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package forcomp | |
import common._ | |
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 | |
/** Converts the word into its character occurence 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. | |
*/ | |
def wordOccurrences(w: Word): Occurrences = w.toLowerCase.groupBy(x => x).mapValues(x => x.length).toList.sorted | |
/** Converts a sentence into its character occurrence list. */ | |
def sentenceOccurrences(s: Sentence): Occurrences = { | |
wordOccurrences(s.flatten.mkString) | |
} | |
/** 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(x => wordOccurrences(x)) | |
} | |
/** 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] = { | |
val ocs : List[Occurrences] = occurrences.map( x => (for(i <- 1 until (x._2+1)) yield (x._1,i)).toList) | |
ocs.foldRight(List[Occurrences](Nil))((x,y) => y ++ (for(i <- x; j <- y) yield (i :: j))) | |
} | |
/** 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 (p1,p2) = x.partition(a => y.exists(b => a._1 == b._1)) | |
val diffed = for( (a,b) <- p1.zip(y) if a._2 != b._2) yield (a._1,a._2-b._2) | |
(p2 ++ diffed).sorted | |
} | |
/** 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] = { | |
sentenceAnagramsInner(sentenceOccurrences(sentence)) | |
} | |
def sentenceAnagramsInner(o: Occurrences): List[Sentence] = { | |
if (o.isEmpty) { | |
List(Nil) | |
} else { | |
val combs = combinations(o) | |
for (i <- combs if dictionaryByOccurrences.keySet(i); | |
j <- dictionaryByOccurrences(i); | |
s <- sentenceAnagramsInner(subtract(o, i))) yield {j :: s} | |
} | |
} | |
} |
I have same code as yours but it is not working!
def sentenceAnagrams(sentence: Sentence): List[Sentence] = {
def anagramsForOccurrences(occurrences: Occurrences): List[Sentence] = {
if(occurrences.isEmpty) List(Nil)
else
for {
combo <- combinations(occurrences)
if dictionaryByOccurrences.contains(combo)
anagram <- dictionaryByOccurrences(combo)
restAnagram <- anagramsForOccurrences(subtract(occurrences, combo))
} yield { anagram :: restAnagram }
}
anagramsForOccurrences(sentenceOccurrences(sentence))
}
Any help would be great.
Thanks!
For subtract method you can just write x.filterNot(y.contains)
perfect
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Shouldn't you use
to
sequence instead ofuntil
in1 until (x._2+1)
?