"Comparison method violates its general contract!"

Question

Can someone explain me in simple terms, why does this code throw an exception, "Comparison method violates its general contract!", and how do I fix it?

private int compareParents(Foo s1, Foo s2) {
    if (s1.getParent() == s2) return -1;
    if (s2.getParent() == s1) return 1;
    return 0;
}

Answer 1

Your comparator is not transitive.

Let A be the parent of B, and B be the parent of C. Since A > B and B > C, then it must be the case that A > C. However, if your comparator is invoked on A and C, it would return zero, meaning A == C. This violates the contract and hence throws the exception.

It's rather nice of the library to detect this and let you know, rather than behave erratically.

One way to satisfy the transitivity requirement in compareParents() is to traverse the getParent() chain instead of only looking at the immediate ancestor.

Answer 2

Just because this is what I got when I Googled this error, my problem was that I had

if (value < other.value)
  return -1;
else if (value >= other.value)
  return 1;
else
  return 0;

the value >= other.value should (obviously) actually be value > other.value so that you can actually return 0 with equal objects.

Answer 3

The violation of the contract often means that the comparator is not providing the correct or consistent value when comparing objects. For example, you might want to perform a string compare and force empty strings to sort to the end with:

if ( one.length() == 0 ) {
    return 1;                   // empty string sorts last
}
if ( two.length() == 0 ) {
    return -1;                  // empty string sorts last                  
}
return one.compareToIgnoreCase( two );

But this overlooks the case where BOTH one and two are empty - and in that case, the wrong value is returned (1 instead of 0 to show a match), and the comparator reports that as a violation. It should have been written as:

if ( one.length() == 0 ) {
    if ( two.length() == 0 ) {
        return 0;               // BOth empty - so indicate
    }
    return 1;                   // empty string sorts last
}
if ( two.length() == 0 ) {
    return -1;                  // empty string sorts last                  
}
return one.compareToIgnoreCase( two );

Answer 4

Even if your compareTo is holds transitivity in theory, sometimes subtle bugs mess things up... such as floating point arithmetic error. It happened to me. this was my code:

public int compareTo(tfidfContainer compareTfidf) {
    //descending order
    if (this.tfidf > compareTfidf.tfidf)
        return -1;
    else if (this.tfidf < compareTfidf.tfidf)
        return 1;
    else
        return 0;

}   

The transitive property clearly holds, but for some reason I was getting the IllegalArgumentException. And it turns out that due to tiny errors in floating point arithmetic, the round-off errors where causing the transitive property to break where they shouldn't! So I rewrote the code to consider really tiny differences 0, and it worked:

public int compareTo(tfidfContainer compareTfidf) {
    //descending order
    if ((this.tfidf - compareTfidf.tfidf) < .000000001)
        return 0;
    if (this.tfidf > compareTfidf.tfidf)
        return -1;
    else if (this.tfidf < compareTfidf.tfidf)
        return 1;
    return 0;
}   

Answer 5

In our case were were getting this error because we had accidentally flipped the order of comparison of s1 and s2. So watch out for that. It was obviously way more complicated than the following but this is an illustration:

s1 == s2   
    return 0;
s2 > s1 
    return 1;
s1 < s2 
    return -1;

Answer 6

Editing VM Configuration worked for me.

-Djava.util.Arrays.useLegacyMergeSort=true

Answer 7

I've seen this happen in a piece of code where the often recurring check for null values was performed:

if(( A==null ) && ( B==null )
  return +1;//WRONG: two null values should return 0!!!

Answer 8

In my case I was doing something like the following:

if (a.someField == null) {
    return 1;
}

if (b.someField == null) {
    return -1;
}

if (a.someField.equals(b.someField)) {
    return a.someOtherField.compareTo(b.someOtherField);
}

return a.someField.compareTo(b.someField);

What I forgot to check was when both a.someField and b.someField are null.

Answer 9

Java does not check consistency in a strict sense, only notifies you if it runs into serious trouble. Also it does not give you much information from the error.

I was puzzled with what's happening in my sorter and made a strict consistencyChecker, maybe this will help you:

/**
 * @param dailyReports
 * @param comparator
 */
public static <T> void checkConsitency(final List<T> dailyReports, final Comparator<T> comparator) {
  final Map<T, List<T>> objectMapSmallerOnes = new HashMap<T, List<T>>();

  iterateDistinctPairs(dailyReports.iterator(), new IPairIteratorCallback<T>() {
    /**
     * @param o1
     * @param o2
     */
    @Override
    public void pair(T o1, T o2) {
      final int diff = comparator.compare(o1, o2);
      if (diff < Compare.EQUAL) {
        checkConsistency(objectMapSmallerOnes, o1, o2);
        getListSafely(objectMapSmallerOnes, o2).add(o1);
      } else if (Compare.EQUAL < diff) {
        checkConsistency(objectMapSmallerOnes, o2, o1);
        getListSafely(objectMapSmallerOnes, o1).add(o2);
      } else {
        throw new IllegalStateException("Equals not expected?");
      }
    }
  });
}

/**
 * @param objectMapSmallerOnes
 * @param o1
 * @param o2
 */
static <T> void checkConsistency(final Map<T, List<T>> objectMapSmallerOnes, T o1, T o2) {
  final List<T> smallerThan = objectMapSmallerOnes.get(o1);

  if (smallerThan != null) {
    for (final T o : smallerThan) {
      if (o == o2) {
        throw new IllegalStateException(o2 + "  cannot be smaller than " + o1 + " if it's supposed to be vice versa.");
      }
      checkConsistency(objectMapSmallerOnes, o, o2);
    }
  }
}

/**
 * @param keyMapValues 
 * @param key 
 * @param <Key> 
 * @param <Value> 
 * @return List<Value>
 */ 
public static <Key, Value> List<Value> getListSafely(Map<Key, List<Value>> keyMapValues, Key key) {
  List<Value> values = keyMapValues.get(key);

  if (values == null) {
    keyMapValues.put(key, values = new LinkedList<Value>());
  }

  return values;
}

/**
 * @author Oku
 *
 * @param <T>
 */
public interface IPairIteratorCallback<T> {
  /**
   * @param o1
   * @param o2
   */
  void pair(T o1, T o2);
}

/**
 * 
 * Iterates through each distinct unordered pair formed by the elements of a given iterator
 *
 * @param it
 * @param callback
 */
public static <T> void iterateDistinctPairs(final Iterator<T> it, IPairIteratorCallback<T> callback) {
  List<T> list = Convert.toMinimumArrayList(new Iterable<T>() {

    @Override
    public Iterator<T> iterator() {
      return it;
    }

  });

  for (int outerIndex = 0; outerIndex < list.size() - 1; outerIndex++) {
    for (int innerIndex = outerIndex + 1; innerIndex < list.size(); innerIndex++) {
      callback.pair(list.get(outerIndex), list.get(innerIndex));
    }
  }
}

Answer 10

If compareParents(s1, s2) == -1 then compareParents(s2, s1) == 1 is expected. With your code it's not always true.

Specifically if s1.getParent() == s2 && s2.getParent() == s1. It's just one of the possible problems.

Answer 11

In my case, it was an infinite sort. That is, at first the line moved up according to the condition, and then the same line moved down to the same place. I added one more condition at the end that unambiguously established the order of the lines.

Answer 12

So The main reason this happens is when the compare or compareTo method is not following the basic contract of comparison which is transitivity.

`If A>B And B>C then A has to be greater than A has to greater than C`

Please don't directly jump at using

` -Djava.util.Arrays.useLegacyMergeSort=true `

It will make your sorting logic buggy.

`   
    Foo s1 = new Foo();
    Foo s2 = new Foo();
    Foo s3 = new Foo();

    s1.setParent(s2);
    s2.setParent(s3);
    s3.setParent(s1);

    compareParents(s1, s2); // returns -1
    compareParents(s2, s3); // returns 1
    compareParents(s1, s3); // returns 1
`

In these cases your compareParents method will break the transitive property.

Answer 13

You can't compare object data like this:s1.getParent() == s2 - this will compare the object references. You should override equals function for Foo class and then compare them like this s1.getParent().equals(s2)

Answer 14

I faced the same issue and I solved it.

//This this your code

private int compareParents(Foo s1, Foo s2) {
    if (s1.getParent() == s2) return -1;
    if (s2.getParent() == s1) return 1;
    return 0;
}

The violation is comparing different things with each other.

//acceptable
compare between s1.getParent() and s2.getParent()

//acceptable
compare between s1 and s2

//NOT acceptable
compare between s1 and s2.getParent()

//NOT acceptable
compare between s1.getParent() and s2

In my code, I wanted to sort addresses by their coordination. In the comparator, I compared between X and Y (by mistake), instead of X and X.

//My code:
    private void sortBasedOnX(){
        //addresses is a list of addresses where each address has X and Y
        addresses.sort((o1, o2) -> {

            String a = o1.getAddress().getX(); 
            String b = o2.getAddress().getY(); //<-- this is supposed to be getX

            return Integer.parseInt(a)-Integer.parseInt(b);
        });
    }

//acceptable
compare between o1.getAddress().getX() and o1.getAddress().getX()

//acceptable
compare between o1.getAddress().getY() and o1.getAddress().getY()

//NOT acceptable
compare between o1.getAddress().getX() and o1.getAddress().getY()

//NOT acceptable
compare between o1.getAddress().getX() and o1.getAddress()

//NOT acceptable
compare between o1.getAddress().getX() and o1