combinatorials in scala; last week's assignment for progfun1

Anagrams.scala

package forcomp

import java.io.Serializable


object Anagrams {

  /** A word is simply a `String`. */
  type Word = String

  /** A sentence is a `List` of words. */
  type Sentence = List[Word]

  /** `Occurrences` is a `List` of pairs of characters and positive integers saying
    *  how often the character appears.
    *  This list is sorted alphabetically w.r.t. to the character in each pair.
    *  All characters in the occurrence list are lowercase.
    *
    *  Any list of pairs of lowercase characters and their frequency which is not sorted
    *  is **not** an occurrence list.
    *
    *  Note: If the frequency of some character is zero, then that character should not be
    *  in the list.
    */
  type Occurrences = List[(Char, Int)]

  /** The dictionary is simply a sequence of words.
    *  It is predefined and obtained as a sequence using the utility method `loadDictionary`.
    */
  val dictionary: List[Word] = loadDictionary.filter(word => word.forall(char => char.isLetter))

  /** Converts the word into its character occurrence list.
    *
    *  Note: the uppercase and lowercase version of the character are treated as the
    *  same character, and are represented as a lowercase character in the occurrence list.
    *
    *  Note: you must use `groupBy` to implement this method!
    */
  //  def wordOccurrences(word: Word): Occurrences = word.toLowerCase.toList.groupBy(char => char).toList.map{case (a , b) => (a, b.length)}.sorted
  def wordOccurrences(word: Word): Occurrences =
  for{
    (char, occList) <- word.toLowerCase.groupBy(char => char).toList.sorted
  } yield (char, occList.length)

  /** Converts a sentence into its character occurrence list. */
  def sentenceOccurrences(s: Sentence): Occurrences = wordOccurrences(s.foldLeft("")(_ ++ _))

  /** The `dictionaryByOccurrences` is a `Map` from different occurrences to a sequence of all
    *  the words that have that occurrence count.
    *  This map serves as an easy way to obtain all the anagrams of a word given its occurrence list.
    *
    *  For example, the word "eat" has the following character occurrence list:
    *
    *     `List(('a', 1), ('e', 1), ('t', 1))`
    *
    *  Incidentally, so do the words "ate" and "tea".
    *
    *  This means that the `dictionaryByOccurrences` map will contain an entry:
    *
    *    List(('a', 1), ('e', 1), ('t', 1)) -> Seq("ate", "eat", "tea")
    *
    */
  lazy val dictionaryByOccurrences: Map[Occurrences, List[Word]] =
    (dictionary groupBy wordOccurrences).withDefaultValue(List())

  /** Returns all the anagrams of a given word. */
  def wordAnagrams(word: Word): List[Word] = dictionaryByOccurrences(wordOccurrences(word))

  /** Returns the list of all subsets of the occurrence list.
    *  This includes the occurrence itself, i.e. `List(('k', 1), ('o', 1))`
    *  is a subset of `List(('k', 1), ('o', 1))`.
    *  It also include the empty subset `List()`.
    *
    *  Example: the subsets of the occurrence list `List(('a', 2), ('b', 2))` are:
    *
    *    List(
    *      List(),
    *      List(('a', 1)),
    *      List(('a', 2)),
    *      List(('b', 1)),
    *      List(('a', 1), ('b', 1)),
    *      List(('a', 2), ('b', 1)),
    *      List(('b', 2)),
    *      List(('a', 1), ('b', 2)),
    *      List(('a', 2), ('b', 2))
    *    )
    *
    *  Note that the order of the occurrence list subsets does not matter -- the subsets
    *  in the example above could have been displayed in some other order.
    */
  def combinations(occurrences: Occurrences): List[Occurrences] = occurrences match {
    case List() => List(List())
    case Nil => List(List())
    case (char, occNumber) :: xs =>
      for {
        subset <- combinations(xs)
        occurrence <- 0 to occNumber
      } yield if (occurrence == 0) subset else (char, occurrence) :: subset
  }

  /** Subtracts occurrence list `y` from occurrence list `x`.
    *
    *  The precondition is that the occurrence list `y` is a subset of
    *  the occurrence list `x` -- any character appearing in `y` must
    *  appear in `x`, and its frequency in `y` must be smaller or equal
    *  than its frequency in `x`.
    *
    *  Note: the resulting value is an occurrence - meaning it is sorted
    *  and has no zero-entries.
    */
  def subtract(x: Occurrences, y: Occurrences): Occurrences = {
      val yMap = y.toMap
      for {
        (char, occ) <- x
        resultingOcc = occ - yMap.getOrElse(char, 0)
        if (resultingOcc > 0)
      } yield (char, resultingOcc)
  }


  /** Returns a list of all anagram sentences of the given sentence.
    *
    *  An anagram of a sentence is formed by taking the occurrences of all the characters of
    *  all the words in the sentence, and producing all possible combinations of words with those characters,
    *  such that the words have to be from the dictionary.
    *
    *  The number of words in the sentence and its anagrams does not have to correspond.
    *  For example, the sentence `List("I", "love", "you")` is an anagram of the sentence `List("You", "olive")`.
    *
    *  Also, two sentences with the same words but in a different order are considered two different anagrams.
    *  For example, sentences `List("You", "olive")` and `List("olive", "you")` are different anagrams of
    *  `List("I", "love", "you")`.
    *
    *  Here is a full example of a sentence `List("Yes", "man")` and its anagrams for our dictionary:
    *
    *    List(
    *      List(en, as, my),
    *      List(en, my, as),
    *      List(man, yes),
    *      List(men, say),
    *      List(as, en, my),
    *      List(as, my, en),
    *      List(sane, my),
    *      List(Sean, my),
    *      List(my, en, as),
    *      List(my, as, en),
    *      List(my, sane),
    *      List(my, Sean),
    *      List(say, men),
    *      List(yes, man)
    *    )
    *
    *  The different sentences do not have to be output in the order shown above - any order is fine as long as
    *  all the anagrams are there. Every returned word has to exist in the dictionary.
    *
    *  Note: in case that the words of the sentence are in the dictionary, then the sentence is the anagram of itself,
    *  so it has to be returned in this list.
    *
    *  Note: There is only one anagram of an empty sentence.
    */
  def sentenceAnagrams(sentence: Sentence): List[Sentence] = getAnagrams(sentenceOccurrences(sentence))

  def getAnagrams(occs: Occurrences): List[Sentence] = occs match {
    case Nil => List(List())
    case default =>
      for {
        comb <- combinations(occs)
        restOcc = subtract(occs, comb)
        word <- dictionaryByOccurrences(comb)
        sentence <- getAnagrams(restOcc)
      } yield word :: sentence
  }
}