I am writing a Math Quiz app for my daughter in xcode/swift.Specifically, I want to produce a question that will contain at least one negative number to be added or subtracted against a second randomly generated number.Cannot be two positive numbers.
i.e.
What is (-45) subtract 12?
What is 23 Minus (-34)?
I am struggling to get the syntax right to generate the numbers, then decide if the said number will be a negative or positive.
Then the second issue is randomizing if the problem is to be addition or subtraction.
It's possible to solve this without repeated number drawing. The idea is to:
Draw a random number, positive or negative
If the number is negative: Draw another number from the same range and return the pair.
If the number is positive: Draw the second number from a range constrained to negative numbers.
Here's the implementation:
extension CountableClosedRange where Bound : SignedInteger {
/// A property that returns a random element from the range.
var random: Bound {
return Bound(arc4random_uniform(UInt32(count.toIntMax())).toIntMax()) + lowerBound
}
/// A pair of random elements where always one element is negative.
var randomPair: (Bound, Bound) {
let first = random
if first >= 0 {
return (first, (self.lowerBound ... -1).random)
}
return (first, random)
}
}
Now you can just write...
let pair = (-10 ... 100).randomPair
... and get a random tuple where one element is guaranteed to be negative.
Here's my attempt. Try running this in a playground, it should hopefully get you the result you want. I hope I've made something clean enough...
//: Playground - noun: a place where people can play
import Cocoa
let range = Range(uncheckedBounds: (-50, 50))
func generateRandomCouple() -> (a: Int, b: Int) {
// This function will generate a pair of random integers
// (a, b) such that at least a or b is negative.
var first, second: Int
repeat {
first = Int(arc4random_uniform(UInt32(range.upperBound - range.lowerBound))) - range.upperBound
second = Int(arc4random_uniform(UInt32(range.upperBound - range.lowerBound))) - range.upperBound
}
while (first > 0 && second > 0);
// Essentially this loops until at least one of the two is less than zero.
return (first, second)
}
let couple = generateRandomCouple();
print("What is \(couple.a) + (\(couple.b))")
// at this point, either of the variables is negative
// I don't think you can do it in the playground, but here you would read
// her input and the expected answer would, naturally, be:
print(couple.a + couple.b)
In any case, feel free to ask for clarifications. Good luck !
Related
I want to get one decimal place of a double in Dart. I use the toStringAsFixed() method to get it, but it returns a round-up value.
double d1 = 1.151;
double d2 = 1.150;
print('$d1 is ${d1.toStringAsFixed(1)}');
print('$d2 is ${d2.toStringAsFixed(1)}');
Console output:
1.151 is 1.2
1.15 is 1.1
How can I get it without a round-up value? Like 1.1 for 1.151 too. Thanks in advance.
Not rounding seems highly questionable to me1, but if you really want to truncate the string representation without rounding, then I'd take the string representation, find the decimal point, and create the appropriate substring.
There are a few potential pitfalls:
The value might be so large that its normal string representation is in exponential form. Note that double.toStringAsFixed just returns the exponential form anyway for such large numbers, so maybe do the same thing.
The value might be so small that its normal string representation is in exponential form. double.toStringAsFixed already handles this, so instead of using double.toString, use double.toStringAsFixed with the maximum number of fractional digits.
The value might not have a decimal point at all (e.g. NaN, +infinity, -infinity). Just return those values as they are.
extension on double {
// Like [toStringAsFixed] but truncates (toward zero) to the specified
// number of fractional digits instead of rounding.
String toStringAsTruncated(int fractionDigits) {
// Require same limits as [toStringAsFixed].
assert(fractionDigits >= 0);
assert(fractionDigits <= 20);
if (fractionDigits == 0) {
return truncateToDouble().toString();
}
// [toString] will represent very small numbers in exponential form.
// Instead use [toStringAsFixed] with the maximum number of fractional
// digits.
var s = toStringAsFixed(20);
// [toStringAsFixed] will still represent very large numbers in
// exponential form.
if (s.contains('e')) {
// Ignore values in exponential form.
return s;
}
// Ignore unrecognized values (e.g. NaN, +infinity, -infinity).
var i = s.indexOf('.');
if (i == -1) {
return s;
}
return s.substring(0, i + fractionDigits + 1);
}
}
void main() {
var values = [
1.151,
1.15,
1.1999,
-1.1999,
1.0,
1e21,
1e-20,
double.nan,
double.infinity,
double.negativeInfinity,
];
for (var v in values) {
print(v.toStringAsTruncated(1));
}
}
Another approach one might consider is to multiply by pow(10, fractionalDigits), use double.truncateToDouble, divide by the power-of-10 used earlier, and then use .toStringAsFixed(fractionalDigits). That could work for human-scaled values, but it could generate unexpected results for very large values due to precision loss from floating-point arithmetic. (This approach would work if you used package:decimal instead of double, though.)
1 Not rounding seems especially bad given that using doubles to represent fractional base-10 numbers is inherently imprecise. For example, since the closest IEEE-754 double-precision floating number to 0.7 is 0.6999999999999999555910790149937383830547332763671875, do you really want 0.7.toStringAsTruncated(1) to return '0.6' instead of '0.7'?
I translated a DTW matlab function to Swift. The code looks as follows:
private func dtw(x1 : [Double], x2 : [Double]) -> Double {
let n1 = x1.count;
let n2 = x2.count;
var table = [[Double]](repeating: [Double](repeating: 0, count: n2 + 1), count: 2);
table[0][0] = 0;
for i in 1...n2 { table[0][i] = Double.infinity }
for i in 1 ... n1 {
table[1][0] = Double.infinity;
for j in 1 ... n2 {
let cost = abs(x1[i - 1] - x2[j - 1]);
var min = table[0][j - 1];
if (min > table[0][j]) {
min = table[0][j];
}
if (min > table[1][j - 1]) { min = table[1][j - 1]; }
table[1][j] = cost + min;
}
let swap = table[0];
table[0] = table[1];
table[1] = swap;
}
return table[0][n2];
}
This function takes an average of 16 ms to complete on an iPhone 11. For my use case, this is very slow. I want to investigate ways to improve speed. I recently read these two articles : DTW in Swift Orailly and Parallel programming with Swift. In the first article, there is a good quote:
Our implementation of DTW is naïve, and can be accelerated using parallel computing. To calculate the new row/column in a distance matrix, you don't need to wait until the previous one is finished; you only need it to be filled one cell ahead of your row/column
This would make the for j in 1 ... n2 { for loop an ideal candidate. ( I think ) Looking at the code, only these two operations should be thread-safe due to the read / write:
table[1][j - 1]
table[1][j]
The problem I am currently experiencing in introducing parallel computing ( from article 2 ) is that I cannot figure out how to tell swift run everything in parallel, except when I come to the two below lines, as they depend on their predocessor:
if (min > table[1][j - 1]) { min = table[1][j - 1]; }
table[1][j] = cost + min;
I suspect I could solve this issue with DispatchQueue.concurrentPerform and an NSLock(), if I implemented it correctly. ( I have not ) It could also be the wrong tool of choice, yielding me back to my question:
What can I do, to improve the speed of my DTW function where the only constraint in performing a task is that the previous execution in an array had to have completed ( parallelization, concurrency, etc. ) A code example would go a long way.
Your first problem is that you're creating an array of arrays. This is not an efficient data structure, and is not a "2 dimensional array" in the way most people mean (i.e a matrix). It is an array made up of other arrays, all of which can have arbitrary sizes, and this can be very expensive to mutate. As a rule, if you want a matrix, you should back it with a flat array and use multiplication to find its offsets, particularly if you're mutating it. Instead of table[i][j] you would use table[i * width + j].
But in your case it's even easier, since there are exactly two rows. So you don't a multi-dimensional array at all. You can just use two variables, and it'll be much more efficient. (In my tests, just making this change is about 30% faster than the original code.)
The major thing that slows you down is contention. You read and write to the same array in the loop. That gets in the way of various reordering and caching optimizations. In particular, it happens here:
if (min > table[1][j - 1]) { min = table[1][j - 1]; }
table[1][j] = cost + min;
If you rewrite that using two row variables rather than an array, it still looks like this:
if (min > row1[j - 1]) { min = row1[j - 1] }
row1[j] = cost + min
This forces the previous write to row1 to be fully completed before the next minimum can be computed, and then requires an array lookup to get the value back. But that's not really necessary. You can just cache the previous value between loops. Doing that means the loop only performs reads on row0 and only performs writes on row1. That's good for memory contention.
Putting those together, I wrote it this way. I changed the offsets to run from 0 rather than 1; it just made the code a little simpler to understand IMO. In my tests, this is about 3x faster than the original code for two arrays of 10k elements each.
func dtw(x1 : [Double], x2 : [Double]) -> Double {
let n1 = x1.count
let n2 = x2.count
var row0 = Array(repeating: Double.infinity, count: n2 + 1)
row0[0] = 0
var row1 = Array(repeating: 0.0, count: n2 + 1)
for i in 0 ..< n1 {
row1[0] = .infinity
// Keep track of the last value so we never have to read from row1.
var lastValue = Double.infinity
for j in 0 ..< n2 {
let cost = abs(x1[i] - x2[j])
// Don't be tempted to use the 3-value version of `min` here. It's much slower.
var minimum = min(row0[j], row0[j + 1])
minimum = min(minimum, lastValue)
lastValue = cost + minimum
row1[j + 1] = lastValue
}
swap(&row0, &row1)
}
return row0[n2];
}
This code is somewhat hard to make parallel, because the operations are not independent. Each row depends on the other rows. The key to good queue-based parallelism is the ability to split up fairly large chunks of independent work, and then efficiently combine them at the end. The cost of coordination will eat your benefits if the work units are too small. In many cases, vectorization (SIMD) is much more efficient than dispatching to multiple queues.
The cost function is independent, and I explored computing it with Accelerate (the main vectorization framework), but this generally made things slower. The compiler is very good at optimizing simple math in loops, and will do quite a lot of vectorizing for you if you let it. Accelerate is best when you need to do an expensive, consistent, and independent computation on a lot of values. And this loop isn't expensive or independent.
I’m trying to make a basic simulation of a 16 bit computer with Swift. The computer will feature
An ALU
2 registers
That’s all. I have enough knowledge to create these parts visually and understand how they work, but it has become increasingly difficult to make larger components with more inputs while using my current approach.
My current approach has been to wrap each component in a struct. This worked early on, but is becoming increasingly difficult to manage multiple inputs while staying true to the principles of computer science.
The primary issue is that the components aren’t updating with the clock signal. I have the output of the component updating when get is called on the output variable, c. This, however, neglects the idea of a clock signal and will likely cause further problems later on.
It’s also difficult to make getters and setters for each variable without getting errors about mutability. Although I have worked through these errors, they are annoying and slow down the development process.
The last big issue is updating the output. The output doesn’t update when the inputs change; it updates when told to do so. This isn’t accurate to the qualities of real computers and is a fundamental error.
This is an example. It is the ALU I mentioned earlier. It takes two 16 bit inputs and outputs 16 bits. It has two unary ALUs, which can make a 16 bit number zero, negate it, or both. Lastly, it either adds or does a bit wise and comparison based on the f flag and inverts the output if the no flag is selected.
struct ALU {
//Operations are done in the order listed. For example, if zx and nx are 1, it first makes input 1 zero and then inverts it.
var x : [Int] //Input 1
var y : [Int] //Input 2
var zx : Int //Make input 1 zero
var zy : Int //Make input 2 zero
var nx : Int //Invert input 1
var ny : Int //Invert input 2
var f : Int //If 0, do a bitwise AND operation. If 1, add the inputs
var no : Int //Invert the output
public var c : [Int] { //Output
get {
//Numbers first go through unary ALUs. These can negate the input (and output the value), return 0, or return the inverse of 0. They then undergo the operation specified by f, either addition or a bitwise and operation, and are negated if n is 1.
var ux = UnaryALU(z: zx, n: nx, x: x).c //Unary ALU. See comments for more
var uy = UnaryALU(z: zy, n: ny, x: y).c
var fd = select16(s: f, d1: Add16(a: ux, b: uy).c, d0: and16(a: ux, b: uy).c).c //Adds a 16 bit number or does a bitwise and operation. For more on select16, see the line below.
var out = select16(s: no, d1: not16(a: fd).c, d0: fd).c //Selects a number. If s is 1, it returns d1. If s is 0, it returns d0. d0 is the value returned by fd, while d1 is the inverse.
return out
}
}
public init(x:[Int],y:[Int],zx:Int,zy:Int,nx:Int,ny:Int,f:Int,no:Int) {
self.x = x
self.y = y
self.zx = zx
self.zy = zy
self.nx = nx
self.ny = ny
self.f = f
self.no = no
}
}
I use c for the output variable, store values with multiple bits in Int arrays, and store single bits in Int values.
I’m doing this on Swift Playgrounds 3.0 with Swift 5.0 on a 6th generation iPad. I’m storing each component or set of components in a separate file in a module, which is why some variables and all structs are marked public. I would greatly appreciate any help. Thanks in advance.
So, I’ve completely redone my approach and have found a way to bypass the issues I was facing. What I’ve done is make what I call “tracker variables” for each input. When get is called for each variable, it returns that value of the tracker assigned to it. When set is called it calls an update() function that updates the output of the circuit. It also updates the value of the tracker. This essentially creates a ‘copy’ of each variable. I did this to prevent any infinite loops.
Trackers are unfortunately necessary here. I’ll demonstrate why
var variable : Type {
get {
return variable //Calls the getter again, resulting in an infinite loop
}
set {
//Do something
}
}
In order to make a setter, Swift requires a getter to be made as well. In this example, calling variable simply calls get again, resulting in a never-ending cascade of calls to get. Tracker variables are a workaround that use minimal extra code.
Using an update method makes sure the output responds to a change in any input. This also works with a clock signal, due to the architecture of the components themselves. Although it appears to act as the clock, it does not.
For example, in data flip-flops, the clock signal is passed into gates. All a clock signal does is deactivate a component when the signal is off. So, I can implement that within update() while remaining faithful to reality.
Here’s an example of a half adder. Note that the tracker variables I mentioned are marked by an underscore in front of their name. It has two inputs, x and y, which are 1 bit each. It also has two outputs, high and low, also known as carry and sum. The outputs are also one bit.
struct halfAdder {
private var _x : Bool //Tracker for x
public var x: Bool { //Input 1
get {
return _x //Return the tracker’s value
}
set {
_x = x //Set the tracker to x
update() //Update the output
}
}
private var _y : Bool //Tracker for y
public var y: Bool { //Input 2
get {
return _y
}
set {
_y = y
update()
}
}
public var high : Bool //High output, or ‘carry’
public var low : Bool //Low output, or ‘sum’
internal mutating func update(){ //Updates the output
high = x && y //AND gate, sets the high output
low = (x || y) && !(x && y) //XOR gate, sets the low output
}
public init(x:Bool, y:Bool){ //Initializer
self.high = false //This will change when the variables are set, ensuring a correct output.
self.low = false //See above
self._x = x //Setting trackers and variables
self._y = y
self.x = x
self.y = y
}
}
This is a very clean way, save for the trackers, do accomplish this task. It can trivially be expanded to fit any number of bits by using arrays of Bool instead of a single value. It respects the clock signal, updates the output when the inputs change, and is very similar to real computers.
I want to write an algorithm which allows me to rescale numbers to between 0 and 1. This means if I pass 25, 100, 500 then it should generate a new scale and represent those numbers on a scale of 0 to 1.
Here is what I have which is incorrect and does not make sense.
height: item.height/item.height * 20
Pass in the numbers in an array.
Loop through the numbers and find the max.
Map the array of integers to an array of Doubles, each one being the value from the source array, divided by the max.
Try to write that code. If you have trouble, update your question with your attempt and tell us what's going wrong.
EDIT:
Your answer shows how to print your resulting scaled values, but you implied that you actually want to create a new array containing the scaled values. For that you could use a function like this:
func scaleArray(_ sourceArray: [Int]) -> [Double] {
guard let max = sourceArray.max() else {
return [Double]()
}
return sourceArray.map {
return Double($0)/Double(max)
}
}
Edit #2:
Here is code that would let you test the above:
func scaleAndPrintArray(_ sourceArray: [Int]) {
let scaledArray = scaleArray(sourceArray)
for index in 0..<sourceArray.count {
print(String(format: "%3d", sourceArray[index]), String(format: "%0.5f",scaledArray[index]))
}
}
for arrayCount in 1...5 {
let arraySize = Int(arc4random_uniform(15)) + 5
var array = [Int]()
for _ in 1..<arraySize {
array.append(Int(arc4random_uniform(500)))
}
scaleAndPrintArray(array)
if arrayCount < 5 {
print("-----------")
}
}
(Sorry but I don't know swift)
If you're wanting to create a linear scale, a linear equation is y(x) = m*x + c. You wish the output to range from 0 to 1 when the input ranges from the minimum value to the maximum (your question is ambiguous, maybe you may wish to lock y(0) to 0).
y(0) = min
y(1) = max
therefore
c = min
m = max - min
and to find the value of any intervening value
y = m*x + c
Say I have two arrays:
var arrayOne = ["Hi", "Hello", "Hey", "Howdy"]
var arrayOne = ["Hi", "Hello", "Hey", "Not Howdy"]
What could I do to compare how similar the array elements are? As in a function that would return 75% Because the first three elements are the same but the last element are not. The arrays I'm using in my project are strings but they will almost entirely match except for a few elements. I need to see What percent the differences are. Any ideas?
let arrayOne = ["Hi", "Hello", "Hey", "Howdy"]
let arrayTwo = ["Hi", "Hello", "Hey", "Not Howdy"]
var matches = 0
for (index, item) in enumerate(arrayOne) {
if item == arrayTwo[index] {
matches++
}
}
Double(matches) / Double(arrayOne.count) // 0.75
Both of these algorithms use the idea that if you have two different length arrays, the highest similarity you can have is short length / long length, meaning that the difference in the array lengths are counted as not matching.
You could add all of the terms to a set and then make your percentage the size of the set / length of longest array.
You could sort both arrays and then do a loop with an index variable for each array and compare the values at the two indices, advancing the index for the array that has the "lower" value in the comparison, or increment a counter if they are equivalent. Your percentage would be the counter / length of longest array.
One thing to think about though is how you want to measure similarity in weird cases. Suppose you have two arrays: [1, 2, 3, 4, 5] and [1, 1, 1, 1, 1]. I don't know whether you would want to say they are completely similar, since all of the elements in the second array are in the first array, or if they only have a similarity of 20% because once the 1 in the first array is "used", it can't be used again.
Just some thoughts.
maybe something like this? (written off top of my head so havent checked if it actually compiles)
var arrayOne = ["Hi", "Hello", "Hey", "Howdy"]
var arrayTwo = ["Hi", "Hello", "Hey", "Not Howdy"]
var matches = 0
for i in 0...arrayOne.count { //assuming the arrays are always the same length
if arrayOne[i] == arrayTwo[i]{
matches++
}
}
var percent = matches / arrayOne.count
A good way to measure the similarity of 2 arrays is to iterate all elements of an array, and keep a cursor on the 2nd array, such that at any time the current element of the iterated array is not greater than the element at the cursor position.
As you may argue, this algorithm require elements to be comparable, and as such it works if the arrays type implements the Comparable interface.
I've worked on a generic function that perform that calculation, here it is:
func compare<T: Comparable>(var lhs: [T], var rhs: [T]) -> (matches: Int, total: Int) {
lhs.sort { $0 < $1 } // Inline sort
rhs.sort { $0 < $1 } // Inline sort
var matches = 0
var rightSequence = SequenceOf(rhs).generate()
var right = rightSequence.next()
for left in lhs {
while right != nil && left > right {
right = rightSequence.next()
}
if left == right {
++matches
right = rightSequence.next()
}
}
return (matches: matches, total: max(lhs.count, rhs.count))
}
Let me say that the implementation can probably be optimized, but my goal here is to show the algorithm, not to provide its best implementation.
The first thing to do is to obtain a sorted version of each of the 2 arrays - for simplicity, I have declared both parameters as var, which allows me to edit them, leaving all changes in the local scope. That's way I am using in-place sort.
A sequence on the 2nd array is created, called rightSequence, and the first element is extracted, copied into the right variable.
Then the first array is iterated over - for each element, the sequence is advanced to the next element until the left element is not greater than the right one.
Once this is done, left and right are compared for equality, in which case the counter of matches is incremented.
The algorithm works for arrays having repetitions, different sizes, etc.