CS145 Assignment #4

Due Wednesday, Oct. 28, 1998

Step 4 of Your PDA

1. (15 pts.) Write five queries on your PDA database, using the select-from-where construct of SQL. To receive full credit, all but perhaps one of your queries should exhibit some interesting feature of SQL: queries over more than one relation, or subqueries, for example. We suggest that you experiment with your SQL commands on a small database (e.g., your hand-created database), before running them on the large database that you loaded in PDA part 3. Initial debugging is much easier when you're operating on small amounts of data. Once you're confident that your queries are working, run them on your complete database. If you discover that most or all of your ``interesting'' queries return an empty answer on your large database, check whether you followed the instructions in Assignment #3 for generating data values that join properly. You will need to modify your data generator accordingly. Turn in a copy of all of your SQL queries, along with a script illustrating their execution. Your script should be sufficient to convince us that your commands run successfully. Please do not, however, turn in query results that are thousands (or hundreds of thousands) of lines long!
2. (15 pts.) Write five data modification commands on your PDA database. Most of these commands should be ``interesting,'' in the sense that they involve some complex feature, such as inserting the result of a query, updating several tuples at once, or deleting a set of tuples that is more than one but less than all the tuples in a relation. As for the queries in (1), you might want to try out your commands on small data before trying it on your full database. Hand in a script that shows your modification commands running in a convincing fashion.
3. (10 pts.) Create two views on top of your database schema. Show your CREATE VIEW statements and the response of the system. Also, show a query involving each view and the system response (but truncate the response if there are more than a few tuples produced). Finally, show a script of what happens when you try to update your view, say by inserting a new tuple into it. Are either of your views updatable? Why or why not? (Updatable views are discussed in Section 5.8.4 of the text. Essentially, a view is updatable if it is a selection on one base table.)
4. (20 pts.) In part (1) you probably discovered that some queries run very slowly over your large database. An important technique for improving the performance of queries is to create indexes. An index on an attribute A of relation R allows the database to quickly find all tuples in R with a given value for attribute A. This index is useful if a value of A is specified by your query (in the where-clause). It may also be useful if A is involved in a join that equates it to some other attribute. For example, in the query

        SELECT Bars.address
        FROM Drinkers, Bars
        WHERE   Drinkers.name = 'joe'
            AND Drinkers.frequents = Bars.name;
   

   we might use an index on Drinkers.name to help us find the tuple for drinker Joe quickly. We might also like an index on Bars.name, so we can take all the bars Joe frequents and quickly find the tuples for those bars to read their addresses.

   In Oracle, you can get an index by the command:

        CREATE INDEX <IndexName> ON <RelName>(<Attribute List>)
            TABLESPACE indexes;
   

   Note:
   • The text on the second line (``tablespace indexes'') is unique to the Stanford Oracle installation and is designed to get our indexes onto a second disk used only for that purpose, so one disk can be retrieving data while the other is using an index.

   • Oracle creates indexes automatically when you declare an attribute(s) PRIMARY KEY or UNIQUE. Thus, some indexes may exist before you create them and may also be making certain queries run faster than than they would with no indexes at all. Be sure to take this factor into account when trying to explain differences in running times.

   If the attribute list contains more than one attribute, then the index requires values for all the listed attributes to find a tuple. That situation might be helpful if the attributes together form a key, for example. An illustration of the CREATE INDEX command is

        CREATE INDEX DrinkerInd ON Drinkers(name)
            TABLESPACE indexes;
        CREATE INDEX BarInd ON Bars(name)
            TABLESPACE indexes;
   

   which creates the two indexes mentioned above. To get rid of an index, you can say DROP INDEX followed by the name of the index. Notice that each index must have a name, even though we only refer to the name if we want to drop the index.

   Create at least two useful indexes for your PDA. Run your queries from part (1) on your large database with the indexes and without the indexes. To time your commands, you may issue the following commands to sqlplus:

   1. TIMING START <TimerName>; starts your timer. Give it whatever name you wish.
   2. TIMING SHOW; prints the current wall-clock time of your current timer.
   (There is a way to switch among timers, which is why they are named, but we shall not use this feature.)
   3. TIMING STOP; prints the current time of your timer and stops it.

   Naturally these times may be affected by external factors such as system load, etc. Still, you should see a dramatic difference between the execution times with indexes and the times without. Turn in a script showing your commands to create indexes, and showing the relative times of query execution with and without indexes.

Problem Set

Movies(Title, Year, filmType, DirectedBy)
Stars(Title, Year, StarName, Salary)
Oscars(Title, Year, OscarAwardType, AwardedTo)
BoxOffice(Title, Year, City, GrossSales)
Ratings(Title, Year, Reviewer, Score)

The Movies relation contains the list of all movies. The value of the filmType attribute may be 'horror', 'romance', 'thriller' etc. (Title, Year) form a key in the Movies relation. The Salary attribute in the Stars relation is the amount paid to StarName for working in the movie: (Title, Year). The Oscars relation has a tuple for each Oscar awarded to a movie. The only awards of interest to us are the Best Actor, Best Actress, and Best Director awards. A typical record in the Oscars relation would look like

(Cleopatra, 1963, Best Actor, Rex Harrison).

The BoxOffice relation records for each movie its gross sales at the box office for each city. Also, each movie is rated at a scale of 1-10 by many reviewers. The rating information is kept in the relation Ratings. Write SQL statements to produce tables that contain the following information. You may use views to store intermediate results.

1. (5 pts.) A table containing for each year, the movie that has the maximum TotalSalaryCost (total salary paid to stars).

2. (5 pts.) A table containing total gross sales of all movies that won some Oscar award in the 1960s.

3. (5 pts.) A table containing, for each director, the total number of his/her movies that have won an Oscar.

4. (5 pts.) Above question, except that consider only those movies that were rated 7 or above by at least two reviewers. You are REQUIRED to use views to answer this one.

5. (5 pts.) A table containing, for each star, the total number of movies for which he/she has won an Oscar AND that have won at least two Oscars. You need not include any items for which this count is zero.

6. (5 pts.) A table containing the list of all directors who received the "Best Director" award, even though the movie for which they received the award had an average rating of less than 5.

   DATA MODIFICATION

7. (5 pts.) Write an UPDATE statement that changes the filmType of a movie that has won an Oscar to 'OscarWinner'.

8. (5 pts.) Write an INSERT statement to add, for each movie, one more rating score, equal to the average rating for that movie, to the relation Ratings. Let the reviewer for the newly added rating be 'AverageScore'.