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Hashing Part One Reaching for the Perfect Search Most of this material stolen from "File Structures" by Folk, Zoellick and Riccardi.

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Presentation on theme: "Hashing Part One Reaching for the Perfect Search Most of this material stolen from "File Structures" by Folk, Zoellick and Riccardi."— Presentation transcript:

1 Hashing Part One Reaching for the Perfect Search Most of this material stolen from "File Structures" by Folk, Zoellick and Riccardi

2  Text File v. Binary File  Unordered Binary File ◦ average search takes N/2 file operations  Ordered Binary File ◦ average search takes Log 2 N file operations ◦ but keeping the data file sorted is costly  Indexed File ◦ average search takes 3 or 4 file operations  Perfect Search ◦ search time = 1file read

3  Definition: o a magic black box that converts a key to the file address of that record NameField1Field2... Dannelly Hash Function

4  Example Hashing Function: o Key = Customer's Name o Function = 1 st letter x 2 nd letter, then use rightmost 4 letters. Name ascii product RRN BALL 66x65 = 4290 290 LOWELL 76x79 = 6004 004 TREE 84x82 = 6888 888 OLIVIER 79x76 = 6004 004

5  Definition: ◦ When two or more keys hash to the same address.  Minimizing the Number of Collisions: 1)pick a hash function that avoids collisions, i.e. one with a seemingly random distribution ◦ e.g. our previous function is terrible because letters like "E" and "L" occur frequently, while no one's name starts with "XZ". 2)spread out the records ◦ 300 records in a file with space for 1000 records will have many fewer collisions than 300 records in a file with capacity of 400

6  Our objective is to muddle the relationship between the keys and the addresses.  Good Ideas:  use both addition and multiplication  avoid integer overflow  so mix in some subtraction and division too  divide by prime numbers

7 1. pad the name with spaces 2. fold and add pairs of letters 3. mod by a prime after each add 4. divide sum by file size  Example: Key="LOWELL" and file size = 1,000 L O W E L L 76 79 | 87 69 | 76 76 | 32 32 | 32 32 | 32 32 7679 + 8769 = 16,448 % 19,937 = 16,448 16448 + 7676 = 24,124 % 19,937 = 4,187 4187 + 3232 = 7,419 % 19,937 = 7,419 7419 + 3232 = 10,651 % 19,937 = 10,651 10651 + 3232 = 13,883 % 19,937 = 13,833 833 13833 % 1000 = 833 Why 19,937 ? 19,937 is the largest prime that insures the next add will not cause integer overflow.

8  The simplest hash function for a string is "add up all the characters, then divide by filesize"  For example, ◦ filesize = 100 records ◦ key = "pen" ◦ address = ( 16 + 5 + 14 ) % 100 = 35 1.Find another word with the same mapping 2.Give an improvement to this hash function abcdefghijklmnopqrstuvwxyz 1234567891011121314151617181920212223242526

9  The optimal hash function for a set of keys: 1.will evenly distribute the keys across the address space, and 2.every address has a equal chance of being used.  Uniform distribution is nearly impossible. Good MappingPoor Mapping key address 11 22 A 3A3 B 4B4 C 5C5 D 6D6 E 7E7 88 9910

10  Suppose we have a file of 10,000 records, finding a hash function that will take our 10,000 keys and yield 10,000 different addresses is essentially impossible.  So, our 10,000 records are stored in a larger file.  How much larger than 10,000? o 10,500? o 12,000? o 50,000?  It Depends ◦ larger datafile:  more empty (wasted) space  fewer collisions

11  Even with a very good hash function, collisions will occur.  We must have an algorithm to locate alternative addresses.  Example, ◦ Suppose "dog" and "cat" both hash to location 25. ◦ If we add "dog" first, then dog goes in location 25. ◦ If we later add "cat", where does it go? ◦ Same idea for searching. If cat is supposed to be at 25 but dog is there, where do we look next?

12  "Linear Probing" or "Progressive Overflow"  When a key maps to address already in use, just try the next one. If that one is in use, try the next one. yadda yadda  Easy to implement.  Usually works well, especially with a non- dense file and a good hash function.  Can lead to clumps of records.

13  Assume these keys map to these addresses: 1.adams = 20 2.bates = 22 3.cole = 20 4.dean = 21 5.evans = 23  Where will each record be placed if inserted in that order?  Using linear probing, how many file accesses for each?

14  How many collisions is acceptable? ◦ Analysis: packing density v probing length  Is there a collision resolution algorithm better than linear probing? ◦ buckets


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