## Optimizing Hash Functions For a Perfect Map

by Jack Shirazi01/25/2001

`Map`

s that are hash tables are normally implemented
using a *hash function* which maps a data item (the key) into
an indexing table. The hash function takes the key and uses some
algorithm to convert it to an index value into an array. For example,
the hash function from the Java SDK `HashMap`

class:

`int index = (key.hashCode() & 0x7FFFFFFF) % table.length;`

This hash function extracts some bits from the hashcode of the key
object (to quickly convert the hashcode to some positive number), and
then uses the remainder, after division by the table length, to get an
index that fits into the `HashMap`

internal array. A hash
function cannot generally guarantee every data item maps to a
different index, so the structure of a hash table is complicated by
needing to maintain a list at every index, allowing multiple objects
to be held at any index. Having multiple objects at the same index is
inefficient since the collection at that index must be searched every
time an element from that collection is required.

A *perfect* hash function is one that maps every key to a
different index. `Map`

implementations with perfect hash
functions are much simpler than using generic hash functions; they
consist of two arrays, one for the keys and one to hold all
corresponding values. The array elements do not need to hold
collections since only one object is ever located at each index,
making processing easier and faster. (Note that the hash function does
not have to map a key into every index, i.e. some indexes can be
empty. A hash function which maps each key to a unique index
*and* leaves no empty entries in the table is called a
*minimal perfect hash function*). But how can you get a perfect
hash function?

### An interesting problem

Recently I came across an interesting performance problem. The
application used a number of objects, which were at core
`Map`

s with known key objects (not
`String`

s). There were several kinds of `Map`

s,
and each kind was allowed to contain only one specific set of key
objects. For example, suppose the key objects were defined simply
as

```
public class Key {
int id;
public Key(int id) {this.id = id;}
}
```

Then the first kind of `Map`

might be allowed to have
keys with ids of 5, 10, 11, and 27, while another kind of
`Map`

must contain only keys with ids 10, 11 and 15,
etc. There could be multiple instances of each kind of map, but every
instance could only contain its defined keys.

In this application these `Map`

objects were accessed
and updated often. The key lookup needed to be as fast as
possible. Note the information we have about the `Map`

data: the key set was restricted and known before construction for
every `Map`

, and the keys had associated numerical values
which were unique to each key. Because the key data was restricted and
known, my first thought was that the these `Map`

s were
ideal for optimization using perfect hash functions.

### Optimizing the Maps

Creating optimized classes for these `Map`

s is
straightforward except for the hash function. To keep the code clear
and optimized for `Key`

object keys, I won't implement the
`Map`

interface precisely -- converting the following class
to a standard implementation of `Map`

is easy enough.

```
public class PerfectKeyHashMap
{
Key[] keys;
Object[] values;
public void put(Key key, Object value)
{
//The following line is the currently unknown hash function
int index = someHashFunctionOn(key);
keys[index] = key;
values[index] = values;
}
public Object get(Key key)
{
//The following line is the currently unknown hash function
int index = someHashFunctionOn(key);
//keys assumed to be entered on map creation with null values.
//no validation done on key, but if needed, that could be
//if(keys[index].id != key.id) throw new Exception
//("invalid key");
return values[index];
}
...
}
```

Now how do we get our perfect hash function? We want it to be quick
so we should use as few simple operations as possible. The
`HashMap`

function manipulates bits and uses the remainder
operator. Remainder looks like it could be quite a useful operation
for our hash function. It will help map arbitrary numbers into a small
table, and it can be used with a variable argument. So let's try
it.

### Generating a perfect hash function

We have a known maximum number of keys in the hashtable. We also know that we want each of those keys to map to a different index. So clearly our remainder operator argument must be a number greater than or equal to the number of keys we have. Let's try out some tests to get a feel for the data. Using some examples, including the ones we had earlier, what would work using the remainder operator? We'll start with the list of key values and apply the remainder operator to each value. Then we use the argument that starts with the size of the key set, which increments by one each time until we get a result set of unique keys (the repeated values in each column are marked with a *):

Table 1: Finding the hash function for the set {5,10,11,27} | ||||

keys 1 |
%4 |
%5 |
%6 |
%7 |

5 |
1 |
0* |
5* |
5 |

10 |
2 |
0* |
4 |
3 |

11 |
3* |
1 |
5* |
4 |

27 |
3* |
2 |
3 |
6 |

Table 2: Finding the hash function for the set {10,11,15} | |

keys 2 |
%3 |

10 |
1 |

11 |
2 |

15 |
0 |

Table 3: Finding the hash function for the set {1,55,300,1095,1111} | ||||||||||

keys 3 |
%5 |
%6 |
%7 |
%8 |
%9 |
%10 |
%11 |
%12 |
%13 |
%14 |

300 |
0 |
0 |
6* |
4 |
3 |
0 |
3 |
0 |
1 |
6 |

1095 |
0 |
3 |
3 |
7* |
6 |
5* |
6 |
3 |
3* |
3 |

55 |
0 |
1* |
6* |
5 |
1* |
5* |
0* |
7* |
3* |
13 |

1 |
1* |
1* |
1 |
1 |
1* |
1 |
1 |
1 |
1 |
1 |

1111 |
1* |
1* |
5 |
7* |
4 |
1 |
0* |
7* |
6 |
5 |

For each of these three examples, we eventually gained a set of indexes which were unique and would map to a fairly small table. This method of obtaining indexes by successively increasing the remainder argument will definitely lead to a set of unique value for any set of non-repeated keys. This is because the value of one plus the largest key in the set for the remainder operator argument is guaranteed to produce a unique set of integers. For the last example this argument would be 1112 as shown in this table:

Table 4: Showing that a hash function producing a unique set is always obtainable | |

keys 3 |
%1112 |

300 |
300 |

1095 |
1095 |

55 |
55 |

1 |
1 |

1111 |
1111 |

So our perfect hash map needs an extra instance variable to hold the remainder appropriate for its set of keys. The hash function is just

` int index = key.id % remainder;`

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