Carried on: bags

(This is part 3 of a series on the bag/multiset data structure. Check out parts 1 and 2.)

Improving the hash function

So our bag works per our tests. Let's take a look at the naïve implementation of hashing, at least for strings:

// This is the equivalent of `impl`ing a trait in Rust.
type String = Hashable;

String.prototype.hashCode = function(): number {
  return [...this].reduce((hash, character) => {
    hash = (hash << 5) - hash + character.charCodeAt(0);
    return hash & hash;
  }, 0);
String.prototype.minimumHashValue = -0x80000000;
String.prototype.maximumHashValue = 0x7fffffff;

export default String;

One bit of good news is this: this is not baked in to the bag's implementation. We already discussed how it was desirable from an Open–Closed Principle point of view to make this the responsibility of each type that would like to be stored in a bag.

I don't know enough about hashing algorithms to eyeball it in order to evaluate whether this one is collision-resistant in common applications for strings. But we can test it out empirically using a large SCOWL dictionary.

With our hashCode() ported from Java per Stack Overflow, let's bucket our 657,769 dictionary entries in 100 buckets and count how many values are in each bucket. The mean is, of course, about 6,578. But the median is 5,830 and the standard deviation is about 3,090. The smallest bucket has 4,677 entries while the largest has 26,835 entries.

Not great. Let's try this with the MurmurHash3 algorithm. I will simply steal the minified JavaScript from here.

Now we have a median of about 6,571, a standard deviation of about 72, and a minimum and maximum of about 6,424 and 6,781, respectively. That seems better.

And this reinforces our decision to have the data we want to put in our bag extend Hashable: We could test this particular hashing algorithm against a dictionary of English words without worrying about the behavior for other kinds of data. For other data types, some other means of hashing might be better.

Improving load factor

If you have 100 buckets and 75 entries, the load factor is the ratio of those two, 0.75. Different languages use slightly different values, but 0.75 to at most about 0.90 appear to be the maximum load factors allowable for best performance.

One of those surprising statistical factoids is that if you're hosting a party and want to ensure a 50% chance that two guests share the same birthday, you need only invite 23 people; with 70 people, there's a 99.9% chance of a shared birthday (in each case assuming birthdays are uniformly distributed).1 And so it is with hash tables, where collisions become a problem sooner than you'd think.

To make sure load factor is in good shape, we need to pick a good starting value, then resize our hashtable when the load factor gets too high.

Buckets based on modulus

My naïve implementation added unneeded complexity by basing what bucket chosen on calculating the percentile of the hash in the range of possible hash values and multiplying that by the number of buckets. That's more operations than are needed. Using hashResult % numBuckets should distribute the result evenly across the hash buckets. This also lets us simplify our Hashable implementation.

Linked lists

My naïve implementation uses an array of buckets, with each bucket itself an array.2 The usual implementation is to use a linked list. It is a bit unclear from my reading whether that's desirable per se, or if it is simply because Donald Knuth said to do it that way in an example.3 I think it's because linked lists don't do spooky things like reallocate unexpectedly, and by the time a bucket's contents would need to reallocate, we should be re-hashing.

Open addressing

Why it's good

Another way of building a hashtable is to use open addressing.

Remember how our buckets hold linked lists or arrays rather than the values themselves? That's called “chaining.” Doing that means you have to go to some other place in the computer's memory, and so can't take advantage of “locality of reference.” That term means that computers are clever about caching things and that accessing something close in memory to something else you recently accessed can be more efficient if the second thing happened to be cached when you looked up the first thing.

So in practice, it is often faster to store the data directly in the array of buckets. That is, you can have the buckets that make up the array actually hold the values rather than holding pointers to linked lists or other arrays.

Dealing with collisions

But this raises its own problem. Remember that with an array or linked list, collisions are easy to deal with: we just add a new value to the collection stored in each bucket. But when there's just that one array, we have to find a different place to stick the value that had a collision, and we have to think about how we'll delete that value.



We want to insert Z, which hashes to 20.

Current Value     | A  | Q  | M  | empty |
Bucket            | 20 | 21 | 22 | 23    |
Standard insertion

The usual approach is to put the value in the next empty slot. Then when it's time to go looking for a value, you hash it, go to the appropriate bucket, and if it's not in that bucket, keep going along the array till you find the value or an empty bucket. In other words, go to the bucket where you'd expect the value to be if there wasn't a collision, then check the buckets following that one, where you might have put it if there was a collision. Finding an empty slot means you didn't stick the value in the next available slot after a collision—because you wouldn't have left a gap in that case.


Current Value     | A  | Q  | M  | Z  |
Bucket            | 20 | 21 | 22 | 23 |
Robin Hood hashing

You can also do what's called Robin Hood hashing. In short, it means that if you have a collision, as you look for an empty spot for your value, you keep checking which value is more out of place—the one you're inserting or the one that's already there. We rob from the rich to give to the poor, which is where the name comes from.

Here's the same example, but with some extra information:


We want to insert Z, which hashes to 20.

Current Value            | A  | Q  | M  | empty |
Correct bucket           | 19 | 19 | 21 | N/A   |
How out of place?        | 1  | 2  | 1  | N/A   |
How far off would Z be?  | 0  | 1  | 2  | 3     |
Bucket                   | 20 | 21 | 22 | 23    |

We are trying to insert the value Z, which hashes to 20. Using Robin Hood hashing would have us try to insert the value in Bucket 20. But that bucket is occupied. And its current value A really should be in Bucket 19, so it's already one bucket out of place. Meanwhile, Z is zero buckets out of place here at Bucket 20, so we aren't going to exchange Z and A. A is “poorer” because it is more out of place than Z is. So we don't rob from the poor to pay the rich here.

Now we come to Bucket 21, which contains Q, which is two buckets out of place. Z is now one bucket out of place. No exchange there: again, it would be robbing from the poor to pay the rich.

We come next to Bucket 22, which contains M, which is one bucket out of place. But now Z is two buckets out of place. Let's put Z in Bucket 22 and carry M forward. We rob from the richer M, which is only one bucket out of place, to pay Z, which is two buckets out of place.

Finally, we get to Bucket 23, which is empty, so M goes there. The final result is this:


Current Value     | A  | Q  | Z  | M  |
Correct bucket    | 19 | 19 | 20 | 21 |
How out of place? | 1  | 2  | 2  | 2  |
Bucket            | 20 | 21 | 22 | 23 |

By following this algorithm, values will not wind up wildly out of place, thereby reducing the search cost following a collision, at little extra cost (either way you must probe to the first empty spot; but by switching the values, you make searches faster on average).


There's still one hitch (and bear with me till the example in the next paragraph): what if you insert a value, then have a collision and insert another value in the next available bucket, but then delete the first value (the one in its proper bucket)? Or, for that matter, delete any value that displaced another value because of a collision?

Imagine B hashes to Bucket 22. Then H also hashes to Bucket 22, so it goes in Bucket 23.


We want to delete B, which hashes to 22.

Current Value     | B  | H  |
Bucket            | 22 | 23 |

Now you delete B. Finally, you check to see if H is in the hashtable. It hashes to bucket 22, but that bucket is empty. So you incorreclty conclude that H isn't in the hashtable, and report that back to the user.

                    ↓ Problem!
Current Value     | empty  | H  |
Bucket            | 22     | 23 |

There are two approaches often used in this situation.


First, you might replace the deleted entry with a tombstone, a special flag that means “I deleted something here.” So you treat that bucket exactly as you would a bucket that was filled with a value that didn't match the one we are searching for: it's not empty, and so you look at the next one. This is easy, but it means load factor doesn't go down when you delete values: you might have a hashtable with lots of tombstones just wasting space.


Current Value     | TOMBSTONE  | H  |
Bucket            | 22         | 23 |
Left shift

Alternatively, you might shift values backward to fill in the gap. This avoids using tombstones, but adds to the time complexity of deletion.


Current Value     | H  | empty  |
Bucket            | 22 | 23     |


So we have discussed some fixes here that will improve our implementation:

I will implement those changes and describe updating the code in my next post.

  1. Wikipedia, Birthday problem, (last visited Jan. 10, 2020). ↩︎

  2. When I say the array is “in” the bucket, I actually mean that the bucket holds a pointer to the array, which is on the heap. So it's really a pointer that is “in” the array. ↩︎

  3. Roman Leventov, Answer, Why are we using linked list to address collisions in hash tables?, Stack Overflow, (May 14, 2015). ↩︎