• Hash Table Collision Handling


    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 factorn/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 = (ke) 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(ke) 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

  • 相关阅读:
    Unity3d在各个平台读取Sqlite3数据库
    UI
    Could..... not preload global game manager
    Asset Store 下载的package存在什么地方?
    NDK下载地址
    UGUI富文本
    Unity播放视频
    让IIS支持无后缀名访问
    PC Android IOS资料同步更新
    PHP memcached 扩展的安装
  • 原文地址:https://www.cnblogs.com/grainy/p/7244183.html
Copyright © 2020-2023  润新知