- 5.3.1 Hash Codes for Primitive Data Types
- 5.3.2 Hash Codes for Compound Objects
- 5.3.3 Hash Codes for Arrays and Strings

The hash tables discussed in the previous section are used to associate data with integer keys consisting of bits. In many cases, we have keys that are not integers. They may be strings, objects, arrays, or other compound structures. To use hash tables for these types of data, we must map these data types to -bit hash codes. Hash code mappings should have the following properties:

- If
and
are equal, then
and
are equal.
- If and are not equal, then the probability that should be small (close to ).

The first property ensures that if we store in a hash table and later look up a value equal to , then we will find --as we should. The second property minimizes the loss from converting our objects to integers. It ensures that unequal objects usually have different hash codes and so are likely to be stored at different locations in our hash table.

Small primitive data types like , , , and are usually easy to find hash codes for. These data types always have a binary representation and this binary representation usually consists of or fewer bits. (For example, in Java, is an 8-bit type and is a 32-bit type.) In these cases, we just treat these bits as the representation of an integer in the range . If two values are different, they get different hash codes. If they are the same, they get the same hash code.

A few primitive data types are made up of more than bits, usually bits for some constant integer . (Java's and types are examples of this with .) These data types can be treated as compound objects made of parts, as described in the next section.

5.3.2 Hash Codes for Compound Objects

For a compound object, we want to create a hash code by combining the individual hash codes of the object's constituent parts. This is not as easy as it sounds. Although one can find many hacks for this (for example, combining the hash codes with bitwise exclusive-or operations), many of these hacks turn out to be easy to foil (see Exercises 5.4-5.6). However, if one is willing to do arithmetic with bits of precision, then there are simple and robust methods available. Suppose we have an object made up of several parts whose hash codes are . Then we can choose mutually independent random -bit integers and a random -bit odd integer and compute a hash code for our object with

int hashCode() { long[] z = {0x2058cc50L, 0xcb19137eL, 0x2cb6b6fdL}; // random long zz = 0xbea0107e5067d19dL; // random long h0 = x0.hashCode() & ((1L<<32)-1); // unsigned int to long long h1 = x1.hashCode() & ((1L<<32)-1); long h2 = x2.hashCode() & ((1L<<32)-1); return (int)(((z[0]*h0 + z[1]*h1 + z[2]*h2)*zz) >>> 32); }The following theorem shows that, in addition to being straightforward to implement, this method is provably good:

where

since each of and is at most , so their product is at most . By assumption, , so (5.5) has at most one solution in . Therefore, since and are independent ( are mutually independent), the probability that we select so that is at most .

The final step of the hash function is to apply multiplicative hashing to reduce our -bit intermediate result to a -bit final result . By Theorem 5.3, if , then .

To summarize,

5.3.3 Hash Codes for Arrays and Strings

The method from the previous section works well for objects that have a fixed, constant, number of components. However, it breaks down when we want to use it with objects that have a variable number of components since it requires a random -bit integer for each component. We could use a pseudorandom sequence to generate as many 's as we need, but then the 's are not mutually independent, and it becomes difficult to prove that the pseudorandom numbers don't interact badly with the hash function we are using. In particular, the values of and in the proof of Theorem 5.3 are no longer independent.

A more rigorous approach is to base our hash codes on polynomials over prime fields. This method is based on the following theorem, which says that polynomials over prime fields behave pretty-much like usual polynomials:

To use Theorem 5.4, we hash a sequence of integers with each using a random integer via the formula

Note the extra term at the end of the formula. It helps to think of as the last element, , in the sequence . Note that this element differs from every other element in the sequence (each of which is in the set ). We can think of as an end-of-sequence marker.

The following theorem, which considers the case of two sequences of the same length, shows that this hash function gives a good return for the small amount of randomization needed to choose :

Since , this polynomial is non-trivial. Therefore, by Theorem 5.4, it has at most solutions in . The probability that we pick to be one of these solutions is therefore at most .

Note that this hash function also deals with the case in which two sequences have different lengths, even when one of the sequences is a prefix of the other. This is because this function effectively hashes the infinite sequence

The following example code shows how this hash function is applied to an object that contains an array, , of values:

int hashCode() { long p = (1L<<32)-5; // prime: 2^32 - 5 long z = 0x64b6055aL; // 32 bits from random.org int z2 = 0x5067d19d; // random odd 32 bit number long s = 0; long zi = 1; for (int i = 0; i < x.length; i++) { long xi = (x[i].hashCode() * z2) >>> 1; // reduce to 31 bits s = (s + zi * xi) % p; zi = (zi * z) % p; } s = (s + zi * (p-1)) % p; return (int)s; }

The above code sacrifices some collision probability for implementation convenience. In particular, it applies the multiplicative hash function from Section 5.1.1, with to reduce to a 31-bit value. This is so that the additions and multiplications that are done modulo the prime can be carried out using unsigned 63-bit arithmetic. This means that the probability of two different sequences, the longer of which has length , having the same hash code is at most

opendatastructures.org