Skip to content

Instantly share code, notes, and snippets.

@carloxavier
Last active October 10, 2016 22:55
Show Gist options
  • Save carloxavier/fc76adccdd855c0d20fcaed0ff4b4183 to your computer and use it in GitHub Desktop.
Save carloxavier/fc76adccdd855c0d20fcaed0ff4b4183 to your computer and use it in GitHub Desktop.
combinatorials in scala; last week's assignment for progfun1
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
}
}
Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment