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Hashing Part Two Better Collision Resolution Small parts of this material stolen from "File Organization and Access" by Austing and Cassel.

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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:

1 Hashing Part Two Better Collision Resolution Small parts of this material stolen from "File Organization and Access" by Austing and Cassel

2  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

3  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

4  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

5  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

6 1.adams = 20 2.bates = 22 3.cole = 20 4.dean = 21 5.evans = 23 Linear Probing AddressData 20Adams 21Cole 22Bates 23Dean 24Evans Random Probing, R = 5 AddressData 20Adams 21Dean 22Bates 23Evans 24 25Cole But what if 25 and 30 already had keys directly hashed to those locations? Cole would be at probes away.

7  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

8  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

9  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

10  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

11  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

12  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

13  Given ◦ N = 1000 ◦ r = 1000  Probability that a given address will have exactly one, two, or three keys hashed to it: p(1) = p(2) = p(3) = 0.061

14  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) = 10000x = ,000 x p(2) = 10000x = ,000 x p(3) = 10000x = 613  So, 1839 keys will collide once and 613 will collide at least twice.  Many of those collisions will disrupt probing.

15  Given ◦ r = 500 ◦ N = 1000 ◦ one record per address  Addresses with exact one record? N x p(1) = 1000 x = 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 x 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

16 Packing Density (%) Synonyms as percent of records

17  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

18  Thursday April 14 ◦ No Class  Tuesday April 19 ◦ B-Trees  Thursday April 21 ◦ Review


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