Is there an efficient method to generate all possible arrays of booleans with a given number of "true" values?
Right now I'm incrementing a number and checking if its binary representation has the given number of 1s (and if so, adding that array). But this becomes extremely slow for larger givens.
This is the kind of input-output that I'm looking for:
(length: 4, trues: 2) -> [[1,1,0,0],[1,0,1,0],[0,1,1,0],[1,0,0,1],[0,1,0,1],[0,0,1,1]]
The trouble is doing it in less than O(2^N), and so that they're ordered as the little endian binary representations would be.
If it helps the length would be a fixed number at compile time (currently it's 64). I wrote it as an input because I might have to increase it to 128, but it won't vary during runtime.
You can define a recursive solution to this problem.
fn solution(length: u32, trues: u32) -> Vec<Vec<bool>>;
How do we formulate this function recursively? Let's think about the last element of the output arrays. If that last element is false, then the amount of true elements in the 0..length-1 elements must be trues. If that last element is true, then the amount of true elements in the 0..length-1 elements must be trues-1.
So we can just answer the problem for (length-1, trues) and then extend them all with false, and answer the problem for (length-1, trues-1) and extend them all with true, and then we can combine the result (putting the ends-with-true case first for little endian ordering). Adding in some base cases, we get the following code:
fn solution(length: u32, trues: u32) -> Vec<Vec<bool>> {
if trues > length {
// no candidate arrays exist
return vec![];
}
if length == 0 {
// one array exists: the empty array
return vec![vec![]];
}
if trues == 0 {
// one array exists: the all-false array
return vec![vec![false; length as usize]];
}
let mut result = Vec::new();
for mut ones in solution(length-1, trues-1) {
ones.push(true);
result.push(ones);
}
for mut zeroes in solution(length-1, trues) {
zeroes.push(false);
result.push(zeroes);
}
result
}
If the time complexity for solution(L, T) is O(S(L,T)), then the time complexity of solution(L,T) can be recursively expressed as O(S(L-1,T-1) + S(L-1,T)), with the base cases S(0,0) = 1, S(L,L+1)=1, S(L,0) = L. This is the best achievable time complexity, since S(L,T) = S(L-1,T-1) + S(L-1,T) is also the recursive formula for how many arrays of length L and true-count T exist. This is also the same recurrence formula as the binomial recurrence equation, but the base cases are different. I will leave computing the time complexity as an exercise to the reader, and there are a couple of trivial optimizations to the above code that can be done to bring down the computation time further.
Related
Suppose I have a Supply, Channel, IO::Handle, or similar stream-like source of text, and I want to scan it for substrings matching a regex. I can't be sure that matching substrings do not cross chunk boundaries. The total length is potentially infinite and cannot be slurped into memory.
One way this would be possible is if I could instantiate a regex matching engine and feed it chunks of text while it maintains its state. But I don't see any way to do that -- I only see methods to run the match engine to completion.
Is this possible?
After some more searching, I may have answered my own question. Specifically, it seems Seq.comb is capable of combining chunks and lazily processing them:
my $c = supply {
whenever Supply.interval(1.0) -> $v {
my $letter = do if ($v mod 2 == 0) { "a" } else { "b" };
my $chunk = $letter x ($v + 1);
say "Pushing {$chunk}";
emit($chunk);
}
};
my $c2 = $c.comb(/a+b+/);
react {
whenever $c2 -> $v {
say "Got {$v}";
}
}
See also the concurrency features used to construct this example.
I would like to generate random numbers from a union of ranges in Kotlin. I know I can do something like
((1..10) + (50..100)).random()
but unfortunately this creates an intermediate list, which can be rather expensive when the ranges are large.
I know I could write a custom function to randomly select a range with a weight based on its width, followed by randomly choosing an element from that range, but I am wondering if there is a cleaner way to achieve this with Kotlin built-ins.
Suppose your ranges are nonoverlapped and sorted, if not, you could have some preprocessing to merge and sort.
This comes to an algorithm choosing:
O(1) time complexity and O(N) space complexity, where N is the total number, by expanding the range object to a set of numbers, and randomly pick one. To be compact, an array or list could be utilized as the container.
O(M) time complexity and O(1) space complexity, where M is the number of ranges, by calculating the position in a linear reduction.
O(M+log(M)) time complexity and O(M) space complexity, where M is the number of ranges, by calculating the position using a binary search. You could separate the preparation(O(M)) and generation(O(log(M))), if there are multiple generations on the same set of ranges.
For the last algorithm, imaging there's a sorted list of all available numbers, then this list can be partitioned into your ranges. So there's no need to really create this list, you just calculate the positions of your range s relative to this list. When you have a position within this list, and want to know which range it is in, do a binary search.
fun random(ranges: Array<IntRange>): Int {
// preparation
val positions = ranges.map {
it.last - it.first + 1
}.runningFold(0) { sum, item -> sum + item }
// generation
val randomPos = Random.nextInt(positions[ranges.size])
val found = positions.binarySearch(randomPos)
// binarySearch may return an "insertion point" in negative
val range = if (found < 0) -(found + 1) - 1 else found
return ranges[range].first + randomPos - positions[range]
}
Short solution
We can do it like this:
fun main() {
println(random(1..10, 50..100))
}
fun random(vararg ranges: IntRange): Int {
var index = Random.nextInt(ranges.sumOf { it.last - it.first } + ranges.size)
ranges.forEach {
val size = it.last - it.first + 1
if (index < size) {
return it.first + index
}
index -= size
}
throw IllegalStateException()
}
It uses the same approach you described, but it calls for random integer only once, not twice.
Long solution
As I said in the comment, I often miss utils in Java/Kotlin stdlib for creating collection views. If IntRange would have something like asList() and we would have a way to concatenate lists by creating a view, this would be really trivial, utilizing existing logic blocks. Views would do the trick for us, they would automatically calculate the size and translate the random number to the proper value.
I implemented a POC, maybe you will find it useful:
fun main() {
val list = listOf(1..10, 50..100).mergeAsView()
println(list.size) // 61
println(list[20]) // 60
println(list.random())
}
#JvmName("mergeIntRangesAsView")
fun Iterable<IntRange>.mergeAsView(): List<Int> = map { it.asList() }.mergeAsView()
#JvmName("mergeListsAsView")
fun <T> Iterable<List<T>>.mergeAsView(): List<T> = object : AbstractList<T>() {
override val size = this#mergeAsView.sumOf { it.size }
override fun get(index: Int): T {
if (index < 0 || index >= size) {
throw IndexOutOfBoundsException(index)
}
var remaining = index
this#mergeAsView.forEach { curr ->
if (remaining < curr.size) {
return curr[remaining]
}
remaining -= curr.size
}
throw IllegalStateException()
}
}
fun IntRange.asList(): List<Int> = object : AbstractList<Int>() {
override val size = endInclusive - start + 1
override fun get(index: Int): Int {
if (index < 0 || index >= size) {
throw IndexOutOfBoundsException(index)
}
return start + index
}
}
This code does almost exactly the same thing as short solution above. It only does this indirectly.
Once again: this is just a POC. This implementation of asList() and mergeAsView() is not at all production-ready. We should implement more methods, like for example iterator(), contains() and indexOf(), because right now they are much slower than they could be. But it should work efficiently already for your specific case. You should probably test it at least a little. Also, mergeAsView() assumes provided lists are immutable (they have fixed size) which may not be true.
It would be probably good to implement asList() for IntProgression and for other primitive types as well. Also you may prefer varargs version of mergeAsView() than extension function.
As a final note: I guess there are libraries that does this already - probably some related to immutable collections. But if you look for a relatively lightweight solution, it should work for you.
I've just started learning programming and I'm having a problem understanding a piece of code from a tutorial. Could anyone explain what the Kotlin code below does?
Thank you
fun f(i:Int, list:MutableList<Int>) : Boolean {
for (number in list) {
if (i % number == 0) {
return false
}
}
return true
}
fun main(args:Array<String>) {
val result = mutableListOf<Int>()
for (number in 2..100) {
if (f(number, result)) {
result.append(number)
}
print(result.joinToString())
}
The main method creates a new list of integers. In a loop from 2 to 200 it calls the function f with current number of the loop (number) and the list created.
The function checks if the number handed over can be divided by any number in the list. In case it can be divided, false is returned, else true.
If the number couldn't be divided then the number is stored inside the list.
So it is a simple algorithm to find prime numbers. The list stores all so far found prime numbers. And the function checks if the number can be divided by any of the prime numbers.
f(...) checks whether i divides with any number in the list - if so, returns false.
main(..) loops through all numbers from 2..100 and adds numbers that don't divide with any number previously added to the list.
Basically, it will print all prime numbers between 2..100
I am pretty confused with both functions fold() and reduce() in Kotlin, can anyone give me a concrete example that distinguishes both of them?
fold takes an initial value, and the first invocation of the lambda you pass to it will receive that initial value and the first element of the collection as parameters.
For example, take the following code that calculates the sum of a list of integers:
listOf(1, 2, 3).fold(0) { sum, element -> sum + element }
The first call to the lambda will be with parameters 0 and 1.
Having the ability to pass in an initial value is useful if you have to provide some sort of default value or parameter for your operation. For example, if you were looking for the maximum value inside a list, but for some reason want to return at least 10, you could do the following:
listOf(1, 6, 4).fold(10) { max, element ->
if (element > max) element else max
}
reduce doesn't take an initial value, but instead starts with the first element of the collection as the accumulator (called sum in the following example).
For example, let's do a sum of integers again:
listOf(1, 2, 3).reduce { sum, element -> sum + element }
The first call to the lambda here will be with parameters 1 and 2.
You can use reduce when your operation does not depend on any values other than those in the collection you're applying it to.
The major functional difference I would call out (which is mentioned in the comments on the other answer, but may be hard to understand) is that reduce will throw an exception if performed on an empty collection.
listOf<Int>().reduce { x, y -> x + y }
// java.lang.UnsupportedOperationException: Empty collection can't be reduced.
This is because .reduce doesn't know what value to return in the event of "no data".
Contrast this with .fold, which requires you to provide a "starting value", which will be the default value in the event of an empty collection:
val result = listOf<Int>().fold(0) { x, y -> x + y }
assertEquals(0, result)
So, even if you don't want to aggregate your collection down to a single element of a different (non-related) type (which only .fold will let you do), if your starting collection may be empty then you must either check your collection size first and then .reduce, or just use .fold
val collection: List<Int> = // collection of unknown size
val result1 = if (collection.isEmpty()) 0
else collection.reduce { x, y -> x + y }
val result2 = collection.fold(0) { x, y -> x + y }
assertEquals(result1, result2)
Another difference that none of the other answers mentioned is the following:
The result of a reduce operation will always be of the same type (or a super type) as the data that is being reduced.
We can see that from the definition of the reduce method:
public inline fun <S, T : S> Iterable<T>.reduce(operation: (acc: S, T) -> S): S {
val iterator = this.iterator()
if (!iterator.hasNext()) throw UnsupportedOperationException("Empty collection can't be reduced.")
var accumulator: S = iterator.next()
while (iterator.hasNext()) {
accumulator = operation(accumulator, iterator.next())
}
return accumulator
}
On the other hand, the result of a fold operation can be anything, because there are no restrictions when it comes to setting up the initial value.
So, for example, let us say that we have a string that contains letters and digits. We want to calculate the sum of all the digits.
We can easily do that with fold:
val string = "1a2b3"
val result: Int = string.fold(0, { currentSum: Int, char: Char ->
if (char.isDigit())
currentSum + Character.getNumericValue(char)
else currentSum
})
//result is equal to 6
reduce - The reduce() method transforms a given collection into a single result.
val numbers: List<Int> = listOf(1, 2, 3)
val sum: Int = numbers.reduce { acc, next -> acc + next }
//sum is 6 now.
fold - What would happen in the previous case of an empty list? Actually, there’s no right value to return, so reduce() throws a RuntimeException
In this case, fold is a handy tool. You can put an initial value by it -
val sum: Int = numbers.fold(0, { acc, next -> acc + next })
Here, we’ve provided initial value. In contrast, to reduce(), if the collection is empty, the initial value will be returned which will prevent you from the RuntimeException.
Simple Answer
Result of both reduce and fold is "a list of items will be transformed into a single item".
In case of fold,we provide 1 extra parameter apart from list but in case of reduce,only items in list will be considered.
Fold
listOf("AC","Fridge").fold("stabilizer") { freeGift, itemBought -> freeGift + itemBought }
//output: stabilizerACFridge
In above case,think as AC,fridge bought from store & they give stabilizer as gift(this will be the parameter passed in the fold).so,you get all 3 items together.Please note that freeGift will be available only once i.e for the first iteration.
Reduce
In case of reduce,we get items in list as parameters and can perform required transformations on it.
listOf("AC","Fridge").reduce { itemBought1, itemBought2 -> itemBought1 + itemBought2 }
//output: ACFridge
The difference between the two functions is that fold() takes an initial value and uses it as the accumulated value on the first step, whereas the first step of reduce() uses the first and the second elements as operation arguments on the first step.
When normally using a for-in-loop, the counter (in this case number) is a constant in each iteration:
for number in 1...10 {
// do something
}
This means I cannot change number in the loop:
for number in 1...10 {
if number == 5 {
++number
}
}
// doesn't compile, since the prefix operator '++' can't be performed on the constant 'number'
Is there a way to declare number as a variable, without declaring it before the loop, or using a normal for-loop (with initialization, condition and increment)?
To understand why i can’t be mutable involves knowing what for…in is shorthand for. for i in 0..<10 is expanded by the compiler to the following:
var g = (0..<10).generate()
while let i = g.next() {
// use i
}
Every time around the loop, i is a freshly declared variable, the value of unwrapping the next result from calling next on the generator.
Now, that while can be written like this:
while var i = g.next() {
// here you _can_ increment i:
if i == 5 { ++i }
}
but of course, it wouldn’t help – g.next() is still going to generate a 5 next time around the loop. The increment in the body was pointless.
Presumably for this reason, for…in doesn’t support the same var syntax for declaring it’s loop counter – it would be very confusing if you didn’t realize how it worked.
(unlike with where, where you can see what is going on – the var functionality is occasionally useful, similarly to how func f(var i) can be).
If what you want is to skip certain iterations of the loop, your better bet (without resorting to C-style for or while) is to use a generator that skips the relevant values:
// iterate over every other integer
for i in 0.stride(to: 10, by: 2) { print(i) }
// skip a specific number
for i in (0..<10).filter({ $0 != 5 }) { print(i) }
let a = ["one","two","three","four"]
// ok so this one’s a bit convoluted...
let everyOther = a.enumerate().filter { $0.0 % 2 == 0 }.map { $0.1 }.lazy
for s in everyOther {
print(s)
}
The answer is "no", and that's a good thing. Otherwise, a grossly confusing behavior like this would be possible:
for number in 1...10 {
if number == 5 {
// This does not work
number = 5000
}
println(number)
}
Imagine the confusion of someone looking at the number 5000 in the output of a loop that is supposedly bound to a range of 1 though 10, inclusive.
Moreover, what would Swift pick as the next value of 5000? Should it stop? Should it continue to the next number in the range before the assignment? Should it throw an exception on out-of-range assignment? All three choices have some validity to them, so there is no clear winner.
To avoid situations like that, Swift designers made loop variables in range loops immutable.
Update Swift 5
for var i in 0...10 {
print(i)
i+=1
}