I want to compute the cartesian product of an arbitrary number of nonempty sets in Java.
I\'ve wrote that iterative code...
public s
Here's an iterative, lazy implementation I wrote. The interface is very similar to Google's Sets.cartesianProduct, but it's a bit more flexible: it deals in Iterables instead of Sets. This code and its unit tests are at 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()";
}
}
}
}
I believe this is correct. It is not seeking efficiency, but a clean style through recursion and abstraction.
The key abstraction is to introduce a simple Tuple
class. This helps the generics later:
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;
}
}
With this class, we can write a class like so:
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;
}
}
I have a decent example of this working, but it is omitted for brevity.
You might be interested in Another question about cartesian products (edit: removed to conserve hyperlinks, search for the tag cartesian products). That answer has a nice recursive solution that I'd be hard pressed to improve on. Do you specifically want an iterative solution instead of recursive solution?
EDIT:
After looking at another iterative solution on stack overflow in perl and a clean explanation , here is another solution:
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;
}
I wrote an recursive cartesian product algorithm for table of Strings. You can modify it to have sets istead. Below is the algorithm. It's also explained in my 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;
}}
The following answer uses iteration and not recursion. It uses the same Tuple
class from my previous answer.
It is a separate answer because IMHO both are valid, different approaches.
Here is the new main class:
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;
}
}
Here is a lazy iterator approach that uses a function to produce an appropriate output type.
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;
}
};
}
};
}