Is it possible to reasonably emulate yield-syntax, perhaps with help of Java 8?

Question

I was experimenting with this question today, from Euler Problems:

A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.

Find the largest palindrome made from the product of two 3-digit numbers.

I thought about it and it can of course be done with for-loops, however I want to use Java 8 as it opens up new options.

However first of all, I do not know how to generate an IntStream that produces such elements, so I still ended up using normal for-loops:

public class Problem4 extends Problem<Integer> {
    private final int digitsCount;

    private int min;
    private int max;

    public Problem4(final int digitsCount) {
        this.digitsCount = digitsCount;
    }

    @Override
    public void run() {
        List<Integer> list = new ArrayList<>();
        min = (int)Math.pow(10, digitsCount - 1);
        max = min * 10;

        for (int i = min; i < max; i++) {
            for (int j = min; j < max; j++) {
                int sum = i * j;
                if (isPalindrome(sum)) {
                    list.add(sum);
                }
            }
        }

        result = list.stream().mapToInt(i -> i).max().getAsInt();
    }

    private boolean isPalindrome(final int number) {
        String numberString = String.valueOf(number);
        String reversed = new StringBuilder(numberString).reverse().toString();
        return (numberString.equals(reversed));
    }

    @Override
    public String getName() {
        return "Problem 4";
    }
}

As you can see I might be a bit lazy, bit really the IntStream::max is a very nice method and I think it is better to use that, as to write it yourself.

Here comes the issue though, I need to have a list now to be able to obtain the maximum in this manner, which means I need to store data, where I really should not do so.

So, the question now, would it be possible to implement this in Java 8?

for (int i = min; i < max; i++) {
    for (int j = min; j < max; j++) {
        yield i * j;
    }
}

And then out of that method create an PrimitiveIterator.OfInt (unboxes version of Iterator<Integer>, or create an IntStream directly?
Then getting the answer with streamFromYield.filter(this::isPalindrome).max().getAsInt() would be really easy to implement.

Lastly, I know this question has been asked before, however the last time is already quite a bit ago and now Java 8 is going to happen very soon, where they have added as big concept Stream<T> and the new language construct, called lambdas.
So making such code may be very different now than when people were making it for Java 6 or 7.

Answer 1

Well, I think we've gotten carried away using the Streams API from the "outside," using flatMap, optimizing the palindrome-finding algorithm, etc. See answers from Boris the Spider and assylias. However, we've sidestepped the original question of how to write a generator function using something like Python's yield statement. (I think the OP's nested-for example with yield was using Python.)

One of the problems with using flatMap is that parallel splitting can only occur on the outermost stream. The inner streams (returned from flatMap) are processed sequentially. We could try to make the inner streams also parallel, but they'd possibly compete with the outer ones. I suppose nested splitting could work, but I'm not too confident.

One approach is to use the Stream.generate or (like assylias' answer) the Stream.iterate functions. These create infinite streams, though, so an external limit must be supplied to terminate the stream.

It would be nice if we could create a finite but "flattened" stream so that the entire stream of values is subject to splitting. Unfortunately creating a stream is not nearly as convenient as Python's generator functions. It can be done without too much trouble, though. Here's an example that uses the StreamSupport and AbstractSpliterator classes:

class Generator extends Spliterators.AbstractIntSpliterator {
    final int min;
    final int max;
    int i;
    int j;

    public Generator(int min, int max) {
        super((max - min) * (max - min), 0);
        this.min = min;
        this.max = max;
        i = min;
        j = min;
    }

    public boolean tryAdvance(IntConsumer ic) {
        if (i == max) {
            return false;
        }
        ic.accept(i * j);
        j++;
        if (j == max) {
            i++;
            j = min;
        }
        return true;
    }
}

public static void main(String[] args) {
    Generator gen = new Generator(100, 1000);
    System.out.println(
        StreamSupport.intStream(gen, false)
            .filter(i -> isPalindrome(i))
            .max()
            .getAsInt());
}

Instead of having the iteration variables be on the stack (as in the nested-for with yield approach) we have to make them fields of an object and have the tryAdvance increment them until the iteration is complete. Now, this is the simplest form of a spliterator and it doesn't necessarily parallelize well. With additional work one could implement the trySplit method to do better splitting, which in turn would enable better parallelism.

The forEachRemaining method could be overridden, and it would look almost like the nested-for-loop-with-yield example, calling the IntConsumer instead of yield. Unfortunately tryAdvance is abstract and therefore must be implemented, so it's still necessary to have the iteration variables be fields of an object.

Answer 2

How about looking at it from another direction:

You want a Stream of [100,1000), and for each element of that Stream you want another Stream of that element multiplied by each of [100, 1000). This is what flatMap is for:

public static void main(final String[] args) throws Exception {
    OptionalInt max = IntStream.range(100, 1000).
            flatMap((i) -> IntStream.range(i, 1000).map((j) -> i * j)).
            unordered().
            parallel().
            filter((i) -> {
                String forward = Integer.toString(i);
                String backward = new StringBuilder(forward).reverse().toString();
                return forward.equals(backward);
            }).
            max();
    System.out.println(max);
}

Not sure if getting a String and then the reverse is the most efficient way to detect palindromes, off the top of my head this would seem to be faster:

final String asString = Integer.toString(i);
for (int j = 0, k = asString.length() - 1; j < k; j++, k--) {
    if (asString.charAt(j) != asString.charAt(k)) {
        return false;
    }
}
return true;

It gives the same answer but I haven't put it under an rigorous testing... Seems to be about 100ms faster on my machine.

Also not sure this problem is big enough for unordered().parallel() - removing that gives a little boost to speed too.

Was just trying to demonstrate the capabilities of the Stream API.

EDIT

As @Stuart points out in the comments, as multiplication is commutative, we only need to IntStream.range(i, 1000) in the sub-stream. This is because once we check a x b we don't need to check b x a. I have updated the answer.

Answer 3

There always have been ways to emulate that overrated yield feature, even without Java 8. Basically it is about storing the state of an execution, i.e. the stack frame(s), which can be done by a thread. A very simple implementation could look like this:

import java.util.Iterator;
import java.util.NoSuchElementException;

public abstract class Yield<E> implements Iterable<E> {
  protected interface Flow<T> { void yield(T item); }
  private final class State implements Runnable, Iterator<E>, Flow<E> {
    private E nextValue;
    private boolean finished, value;

    public synchronized boolean hasNext() {
      while(!(value|finished)) try { wait(); } catch(InterruptedException ex){}
      return value;
    }
    public synchronized E next() {
      while(!(value|finished)) try { wait(); } catch(InterruptedException ex){}
      if(!value) throw new NoSuchElementException();
      final E next = nextValue;
      value=false;
      notify();
      return next;
    }
    public void remove() { throw new UnsupportedOperationException(); }
    public void run() {
      try { produce(this); }
      finally {
        synchronized(this) {
          finished=true;
          notify();
        }
      }
    }
    public synchronized void yield(E next) {
      while(value) try { wait(); } catch(InterruptedException ex){}
      nextValue=next;
      value=true;
      notify();
    }
  }

  protected abstract void produce(Flow<E> f);

  public Iterator<E> iterator() {
    final State state = new State();
    new Thread(state).start();
    return state;
  }
}

Once you have such a helper class, the use case will look straight-forward:

// implement a logic the yield-style
Iterable<Integer> y=new Yield<Integer>() {
  protected void produce(Flow<Integer> f) {

    for (int i = min; i < max; i++) {
      for (int j = min; j < max; j++) {
          f.yield(i * j);
      }
    }

  }
};

// use the Iterable, e.g. in a for-each loop
int maxPalindrome=0;
for(int i:y) if(isPalindrome(i) && i>maxPalindrome) maxPalindrome=i;
System.out.println(maxPalindrome);

The previous code didn’t use any Java 8 features. But it will allow using them without the need for any change:

// the Java 8 way
StreamSupport.stream(y.spliterator(), false).filter(i->isPalindrome(i))
  .max(Integer::compare).ifPresent(System.out::println);

Note that the Yield support class above is not the most efficient implementation and it doesn’t handle the case if an iteration is not completed but the Iterator abandoned. But it shows that such a logic is indeed possible to implement in Java (while the other answers convincingly show that such a yield logic is not necessary to solve such a problem).

Answer 4

I'll give it a go. Version with a loop then with a stream. Although I start from the other end so it's easier because I can limit(1).

public class Problem0004 {

    public static void main(String[] args) {
        int maxNumber = 999 * 999;
        //with a loop
        for (int i = maxNumber; i > 0; i--) {
            if (isPalindrome(i) && has3DigitsFactors(i)) {
                System.out.println(i);
                break;
            }
        }
        //with a stream
        IntStream.iterate(maxNumber, i -> i - 1)
                .parallel()
                .filter(i -> isPalindrome(i) && has3DigitsFactors(i))
                .limit(1)
                .forEach(System.out::println);
    }

    private static boolean isPalindrome(int n) {
        StringBuilder numbers = new StringBuilder(String.valueOf(n));
        return numbers.toString().equals(numbers.reverse().toString());
    }

    private static boolean has3DigitsFactors(int n) {
        for (int i = 999; i > 0; i--) {
            if (n % i == 0 && n / i < 1000) {
                return true;
            }
        }
        return false;
    }
}