Advanced use of hash codes in Java: some brief statistics
To understand some advanced uses of hash maps that we're going to introduce, it's worth having a brief overview of some
of the statistics involved.
Collisions: chance of repeated hash codes
A collision is when two items end up with the same hash code, or with
the same index in a hash table of a given size. If we take n hash codes from a possible
range of d, then a rough formula for working out the number of expected collisions
is as follows:
(For a slightly more accurate calculation, you can also use the collision calculator opposite.
This uses a more accurate formula discussed below.)
The above approximation works well when d (the number of possible hash codes) is fairly
large– say, at least in the hundreds– and n is much smaller than d (say, at most half,
but ideally within an order of magnitude of the square root of d).
In practice, this approximation can be stated in easier terms if you're dealing with hash codes and hash
tables that are powers of two, as is often the case:
- Operating with x-bit hash codes, if we take
20.5x + 1 codes from this range, we'll expect 2 collisions.
- Doubling the number of items quadruples the number of collisions.
For example, say we are using 32-bit hash codes (so x is 32). This is the normal range
for hash codes of Java objects. This means that if we take the hash codes of 217 random
objects, we would on average expect 2 collisions: that is, 2 different hash codes each shared by more
than one object1. 217 is just over 130,000 objects. In other words:
- With 32-bit keys, the chance of having objects sharing the same hash code is fairly small,
but needs to be considered when we start having a moderately large number of objects.
Even with a good hash function, if we put a few thousand objects into a hash table
(or across a number of hash tables) then sooner or later, two objects will
have the same hash code. So in practice for a general-purpose hash table such as Java's
HashMap, the implementation has to store the keys to check that the mapping we're
retrieving really is for the object sought. (Of course, these statistics assume a perfect hash
function; in practice, things are a little worse.)
Collisions in a hash table
Now we can generally apply the above formula in some cases to work out the number of
actual collisions– that is,
the number of buckets with more than one item– that we expect in a hash table of a given size.
With typically-sized hash tables, the number of expected collisions becomes considerable.
For example, if we store 1000 objects in a hash table of size 211 (2048), then we
will expect 10002/4096 = over 200 collisions. Of course, upping the number of buckets in
the hash table reduces the number of collisions: with 1000 objets in a hash table of size 216,
we'd get fewer than 10 collisions (less than 1%). But usually, having a hash table so much bigger than
the number of objects is extremely wasteful. So even if we store only the hash codes instead
of the full keys in a hash table (a case we'll consider in a moment),
we would still generally have to compare the hash code:
bucket index alone isn't generally enough to confirm a match.
Where the number of items chosen approaches the hsah table size, the above approximation
doesn't work so well (although in fact, it tends to overestimate the number of collisions).
A better approximation, from which the above is derived2, is as follows:
On the next page, we exploit the above statistics to consider storing
only the hash code in a hash map or hash set.
1. In all our calculations, we generally assume for the sake of argument that every collision involves
a pair of objects. In reality, of course, it could be that more than 2 objects share the same
hash code; our calculations would count this case as several collisions, when from one point of view
it's still a single collision. This turns out not to matter a great deal: the calculations we give here
are still a good guide in practice.
2. The derivation essentially consists of applying the binomial expansion to the term
(1 - 1/d)n. The first three terms of the binomial expansion are
1 - n/d + n(n-1)/2d2 + .... Plugging these back into the formula
gives the approximation n(n-1)/2d, which is of course close to n2/2d.