I am pretty Newbie to programming. And I am trying to pile up the random blocks dynamically till it hits the upper frame. But it seems that Swift doesn't let me to do so. Did I miss anything please? Any input are appreciated.
let blocks =[block1,block2,block3,block4,block5,block6,block7,block8,block9,block10,block11,block12]
var block:SKSpriteNode!
let blockX:Double = 0.0
var blockY:Double = -(self.size.height/2)
repeat{
block = blocks.randomBlock()
block.zPosition = 2
block.position = CGPoint(x:blockX, y:blockY)
block.size.height = 50
block.size.width = 50
self.addChild(block)
blockY += 50
} while( block.position.y < self.size.height)
extension Array {
func randomBlock()-> Element {
let randint = Int(arc4random_uniform(UInt32(self.count)))
return self[randint]
}
}
you need to have someway of tracking which blocks have been selected and ensure that they don't get selected again. The method below uses an array to store the indexes of selected blocks and then uses recursion to find a cycle through until an unused match is found.
private var usedBlocks = [Int]()
func randomBlock() -> Int {
guard usedBlocks.count != blocks.count else { return -1 }
let random = Int(arc4random_uniform(UInt32(blocks.count)))
if usedBlocks.contains(random) {
return randomBlock()
}
usedBlocks.append(random)
return random
}
in your loop change your initializer to
let index = randomBlock()
if index > -1 {
block = blocks[index]
block.zPosition = 2
block.position = CGPoint(x:blockX, y:blockY)
}
remember that if you restart the game or start a new level, etc. you must clear all of the objects from usedBlocks
usedBlocks.removeAll()
I have a function in Swift that computes the hamming distance of two strings and then puts them into a connected graph if the result is 1.
For example, read to hear returns a hamming distance of 2 because read[0] != hear[0] and read[3] != hear[3].
At first, I thought my function was taking a long time because of the quantity of input (8,000+ word dictionary), but I knew that several minutes was too long. So, I rewrote my same algorithm in Java, and the computation took merely 0.3s.
I have tried writing this in Swift two different ways:
Way 1 - Substrings
extension String {
subscript (i: Int) -> String {
return self[Range(i ..< i + 1)]
}
}
private func getHammingDistance(w1: String, w2: String) -> Int {
if w1.length != w2.length { return -1 }
var counter = 0
for i in 0 ..< w1.length {
if w1[i] != w2[i] { counter += 1 }
}
return counter
}
Results: 434 seconds
Way 2 - Removing Characters
private func getHammingDistance(w1: String, w2: String) -> Int {
if w1.length != w2.length { return -1 }
var counter = 0
var c1 = w1, c2 = w2 // need to mutate
let length = w1.length
for i in 0 ..< length {
if c1.removeFirst() != c2.removeFirst() { counter += 1 }
}
return counter
}
Results: 156 seconds
Same Thing in Java
Results: 0.3 seconds
Where it's being called
var graph: Graph
func connectData() {
let verticies = graph.canvas // canvas is Array<Node>
// Node has key that holds the String
for vertex in 0 ..< verticies.count {
for compare in vertex + 1 ..< verticies.count {
if getHammingDistance(w1: verticies[vertex].key!, w2: verticies[compare].key!) == 1 {
graph.addEdge(source: verticies[vertex], neighbor: verticies[compare])
}
}
}
}
156 seconds is still far too inefficient for me. What is the absolute most efficient way of comparing characters in Swift? Is there a possible workaround for computing hamming distance that involves not comparing characters?
Edit
Edit 1: I am taking an entire dictionary of 4 and 5 letter words and creating a connected graph where the edges indicate a hamming distance of 1. Therefore, I am comparing 8,000+ words to each other to generate edges.
Edit 2: Added method call.
Unless you chose a fixed length character model for your strings, methods and properties such as .count and .characters will have a complexity of O(n) or at best O(n/2) (where n is the string length). If you were to store your data in an array of character (e.g. [Character] ), your functions would perform much better.
You can also combine the whole calculation in a single pass using the zip() function
let hammingDistance = zip(word1.characters,word2.characters)
.filter{$0 != $1}.count
but that still requires going through all characters of every word pair.
...
Given that you're only looking for Hamming distances of 1, there is a faster way to get to all the unique pairs of words:
The strategy is to group words by the 4 (or 5) patterns that correspond to one "missing" letter. Each of these pattern groups defines a smaller scope for word pairs because words in different groups would be at a distance other than 1.
Each word will belong to as many groups as its character count.
For example :
"hear" will be part of the pattern groups:
"*ear", "h*ar", "he*r" and "hea*".
Any other word that would correspond to one of these 4 pattern groups would be at a Hamming distance of 1 from "hear".
Here is how this can be implemented:
// Test data 8500 words of 4-5 characters ...
var seenWords = Set<String>()
var allWords = try! String(contentsOfFile: "/usr/share/dict/words")
.lowercased()
.components(separatedBy:"\n")
.filter{$0.characters.count == 4 || $0.characters.count == 5}
.filter{seenWords.insert($0).inserted}
.enumerated().filter{$0.0 < 8500}.map{$1}
// Compute patterns for a Hamming distance of 1
// Replace each letter position with "*" to create patterns of
// one "non-matching" letter
public func wordH1Patterns(_ aWord:String) -> [String]
{
var result : [String] = []
let fullWord : [Character] = aWord.characters.map{$0}
for index in 0..<fullWord.count
{
var pattern = fullWord
pattern[index] = "*"
result.append(String(pattern))
}
return result
}
// Group words around matching patterns
// and add unique pairs from each group
func addHamming1Edges()
{
// Prepare pattern groups ...
//
var patternIndex:[String:Int] = [:]
var hamming1Groups:[[String]] = []
for word in allWords
{
for pattern in wordH1Patterns(word)
{
if let index = patternIndex[pattern]
{
hamming1Groups[index].append(word)
}
else
{
let index = hamming1Groups.count
patternIndex[pattern] = index
hamming1Groups.append([word])
}
}
}
// add edge nodes ...
//
for h1Group in hamming1Groups
{
for (index,sourceWord) in h1Group.dropLast(1).enumerated()
{
for targetIndex in index+1..<h1Group.count
{ addEdge(source:sourceWord, neighbour:h1Group[targetIndex]) }
}
}
}
On my 2012 MacBook Pro, the 8500 words go through 22817 (unique) edge pairs in 0.12 sec.
[EDIT] to illustrate my first point, I made a "brute force" algorithm using arrays of characters instead of Strings :
let wordArrays = allWords.map{Array($0.unicodeScalars)}
for i in 0..<wordArrays.count-1
{
let word1 = wordArrays[i]
for j in i+1..<wordArrays.count
{
let word2 = wordArrays[j]
if word1.count != word2.count { continue }
var distance = 0
for c in 0..<word1.count
{
if word1[c] == word2[c] { continue }
distance += 1
if distance > 1 { break }
}
if distance == 1
{ addEdge(source:allWords[i], neighbour:allWords[j]) }
}
}
This goes through the unique pairs in 0.27 sec. The reason for the speed difference is the internal model of Swift Strings which is not actually an array of equal length elements (characters) but rather a chain of varying length encoded characters (similar to the UTF model where special bytes indicate that the following 2 or 3 bytes are part of a single character. There is no simple Base+Displacement indexing of such a structure which must always be iterated from the beginning to get to the Nth element.
Note that I used unicodeScalars instead of Character because they are 16 bit fixed length representations of characters that allow a direct binary comparison. The Character type isn't as straightforward and take longer to compare.
Try this:
extension String {
func hammingDistance(to other: String) -> Int? {
guard self.characters.count == other.characters.count else { return nil }
return zip(self.characters, other.characters).reduce(0) { distance, chars in
distance + (chars.0 == chars.1 ? 0 : 1)
}
}
}
print("read".hammingDistance(to: "hear")) // => 2
The following code executed in 0.07 secounds for 8500 characters:
func getHammingDistance(w1: String, w2: String) -> Int {
if w1.characters.count != w2.characters.count {
return -1
}
let arr1 = Array(w1.characters)
let arr2 = Array(w2.characters)
var counter = 0
for i in 0 ..< arr1.count {
if arr1[i] != arr2[i] { counter += 1 }
}
return counter
}
After some messing around, I found a faster solution to #Alexander's answer (and my previous broken answer)
extension String {
func hammingDistance(to other: String) -> Int? {
guard !self.isEmpty, !other.isEmpty, self.characters.count == other.characters.count else {
return nil
}
var w1Iterator = self.characters.makeIterator()
var w2Iterator = other.characters.makeIterator()
var distance = 0;
while let w1Char = w1Iterator.next(), let w2Char = w2Iterator.next() {
distance += (w1Char != w2Char) ? 1 : 0
}
return distance
}
}
For comparing strings with a million characters, on my machine it's 1.078 sec compared to 1.220 sec, so roughly a 10% improvement. My guess is this is due to avoiding .zip and the slight overhead of .reduce and tuples
As others have noted, calling .characters repeatedly takes time. If you convert all of the strings once, it should help.
func connectData() {
let verticies = graph.canvas // canvas is Array<Node>
// Node has key that holds the String
// Convert all of the keys to utf16, and keep them
let nodesAsUTF = verticies.map { $0.key!.utf16 }
for vertex in 0 ..< verticies.count {
for compare in vertex + 1 ..< verticies.count {
if getHammingDistance(w1: nodesAsUTF[vertex], w2: nodesAsUTF[compare]) == 1 {
graph.addEdge(source: verticies[vertex], neighbor: verticies[compare])
}
}
}
}
// Calculate the hamming distance of two UTF16 views
func getHammingDistance(w1: String.UTF16View, w2: String.UTF16View) -> Int {
if w1.count != w2.count {
return -1
}
var counter = 0
for i in w1.startIndex ..< w1.endIndex {
if w1[i] != w1[i] {
counter += 1
}
}
return counter
}
I used UTF16, but you might want to try UTF8 depending on the data. Since I don't have the dictionary you are using, please let me know the result!
*broken*, see new answer
My approach:
private func getHammingDistance(w1: String, w2: String) -> Int {
guard w1.characters.count == w2.characters.count else {
return -1
}
let countArray: Int = w1.characters.indices
.reduce(0, {$0 + (w1[$1] == w2[$1] ? 0 : 1)})
return countArray
}
comparing 2 strings of 10,000 random characters took 0.31 seconds
To expand a bit: it should only require one iteration through the strings, adding as it goes.
Also it's way more concise š.
Gives an Index out of range eror.
Is there a syntax error or logic ?
func generateGameBoard()->([Int]){
var gboard = [Int]();
var i : Int = 0;
for(i=0;i<8;i++){
gboard[i]=1;
}
return gboard;
}
}
Dont you notice error in your code. You create an empty array and then ask index for 0 ..< 8 which is invalid. You should really use count to iterate over the contents.
func generateGameBoard()->([Int]){
var gboard = [Int]();
for i in 0 ..< gboard.count {
gboard[i]=1;
}
return gboard;
}
var gboard = [Int](); // you are creating an empty array here.
you need to append value in array
like
gboard.append(1) instead of gboard[i]=1;
and c style for loop and ++ opeartor will not use in next versions of swift.
You should also get ready for swift 3 and update the for loop part. It won't compile as it stand now in swift 3. You have to change it to:
for i in 0..<8 {
}
I'm new to Swift and pretty new to programming. I'm not sure if it's because it's 2:34 am but can somebody help me identify the difference between these two loops?
The first one is resulting in the values I want yet uses (as the Swift documentation explains) a 'traditional C loop' where the one after is using what seems like a Swift preferred 'for index' loop (which I personally like as it is clearer to me), the problem is that this loop returns every number rather than meeting the conditional.
func findLargestPrimeFactor(number: Int) {
for var i = 2; i < number; i += i {
if number/i % 1 > 0 {
} else {
print(i);
}
}
}
findLargestPrimeFactor(13195);
Below is returning every number to 13195
func findLargestPrimeFactor(number: Int) {
for i in 2...number {
if number/i % 1 > 0 {
} else {
print(i);
}
}
}
findLargestPrimeFactor(13195);
I've realised how stupid my mistakes were, first of all I have no idea why I was doing i += i, that just didn't make sense at all.
Also, my conditional was never being met because I declared it as an Int when really I needed a Float because I care about remainders!
Edit: I've updated my code yet again (taken everyones feedback into consideration), I feel a bit better with this and will continue working on the solution.
Thanks to everyone for your help!
Updated code
func findLargestPrimeFactor(number: Int) {
for i in 2...number {
var n = Float(number)
var count = Float(i)
if n/count % 1 > 0 {
print("not divisible")
} else {
print(i)
}
}
}
findLargestPrimeFactor(13195);
In Swift, I need to create a simple for-condition-increment loop with all the multiples of 3 from 3-100. So far I have:
var multiplesOfThree: [String] = []
for var counter = 0; counter < 30; ++counter {
multiplesOfThree.append("0")
if counter == 3 {
multiplesOfThree.append("3")
} else if counter == 6 {
multiplesOfThree.append("6")
} else if counter == 9 {
multiplesOfThree.append("9")
}
println("Adding \(multiplesOfThree[counter]) to the Array.")
}
I would like to replace all the if and else if statements with something like:
if (index %3 == 0)
but Iām not sure what the proper syntax would be? Also, if I have a single IF statement do I need a .append line to add to the Array?
You are very much on the right track. A few notes:
Swift provides a more concise way to iterate over a fixed number of integers using the ..< operator (an open range operator).
Your if statement with the modulus operator is exactly correct
To make a string from an Int you can use \(expression) inside a string. This is called String Interpolation
Here is the working code:
var multiplesOfThree: [String] = []
for test in 0..<100 {
if (test % 3 == 0) {
multiplesOfThree.append("\(test)")
}
}
However, there is no reason to iterate over every number. You can simply continue to add 3 until you reach your max:
var multiplesOfThree: [String] = []
var multiple = 0
while multiple < 100 {
multiplesOfThree.append("\(multiple)")
multiple += 3
}
As rickster pointed out in the comments, you can also do this in a more concise way using a Strided Range with the by method:
var multiplesOfThree: [String] = []
for multiple in stride(from: 0, to: 100, by: 3) {
multiplesOfThree.append("\(multiple)")
}
Getting even more advanced, you can use the map function to do this all in one line. The map method lets you apply a transform on every element in an array:
let multiplesOfThree = Array(map(stride(from: 0, to: 100, by: 3), { "\($0)" }))
Note: To understand this final code, you will need to understand the syntax around closures well.