1. Introduction to Computer Systems
Lecturer: Steve Maybank
Department of Computer Science and Information Systems
Spring 2017
Week 7b: File management, indexing and hashing
21 February 2017

2. Files and Records File: unit for the storage of information.
Logical record: natural division of the information in a file, i.e. natural given the meaning of the data in the file.
21 February 2017
Brookshear, Section 9.5

3. Example of a Logical Record
In many files each record contains a unique key.
Format of the records:
text (a sequence of
characters)
Name field
Id number (key)
Kimberly Ann Dawson
385172
Example of a record
21 February 2017
Brookshear, Section 9.5

4. Sequential Files Records are accessed in a fixed order.
On a sequential storage medium such as a CD, the records can be stored in the same order as the access order.
Reading the file is efficient if the storage order is also the access order.
21 February 2017
Brookshear, Section 9.5

5. Examples of Sequential Files
Audio files
Video files
Text files, eg. programs or documents (in fact most files produced on a PC).
21 February 2017
Brookshear, Section 9.5

6. Segmentation of a Text File Into Records
fixed length blocks of 25 characters …
……. …
file K i m b e r l y A n D a w s o 3 8 5 1 7 2
The records are read in turn.
Sentinel: special record marking end of file (EOF).
21 February 2017
Brookshear, Section 9.5

7. Merging of Two Sequential Files
Input files
Output file B D
Stage 1 A A C E F B D A B
Stage 2 A C E F
21 February 2017
Brookshear, Section 9.5

8. Procedure for Merging Two Sequential Files
def MergeFiles(inputFileA, inputFileB, OutputFile) :
if (both input files at EOF) :
Stop, with OutputFile empty
if (inputFileA not at EOF) :
Declare its first record to be its current record
if (inputFileB not at EOF)
while (neither input file at EOF) :
Put the current record with the smaller key in OutputFile
if (that current record is the last record in its corresponding input file) :
Declare that input file to be at EOF)
else :
Declare the next record in that input file to be the file’s current record
Starting with the current record in the input file that is not at EOF,
copy the remaining records to OutputFile.
21 February 2017
Brookshear, Section 9.5

9. Random Accessing of Records
It is inefficient to access randomly chosen records from a sequential file.
For efficient random access use an indexed file or a hash file
21 February 2017
Brookshear, Section 9.5

10. Indexed Files Each record is identified by a unique key.
The file has an index which contains the keys, and for each key the address of the corresponding record.
Problem: the index has to be maintained
21 February 2017
Brookshear, Section 9.5

11. Example of an Indexed File
key address
Record for Brown
Brown P1
Johnson P2
Jones P3
Smith P4
Watson P5
Indexed file
Index
21 February 2017
Brookshear, Section 9.5

12. Hash Files Each record is identified by a unique key.
The file has a function which takes a key as input and computes the address of the corresponding record.
More efficient than a sequential file if records are accessed randomly.
21 February 2017
Brookshear, Section 9.5

13. Construction of a Hash File
Memory
New record r4, key k4. r1 r3
Use k4 to compute
an address in memory
and store r4 at that
address. r2 …
21 February 2017
Brookshear, Section 9.5

14. Example of a Hash File Key Memory address 1203 40 467 41 89 37
The hash function
key |-> key%11+36
Key
Memory address
1203 40
467 41 89 37
replaces the index
21 February 2017
Brookshear, Section 9.5

15. Terminology for Hash Files
Bucket: section of the storage area corresponding to a hash function value.
Load factor: ratio of number of records in file to maximum number of records that can be stored. Load factors <=50% are recommended.
Collision: the hash function takes the same value on two different keys.
21 February 2017
Brookshear, Section 9.5

16. Nature of a Hash Function
constant on
strips.
Disk: large
set of keys • • •
• records in
current file
(4 records,
1 collision) •
21 February 2017
Birkbeck College

17. General Result For “reasonable” numbers of buckets and records,
The probability of at least one collision is high.
The probability that any bucket will be the site of many collisions is low.
21 February 2017
Brookshear, Section 9.5

18. Examples of Hash Functions
Mid square: compute (key * key) and set bucket number = middle digits.
Extraction: select digits from certain positions within the key.
Divide key by number of buckets and use the remainder.
21 February 2017
Birkbeck College

19. Hash Function Requirements
Easily and quickly computed.
Values evenly spread over the bucket numbers with few collisions
21 February 2017
Brookshear, Section 9.5

20. Pros and Cons of Hash Functions
Pro: no need to maintain an index.
Cons:
Requires a method of dealing with collisions.
Inefficient use of storage (load factor <=50% recommended).
21 February 2017
Brookshear, Section 9.5

21. Place Two Records in Three Buckets
R1, R2 R1 R2
All nine configurations have
the same probability
The sum of the different
probabilities is 1 R1
 R2 R2 R1
R1, R2 R1
 R2 R2
 R1 R2
 R1
 R1,R2
21 February 2017
Birkbeck College
Number of buckets: 3
Number of records: 2
Number of ways of storing 2 records in 3 buckets = 3x3
Number of ways of storing 2 records in 3 buckets without collisions = 6
Probability of no collision = 6/3x3=2/3

22. Example: Probability of a Collision
Number of buckets: 3
Number of records: 2
Number of ways of storing 2 records in 3 buckets = 3x3
Number of ways of storing 2 records in 3 buckets without collisions = 6
Probability of no collision = 6/3x3=2/3
21 February 2017
Birkbeck College

23. Birthday Problem
A room contains n people. What is the least value of n such that the probability that two people have the same birthday is greater than ½?
A hash file has n records. There are 365 possible values for the hash function. What is the least value of n such that the probability that two records have the same hash value is greater than ½?
21 February 2017
Birkbeck College

24. 2003 Examination, Qu. 15
Suppose that a hashed file is constructed using the ‘divide’ hash function with 5 storage buckets. For each of the following key field values, 10, 23 and 41, identify the bucket in which the record with that key value is placed.
21 February 2017
Birkbeck College