4
\$\begingroup\$

I have tried my hands at "functionalising" a toy problem I had: find the expected number of throws of a six sided die until all sides have been seen (the answer is 14.7)

My starting point is my imperative solution

import scala.collection.mutable.SortedSet
import scala.util.Random

def dice(nsim:Int) : Double = {
    val nsides = 6
    val r = Random
    val throws = SortedSet[Int]()
    var nthrows = 0
    var res = 0.0

    for (i <- 1 to nsim) {
        throws.clear
        nthrows = 0
        while ( throws.size != nsides) {
            nthrows += 1
            throws += r.nextInt(nsides) + 1
        }
        res += nthrows
    }
    res / nsim
}

dice(10000) // ~14.7

I initialise an empty set, and keep adding my thrown dice until i have seen all sides (the set length equals six), while keeping track of how many die I have thrown. The expected number is just the sum of all these throws divided by the number of simulations (repetitions) I perform: approx 14.7

And here is my functional attempt (almost half a day went into this, including plenty of googling, outofmemory errors etc) - and I am not sure if this is considered good.

def throwdie(nsides:Int) : Int = {
    val r = Random
    r.nextInt(nsides) + 1 
}

def nthrows(seen:SortedSet[Int], count:Int, nsides:Int) : Int = {
    if (seen.size == nsides)
        return count
    return nthrows(seen + throwdie(nsides), count + 1, nsides)
}

def fdice(nsim:Int, nsides:Int) : Double = {
    val runs = Iterator.fill(nsim)(SortedSet[Int]())
    runs.map( x=> nthrows(x, 0, nsides) ).reduceLeft(_+_) * 1.0 / nsim
}

fdice(10000, 6) // ~14.7

explanation

So I start with a function throwdie that just returns the outcome of rolling a single die.

Next I have tried to rewrite the while loop using recursion. I came up with nthrows which takes an empty set, a zero count, and the number of die sides, and returns the number of throws until all sides have been seen for one "simulation". I use a mutable data structure here, which as I understand is a nono in functional programming, but it is all I came up with.

Next I run the simulation in fdice where I fill an Iterator with nsim instances of the empty SortedSet, and apply nthrows to all of these and sum the result.

\$\endgroup\$
3
\$\begingroup\$

Let me first present my solution and then comment below: Let me point the code is more verbose than needed, in order to be more descriptive.

import scala.annotation.tailrec
import scala.collection.SortedSet
import scala.util.Random

//=========Neat reusable methods=============

//Infinite iterator of the results of a dice throw
def diceResultIterator(sidesCount: Int): Iterator[Int] = {

  case class Dice(rnd: Random, sidesCount: Int, upside: Int)
  //Gets a dice, returns a dice, functional "mutation"! For more, search State monad
  def roll(initDice: Dice) = {
    val newSide = initDice.rnd.nextInt(initDice.sidesCount) + 1
    Dice(initDice.rnd, initDice.sidesCount, newSide)
  }
  val someDice = Dice(new Random, sidesCount, 1)
  val diceIterator = Iterator.iterate(someDice)(roll)
  diceIterator.map(_.upside)
}

//Infinite iterator of the results of the given function
def simulateManyTimes[T](f: => T) = Iterator.iterate(f)(_ => f)

//A usual general function to calculate and average
def average(values: Iterable[Int]): Option[Double] = {
  val (total, len) = values.foldLeft((0, 0)) { case ((sum, cnt), value) => (sum + value, cnt + 1) }
  if (len > 0) Some(total.toDouble / len) else None
}
def averageResultsOfIntSimulations(f: => Int, repeat: Int) = {
  average(simulateManyTimes(f).take(repeat).toIterable)
}
//===================Problem specific part======================

//Similar to OP method, just encapsulated the recursion
def countThrowsUntilAllSidesSeen(diceIter: Iterator[Int], sidesCount: Int): Int = {
  @tailrec
  def recursiveHelper(seen: SortedSet[Int], count: Int): Int = {
    if (seen.size == sidesCount)
      count
    else
      recursiveHelper(seen + diceIter.next(), count + 1)
  }
  recursiveHelper(SortedSet[Int](), 0)
}
def simulationWrapper(sides: Int, simRepeat: Int) = {
  averageResultsOfIntSimulations(countThrowsUntilAllSidesSeen(diceResultIterator(sides), sides), simRepeat).get
}

simulationWrapper(6, 1000000)

Generally, creating a new Randomfor each throw is a bad practice since it may produce less random results than expected. My proposed solution reuses the same Random in a functional way for each simulation.

I am in favor of the school that says code should be self documenting, and your second method tries to fit too many things in one place, adding obscurity.

val runs = Iterator.fill(nsim)(SortedSet[Int]())
runs.map( x=> nthrows(x, 0, nsides) ).reduceLeft(_+_) * 1.0 / nsim

For example there is no good reason why you would fill an iterator with SortedSets then use then with map, instead of just runs.map( _ => nthrows(SortedSet[Int](), 0, nsides)) Counter-intuitive code is bad code. In production, the dev next to you would certainly tap your shoulder to ask you to explain your code, generally not too happy that he had to do so.

As general advice, tune the code to your audience, like with all forms of communication. For example, the diceResultIterator in my solution is purposefully verbose, to make the concept clearer. When you are comfortable with it, you can see that it is rather trivially refactored to ~half the lines. Scala has a trap to chain operations into obscurity, be mindful about this ;)

When you see recursiveHelper inside countThrowsUntilAllSidesSeenwith arguments seen: SortedSet[Int], count: Int, it's purpose is already quite clear, so the most important piece of information was to say that it is there just to encapsulate recursion.

Lastly, I added explicit return type is the functions that I felt it added clarity. Think of them like training wheels: Just remove them when they feel useless.

\$\endgroup\$
3
\$\begingroup\$

Your solution is already quite good. However, I made some improvements:

import scala.annotation.tailrec
import scala.util.Random

def countThrows(nsides: Int): Int = {
  def throwDie = Random.nextInt(nsides) + 1

  @tailrec
  def go(sidesSeen: Set[Int], nthrows: Int): Int =
    if (sidesSeen.size == nsides) nthrows
    else go(sidesSeen + throwDie, nthrows + 1)

  go(Set(), 0)
}

def calcAverage(nsides: Int, nruns: Int): Double = {
  val results = (1 to nruns).map(_ => countThrows(nsides))
  results.sum * 1.0 / nruns
}

Use tail recursion for helper methods. Inside countThrows, I introduced a simple tail recursive helper method. However, this method has two "artificial" parameters that the caller of countThrows should not have to worry about. You can see that this simplifies calcAverage because you don't have to provide the initial empty sets.

Use immutable data structures. There is no need for a mutable (or ordered) set here. Since the set is only passed along to the recursive call when it is updated, it can be immutable.

Define functions in one line where appropriate. Scala allows you to be very concise and brief. This is certainly a power that can be abused. However, it is perfectly reasonable to write the definition of throwDie in one line. When defined inside the context it is used (countThrows) you can even omit the explicit parameter.

Don't use return. You are already omitting it sometimes. You don't need it anywhere.

Additional Notes

  • Functional random number generation There is one thing not 100% "functional" in the above solution: Random.nextInt(nsides). This call has a side effect. It updates the internal state of the Random object. You could pass this state along but it adds a lot of complexity and I doubt it is necessary in your case. @Billy-Lab's solution is also not optimal because a new Random is created for every new run. Also, Random is not immutable, so it is also not a 100% functional solution. I decided to go for simplicity here. How to treat random number generation in a purely functionial way is covered in the really excellent book Functional Programming in Scala.

  • You can omit the set for a more efficient solution. If four sides of the die have been seen, it is apparently irrelevant which four sides with regard to the probability that the next throw yields one of the unseen sides. So you can just keep track of the number of sides having been seen:

    def go(nseen: Int, nthrows: Int): Int =
      if (nseen == nsides) nthrows
      else go(if (throwDie > nseen) nseen+1 else nseen, nthrows + 1)
    
\$\endgroup\$
  • 1
    \$\begingroup\$ Heya there. I've rejected your edit suggestion on a javascript question. Since the dialog doesn't allow me to properly outline my reasoning I took the liberty to do so here. In short Code Review strives to review the code as-is. For that reason it's generally a bad idea to edit code in a question and as such the official policy is not allowing such edits If you have further questions feel free to join me in Code Review Chat. \$\endgroup\$ – Vogel612 Jan 29 '16 at 21:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.