Presentation on theme: "Hashing Part Two Better Collision Resolution Small parts of this material stolen from "File Organization and Access" by Austing and Cassel."— Presentation transcript:
Hashing Part Two Better Collision Resolution Small parts of this material stolen from "File Organization and Access" by Austing and Cassel
Hash function converts key to file address Collision is when two or more keys hash to the same address Collision Avoidance o Good Hash Function spreads out the keys evenly along the whole address space o Non-Dense File decreases chance of collisions and decreases probes after a collision
Very simple collision resolution if H(key) = A, and A is already used, try A+1, then A+2, etc Advantages easy to implement guaranteed to use all addresses Disadvantages clustering / clumping
Given the following hashes and linear probing: 1.adams = 20 2.bates = 22 3.cole = 20 4.dean = 21 5.evans = 23 Result of either ◦ poor hash function ◦ dense file AddressData 20Adams 21Cole 22Bates 23Dean 24Evans 25 26
Instead of adding 1, spread out by random amount True random would not work. Instead use pseudo-random. While A is in use A = (A + R) mod T A = address R = prime T = Table Size
1.adams = 20 2.bates = 22 3.cole = 20 4.dean = 21 5.evans = 23 Linear Probing AddressData 20Adams 21Cole 22Bates 23Dean 24Evans 25 26 Random Probing, R = 5 AddressData 20Adams 21Dean 22Bates 23Evans 24 25Cole 26 27 28 But what if 25 and 30 already had keys directly hashed to those locations? Cole would be at 35 -- 4 probes away.
Assuming a better hash function and less dense file are not options... And assuming linear and random probing lead to coalesced lists... Chaining : maintain a linked list of collisions, one head per address ◦ Example, after addition of Adams and Cole, and R=5: 19 : null 20 : 35 -> null 21 : null Advantage: Faster at resolving collisions Disadvantage : Space
File Read Time = seek time + latency + data read time Smallest Readable Portion = 1 cluster = 4KB (usually) To access portion of a file, most of the time is in seek time and latency, not read time ◦ so, number of file reads is more important than size of reads, until size gets really big SO... reading a few records from a file takes no more time than reading just one record
Given, collisions will occur... Why not just read 2, or 3, or 4 records instead of just 1 on each read operation? "Bucket" - a group of records at the same address "Hash File of Buckets" - hashed keys collide to small arrays of records in the data file
use avg collisions and stddev? if 1000 records and 200 addresses ◦ then avg is 5.0 ◦ but stddev might be 1.0 start by determining how many records can fit in one or more disk clusters then design a good hash function to match that address space
Advantages: Can achieve relatively fast access ◦ Remember, the hash function tells us where the record is located, so only 1 read operation. And even with collisions, the list of possible records is read into memory, which searches fast. ◦ Search Time = time to read bucket + time to search the array Disadvantages: What do we do when the bucket is full? ◦ solutions are similar to collision resolution ◦ we end up reading multiple sets of records
Collisions will happen! Poisson Function: ◦ p(x) gives the probability that a given address will have had x records assigned to it. (r/N) x e -(r/N) p(x) = --------------- x! N = number of available addresses r = number of records to be stored x = number of records assigned to a given address
Given ◦ N = 1000 ◦ r = 1000 Probability that a given address will have exactly one, two, or three keys hashed to it: p(1) = 0.368 p(2) = 0.184 p(3) = 0.061
Given ◦ N = 10,000 ◦ R = 10,000 How many addresses should have one, two, or three keys hashed to them? 10,000 x p(1) = 10000x0.3679 = 3679 10,000 x p(2) = 10000x0.1839 = 1839 10,000 x p(3) = 10000x0.0613 = 613 So, 1839 keys will collide once and 613 will collide at least twice. Many of those collisions will disrupt probing.
Given ◦ r = 500 ◦ N = 1000 ◦ one record per address Addresses with exact one record? N x p(1) = 1000 x 0.303 = 303 How many overflow records? 1 x N x p(2) + 2 x N x p(3) + 3 x N x p(4) +... = N x [1 x p(2) + 2 x p(3) + 3 x p(4)] = 1000 x [ 1 x 0.076 + 2 x 0.013 + 3 x 0.002] = 107 Percentage of Records NOT stored at home address 107 / 500 = 21.4% Records that never collide = 303 Records that cannot go at their home = 107 Records at their home, but cause collisions = 90 Total = 500
Packing Density (%) Synonyms as percent of records 104.8 3013.6 5021.4 7028.1 10036.8
We must balance many factors: file size e.g., wasted space in hashed files e.g., extra space for index files disk access times available memory frequency of additions and deletions compared to searches Best Solution of All? probably a combination of indexed files, hashing, and buckets
Thursday April 14 ◦ No Class Tuesday April 19 ◦ B-Trees Thursday April 21 ◦ Review
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