CS136, Lecture 32

    1. Hashing functions
      1. Selecting hashing functions
        1. Digit selection
        2. Division
        3. Mid-Square:
        4. Folding
        5. String-valued keys

Hashing functions

The table implementation of an array with keys as the subscripts and values as contents makes sense. Nevertheless there are some important restrictions on the use of this representation of a table.

This implementation assumes that the data has a key which is of a restricted type (some enumerated type in Pascal, integers in Java), which is not always the case.

Note also that the size requirements for this implementation could be prohibitive.

Ex. If the array held 2000 student records indexed by social security number it would be declared as ARRAY[0..999,999,999]

What if most of entries are empty? If we use a smaller array then all elements will still fit.

Suppose we have a lot of data elements of type EltType and a set of locations in which we could store data elements.

Consider a function H: EltType -> Location with the properties

  1. H(elt) can be computed quickly

  2. If elt1 <> elt2 then H( elt1) <> H( elt2). (H is one-to-one function)

This is called a perfect hashing function. Unfortunately, they are difficult to find unless you know all possible entries to the table in advance. This is not often the case.

Instead we use something that behaves well, but not necessarily perfectly:
i.e., keep property 1, but may give up 2.

The goal is to scatter elements through the array randomly so that they won't bump into each other.

Define a function H: Keys -> Addresses, and call H(element.key) the home address of element.

Of course now we can't list elements easily in any kind of order, but hopefully we can find them in time O(1).

Note that each entry in the table will need to include the actual key, since several different keys will likely get mapped to the same subscript.

Hash functions and hashtables are important enough that Hashtable is provided by Java as a utility class (in java.utility.* - see section 12.7 in Arnold & Gosling) and the method

        public int hashCode()
is included in Object. Can use built-in hashCode or redefine for your objects.

Most important property is that for all o and o', if o.equals(o') then must have o.hashCode() = o'.hashCode(). Otherwise hashTable algorithms won't work!

There are two problems to look at:

  1. What are good hashing functions?

  2. What do we do when two different elements get sent to same home address?

Selecting hashing functions

The following quote should be memorized by any trying to design a hashing function: "A given hash function must always be tried on real data in order to find out whether it is effective or not." Data which has certain regularities can completely destroy the usefulness of any hashing function!

Here are some sample Hashing functions.

Presume for the moment that the keys are numbers.

a. Digit selection

Choose digits from certain positions of key (e.g. last 3 digits of SS#).

Unfortunately it is easy to get a biased sample. We can carefully analyze keys to see which will work best. We must watch out for patterns - they should generate all possible table positions. (For example the first digits of SS#'s reflect the region in which they were assigned and hence usually would work poorly as a hashing function.)

b. Division

Let H(key) = key % TableSize.

This is very efficient and often gives good results if the TableSize is chosen properly.

If it is chosen poorly then you can get very poor results. If TableSize = 28 = 256 and the keys are integer ASCII equivalent of two letter pairs, i.e. Key(xy) = 28 * (int)x + (int)y, then all pairs starting ending with the same letter get mapped to same address. Similar problems arise with any power of 2.

The best bet seems to be to let the TableSize be a prime number.

In practice if no divisor of the TableSize is less than 20, the hash function seems to be OK. (Text uses 997 in the sample program)

c. Mid-Square:

In this algorithm you square the key and then select certain bits. Usually the middle half of the bits is taken. The mixing provided by the multiplication ensures that all digits are used in the computation of the hash code.

Example: Let the keys range between 1 and 32000 and let the TableSize be 2048 = 211.

Square the Key and remove the middle 11 bits. (Grabbing certain bits of a word is easy to do using shift operators in assembly language or can be done using the div and mod operators using powers of two.)

In general r bits gives a table of size 2r.

d. Folding

Break the key into pieces (sometimes reversing alternatie chunks) and add them up.

This is often used if the key is too big. E.g., If the keys are Social security numbers, the 9 digits will generally not fit into an integer. Break it up into three pieces - the 1st digit, the next 4, and then the last 4. Then add them together.

Now you can do arithmetic on them.

This technique is often used in conjunction with other methods (e.g. division)

e. String-valued keys

We can use a formula like - Key(xy) = 28 * (int)x + (int)y to convert from alphabetic keys to ASCII equivalents. This is often used in combination with folding (for the rest of the string) and division.

If you use longints to hold the numbers, then you can get 4 letters into one number in this way. If they are all alphabetic (no special characters), then you can subtract (int)'a' from each ASCII code in order to reduce the size of the keys.

Here is a very simple-minded hash code for strings: Add together the ordinal equivalent of all letters and take the remainder mod tableSize.

Problem: Words with same letters get mapped to same places:

miles, slime, smile

This would be much improved if you took the letters in pairs before division.

Nevertheless, for simplicity we adopt this simple-minded (and thus relatively useless) hash function for the following discussion.

Here is a function which adds up ord of letters and then mod tableSize:

hash = 0;
for (int charNo = 0; charNo < word.length(); charNo++) 
    hash = hash + (int)(word.charAt(CharNo));
hash = hash % tableSize;  (* gives 0 <= hash < tableSize *)

Code is only a little more complex to multiply each succeeding character by 2*8.

Efficient way using Horner's rule:

hash = 0;
for (in CharNo = word.length()-1;charNo >= 0; charNo--)
    hash = (256*hash + (int)(word.charAt(CharNo))) % tableSize;

Notice we mod by tableSize each time we update hash to prevent overflows.

Efficient way of calculating uses only word.length() multiplications, while normal way would involve O(word.length()2) multiplications.