Producto cartesiano iterativo en Java

Question

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

Me he escrito que el 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 me pareció bastante poco elegante. ¿Alguien tiene una mejor solución aún iterativa? ¿Una solución que utiliza algún maravilloso enfoque funcional? De lo contrario ... sugerencia sobre cómo mejorarlo? ¿Errores?

Answer 1

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

Tenga en cuenta que es posible que no necesita una utilidad tal si el número de juegos que está cartesiano-producting es un número fijo conocido en tiempo de compilación -. Usted podría utilizar ese número de bucles for anidados

EDITAR: el código se lanza ahora.

Sets.cartesianProduct()

creo que seas muy feliz con él. Solo crea las listas individuales cuando las solicite; no llena la memoria con todos los MxNxPxQ de ellos.

Si desea inspeccionar la fuente, es here at line 727.

¡Disfrútalo!

Answer 2

Te puede interesar Otra pregunta sobre los productos cartesianos (corregir: eliminar para conservar hipervínculos, buscar la etiqueta de productos cartesianos). Esa respuesta tiene una buena solución recursiva que sería difícil mejorar. ¿Desea específicamente una solución iterativa en lugar de una solución recursiva?

---

EDIT:

Después de buscar en otra solución iterativa de desbordamiento de pila en Perl y a clean explanation, aquí es 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 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 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 de esta manera:

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 ejemplo digno de este trabajo, pero se omite para ser breve.

Answer 4

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

Es una respuesta separada porque en mi humilde opinión ambos son enfoques válidos y 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 5

Aquí hay un enfoque de iterador diferido 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 6

Escribí un algoritmo recursivo del producto cartesiano para la tabla de cadenas. Puede modificarlo para tener conjuntos en el lugar. A continuación está 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 7

He aquí una aplicación iterativa, perezoso escribí. La interfaz es muy similar al Google Sets.cartesianProduct de Google, pero es un poco más flexible: se trata de Iterables en lugar de Sets. Este código y sus pruebas unitarias están 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:jmkristian@gmail.com">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 8

-Índice de solución basada en

Trabajar con los índices es una alternativa simple que es rápido y eficiente en la memoria y puede manejar cualquier número de conjuntos. La implementación de Iterable permite un uso sencillo en un bucle for-each. 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]]); 
    } 
} 

}

Answer 9

Usando Google Guava 19 y Java 8 es muy simple:

Digamos que tienes la lista de todas las matrices que desea 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] 
]