Simple Swift Fibonacci program crashing (Project Euler 2) - swift

I am trying to solve the second problem on Project Euler. The problem is as follows:
Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
I think I've written a solution, but when I try to run my code it crashes my Swift playground and gives me this error message:
Playground execution aborted: Execution was interrupted, reason: EXC_BAD_INSTRUCTION (code=EXC_I386_INVOP, subcode=0x0)
var prev = 0
var next = 1
var num = 0
var sum = 0
for var i = 1; i < 400; i++ {
num = prev + next
if next % 2 == 0 {
sum += next
}
prev = next
next = num
}
print(sum)
The weird thing is, if I set the counter on my loop to less than 93, it works fine. Explicitly setting the variable names to Double does not help. Anyone know what's going on here?

There is nothing weird about this at all. Do you know how large the 400 fibonacci number is?
176023680645013966468226945392411250770384383304492191886725992896575345044216019675
Swift Int64 or UInt64 simply cannot handle that large of a number. The later can go up to 18446744073709551615 at max - not even close.
If you change your variables to be doubles it works but will be inaccurate:
var prev : Double = 0
var next : Double = 1
var num : Double = 0
var sum : Double = 0
will yield
2.84812298108489e+83
which is kind of close to the actual value of
1.76e+83
Luckily you do not need to get values that big. I would recommend not writing a for loop but a while loop that calculates the next fibonacci number until the break condition is met whose values do not exceed four million.

The Fibonacci numbers become very large quickly. To compute large Fibonacci numbers, you need to implement some kind of BigNum. Here is a version the makes a BigNum that is implemented internally as an array of digits. For example, 12345 is implemented internally as [1, 2, 3, 4, 5]. This makes it easy to represent arbitrarily large numbers.
Addition is implemented by making the two arrays the same size, then map is used to add the elements, finally the carryAll function restores the array to single digits.
For example 12345 + 67:
[1, 2, 3, 4, 5] + [6, 7] // numbers represented as arrays
[1, 2, 3, 4, 5] + [0, 0, 0, 6, 7] // pad the shorter array with 0's
[1, 2, 3, 10, 12] // add the arrays element-wise
[1, 2, 4, 1, 2] // perform carry operation
Here is the implementation of BigNum. It is also CustomStringConvertible which makes it possible to print the result as a String.
struct BigNum: CustomStringConvertible {
var arr = [Int]()
// Return BigNum value as a String so it can be printed
var description: String { return arr.map(String.init).joined() }
init(_ arr: [Int]) {
self.arr = carryAll(arr)
}
// Allow BigNum to be initialized with an `Int`
init(_ i: Int = 0) {
self.init([i])
}
// Perform the carry operation to restore the array to single
// digits
func carryAll(_ arr: [Int]) -> [Int] {
var result = [Int]()
var carry = 0
for val in arr.reversed() {
let total = val + carry
let digit = total % 10
carry = total / 10
result.append(digit)
}
while carry > 0 {
let digit = carry % 10
carry = carry / 10
result.append(digit)
}
return result.reversed()
}
// Enable two BigNums to be added with +
static func +(_ lhs: BigNum, _ rhs: BigNum) -> BigNum {
var arr1 = lhs.arr
var arr2 = rhs.arr
let diff = arr1.count - arr2.count
// Pad the arrays to the same length
if diff < 0 {
arr1 = Array(repeating: 0, count: -diff) + arr1
} else if diff > 0 {
arr2 = Array(repeating: 0, count: diff) + arr2
}
return BigNum(zip(arr1, arr2).map { $0 + $1 })
}
}
// This function is based upon this question:
// https://stackoverflow.com/q/52975875/1630618
func fibonacci(to n: Int) {
guard n >= 2 else { return }
var array = [BigNum(0), BigNum(1)]
for i in 2...n {
array.append(BigNum())
array[i] = array[i - 1] + array[i - 2]
print(array[i])
}
}
fibonacci(to: 400)
Output:
1
2
3
5
8
...
67235063181538321178464953103361505925388677826679492786974790147181418684399715449
108788617463475645289761992289049744844995705477812699099751202749393926359816304226
176023680645013966468226945392411250770384383304492191886725992896575345044216019675

Related

Calling different extent of randomness of arc4random in Swift?

This might be rather stupid question. I would like to know if different nuances/extent of randomness would be possible using arc4random_uniform in Swift. Here's an example:
let number = arc4random_uniform(10) + 1
print(number)
In this case, a number will be printed randomly from 1 to 10. But is there a way that I can repeat the random result, 2 to 3 times? The result would be something like this:
1, 1, 6, 6, 6, 3, 3, 8, 8, 9, 9, 9 ...
// 1) Randomly selected and 2) repeated 2 to 3 times randomly.
Perhaps I might use two arc4random_uniform functions together, but cannot express them properly. Would be much appreciated if you could give me some suggestions. <3
In order to do this, you will need to generate two values: your random value and a repeatCount. Also, you'll need to remember both of those values so that you can repeat the value. You can do this with a custom class:
class RandomWithRepeats {
var range: ClosedRange<Int>
var repeatRange: ClosedRange<Int>
var repeatCount = 0
var value = 0
init(range: ClosedRange<Int>, repeatRange: ClosedRange<Int>) {
self.range = range
self.repeatRange = repeatRange
}
// generate a random number in a range
// Just use Int.random(in:) with Swift 4.2 and later
func random(in range: ClosedRange<Int>) -> Int {
return Int(arc4random_uniform(UInt32(range.upperBound - range.lowerBound + 1))) + range.lowerBound
}
func nextValue() -> Int {
// if repeatCount is 0, its time to generate a new value and
// a new repeatCount
if repeatCount == 0 {
// For Swift 4.2, just use Int.random(in:) instead
value = self.random(in: range)
repeatCount = self.random(in: repeatRange)
}
repeatCount -= 1
return value
}
}
Example:
let rand = RandomWithRepeats(range: 1...10, repeatRange: 2...3)
// generate 20 random repeated numbers
for _ in 1...20
{
print(rand.nextValue(), terminator: " ")
}
6 6 6 8 8 8 10 10 10 2 2 9 9 5 5 8 8 8 5 5
With regards to the nuances of random number generators: have a look at GKRandomSource.
What you're doing here is not really making something less random, or modifying the parameters in the random number generator. You're simply applying an operation (with one random parameter) to a collection of random integers.
extension Collection {
func duplicateItemsRandomly(range: CountableClosedRange<Int>) -> [Element] {
return self.reduce(into: [Element](), { (acc, element) in
let distance = UInt32(range.upperBound - range.lowerBound + 1)
let count = Int(arc4random_uniform(distance) + UInt32(range.lowerBound))
let result = Array.init(repeating: element, count: count)
acc.append(contentsOf: result)
})
}
}
let sequence = [1, 6, 3, 8, 9]
sequence.duplicateItemsRandomly(range: 2...3)
// [1, 1, 6, 6, 6, 3, 3, 3, 8, 8, 8, 9, 9, 9]
P.S: If you're writing this code in Swift 4.2, please use Int.random(in:).
I'd suggest a custom Sequence:
class RepeatingRandomSequence : Sequence {
let rangeLow, rangeSpan : UInt32
let repeatLow, repeatSpan : UInt32
init(range:Range<UInt32>, count:Range<UInt32>) {
rangeLow = range.lowerBound
rangeSpan = range.upperBound - range.lowerBound + 1
repeatLow = count.lowerBound
repeatSpan = count.upperBound - count.lowerBound + 1
}
func makeIterator() -> AnyIterator<UInt32> {
var count : UInt32 = 0
var value : UInt32 = 0
return AnyIterator {
if(count <= 0) {
count = arc4random_uniform(self.repeatSpan) + self.repeatLow
value = arc4random_uniform(self.rangeSpan) + self.rangeLow
}
defer { count = count - 1 }
return value
}
}
}
let sequence = RepeatingRandomSequence(range: 0..<10, count: 2..<3)
let randoms = sequence.makeIterator()
Note that the iterator, randoms now generates an endless sequence of random numbers using randoms.next() Since the sequence is endless, many things aren't particularly useful, like sort, map, etc. You could however use it like:
for value in random {
print(value)
if(value == 9) { // or any other termination condition
break
}
}
Or more conventionally, as:
(0..<10).forEach { _ in
print(String(describing: random.next()))
}

Swift 4 - Accurately getting large factorials in playgrounds [duplicate]

I have written this function to return the factorial of a given number
func factorial(_ n: Int) -> Int {
if n == 0 {
return 1
}
else {
return n * factorial(n - 1)
}
}
print( factorial(20) ) // 2432902008176640000
Works as it should, as long the given number does not exceed 20, because then the result becomes too high!
How can I circumvent this limit and thus calculate the factorial of higher numbers?
I have searched around and found some bignum libraries for Swift. I'm doing this to learn and be familiar with Swift, therefore I want to figure this out on my own.
Here's an approach that will let you find very large factorials.
Represent large numbers as an array of digits. For instance 987 would be [9, 8, 7]. Multiplying that number by an integer n would require two steps.
Multiply each value in that array by n.
Perform a carry operation to return a result that is again single digits.
For example 987 * 2:
let arr = [9, 8, 7]
let arr2 = arr.map { $0 * 2 }
print(arr2) // [18, 16, 14]
Now, perform the carry operation. Starting at the one's digit, 14 is too big, so keep the 4 and carry the 1. Add the 1 to 16 to get 17.
[18, 17, 4]
Repeat with the ten's place:
[19, 7, 4]
And then with the hundred's place:
[1, 9, 7, 4]
Finally, for printing, you could convert this back to a string:
let arr = [1, 9, 7, 4]
print(arr.map(String.init).joined())
1974
Applying that technique, here is a carryAll function that performs the carry operation, and a factorial that uses it to calculate very large factorials:
func carryAll(_ arr: [Int]) -> [Int] {
var result = [Int]()
var carry = 0
for val in arr.reversed() {
let total = val + carry
let digit = total % 10
carry = total / 10
result.append(digit)
}
while carry > 0 {
let digit = carry % 10
carry = carry / 10
result.append(digit)
}
return result.reversed()
}
func factorial(_ n: Int) -> String {
var result = [1]
for i in 2...n {
result = result.map { $0 * i }
result = carryAll(result)
}
return result.map(String.init).joined()
}
print(factorial(1000))
402387260077093773543702433923003985719374864210714632543799910429938512398629020592044208486969404800479988610197196058631666872994808558901323829669944590997424504087073759918823627727188732519779505950995276120874975462497043601418278094646496291056393887437886487337119181045825783647849977012476632889835955735432513185323958463075557409114262417474349347553428646576611667797396668820291207379143853719588249808126867838374559731746136085379534524221586593201928090878297308431392844403281231558611036976801357304216168747609675871348312025478589320767169132448426236131412508780208000261683151027341827977704784635868170164365024153691398281264810213092761244896359928705114964975419909342221566832572080821333186116811553615836546984046708975602900950537616475847728421889679646244945160765353408198901385442487984959953319101723355556602139450399736280750137837615307127761926849034352625200015888535147331611702103968175921510907788019393178114194545257223865541461062892187960223838971476088506276862967146674697562911234082439208160153780889893964518263243671616762179168909779911903754031274622289988005195444414282012187361745992642956581746628302955570299024324153181617210465832036786906117260158783520751516284225540265170483304226143974286933061690897968482590125458327168226458066526769958652682272807075781391858178889652208164348344825993266043367660176999612831860788386150279465955131156552036093988180612138558600301435694527224206344631797460594682573103790084024432438465657245014402821885252470935190620929023136493273497565513958720559654228749774011413346962715422845862377387538230483865688976461927383814900140767310446640259899490222221765904339901886018566526485061799702356193897017860040811889729918311021171229845901641921068884387121855646124960798722908519296819372388642614839657382291123125024186649353143970137428531926649875337218940694281434118520158014123344828015051399694290153483077644569099073152433278288269864602789864321139083506217095002597389863554277196742822248757586765752344220207573630569498825087968928162753848863396909959826280956121450994871701244516461260379029309120889086942028510640182154399457156805941872748998094254742173582401063677404595741785160829230135358081840096996372524230560855903700624271243416909004153690105933983835777939410970027753472000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
You can use this library:
BigInt
Install it using CocoaPods:
pod 'BigInt'
Then you can use it like this:
import BigInt
func factorial(_ n: Int) -> BigInt {
if n == 0 {
return 1
}
else {
return BigInt(n) * factorial(n - 1)
}
}
print( factorial(50) ) // 30414093201713378043612608166064768844377641568960512000000000000

Speeding up Swift CodeFight Challenge

Per Codefighters:
Note: Write a solution with O(n) time complexity and O(1) additional space complexity, since this is what you would be asked to do during a real interview.
Given an array a that contains only numbers in the range from 1 to a.length, find the first duplicate number for which the second occurrence has the minimal index. In other words, if there are more than 1 duplicated numbers, return the number for which the second occurrence has a smaller index than the second occurrence of the other number does. If there are no such elements, return -1.
Example
For a = [2, 3, 3, 1, 5, 2], the output should be firstDuplicate(a) = 3.
There are 2 duplicates: numbers 2 and 3. The second occurrence of 3 has a smaller index than than second occurrence of 2 does, so the answer is 3.
For a = [2, 4, 3, 5, 1], the output should be firstDuplicate(a) = -1.
So here is what I came up with. It works but fails on the final test because it ran over 4000ms. I am stuck to what else I can do. Any Ideas to improve speed?
func firstDuplicate(a : [Int]) -> Int {
var duplicateIndexArray = [Int]()
for firstIndex in 0..<a.count {
for secondIndex in 0..<a.count {
if a[firstIndex] == a[secondIndex] && firstIndex != secondIndex {
print(firstIndex, secondIndex)
if !(duplicateIndexArray.contains(firstIndex)){
duplicateIndexArray.append(secondIndex)
break
}
}
}
}
// Check for duplicacy
if duplicateIndexArray.count > 0 {
print(duplicateIndexArray)
return a[duplicateIndexArray.min()!]
}
return -1
}
The O(n) time part is easy, but the O(1) additional space is a bit tricky. Usually, a hash set (or bit array in your case) can be used to check if a number occurred more than once, but that requires O(n) additional space. For O(1) additional space, we can use the source array itself as a bit array by making some of the numbers in it negative.
For example if the first number in the array is 3, then we make the number at position 3-1 negative. If one of the other numbers in the array is also 3, we can check if the number at position 3-1 is negative.
I don't have any experience with Swift, so I'll try to write a function in pseudocode:
function firstDuplicate(a)
result = -1
for i = 0 to a.count - 1
if a[abs(a[i])-1] < 0 then
result = a[i]
exit for loop
else
a[abs(a[i])-1] = -a[abs(a[i])-1]
// optional restore the negative numbers back to positive
for i = 0 to a.count - 1
if a[i] < 0 then
a[i] = -a[i]
return result
Replace this line
for secondIndex in 0..<a.count
with
for secondIndex in firstIndex..<a.count
There is no requirement of double checking
So Your Final code is
func firstDuplicate(a : [Int]) -> Int {
var duplicateIndexArray = [Int]()
for firstIndex in 0..<a.count {
for secondIndex in firstIndex..<a.count {
if a[firstIndex] == a[secondIndex] && firstIndex != secondIndex {
print(firstIndex, secondIndex)
if !(duplicateIndexArray.contains(firstIndex))
{
duplicateIndexArray.append(secondIndex)
break
}
}
}
}
// Check for duplicacy
if duplicateIndexArray.count > 0
{
print(duplicateIndexArray)
return a[duplicateIndexArray.min()!]
}
return -1
}
func firstDuplicate(input: [Int]) -> Int{
var map : [String : Int] = [:]
var result = -1
for i in 0 ..< input.count {
if map["\(input[i])"] != nil {
result = i
break
}
else {
map["\(input[i])"] = i
}
}
return result
}

does Swift array.count get evaluated each time in a loop

I suppose this is a question that must get asked for every language, but when you write for example:
while i < array.count {
...
}
does array.count get evaluated each time the loop runs? Is it better to store it in a let constant before running the loop like this?
let length = array.count
while i < length {
...
}
Array gets the count property because it conforms to Collection. The documentation for count in Collection states
Complexity: O(1) if the collection conforms to RandomAccessCollection; otherwise, O(n), where n is the length of the collection.
Source
Since Array also conforms to RandomAccessCollection, it is a constant time operation to get the count of the array. There shouldn't be any major performance difference between getting it once at the start vs every loop iteration.
while loops (and do while loops) have their predicates evaluated on each iteration.
for loops evaluate the sequences once.
Here's is a demonstration:
var array: [Int]
print("Test Case 1 - while i < array.count")
array = [1, 2, 3, 4, 5, 6]
var i = 0
while i < array.count {
print(array[i])
if i < 3 { array.append(123) }
i += 1
}
print("\r\nTest Case 2 - for i in array.indices")
array = [1, 2, 3, 4, 5, 6]
for i in array.indices {
print(array[i])
if i < 3 { array.append(123) }
}
print("\r\nTest Case 3 - for i in 0 ..< array.count")
array = [1, 2, 3, 4, 5, 6]
for i in 0 ..< array.count {
print(array[i])
if i < 3 { array.append(123) }
}
Test Case 1 - while i < array.count
1
2
3
4
5
6
123
123
123
Test Case 2 - for i in array.indices
1
2
3
4
5
6
Test Case 3 - for i in 0 ..< array.count
1
2
3
4
5
6
Yes it's evaluated on each iteration.
Assigning to a constant will be slightly more performant. However with all of the optimisations in a modern compiler I wouldn't bother. Unless the loop count is going to be humongous.
Careful with for loops like the following. Since elems.count is part of a range that gets constructed once at the top of the loop, it is evaluated exactly once. The following code will die when i = 4:
var elems = [1, 2, 3, 4, 5, 6]
for i in 0 ..< elems.count {
if i % 2 == 0 { // remove even elements
elems.remove(at: i)
}
}
The array.count in your while does indeed get evaluated each time the condition is evaluated.
Yes, it gets called every time
Let's run a simple test, first of all we need the following Array Extension
extension Array {
var myCustomCount: Int {
print("myCustomCount")
return self.count
}
}
And then we can try this code
let nums = [1, 2, 3, 4, 5]
var i = 0
while i < nums.myCustomCount {
i += 1
}
The output is
myCustomCount
myCustomCount
myCustomCount
myCustomCount
myCustomCount
myCustomCount

How do I insert an item repeatedly between every two elements of an array?

I have this code:
let items = [1, 2, 3]
let sep = 0
I want to insert sep between every two elements of items to get a result similar to this:
newItems = [1, 0, 2, 0, 3]
Is there a concise way to do this through functional programming in Swift? Something similar to String.join(), but for arrays.
(Note: The answer has been updated for Swift 3 and later with the help of Brandon's answer and ober's answer).
This does the trick:
let items = [1, 2, 3]
let sep = 0
let newItems = Array(items.map { [$0] }.joined(separator: [sep]))
print(newItems) // [1, 0, 2, 0, 3]
items.map { [ $0 ] } creates an array of single-element arrays, and joined(separator: [sep]) then interposes the separator and concatenates
the elements. The result is a JoinedSequence from which we can create an Array.
As it turns out (benchmarks below) it is quite expensive to create many temporary arrays. This can be avoided by using “single-element collections”:
let newItems = Array(items.map(CollectionOfOne.init).joined(separator: CollectionOfOne(sep)))
Another possible solution is
let newItems = (0 ..< 2 * items.count - 1).map { $0 % 2 == 0 ? items[$0/2] : sep }
which maps even indices to the corresponding element of items,
and odd indices to the separator. This turns out to be the fastest solution for large arrays.
Benchmark: With items containing 1,000,000 elements, compiled in Release mode on a 2.3 GHz Intel Core i7 MacBook Pro, I measured the following approximate execution times:
First method (map + joined with arrays): 0.28 seconds.
Second method (map + joined with CollectionOfOne): 0.035 seconds.
Third method (using only map): 0.015 seconds.
converted to swift 5
extension Array {
func insert(separator: Element) -> [Element] {
(0 ..< 2 * count - 1).map { $0 % 2 == 0 ? self[$0/2] : separator }
}
}
Here is the original answer converted to Swift 3/4
let items = [1, 2, 3]
let sep = 0
let newItems = Array(items.map { [$0] }.joined(separator: [sep]))
print(newItems) // [1, 0, 2, 0, 3]