Producto cartesiano iterativo en Java

algorithm cartesian-product

Question

Quiero calcular el producto cartesiano de un número arbitrario de conjuntos no vacíos en Java.

He escrito ese código iterativo ...

public static <T> List<Set<T>> cartesianProduct(List<Set<T>> list) {
List<Iterator<T>> iterators = new ArrayList<Iterator<T>>(list.size());
List<T> elements = new ArrayList<T>(list.size());
List<Set<T>> toRet = new ArrayList<Set<T>>();
for (int i = 0; i < list.size(); i++) {
iterators.add(list.get(i).iterator());
elements.add(iterators.get(i).next());
}
for (int j = 1; j >= 0;) {
toRet.add(Sets.newHashSet(elements));
for (j = iterators.size()-1; j >= 0 && !iterators.get(j).hasNext(); j--) {
iterators.set(j, list.get(j).iterator());
elements.set(j, iterators.get(j).next());
}
elements.set(Math.abs(j), iterators.get(Math.abs(j)).next());
}
return toRet;
}

... pero lo encontré bastante poco elegante. ¿Alguien tiene una solución mejor, todavía iterativa? ¿Una solución que usa algún enfoque funcional maravilloso? De lo contrario ... ¿Sugerencia sobre cómo mejorarla? Errores?

Answer 1

Aquí hay un enfoque de iterador perezoso que usa una función para producir un tipo de salida apropiado.

 public static <T> Iterable<T> cartesianProduct(
final Function<Object[], T> fn, Object[]... options) {
final Object[][] opts = new Object[options.length][];
for (int i = opts.length; --i >= 0;) {
// NPE on null input collections, and handle the empty output case here
// since the iterator code below assumes that it is not exhausted the
// first time through fetch.
if (options[i].length == 0) { return Collections.emptySet(); }
opts[i] = options[i].clone();
}
return new Iterable<T>() {
public Iterator<T> iterator() {
return new Iterator<T>() {
final int[] pos = new int[opts.length];
boolean hasPending;
T pending;
boolean exhausted;
public boolean hasNext() {
fetch();
return hasPending;
}
public T next() {
fetch();
if (!hasPending) { throw new NoSuchElementException(); }
T out = pending;
pending = null; // release for GC
hasPending = false;
return out;
}
public void remove() { throw new UnsupportedOperationException(); }
private void fetch() {
if (hasPending || exhausted) { return; }
// Produce a result.
int n = pos.length;
Object[] args = new Object[n];
for (int j = n; --j >= 0;) { args[j] = opts[j][pos[j]]; }
pending = fn.apply(args);
hasPending = true;
// Increment to next.
for (int i = n; --i >= 0;) {
if (++pos[i] < opts[i].length) {
for (int j = n; --j > i;) { pos[j] = 0; }
return;
}
}
exhausted = true;
}
};
}
};
}

Answer 2

Aquí hay una implementación iterativa y perezosa que escribí. La interfaz es muy similar a la de Sets.cartesianProduct de Google, pero es un poco más flexible: se trata de Iterables en lugar de Sets. Este código y sus pruebas de unidad se encuentran en https://gist.github.com/1911614 .

/* Copyright 2012 LinkedIn Corp.
Licensed under the Apache License, Version 2.0 (the "License");
you may not use this file except in compliance with the License.
You may obtain a copy of the License at
http://www.apache.org/licenses/LICENSE-2.0
Unless required by applicable law or agreed to in writing, software
distributed under the License is distributed on an "AS IS" BASIS,
WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
See the License for the specific language governing permissions and
limitations under the License.
*/
import com.google.common.base.Function;
import com.google.common.collect.Iterables;
import java.lang.reflect.Array;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Iterator;
import java.util.List;
import java.util.NoSuchElementException;
/**
* Implements the Cartesian product of ordered collections.
*
* @author <a href="mailto:[email protected]">John Kristian</a>
*/
public class Cartesian {
/**
* Generate the <a href="http://en.wikipedia.org/wiki/Cartesian_product">Cartesian
* product</a> of the given axes. For axes [[a1, a2 ...], [b1, b2 ...], [c1, c2 ...]
* ...] the product is [[a1, b1, c1 ...] ... [a1, b1, c2 ...] ... [a1, b2, c1 ...] ...
* [aN, bN, cN ...]]. In other words, the results are generated in same order as these
* nested loops:
*
* <pre>
* for (T a : [a1, a2 ...])
* for (T b : [b1, b2 ...])
* for (T c : [c1, c2 ...])
* ...
* result = new T[]{ a, b, c ... };
* </pre>
*
* Each result is a new array of T, whose elements refer to the elements of the axes. If
* you prefer a List, you can call asLists(product(axes)).
* <p>
* Don''t change the axes while iterating over their product, as a rule. Changes to an
* axis can affect the product or cause iteration to fail (which is usually bad). To
* prevent this, you can pass clones of your axes to this method.
* <p>
* The implementation is lazy. This method iterates over the axes, and returns an
* Iterable that contains a reference to each axis. Iterating over the product causes
* iteration over each axis. Methods of each axis are called as late as practical.
*/
public static <T> Iterable<T[]> product(Class<T> resultType,
Iterable<? extends Iterable<? extends T>> axes) {
return new Product<T>(resultType, newArray(Iterable.class, axes));
}
/** Works like product(resultType, Arrays.asList(axes)), but slightly more efficient. */
public static <T> Iterable<T[]> product(Class<T> resultType, Iterable<? extends T>... axes) {
return new Product<T>(resultType, axes.clone());
}
/**
* Wrap the given arrays in fixed-size lists. Changes to the lists write through to the
* arrays.
*/
public static <T> Iterable<List<T>> asLists(Iterable<? extends T[]> arrays) {
return Iterables.transform(arrays, new AsList<T>());
}
/**
* Arrays.asList, represented as a Function (as used in Google collections).
*/
public static class AsList<T> implements Function<T[], List<T>> {
@Override
public List<T> apply(T[] array) {
return Arrays.asList(array);
}
}
/** Create a generic array containing references to the given objects. */
private static <T> T[] newArray(Class<? super T> elementType, Iterable<? extends T> from) {
List<T> list = new ArrayList<T>();
for (T f : from)
list.add(f);
return list.toArray(newArray(elementType, list.size()));
}
/** Create a generic array. */
@SuppressWarnings("unchecked")
private static <T> T[] newArray(Class<? super T> elementType, int length) {
return (T[]) Array.newInstance(elementType, length);
}
private static class Product<T> implements Iterable<T[]> {
private final Class<T> _resultType;
private final Iterable<? extends T>[] _axes;
/** Caution: the given array of axes is contained by reference, not cloned. */
Product(Class<T> resultType, Iterable<? extends T>[] axes) {
_resultType = resultType;
_axes = axes;
}
@Override
public Iterator<T[]> iterator() {
if (_axes.length <= 0) // an edge case
return Collections.singleton(newArray(_resultType, 0)).iterator();
return new ProductIterator<T>(_resultType, _axes);
}
@Override
public String toString() {
return "Cartesian.product(" + Arrays.toString(_axes) + ")";
}
private static class ProductIterator<T> implements Iterator<T[]> {
private final Iterable<? extends T>[] _axes;
private final Iterator<? extends T>[] _iterators; // one per axis
private final T[] _result; // a copy of the last result
/**
* The minimum index such that this.next() will return an array that contains
* _iterators[index].next(). There are some special sentinel values: NEW means this
* is a freshly constructed iterator, DONE means all combinations have been
* exhausted (so this.hasNext() == false) and _iterators.length means the value is
* unknown (to be determined by this.hasNext).
*/
private int _nextIndex = NEW;
private static final int NEW = -2;
private static final int DONE = -1;
/** Caution: the given array of axes is contained by reference, not cloned. */
ProductIterator(Class<T> resultType, Iterable<? extends T>[] axes) {
_axes = axes;
_iterators = Cartesian.<Iterator<? extends T>> newArray(Iterator.class, _axes.length);
for (int a = 0; a < _axes.length; ++a) {
_iterators[a] = axes[a].iterator();
}
_result = newArray(resultType, _iterators.length);
}
private void close() {
_nextIndex = DONE;
// Release references, to encourage garbage collection:
Arrays.fill(_iterators, null);
Arrays.fill(_result, null);
}
@Override
public boolean hasNext() {
if (_nextIndex == NEW) { // This is the first call to hasNext().
_nextIndex = 0; // start here
for (Iterator<? extends T> iter : _iterators) {
if (!iter.hasNext()) {
close(); // no combinations
break;
}
}
} else if (_nextIndex >= _iterators.length) {
// This is the first call to hasNext() after next() returned a result.
// Determine the _nextIndex to be used by next():
for (_nextIndex = _iterators.length - 1; _nextIndex >= 0; --_nextIndex) {
Iterator<? extends T> iter = _iterators[_nextIndex];
if (iter.hasNext()) {
break; // start here
}
if (_nextIndex == 0) { // All combinations have been generated.
close();
break;
}
// Repeat this axis, with the next value from the previous axis.
iter = _axes[_nextIndex].iterator();
_iterators[_nextIndex] = iter;
if (!iter.hasNext()) { // Oops; this axis can''t be repeated.
close(); // no more combinations
break;
}
}
}
return _nextIndex >= 0;
}
@Override
public T[] next() {
if (!hasNext())
throw new NoSuchElementException("!hasNext");
for (; _nextIndex < _iterators.length; ++_nextIndex) {
_result[_nextIndex] = _iterators[_nextIndex].next();
}
return _result.clone();
}
@Override
public void remove() {
for (Iterator<? extends T> iter : _iterators) {
iter.remove();
}
}
@Override
public String toString() {
return "Cartesian.product(" + Arrays.toString(_axes) + ").iterator()";
}
}
}
}

Answer 3

Creo que esto es correcto. No busca la eficiencia, sino un estilo limpio a través de la recursión y la abstracción.

La abstracción clave es introducir una clase simple de Tuple . Esto ayuda a los genéricos más tarde:

class Tuple<T> {
private List<T> list = new ArrayList<T>();
public void add(T t) { list.add(t); }
public void addAll(Tuple<T> subT) {
for (T t : subT.list) {
list.add(t);
}
}
public String toString() {
String result = "(";
for (T t : list) { result += t + ", "; }
result = result.substring(0, result.length() - 2);
result += " )";
return result;
}
}

Con esta clase, podemos escribir una clase así:

public class Example {
public static <T> List<Tuple<T>> cartesianProduct(List<Set<T>> sets) {
List<Tuple<T>> tuples = new ArrayList<Tuple<T>>();
if (sets.size() == 1) {
Set<T> set = sets.get(0);
for (T t : set) {
Tuple<T> tuple = new Tuple<T>();
tuple.add(t);
tuples.add(tuple);
}
} else {
Set<T> set = sets.remove(0);
List<Tuple<T>> subTuples = cartesianProduct(sets);
System.out.println("TRACER size = " + tuples.size());
for (Tuple<T> subTuple : subTuples) {
for (T t : set) {
Tuple<T> tuple = new Tuple<T>();
tuple.addAll(subTuple);
tuple.add(t);
tuples.add(tuple);
}
}
}
return tuples;
}

}

Tengo un buen ejemplo de este trabajo, pero se omite por brevedad.

Answer 4

Es posible que le interese Otra pregunta sobre productos cartesianos (edición: eliminado para conservar los hipervínculos, busque los productos cartesianos de la etiqueta). Esa respuesta tiene una buena solución recursiva que me costaría mucho mejorar. ¿Desea específicamente una solución iterativa en lugar de una solución recursiva?

EDITAR:

Después de ver otra solución iterativa en el desbordamiento de pila en perl y una explicación clara , aquí hay otra solución:

public static <T> List<Set<T>> uglyCartesianProduct(List<Set<T>> list) {
List<Iterator<T>> iterators = new ArrayList<Iterator<T>>(list.size());
List<T> elements = new ArrayList<T>(list.size());
List<Set<T>> toRet = new ArrayList<Set<T>>();
for (int i = 0; i < list.size(); i++) {
iterators.add(list.get(i).iterator());
elements.add(iterators.get(i).next());
}
for(int i = 0; i < numberOfTuples(list); i++)
{
toRet.add(new HashSet<T>());
}
int setIndex = 0;
for (Set<T> set : list) {
int index = 0;
for (int i = 0; i < numberOfTuples(list); i++) {
toRet.get(index).add((T) set.toArray()[index % set.size()]);
index++;
}
setIndex++;
}
return toRet;
}
private static <T> int numberOfTuples(List<Set<T>> list) {
int product = 1;
for (Set<T> set : list) {
product *= set.size();
}
return product;
}

Answer 5

Escribí un algoritmo de producto cartesiano recursivo para la tabla de cadenas. Puedes modificarlo para tener conjuntos istead. A continuación se muestra el algoritmo. También se explica en mi article

public class Main {
public static void main(String[] args) {
String[] A = new String[]{ "a1", "a2", "a3" };
String[] B = new String[]{ "b1", "b2", "b3" };
String[] C = new String[]{ "c1" };
String[] cp = CartesianProduct(0, A, B, C);
for(String s : cp) {
System.out.println(s);
}
}
public static String[] CartesianProduct(int prodLevel, String[] res, String[] ...s) {
if(prodLevel < s.length) {
int cProdLen = res.length * s[prodLevel].length;
String[] tmpRes = new String[cProdLen];
for (int i = 0; i < res.length; i++) {
for (int j = 0; j < s[prodLevel].length; j++) {
tmpRes[i * res.length + j] = res[i] + s[prodLevel][j];
}
}
res = Main.CartesianProduct(prodLevel + 1, tmpRes, s);
}
return res;
}}

Answer 6

He escrito una solución que no requiere que llenes una gran colección en la memoria. Desafortunadamente, el código requerido es de cientos de líneas. Es posible que tenga que esperar hasta que aparezca en el proyecto Guava ( http://guava-libraries.googlecode.com ), que espero sea para finales de año. Lo siento. :(

Tenga en cuenta que es posible que no necesite dicha utilidad si el número de conjuntos que está produciendo cartesiano es un número fijo conocido en el momento de la compilación; podría usar ese número de bucles anidados.

EDITAR: el código se libera ahora.

Sets.cartesianProduct()

Creo que estarás muy feliz con eso. Solo crea las listas individuales cuando las pides; no llena la memoria con todos los MxNxPxQ de ellos.

Si desea inspeccionar la fuente, está aquí en la línea 727 .

¡Disfrutar!

Answer 7

La siguiente respuesta usa iteración y no recursión. Utiliza la misma clase Tuple de mi respuesta anterior.

Es una respuesta separada porque en mi humilde opinión ambos enfoques son válidos, diferentes.

Aquí está la nueva clase principal:

public class Example {
public static <T> List<Tuple<T>> cartesianProduct(List<Set<T>> sets) {
List<Tuple<T>> tuples = new ArrayList<Tuple<T>>();
for (Set<T> set : sets) {
if (tuples.isEmpty()) {
for (T t : set) {
Tuple<T> tuple = new Tuple<T>();
tuple.add(t);
tuples.add(tuple);
}
} else {
List<Tuple<T>> newTuples = new ArrayList<Tuple<T>>();
for (Tuple<T> subTuple : tuples) {
for (T t : set) {
Tuple<T> tuple = new Tuple<T>();
tuple.addAll(subTuple);
tuple.add(t);
newTuples.add(tuple);
}
}
tuples = newTuples;
}
}
return tuples;
}
}

Answer 8

Usar Google Guava 19 y Java 8 es muy simple:

Digamos que tienes la lista de todos los arrays que deseas asociar ...

public static void main(String[] args) {
List<String[]> elements = Arrays.asList(
new String[]{"John", "Mary"},
new String[]{"Eats", "Works", "Plays"},
new String[]{"Food", "Computer", "Guitar"}
);
// Create a list of immutableLists of strings
List<ImmutableList<String>> immutableElements = makeListofImmutable(elements);
// Use Guava''s Lists.cartesianProduct, since Guava 19
List<List<String>> cartesianProduct = Lists.cartesianProduct(immutableElements);
System.out.println(cartesianProduct);
}

El método para hacer la lista de listas inmutables es el siguiente:

/**
* @param values the list of all profiles provided by the client in matrix.json
* @return the list of ImmutableList to compute the Cartesian product of values
*/
private static List<ImmutableList<String>> makeListofImmutable(List<String[]> values) {
List<ImmutableList<String>> converted = new LinkedList<>();
values.forEach(array -> {
converted.add(ImmutableList.copyOf(array));
});
return converted;
}

La salida es la siguiente:

[
[John, Eats, Food], [John, Eats, Computer], [John, Eats, Guitar],
[John, Works, Food], [John, Works, Computer], [John, Works, Guitar],
[John, Plays, Food], [John, Plays, Computer], [John, Plays, Guitar],
[Mary, Eats, Food], [Mary, Eats, Computer], [Mary, Eats, Guitar],
[Mary, Works, Food], [Mary, Works, Computer], [Mary, Works, Guitar],
[Mary, Plays, Food], [Mary, Plays, Computer], [Mary, Plays, Guitar]
]

Answer 9

Solución basada en índices

Trabajar con los índices es una alternativa simple que es rápida y eficiente en memoria y puede manejar cualquier número de conjuntos. Implementar Iterable permite un uso fácil en un bucle para cada uno. Vea el método #main para un ejemplo de uso.

public class CartesianProduct implements Iterable<int[]>, Iterator<int[]> {
private final int[] _lengths;
private final int[] _indices;
private boolean _hasNext = true;
public CartesianProduct(int[] lengths) {
_lengths = lengths;
_indices = new int[lengths.length];
}
public boolean hasNext() {
return _hasNext;
}
public int[] next() {
int[] result = Arrays.copyOf(_indices, _indices.length);
for (int i = _indices.length - 1; i >= 0; i--) {
if (_indices[i] == _lengths[i] - 1) {
_indices[i] = 0;
if (i == 0) {
_hasNext = false;
}
} else {
_indices[i]++;
break;
}
}
return result;
}
public Iterator<int[]> iterator() {
return this;
}
public void remove() {
throw new UnsupportedOperationException();
}
/**
* Usage example. Prints out
*
* <pre>
* [0, 0, 0] a, NANOSECONDS, 1
* [0, 0, 1] a, NANOSECONDS, 2
* [0, 0, 2] a, NANOSECONDS, 3
* [0, 0, 3] a, NANOSECONDS, 4
* [0, 1, 0] a, MICROSECONDS, 1
* [0, 1, 1] a, MICROSECONDS, 2
* [0, 1, 2] a, MICROSECONDS, 3
* [0, 1, 3] a, MICROSECONDS, 4
* [0, 2, 0] a, MILLISECONDS, 1
* [0, 2, 1] a, MILLISECONDS, 2
* [0, 2, 2] a, MILLISECONDS, 3
* [0, 2, 3] a, MILLISECONDS, 4
* [0, 3, 0] a, SECONDS, 1
* [0, 3, 1] a, SECONDS, 2
* [0, 3, 2] a, SECONDS, 3
* [0, 3, 3] a, SECONDS, 4
* [0, 4, 0] a, MINUTES, 1
* [0, 4, 1] a, MINUTES, 2
* ...
* </pre>
*/
public static void main(String[] args) {
String[] list1 = { "a", "b", "c", };
TimeUnit[] list2 = TimeUnit.values();
int[] list3 = new int[] { 1, 2, 3, 4 };
int[] lengths = new int[] { list1.length, list2.length, list3.length };
for (int[] indices : new CartesianProduct(lengths)) {
System.out.println(Arrays.toString(indices) //
+ " " + list1[indices[0]] //
+ ", " + list2[indices[1]] //
+ ", " + list3[indices[2]]);
}
}

}