Project Euler #17

Project Euler presents problem 17:

If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total. If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used?

import scalaz.syntax.applicative._
import scalaz.std.option._
import scala.util.matching.Regex
import scala.annotation.tailrec

object Problem17 {

val oneDigit: Regex = "^[1-9]$".r val twoDigit: Regex = "^[1-9][0-9]$".r
val threeDigit: Regex = "^[1-9][0-9][0-9]$".r val fourDigit: Regex = "^[1-9][0-9][0-9][0-9]$".r

private def isOneToFourDigitNum(x: String): Boolean =
x.matches(oneDigit.toString) |
x.matches(twoDigit.toString) |
x.matches(threeDigit.toString) |
x.matches(fourDigit.toString)

// If all the numbers from 1 to 1000 (one thousand) inclusive were written out
// in words, how many letters would be used?
def runProblem: Option[Int] = {
val oneToThousand: List[Int] = (1 to 1000).toList
val numberList: List[String] = oneToThousand.map(_.toString)
val wordLengths: List[Option[Int]] = numberList.map(getLengthOfMaybeNumberWord)
wordLengths match {
case Nil        => None
case x :: xs    => xs.foldLeft[Option[Int]](x) {
(acc: Option[Int], elem: Option[Int]) => ^(acc, elem)(_ + _)
}
}
}

def getLengthOfMaybeNumberWord(number: String): Option[Int]= {
if (isOneToFourDigitNum(number)) {
val numberAsWord: Option[String] = getNumberAsWord(number)
val wordFiltered: Option[String] = numberAsWord.flatMap(a => Some(a.filter(x => x != ' ').filter(x => x != '-')))
wordFiltered.map(_.length)
}
else None
}

def getNumberAsWord(num: String): Option[String] = {
@tailrec
def go(numbers: String)(acc: Option[String]): Option[String] = numbers.toList match {
case Nil => acc
case a :: b :: c :: d :: Nil =>  {
val rest: String = b.toString + c.toString + d.toString
val thousandWord = ^(convertSingleDigitOnes(a), Some(" thousand"))(_ ++ _)
val newAcc = ^(acc, thousandWord)(_ + _)
go(rest)(newAcc)
}
case b :: c :: d :: Nil => {
val rest: String = c.toString + d.toString
val hundredWord = convertSingleDigitHundred(b)
val newAcc = ^(acc, hundredWord)(_ + _)
go(rest)(newAcc)
}
case '0' :: '0' :: Nil => acc
case c :: d :: Nil     => acc match {
case Some("") => {
val twoDigitsWord = convertTwoDigits(c)(d)
val newAcc = ^(acc, twoDigitsWord)(_ + _)
go("")(newAcc)
}
case Some(_: String) => {
val twoDigitsWord: Option[String] = convertTwoDigits(c)(d)
val addingAnd: Option[String] = ^(Some(" and "), twoDigitsWord)(_ + _)
val newAcc = ^(acc, addingAnd)(_ + _)
go("")(newAcc)
}
case None => None

}
case d :: Nil          => convertSingleDigitOnes(d)
case _                 => None
}
go(num)(Some(""))
}

private def convertTwoDigits(tens: Char)(ones: Char): Option[String] = (tens, ones) match {
case ('0', _)  => convertSingleDigitOnes(ones)
case ('1', _)  => convertTensWithOne(ones)
case (_, '0')  => convertTens(tens)
case (_, _)    => ^(convertTens(tens).map(_ ++ "-"), convertSingleDigitOnes(ones))(_ + _)
}

// Converts a 2-digit number to its word equivalent *where* "1" is in the tens column
// Examples: 15, 12, 18, etc.
private def convertTensWithOne(ones: Char): Option[String] = ones match {
case '0' => Some("ten")
case '1' => Some("eleven")
case '2' => Some("twelve")
case '3' => Some("thirteen")
case '4' => Some("fourteen")
case '5' => Some("fifteen")
case '6' => Some("sixteen")
case '7' => Some("seventeen")
case '8' => Some("eighteen")
case '9' => Some("nineteen")
case _    => None
}

// Given a single Digit in the Ones column, return its corresponding word
// Example: f(1) -> "one", f(2) -> "two"
private def convertSingleDigitOnes(x: Char): Option[String] = x match {
case '1' => Some("one")
case '2' => Some("two")
case '3' => Some("three")
case '4' => Some("four")
case '5' => Some("five")
case '6' => Some("six")
case '7' => Some("seven")
case '8' => Some("eight")
case '9' => Some("nine")
case '0' => Some("zero")
case  _  => None
}

// Given a single Digit in the Ones column, return its corresponding word
// Example: f(1) -> "one", f(2) -> "two"
private def convertSingleDigitHundred(x: Char): Option[String] = x match {
case '1' => Some("one hundred")
case '2' => Some("two hundred")
case '3' => Some("three hundred")
case '4' => Some("four hundred")
case '5' => Some("five hundred")
case '6' => Some("six hundred")
case '7' => Some("seven hundred")
case '8' => Some("eight hundred")
case '9' => Some("nine hundred")
case '0' => Some("")
case  _  => None
}

// Given a single Digit in the Tens column, return its corresponding word.
// Example: f(2) -> "twenty", f(5) = "fifty"
// Due to 1's uniqueness, i.e. 15 = "fifteen", not "ten-five", it's handled in
// convertTensWithOne.
private def convertTens(x: Char): Option[String] = x match {
case '2' => Some("twenty")
case '3' => Some("thirty")
case '4' => Some("forty")
case '5' => Some("fifty")
case '6' => Some("sixty")
case '7' => Some("seventy")
case '8' => Some("eighty")
case '9' => Some("ninety")
case _   => None
}
}

Not to be a pain here, but this code is an example of how to complicate something that is reasonably simple.

First, converting the numbers to strings and then using regular expressions is just wrong. Numbers should be dealt with as numbers. Changing them to strings will do nothing but add awkwardness and complication to the algorithm.

Second, it is not necessary to actually create the string in order to calculate it's length.

Third, the method runProblem is a perfect example of when NOT to use a list because a plain, ordinary loop will work just fine, and use far less memory and CPU time.

• Hi Donald - thank you for this criticism! You make good points that I will incorporate into an improved version. Sep 22 '14 at 14:03
• I have been guilty of overcomplicating things too - keeping things simple isn't easy. I look forward to seeing the improved version. Sep 22 '14 at 15:25

For Euler problem number 17 there is no reason to utilize Option. At a high level using Option implies that there are missing values in the domain/codomain of your function.

To put it another way, as long as we can trust that (1 to 1000) will produce an ordered sequence of integers from 1 to 1000 we can drop the use of Option and just focus on the logic of the problem.

It is rare as a programmer that we get such an explicit definition of what our program input will be, so we may as well take advantage when we can :) The advantage in this case is being able to not use Option and thus allowing our code to be much more concise.

Below I've coded up a solution to 17. I partially obfuscated the code in case you (or whomever might be reading this) didn't want a direct answer to the question. I can give proper variable and function names if you'd like me to.

val ones = Array(0, "one".length, 3, 5, 4, 4, 3, 5, 5, "nine".length)
val teens = Array("ten".length, 6, 6, 8, 8, 7, 7, 9, 8, "nineteen".length)
val tens = Array(0, 0, "twenty".length, 6, 5, 5, 5, 7, 6, "ninety".length)

def f1(i: Int): Int = i match {
case 0 => 0
case _ => (ones(i) + 10) * 100 - 3
}

def f0(i: Int, j: Int): Int = i match {
case 1 => teens(j)
case _ => tens(i) + ones(j)
}

val fooMatrix = Vector.tabulate(10, 10)(f0(_,_))
val matrixSum = fooMatrix map (row => row.sum) reduce (_ + _)
val hundredzz = Vector.tabulate(10)(f1(_))

(hundreadzz :\ 0)(matrixSum + _ + _) + "onethousand".length