moment
We start from 0 ... upto our range where we need prime numbers upto (with little modification it can be changed if the range starts from other number beside 0)
Remember, for a number to be Prime it should have 2 divisor means should be only completely divisible by 2 numbers. First is 1 and second is itself. (Completely divisible means having remainder 0)
The counter variable is used to keep count of how many numbers divide the current number being worked on.
Since 1 is only has 1 Divisor itself hence its not a Prime number so we start from number > 1.
First as soon as we get in, we add the current number being checked into the number array to keep track of numbers being used
We run for loop to on number array and check if the Current Number (which in our case will always be New and Greater then previous ones) when divided by numbers in numArr leaves a remainder of 0.
If Remainder is 0, we add 1 to the counter.
Since we are already ignoring 1, the max number of counter for a prime number should be 1 which means only divisible by itself (only because we are ignoring it being divisible by 1)
Hence if counter is equal to 1, it confirms that the number is prime and we add it to the primeNumArr
And that's it. This will give you all prime numbers within your range.
PS: This code is written on current version of swift
Optimised with less number of loops
Considered below conditions
Even Number can not be prime number expect 2 so started top loop form 3 adding 2
Any prime number can not multiplier of even number expect 2 so started inner loop form 3 adding 2
Maximum multiplier of any number if half that number
var primeNumbers:[Int] = [2]
for index in stride(from: 3, to: 100, by: 2) {
var count = 0
for indexJ in stride(from: 3, to: index/2, by: 2) {
if index % indexJ == 0 {
count += 1
}
if count == 1 {
break
}
}
if count == 0 {
primeNumbers.append(index)
}
}
print("primeNumbers ===", primeNumbers)
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)
}
}
I want to write a program that can find the N-th number,which only contains factor 2 , 3 or 5.
def method3(n:Int):Int = {
var q2 = mutable.Queue[Int](2)
var q3 = mutable.Queue[Int](3)
var q5 = mutable.Queue[Int](5)
var count = 1
var x:Int = 0
while(count != n){
val minVal = Seq(q2,q3,q5).map(_.head).min
if(minVal == q2.head){
x = q2.dequeue()
q2.enqueue(2*x)
q3.enqueue(3*x)
q5.enqueue(5*x)
}else if(minVal == q3.head){
x = q3.dequeue()
q3.enqueue(3*x)
q5.enqueue(5*x)
}else{
x = q5.dequeue()
q5.enqueue(5*x)
}
count+=1
}
return x
}
println(method3(1000))
println(method3(10000))
println(method3(100000))
The results
51200000
0
0
When the input number gets larger , I get 0 from the function.
But if I change the function to
def method3(n:Int):Int = {
...
q5.enqueue(5*x)
}
if(x > 1000000000) println(('-',x)) //note here!!!
count+=1
}
return x
}
The results
51200000
(-,1006632960)
(-,1007769600)
(-,1012500000)
(-,1019215872)
(-,1020366720)
(-,1024000000)
(-,1025156250)
(-,1033121304)
(-,1036800000)
(-,1048576000)
(-,1049760000)
(-,1054687500)
(-,1061683200)
(-,1062882000)
(-,1073741824)
0
.....
So I don't know why the result equals to 0 when the input number grows larger.
An Int is only 32 bits (4 bytes). You're hitting the limits of what an Int can hold.
Take that last number you encounter: 1073741824. Multiply that by 2 and the result is negative (-2147483648). Multiply it by 4 and the result is zero.
BTW, if you're working with numbers "which only contains factor 2, 3 or 5", in other words the numbers 2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 15, ... etc., then the 1,000th number in that sequence shouldn't be that big. By my calculations the result should only be 1365.
After hours of Googling, I'm still at a standstill. I would appreciate if someone would point out the error in my formula or coding choice. Please keep in mind I'm new to Swift. I'm not used to non C-style for loops.
if textField.text != "" {
input = Double(textField.text!)! // parse input
// return if number less than 2 entered
if input < 2 {
resultLabel.text = "Enter a number greater than or equal to 2."
return;
}
// get square root of input and parse to int
inputSquared = Int(sqrt(input));
// loop from 2 to input iterating by 1
for i in stride(from: 2, through: input, by: 1) {
if inputSquared % Int(i) == 0 {
resultLabel.text = "\(Int(input)) is not a prime number."
}
else {
resultLabel.text = "\(Int(input)) is a prime number!"
}
}
}
I didn't know the formula on how to find a prime number. After looking up multiple formulas I have sorta settled on this one. Every result is a prime number, however. So my if condition is wrong. I just don't know how to fix it.
Check my algorithm.It works.But I'm not sure this is an effective algorithm for prime number
var input:Int = 30
var isPrime:Bool = true
if(input == 2){
print("Input value 2 is prim number")
}
else if(input < 2){
print("Input value must greater than 2")
}
else{
for i in 2...input-1{
if((input%i) == 0){
isPrime = false
break;
}
}
if(isPrime){
print("Your Input Value \(input) is Prime!")
}
}
A couple of solutions that work have been posted, but none of them explain why yours doesn't. Some of the comments get close, however.
Your basic problem is that you take the square root of input, then iterate from 2 to the input checking if the integer part of the square root is divisible by i. You got that the wrong way round. You need to iterate from 2 to the square root and check that the input is divisible by i. If it is, you stop because input is not prime. If you get to the end without finding a divisor, you have a prime.
try this code in playground you will get this better idea and try to use playground when you try the swift as you are not familiar with swift playground is best.
let input = 13 // add your code that take value from textfield
var prime = 1
// throw error less than 2 entered
if input < 2 {
assertionFailure("number should be 2 or greater")
}
// loop from 2 to input iterating by 1
for i in stride(from: 2, to: input, by: 1) {
if input % i == 0{
prime = 0
}
}
if prime == 1 {
print("\(input) number is prime")
} else {
print("\(input) number is not prime")
}
Is there a more elegant way to write this? I dont want to use a for loop
if i==1 || i==6 || i==11 || i==16 || i==21 || i==26 || i==31 || i==36
function
end
basically i is the index of a vector, after each fifth element of this vector (starting with the first ) a specific function is applied. i starts with 1 and it increments after the if statement and just if it equals these values of the if condition the if statement is valid
EDITTED FOR MATLAB CODE OF MODULO
output = mod(input, 5); //this will output 1 if it is 1, 5, 11, 16
//input is your 1, 5, 11, 16 etc
//output is the result of modulo. else it is 0, 2, 3, 4
if(output == 1)
[previous answer]
i forgot how to write this in matlab but with your values, put it this way.
if(number%5==1)
any input 1 or 6 or 11 or any else that can add 5 to it, you'll end up in 1. else it will return false
I am trying to find the odd numbers and a multiple of 7 between a 1 to 100 and append them into an array. I have got this far:
var results: [Int] = []
for n in 1...100 {
if n / 2 != 0 && 7 / 100 == 0 {
results.append(n)
}
}
Your conditions are incorrect. You want to use "modular arithmetic"
Odd numbers are not divisible by 2. To check this use:
if n % 2 != 0
The % is the mod function and it returns the remainder of the division (e.g. 5 / 2 is 2.5 but integers don't have decimals, so the integer result is 2 with a remainder of 1 and 5 / 2 => 2 and 5 % 2 => 1)
To check if it's divisible by 7, use the same principle:
if n % 7 == 0
The remainder is 0 if the dividend is divisible by the divisor. The complete if condition is:
if n % 2 != 0 && n % 7 == 0
You can also use n % 2 == 1 because the remainder is always 1. The result of any mod function, a % b, is always between 0 and b - 1.
Or, using the new function isMultiple(of:, that final condition would be:
if !n.isMultiple(of: 2) && n.isMultiple(of: 7)
Swift 5:
Since Swift 5 has been released, you could use isMultiple(of:) method.
In your case, you should check if it is not multiple of ... :
if !n.isMultiple(of: 2)
Swift 5 is coming with isMultiple(of:) method for integers , so you can try
let res = Array(1...100).filter { !$0.isMultiple(of:2) && $0.isMultiple(of:7) }
Here is an efficient and concise way of getting the odd multiples of 7 less than or equal to 100 :
let results: [Int] = Array(stride(from: 7, through: 100, by: 14))
You can also use the built-in filter to do an operation on only qualified members of an array. Here is how that'd go in your case for example
var result = Array(1...100).filter { (number) -> Bool in
return (number % 2 != 0 && number % 7 == 0)
}
print(result) // will print [7, 21, 35, 49, 63, 77, 91]
You can read more about filter in the doc but here is the basics: it goes through each element and collects elements that return true on the condition. So it filters the array and returns what you want