Hash Table Collision Handling

标签：proc   val   linear   ring   not   arch   ati   repair   item   

Two basic methods; separate chaining and open address.

Separate Chain

Hangs an additional data structure off of the buckets.  For example the bucket array becomes an array of link list.  So to find an item we first go to the bucket then compare keys..  This is a popular method, and if link list is used the hash never fills up.

Illustrate

load factor, f = n/N where n is number of items stored in the hash table. Like for the load factor to be less then 1.

The cost for get(k) is on average O(n/N)

Open Addressing

The problem with separate chaining is that the data structure can grow with out bounds.  Sometimes this is not appropriate because of finite storage, for example in embedded processors.

Open addressing does not introduce a new structure.  If a collision occurs then we look for availability in the next spot generated by an algorithm. Open Addressing is generally used where storage space is a premium, i.e. embedded processors. Open addressing not necessarily faster then separate chaining.

Methods for Open Addressing:

1. Linear Probing:

   We try to insert Item = (k, e) into bucket A[i] and find it full so the next bucket we try is:

   A[(i + 1) mod N]

   then try A[(i + 1) mod N], etc.

   Illustrate with 11 buckets: Note the probing is linear.

   Note the hash table can be filled up.

   Also what to do if we remove an Item.  Should repair the array A but this is too costly.  Instead we mark the bucket as available/deactivated.  Then the next use of findElement(k) would skip over the available/deactivated bucket.  insertItem(k, e) would insert into a available/deactivated.

   Clustering slows down searches. 

2. Quadratic Probing:

   A[ (i + f(j) )mod N]  where j = 0, 1, 2, ... and f(j) = j²

   Helps avoids clustering.  Secondary clustering can occur. We can imagine a more complicated function for f. 

3. Double Hashing:

   Use a second hash function h‘.

   A[ (i + f(j) )mod N]   where  f(j) = j*h‘(k) should not evaluate to zero. Example: h‘(k) = q - (k mod q).  Note that still i = h(k).

http://www.csl.mtu.edu/cs2321/www/newLectures/17_Hash_Tables_Collisions.html

Hash Table Collision Handling

标签：proc   val   linear   ring   not   arch   ati   repair   item   

原文地址：http://www.cnblogs.com/grainy/p/7244183.html