Trampolining scalaz' Monad.whileM_ to prevent stack overflow - monads

I'm using scalaz' Monad.whileM_ to implement a while loop in a functional way as follows:
object Main {
import scalaz._
import Scalaz._
import scala.language.higherKinds
case class IState(s: Int)
type IStateT[A] = StateT[Id, IState, A]
type MTransT[S[_], A] = EitherT[S, String, A]
type MTrans[A] = MTransT[IStateT, A]
def eval(k: Int): MTrans[Int] = {
for {
state <- get[IState].liftM[MTransT]
_ <- put(state.copy(s = (state.s + 1) % k)).liftM[MTransT]
} yield (k + 1)
}
def evalCond(): MTrans[Boolean] = {
for {
state <- get[IState].liftM[MTransT]
} yield (state.s != 0)
}
def run() = {
val k = 10
eval(k).whileM_(evalCond()).run(IState(1))
}
}
While this works for small k, it results in a StackOverflow error for large k (e.g. 1000000). Is there a way to trampoline whileM_ or is there a better way to be stack safe?

Use scalaz.Free.Trampoline instead of scalaz.Id.Id.
type IStateT[A] = StateT[Trampoline, IState, A]
The state operations used here return State[S, A] which is just an alias for StateT[Id, S, A]. You need to use the lift[M[_]] function defined on StateT to lift StateT[Id, S, A] to StateT[Trampoline, S, A].
def eval(k: Int): MTrans[Int] = {
for {
state <- get[IState].lift[Trampoline].liftM[MTransT]
_ <- put(state.copy(s = (state.s + 1) % k)).lift[Trampoline].liftM[MTransT]
} yield (k + 1)
}
def evalCond(): MTrans[Boolean] = {
for {
state <- get[IState].lift[Trampoline].liftM[MTransT]
} yield (state.s != 0)
}
Finally, calling .run(IState(1)) now results in Trampoline[(IState, String \/ Unit)]. You must additionally run this as well.
eval(k).whileM_(evalCond()).run(IState(1)).run

Related

Function always return Nil

I am trying to resolve some anagrams assignments. And I can't figure out the problem behind getting always a List() when running my sentenceAnagrams function. Any Help !
type Word = String
type Sentence = List[Word]
type Occurrences = List[(Char, Int)]
def combinations(occurrences: Occurrences): List[Occurrences] = occurrences match {
case Nil => List(Nil)
case x :: xs => (for {z <- combinations(xs); i <- 1 to x._2} yield (x._1, i) :: z).union(combinations(xs))
}
def subtract(x: Occurrences, y: Occurrences): Occurrences = {
if (y.isEmpty) x
else {
val yMap = y.toMap withDefaultValue 0
x.foldLeft(x) { (z, i) => if (combinations(x).contains(y)) {
val diff = i._2 - yMap.apply(i._1)
if (diff > 0) z.toMap.updated(i._1, diff).toList else z.toMap.-(i._1).toList
} else z
}
}}
--
def sentenceAnagrams(sentence: Sentence): List[Sentence] = {
def sentenceAnag(occ: Occurrences): List[Sentence] =
if (occ.isEmpty) List(List())
else (for {
comb <- combinations(occ)
word <- (dictionaryByOccurrences withDefaultValue List()).apply(comb)
otherSentence <- sentenceAnag(subtract(occ, comb))
} yield word :: otherSentence).toList
sentenceAnag(sentenceOccurrences(sentence))
}

Scala: Find all strings of length up to n in regular language

I have a (possibly infinite) regular language which I describe with a regular expression. From this regular language I want to obtain all strings of length up to n, using scala. Some quick googling tells me there are some libraries out there that can help me. Before using an external library I want to know if this is something that is easy (as in something a decent programmer can implement in under 15 minutes) to do myself in Scala. If not, are there some good libraries that you can recommend for this?
To make what I want more concrete. Suppose I have the language A*B* and my n is 3, I then want the strings "", "A", "B", "AA", "AB", "BB", "AAA", "AAB", "ABB", "BBB".
Answer
Edits
26Nov, 4:30pm - added iterator-based version to reduce runtime and memory consumption. Seq-based version of canonic is at the bottom under (1)
26Nov, 2:45pm - added working seq-based version for canonic, non working old version of canonic is at the bottom (2)
Approach
Canonically generate all words possible for a given alphabet up to length n.
Filter the generated words by a regular expression (your regular language in that case)
Code
object SO {
import scala.annotation.tailrec
import scala.collection.{AbstractIterator, Iterator}
import scala.util.matching.Regex
def canonic(alphabet: Seq[Char], n: Int): Iterator[String] =
if (n < 0) Iterator.empty
else {
val r: IndexedSeq[Iterator[String]] = for (i <- 1 to n)
yield new CanonicItr(alphabet, i)
r.reduce(_ ++ _)
}
private class CanonicItr(alphabet: Seq[Char], width: Int) extends AbstractIterator[String] {
val aSize = alphabet.size
val alph = alphabet.toVector
val total = aSizePower(width)
println("total " + total)
private var pos = 0L
private def aSizePower(r: Int): Long = scala.math.pow(aSize, r).toLong
def stringFor(id: Long): String = {
val r = for {
i <- (0 until width).reverse
// (738 / 10^0) % 10 = 8
// (738 / 10^1) % 10 = 3
// (738 / 10^2) % 10 = 7
charIdx = ((id / (aSizePower(i))) % aSize).toInt
} yield alph(charIdx)
r.mkString("")
}
override def hasNext: Boolean = pos < total
override def next(): String = {
val s = stringFor(pos)
pos = pos + 1
s
}
}
def main(args: Array[String]): Unit = {
// create all possible words with the given alphabet
val canonicWordSet = canonic(Seq('a', 'b', 'c'), 8)
// formal regular language definition
val languageDef: Regex = "a*b".r
// determine words of language by filtering the canocic set.
val wordsOfLanguage = canonicWordSet.filter(word => languageDef.pattern.matcher(word).matches)
println(wordsOfLanguage.toList)
}
}
1) Working version of canonic but with high memory requirements
object SO {
import scala.util.matching.Regex
/**
* Given a sequence of characters (e.g. Seq('a', 'b', 'c') )
* generates all combinations up to lneght of n (incl.).
*
* #param alphabet sequence of characters
* #param n is the max length
* #return all combinations of up to length n.
*/
def canonic(alphabet:Seq[Char], n: Int): Seq[String] = {
def combination( input: Seq[String], chars: Seq[Char]) = {
for {
i <- input
c <- chars
} yield (i+c)
}
#tailrec
def rec(left: Int, current: Seq[String], accum: Seq[String] ) : Seq[String] = {
left match {
case 0 => accum
case _ => {
val next = combination( current, alphabet )
rec( left-1, next, accum ++ next )
}
}
}
rec(n, Seq(""), Seq(""))
}
def main(args: Array[String]) : Unit = {
// create all possible words with the given alphabet
val canonicWordSet= canonic( Seq('a', 'b', 'c'), 3)
// formal regular language definition
val languageDef: Regex = "a*b".r
// determine words of language by filtering the canocic set.
val wordsOfLanguage = canonicWordSet.filter( word => languageDef.pattern.matcher(word).matches )
println( wordsOfLanguage.toList )
}
}
2) Non working version of canonic not working correctly
def canonic(alphabet:Seq[Char], n: Int): Iterator[String] = {
for {
i <- (0 to n).iterator
combi <- alphabet.combinations(i).map(cs => cs.mkString)
} yield combi
}
I have not completely understood your meaning, is this OK?
scala> def generator(chars: Seq[Char], n: Int): Iterator[String] =
| (0 to n).iterator flatMap (i => (chars flatMap (_.toString*i) mkString) combinations i)
generator: (chars: Seq[Char], n: Int)Iterator[String]
scala>
scala> generator("AB", 3) toList
res0: List[String] = List("", A, B, AA, AB, BB, AAA, AAB, ABB, BBB)
scala> generator("ABC", 3) toList
res1: List[String] = List("", A, B, C, AA, AB, AC, BB, BC, CC, AAA, AAB, AAC, ABB, ABC, ACC, BBB, BBC, BCC, CCC)

Update the values of a list with their absolute values

Newbie to scala.
I am trying to make this code to work for a few hours now . It is intended to update the List[Int](list of integers) with absolute values of the integers.
Took a long time to figure out that List is immutable, so found that ListBuffer can be the saviour, but eventually in returning it back into the List form is seeing some issue i guess.
def f (arr:List[Int]) : List[Int] =
{
val list = new scala.collection.mutable.ListBuffer[Int]();
val len = arr.length;
for ( i <- 0 to len)
{
if(arr(i) < 0)
{
list.append((-1)*arr(i)) ;
}
else
{
list.append(arr(i));
}
}
return list.toList;
}
which is giving this error:
java.lang.IndexOutOfBoundsException: 12
at scala.collection.LinearSeqOptimized$class.apply(LinearSeqOptimized.scala:52)
at scala.collection.immutable.List.apply(List.scala:84)
at Solution$.f(Solution.scala:7)
at Solution$delayedInit$body.apply(Solution.scala:23)
at scala.Function0$class.apply$mcV$sp(Function0.scala:40)
at scala.runtime.AbstractFunction0.apply$mcV$sp(AbstractFunction0.scala:12)
at scala.App$$anonfun$main$1.apply(App.scala:71)
at scala.App$$anonfun$main$1.apply(App.scala:7...
Not getting what's wrong here.
The best way is to use Scala functions like #senia suggested in comments. For example:
val res = list map math.abs
But if you want to fix your code just replace to with until. You are getting off by one error:
def f (arr:List[Int]) : List[Int] =
{
val list = new scala.collection.mutable.ListBuffer[Int]();
val len = arr.length;
for ( i <- 0 until len)
{
if(arr(i) < 0)
{
list.append((-1)*arr(i)) ;
}
else
{
list.append(arr(i));
}
}
return list.toList;
}
Here is the difference between until and to:
1 to 3
// Range(1, 2, 3)
1 until 3
// Range(1, 2)
You can also remove return, ; and even braces { used with if/else.
Yet another version using a for comprehension that avoids indexing,
def f (arr:List[Int]) : List[Int] =
{
val list = new scala.collection.mutable.ListBuffer[Int]();
for {
a <- arr
sign = if (a < 0) -1 else 1
} list.append(sign * a)
return list.toList;
}
As mentioned above, the return may be omitted.
You can try using case statements for more neat syntax :
def f(arr:List[Int]):List[Int] = {
val list = scala.collection.mutable.ListBuffer[Int]()
arr.foreach{
x =>
x match {
case _ if (x <0) => list+= (x*(-1))
case _ => list +=x
}
}
list.toList
}
Looks like you were trying to solve the challenge from here. Probably you may want to use more functional approach with recursion and immutable List.
def f(arr: List[Int]): List[Int] = arr match {
case Nil => Nil
case x :: rest => java.lang.Math.abs(x) :: f(rest)
}
Beginner friendly: this is how I wrote it
def f(arr: List[Int]) : List[Int] = {
var list = new scala.collection.mutable.ArrayBuffer[Int]();
// var len = arr.length;
for(i <-0 until arr.length) {
list.append( math.abs(arr(i)));
}
return list.toList; }
I haven't done any time complexity analysis but it's the most straightforward for beginners to understand. Also, it passes all the tests on hackerrank
def f (arr: List[Int]) : List[Int] = {
arr.map {
case i if 0 > i => i * -1
case i => i
}
}

Deep-reverse of nested lists in Scala

I'd like to reverse a list of lists, recursively, in Scala.
I've written deep list reverses in Python like this:
def deepReverse(items):
if type(items) == list:
return [deepReverse(item) for item in reversed(items)]
else:
return items
How would I do the equivalent in Scala? The problem isn't the algorithm - it's the type stuff, which I'm newer on.
I need the function to take a list of [T], or a List[List[T]], or a list of T's and lists of Ts, to any arbitrary depth. I tried making a case class to do that based on an example I'd seen elsewhere. I don't want a function that just returns Any and accepts Any; that feels like cheating.
case class NL[+T](val v : Either[List[NL[T]],T])
Still, I couldn't quite get my types to balance out. I'm new to Scala, but I figured it'd be a perfect opportunity to mess with recursion and typing.
It's actually not too hard to write a version of the type class approach that sschaef proposes that will work for arbitrarily nested lists:
trait Reverser[C] {
def reverse(xs: C): C
}
implicit def rev[A](implicit ev: Reverser[A] = null) = new Reverser[List[A]] {
def reverse(xs: List[A]) =
Option(ev).map(r => xs map r.reverse).getOrElse(xs).reverse
}
def deepReverse[A](xs: A)(implicit ev: Reverser[A]): A = ev.reverse(xs)
The implicit argument ev in our rev method is evidence that A itself is reversable, and if ev is null that means it's not. If we have this evidence that A is reversable, we use it to reverse the elements of our List[A] (this is what the map is doing), and then we reverse the list itself. If we don't have this evidence (the getOrElse case), we can just reverse the list.
We could write rev a little less concisely (but possibly more performantly) like this:
implicit def rev[A](implicit ev: Reverser[A] = null) = if (ev == null) {
new Reverser[List[A]] {
def reverse(xs: List[A]) = xs.reverse
}
} else {
new Reverser[List[A]] {
def reverse(xs: List[A]) = (xs map ev.reverse).reverse
}
}
To test either of these two versions, we can write the following:
scala> deepReverse(List.tabulate(3)(identity))
res0: List[Int] = List(2, 1, 0)
scala> deepReverse(List.tabulate(2,3) { case (a, b) => a + b })
res1: List[List[Int]] = List(List(3, 2, 1), List(2, 1, 0))
scala> deepReverse(List.tabulate(2, 3, 4, 5, 6) {
| case (a, b, c, d, e) => a + b + c + d + e
| }).head.head.head.head
res2: List[Int] = List(15, 14, 13, 12, 11, 10)
As expected.
I should add that the following is a more common idiom for getting the implicits right in a case like this:
trait ReverserLow {
implicit def listReverser[A] = new Reverser[List[A]] {
def reverse(xs: List[A]) = xs.reverse
}
}
object ReverserHigh extends ReverserLow {
implicit def nestedListReverser[A](implicit ev: Reverser[A]) =
new Reverser[List[A]] {
def reverse(xs: List[A]) = xs.map(ev.reverse).reverse
}
}
import ReverserHigh._
If we'd just written listReverser and nestedListReverser at the same level, we'd get the following error when we try to reverse a list of lists:
scala> deepReverse(List.tabulate(2, 3)(_ + _))
<console>:12: error: ambiguous implicit values:
both method listReverser...
and method nestedListReverser...
match expected type Reverser[List[List[Int]]]
deepReverse(List.tabulate(2, 3)(_ + _))
The standard approach to prioritizing the two is to put the lower priority implicit in a trait (WhateverLow) and the other in an object (WhateverHigh) that extends that trait. In a fairly simple case like this, though, it's more concise (and clearer, to my eye) to use the default argument trick in my rev method above. But you're more likely to see the other version in other people's code.
If you wanna have this really typesafe then the typeclass pattern is your friend:
object Reverse extends App {
trait Reverser[C] {
def reverse(xs: C): C
}
implicit def List1Reverser[A] = new Reverser[List[A]] {
def reverse(xs: List[A]) =
xs.reverse
}
implicit def List2Reverser[A] = new Reverser[List[List[A]]] {
def reverse(xs: List[List[A]]) =
xs.map(_.reverse).reverse
}
implicit def List3Reverser[A] = new Reverser[List[List[List[A]]]] {
def reverse(xs: List[List[List[A]]]) =
xs.map(_.map(_.reverse).reverse).reverse
}
def deepReverse[A](xs: A)(implicit rev: Reverser[A]): A =
rev.reverse(xs)
val xs = List(1,2)
val xxs = List(List(1,2),List(1,2),List(1,2))
val xxxs = List(List(List(1,2),List(1,2)),List(List(1,2),List(1,2)),List(List(1,2),List(1,2)))
println(deepReverse(xs))
println(deepReverse(xxs))
println(deepReverse(xxxs))
}
The only problem with this is that you need a typeclass for each nested list type.

Scala: extracting a repeated value from a list

I have often the need to check if many values are equal and in case extract the common value. That is, I need a function that will work like follows:
extract(List()) // None
extract(List(1,2,3)) // None
extract(List(2,2,2)) // Some(2)
Assuming one has a pimp that will add tailOption to seqs (it is trivial to write one or there is one in scalaz), one implementation looks like
def extract[A](l: Seq[A]): Option[A] = {
def combine(s: A)(r: Seq[A]): Option[A] =
r.foldLeft(Some(s): Option[A]) { (acc, n) => acc flatMap { v =>
if (v == n) Some(v) else None
} }
for {
h <- l.headOption
t <- l.tailOption
res <- combine(h)(t)
} yield res
}
Is there something like that - possibly more general - already in Scalaz, or some simpler way to write it?
This seems like a really complicated way to write
def extract[A](l:Seq[A]):Option[A] = l.headOption.flatMap(h =>
if (l.tail.forall(h==)) Some(h) else None)
You don't need tailOption, since the anonymous function that gets passed as an argument to flatMap is only executed if l is not empty.
If you only want to delete duplicates toSet is enough:
def equalValue[A](xs: Seq[A]): Option[A] = {
val set = xs.toSet
if (set.size == 1) Some(set.head) else None
}
scala> equalValue(List())
res8: Option[Nothing] = None
scala> equalValue(List(1,2,3))
res9: Option[Int] = None
scala> equalValue(List(2,2,2))
res10: Option[Int] = Some(2)
This is a fluent solution
yourSeq.groupBy(x => x) match {case m if m.size==1 => m.head._1; case _ => None}
You could use a map to count the number of occurrences of each element in the list and then return only those that occur more than once:
def extract[T](ts: Iterable[T]): Iterable[T] = {
var counter: Map[T, Int] = Map()
ts.foreach{t =>
val cnt = counter.get(t).getOrElse(0) + 1
counter = counter.updated(t, cnt)
}
counter.filter(_._2 > 1).map(_._1)
}
println(extract(List())) // List()
println(extract(List(1,2,3))) // List()
println(extract(List(2,2,2))) // List(2)
println(extract(List(2,3,2,0,2,3))) // List(2,3)
You can also use a foldLeft instead of foreach and use the empty map as the initial accumulator of foldLeft.