ritengo che la soluzione mio è meglio di altri
def withReplacements(chars: String, n: Int) {
def internal(path: String, acc: List[String]): List[String] = {
if (path.length == n) path :: acc else
chars.toList.flatMap {c => internal(path + c, acc)}
}
val res = internal("", Nil)
println("there are " + res.length + " " + n + "-permutations with replacement for " + chars + " = " + res)
} //> withReplacements: (chars: String, n: Int)Unit
def noReplacements(chars: String, n: Int) {
//val set = chars.groupBy(c => c).map {case (c, list) => (c -> list.length)}.toList
import scala.collection.immutable.Queue
type Set = Queue[Char]
val set = Queue[Char](chars.toList: _*)
type Result = Queue[String]
// The idea is that recursions will scan the set with one element excluded.
// Queue was chosen to implement the set to enable excluded element to bubble through it.
def internal(set: Set, path: String, acc: Result): Result = {
if (path.length == n) acc.enqueue(path)
else
set.foldLeft(acc, set.dequeue){case ((acc, (consumed_el, q)), e) =>
(internal(q, consumed_el + path, acc), q.enqueue(consumed_el).dequeue)
}. _1
}
val res = internal(set, "", Queue.empty)
println("there are " + res.length + " " + n + "-permutations without replacement for " + set + " = " + res)
} //> noReplacements: (chars: String, n: Int)Unit
withReplacements("abc", 2) //> there are 9 2-permutations with replacement for abc = List(aa, ab, ac, ba,
//| bb, bc, ca, cb, cc)
noReplacements("abc", 2) //> there are 6 2-permutations without replacement for Queue(a, b, c) = Queue(b
//| a, ca, cb, ab, ac, bc)
noReplacements("abc", 3) //> there are 6 3-permutations without replacement for Queue(a, b, c) = Queue(c
//| ba, bca, acb, cab, bac, abc)
withReplacements("abc", 3) //> there are 27 3-permutations with replacement for abc = List(aaa, aab, aac,
//| aba, abb, abc, aca, acb, acc, baa, bab, bac, bba, bbb, bbc, bca, bcb, bcc,
//| caa, cab, cac, cba, cbb, cbc, cca, ccb, ccc)
// you can run with replacements (3 chars, n = 4) but noReplacements will fail for obvious reason -- you cannont combine 3 chars to produce 4
withReplacements("abc", 4) //> there are 81 4-permutations with replacement for abc = List(aaaa, aaab, aaa
//| c, aaba, aabb, aabc, aaca, aacb, aacc, abaa, abab, abac, abba, abbb, abbc,
//| abca, abcb, abcc, acaa, acab, acac, acba, acbb, acbc, acca, accb, accc, baa
//| a, baab, baac, baba, babb, babc, baca, bacb, bacc, bbaa, bbab, bbac, bbba,
//| bbbb, bbbc, bbca, bbcb, bbcc, bcaa, bcab, bcac, bcba, bcbb, bcbc, bcca, bcc
//| b, bccc, caaa, caab, caac, caba, cabb, cabc, caca, cacb, cacc, cbaa, cbab,
//| cbac, cbba, cbbb, cbbc, cbca, cbcb, cbcc, ccaa, ccab, ccac, ccba, ccbb, ccb
//| c, ccca, cccb, cccc)
(1 to 3) foreach (u => noReplacements("aab", u))//> there are 3 1-permutations without replacement for Queue(a, a, b) = Queue(a
//| , a, b)
//| there are 6 2-permutations without replacement for Queue(a, a, b) = Queue(a
//| a, ba, ba, aa, ab, ab)
//| there are 6 3-permutations without replacement for Queue(a, a, b) = Queue(b
//| aa, aba, aba, baa, aab, aab)
Questi sono gli stessi 3 linee di codice concatenazioni ma sono supportate lunghezze variabili e permutazione elenco vengono eliminati.
Ho reso il secondo più ideologico (in modo da impedire l'unione di accumulatori di mappe piane, che rende anche più ricurvo in coda) e esteso in permutazioni multiset, in modo che si possa dire che "aab", "aba "e" baa "sono permutazioni (l'una dell'altra). L'idea è che la lettera "a" sia sostituibile due volte anziché infinitamente (con sostituzione) o disponibile solo una volta (senza sostituzioni). Quindi, hai bisogno di un contatore, che ti dice quante volte ogni lettera è disponibile per la sostituzione.
// Rewrite with replacement a bit to eliminate flat-map merges.
def norep2(chars: String, n: Int/* = chars.length*/) {
import scala.collection.immutable.Queue
type Set = Queue[Char]
val set = Queue[Char](chars.toList: _*)
type Result = Queue[String]
def siblings(set: (Char, Set), offset: Int, path: String, acc: Result): Result = set match {case (bubble, queue) =>
val children = descend(queue, path + bubble, acc) // bubble was used, it is not available for children that will produce combinations in other positions
if (offset == 0) children else siblings(queue.enqueue(bubble).dequeue, offset - 1, path, children) // siblings will produce different chars at the same position, fetch next char for them
}
def descend(set: Set, path: String, acc: Result): Result = {
if (path.length == n) acc.enqueue(path) else siblings(set.dequeue, set.size-1, path, acc)
}
val res = descend(set, "", Queue.empty)
println("there are " + res.length + " " + n + "-permutations without replacement for " + set + " = " + res)
} //> norep2: (chars: String, n: Int)Unit
assert(norep2("abc", 2) == noReplacements("abc", 2))
//> there are 6 2-permutations without replacement for Queue(a, b, c) = Queue(a
//| b, ac, bc, ba, ca, cb)
//| there are 6 2-permutations without replacement for Queue(a, b, c) = Queue(b
//| a, ca, cb, ab, ac, bc)
assert(norep2("abc", 3) == noReplacements("abc", 3))
//> there are 6 3-permutations without replacement for Queue(a, b, c) = Queue(a
//| bc, acb, bca, bac, cab, cba)
//| there are 6 3-permutations without replacement for Queue(a, b, c) = Queue(c
//| ba, bca, acb, cab, bac, abc)
def multisets(chars: String, n: Int/* = chars.length*/) {
import scala.collection.immutable.Queue
type Set = Queue[Bubble]
type Bubble = (Char, Int)
type Result = Queue[String]
def siblings(set: (Bubble, Set), offset: Int, path: String, acc: Result): Result = set match {case ((char, avail), queue) =>
val children = descend(if (avail - 1 == 0) queue else queue.enqueue(char -> {avail-1}), path + char, acc) // childern can reuse the symbol while if it is available
if (offset == 0) children else siblings(queue.enqueue((char, avail)).dequeue, offset - 1, path, children)
}
def descend(set: Set, path: String, acc: Result): Result = {
if (path.length == n) acc.enqueue(path) else siblings(set.dequeue, set.size-1, path, acc)
}
val set = Queue[Bubble]((chars.toList groupBy (c => c) map {case (k, v) => (k, v.length)}).toList: _*)
val res = descend(set, "", Queue.empty)
println("there are " + res.length + " multiset " + n + "-permutations for " + set + " = " + res)
} //> multisets: (chars: String, n: Int)Unit
assert(multisets("abc", 2) == norep2("abc", 2)) //> there are 6 multiset 2-permutations for Queue((b,1), (a,1), (c,1)) = Queue(
//| ba, bc, ac, ab, cb, ca)
//| there are 6 2-permutations without replacement for Queue(a, b, c) = Queue(a
//| b, ac, bc, ba, ca, cb)
assert(multisets("abc", 3) == norep2("abc", 3)) //> there are 6 multiset 3-permutations for Queue((b,1), (a,1), (c,1)) = Queue(
//| bac, bca, acb, abc, cba, cab)
//| there are 6 3-permutations without replacement for Queue(a, b, c) = Queue(a
//| bc, acb, bca, bac, cab, cba)
assert (multisets("aaab", 2) == multisets2("aaab".toList, 2))
//> there are 3 multiset 2-permutations for Queue((b,1), (a,3)) = Queue(ba, ab,
//| aa)
//| there are 3 multiset 2-permutations for Queue((b,1), (a,3)) = List(List(a,
//| a), List(b, a), List(a, b))
multisets("aab", 2) //> there are 3 multiset 2-permutations for Queue((b,1), (a,2)) = Queue(ba, ab,
//| aa)
multisets("aab", 3) //> there are 3 multiset 3-permutations for Queue((b,1), (a,2)) = Queue(baa, ab
//| a, aab)
norep2("aab", 3) //> there are 6 3-permutations without replacement for Queue(a, a, b) = Queue(a
//| ab, aba, aba, aab, baa, baa)
Come generalizzazione, è possibile ottenere con/senza sostituzioni utilizzando la funzione multiset. Per esempio,
//take far more letters than resulting permutation length to emulate withReplacements
assert(multisets("aaaaabbbbbccccc", 3) == withReplacements("abc", 3))
//> there are 27 multiset 3-permutations for Queue((b,5), (a,5), (c,5)) = Queue
//| (bac, bab, baa, bcb, bca, bcc, bba, bbc, bbb, acb, aca, acc, aba, abc, abb,
//| aac, aab, aaa, cba, cbc, cbb, cac, cab, caa, ccb, cca, ccc)
//| there are 27 3-permutations with replacement for abc = List(aaa, aab, aac,
//| aba, abb, abc, aca, acb, acc, baa, bab, bac, bba, bbb, bbc, bca, bcb, bcc,
//| caa, cab, cac, cba, cbb, cbc, cca, ccb, ccc)
//take one letter of each to emulate withoutReplacements
assert(multisets("aaaaabbbbbccccc", 3) == noReplacements("abc", 3))
//> there are 27 multiset 3-permutations for Queue((b,5), (a,5), (c,5)) = Queue
//| (bac, bab, baa, bcb, bca, bcc, bba, bbc, bbb, acb, aca, acc, aba, abc, abb,
//| aac, aab, aaa, cba, cbc, cbb, cac, cab, caa, ccb, cca, ccc)
//| there are 6 3-permutations without replacement for Queue(a, b, c) = Queue(c
//| ba, bca, acb, cab, bac, abc)
If you are more interested about permutations, you may look at
Questo dovrebbe probabilmente essere codereview.SE. – Raphael
@Raphael Ho dovuto google così qui è per il pigro http://codereview.stackexchange.com/ – simbo1905
Penso che l'OP va bene per SO. Le persone hanno bisogno di vedere cosa potrebbero fare gli altri con alcuni problemi, le permutazioni qui, sulla loro strada per migliorare la programmazione di Scala. – lcn