Re: relation algebra query

From: --CELKO-- <71062.1056_at_compuserve.com>
Date: 6 May 2002 09:44:15 -0700
Message-ID: <c0d87ec0.0205060844.5f19651f_at_posting.google.com>


>> student(sid, sname)
 subject(subid, subname)
 enrolled(subid, sid, year)  

how would you write the relational algebra for finding the subject name which is taken by all students in a particular year. ie. only subjects which have all students are displayed. <<

This is a relational division of enrolled by the join of student and subject. You just got the algebr, but perhaps the SQL will help, so let me do one of my cut & paste jobs:

Relational division is one of the eight basic operations in Codd's relational algebra. The idea is that a divisor table is used to partition a dividend table and produce a quotient or results table. The quotient table is made up of those values of one column for which a second column had all of the values in the divisor.

This is easier to explain with an example. We have a table of pilots and the planes they can fly (dividend); we have a table of planes in the hangar (divisor); we want the names of the pilots who can fly every plane (quotient) in the hangar. To get this result, we divide the PilotSkills table by the planes in the hangar.

CREATE TABLE PilotSkills
(pilot CHAR(15) NOT NULL,
 plane CHAR(15) NOT NULL,
 PRIMARY KEY (pilot, plane));

PilotSkills
pilot plane



'Celko' 'Piper Cub'
'Higgins' 'B-52 Bomber'
'Higgins' 'F-14 Fighter'
'Higgins' 'Piper Cub'
'Jones' 'B-52 Bomber'
'Jones' 'F-14 Fighter'
'Smith' 'B-1 Bomber'
'Smith' 'B-52 Bomber'
'Smith' 'F-14 Fighter'
'Wilson' 'B-1 Bomber'
'Wilson' 'B-52 Bomber'
'Wilson' 'F-14 Fighter'
'Wilson' 'F-17 Fighter'

CREATE TABLE Hangar
(plane CHAR(15) NOT NULL PRIMARY KEY);

Hangar
plane



'B-1 Bomber'
'B-52 Bomber'
'F-14 Fighter'

PilotSkills DIVIDED BY Hangar
pilot



'Smith'
'Wilson'

In this example, Smith and Wilson are the two pilots who can fly everything in the hangar. Notice that Higgins and Celko know how to fly a Piper Cub, but we don't have one right now. In Codd's original definition of relational division, having more rows than are called for is not a problem.

The important characteristic of a relational division is that the CROSS JOIN (Cartesian product) of the divisor and the quotient produces a valid subset of rows from the dividend. This is where the name comes from, since the CROSS JOIN acts like a multiplication operator.

Division with a Remainder

There are two kinds of relational division. Division with a remainder allows the dividend table to have more values than the divisor, which was Codd's original definition. For example, if a pilot can fly more planes than just those we have in the hangar, this is fine with us. The query can be written in SQL-89 as

SELECT DISTINCT pilot
  FROM PilotSkills AS PS1
 WHERE NOT EXISTS

       (SELECT *
          FROM Hangar
         WHERE NOT EXISTS 
               (SELECT *
                  FROM PilotSkills AS PS2
                 WHERE (PS1.pilot = PS2.pilot)
                   AND (PS2.plane = Hangar.plane)));

The quickest way to explain what is happening in this query is to imagine an old World War II movie where a cocky pilot has just walked into the hangar, looked over the fleet, and announced, "There ain't no plane in this hangar that I can't fly!" We are finding the pilots for whom there does not exist a plane in the hangar for which they have no skills. The use of the NOT EXISTS() predicates is for speed. Most SQL systems will look up a value in an index rather than scan the whole table. The SELECT * clause lets the query optimizer choose the column to use when looking for the index.

This query for relational division was made popular by Chris Date in his textbooks, but it is not the only method nor always the fastest. Another version of the division can be written so as to avoid three levels of nesting. While it is not original with me, I have made it popular in my books.

 SELECT PS1.pilot
   FROM PilotSkills AS PS1, Hangar AS H1   WHERE PS1.plane = H1.plane
  GROUP BY PS1.pilot
 HAVING COUNT(PS1.plane) = (SELECT COUNT(plane) FROM Hangar);

There is a serious difference in the two methods. Burn down the hangar, so that the divisor is empty. Because of the NOT EXISTS() predicates in Date's query, all pilots are returned from a division by an empty set. Because of the COUNT() functions in my query, no pilots are returned from a division by an empty set.

In the sixth edition of his book, INTRODUCTION TO DATABASE SYSTEMS (Addison-Wesley; 1995 ;ISBN 0-201-82458-2), Chris Date defined another operator (DIVIDEBY ... PER) which produces the same results as my query, but with more complexity.

Exact Division

The second kind of relational division is exact relational division. The dividend table must match exactly to the values of the divisor without any extra values.

SELECT PS1.pilot
  FROM PilotSkills AS PS1

       LEFT OUTER JOIN
       Hangar AS H1
       ON PS1.plane = H1.plane

 GROUP BY PS1.pilot
HAVING COUNT(PS1.plane) = (SELECT COUNT(plane) FROM Hangar)

   AND COUNT(H1.plane) = (SELECT COUNT(plane) FROM Hangar);

This says that a pilot must have the same number of certificates as there planes in the hangar and these certificates all match to a plane in the hangar, not something else. The "something else" is shown by a created NULL from the LEFT OUTER JOIN.

Please do not make the mistake of trying to reduce the HAVING clause with a little algebra to:

 HAVING COUNT(PS1.plane) = COUNT(H1.plane)

because it does not work; it will tell you that the hangar has (n) planes in it and the pilot is certified for (n) planes, but not that those two sets of planes are equal to each other.

Note on Performance

The nested EXISTS() predicates version of relational division was made popular by Chris Date's textbooks, while the author is associated with popularizing the COUNT(*) version of relational division. The Winter 1996 edition of DB2 ON-LINE MAGAZINE
(http://www.db2mag.com/96011ar:htm) had an article entitled "Powerful SQL:Beyond the Basics" by Sheryl Larsen which gave the results of testing both methods. Her conclusion for DB2 was that the nested EXISTS() version is better when the quotient has less than 25% of the dividend table's rows and the COUNT(*) version is better when the quotient is more than 25% of the dividend table. Received on Mon May 06 2002 - 18:44:15 CEST

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