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Material for Database Class.
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List of commits:
Subject Hash Author Date (UTC)
Working on exercises. 42549c8d28249f1ecf85dd21f6d004ae63c533ad aubert@math.cnrs.fr 2018-05-19 05:43:04
Working on incorporating the exercises. f32148faeb8b11425fe9a57770db9bc85ae02c34 aubert@math.cnrs.fr 2018-05-19 04:51:37
working on sum for exo. 5fb55d2c31b06161ade3a7cae707485d8c02447a aubert@math.cnrs.fr 2018-05-18 21:12:56
Adding various small notes. 5eac6e0b19441fca9a3f70f239d5cbbc4d769976 aubert@math.cnrs.fr 2018-05-17 20:37:14
First pass over! ea1e4f6ddabd91287d10384dc8f28565a762526b aubert@math.cnrs.fr 2018-05-17 19:26:28
Working on notes on normal forms. 58bfcedc6bd6336592a5f93a43ffc0aaf64fa8f6 aubert@math.cnrs.fr 2018-05-16 16:40:12
Initial commit, draft of lecture notes. 200d2739bca881d60c7c9381b16f5a4d6384ff29 aubert@math.cnrs.fr 2018-05-16 15:35:36
Commit 42549c8d28249f1ecf85dd21f6d004ae63c533ad - Working on exercises.
Author: aubert@math.cnrs.fr
Author date (UTC): 2018-05-19 05:43
Committer name: aubert@math.cnrs.fr
Committer date (UTC): 2018-05-19 05:43
Parent(s): f32148faeb8b11425fe9a57770db9bc85ae02c34
Signer:
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Tree: e2e4b6afe443ac72580a52fc53aef39dbe8240fb
File Lines added Lines deleted
exercises/sum.md 3101 0
exercises/sum.tex 0 2783
notes/00_Questions_Notes.md 14 0
File exercises/sum.md added (mode: 100644) (index 0000000..d4438fe)
1 ---
2 documentclass: scrreprt
3 papersize: letter
4 link-citations: true
5 title: CSCI 3410 - Database Systems
6 subtitle: Lecture Notes (Draft)
7 author: Clément Aubert
8 institute: Augusta University
9 dir: ltr
10 lang: en
11 numbersections: true
12 keywords:
13 - Computer Science
14 - Database
15 - MySQL programming
16 header-includes:
17 - \usepackage[table]{xcolor}
18 - \setromanfont[Ligatures={Common,TeX}]{Linux Libertine O}
19 - \setmainfont[Ligatures={Common,TeX}]{Linux Libertine O}
20 - \setmonofont{Latin Modern Mono Light}
21 - \setmonofont[SmallCapsFont={Latin Modern Mono Caps}]{Latin Modern Mono Light}
22 - \usepackage{xunicode}
23 - \usepackage{fvextra}
24 - \DefineVerbatimEnvironment{Highlighting}{Verbatim}{breaklines,commandchars=\\\{\}}
25 - \definecolor{lgray}{HTML}{e6e7e8}
26 ---
27
28 <!--
29
30 pandoc sum.md --pdf-engine=xelatex --toc --filter pandoc-numbering --top-level-division=chapter -M date="`date "+%B %e, %Y"`" -o sum.pdf
31
32 pandoc sum.md --pdf-engine=xelatex --toc --filter pandoc-numbering --top-level-division=chapter -M date="`date "+%B %e, %Y"`" -s -o sum.tex && xelatex sum.tex
33
34 -->
35
36 Exercise +.#
37
38 What does it mean to say that `SQL` is at the same time a "data definition language" and a "data manipulation language"?
39
40
41 Solution +.#
42
43 It can specify the conceptual and internal schema, \textbf{and} it can manipulate the data.
44
45
46 Exercise +.#
47
48 Name three kind of objects (for lack of a better word) a `CREATE` statement can create.
49
50
51 Solution +.#
52
53 Database (schema), table, view, assertion, trigger, etc.
54
55
56 Exercise +.#
57
58 Write a `SQL` statement that adds a primary key constraint to an attribute named `Id` in an already existing table named `STAFF`.
59
60
61 Solution +.#
62
63 `ALTER TABLE STAFF ADD PRIMARY KEY(Id);`
64
65
66 Exercise +.#
67
68 Complete the following table wo different examples when asked for examples:
69
70 \begin{tabular}{m{6cm} | m{6cm} }
71 \hline
72 \rowcolor{lgray}
73 Data Type & Examples \\
74 \hline
75 Int & $4$, $-32$\\
76 \hline
77 Char(4) & %\blank[width=1.9em]{'trai', 'plol'}
78 \\
79 \hline
80 % \blank[width=1.9em]{VarChar(10)}
81 & 'Train', 'Michelle'\\
82 \hline
83 Bit(4) & % \blank[width=1.9em]{ B'1010', B'0101'}
84 \\
85 \hline
86 % \blank[width=1.9em]{Boolean}
87 & TRUE, UNKNOWN\\
88 \hline
89 \end{tabular}
90
91
92
93 Exercise +.#
94
95 Explain what the following SQL statement does
96
97 ```mysql
98 CREATE SCHEMA Faculty;
99 ```
100
101 Solution +.#
102
103 It creates a schema, i.e., a database, named "Faculty".
104
105
106 Exercise +.#
107
108 If I want to enter January 21, 2016, as a value for an attribute with the `DATE` datatype, what value should I enter?
109
110
111 Solution +.#
112
113 `DATE`'2016-01-21'
114
115
116 Exercise +.#
117
118 Write a statement that inserts the values "Thomas" and "$4$" into the table `TRAINS`.
119
120
121 Solution +.#
122
123 ```sql
124 INSERT INTO TRAINS VALUES('Thomas', 4);
125 ```
126
127
128 Exercise +.#
129
130 If `PkgName` is a primary key, what can you tell about the number of rows returned by the following statement?
131 ```sql
132 SELECT * FROM MYTABLE WHERE PkgName = 'MySQL';
133 ```
134
135
136 Solution +.#
137
138 Yes.
139
140
141 Exercise +.#
142
143 What is the difference between an implicit, an explicit, and a semantic constraint?
144
145
146 Solution +.#
147
148 The textbook reads, pp. 67 - 69:
149 \begin{enumerate}
150 \item Constraints that are inherent in the data model. We call these inherent model-based constraints or implicit constraints.
151 \item Constraints that can be directly expressed in schemas of the data model, typically by specifying them in the DDL (data definition language, see Section 2.3.1). We call these schema-based constraints or explicit constraints.
152 \item Constraints that cannot be directly expressed in the schemas of the data model, and hence must be expressed and enforced by the application programs. We call these application-based or semantic constraints or business rules.
153 \end{enumerate}
154
155
156 Exercise +.#
157
158 If you want every tuple referencing the tuple you're about to delete to be deleted as well, what mechanism should you use?
159
160
161 Solution +.#
162
163 We should use a referential triggered action clause, `ON DELETE CASCADE`.
164
165
166 Exercise +.#
167
168 If a database designer is using the `ON UPDATE SET NULL` for a foreign key, what mechanism is he implementing (i.e., describe how the database will react a certain operation)?
169
170
171 Solution +.#
172
173 If the referenced tuple is updated, then the attribute of the referencing relation are set to NULL.
174
175
176 Exercise +.#
177
178 If the following is part of the design of a table:
179
180 ```sqlmysql
181 FOREIGN KEY (DptNumber) REFERENCES DEPARTMENT(DptNumber) ON DELETE SET DEFAULT ON UPDATE CASCADE
182 ```
183
184 What happen to the row whose foreign key `DptNumber` is set to $3$ if
185 \begin{enumerate}
186 \item the row in the `DEPARTEMENT` table with primary key `DptNumber` set to \(3\) is deleted?
187 \item the row in the `DEPARTEMENT` table with primary key `DptNumber` set to \(3\) as its value for `DptNumber` updated to \(5\)?
188 \end{enumerate}
189
190
191
192 Solution +.#
193
194 The department number is set to the default value.
195
196 The department number is updated accordingly.
197
198
199 Exercise +.#
200
201 If the following is part of the design of a `WORKER` table:
202
203 ```sqlmysql
204 FOREIGN KEY WORKER(DptNumber) REFERENCES DEPARTMENT(DptNumber) ON UPDATE CASCADE
205 ```
206
207 Admitting there is a row in the `WORKER` table whose foreign key `DptNumber` is set to $3$, what would happen if
208 \begin{enumerate}
209 \item We tried to delete the row in the `DEPARTEMENT` table with primary key `DptNumber` set to \(3\)?
210 \item We tried to change the value of `DptNumber` of the row in the `DEPARTEMENT` table with primary key `DptNumber` set to \(3\)?
211 \end{enumerate}
212
213
214
215 Exercise +.#
216
217 Given a relation `TOURIST(Name, EntryDate, Address)`, write a `SQL` statement printing the name and address of all the tourists who entered the territory after the 15 September, 2012.
218
219
220 Solution +.#
221
222 ```sqlmysql
223 SELECT Name, Address
224 FROM TOURIST
225 WHERE EntryDate > DATE'2012-09-15';
226 ```
227
228
229
230
231 Exercise +.#
232
233 What is the fully qualified name of an attribute? Give an example.
234
235
236 Solution +.#
237
238 The name of the relation with the name of its schema and a period beforehand. myDB.MYTABLE, EMPLOYEE.name, etc.
239
240
241 % Exercise +.#
242
243 If `DEPARTMENT` is a database, what is `DEPARTMENT.*`?}
244
245
246 Solution +.#
247
248 All the tables in that database.
249
250
251 Exercise +.#
252
253 What is a multi-set? What does it mean to say that SQL treats tables as multisets?
254
255
256 Solution +.#
257
258 A set where the same value can occur twice.
259 The same tuple can occur twice in a table.
260
261
262 Exercise +.#
263
264 What is the difference between those two queries?
265 ```mySQL
266 SELECT ALL * FROM MYTABLE;
267 ```
268 and
269 ```mySQL
270 SELECT DISTINCT * FROM MYTABLE;
271 ```
272 How are the results the same? How are they different?
273
274
275 Solution +.#
276
277 All will print the duplicate, distinct will eliminate them.
278
279 Problem +.#
280
281 *This exercise assume you successfully completed Problem~2 of Homework 1.*
282
283 1. Connect to your MySQL DBMS as `testuser`:
284 - **In windows**, open a command prompt (search for "cmd") and type
285
286 ```bat
287 cd "C:\Program Files\MySQL\MySQL Server 5.7\bin"
288 ```
289 - **In Linux**, open a shell (as a normal user)
290
291 Then, in both cases, type
292
293 ```plain
294 mysql -u testuser -p
295 ```
296
297 and enter the password `password`.
298 If you are prompted with a message
299
300 ```bash
301 ERROR 1045 (28000): Access denied for user 'testuser'@'localhost' (using password: YES)
302 ```
303
304 then you probably typed the wrong password.
305 Otherwise, you should see a welcoming message from MySQL / MariaDB and a prompt.
306 2. Create a new database called `HW_2Q1`
307
308 ```sqlmysql
309 CREATE DATABASE HW_2Q1;
310 ```
311 3. List the databases
312
313 ```sqlmysql
314 SHOW DATABASES;
315 ``` Make sure `HW_2Q1` is a part of it.
316 4. Remove ("drop") the database, using
317
318 ```sqlmysql
319 DROP DATABASE HW_2Q1;
320 ```
321 In the future, we will refer to the commands \ref{item-connect}.\ and \ref{item-create}.\ as "log-in as `testuser`" and create a database `HW_2Q1`.
322
323
324
325 Problem +.#
326
327 The goal of this problem is to learn where to find the documentation of your DBMS, and to have a first look at the syntax of SQL commands.
328
329 You can consult your textbook, Table~5.2, p. 140 (6th edition) or Table~7.2, p. 235 (7th edition), for a very quick summary of the most common commands.
330 Make sure you are familiar with the Backus–Naur form (BNF) notation commonly used:
331 \begin{itemize}
332 \item non-terminal symbols (i.e., variables, parameters) are enclosed in angled brackets,
333 ```plain
334 <...>
335 ```
336
337 \item optional parts are shown in square brackets,
338 ```plain
339 [...]
340 ```
341
342 \item repetitons are shown in braces
343 ```plain
344 {...}
345 ```
346
347 \item alternatives are shown in parenthesis and separated by vertical bars,
348
349 ```plain
350 (...|...|...)
351 ```
352
353 \end{itemize}
354
355 The most complete lists of commands are probably at
356 \begin{itemize}
357 \item \url{https://mariadb.com/kb/en/library/sql-statements/} and
358 \item \url{https://dev.mysql.com/doc/refman/5.7/en/sql-syntax.html}
359 \end{itemize}
360 Those are the commands implemented in the DBMS you are actually using.
361 Since there are small variations from one implementation to the other, it's better to take one of this link as a reference in the future.
362
363
364 Problem +.#
365
366 \emph{This exercise, and the following ones, assume you successfully completed Problem~\ref{setup}.}
367
368 Log in as `testuser` and create a database `HW_2Q3`.
369
370 1. We will first set-up our meta-data (i.e., the schema, the structure where our tables will be, and the tables themselves).
371
372 Let us first tell MySQL that we want to use that database
373 ```sqlmysql
374 USE HW_2Q3;
375 ```
376 and ask what it contains
377 ```sqlmysql
378 SHOW TABLES;
379 ```
380 Now, create a first table named `NAME`
381
382 ```sqlmysql
383 CREATE TABLE NAME(
384 FName VARCHAR(15),
385 LName VARCHAR(15),
386 Id INT,
387 PRIMARY KEY(Id)
388 );
389 ```
390
391
392 Note that the `SQL` syntax and your DBMS are completely fine with your statement spreading over multiple lines.
393 Let us create a second table, named `ADDRESS`
394
395 ```sqlmysql
396 CREATE TABLE ADDRESS(
397 StreetName VARCHAR(15),
398 Number INT,
399 Habitants INT,
400 PRIMARY KEY(StreetName, Number)
401 );
402 ```
403
404
405 To make sure those two tables were actually created, we can use
406
407 ```sqlmysql
408 SHOW TABLES;
409 ```
410 But how to make sure that you entered the attributes correctly?
411 One way to make sure is to enter the command
412
413 ```sqlmysql
414 show create table name;
415 ```
416
417 ```sqlmysql
418 DESC ADDRESS;
419 ```
420 and to examine carefully the message printed. You should read
421
422 ```sqlmysql
423 +------------+-------------+------+-----+---------+-------+
424 | Field | Type| Null | Key | Default | Extra |
425 +------------+-------------+------+-----+---------+-------+
426 | StreetName | varchar(15) | NO | PRI | NULL| |
427 | Number | int(11) | NO | PRI | NULL| |
428 | Habitants | int(11) | YES | | NULL| |
429 +------------+-------------+------+-----+---------+-------+
430 ```
431
432 If you believe there is a mistake, you can erase ("drop") the \emph{table} (not the whole database, as we did in Problem~1) using
433
434 ```sqlmysql
435 DROP TABLE ADDRESS;
436 ```
437 and then re-create it. Of course, you can do the same for the `NAME` table.
438
439 Now, let us add a foreign key to the `ADDRESS` table:
440
441 ```sqlmysql
442
443 ALTER TABLE ADDRESS
444 ADD FOREIGN KEY (Habitants)
445 REFERENCES NAME(Id);
446 ```
447
448 And observe the modification:
449 ```sqlmysql
450 DESC ADDRESS;
451 ```
452
453
454
455 2. This second part is about data, i.e., filling our tables with actual information.
456 We begin by adding some data in the `NAME` table:
457 ```sqlmysql
458 INSERT INTO NAME VALUES ('Barbara', 'Liskov', 003);
459 INSERT INTO NAME VALUES ('Tuong Lu', 'Kim', 004);
460 INSERT INTO NAME VALUES ('Samantha', NULL, 080);
461 ```
462
463
464 To display the data we just inserted, we can use:
465 ```sqlmysql
466 SELECT * FROM NAME;
467 ```
468
469
470 Do you notice anything regarding the values we entered for the `Id` attribute?
471
472 We can add some data into the `ADDRESS` table (using a different syntax):
473 ```sqlmysql
474 INSERT INTO ADDRESS (StreetName, Number, Habitants)
475 VALUES
476 ('Armstrong Drive', 10019, 003),
477 ('North Broad St.', 23, 004),
478 ('Robert Lane', 120, NULL);
479 ```
480
481
482 And let's use our first update command:
483 ```sqlmysql
484 UPDATE ADDRESS SET Habitants = 3 WHERE Number = 120;
485 ```
486
487
488
489 3.
490 Now, do the following:
491 1. Have a look at the formal definition of the
492 ```sqlmysql
493 ALTER
494 ```
495 command, at \url{https://dev.mysql.com/doc/refman/5.7/en/alter-table.html} or \url{https://mariadb.com/kb/en/library/alter-table/}.
496 2. Draw the relations corresponding to that database, including the domains and primary, as well as foreign, keys.
497 3. Write a
498
499 ```sqlmysql
500 SELECT
501 ```
502 statement that returns the
503 ```sqlmysql
504 Id
505 ```
506 number of the person whose first name is "Samantha".
507 % SELECT Id FROM NAME WHERE FName = 'Samantha';
508 4. Write a statement that violate the entity integrity constraint. What is the error message returned?
509 % ERROR 1062 (23000): Duplicate entry '80' for key 'PRIMARY'
510 5. Execute an `UPDATE` statement that violate the referential integrity constraint. What is the error message returned?
511 % ERROR 1452 (23000): Cannot add or update a child row: a foreign key constraint fails (`HW_2Q3`.`ADDRESS`, CONSTRAINT `fk_id` FOREIGN KEY (`Habitants`) REFERENCES `name` (`Id`))
512 6. Write a statement that violate another kind of constraint. Explain what constraint you are violating, and explain the error message.
513 % ERROR 1136 (21S01): Column count doesn't match value count at row 1
514
515 Problem +.#
516 Log in as `testuser` and create a database `HW_2Q4`. Tell MySQL that you want to use that database, and create a table (I will assume you named it `EXAMPLE` in the following, but you are free to name it the way you want) with at least two attributes that have different data types. Don't declare a primary key yet. Answer the following:
517 \begin{enumerate}
518 \item Add a tuple to your table using
519 ```mySQL
520 INSERT INTO EXAMPLE VALUES(X, Y);
521 ```
522 where "X" and "Y" are values with the right type.
523 Try to add this tuple again. What do you observe? (You can use
524 ```mySQL
525 SELECT * FROM EXAMPLE;
526 ```
527 to observe what is stored in this table.)
528 \item Alter your table to add a primary key, using
529 ```mySQL
530 ALTER TABLE EXAMPLE ADD PRIMARY KEY (Attribute);
531 ``` where `Attribute` is the name of the attribute you want to be a primary key. What do you observe?
532 \item Empty your table using```sqlmysql
533 DELETE FROM EXAMPLE;
534 ``` and alter your table to add a primary key, using the command we gave at the previous step. What do you observe?
535 \item Try to add the same tuple twice. What do you observe?
536 %\item Now, create \emph{another table} (just one attribute and no primary key is fine) and try to create a foreign key pointing to the attribute in `EXAMPLE` that is not the primary key.
537 % What do you observe?
538 % => Messy, they get an error message if the attribute is not the primary key or if the datatype isn't the same, or if the command is not properly formatted... The error message is too vague for them to learn anything.
539 \end{enumerate}
540
541
542 Solution +.#
543
544
545 ```sqlmysql
546 CREATE TABLE EXAMPLE(
547 X VARCHAR(15),
548 Y INT);
549 ```
550
551
552
553 ```sqlmysql
554 INSERT INTO EXAMPLE VALUES('Train', 4);
555 ```
556
557
558 OK to insert same tuple twice.
559
560 ```sqlmysql
561 ALTER TABLE EXAMPLE ADD PRIMARY KEY (X);
562 ERROR 1062 (23000): Duplicate entry 'Train' for key 'PRIMARY'
563 ```
564
565
566
567
568 %%%%% 03
569
570
571 Exercise +.#
572
573 Describe what the star do in the statement `SELECT ALL * FROM MYTABLE;`.
574
575
576 Solution +.#
577
578 It selects all the attributes.
579
580
581
582 Exercise +.#
583
584 What is wrong with the statement `SELECT * WHERE Name = 'CS' FROM DEPARTMENT;`?
585
586
587 Solution +.#
588
589 You can't have the `WHERE` before `FROM`.
590
591
592
593 Exercise +.#
594
595 When is it useful to use a select-project-join query?
596
597
598 Solution +.#
599
600 We use those query that projects on attributes using a selection and join conditions when we need to look for information gathered in multiple tables.
601
602
603 Exercise +.#
604
605 When is a tuple variable useful?
606
607
608 Solution +.#
609
610 It makes the distinction between two different copies of the same relation, it is useful when we want to select a tuple in a relation that is in a particular relation with a tuple in the same relation.
611
612 cf. \url{https://stackoverflow.com/a/7698796/2657549}:
613 They are useful for saving typing, but there are other reasons to use them:
614 \begin{itemize}
615 \item If you join a table to itself you must give it two different names otherwise referencing the table would be ambiguous.
616 \item It can be useful to give names to derived tables, and in some database systems it is required... even if you never refer to the name.
617 \end{itemize}
618
619
620 Exercise +.#
621
622 Write a query that changes the name of the professor whose login is 'caubert' to 'Hugo Pernot' in the table `PROF`.
623
624
625 Solution +.#
626
627 `UPDATE PROF SET Name = 'Hugo Pernot' WHERE Login = 'caubert';`
628
629
630 Exercise +.#
631
632 Can an `UPDATE` statement have a `WHERE` condition using an attribute that isn't the primary key? If no, justify, if yes, tell what could happen.
633
634
635 Solution +.#
636
637 Yes, we could update more than one tuple at a time.
638
639
640 Exercise +.#
641
642 What is a multi-set?
643 What does it mean to say that `SQL` treats tables as multisets?
644
645
646 Solution +.#
647
648 A set where the same value can occur twice.
649 The same tuple can occur twice in a table.
650
651
652 Exercise +.#
653
654 What is the difference between those two queries?
655 ```mySQL
656 SELECT ALL * FROM MYTABLE;
657 ```
658 and
659 ```mySQL
660 SELECT DISTINCT * FROM MYTABLE;
661 ```
662 How are the results the same? How are they different?
663
664
665 Solution +.#
666
667 `ALL` will print the duplicate, `DISTINCT` will eliminate them.
668
669
670 Exercise +.#
671
672 What are the possible meanings or interpretations for a `NULL` value?
673
674
675 Solution +.#
676
677 Unknown value, unavailable / withheld, N/A.
678
679
680
681 Exercise +.#
682
683 What are the values of the following expressions (i.e., do they evaluate to `TRUE`, `FALSE`, or `UNKNOWN`)?
684
685 `TRUE AND FALSE`\hspace{2.2em}
686 `TRUE AND UNKNOWN`\hspace{2.2em}
687 `NOT UNKNOWN`\hspace{2.2em}
688 `FALSE OR UNKNOWN`
689
690
691
692 Solution +.#
693
694 `FALSE`, `UNKNOWN`, `UNKNOWN`, `FALSE`
695
696
697 Exercise +.#
698
699 What comparison expression should you use to test if a value is different from `NULL`?
700
701
702 Solution +.#
703
704 `IS NOT`
705
706
707 Exercise +.#
708
709 Explain this query:
710
711 ```sqlmysql
712 SELECT Login
713 FROM PROF
714 WHERE Department IN ( SELECT Major
715 FROM STUDENT
716 WHERE Login = 'jrakesh');
717 ```
718
719
720 Solution +.#
721
722 It list the login of the professors teaching in the department where the student whose login is "jrakesh" is majoring.
723
724
725 Exercise +.#
726
727 What is wrong with this query?
728
729 ```sqlmysql
730 SELECT Name
731 FROM STUDENT
732 WHERE Login IN ( SELECT Code
733 FROM Department
734 WHERE head = 'aturing');
735 ```
736
737
738
739
740
741 Solution +.#
742
743 It tries to find a `Login` in a `Code`!
744
745
746 Exercise +.#
747
748 Write a query that returns the number of row (i.e., of entries, of tuples) in a table named `BOOK`.
749
750
751 Solution +.#
752
753 `SELECT COUNT(*) FROM BOOK;`
754
755
756 Exercise +.#
757
758
759 Consider the following `SQL` code:
760
761 ```sqlmysql
762 CREATE TABLE COMPUTER(
763 Id VARCHAR(20) PRIMARY KEY,
764 Model VARCHAR(20)
765 );
766 CREATE TABLE PRINTER(
767 Id VARCHAR(20) PRIMARY KEY,
768 Model VARCHAR(20)
769 );
770 CREATE TABLE CONNEXION(
771 Computer VARCHAR(20),
772 Printer VARCHAR(20),
773 PRIMARY KEY(Computer, Printer),
774 FOREIGN KEY (Computer) REFERENCES COMPUTER(Id),
775 FOREIGN KEY (Printer) REFERENCES PRINTER(Id)
776 );
777
778 INSERT INTO COMPUTER VALUES
779 ('A', 'DELL A'),
780 ('B', 'HP X'),
781 ('C', 'ZEPTO D'),
782 ('D', 'MAC Y');
783 INSERT INTO PRINTER VALUES
784 ('12', 'HP-140'),
785 ('13', 'HP-139'),
786 ('14', 'HP-140'),
787 ('15', 'HP-139');
788 INSERT INTO CONNEXION VALUES
789 ('A', '12'),
790 ('A', '13'),
791 ('B', '13'),
792 ('C', '14');
793 ```
794
795
796 Write a query that returns … (in parenthesis, the values returned \emph{in this set-up}, but you have to be general)
797
798 1. … the number of computers connected to the printer whose id is '13' (i.e., \(2\) ).
799 2. … the number of different models of printers (i.e., \(2\)).
800 3. … the model(s) of the printer connected to the computer whose id is 'A' (i.e., 'HP-140' and 'HP-139').
801 4. … the id of the computer(s) not connected to any printer (i.e., 'D').
802
803
804
805 Exercise +.#
806 Write a query that returns the sum of all the values stored in the `Pages` attribute of a `BOOK` table.
807
808
809 Solution +.#
810
811 `SELECT SUM(Pages)FROM BOOK;`
812
813
814
815 Exercise +.#
816 Write a query that adds a `Pages` attribute of type `INT` into a (already existing) `BOOK` table.
817
818
819 Solution +.#
820
821 `ALTER TABLE BOOK ADD COLUMN Pages INT;`
822
823
824 Exercise +.#
825
826 Write a query that removes the default value for a `Pages` attribute in a `BOOK` table.
827
828
829 Solution +.#
830
831 `ALTER TABLE BOOK ALTER COLUMN Pages DROP DEFAULT;`
832
833 Problem +.#
834
835 Create the `PROF`, `STUDENT`, `DEPARTMENT` and `GRADE` tables as during the lecture.
836 Populate them with some data.
837
838 1. Create a `LECTURE` table as follows:
839 \begin{itemize}
840 \item It should have four attributes, `Name`, `Instructor`, `Code`, and \texttt{Year}, of type `VARCHAR(25)` for the two first, `YEAR(4)`, and `CHAR(5)`.
841 \item The \texttt{Year} and `Code` attributes should be the primary key (yes, have \emph{two} attributes be the primary key).
842 \item The `Instructor` attribute should be a foreign key referencing the `Login` attribute in `PROF`.
843 \end{itemize}
844 2. Populate the `LECTURE` table with some made-up data.
845 3. Add two columns to the `GRADE` table, using
846 ```sqlmysql
847 ALTER TABLE GRADE
848 ADD COLUMN LectureCode CHAR(5),
849 ADD COLUMN LectureYear YEAR(4);
850 ```
851
852 4. Use `DESCRIBE` and `SELECT` to observe the schema of the `GRADE` table and its rows (i.e., tuples).
853 Is that what you would have expected?
854 5. Add a foreign key in `GRADE`, using
855
856 ```sqlmysql
857 ALTER TABLE GRADE
858 ADD FOREIGN KEY (LectureYear, LectureCode)
859 REFERENCES LECTURE(Year, Code);
860 ```
861
862 6. Update the tuples in `GRADE` with some made-up data: now, every row should contain, on top of a login and a grade, a lecture year and a lecture code.
863 7. Write `SELECT` statements answering the following questions (where XXX and YY should be relevant values considering your data):
864 1. "I taught class XXX in YYYY, could you give me the logins and grades of the students who took it?"
865 2. "Could you list the instructors who taught in YYYY?" (and, please, avoid duplicates)
866 3. "Could your list the name and grade of all the student who took class XXX (no matter the year)?"
867 4. "Could you tell me which years was the class XXX taught?"
868 5. "Could you list the classes taught the same year as class XXX?"
869 6. "Could you print the name of the students who registered after Ava Alyx?"
870 7. "How many departments' heads are teaching this year?"
871
872
873 %Solution +.#
874
875 ```sqlmysql
876 CREATE TABLE LECTURE(
877 Name VARCHAR(25),
878 Instructor VARCHAR(25),
879 Year YEAR(4),
880 Code CHAR(5),
881 PRIMARY KEY(Year, Code),
882 FOREIGN KEY (Instructor) REFERENCES PROF(Login)
883 );
884
885 INSERT INTO LECTURE VALUES
886 ('Intro to CS', 'caubert', 2017, '1304'),
887 ('Intro to Algebra', 'perdos', 2017, '1405'),
888 ('Intro to Cyber', 'aturing', 2017, '1234');
889
890 ALTER TABLE GRADE ADD COLUMN LectureCode CHAR(5),
891 ADD COLUMN LectureYear YEAR(4);
892
893 DESCRIBE GRADE;
894
895 SELECT * FROM GRADE;
896
897 ALTER TABLE GRADE ADD FOREIGN KEY (LectureYear, LectureCode) REFERENCES LECTURE(Year, Code);
898
899 UPDATE GRADE SET LectureCode = '1304', LectureYear = 2017 WHERE Login = 'jrakesh' AND Grade = '2.85';
900 UPDATE GRADE SET LectureCode = '1405', LectureYear = 2017 WHERE Login = 'svlatka' OR (Login = 'jrakesh' AND Grade = '3.85');
901 UPDATE GRADE SET LectureCode = '1234', LectureYear = 2017 WHERE Login = 'aalyx' OR Login = 'cjoella';
902
903 SELECT Login, Grade FROM Grade WHERE Lecturecode='1304' and LectureYear = '2017';
904
905 SELECT DISTINCT Instructor FROM LECTURE WHERE year = 2017;
906
907 SELECT Name, Grade FROM STUDENT, GRADE WHERE GRADE.LectureCode = 1405 AND STUDENT.Login = GRADE.Login;
908
909 SELECT Year FROM Lecture WHERE Code = '1234';
910
911 SELECT Name FROM LECTURE WHERE Year IN (SELECT Year FROM LECTURE WHERE CODE = '1234');
912
913 SELECT B.name FROM STUDENT AS A, STUDENT AS B WHERE A.Name = 'Ava Alyx' AND A.Registered > B.Registered;
914
915 SELECT COUNT(DISTINCT PROF.Name) AS 'Head Teaching This Year' FROM LECTURE, DEPARTMENT, PROF WHERE Year = 2017 AND Instructor = Head AND Head = PROF.Login;
916 ```
917
918
919
920
921
922
923
924
925 <!--
926 OK TATA UP TO HERE
927 -->
928
929
930 %%%%% 04
931
932
933 Exercise +.#
934
935 What could be the decomposition of an attribute used to store an email address? When could that be useful?
936
937
938 Solution +.#
939
940 Name / extension.
941 To have statistics about the extensions, to sort the username by length, etc.
942
943
944 Exercise +.#
945
946 Draw the ER diagram for a "COMPUTER" entity that has one multivalued attribute "Operating\_System", a composite attribute "Devices" (decomposed into "Keyboard" and "Mouse") and an "Id" key attribute.
947
948
949 Solution +.#
950
951 \begin{tikzpicture}[node distance=7em]
952 \node[entity] (computer) {COMPUTER};
953 \node[attribute] (pid) [left of=computer] {\key{Id}} edge (computer);
954 \node[multi attribute] (os) [above of=computer] {Operating\_System} edge (computer);
955 \node[attribute] (devices) [right of=computer] {Devices} edge (computer);
956 \node[attribute] (mouse) [above right of=devices] {Mouse} edge (devices);
957 \node[attribute] (keyboard) [right of=devices] {Keyboard} edge (devices);
958 \end{tikzpicture}
959
960
961 Exercise +.#
962
963 Draw the ER diagram for a "CELLPHONE" entity that has a composite attribute "Plan" (decomposed into "Carrier" and "Price"), a "Mobile\_identification\_number" key attribute, and a multi-valued "App\_Installed" attribute.
964
965
966 Solution +.#
967
968 \begin{tikzpicture}[node distance=7em]
969 \node[entity] (cellphone) {CELLPHONE};
970 \node[attribute] (mid) [left of=cellphone] {\key{MIN}} edge (cellphone);
971 \node[multi attribute] (app) [above of=cellphone] {App\_Installed} edge (cellphone);
972 \node[attribute] (plan) [right of=cellphone] {Plan} edge (cellphone);
973 \node[attribute] (carrier) [above right of=plan] {Carrier} edge (plan);
974 \node[attribute] (price) [right of=plan] {Price} edge (plan);
975 \end{tikzpicture}
976
977
978
979 Exercise +.#
980
981 Name one difference between a primary key in the relational model, and a key attribute in the ER model.
982
983
984 Solution +.#
985
986 There can be more than one key in the ER model.
987
988
989 Exercise +.#
990
991 What is the difference between an entity type and a weak entity type?
992
993
994 Solution +.#
995
996 The weak entity type doesn't have a primary key.
997
998
999 Exercise +.#
1000
1001 What is the degree of a relationship type?
1002
1003
1004 Solution +.#
1005
1006 The number of participating entity types.
1007
1008
1009 Exercise +.#
1010
1011 What is a self-referencing, or recursive, relationship type? Give two examples.
1012
1013
1014 Solution +.#
1015
1016 A relationship type where the same entity type participates more than once.
1017 On rooms, "is to the left of", on persons, "is married to".
1018
1019
1020 Exercise +.#
1021
1022 What does it mean for a binary relationship type "Owner" between entity types "Person" and "Computer" to have a cardinality ration $M:N$?
1023
1024
1025 Solution +.#
1026
1027 That a person can own multiple computers, and that a computer can have multiple owners.
1028
1029
1030 Exercise +.#
1031
1032 What are the two possible structural constraints on a relationship type?
1033
1034
1035 Solution +.#
1036
1037 Cardinality ration and participation constraints.
1038
1039
1040 Exercise +.#
1041
1042 Under what condition(s) can an attribute of a binary relationship type be migrated to become an attribute of one of the participating entity type?
1043
1044
1045 Solution +.#
1046
1047 When the cardinality is $1 : N$, $1 : 1$ or $N : 1$.
1048
1049
1050 Exercise +.#
1051
1052 What is a partial key?
1053
1054
1055 Solution +.#
1056
1057 For a weak entity attribute, it is the attribute that can uniquely identify weak entites that are related to the same owner entity.
1058
1059
1060 Exercise +.#
1061
1062 For the following binary relationships, suggest cardinality ratios based on the common-sense meaning of the entity types.
1063
1064 {\centering
1065
1066 \begin{tabular}{l c l}
1067 \textbf{Entity 1} & \textbf{Cardinality Ratio} & \textbf{Entity 2}\\
1068 STUDENT & & MAJOR\\ % 1 : N
1069 CAR & & TAG \\ % 1 : 1
1070 INSTRUCTOR & & LECTURE \\ % N :1
1071 INSTRUCTOR & & OFFICE \\ % 1 : N
1072 COMPUTER & & OPERATING\_SYSTEM % N : M
1073 \end{tabular}
1074
1075 }
1076
1077
1078 Solution +.#
1079
1080 N:1 , 1 : 1, 1 : N , 1:N, M : N
1081
1082
1083 Exercise +.#
1084
1085 Give an example of a binary relationship type of cardinality $1 : N$.
1086
1087
1088 Solution +.#
1089
1090 SUPERVISION, a recursive relationship on EMPLOYEE.
1091
1092
1093 Exercise +.#
1094
1095 Give an example of a binary relationship type of cardinality $N:1$, and draw the corresponding diagram (you don't have to include details on the participating entity types).
1096
1097
1098 Solution +.#
1099
1100 \begin{tikzpicture}[node distance=7em]
1101 \node[relationship] (contains){BELONGS};
1102 \node[entity] (burger) [left = 1cm of contains] {HAND} edge node[above] {$N$} (contains);
1103 \node[entity] (ingredient) [right = 1cm of contains] {PERSON} edge node[above]{$1$} (contains);
1104 \end{tikzpicture}
1105
1106
1107
1108 Exercise +.#
1109
1110 Draw an ER diagram with a single entity type, with two stored attributes and one derived attribute. In your answer, it should be clear that the value for the derived attribute will always be obtained from the value(s) for the other attribute(s).
1111
1112
1113 Solution +.#
1114
1115
1116 \begin{tikzpicture}[node distance=7em]
1117 \node[entity] (person) {PERSON};
1118 \node[attribute] (ssn) [left of=person] {\key{SSN}} edge (person);
1119 \node[attribute] (dob) [above of=person] {Date\_Of\_Birth} edge (person);
1120 \node[derived attribute] (age) [right of=person] {Age} edge (person);
1121 \end{tikzpicture}
1122
1123
1124
1125 Exercise +.#
1126
1127 Draw an ER diagram expressing the total participation of an entity type "BURGER" in a binary relation "CONTAINS" between "BURGER" and "INGREDIENT".
1128 What would be the ratio of such a relation?
1129
1130
1131 Solution +.#
1132
1133 \begin{tikzpicture}[node distance=7em]
1134 \node[relationship] (contains){CONTAINS};
1135 \node[entity] (burger) [left = 1cm of contains] {BURGER} edge[total] node[above] {$N$} (contains);
1136 \node[entity] (ingredient) [right = 1cm of contains] {INGREDIENT} edge node[above]{$M$} (contains);
1137 \end{tikzpicture}
1138
1139
1140 Exercise +.#
1141
1142 Convert the following ER diagram into a relational model:
1143
1144 \begin{tikzpicture}[node distance=7em]
1145 \node[relationship] (contains){STAYS\_AT};
1146 \node[entity] (person) [above = 1cm of contains] {PERSON} edge node[right] {$N$} (contains);;
1147 \node[attribute] (ssn) [left of=person] {\key{SSN}} edge (person);
1148 \node[attribute] (dob) [above = .5cm of person] {Date\_Of\_Birth} edge (person);
1149 \node[entity] (burger) [below of= contains] {PLACE} edge node[right] {$1$} (contains);
1150 \node[attribute] (dob) [below = .5cm of burger] {\key{Address}} edge (burger);
1151 \node[attribute] (ssn) [left of=burger] {Rooms} edge (burger);
1152 \end{tikzpicture}
1153
1154
1155 Solution +.#
1156
1157 \begin{tikzpicture}[node distance=7em]
1158 \node[relationship] (contains){CONTAINS};
1159 \node[entity] (burger) [left = 1cm of contains] {BURGER} edge[total] node[above] {$N$} (contains);
1160 \node[entity] (ingredient) [right = 1cm of contains] {INGREDIENT} edge node[above]{$M$} (contains);
1161 \end{tikzpicture}
1162
1163
1164 Exercise +.#
1165
1166 Why do weak entity type have a total participation constraint?
1167
1168
1169 Solution +.#
1170
1171 Otherwise, we couldn't identify entities in it without owner entity.
1172
1173 Problem +.#
1174
1175
1176 Consider the ER schema for the MOVIES database in Figure~\ref{fig:movies}.
1177
1178 \begin{figure}
1179 \begin{tikzpicture}
1180 \node (a) at (0,0) {\includegraphics[page=1,width=1\textwidth]{img/te.pdf}};
1181 \fill[white] (-8.3, 6.5) rectangle (-3.5, 4);
1182 \end{tikzpicture}
1183 \caption{An ER diagram for a MOVIES database schema.} \label{fig:movies}
1184 \end{figure}
1185
1186 Assume that MOVIES is a populated database.
1187 ACTOR is used as a gender-netral term.
1188 Given the constraints shown in the ER schema, respond to the following statements with \emph{True} or \emph{False}.
1189 Justify each answer.
1190
1191 \begin{enumerate}
1192 \item There are no actors in this database that have been in no movies.
1193 \item There might be actors who have acted in more than ten movies.
1194 \item Some actors could have done a lead role in multiple movies.
1195 \item A movie can have only a maximum of two lead actors.
1196 \item Every director have to have been an actor in some movie.
1197 \item No producer has ever been an actor.
1198 \item A producer cannot be an actor in some other movie.
1199 \item There could be movies with more than a dozen actors.
1200 \item Producers can be directors as well.
1201 \item A movie can have one director and one producer.
1202 \item A movie can have one director and several producers.
1203 \item There could be some actors who have done a lead role, directed a movie, and produced some movie.
1204 \item It is impossible for a director to play in the movie (s)he directed.
1205 \end{enumerate}
1206
1207
1208
1209 Solution +.#
1210
1211 \begin{enumerate}
1212 \item true
1213 \item true
1214 \item true
1215 \item true
1216 \item false
1217 \item false
1218 \item false
1219 \item true
1220 \item true
1221 \item true
1222 \item true
1223 \item true
1224 \item false
1225 \end{enumerate}
1226
1227
1228 Problem +.#
1229
1230 Draw the ER diagram for the following situation:
1231 A car-insurance company wants to have a database of accidents.
1232 An accident involves cars, drivers, and it has several aspects: the moment and place where it took place, the amount of damages, and a (unique) report number.
1233 A car has a license, a model, a year, and an owner.
1234 A driver has an id, an age, a name, and an address.
1235
1236 One of the interesting choice is: should "accident" be an entity type or a relationship type?
1237
1238 Solution +.#
1239
1240 \includegraphics[page=1,width=1\textwidth]{img/p.pdf}
1241
1242 OR
1243
1244
1245
1246 \begin{tikzpicture}[node distance=7em]
1247 \node[entity] (person) {Person};
1248 \node[attribute] (pid) [left of=person] {\key{id}} edge (person);
1249 \node[attribute] (name) [above left of=person] {Name} edge (person);
1250 \node[attribute] (address) [below left of=person] {address} edge (person);
1251
1252 \node[relationship] (owns) [above right of=person] {Owns} edge node[above, sloped]{$M$} (person);
1253 \node[entity] (car) [above right of=owns] {Car} edge node[above, sloped]{$N$} (owns);
1254 \node[attribute] (licence) [above left of=car] {\key{licence}} edge (car);
1255 \node[attribute] (model) [above of =car] {model} edge (car);
1256 \node[attribute] (year) [above right of =car] {year} edge (car);
1257
1258 \node[relationship] (accident) [right = 5cm of person] {Accident} edge[total] node[above, sloped]{$M$} (person);
1259 \node[attribute] (report_number) [below of = accident] {\key{Report Number}} edge (accident);
1260 \node[attribute] (time) [right of = accident] {Time} edge (accident);
1261 \node[attribute] (place) [below right of = accident] {Place} edge (accident);
1262 \node[attribute] (damage_amount) [above right of = accident] {Damage\_Amount} edge (accident);
1263
1264 \draw (car) edge[total] node[above, sloped]{$N$} (accident);
1265 \end{tikzpicture}
1266
1267
1268
1269 Problem +.#
1270
1271
1272 Apply the ER-to-Relation mapping to your ER diagram from the previous problem.
1273
1274
1275 %%% 05
1276
1277
1278 Exercise +.#
1279
1280 What is insertion anomaly? Give an example.
1281
1282
1283 Solution +.#
1284
1285
1286 When you have to invent a primary key or add a lot of `NULL` value to be able to add a tuple.
1287 I want to add a room in my DB, but the only place where rooms are listed are as an attribute on a Instructor table, so I have to "fake" an instructor to add a room.
1288
1289
1290 Exercise +.#
1291
1292 Why should we avoid attributes whose value will often be `NULL`?
1293 Can the usage of `NULL` be completely avoided?
1294
1295
1296 Solution +.#
1297
1298 Because they waste space, and because they are ambiguous (N/A, or unknown, or not communicated?).
1299 No, it is necessary sometimes.
1300
1301
1302 Exercise +.#
1303
1304 Consider the following relation:
1305
1306 \begin{tabular}{p{3em} l}
1307 & PROF(\underline{SSN}, Name, Department, Bike\_brand)
1308 \end{tabular}
1309
1310 Why is it a poor design to have a "Bike\_brand" attribute in such a relation?
1311 How should we store this information?
1312
1313
1314 Solution +.#
1315
1316 Because it will be `NULL` most of the time.
1317 In a separate table.
1318
1319
1320 Exercise +.#
1321
1322 Consider the following relation:
1323
1324 \begin{tabular}{p{3em} l}
1325 & STUDENT(\underline{SSN}, Name, \(\hdots\), Sibling\_On\_Campus)
1326 \end{tabular}
1327
1328 Why is it a poor design to have a "Sibling\_On\_Campus" attribute in such a relation?
1329 How should we store this information?
1330
1331
1332 Solution +.#
1333
1334 Because it will be `NULL` most of the time, and because students could have more than one sibling on campus.
1335 In a separate table.
1336
1337
1338 Exercise +.#
1339
1340 Consider the following relational database schema:
1341
1342 \begin{tabular}{p{3em} l}
1343 & STUDENT(\underline{Login}, Name, \(\hdots\), Major, Major\_Head)\\
1344 & DEPARTMENT(\underline{Code}, Name, Major\_Head)
1345 \end{tabular}
1346
1347 Assuming that "Major" is a foreign key referencing "DEPARTMENT.Code", what is the problem with that schema? How could you address it?
1348
1349
1350 Solution +.#
1351
1352 Major\_Head will give update anomalies.
1353 By putting the Head of the department in the DEPARTMENT relation only, i.e., removing it from STUDENT.
1354
1355
1356 Exercise +.#
1357
1358 Consider the relation \(R(A , B, C, D, E, F)\) and the following functional dependencies:
1359 \begin{enumerate}
1360 \item \(F \to \{D, C\}, D \to \{B, E\}, \{B, E\} \to A\)
1361 \item \(\{A, B\} \to \{C, D\}, \{B, E\} \to F\)
1362 \item \(A \to \{C, D\}, E \to F, D \to B\)
1363 \end{enumerate}
1364 For each set of functional dependency, give a key for \(R\). We want a key, so it has to be minimal.
1365
1366
1367 Solution +.#
1368
1369 \begin{enumerate}
1370 \item \(F\)
1371 \item \(\{A, B, E\}\)
1372 \item \(\{A, E\}\)
1373 \end{enumerate}
1374
1375
1376 \begin{exercise}[tags={none}]
1377 Consider the relation \(R(A, B, C, D, E, F)\) and the following functional dependencies:
1378 \[ A \to \{D, E\}, D \to \{B, F\}, \{B, E\} \to A, \{A,C\} \to \{B, D, F\}, A \to F\]
1379 Answer the following:
1380 \begin{enumerate}
1381 \item How many candidate keys is there? List them.
1382 \item How many transitive dependencies can you find? Give them and justify them.
1383 \end{enumerate}
1384
1385
1386 Solution +.#
1387
1388 \begin{enumerate}
1389 \item \(\{A, C\}\),
1390 \item \(A \to F\) by \(A \to D, D\to F\).
1391 \end{enumerate}
1392
1393
1394 Exercise +.#
1395
1396 Consider the relation \(R(A, B, C, D)\) and answer the following:
1397 \begin{enumerate}
1398 \item If \(\{A, B\}\) is the only key, is \(\{A, B\} \to \{C,D\}, \{B, C\} \to D\) a 2NF? List the nonprime attributes and justify.
1399 \item If \(\{A, B, C\}\) is the only key, is \(A \to \{B, D\}, \{A, B, C\} \to D\) a 2NF? List the nonprime attributes and justify.
1400 \end{enumerate}
1401
1402
1403 Solution +.#
1404
1405 \(1.\) Yes. \(C\) and \(D\) are non prime, and they fully depend on \(\{A,B\}\).
1406
1407 \(2.\) No. \(D\) is the only non prime, and it depends only on \(A\).
1408
1409
1410 Exercise +.#
1411
1412 Consider the relation \(R(A, B, C, D, E, F)\) with candidate keys \(\{A, B\}\) and \(C\). Answer the following:
1413 \begin{enumerate}
1414 \item What are the prime attributes in \(R\)?
1415 \item Is \(\{C,D\} \to E\) a fully functional dependency?
1416 \item Write a set of functional dependencies containing at least one transitive depency, and justify your answer.
1417 \end{enumerate}
1418
1419
1420 Solution +.#
1421
1422 \begin{enumerate}
1423 \item\(A,B,C\)
1424 \item No, because we can remove \(D\),
1425 \item \(A \to D\), \(D \to E\) and \(A \to E\)
1426 \end{enumerate}
1427
1428
1429 Exercise +.#
1430
1431 Consider the relation \(R(A , B, C, D, E)\) and the following functional dependencies:
1432 \begin{enumerate}
1433 \item \(C \to D, \{C, B\} \to A, A \to \{B, C, D\}, B \to E\)
1434 \item \(\ A \to \{C, D\}, C \to B, D \to E, \{E, C\} \to A\)
1435 %\item \(\{A, B\} \to D, D \to \{B, C\}, E \to C\)
1436 \end{enumerate}
1437 For each one, give one candidate key for \(R\).%two candidate keys for \(R\).
1438
1439
1440 Solution +.#
1441
1442 \begin{enumerate}
1443 \item \(\{B, C\}\), \(A\)
1444 \item \(A\), \(\{C,E\}\),
1445 \item \(\{A, D, E\}\), \(\{A, B, E\}\)
1446 \end{enumerate}
1447
1448
1449 Exercise +.#
1450
1451 Consider the relation \(R(A, B, C, D, E)\) and answer the following:
1452 \begin{enumerate}[topsep=0pt,itemsep=10em,partopsep=1ex,parsep=1ex]
1453 \item If \(\{A, B\}\) is the primary key, is \(B \to E, C \to D\) a 2NF?% List the nonprime attributes and justify.
1454 \item If \(\{A\}\) is the primary key, is \(B \to C, B \to D\) a 2NF?% List the nonprime attributes and justify.
1455 \end{enumerate}
1456
1457
1458 Solution +.#
1459
1460 \(1.\) No. \(C, D, E\), and \(E\) has a partial relation to \(B\)
1461
1462 \(2.\) Yes. Since the primary key is a singleton, it is obvious.
1463
1464
1465 Exercise +.#
1466
1467 Consider the relation \(R(A, B, C, D, E, F)\), and let \(\{B, D\}\) be the primary key, and have additionnaly the functional dependencies \(\{A, D\} \to E, C \to F\).
1468 This relation is not in 3NF, can you tell why?% List the nonprime attributes and justify.
1469 %\item If \(A\) is the primary key, is \(\{A, B\} \to C, \{A, B, C\} \to E\) a 3NF?% List the nonprime attributes and justify.
1470 \end{enumerate}
1471
1472
1473 Solution +.#
1474
1475 \(1.\) No. \(A, C, E, F\) are non-prime, \(\{B, D\} \to C \to F\) breaks the 3NF.
1476
1477 \(2.\) Yes. \(B, C, E, F\), \(\{A,B\}\) and \(\{A, B, C\}\) are not strict subset of the primary key.
1478
1479
1480 Exercise +.#
1481
1482 Consider the relation \(R(A, B, C, D)\) and answer the following:
1483 \begin{enumerate}
1484 \item If \(A\) is the only key, is \(A \to \{B,C,D\}, \{A, B\} \to C, \{B, C\} \to D\) a 3NF? List the nonprime attributes and justify.
1485 \item If \(B\) is the only key, is \(B \to \{A, C, D\}, A \to \{C, D\}, \{A, C\} \to D\) a 3NF? List the nonprime attributes and justify.
1486 \end{enumerate}
1487
1488
1489 Solution +.#
1490
1491 \(1.\) No. \(B\), \(C\) and \(D\) are non prime, \(A \to \{B,C\} \to D\) breaks the 3NF.
1492
1493 \(2.\) No. \(A\), \(B\) and \(D\) are non prime, \(B \to \{A,C\} \to D\) breaks the 3NF.
1494
1495
1496 \begin{problem}[tags={none}]
1497 Consider the following relation, and its functional dependencies:
1498
1499 \begin{tabular}{l}
1500 CAR\_SALE(\underline{Car\_no}, Date\_sold, \underline{Salesman\_no}, Commission, Discount\_amt)
1501 \end{tabular}
1502
1503 \begin{tabular}{rcl}
1504 \{Car\_no, Salesman\_no\} & \(\to\)& \{Date\_sold, Commission, Discount\_amt\}\\
1505 Date\_sold & \(\to\)& Discount\_amt\\
1506 Salesman\_no & \(\to\)& Commission
1507 \end{tabular}
1508
1509 Based on the given primary key, is this relation in 1NF, 2NF, or 3NF? Why or why not? Normalize it completely. Then draw an ER diagram for the resulting schema.
1510
1511
1512 Solution +.#
1513
1514 This relation satisfies 1NF but not 2NF (Car\_no \(\to\) Date\_sold and Salesman\_no \(\to\) Commission so these two attributes are not fully functional dependent on the primary key) and not 3NF.
1515
1516 To normalize,
1517
1518 2NF:
1519 \begin{tabular}{l}
1520 Car\_Sale1(Car\_no, Date\_sold, Discount\_amt)\\
1521 Car\_Sale2(Car\_no, Salesman\_no)\\
1522 Car\_Sale3(Salesman\_no,Commission)
1523 \end{tabular}
1524 3NF:
1525 \begin{tabular}{c}
1526 Car\_Sale1-1(Car\_no, Date\_sold)\\
1527 Car\_Sale1-2(Date\_sold, Discount\_amt)\\
1528 Car\_Sale2(Car\_no, Salesman\_no)\\
1529 Car\_Sale3(Salesman\_no,Commission)
1530 \end{tabular}
1531
1532
1533 Problem +.#
1534
1535 This problem asks you to convert business statements into dependencies.
1536 Consider the following relation:
1537 \begin{center}
1538 BIKE(Serial\_no, Manufacturer, Model, Batch, Wheel\_size, Retailer)
1539 \end{center}
1540
1541 Each tuple in the relation BIKE contains information about a bike with a serial number, made by a manufacturer, with a particular model number, released in a certain batch, which has a certain wheel size, and is sold by a certain retailer.
1542 \begin{itemize}
1543 \item Write each of the following dependencies as a functional dependency (I give you the first one as an example):
1544
1545 \begin{enumerate}
1546 \item A retailer can't have two bikes of the same model from different batches.
1547
1548 \emph{solution:} \{Retailer, Model\} \(\to\) Batch
1549 \item The manufacturer and serial number uniquely identifies the bike and where it is sold.
1550 \item A model number is registered by a manufacturer and therefore can't be
1551 used by another manufacturer.
1552 \item All bikes in a particular batch are of the same model.
1553 \item All bikes of a certain model have
1554 the same wheel size.
1555 \end{enumerate}
1556 \item Based on those statements, what could be a key for this relation?
1557 \item Assuming all those functional dependencies hold, and taking the primary key you identified at the previous step, what is the degree of normality of this relation? Justify your answer.
1558 \end{itemize}
1559
1560
1561 Solution +.#
1562
1563 \begin{itemize}
1564 \item
1565 \begin{enumerate}
1566 \item \{ Manufacturer, Serial\_no \} \(\to\) \{ Model, Batch, Wheel\_size, Retailer\}
1567 \item Model \(\to\) Manufacturer
1568 \item Batch \(\to\) Model
1569 \item \{Model, Manufacturer\} \(\to\) Wheel\_size
1570 \end{enumerate}
1571 \item \{Manufacturer, Serial\_no \}
1572 \item If every attribute is atomic, it is in second nf.
1573 \{ Manufacturer, Serial\_no \} \(\to\) Batch \(\to\)Model breaks the 3NF.
1574
1575 \end{itemize}
1576
1577
1578
1579 Problem +.#
1580
1581 Consider the relations \(R\) and \(T\) below, and their functional dependencies (on top of the one induced by the primary keys):
1582
1583 \begin{tabular}{l}
1584 R(\underline{EventId}, \underline{Email}, Time, Date, Location, Status)\\
1585 T(\underline{Invno}, Subtotal, Tax, Total, Email, Lname, Fname, Phone)
1586 \end{tabular}
1587
1588 \begin{tabular}{r c l}
1589 \{EventId, Email\} & \(\to\) & Status\\
1590 EventId & \(\to\) & \{Time, Date, Location\}\\
1591 Invno & \(\to\) & \{Subtotal, Tax, Total, Email\}\\
1592 Email & \(\to\) & \{Fname, Lname, Phone\}
1593 \end{tabular}
1594
1595 Normalize the relations to 2NF and 3NF. Show all relations at each stage (2NF and 3NF) of the normalization process.
1596
1597
1598 Solution +.#
1599
1600 TO BE WRITTEN
1601
1602
1603
1604 Problem +.#
1605
1606 Consider the following relation for published books:
1607 \begin{center}
1608 BOOK (Book\_title, Book\_type, Author\_name, List\_price, Author\_affil, Publisher)
1609 \end{center}
1610
1611 Suppose we have the following dependencies:
1612
1613 \begin{center}
1614 \begin{tabular}{r c l}
1615 Book\_title & \(\to\) & \{ Publisher, Book\_type \}\\
1616 Book\_type & \(\to\) & List\_price\\
1617 Author\_name & \(\to\) & Author\_affil
1618 \end{tabular}
1619 \end{center}
1620
1621 \begin{itemize}
1622 \item What would be a suitable key for this relation?
1623 \item How could this relation not be in first normal form? Explain your answer.
1624 \item This relation is not in second normal form: explain why and normalize it.
1625 \item Is the relations you obtained at the previous step in third normal form? Explain why, and normalize them if needed.
1626 \end{itemize}
1627
1628
1629
1630 Solution +.#
1631
1632 \begin{itemize}
1633 \item \{Book Title, Author Name\}
1634 \item If an attribute is composite or multi-valued.
1635 \item Because of \{ Book\_title \}\(\to\) \{ Publisher, Book\_type \}.
1636 We can normalize it as
1637 (Book Title, Publisher, Book Type, List Price), (Author Name, Author Affiliation), (Author Name, Book Title).
1638 \item Because of \{Book title\} \(\to\) \{ Book\_type\} \(\to\) \{ List\_price\}\\
1639 (Book Title, Publisher, Book Type) and (Book Type, List Price), (Author Name, Author Affiliation), (Author Name, Book Title).
1640
1641 \end{itemize}
1642
1643
1644 %%% 06
1645
1646
1647 Exercise +.#
1648
1649 Consider the relation \(R(A, B, C, D, E)\) and the functional dependencies \( \{A, B\} \to C, B \to D, C \to E\).
1650 Answer the following:
1651
1652 \begin{enumerate}
1653 \item \(A\) by itself is not a primary key, but what is the only key that contains \(A\)?
1654 \item List the non-prime attributes.
1655 \item This relation is not in 2NF: what transformation can you operate to obtain a 2NF?
1656 \item One of the relation you obtained at the previous step is likely not to be in 3NF.
1657 Can you normalize it?
1658 If yes, how?
1659 \end{enumerate}
1660
1661
1662 Solution +.#
1663
1664 \(\{A, B\}\),
1665
1666 \(C, D, E\),
1667
1668 \(R_1(A, B, C, E)\) and \(R_2(B, D)\)
1669
1670 \(R_1(A, B, C)\), \(R_2(C, E)\) and \(R_3(B, D)\)
1671
1672
1673
1674 Exercise +.#
1675
1676 What are the two different categories of U.M.L. diagram?
1677
1678
1679 Solution +.#
1680
1681 Behaviour and structure
1682
1683
1684 Exercise +.#
1685
1686 Can a \texttt{C++} developer working on Linux and a \texttt{Java} developer working on MacOS use the same class diagram as a basis to write their programs?
1687 Justify your answer.
1688
1689
1690 Solution +.#
1691
1692 Yes, U.M.L. diagram is language-independent and platform-independent.
1693
1694
1695 Exercise +.#
1696
1697 What kind of diagram should we use if we want to \(\hdots\)
1698 \begin{enumerate}
1699 \item describe the functional behavior of the system as seen by the user?
1700 \item capture the flow of messages in a software?
1701 \item represent the workflow of actions of an user?
1702 \end{enumerate}
1703
1704
1705 Solution +.#
1706
1707 \begin{enumerate}
1708 \item Use-case
1709 \item Sequence diagram
1710 \item Activity diagram
1711 \end{enumerate}
1712
1713
1714
1715 Exercise +.#
1716
1717 Name two reasons why one would want to use a U.M.L. class diagram over an E.-R. diagram to represent a conceptual schema.
1718
1719
1720 Solution +.#
1721
1722 To use direction for association, to have a common language with someone less knowledgeable of other diagrammatic notations.
1723 For the concept of integration.
1724
1725
1726
1727
1728 Exercise +.#
1729
1730 Consider the following diagram:
1731
1732 \begin{tikzpicture}[scale = 0.8]
1733 \begin{class}[text width=7cm]{Flight}{0,0}
1734 \attribute{flightNumber : Integer}
1735 \attribute{departureTime : Date}
1736 \attribute{flightDuration : Minutes}
1737 \attribute{departingAirport : String}
1738 \attribute{arrivingAirport : String}
1739 \operation{delayFlight(numberOfMinutes : Minutes)}
1740 \operation{getArrivalTime( ) : Date}
1741 \end{class}
1742 \begin{class}{Plane}{11, -0.8}
1743 \attribute{airPlaneType : String}
1744 \attribute{maximumSpeed : MPH}
1745 \attribute{maximumDistance : Miles}
1746 \attribute{tailID : String}
1747 \end{class}
1748 \association{Plane}{}{0..1}{Flight}{0..*}{}
1749 \end{tikzpicture}
1750
1751 Give the number of attributes for both classes, and suggest two operations for the class that doesn't have any.
1752 Discuss the multiplicities: why did the designer picked those values?
1753
1754
1755 Solution +.#
1756
1757 \(5\) and \(4\).
1758
1759 getLastFlightNumber() : Integer
1760
1761 stMaximumSpeed(MPH) : void
1762
1763 A flight could be assigned to no plane, and a plane could not be assigned to a flight.
1764
1765 A plane can be assigned to multiple (or no) flights, but a flight must have at most one plane (and could have none).
1766
1767
1768 Exercise +.#
1769
1770 Briefly explain the difference between an aggregation and a composition association.
1771
1772
1773 Solution +.#
1774
1775 Aggregation: associated class can have an existence of its own.
1776
1777 Composition association: class doesn't exist without the association.
1778
1779
1780 Exercise +.#
1781
1782 How is generalization (or inheritance) represented in a U.M.L. class diagram?
1783 Why is such a concept useful?
1784
1785
1786 Solution +.#
1787
1788 \tikz{\draw[>=open triangle 60,->] (0,0) to (1,0);}
1789 Because it avoids redundancy.
1790
1791
1792 Exercise +.#
1793
1794 Convert the following E.R. diagram into a U.M.L. class diagram:
1795
1796 \begin{tikzpicture}[node distance=8em]
1797 \node[entity] (person) {PILOT};
1798 \node[attribute] (pid) [left of=person] {\key{ID}} edge (person);
1799 \node[attribute] (name) [above left of=person] {Name} edge (person);
1800 \node[attribute] (phone) [above of=person] {Experience} edge (person);
1801
1802 \node[relationship] (drives) [below of=person] {ASSIGNED\_TO} edge node[right, pos=0.1] {$N$} (person);
1803 \node[entity] (plane) [below of=drives] {PLANE} edge [total] node[right, pos=0.7] {$1$} (drives);
1804 \node[attribute] (make) [left of=plane] {\key{TailID}} edge (plane);
1805 \node[attribute] (maxS) [below left of =plane] {MaxSpeed} edge (plane);
1806 \node[attribute] (airPlaneType) [below of = plane] {AirPlaneType} edge (plane);
1807 \end{tikzpicture}
1808
1809 Problem +.#
1810
1811 \label{probMYSQLW}
1812 Consider the following E.R. schema for the CAR\_INFO database:
1813
1814 \begin{tikzpicture}[node distance=8em]
1815 \node[entity] (person) {PERSON};
1816 \node[attribute] (pid) [left of=person] {\key{ID}} edge (person);
1817 \node[attribute] (name) [above left of=person] {Name} edge (person);
1818 \node[multi attribute] (phone) [above of=person] {Phone} edge (person);
1819 \node[attribute] (address) [above right of=person] {Address} edge (person);
1820 \node[attribute] (street) [above right of=address] {Street} edge (address);
1821 \node[attribute] (city) [right of=address] {City} edge (address);
1822 %\node[derived attribute] (age) [right of=person] {Age} edge (person);
1823
1824 \node[relationship] (drives) [below right of=person] {DRIVES} edge node[above, pos=0.1] {$1$} (person);
1825 \node[entity] (car) [below left of=drives] {CAR} edge node[above, pos=0.7] {$1$} (drives);
1826 \node[attribute] (make) [left of=car] {Make} edge (car);
1827 \node[attribute] (year) [below left of =car] {Year} edge (car);
1828 \node[attribute] (brand) [below of =car] {Brand} edge (car);
1829
1830 \node[relationship] (seats) [below left of=person] {SEATS\_IN} edge node[above, pos=0.1] {$N$} (person);
1831 \draw (seats) edge node[above, pos=0.3] {$1$} (car);
1832 \node[attribute] (position) [left of=seats] {Position} edge (seats);
1833
1834 \node[ident relationship] (insured) [right of=car] {INSURED} edge node[above, pos=0.3] {$1$} (car);
1835 \node[weak entity] (insurance) [right = 1cm of insured] {CAR\_INSURANCE} edge[total] node[above, pos=0.7] {$N$} (insured);
1836 \node[attribute] (amount) [above of =insurance] {Covered Amount} edge (insurance);
1837 \node[attribute] (policy) [above right of =insurance] {Policy Number} edge (insurance);
1838 \node[attribute] (company) [below right of =insurance] {Company Name} edge (insurance);
1839 \end{tikzpicture}
1840
1841 Note that a car can have at most one driver, \(N\) passengers, \(N\) insurances, and that car insurances exist only if they are "tied up" to a car (i.e., they are weak entities, and their identifying relationship is called "Insured").
1842
1843 \begin{enumerate}
1844 \item Find the key attribute for "Car", and the partial key for "Car Insurance". If you can't think of any, add a dummy attribute and make it be the key.
1845 \item Convert that E.-R. diagram to a relational database schema.
1846 \item Convert the E.-R. diagram to a U.M.L. class diagram. Comparing Figure 7.16 with Figure 7.2 from your textbook should guide you.
1847 \end{enumerate}
1848
1849
1850
1851 Solution +.#
1852
1853 For "Car", we need to create an attribute, like "vin".
1854 For "Car Insurance", "Policy Number" is perfect.
1855
1856 \colorlet{lightgray}{gray!20}
1857
1858 \begin{tikzpicture}[relation/.style={rectangle split, rectangle split parts=#1, rectangle split part align=base, draw, anchor=center, align=center, text height=3mm, text centered}]\hspace*{-0.3cm}
1859
1860 % RELATIONS
1861
1862 \node (phonetitle) {\textbf{PHONE}};
1863
1864 \node [relation=2, rectangle split horizontal, rectangle split part fill={lightgray!50}, anchor=north west, below=0.6cm of phonetitle.west, anchor=west] (phone)
1865 {\underline{id}%
1866 \nodepart{two} \underline{number}
1867 };
1868
1869 \node [below=1.3cm of phone.west, anchor=west] (persontitle) {\textbf{PERSON}};
1870
1871 \node [relation=6, rectangle split horizontal, rectangle split part fill={lightgray!50}, below=0.6cm of persontitle.west, anchor=west] (person)
1872 {\underline{id}%
1873 \nodepart{two} Name
1874 \nodepart{three} Street
1875 \nodepart{four} City
1876 \nodepart{five} Seat
1877 \nodepart{six}{Position}
1878 };
1879
1880 \node [below=1.1cm of person.west, anchor=west] (cartitle) {\textbf{CAR}};
1881
1882 \node [relation=5, rectangle split horizontal, rectangle split part fill={lightgray!50}, anchor=north west, below=0.6cm of cartitle.west, anchor=west] (car)
1883 {\underline{Vin}%
1884 \nodepart{two} Make
1885 \nodepart{three} Year
1886 \nodepart{four} Brand
1887 \nodepart{five} Driver
1888 };
1889
1890 \node [below=1.4cm of car.west, anchor=west] (carinsurancetitle) {\textbf{CAR INSURANCE}};
1891
1892 \node [relation=4, rectangle split horizontal, rectangle split part fill={lightgray!50}, anchor=north west, below=0.6cm of carinsurancetitle.west, anchor=west] (carinsurance)
1893 {\underline{Policy Number}%
1894 \nodepart{two} Covered Amount
1895 \nodepart{three} Company Name
1896 \nodepart{four} Insured Car
1897 };
1898
1899 % FOREIGN KEYS
1900
1901 \draw[-latex] (phone.one south) -- ++(0,-0.2) --++(7,0) --++(0, -2) -| ($(person.one south) + (0.2,0)$);
1902 %\draw[-latex] (person.five south) -- ++(0,-0.2) --++(3,0) --++(0, -2) -| ($(car.one south) + (-0.2,0)$);
1903 \draw[-latex] (carinsurance.four south) -- ++(0,-0.2) --++(2,0) --++(0, 1.75) -| ($(car.one south) + (0.1,0)$);
1904 \draw[-latex] (car.five south) -- ++(0,-0.2) --++(2,0) --++(0, 1.5) -| ($(person.one south) + (-0.1,0)$);
1905 \end{tikzpicture}
1906
1907
1908 \begin{tikzpicture}[scale = 0.8]
1909 \node (person) at (0,0){\begin{tabular}{| r l |}
1910 \hline
1911 \multicolumn{2}{| c |}{\textbf{Person}}\\
1912 \hline
1913 id &: String\\
1914 name &: String\\
1915 address &: Street\\
1916 & City\\
1917 \hline
1918 getAge()&: Int\\
1919 \hline
1920 \end{tabular}
1921 };
1922 \node (phone) at (8, 1){\begin{tabular}{| r l |}
1923 \hline
1924 \multicolumn{2}{| c |}{\textbf{Phone}}\\
1925 \hline
1926 number &: String\\
1927 \hline
1928 \end{tabular}
1929 };
1930 \draw[open diamond-] (person) -- node[above, pos=0.1]{\(0..\star\)} node[above, pos=0.9]{\(1..1\)} (phone);
1931 \node (car) at (8, -3){\begin{tabular}{| r l |}
1932 \hline
1933 \multicolumn{2}{| c |}{\textbf{Car}}\\
1934 \hline
1935 vin &: String\\
1936 make &: String\\
1937 year &: Year\\
1938 brand &: String\\
1939 \hline
1940 \end{tabular}
1941 };
1942 \draw[-] ($(person)+(2.5,-0.5)$) -- node[above, pos=0.1]{\(0..1\)} node[above, pos=0.5]{drives} node[above, pos=0.9]{\(0..1\)} (car);
1943 %\draw[-] (person) -- node[below, pos=0.1]{\(0..1\)} node[below, pos=0.9]{\(0..4\)} ($(car)+(-2.5, 0.5)$);
1944 \node (seat) at (3.5, -5){\begin{tabular}{| r l |}
1945 \hline
1946 \multicolumn{2}{| c |}{\textbf{Seats In}}\\
1947 \hline
1948 position &: String\\
1949 \hline
1950 \end{tabular}
1951 };
1952 \draw[dashed] (seat) -- (3.5, -1.6);
1953
1954 \node (insu) at (14.3, -2){\begin{tabular}{| r l |}
1955 \hline
1956 \multicolumn{2}{| c |}{\textbf{Car Insurance}}\\
1957 \hline
1958 policy number &: String\\
1959 covered amount &: int\\
1960 compagny name &: String\\
1961 \hline
1962 \end{tabular}
1963 };
1964
1965 \draw[diamond -] (car) -- node[above, pos=0.1]{\(0..\star\)} node[above, pos=0.9]{\(1\)} (insu);
1966 \end{tikzpicture}
1967
1968
1969
1970 Problem +.#
1971
1972 In this exercise, we will install and explore the basic functionalities of MySQL Workbench, which is a cross-platform, open-source, and free graphical interface for database design.
1973 \begin{enumerate}
1974 \item Install MySQL Workbench: use your package manager, or download the binaries from \url{https://dev.mysql.com/downloads/workbench/}.
1975 \item Once installed, execute the software. Under the panel "MySQL Connections", you should see your local installation listed as "Local instance 3306". Click on the top-right corner of that box, and then on "Edit Connections". Alternatively, click on "Database", on "Manage Connections", and then on "Local instance 3306".
1976 \item Check that all the parameters are correct. Normally, you only have to change the name of the user to "testuser", and leave the rest as it is. Click on "Test the connection", and enter your password (which should be "password") when prompted. If you receive a warning about "Incompatible/nonstandard server version or connection protocol detected", click on "Continue anyway".
1977 \item Now, click on the box "Local instance 3306", and enter your password. A new tab appears, you can see the list of schemas in the bottom part of the left panel.
1978 \item Click on "Database", and then on "Reverse Engineering" (or hit \texttt{ctrl} + \texttt{r}), click on "next", enter your password, and click on "next". You should see the list of the schemas stored in your database. Select one (any one, we are just exploring the functionalities at that point), click on "next", and then click on "execute", "next", and "close".
1979 \item You're back on the previous view, but you should now see "E.E.R. diagram" on the top of the middle panel. Click on "E.E.R. diagram" twice, scroll down if needed, and you should see the E.E.R. diagram.
1980 \item This diagram isn't exaclty an E.-R. diagram, and it's not a U.M.L. diagram either. Yet, you should still be able to understand parts of it, and should try to modify it. Make some relations mandatory, change their name, add an attribute, change the name of another, insert a couple of elements in an entity, add a row in a table, etc. Make sure you understand the meaning of the lines between the entities.
1981 \item Once you're done, try to "Forward Engineer" by hitting "Ctrl" + "G". Click on "next" twice, enter your password, click on lick on "next" once more, and you should see the \texttt{SQL} code needed to produce the table you just designed using the graphical tool.
1982 \end{enumerate}
1983
1984
1985 Problem +.#
1986
1987 \emph{This problem requires you to have successfully completed Problem~\ref{probMYSQLW} and Problem~\ref{probERtoREL}.}
1988
1989 Using the relational database schema you obtained in question~\ref{ERtoREL} of Problem~\ref{probERtoREL}, write the \texttt{SQL} implementation of that database.
1990 Then, using MySQL Workbench, use the "Reverse Engineering" function to obtain a E.E.R. diagram of your database, and compare it with the U.M.L. diagram you draw in question~\ref{ERtoUML} of Problem~\ref{probERtoREL}.
1991 Apart from the difference inherent to the nature of the diagram (i.e., U.M.L. Vs E.E.R.), how are they the same? How do they differ? Is the automated tool as efficient and accurate as you are?
1992
1993
1994 Solution +.#
1995
1996 \inputminted{sql}{include/sol.sql}
1997 \includegraphics{include/mysql_workbench_drawing-crop}
1998
1999
2000 %%% 07
2001
2002
2003 Exercise +.#
2004
2005 What are the technologies that makes it possible for a Java application to communicate with a DBMS?
2006
2007
2008 Solution +.#
2009
2010 API + driver
2011
2012
2013 Exercise +.#
2014
2015 What JDBC method do you call to get a connection to a database?
2016
2017 Solution +.#
2018
2019 DriverManager.getConnection()
2020
2021
2022 Exercise +.#
2023
2024 Briefly explain what the \texttt{next()} method from the \texttt{ResultSet} class does, and what is its return type.
2025
2026 Solution +.#
2027
2028 It checks if there is data to read, and if move the cursor reads it.
2029 It returns a Boolean.
2030
2031
2032 Exercise +.#
2033
2034 How do you submit a `SELECT` statement to the DBMS?
2035
2036
2037 Solution +.#
2038
2039 Using
2040
2041 ```java
2042 .executeQuery(strSelect)
2043 ```
2044
2045
2046 Exercise +.#
2047
2048 Where is a ResultSet object's cursor initially pointing? How do you move the cursor forward in the result set?
2049
2050
2051 Solution +.#
2052
2053 Before the first line.
2054
2055 ```java
2056 next()
2057 ```
2058
2059 method
2060
2061 Exercise +.#
2062
2063 Give three navigation methods provided by `ResultSet`.
2064
2065
2066 Solution +.#
2067
2068 `first()`
2069 `last()`
2070 `next()`
2071 `previous()`
2072 `relative(rows)`
2073 `absolute(row)` method.
2074
2075
2076 Exercise +.#
2077
2078 Explain this JDBC URL format:
2079
2080 ```java
2081 jdbc:mysql://localhost:3306/HW_NewDB?createDatabaseIfNotExist=true&useSSL=true
2082 ```
2083
2084
2085 Solution +.#
2086
2087 Connect to localhost:3306 and create a new database if needed, and use secure connection.
2088
2089
2090 Exercise +.#
2091
2092 In what class is the `getColumnName()} method?
2093
2094
2095 Solution +.#
2096
2097 ResultSetMetaData
2098
2099
2100 Exercise +.#
2101
2102 What is a prepared statement?
2103
2104
2105 Solution +.#
2106
2107 A prepared statement is a feature used to execute the same (or similar) SQL statements repeatedly with high efficiency.
2108
2109
2110 Exercise +.#
2111
2112 In the code below, there are \(5\) errors between line \(11\) and line \(30\). They are \emph{not} subtle Java errors (like misspelling a key word) and do not come from the DBMS (so you should assume that the password is correct, that the database exists, etc.). For each error, highlight it precisely and give a short explanation.
2113
2114 \begin{minted}[linenos, breaklines=true, baselinestretch=1.4]{Java}
2115 import java.sql.*;
2116
2117 public class MyProg{
2118 public static void main(String[] args) {
2119 try (
2120 Connection conn = DriverManager.getConnection("jdbc:mysql://localhost:3306/"
2121 +"HW_TestDB?user=testuser&password=password");
2122 Statement stmt = conn.createStatement();
2123 ) {
2124
2125 /* Errors after this point.*/
2126
2127 String strSelect = "SELECT title WHERE qty > 40 FROM disks;";
2128 ResultSet rset = stmt.executeUpdate(strSelect);
2129
2130 System.out.println("The records selected are: (listed last first):");
2131 rset.last();
2132
2133 while(rset.previous()) {
2134 String title = rset.getDouble("title");
2135 System.out.println(title + "\n");
2136 }
2137
2138 String sss = "SELECT title FROM disks WHERE Price <= ?";
2139 PreparedStatement ps = conn.prepareStatement(sss);
2140 ResultSet result = ps.executeQuery();
2141
2142 conn.close();
2143
2144 /* Errors before this point.*/
2145
2146 } catch(SQLException ex) {
2147 ex.printStackTrace();
2148 }
2149 }
2150 }
2151 ```
2152
2153
2154
2155
2156 %
2157 %METADATA
2158 %
2159 %ResultSet is array, other?
2160 %
2161 %Multiple queries
2162 %
2163 %qet datatype
2164 %
2165 %prepared statemernt.
2166 %
2167 %http://tutorials.jenkov.com/jdbc/resultset.html
2168
2169
2170
2171 This part will mainly hands on, and ask you to read code.
2172 It is important that you learn how to get a working environment for a new technology quickly, for your understanding of this lecture, for the exam, and as a CS major.
2173 I'll assume that you will have successfully completed those tasks by \hwDateNext, so don't wait and let me know if you had difficulties solving them.
2174
2175 **References:**
2176
2177 \begin{itemize}
2178 \item This second part takes some inspiration from \url{https://www.ntu.edu.sg/home/ehchua/programming/java/JDBC_Basic.html}. If you experience troubles, \url{https://www.ntu.edu.sg/home/ehchua/programming/howto/ErrorMessages.html#JDBCErrors} might be a good read.
2179 \item Section 13.3.2 of your textbook is a condensed, but good read. Many textbook on Java includes a part on Databases, cf. for instance the Chapter 16 of \emph{Starting Out with Java: Early Objects} (5th Edition) by Tony Gaddis.
2180 \end{itemize}
2181
2182
2183 Problem +.#
2184
2185 In the archive, navigate to \texttt{code/sql/}, open and read \texttt{HW\_ebookshop.sql}.
2186
2187 Then, open a terminal (or command-line interpreter), navigate to the folder where you stored that file (using `cd`), and type
2188
2189 \begin{minted}{bash}
2190 mysql -u testuser -p < HW_ebookshop.sql
2191 ```
2192
2193
2194
2195 for linux, or (something like)
2196
2197 \begin{minted}[breaklines=true]{bash}
2198 "C:\Program Files\MySQL\MySQL Server 5.7\bin\mysql.exe" -u testuser -p < HW_ebookshop.sql
2199 ```
2200
2201
2202 for Windows.
2203
2204 You just discovered MySQL's batch mode, that perform \emph{series} of instructions from a file.
2205 You can easily make sure that the database and the table were indeed created, and the values inserted.
2206
2207
2208 Problem +.#
2209
2210 This exercise supposes you successfully completed Problem~\ref{qu:batch}.
2211 We will compile and execute your first database application, using Java and MySQL.
2212 \begin{itemize}
2213 \item I will assume that you have MySQL installed and set-up as indicated in Homeworks \#1 and \#2.
2214 \item I will assume that you have Java installed. If not, please refer to \url{http://spots.augusta.edu/caubert/teaching/general/java/} for a simple program and the instructions to compile and execute it.
2215 \item We need to set up the \emph{driver} (or \emph{connector}) to make the java \texttt{sql} API and MySQL communicate. To do so,
2216 \begin{itemize}
2217 \item Go to \url{https://dev.mysql.com/downloads/connector/j/}
2218 \item Click on "Download" in front of "Platform Independent (Architecture Independent), ZIP Archive"
2219 \item Look for the (somewhat hidden) "No thanks, just start my download."
2220 \item You will download a file named "mysql-connector-java-***.zip", where \texttt{***} is the version number.
2221 \item Upon completion of the download, unzip the file, and locate the "mysql-connector-java-***-bin.jar" file.
2222 \item Copy that file in \texttt{code/java/}.
2223 \end{itemize}
2224 \item Open a terminal in that same folder, and compile \texttt{FirstProg.java}, using
2225 \begin{minted}{bash}
2226 javac FirstProg.java
2227 ```
2228
2229
2230 (or an equivalent command for windows).
2231 Normally, nothing will be printed, but a \texttt{FirstProg.class} file will be created.
2232 \item Now, execute that program, using
2233 \begin{minted}{bash}
2234 java -cp .:mysql-connector-java-***-bin.jar FirstProg
2235 ```
2236
2237
2238 in Linux, or
2239 \begin{minted}{bash}
2240 java -cp .;mysql-connector-java-***-bin.jar FirstProg
2241 ```
2242
2243
2244 in Windows.
2245 The \texttt{-cp} option lists the places where java should look for the class used in the program: we are explicitely asking java to use the \texttt{mysql-connector-java-***-bin.jar} executable to execute our \texttt{FirstProg} executable.
2246 Try to execute \texttt{FirstProg} without that flag, and see what happens.
2247 \end{itemize}
2248
2249
2250 Solution +.#
2251
2252 \begin{minted}[breaklines=true]{bash}
2253 $ java FirstProg
2254 java.sql.SQLException: No suitable driver found for jdbc:mysql://localhost:3306/HW_ebookshop
2255 at java.sql.DriverManager.getConnection(DriverManager.java:689)
2256 at java.sql.DriverManager.getConnection(DriverManager.java:247)
2257 at FirstProg.main(FirstProg.java:9)
2258 ```
2259
2260
2261
2262
2263 Problem +.#
2264
2265 There are three other programs in that archive:
2266 \begin{description}
2267 \item[Variation.java] is the program we looked at during class.
2268 \item[NullProg.java] is an attempt to answer the question whenever SQL's `NULL` is JAVA's \texttt{NULL} when it comes to strings.
2269 \item[AdvancedProg.java] is a long and commented program that introduces multiple new tools and ideas.
2270 \end{description}
2271
2272 Read, execute, break, edit, compile, patch, hack and (most importantly) understand those three programs, with, of course, a focus on the last one.
2273
2274
2275
2276 %%% 08 -- Exam 1
2277
2278
2279 Problem +.#
2280 %30
2281 A cinema company wants you to design a relational model for the following set-up:
2282 \begin{itemize}
2283 \item The company has movie stars. Each star has a name, gender, birthday, and unique id.
2284 \item The company has information about movies: title, year, length, genre, and studio. Each movie has a unique id, and features multiple stars.
2285 \item The company owns movie theaters as well. Each theater has a name, address, and a unique id.
2286 \item Furthermore, each theater has a set of auditoriums. Each auditorium has a unique number, and seating capacity.
2287 \item Each theater can schedule movies at show-times.
2288 Each show-time has a unique id, a start time, and is for a specific movie, at a theater auditorium.
2289 \item The company sells tickets for scheduled show-times. Each ticket has a unique ticket id, and a price.
2290 \end{itemize}
2291 %Your relations should be written as follow: relation names in uppercase, coma separated attributes enclosed in parenthesis, primary key underlined, and foreign key(s) detailed on the side (or drawn). That is, for instance,
2292 %"PROF(\underline{Login", Name, Department) ; where Department references the Code attribute in the DEPARTMENT table.}
2293 Complete the following relational model.
2294 If you think there are missing relations, add them, and when the parenthesis isn't closed, close it or add attribute(s).
2295 Also, specify the primary as well as foreign keys.
2296
2297 {
2298 \renewcommand{\arraystretch}{3}%
2299 \large
2300 \hspace{3em}\begin{tabular}{l}
2301 STAR(Name, Gender, Birthday, Id\\
2302 MOVIE(Title\\
2303 FEATURE\_IN(StarId, MovieId)\\
2304 THEATER(Name\\
2305 AUDITORIUM(Id, Capacity, Theater\\
2306 SHOWTIME(
2307 \end{tabular}
2308 }
2309
2310
2311
2312 \begin{problem}[tags={ex}, points ={10}]
2313 Your professor designed the following relational model at some point in his career, to help him organizing his exams and the students grades:
2314
2315 \begin{center}
2316 \begin{tabular}{l | l}
2317 Table Name and Attributes & Example of Value\\
2318 \hline
2319 EXAM(\underline{Number}, Date, Course) & \(\langle 1, \text{'2018-02-14'}, \text{'CSCI3410'}\rangle\)\\
2320 PROBLEM(\underline{Statement}, Points, Length, Exam) & \(\langle \text{'Your professor designed…'}, 10, \text{'00:10:00'}, 1\rangle\) \\
2321 STUDENT\_GRADE(\underline{Login}, \underline{Exam}, Grade) & \(\langle \text{'aalyx'}, 1, 83\rangle\)
2322 \end{tabular}
2323 \end{center}
2324
2325 The idea was to have
2326 \begin{itemize}
2327 \item The EXAM table storing information about exams,
2328 \item One entry per problem in the PROBLEM table, and to associate every problem to an exam (i.e., to have the Exam attribute referencing the Number attribute in the EXAM table),
2329 \item The grade of one student for one particular exam stored in the STUDENT\_GRADE table, where Exam is referencing the Number attribute in the EXAM table.
2330 \end{itemize}
2331 Unfortunately, this design turned out to be terrible.
2332 Describe at least one common and interesting situation where this model would fail to fulfill its purpose, and propose a way to correct the particular problem you identified.
2333
2334
2335
2336
2337
2338 Problem +.#
2339 Look at the SQL code below, and then answer the following questions.
2340
2341 \begin{minted}[linenos]{mysql}
2342 CREATE TABLE TRAIN(
2343 Id VARCHAR(30),
2344 Model VARCHAR(30),
2345 ConstructionYear YEAR(4)
2346 );
2347
2348 CREATE TABLE CONDUCTOR(
2349 Id VARCHAR(20),
2350 Name VARCHAR(20),
2351 ExperienceLevel VARCHAR(20)
2352 );
2353
2354 CREATE TABLE ASSIGNED_TO(
2355 TrainId VARCHAR(20),
2356 ConductorId VARCHAR(20),
2357 Day DATE,
2358 PRIMARY KEY(TrainId, ConductorId)
2359 );
2360 ```
2361
2362
2363
2364 \begin{enumerate}[topsep=0pt,itemsep=10em,partopsep=1ex,parsep=1ex]
2365 \item Modify the `CREATE` statement that creates the `TRAIN` table (l. 1--5), so that `Id` would be declared as the primary key.
2366 You can write only the line(s) that need to change.
2367 \item Write an `ALTER` statement that makes `Id` become the primary key of the `CONDUCTOR` table.
2368 \item Modify the `CREATE` statement that creates the `ASSIGNED_TO` table (l. 13--18), so that it has two foreign keys: `ConductorId` would be referencing the `Id` attribute in `CONDUCTOR` and `TrainId` would be referencing the `Id` attribute in `TRAIN`.
2369 You can write only the line(s) that need to change.
2370
2371 \item Write `INSERT` statements that insert one tuple of your invention in each relation.
2372 Your statements should respect all the constraints (including the ones we added at the previous questions) and result in actual insertions. (Remember that four digits is a valid value for an attribute with the `YEAR(4)` datatype.)
2373 \item Write a statement that sets the value of the `ExperienceLevel` attribute to "Senior" in all the tuples where the `Id` attribute is "GP1029" in the `CONDUCTOR` relation.\item For each of the following questions, write a `SELECT` statement that would answer it:
2374 \begin{enumerate}[topsep=0pt,itemsep=6em,partopsep=1ex,parsep=1ex]
2375 \item "What are the identification numbers of the trains?"
2376 \item "What are the names of the conductors with a \enquote{Senior" experience level?}
2377 \item "What are the construction years of the \enquote{Surfliner" and "Regina" models that we have?}
2378 \item "What is the id of the conductor that was responsible of the train referenced \enquote{K-13" on 2015/12/14?}
2379 \item "What are the models that were ever conducted by the conductor whose id is \enquote{GP1029"?}
2380 %\item "What is the name of the conductor that was responsible of the \enquote{Surfliner" model on October, 25th, 2015?}
2381 \end{enumerate}
2382 \end{enumerate}
2383
2384
2385
2386
2387 \begin{problem}[tags={ex}, bonus-points = {4}, points={40}]
2388 Suppose we have the relational model depicted in \autoref{fig:coffee}, p.~\pageref{fig:coffee}, with the indicated data in it.
2389
2390 In the following, we will assume that this model was implemented in a DBMS (MySQL or MariaDB), the primary keys being the underlined attributes, and the foreign keys as follow:
2391
2392 \begin{tabular}{r c l}
2393 \textbf{FavCoffee} in the \textbf{CUSTOMER} relation & refers to & \textbf{Ref} in the \textbf{COFFEE} relation\\
2394 \textbf{Provider} in the \textbf{SUPPLY} & refers to & \textbf{Name} in the \textbf{PROVIDER} relation\\
2395 \textbf{Coffee} in the \textbf{SUPPLY} & refers to & \textbf{Ref} in the \textbf{COFFEE} relation
2396 \end{tabular}
2397
2398 You will be asked to read and write SQL commands.
2399 You should assume that
2400 \begin{enumerate}
2401 \item Datatype doesn't matter: we use only strings and appropriate numerical datatypes.
2402 \item Every statement respects SQL's syntax (there's no "a semi-colon is missing" trap).
2403 \item None of those commands are actually executed: the data is always in the state of \autoref{fig:coffee}, you are asked to answer "what if" questions.
2404 \end{enumerate}
2405
2406 In questions (\ref{ques:upd}) and (\ref{ques:del}), please, use \textbf{COFFEE}.\(1\) to denote the first tuple in \textbf{COFFEE}, and similarly for other relations and tuples (so that, for instance \textbf{SUPPLY}.\(4\) corresponds to "Johns \& Co., \(221\)").
2407
2408 \begin{enumerate}
2409 \item Determine if the following insertion statements would violate the
2410 the \textbf{E}ntity integrity constraint, % ("primary key cannot be \texttt{NULL" and should be unique})
2411 the \textbf{R}eferential integrity constraint% ("the foreign key must refer to something that exists"),
2412 , if there would be some \textbf{O}ther kind of error% (ignoring the plausability / revelance of inserting that tuple)
2413 , or if it would result in \textbf{S}uccessful insertion.
2414
2415 { \renewcommand{\arraystretch}{2}
2416 \begin{tabular}{l | c | c | c | c |}
2417 & \textbf{E} & \textbf{R} & \textbf{O} & \textbf{S}\\
2418 \hline
2419 `INSERT INTO CUSTOMER VALUES(005, 'Bob Hill', NULL, 001);` & & & & \\ \hline
2420 `INSERT INTO COFFEE VALUES(002, "Peru", "Decaf", 3.00);` & & & & \\ \hline
2421 `INSERT INTO PROVIDER VALUES(NULL, "contact@localcof.com");` & & & & \\ \hline
2422 `INSERT INTO SUPPLY VALUES("Johns & Co.", 121);` & & & & \\ \hline
2423 `INSERT INTO SUPPLY VALUES("Coffee Unl.", 311, 221);` & & & & \\ \hline
2424 \end{tabular}
2425 }
2426
2427 \item Assuming that we cascade update statements, list the tuples modified by the following statements:
2428 \begin{enumerate}[topsep=0pt,itemsep=4em,partopsep=1ex,parsep=1ex]
2429 \item `UPDATE CUSTOMER SET FavCoffee = 001 WHERE CardNo = 001;`
2430 \item `UPDATE COFFEE SET TypeOfRoast = 'Decaf' WHERE Origin = 'Brazil';`
2431 \item `UPDATE PROVIDER SET Name = 'Coffee Unlimited' WHERE Name = 'Coffee Unl.';`
2432 \item `UPDATE COFFEE SET PricePerPound = 10.00 WHERE PricePerPound > 10.00;`
2433 \end{enumerate}
2434
2435 \item Assuming that we cascade delete statements, list the tuples deleted by the following statements:
2436 \begin{enumerate}[topsep=0pt,itemsep=4em,partopsep=1ex,parsep=1ex]
2437 \item `DELETE FROM CUSTOMER WHERE Name LIKE '%S%';`
2438 \item `DELETE FROM COFFEE WHERE Ref = 001;`
2439 \item `DELETE FROM SUPPLY WHERE Provider = 'Coffee Unl.' AND Coffee = 001;`
2440 \item `DELETE FROM PROVIDER WHERE Name = 'Johns & Co.';`
2441 \end{enumerate}
2442
2443 \item %Starting here, assume that there is more data in our table than what was given at the beginning of the problem.
2444 Write queries that answer the following questions (the last two are bonus):
2445 \begin{enumerate}[topsep=0pt,itemsep=6em,partopsep=1ex,parsep=1ex]
2446 \item "What are the origins of your dark coffees?"
2447 \item "What is the reference of Bob's favorite coffee?" (nota: it doesn't matter if you return the favorite coffee of all the Bobs in the database.)
2448 \item "What are the names of the providers who didn't give their email?"
2449 \item "How many coffees does Johns \& co. provide us with?"
2450 \item "What are the names of the providers of my dark coffees?"
2451 \end{enumerate}
2452
2453 \end{enumerate}
2454
2455
2456
2457
2458
2459 \begin{figure}
2460
2461 \begin{multicols}{2}
2462 \textbf{COFFEE}
2463
2464 \begin{tabular}{| c | c | c | c | }
2465 \hline
2466 \rowcolor{gray!50} \underline{\textbf{Ref}} & \textbf{Origin} & \textbf{TypeOfRoast} & \textbf{PricePerPound}\\ \hline
2467 \(001\) & Brazil & Light & 8.90\\
2468 \(121\) & Bolivia & Dark& 7.50\\
2469 \(311\) & Brazil & Medium & 9.00\\
2470 \(221\) & Sumatra & Dark & 10.25\\ \hline
2471 \end{tabular}
2472 \\[2em]
2473 \textbf{CUSTOMER}
2474
2475 \begin{tabular}{| c | c | c | c | }
2476 \hline
2477 \rowcolor{gray!50} \underline{\textbf{CardNo}} & \textbf{Name} & \textbf{Email} & \textbf{FavCoffee}\\ \hline
2478 \(001\) & Bob Hill & b.hill@isp.net & 221\\
2479 \(002\) & Ana Swamp & swampa@nca.edu & 311\\
2480 \(003\) & Mary Sea & brig@gsu.gov & 121\\
2481 \(004\) & Pat Mount & pmount@fai.fr & 121\\ \hline
2482 \end{tabular}
2483 %\\[2em]
2484 %\noindent
2485
2486 \hspace{5em}
2487 \textbf{SUPPLY}
2488
2489 \hspace{5em}
2490 \begin{tabular}{| c | c | }
2491 \hline
2492 \rowcolor{gray!50} \underline{\textbf{Provider}} & \underline{\textbf{Coffee}}\\ \hline
2493 Coffee Unl. & \(001\) \\
2494 Coffee Unl. & \(121\) \\
2495 Coffee Exp. & \(311\) \\
2496 Johns \& Co. & \(221\) \\ \hline
2497 \end{tabular}
2498 \\[2em]
2499
2500 \hspace{5em}
2501 \textbf{PROVIDER}
2502
2503 \hspace{5em}
2504 \begin{tabular}{| c | c | }
2505 \hline
2506 \rowcolor{gray!50} \underline{\textbf{Name}} & \textbf{Email}\\ \hline
2507 Coffee Unl. & bob@cofunl.com \\
2508 Coffee Exp. & pat@coffeex.dk \\
2509 Johns \& Co. & NULL \\ \hline
2510 \end{tabular}
2511 \end{multicols}
2512 \caption{Relational Model and Data for the DB\_COFFEE Database}
2513 \label{fig:coffee}
2514 \end{figure}
2515
2516 %% 09 -- Exam 2
2517
2518
2519 Problem +.#
2520 Consider the relation
2521
2522 \begin{center}
2523 CONTACT(Phone, Call\_center, Email, Zip, Brand, Website)
2524 \end{center}
2525
2526 and the following functional dependencies:
2527
2528 \begin{center}
2529 \begin{tabular}{r c l}
2530 \{Zip, Brand\} & \(\to\) & \{Phone\}\\
2531 \{Brand\} & \(\to\) & \{Email\}\\
2532 \{ Brand\} & \(\to\) & \{Website\}\\
2533 \{Phone\} & \(\to\) & \{Call\_center\}
2534 \end{tabular}
2535 \end{center}
2536
2537 Assume that \{Zip, Brand\} is the primary key.
2538 Normalize this relation to the second normal form, and \emph{then} to the third normal form.
2539 Give the relations, their primary keys, and functional dependencies for both steps.
2540
2541
2542
2543
2544 Problem +.#
2545
2546 Consider the relation
2547
2548 \begin{center}
2549 CONSULTATION(Doctor\_no, Patient\_no, Date, Diagnosis, Treatment, Charge, Insurance)
2550 \end{center}
2551
2552 with the following functional dependencies:
2553
2554 \begin{center}
2555 \begin{tabular}{r c l}
2556 \{Doctor\_no, Patient\_no, Date\} & \(\to\) & \{Diagnosis\}\\
2557 \{Doctor\_no, Patient\_no, Date\} & \(\to\) & \{Treatment\}\\
2558 \{Treatment, Insurance\} & \(\to\) & \{Charge\}\\
2559 \{Patient\_no\} & \(\to\) & \{Insurance\}
2560 \end{tabular}
2561 \end{center}
2562
2563 \begin{enumerate}[topsep=0pt,itemsep=10em,partopsep=1ex,parsep=1ex]
2564 \item The designer decided not to add the functional dependency \{Diagnosis\} $\to$ \{Treatment\}.
2565 Explain what could be the designer's justification, at the level of the mini-world.
2566 \item Identify a primary key for this relation.
2567 \item What is the degree of normalization of this relation?
2568 Normalize it to the third normal form if necessary.
2569 \end{enumerate}
2570
2571
2572
2573 Solution +.#
2574
2575 Answer:
2576 Because there are no partial dependencies, the given relation is in 2NF already. This however is not 3NF because the Charge is a nonkey attribute that is determined by another nonkey attribute, Treatment. We must decompose further:
2577
2578 R (Doctor\_no, Patient\_no, Date, Diagnosis, Treatment)
2579
2580 R1 (Treatment, Charge)
2581
2582 We could further infer that the treatment for a given diagnosis is functionally dependant, but we should be sure to allow the doctor to have some flexibility when prescribing cures.
2583
2584
2585 Problem +.#
2586 A friend of yours want you to review and improve the code for a role-playing game.
2587
2588 The original idea was that each character should have a name, a class (e.g., Bard, Assassin, Druid), a certain amount of experience, a level, one or more weapons (providing bonuses) and can complete quests.
2589 A quest have a name, and rewards the characters that completed it with a certain amount of experience, and sometimes (but rarely) with a special item.
2590
2591 Your friend came up with the code presented in \autoref{lst:RPG}, page~\pageref{lst:RPG}, but there are several problems:
2592
2593 \begin{itemize}
2594 \item As of now, a character can have only one weapon.
2595 All the attempts to "hack" the `CHARACTER` table to add an arbitrary number of weapons ended up creating horrible messes.
2596 \item Every time a character completes a quest, a copy of the quest must be created.
2597 Your friend is not so sure why, but nothing else works.
2598 Also it seems that a character can complete only one quest, but your friend is not so sure about that.
2599 \item It would be nice to be able to store features that are tied to the class, and not to the character, like the bonus they provide and the associated element (e.g., all bards use fire, all assassins use wind, etc.).
2600 But you friend simply can't figure out how to do that.
2601 \end{itemize}
2602
2603 Can you provide a \emph{relational database schema} (no need to write the `SQL` code, but remember to indicate the primary and foreign keys) that would solve all of your friend's troubles?
2604
2605
2606
2607
2608
2609
2610 Problem +.#
2611 This exercise asks you to convert business statements into dependencies.
2612 Consider the following relation:
2613 \begin{center}
2614 KEYBOARD(Manufacturer, Model, Layout, Retail\_Store, Price)
2615 \end{center}
2616
2617 A tuple in the KEYBOARD relation contains information about a computer keyboard: its manufacturer, its model, its layout (AZERTY, QWERTY, etc.), the place where it is sold, and its price.
2618
2619 \begin{enumerate}
2620 \item Write each of the following business statement as a functional dependency:
2621
2622 \begin{enumerate}[topsep=0pt,itemsep=7em,partopsep=1ex,parsep=1ex]
2623 \item A model has a fixed layout.
2624 \item A retail store can't have two different models produced by the same manufacturer.
2625 \end{enumerate}
2626
2627 \item Based on those statements, what could be a key for this relation?
2628 \item Assuming all those functional dependencies hold, and taking the primary key you identified at the previous step, what is the degree of normality of this relation? Justify your answer.
2629 \end{enumerate}
2630
2631
2632
2633
2634 Problem +.#
2635 Consider the E.R. schema of Figure~\ref{fig:erCOUNTRY}, p.~\pageref{fig:erCOUNTRY}.
2636 Map that E.-R. diagram to a relational database schema.
2637
2638
2639
2640 \begin{figure}
2641 %\begin{listing}
2642 \begin{minted}[linenos, breaklines=true]{mysql}
2643 CREATE TABLE CHARACTER(
2644 Name VARCHAR(30) PRIMARY KEY,
2645 Class VARCHAR(30),
2646 XP INT,
2647 LVL INT,
2648 Weapon_Name VARCHAR(30),
2649 Weapon_Bonus INT,
2650 Quest_Completed VARCHAR(30)
2651 );
2652
2653 CREATE TABLE QUEST(
2654 Id VARCHAR(20) PRIMARY KEY,
2655 Completed_By VARCHAR(30),
2656 XP_Gained INT,
2657 Special_Item VARCHAR(20),
2658 FOREIGN KEY (Completed_By) REFERENCES CHARACTER(Name)
2659 );
2660
2661 ALTER TABLE CHARACTER ADD FOREIGN KEY (Quest_Completed) REFERENCES QUEST(Id);
2662 ```
2663
2664
2665 %\end{listing}
2666 \caption{A `SQL` Code for a Role-Playing Game}
2667 \label{lst:RPG}
2668 \end{figure}
2669
2670 \begin{figure}
2671 \begin{tikzpicture}[node distance=8em]
2672 \node[entity] (COUNTRY) {COUNTRY};
2673 \node[attribute] (name) [left of=COUNTRY] {\key{Name}} edge (COUNTRY);
2674 \node[attribute] (population) [below left of=COUNTRY] {Population} edge (COUNTRY);
2675
2676 \node[relationship] (SPEAKS) [below right of=COUNTRY] {SPEAKS} edge node[right, pos=0.4] {$M$} (COUNTRY);
2677 \node[entity] (LANGUAGE) [below right of=SPEAKS] {LANGUAGE} edge node[right, pos=0.5] {$N$} (SPEAKS);
2678 \node[attribute] (code) [left of = LANGUAGE] {\key{Code}} edge (LANGUAGE);
2679 \node[attribute] (symbol) [right of = LANGUAGE] {Name} edge (LANGUAGE);
2680
2681 \node[relationship] (BWF) [below of = LANGUAGE] {B\_W\_F};
2682 \draw (LANGUAGE) to node[left, pos=0.6] {$N$} (BWF.west);
2683 \draw (LANGUAGE) to node[right, pos=0.6] {$M$} (BWF.east);
2684
2685 %\node[relationship] (boundary) [above of = COUNTRY] {S\_B\_W};
2686 %\draw (COUNTRY) to node[left, pos=0.6] {$N$} (boundary.west);
2687 %\draw (COUNTRY) to node[right, pos=0.6] {$M$} (boundary.east);
2688
2689 %\node[relationship] (conversion) [below of =LANGUAGE] {CONVERSION};
2690 %\draw (LANGUAGE) to node[left, pos=0.6] {$N$} (conversion.west);
2691 %\draw (LANGUAGE) to node[right, pos=0.6] {$M$} (conversion.east);
2692
2693 %\node[attribute] (timestamp) [below left of=conversion] {\key{Timestamp}} edge (conversion);
2694 %\node[attribute] (rate) [below right of=conversion] {Exchange\_rate} edge (conversion);
2695
2696 \node[ident relationship] (hasfor) [right of=COUNTRY] {SINGS} edge node[above, pos=0.4] {$1$} (COUNTRY);
2697 \node[weak entity] (anthem) %[right of=hasfor]
2698 [right = 1cm of hasfor] {NATIONAL\_ANTHEM} edge[total] node[above, pos=0.6]{\(M\)} (hasfor);
2699 \node[multi attribute] (color) [right =1cm of anthem] {Creator} edge (anthem);
2700 \node[attribute] (name) [below right of = anthem] {\pkey{Name}} edge (anthem);
2701
2702 \node[relationship] (WIN) [above right of =LANGUAGE] {W\_IN};
2703 \draw (LANGUAGE) to node[left, pos=0.6] {$N$} (WIN);
2704 \draw (anthem) to node[right, pos=0.6] {$M$} (WIN);
2705 \end{tikzpicture}
2706
2707 %"S\_B\_W" stands for "SHARES\_A\_BORDER\_WITH".
2708 "W\_IN" stands for "WRITTEN\_IN", and
2709 "B\_W\_F" stands for "BORROWS\_WORDS\_FROM".
2710 For this relationship, on the left-hand side is the language that borrows a word, and on the right-hand side is the language that provides the loanword.
2711 \caption{E.R. Schema for the COUNTRY\_INFO database}
2712 \label{fig:erCOUNTRY}
2713 \end{figure}
2714
2715 %%% 10 Final
2716
2717
2718 Problem +.#
2719 Look at the relational model from Figure~\ref{fig:RelMod}, p.~\pageref{fig:RelMod}, and "reverse-engineer" it to obtain an E.-R. diagram.
2720
2721
2722
2723
2724 Problem +.#
2725 Answer the following in one or two sentences:
2726
2727 \begin{enumerate}[topsep=0pt,itemsep=12em,partopsep=1ex,parsep=1ex]
2728 \item What is deletion anomaly? Is it a desirable feature?
2729 \item What is polyglot persistence? Is it useful?
2730 \item What does it mean to be "schemaless"? What does it imply?
2731 \item What is denormalization?
2732 When could that be useful?
2733 \item What is the (object-relational) impedance mismatch? Is it an issue that can't be overcome?
2734 \end{enumerate}
2735
2736
2737 Solution +.#
2738
2739 A Delete Anomaly exists when certain attributes are lost because of the deletion of other attributes.
2740
2741 When storing data, it is best to use multiple data storage technologies, chosen based upon the way data is being used by individual applications or components of a single application.
2742
2743
2744
2745
2746
2747 Problem +.#
2748 Consider the following relation:
2749 \begin{center}
2750 FLIGHT(From, \qquad To, \qquad Airline, \qquad Flight\#, \qquad DateHour, \qquad HeadQuarter, \qquad Pilot, \qquad TZDifference)
2751 \end{center}
2752
2753 A tuple in the FLIGHT relation contains information about an airplane flight: the airports of departure and arrival, the airline carrier, the number of the flight, its time of departure, the headquarter of the company chartering the flight, the name of the pilot(s), and the time zone difference between the departure and arrival airports.
2754
2755 The "Pilot" attribute is multi-valued (so that between \(1\) and \(4\) pilot's names can be stored in it).
2756 Given an airline and a flight number, one can determine the departure and arrival airports, as well as the date and hour and the pilot(s).
2757 Given the airline carrier, one can determine the headquarter.
2758 Finally, given the departure and arrival airports, one can determine their time zone difference.
2759
2760 Normalize the "FLIGHT" relation to its third normal form.
2761 You can indicate your steps, justify your reasoning, and indicate the foreign keys if you want to, but don't have to.
2762
2763
2764
2765
2766 Problem +.#
2767 Consider the set of requirements in Figure~\ref{fig:reqUNIV}, p.~\pageref{fig:reqUNIV}.
2768 Draw an E.-R. diagram for that schema.
2769 Specify key attributes of each entity type and structural constraints on each relationship type.
2770 Note any unspecified requirements, and make appropriate assumptions to make the specification complete.
2771
2772
2773
2774 Problem +.#
2775 Consider the UML diagram in Figure~\ref{fig:UML}, p.~\pageref{fig:UML}, and convert it to the relational model.
2776 Don't forget to indicate primary and foreign keys.
2777
2778
2779
2780 \begin{figure}
2781 \colorlet{lightgray}{gray!20}
2782
2783 \begin{tikzpicture}[relation/.style={rectangle split, rectangle split parts=#1, rectangle split part align=base, draw, anchor=center, align=center, text height=3mm, text centered}]\hspace*{-0.3cm}
2784
2785 % RELATIONS
2786
2787 \node (actortitle) {\textbf{ACTOR}};
2788
2789 \node [relation=3, rectangle split horizontal, rectangle split part fill={lightgray!50}, anchor=north west, below=0.6cm of actortitle.west, anchor=west] (actor)
2790 {\underline{Id}%
2791 \nodepart{two} Name
2792 \nodepart{three} Birthdate
2793 };
2794
2795 \node [below=1.3cm of actor.west, anchor=west] (movietitle) {\textbf{MOVIE}};
2796
2797 \node [relation=4, rectangle split horizontal, rectangle split part fill={lightgray!50}, below=0.6cm of movietitle.west, anchor=west] (movie)
2798 {\underline{Title}%
2799 \nodepart{two} Year
2800 \nodepart{three} Length
2801 \nodepart{four} Studio
2802 };
2803
2804
2805 \node [right=8cm of actortitle.west, anchor=west] (actingtitle) {\textbf{ACTING}};
2806
2807 \node [relation=2, rectangle split horizontal, rectangle split part fill={lightgray!50}, below=0.6cm of actingtitle.west, anchor=west] (acting)
2808 {\underline{ActorID}%
2809 \nodepart{two} \underline{MovieTitle}%
2810 };
2811
2812 \node [below=1.3cm of movie.west, anchor=west] (theatretitle) {\textbf{THEATRE}};
2813
2814 \node [relation=5, rectangle split horizontal, rectangle split part fill={lightgray!50}, anchor=north west, below=0.6cm of theatretitle.west, anchor=west] (theatre)
2815 {\underline{Name}%
2816 \nodepart{two} Street
2817 \nodepart{three} City
2818 \nodepart{four} State
2819 \nodepart{five} Zip
2820 };
2821
2822 \node [below=1.3cm of acting.west, anchor=west] (showingtitle) {\textbf{SHOWING}};
2823
2824 \node [relation=4, rectangle split horizontal, rectangle split part fill={lightgray!50}, anchor=north west, below=0.6cm of showingtitle.west, anchor=west] (showing)
2825 {\underline{TheaterName}%
2826 \nodepart{two} \underline{MovieTitle}
2827 \nodepart{three} \underline{Day}
2828 \nodepart{four} \underline{Time}
2829 };
2830
2831
2832 % FOREIGN KEYS
2833
2834 %\draw[-latex] (acting) -- -| (actor.one south);%(actor.one south) -- ++(0,-0.2) --++(7,0) --++(0, -2) -| ($(movie.one south) + (0.2,0)$);
2835 \draw[-latex] ($(showing.two south) + (-0.3,0)$)-- ++(0,-0.3) -| (movie.one south);
2836 \draw[-latex] (showing.one south) -- ++ (0,-2.6) -| (theatre.one south);
2837 \draw[-latex] (acting.one south) -- ++ (0,-0.3) -| (actor.one south);
2838 \draw[-latex] (acting.two south) -- ++ (0,-0.5) -- ++(-3, 0) -- ++(0, -2.1)-| ($(movie.one south) + (-0.25,0)$);
2839 %\draw[-latex] (showing.four south) -- ++(0,-0.2) --++(2,0) --++(0, 1.75) -| ($(theatre.one south) + (0.1,0)$);
2840 %\draw[-latex] (theatre.five south) -- ++(0,-0.2) --++(2,0) --++(0, 1.5) -| ($(movie.one south) + (-0.1,0)$);
2841 \end{tikzpicture}
2842 \caption{Relational Model for a MOVIE Database}
2843 \label{fig:RelMod}
2844 \end{figure}
2845
2846 \begin{figure}
2847 Consider the following requirements for a UNIVERSITY database, used to keep track of students' transcripts.
2848 \begin{enumerate}
2849 \item The university keeps track of each student's name, student number, class (freshman, sophomore, …, graduate), major department, minor department (if any), and degree program (B.A., B.S., …, Ph.D.).
2850 Student number has unique values for each student.
2851 \item Each department is described by a name and has a (unique) department code.
2852 \item Each course has a course name, a course number, credit hours, and is offered by at least one department. The value of course number is unique for each course. A course has at least one section.
2853 \item Each section of a course has an instructor, a semester, a year, and a section number.
2854 The section number distinguishes different sections of the same course that are taught during the same semester/year; its values are 1, 2, 3, …, up to the number of sections taught during each semester. Students can enroll in sections and receive a letter grade, and grade point (0, 1, 2, 3, 4 for F, D, C, B, A, respectively).
2855 \end{enumerate}
2856 \caption{Requirements for a UNIVERSITY database}
2857 \label{fig:reqUNIV}
2858 \end{figure}
2859
2860 \begin{figure}
2861 \begin{tikzpicture}[scale = 0.8]
2862 \node (movie) at (0,0){\begin{tabular}{| r l |}
2863 \hline
2864 \multicolumn{2}{| c |}{\textbf{DRIVER}}\\
2865 \hline
2866 id &: String\\
2867 dob &: Date\\
2868 name &: String\\
2869 address &: Street\\
2870 & City\\
2871 \hline
2872 getAge()&: Int\\
2873 \hline
2874 \end{tabular}
2875 };
2876 \node (cdriver) at (8, 1){\begin{tabular}{| r l |}
2877 \hline
2878 \multicolumn{2}{| c |}{\textbf{COMMERCIAL\_DRIVER}}\\
2879 \hline
2880 Class &: String\\
2881 \hline
2882 \end{tabular}
2883 };
2884 %\draw[open diamond-] (movie) -- node[above, pos=0.1]{\(0..\star\)} node[above, pos=0.9]{\(1..1\)} (cdriver);
2885 \draw[{Triangle[open]}-] (movie) -- %node[above, pos=0.1]{\(0..\star\)} node[above, pos=0.9]{\(1..1\)}
2886 (cdriver);
2887 \node (car) at (9, -2){\begin{tabular}{| r l |}
2888 \hline
2889 \multicolumn{2}{| c |}{\textbf{CAR}}\\
2890 \hline
2891 vin &: String\\
2892 make &: String\\
2893 year &: Year\\
2894 brand &: String\\
2895 \hline
2896 \end{tabular}
2897 };
2898 \draw[-] ($(movie)+(2.5,-0.5)$) -- node[above, pos=0.1]{\(0..*\)} node[above=0.5em]{POSSESSES} node[above, pos=0.9]{\(0..*\)} (car);
2899 %\draw[-] (movie) -- node[below, pos=0.1]{\(0..1\)} node[below, pos=0.9]{\(0..4\)} ($(car)+(-2.5, 0.5)$);
2900 \node (seat) at (3.5, -5){\begin{tabular}{| r l |}
2901 \hline
2902 \multicolumn{2}{| c |}{\textbf{POSSESSION\_HISTORY}}\\
2903 \hline
2904 date of purchase &: Date\\
2905 \hline
2906 \end{tabular}
2907 };
2908 \draw[dashed] (seat) -- (3.5, -1.2);
2909
2910 \node (insu) at (15.5, 2){\begin{tabular}{| r l |}
2911 \hline
2912 \multicolumn{2}{| c |}{\textbf{CAR\_INSURANCE}}\\
2913 \hline
2914 policy number &: String\\
2915 covered amount &: int\\
2916 compagny name &: String\\
2917 \hline
2918 \end{tabular}
2919 };
2920
2921 \draw[diamond -] (car) -- node[below, pos=0.1, right=.2]{\(1..1\)} node[below, pos=0.9, right=0.2]{\(1..\star\)} node[right=0.2]{IS\_COVERED\_BY} (insu);
2922 \end{tikzpicture}
2923 \caption{UML Diagram for the CAR\_POSSESSION Database}
2924 \label{fig:UML}
2925 \end{figure}
2926
2927 \end{document}
2928
2929
2930
2931 Problem +.#
2932
2933 Consider the relation
2934
2935 \begin{center}
2936 CONSULTATION(Doctor\_no, Patient\_no, Date, Diagnosis, Treatment, Charge, Insurance)
2937 \end{center}
2938
2939 with the following functional dependencies:
2940
2941 \begin{center}
2942 \begin{tabular}{r c l}
2943 \{Doctor\_no, Patient\_no, Date\} & \(\to\) & \{Diagnosis\}\\
2944 \{Doctor\_no, Patient\_no, Date\} & \(\to\) & \{Treatment\}\\
2945 \{Treatment, Insurance\} & \(\to\) & \{Charge\}\\
2946 \{Patient\_no\} & \(\to\) & \{Insurance\}
2947 \end{tabular}
2948 \end{center}
2949
2950 \begin{enumerate}[topsep=0pt,itemsep=10em,partopsep=1ex,parsep=1ex]
2951 \item The designer decided not to add the functional dependency \{Diagnosis\} $\to$ \{Treatment\}.
2952 Explain what could be the designer's justification, at the level of the mini-world.
2953 \item Identify a primary key for this relation.
2954 \item What is the degree of normalization of this relation?
2955 Normalize it to the third normal form if necessary.
2956 \end{enumerate}
2957
2958
2959
2960 Solution +.#
2961
2962 Answer:
2963 Because there are no partial dependencies, the given relation is in 2NF already. This however is not 3NF because the Charge is a nonkey attribute that is determined by another nonkey attribute, Treatment. We must decompose further:
2964
2965 R (Doctor\_no, Patient\_no, Date, Diagnosis, Treatment)
2966
2967 R1 (Treatment, Charge)
2968
2969 We could further infer that the treatment for a given diagnosis is functionally dependant, but we should be sure to allow the doctor to have some flexibility when prescribing cures.
2970
2971
2972 Problem +.#
2973 A friend of yours want you to review and improve the code for a role-playing game.
2974
2975 The original idea was that each character should have a name, a class (e.g., Bard, Assassin, Druid), a certain amount of experience, a level, one or more weapons (providing bonuses) and can complete quests.
2976 A quest have a name, and rewards the characters that completed it with a certain amount of experience, and sometimes (but rarely) with a special item.
2977
2978 Your friend came up with the code presented in \autoref{lst:RPG}, page~\pageref{lst:RPG}, but there are several problems:
2979
2980 \begin{itemize}
2981 \item As of now, a character can have only one weapon.
2982 All the attempts to "hack" the `CHARACTER` table to add an arbitrary number of weapons ended up creating horrible messes.
2983 \item Every time a character completes a quest, a copy of the quest must be created.
2984 Your friend is not so sure why, but nothing else works.
2985 Also it seems that a character can complete only one quest, but your friend is not so sure about that.
2986 \item It would be nice to be able to store features that are tied to the class, and not to the character, like the bonus they provide and the associated element (e.g., all bards use fire, all assassins use wind, etc.).
2987 But you friend simply can't figure out how to do that.
2988 \end{itemize}
2989
2990 Can you provide a \emph{relational database schema} (no need to write the `SQL` code, but remember to indicate the primary and foreign keys) that would solve all of your friend's troubles?
2991
2992
2993
2994
2995
2996
2997 Problem +.#
2998 This exercise asks you to convert business statements into dependencies.
2999 Consider the following relation:
3000 \begin{center}
3001 KEYBOARD(Manufacturer, Model, Layout, Retail\_Store, Price)
3002 \end{center}
3003
3004 A tuple in the KEYBOARD relation contains information about a computer keyboard: its manufacturer, its model, its layout (AZERTY, QWERTY, etc.), the place where it is sold, and its price.
3005
3006 \begin{enumerate}
3007 \item Write each of the following business statement as a functional dependency:
3008
3009 \begin{enumerate}[topsep=0pt,itemsep=7em,partopsep=1ex,parsep=1ex]
3010 \item A model has a fixed layout.
3011 \item A retail store can't have two different models produced by the same manufacturer.
3012 \end{enumerate}
3013
3014 \item Based on those statements, what could be a key for this relation?
3015 \item Assuming all those functional dependencies hold, and taking the primary key you identified at the previous step, what is the degree of normality of this relation? Justify your answer.
3016 \end{enumerate}
3017
3018
3019
3020
3021 Problem +.#
3022 Consider the E.R. schema of Figure~\ref{fig:erCOUNTRY}, p.~\pageref{fig:erCOUNTRY}.
3023 Map that E.-R. diagram to a relational database schema.
3024
3025
3026
3027 \begin{figure}
3028 %\begin{listing}
3029 \begin{minted}[linenos, breaklines=true]{mysql}
3030 CREATE TABLE CHARACTER(
3031 Name VARCHAR(30) PRIMARY KEY,
3032 Class VARCHAR(30),
3033 XP INT,
3034 LVL INT,
3035 Weapon_Name VARCHAR(30),
3036 Weapon_Bonus INT,
3037 Quest_Completed VARCHAR(30)
3038 );
3039
3040 CREATE TABLE QUEST(
3041 Id VARCHAR(20) PRIMARY KEY,
3042 Completed_By VARCHAR(30),
3043 XP_Gained INT,
3044 Special_Item VARCHAR(20),
3045 FOREIGN KEY (Completed_By) REFERENCES CHARACTER(Name)
3046 );
3047
3048 ALTER TABLE CHARACTER ADD FOREIGN KEY (Quest_Completed) REFERENCES QUEST(Id);
3049 ```
3050
3051
3052 %\end{listing}
3053 \caption{A `SQL` Code for a Role-Playing Game}
3054 \label{lst:RPG}
3055 \end{figure}
3056
3057 \begin{figure}
3058 \begin{tikzpicture}[node distance=8em]
3059 \node[entity] (COUNTRY) {COUNTRY};
3060 \node[attribute] (name) [left of=COUNTRY] {\key{Name}} edge (COUNTRY);
3061 \node[attribute] (population) [below left of=COUNTRY] {Population} edge (COUNTRY);
3062
3063 \node[relationship] (SPEAKS) [below right of=COUNTRY] {SPEAKS} edge node[right, pos=0.4] {$M$} (COUNTRY);
3064 \node[entity] (LANGUAGE) [below right of=SPEAKS] {LANGUAGE} edge node[right, pos=0.5] {$N$} (SPEAKS);
3065 \node[attribute] (code) [left of = LANGUAGE] {\key{Code}} edge (LANGUAGE);
3066 \node[attribute] (symbol) [right of = LANGUAGE] {Name} edge (LANGUAGE);
3067
3068 \node[relationship] (BWF) [below of = LANGUAGE] {B\_W\_F};
3069 \draw (LANGUAGE) to node[left, pos=0.6] {$N$} (BWF.west);
3070 \draw (LANGUAGE) to node[right, pos=0.6] {$M$} (BWF.east);
3071
3072 %\node[relationship] (boundary) [above of = COUNTRY] {S\_B\_W};
3073 %\draw (COUNTRY) to node[left, pos=0.6] {$N$} (boundary.west);
3074 %\draw (COUNTRY) to node[right, pos=0.6] {$M$} (boundary.east);
3075
3076 %\node[relationship] (conversion) [below of =LANGUAGE] {CONVERSION};
3077 %\draw (LANGUAGE) to node[left, pos=0.6] {$N$} (conversion.west);
3078 %\draw (LANGUAGE) to node[right, pos=0.6] {$M$} (conversion.east);
3079
3080 %\node[attribute] (timestamp) [below left of=conversion] {\key{Timestamp}} edge (conversion);
3081 %\node[attribute] (rate) [below right of=conversion] {Exchange\_rate} edge (conversion);
3082
3083 \node[ident relationship] (hasfor) [right of=COUNTRY] {SINGS} edge node[above, pos=0.4] {$1$} (COUNTRY);
3084 \node[weak entity] (anthem) %[right of=hasfor]
3085 [right = 1cm of hasfor] {NATIONAL\_ANTHEM} edge[total] node[above, pos=0.6]{\(M\)} (hasfor);
3086 \node[multi attribute] (color) [right =1cm of anthem] {Creator} edge (anthem);
3087 \node[attribute] (name) [below right of = anthem] {\pkey{Name}} edge (anthem);
3088
3089 \node[relationship] (WIN) [above right of =LANGUAGE] {W\_IN};
3090 \draw (LANGUAGE) to node[left, pos=0.6] {$N$} (WIN);
3091 \draw (anthem) to node[right, pos=0.6] {$M$} (WIN);
3092 \end{tikzpicture}
3093
3094 %"S\_B\_W" stands for "SHARES\_A\_BORDER\_WITH".
3095 "W\_IN" stands for "WRITTEN\_IN", and
3096 "B\_W\_F" stands for "BORROWS\_WORDS\_FROM".
3097 For this relationship, on the left-hand side is the language that borrows a word, and on the right-hand side is the language that provides the loanword.
3098 \caption{E.R. Schema for the COUNTRY\_INFO database}
3099 \label{fig:erCOUNTRY}
3100 \end{figure}
3101
File exercises/sum.tex deleted (index ba4254b..0000000)
1
2 \begin{exercise}[tags={hw,qz}]
3 What does it mean to say that \sqli{SQL} is at the same time a \enquote{data definition language} and a \enquote{data manipulation language}?
4 \end{exercise}
5
6 \begin{solution}
7 It can specify the conceptual and internal schema, \textbf{and} it can manipulate the data.
8 \end{solution}
9
10 \begin{exercise}[tags={hw}]
11 Name three kind of objects (for lack of a better word) a \sqli{CREATE} statement can create.
12 \end{exercise}
13
14 \begin{solution}
15 Database (schema), table, view, assertion, trigger, etc.
16 \end{solution}
17
18 \begin{exercise}[tags={qz}]
19 Write a \sqli{SQL} statement that adds a primary key constraint to an attribute named \sqli{Id} in an already existing table named \sqli{STAFF}.
20 \end{exercise}
21
22 \begin{solution}
23 \sqli{ALTER TABLE STAFF ADD PRIMARY KEY(Id);}
24 \end{solution}
25
26 \begin{exercise}[tags={hw}]
27 Complete the following table wo different examples when asked for examples:
28
29 {\centering
30
31 \begin{tabular}{m{6cm} | m{6cm} }
32 \hline
33 \rowcolor{lgray}
34 Data Type & Examples \\
35 \hline
36 Int & $4$, $-32$\\
37 \hline
38 Char(4) & \blank[width=1.9em]{'trai', 'plol'}\\
39 \hline
40 \blank[width=1.9em]{VarChar(10)} & 'Train', 'Michelle'\\
41 \hline
42 Bit(4) & \blank[width=1.9em]{ B'1010', B'0101'}\\
43 \hline
44 \blank[width=1.9em]{Boolean} & TRUE, UNKNOWN\\
45 \hline
46 \end{tabular}
47
48 }
49 \end{exercise}
50
51 \begin{exercise}[tags={hw}]
52 Explain what the following SQL statement does
53 \mint{mysql}| CREATE SCHEMA Faculty;|
54 \end{exercise}
55
56 \begin{solution}
57 It creates a schema, i.e., a database, named \enquote{Faculty}.
58 \end{solution}
59
60 \begin{exercise}[tags={hw}]
61 If I want to enter January 21, 2016, as a value for an attribute with the \sqli{DATE} datatype, what value should I enter?
62 \end{exercise}
63
64 \begin{solution}
65 \sqli{DATE}'2016-01-21'
66 \end{solution}
67
68 \begin{exercise}[tags={hw,qz}]
69 Write a statement that inserts the values \enquote{Thomas} and \enquote{$4$} into the table \sqli{TRAINS}.
70 \end{exercise}
71
72 \begin{solution}
73 \mint{sql}| INSERT INTO TRAINS VALUES('Thomas', 4);|
74 \end{solution}
75
76 \begin{exercise}[tags={hw}]
77 If \sqli{PkgName} is a primary key, what can you tell about the number of rows returned by the following statement?
78 \mint{sql}| SELECT * FROM MYTABLE WHERE PkgName = 'MySQL';|
79 \end{exercise}
80
81 \begin{solution}
82 Yes.
83 \end{solution}
84
85 \begin{exercise}[tags={hw}]
86 What is the difference between an implicit, an explicit, and a semantic constraint?
87 \end{exercise}
88
89 \begin{solution}
90 The textbook reads, pp. 67 - 69:
91 \begin{enumerate}
92 \item Constraints that are inherent in the data model. We call these inherent model-based constraints or implicit constraints.
93 \item Constraints that can be directly expressed in schemas of the data model, typically by specifying them in the DDL (data definition language, see Section 2.3.1). We call these schema-based constraints or explicit constraints.
94 \item Constraints that cannot be directly expressed in the schemas of the data model, and hence must be expressed and enforced by the application programs. We call these application-based or semantic constraints or business rules.
95 \end{enumerate}
96 \end{solution}
97
98 \begin{exercise}[tags={hw}]
99 If you want every tuple referencing the tuple you're about to delete to be deleted as well, what mechanism should you use?
100 \end{exercise}
101
102 \begin{solution}
103 We should use a referential triggered action clause, \sqli{ON DELETE CASCADE}.
104 \end{solution}
105
106 \begin{exercise}[tags={hw}]
107 If a database designer is using the \sqli{ON UPDATE SET NULL} for a foreign key, what mechanism is he implementing (i.e., describe how the database will react a certain operation)?
108 \end{exercise}
109
110 \begin{solution}
111 If the referenced tuple is updated, then the attribute of the referencing relation are set to NULL.
112 \end{solution}
113
114 \begin{exercise}[tags={hw}]
115 If the following is part of the design of a table:
116 \mint[breaklines, linenos]{mySQL}|FOREIGN KEY (DptNumber) REFERENCES DEPARTMENT(DptNumber) ON DELETE SET DEFAULT ON UPDATE CASCADE|
117 What happen to the row whose foreign key \sqli{DptNumber} is set to $3$ if
118 \begin{enumerate}
119 \item the row in the \sqli{DEPARTEMENT} table with primary key \sqli{DptNumber} set to \(3\) is deleted?
120 \item the row in the \sqli{DEPARTEMENT} table with primary key \sqli{DptNumber} set to \(3\) as its value for \sqli{DptNumber} updated to \(5\)?
121 \end{enumerate}
122 \end{exercise}
123
124
125 \begin{solution}
126 The department number is set to the default value.
127
128 The department number is updated accordingly.
129 \end{solution}
130
131 \begin{exercise}[tags={qz}]
132 If the following is part of the design of a \sqli{WORKER} table:
133 \mint[breaklines, linenos]{mySQL}|FOREIGN KEY WORKER(DptNumber) REFERENCES DEPARTMENT(DptNumber) ON UPDATE CASCADE|
134 Admitting there is a row in the \sqli{WORKER} table whose foreign key \sqli{DptNumber} is set to $3$, what would happen if
135 \begin{enumerate}
136 \item We tried to delete the row in the \sqli{DEPARTEMENT} table with primary key \sqli{DptNumber} set to \(3\)?
137 \IfTagSetF{hw}{\vspace{4em}}
138 \item We tried to change the value of \sqli{DptNumber} of the row in the \sqli{DEPARTEMENT} table with primary key \sqli{DptNumber} set to \(3\)?
139 \end{enumerate}
140 \end{exercise}
141
142
143 \begin{exercise}[tags={hw,qz}]
144 Given a relation \sqli{TOURIST(Name, EntryDate, Address)}, write a \sqli{SQL} statement printing the name and address of all the tourists who entered the territory after the 15 September, 2012.
145 \end{exercise}
146
147 \begin{solution}
148 \begin{minted}{mySQL}
149 SELECT Name, Address
150 FROM TOURIST
151 WHERE EntryDate > DATE'2012-09-15';
152 \end{minted}
153 \end{solution}
154
155 \begin{exercise}[tags={hw}]
156 What is the fully qualified name of an attribute? Give an example.
157 \end{exercise}
158
159 \begin{solution}
160 The name of the relation with the name of its schema and a period beforehand. myDB.MYTABLE, EMPLOYEE.name, etc.
161 \end{solution}
162
163 % \begin{exercise}[tags={hw}]If \sqli{DEPARTMENT} is a database, what is \sqli{DEPARTMENT.*}?}
164 %
165 % \begin{solution}
166 % All the tables in that database.
167 % \end{solution}
168
169 \begin{exercise}[tags={hw}]
170 What is a multi-set? What does it mean to say that SQL treats tables as multisets?
171 \end{exercise}
172
173 \begin{solution}
174 A set where the same value can occur twice.
175 The same tuple can occur twice in a table.
176 \end{solution}
177
178 \begin{exercise}[tags={hw}]
179 What is the difference between those two queries?
180 \mint{mySQL}| SELECT ALL * FROM MYTABLE;|
181 and
182 \mint{mySQL}| SELECT DISTINCT * FROM MYTABLE;|
183 How are the results the same? How are they different?
184 \end{exercise}
185
186 \begin{solution}
187 All will print the duplicate, distinct will eliminate them.
188 \end{solution}
189
190 \IfTagSetF{hw}{
191 \end{document}
192 }
193
194 \section{Part II --- Programming Exercises}
195
196 This part will help you in assessing your level of understanding of this lecture, and help you to become more familiar with \sqli{SQL}.
197 It constitutes the practical part of this class and shouldn't be overlooked.
198 This time, all the problems require a computer.
199 They might look long, but it is because the instructions are detailed: follow them carefully and the four problems shouldn't take you very long.
200 I'll assume that you will have successfully completed them by the time Homework \#\the\numexpr\hwNum +1\relax{} is released (\hwDateNext), so don't wait and let me know if you had difficulties doing them.
201
202 \begin{problem}[tags={hw}]
203 \label{setup}
204 \emph{This exercise assume you successfully completed Problem~2 of Homework 1.}
205 \begin{enumerate}
206 \item \label{item-connect} Connect to your MySQL DBMS as \sqli{testuser}:
207 \begin{itemize}
208 \item In windows, open a command prompt (search for \enquote{cmd}) and type
209 \mint{bat}| cd "C:\Program Files\MySQL\MySQL Server 5.7\bin"|
210 \item In Linux, open a shell (as a normal user)
211 \end{itemize}
212 Then, in both cases, type \mint{BNF}| mysql -u testuser -p| and enter the password \sqli{password}.
213 If you are prompted with a message
214 \mint[breaklines]{BNF}| ERROR 1045 (28000): Access denied for user 'testuser'@'localhost' (using password: YES)|
215 then you probably typed the wrong password.
216 Otherwise, you should see a welcoming message from MySQL / MariaDB and a prompt.
217 \item \label{item-create} Create a new database called \sqli{HW_2Q1} \mint{mySQL}| CREATE DATABASE HW_2Q1;|
218 \item List the databases \mint{mySQL}| SHOW DATABASES;| Make sure \sqli{HW_2Q1} is a part of it.
219 \item Remove (\enquote{drop}) the database, using \mint{mySQL}| DROP DATABASE HW_2Q1;|
220 \end{enumerate}
221 In the future, we will refer to the commands \ref{item-connect}.\ and \ref{item-create}.\ as \enquote{log-in as \sqli{testuser} and create a database \sqli{HW_2Q1}}.
222
223 \end{problem}
224
225 \begin{problem}[tags={hw}]
226 The goal of this problem is to learn where to find the documentation of your DBMS, and to have a first look at the syntax of SQL commands.
227
228 You can consult your textbook, Table~5.2, p. 140 (6th edition) or Table~7.2, p. 235 (7th edition), for a very quick summary of the most common commands.
229 Make sure you are familiar with the Backus–Naur form (BNF) notation commonly used:
230 \begin{itemize}
231 \item non-terminal symbols (i.e., variables, parameters) are enclosed in angled brackets, \mintinline{BNF}|<...>|
232 \item optional parts are shown in square brackets, \mintinline{BNF}|[...]|
233 \item repetitons are shown in braces \mintinline{BNF}|{...}|
234 \item alternatives are shown in parenthesis and separated by vertical bars, \mintinline{BNF}\(...|...|...)\
235 \end{itemize}
236
237 The most complete lists of commands are probably at
238 \begin{itemize}
239 \item \url{https://mariadb.com/kb/en/library/sql-statements/} and
240 \item \url{https://dev.mysql.com/doc/refman/5.7/en/sql-syntax.html}
241 \end{itemize}
242 Those are the commands implemented in the DBMS you are actually using.
243 Since there are small variations from one implementation to the other, it's better to take one of this link as a reference in the future.
244 \end{problem}
245
246 \begin{problem}[tags={hw}]
247 \emph{This exercise, and the following ones, assume you successfully completed Problem~\ref{setup}.}
248
249 Log in as \sqli{testuser} and create a database \sqli{HW_2Q3}.
250
251 \begin{enumerate}
252 \item We will first set-up our meta-data (i.e., the schema, the structure where our tables will be, and the tables themselves).
253
254 Let us first tell MySQL that we want to use that database
255 \mint{mySQL}| USE HW_2Q3;|
256 and ask what it contains
257 \mint{mySQL}| SHOW TABLES;|
258 Now, create a first table named \sqli{NAME}
259 \begin{minted}{mySQL}
260 CREATE TABLE NAME(
261 FName VARCHAR(15),
262 LName VARCHAR(15),
263 Id INT,
264 PRIMARY KEY(Id)
265 );
266 \end{minted}
267 Note that the \sqli{SQL} syntax and your DBMS are completely fine with your statement spreading over multiple lines.
268 Let us create a second table, named \sqli{ADDRESS}
269 \begin{minted}{mySQL}
270 CREATE TABLE ADDRESS(
271 StreetName VARCHAR(15),
272 Number INT,
273 Habitants INT,
274 PRIMARY KEY(StreetName, Number)
275 );
276 \end{minted}
277 To make sure those two tables were actually created, we can use
278 \mint{mySQL}| SHOW TABLES;|
279 But how to make sure that you entered the attributes correctly?
280 One way to make sure is to enter the command
281 % \mint{mySQL}| show create table name;|
282 \mint{mySQL}| DESC ADDRESS;|
283 and to examine carefully the message printed. You should read
284 \begin{minted}{mySQL}
285 +------------+-------------+------+-----+---------+-------+
286 | Field | Type | Null | Key | Default | Extra |
287 +------------+-------------+------+-----+---------+-------+
288 | StreetName | varchar(15) | NO | PRI | NULL | |
289 | Number | int(11) | NO | PRI | NULL | |
290 | Habitants | int(11) | YES | | NULL | |
291 +------------+-------------+------+-----+---------+-------+
292 \end{minted}
293 If you believe there is a mistake, you can erase (\enquote{drop}) the \emph{table} (not the whole database, as we did in Problem~1) using
294 \mint{mySQL}| DROP TABLE ADDRESS;|
295 and then re-create it. Of course, you can do the same for the \sqli{NAME} table.
296
297 Now, let us add a foreign key to the \sqli{ADDRESS} table:
298
299 \begin{minted}{mySQL}
300 ALTER TABLE ADDRESS
301 ADD FOREIGN KEY (Habitants)
302 REFERENCES NAME(Id);
303 \end{minted}
304 And observe the modification:
305 \begin{minted}{mySQL}
306 DESC ADDRESS;
307 \end{minted}
308
309 \item This second part is about data, i.e., filling our tables with actual information.
310 We begin by adding some data in the \sqli{NAME} table:
311 \begin{minted}{mySQL}
312 INSERT INTO NAME VALUES ('Barbara', 'Liskov', 003);
313 INSERT INTO NAME VALUES ('Tuong Lu', 'Kim', 004);
314 INSERT INTO NAME VALUES ('Samantha', NULL, 080);
315 \end{minted}
316 To display the data we just inserted, we can use:
317 \begin{minted}{mySQL}
318 SELECT * FROM NAME;
319 \end{minted}
320 Do you notice anything regarding the values we entered for the \sqli{Id} attribute?
321
322 We can add some data into the \sqli{ADDRESS} table (using a different syntax):
323 \begin{minted}{mySQL}
324 INSERT INTO ADDRESS (StreetName, Number, Habitants)
325 VALUES
326 ('Armstrong Drive', 10019, 003),
327 ('North Broad St.', 23, 004),
328 ('Robert Lane', 120, NULL);
329 \end{minted}
330 And let's use our first update command:
331 \begin{minted}{mySQL}
332 UPDATE ADDRESS SET Habitants = 3 WHERE Number = 120;
333 \end{minted}
334
335 \item
336 Now, do the following:
337 \begin{enumerate}
338 \item Have a look at the formal definition of the \mintinline{SQL}|ALTER| command, at \url{https://dev.mysql.com/doc/refman/5.7/en/alter-table.html} or \url{https://mariadb.com/kb/en/library/alter-table/}.
339 \item Draw the relations corresponding to that database, including the domains and primary, as well as foreign, keys.
340 \item Write a \mintinline{SQL}{SELECT} statement that returns the \mintinline{SQL}{Id} number of the person whose first name is \enquote{Samantha}.
341 % SELECT Id FROM NAME WHERE FName = 'Samantha';
342 \item Write a statement that violate the entity integrity constraint. What is the error message returned?
343 % ERROR 1062 (23000): Duplicate entry '80' for key 'PRIMARY'
344 \item Execute an \sqli{UPDATE} statement that violate the referential integrity constraint. What is the error message returned?
345 % ERROR 1452 (23000): Cannot add or update a child row: a foreign key constraint fails (`HW_2Q3`.`ADDRESS`, CONSTRAINT `fk_id` FOREIGN KEY (`Habitants`) REFERENCES `name` (`Id`))
346 \item Write a statement that violate another kind of constraint. Explain what constraint you are violating, and explain the error message.
347 % ERROR 1136 (21S01): Column count doesn't match value count at row 1
348 \end{enumerate}
349
350 \end{enumerate}
351 \end{problem}
352
353 \begin{problem}[tags={hw}]Log in as \sqli{testuser} and create a database \sqli{HW_2Q4}. Tell MySQL that you want to use that database, and create a table (I will assume you named it \sqli{EXAMPLE} in the following, but you are free to name it the way you want) with at least two attributes that have different data types. Don't declare a primary key yet. Answer the following:
354 \begin{enumerate}
355 \item Add a tuple to your table using
356 \mint{mySQL}| INSERT INTO EXAMPLE VALUES(X, Y);|
357 where \enquote{X} and \enquote{Y} are values with the right type.
358 Try to add this tuple again. What do you observe? (You can use \mintinline{mySQL}|SELECT * FROM EXAMPLE;| to observe what is stored in this table.)
359 \item Alter your table to add a primary key, using
360 \mint{mySQL}| ALTER TABLE EXAMPLE ADD PRIMARY KEY (Attribute);| where \sqli{Attribute} is the name of the attribute you want to be a primary key. What do you observe?
361 \item Empty your table using \mint{mySQL}| DELETE FROM EXAMPLE;| and alter your table to add a primary key, using the command we gave at the previous step. What do you observe?
362 \item Try to add the same tuple twice. What do you observe?
363 % \item Now, create \emph{another table} (just one attribute and no primary key is fine) and try to create a foreign key pointing to the attribute in \sqli{EXAMPLE} that is not the primary key.
364 % What do you observe?
365 % => Messy, they get an error message if the attribute is not the primary key or if the datatype isn't the same, or if the command is not properly formatted... The error message is too vague for them to learn anything.
366 \end{enumerate}
367 \end{problem}
368
369 \begin{solution}
370 \begin{minted}{SQL}
371 CREATE TABLE EXAMPLE(
372 X VARCHAR(15),
373 Y INT);
374 \end{minted}
375 \begin{minted}{SQL}
376 INSERT INTO EXAMPLE VALUES('Train', 4);
377 \end{minted}
378 OK to insert same tuple twice.
379 \begin{minted}{SQL}
380 ALTER TABLE EXAMPLE ADD PRIMARY KEY (X);
381 ERROR 1062 (23000): Duplicate entry 'Train' for key 'PRIMARY'
382 \end{minted}
383 \end{solution}
384
385 %%%%% 03
386
387
388 \begin{exercise}[tags={hw}]
389 Describe what the star do in the statement \sqli{SELECT ALL * FROM MYTABLE;}.
390 \end{exercise}
391
392 \begin{solution}
393 It selects all the attributes.
394 \end{solution}
395
396
397 \begin{exercise}[tags={hw}]
398 What is wrong with the statement \sqli{SELECT * WHERE Name = 'CS' FROM DEPARTMENT;}?
399 \end{exercise}
400
401 \begin{solution}
402 You can't have the \sqli{WHERE} before \sqli{FROM}.
403 \end{solution}
404
405
406 \begin{exercise}[tags={hw}]
407 When is it useful to use a select-project-join query?
408 \end{exercise}
409
410 \begin{solution}
411 We use those query that projects on attributes using a selection and join conditions when we need to look for information gathered in multiple tables.
412 \end{solution}
413
414 \begin{exercise}[tags={hw}]
415 When is a tuple variable useful?
416 \end{exercise}
417
418 \begin{solution}
419 It makes the distinction between two different copies of the same relation, it is useful when we want to select a tuple in a relation that is in a particular relation with a tuple in the same relation.
420
421 cf. \url{https://stackoverflow.com/a/7698796/2657549}:
422 They are useful for saving typing, but there are other reasons to use them:
423 \begin{itemize}
424 \item If you join a table to itself you must give it two different names otherwise referencing the table would be ambiguous.
425 \item It can be useful to give names to derived tables, and in some database systems it is required... even if you never refer to the name.
426 \end{itemize}
427 \end{solution}
428
429 \begin{exercise}[tags={hw}]
430 Write a query that changes the name of the professor whose login is 'caubert' to 'Hugo Pernot' in the table \sqli{PROF}.
431 \end{exercise}
432
433 \begin{solution}
434 \sqli{UPDATE PROF SET Name = 'Hugo Pernot' WHERE Login = 'caubert';}
435 \end{solution}
436
437 \begin{exercise}[tags={hw}]
438 Can an \sqli{UPDATE} statement have a \sqli{WHERE} condition using an attribute that isn't the primary key? If no, justify, if yes, tell what could happen.
439 \end{exercise}
440
441 \begin{solution}
442 Yes, we could update more than one tuple at a time.
443 \end{solution}
444
445 \begin{exercise}[tags={hw}]
446 What is a multi-set?
447 What does it mean to say that \sqli{SQL} treats tables as multisets?
448 \end{exercise}
449
450 \begin{solution}
451 A set where the same value can occur twice.
452 The same tuple can occur twice in a table.
453 \end{solution}
454
455 \begin{exercise}[tags={hw}]
456 What is the difference between those two queries?
457 \mint{mySQL}| SELECT ALL * FROM MYTABLE;|
458 and
459 \mint{mySQL}| SELECT DISTINCT * FROM MYTABLE;|
460 How are the results the same? How are they different?
461 \end{exercise}
462
463 \begin{solution}
464 \sqli{ALL} will print the duplicate, \sqli{DISTINCT} will eliminate them.
465 \end{solution}
466
467 \begin{exercise}[tags={hw}]
468 What are the possible meanings or interpretations for a \sqli{NULL} value?
469 \end{exercise}
470
471 \begin{solution}
472 Unknown value, unavailable / withheld, N/A.
473 \end{solution}
474
475
476 \begin{exercise}[tags={hw}]
477 What are the values of the following expressions (i.e., do they evaluate to \sqli{TRUE}, \sqli{FALSE}, or \sqli{UNKNOWN})?
478
479 \sqli{TRUE AND FALSE}\hspace{2.2em}
480 \sqli{TRUE AND UNKNOWN}\hspace{2.2em}
481 \sqli{NOT UNKNOWN}\hspace{2.2em}
482 \sqli{FALSE OR UNKNOWN}
483
484 \end{exercise}
485
486 \begin{solution}
487 \sqli{FALSE}, \sqli{UNKNOWN}, \sqli{UNKNOWN}, \sqli{FALSE}
488 \end{solution}
489
490 \begin{exercise}[tags={hw,qz}]
491 What comparison expression should you use to test if a value is different from \sqli{NULL}?
492 \end{exercise}
493
494 \begin{solution}
495 \sqli{IS NOT}
496 \end{solution}
497
498 \begin{exercise}[tags={hw}]
499 Explain this query:
500 \begin{minted}{sql}
501 SELECT Login
502 FROM PROF
503 WHERE Department IN ( SELECT Major
504 FROM STUDENT
505 WHERE Login = 'jrakesh');
506 \end{minted}
507 \end{exercise}
508
509 \begin{solution}
510 It list the login of the professors teaching in the department where the student whose login is \enquote{jrakesh} is majoring.
511 \end{solution}
512
513 \begin{exercise}[tags={hw}]
514 What is wrong with this query?
515
516 \begin{minted}{sql}
517 SELECT Name
518 FROM STUDENT
519 WHERE Login IN ( SELECT Code
520 FROM Department
521 WHERE head = 'aturing');
522 \end{minted}
523
524 \end{exercise}
525
526 \begin{solution}
527 It tries to find a \sqli{Login} in a \sqli{Code}!
528 \end{solution}
529
530 \begin{exercise}[tags={hw}]
531 Write a query that returns the number of row (i.e., of entries, of tuples) in a table named \sqli{BOOK}.
532 \end{exercise}
533
534 \begin{solution}
535 \sqli{SELECT COUNT(*) FROM BOOK;}
536 \end{solution}
537
538 \begin{exercise}[tags={qz}]
539 \vspace{-5em}
540 Consider the following \sqli{SQL} code:
541 \begin{multicols}{2}
542 \begin{minted}[breaklines=true, linenos]{sql}
543 CREATE TABLE COMPUTER(
544 Id VARCHAR(20) PRIMARY KEY,
545 Model VARCHAR(20)
546 );
547 CREATE TABLE PRINTER(
548 Id VARCHAR(20) PRIMARY KEY,
549 Model VARCHAR(20)
550 );
551 CREATE TABLE CONNEXION(
552 Computer VARCHAR(20),
553 Printer VARCHAR(20),
554 PRIMARY KEY(Computer, Printer),
555 FOREIGN KEY (Computer) REFERENCES COMPUTER(Id),
556 FOREIGN KEY (Printer) REFERENCES PRINTER(Id)
557 );
558
559 INSERT INTO COMPUTER VALUES
560 ('A', 'DELL A'),
561 ('B', 'HP X'),
562 ('C', 'ZEPTO D'),
563 ('D', 'MAC Y');
564 INSERT INTO PRINTER VALUES
565 ('12', 'HP-140'),
566 ('13', 'HP-139'),
567 ('14', 'HP-140'),
568 ('15', 'HP-139');
569 INSERT INTO CONNEXION VALUES
570 ('A', '12'),
571 ('A', '13'),
572 ('B', '13'),
573 ('C', '14');
574
575 \end{minted}
576 \end{multicols}
577 Write a query that returns \dots (in parenthesis, the values returned \emph{in this set-up}, but you have to be general)
578 \begin{enumerate}[topsep=0pt,itemsep=7em,partopsep=1ex,parsep=1ex]
579 \item \dots the number of computers connected to the printer whose id is '13' (i.e., \(2\) ).
580 \item \dots the number of different models of printers (i.e., \(2\)).
581 \item \dots the model(s) of the printer connected to the computer whose id is 'A' (i.e., 'HP-140' and 'HP-139').
582 \item \dots the id of the computer(s) not connected to any printer (i.e., 'D').
583 \end{enumerate}
584
585 \end{exercise}
586
587 \begin{exercise}[tags={hw}]
588 Write a query that returns the sum of all the values stored in the \sqli{Pages} attribute of a \sqli{BOOK} table.
589 \end{exercise}
590
591 \begin{solution}
592 \sqli{SELECT SUM(Pages) FROM BOOK;}
593 \end{solution}
594
595
596 \begin{exercise}[tags={hw}]
597 Write a query that adds a \sqli{Pages} attribute of type \sqli{INT} into a (already existing) \sqli{BOOK} table.
598 \end{exercise}
599
600 \begin{solution}
601 \sqli{ALTER TABLE BOOK ADD COLUMN Pages INT;}
602 \end{solution}
603
604 \begin{exercise}[tags={hw}]
605 Write a query that removes the default value for a \sqli{Pages} attribute in a \sqli{BOOK} table.
606 \end{exercise}
607
608 \begin{solution}
609 \sqli{ALTER TABLE BOOK ALTER COLUMN Pages DROP DEFAULT;}
610 \end{solution}
611
612
613 \IfTagSetF{hw}{
614 \end{document}
615 }
616
617 \section{Part II --- Programming Exercises}
618
619 This time, the (only) problem requires a computer and to have access to the code used during the lecture.
620 I'll assume that you will have successfully completed it by the time Homework \#\the\numexpr\hwNum +1\relax{} is released (\hwDateNext), so don't wait and let me know if you had difficulties doing it.
621
622 \begin{problem}[tags={hw}]
623 Create the \sqli{PROF}, \sqli{STUDENT}, \sqli{DEPARTMENT} and \sqli{GRADE} tables as during the lecture.
624 Populate them with some data.
625 \begin{enumerate}
626 \item Create a \sqli{LECTURE} table as follows:
627 \begin{itemize}
628 \item It should have four attributes, \sqli{Name}, \sqli{Instructor}, \sqli{Code}, and \texttt{Year}, of type \sqli{VARCHAR(25)} for the two first, \sqli{YEAR(4)}, and \sqli{CHAR(5)}.
629 \item The \texttt{Year} and \sqli{Code} attributes should be the primary key (yes, have \emph{two} attributes be the primary key).
630 \item The \sqli{Instructor} attribute should be a foreign key referencing the \sqli{Login} attribute in \sqli{PROF}.
631 \end{itemize}
632 \item Populate the \sqli{LECTURE} table with some made-up data.
633 \item Add two columns to the \sqli{GRADE} table, using
634 \begin{minted}{sql}
635 ALTER TABLE GRADE
636 ADD COLUMN LectureCode CHAR(5),
637 ADD COLUMN LectureYear YEAR(4);
638 \end{minted}
639 \item Use \sqli{DESCRIBE} and \sqli{SELECT} to observe the schema of the \sqli{GRADE} table and its rows (i.e., tuples).
640 Is that what you would have expected?
641 \item Add a foreign key in \sqli{GRADE}, using
642 \begin{minted}{mysql}
643 ALTER TABLE GRADE
644 ADD FOREIGN KEY (LectureYear, LectureCode)
645 REFERENCES LECTURE(Year, Code);
646 \end{minted}
647 \item Update the tuples in \sqli{GRADE} with some made-up data: now, every row should contain, on top of a login and a grade, a lecture year and a lecture code.
648 \item Write \sqli{SELECT} statements answering the following questions (where XXX and YY should be relevant values considering your data):
649
650 \begin{enumerate}
651 \item \enquote{I taught class XXX in YYYY, could you give me the logins and grades of the students who took it?}
652 \item \enquote{Could you list the instructors who taught in YYYY?} (and, please, avoid duplicates)
653 \item \enquote{Could your list the name and grade of all the student who took class XXX (no matter the year)?}
654 \item \enquote{Could you tell me which years was the class XXX taught?}
655 \item \enquote{Could you list the classes taught the same year as class XXX?}
656 \item \enquote{Could you print the name of the students who registered after Ava Alyx?}
657 \item \enquote{How many departments' heads are teaching this year?}
658 \end{enumerate}
659 \end{enumerate}
660 \end{problem}
661
662 %\begin{solution}[print=true]
663 % Hey!
664 \IfSolutionPrintT{
665 % Essayer avec print = true?
666 \begin{minted}[breaklines=true, linenos]{sql}
667 CREATE TABLE LECTURE(
668 Name VARCHAR(25),
669 Instructor VARCHAR(25),
670 Year YEAR(4),
671 Code CHAR(5),
672 PRIMARY KEY(Year, Code),
673 FOREIGN KEY (Instructor) REFERENCES PROF(Login)
674 );
675
676 INSERT INTO LECTURE VALUES
677 ('Intro to CS', 'caubert', 2017, '1304'),
678 ('Intro to Algebra', 'perdos', 2017, '1405'),
679 ('Intro to Cyber', 'aturing', 2017, '1234');
680
681 ALTER TABLE GRADE ADD COLUMN LectureCode CHAR(5),
682 ADD COLUMN LectureYear YEAR(4);
683
684 DESCRIBE GRADE;
685
686 SELECT * FROM GRADE;
687
688 ALTER TABLE GRADE ADD FOREIGN KEY (LectureYear, LectureCode) REFERENCES LECTURE(Year, Code);
689
690 UPDATE GRADE SET LectureCode = '1304', LectureYear = 2017 WHERE Login = 'jrakesh' AND Grade = '2.85';
691 UPDATE GRADE SET LectureCode = '1405', LectureYear = 2017 WHERE Login = 'svlatka' OR (Login = 'jrakesh' AND Grade = '3.85');
692 UPDATE GRADE SET LectureCode = '1234', LectureYear = 2017 WHERE Login = 'aalyx' OR Login = 'cjoella';
693
694 SELECT Login, Grade FROM Grade WHERE Lecturecode='1304' and LectureYear = '2017';
695
696 SELECT DISTINCT Instructor FROM LECTURE WHERE year = 2017;
697
698 SELECT Name, Grade FROM STUDENT, GRADE WHERE GRADE.LectureCode = 1405 AND STUDENT.Login = GRADE.Login;
699
700 SELECT Year FROM Lecture WHERE Code = '1234';
701
702 SELECT Name FROM LECTURE WHERE Year IN (SELECT Year FROM LECTURE WHERE CODE = '1234');
703
704 SELECT B.name FROM STUDENT AS A, STUDENT AS B WHERE A.Name = 'Ava Alyx' AND A.Registered > B.Registered;
705
706 SELECT COUNT(DISTINCT PROF.Name) AS 'Head Teaching This Year' FROM LECTURE, DEPARTMENT, PROF WHERE Year = 2017 AND Instructor = Head AND Head = PROF.Login;
707 \end{minted}
708 }
709
710 %%%%% 04
711
712
713 \begin{exercise}[tags={hw}]
714 What could be the decomposition of an attribute used to store an email address? When could that be useful?
715 \end{exercise}
716
717 \begin{solution}
718 Name / extension.
719 To have statistics about the extensions, to sort the username by length, etc.
720 \end{solution}
721
722 \begin{exercise}[tags={hw}]
723 Draw the ER diagram for a \enquote{COMPUTER} entity that has one multivalued attribute \enquote{Operating\_System}, a composite attribute \enquote{Devices} (decomposed into \enquote{Keyboard} and \enquote{Mouse}) and an \enquote{Id} key attribute.
724 \end{exercise}
725
726 \begin{solution}
727 \begin{tikzpicture}[node distance=7em]
728 \node[entity] (computer) {COMPUTER};
729 \node[attribute] (pid) [left of=computer] {\key{Id}} edge (computer);
730 \node[multi attribute] (os) [above of=computer] {Operating\_System} edge (computer);
731 \node[attribute] (devices) [right of=computer] {Devices} edge (computer);
732 \node[attribute] (mouse) [above right of=devices] {Mouse} edge (devices);
733 \node[attribute] (keyboard) [right of=devices] {Keyboard} edge (devices);
734 \end{tikzpicture}
735 \end{solution}
736
737 \begin{exercise}[tags={qz}]
738 Draw the ER diagram for a \enquote{CELLPHONE} entity that has a composite attribute \enquote{Plan} (decomposed into \enquote{Carrier} and \enquote{Price}), a \enquote{Mobile\_identification\_number} key attribute, and a multi-valued \enquote{App\_Installed} attribute.
739 \end{exercise}
740
741 \begin{solution}
742 \begin{tikzpicture}[node distance=7em]
743 \node[entity] (cellphone) {CELLPHONE};
744 \node[attribute] (mid) [left of=cellphone] {\key{MIN}} edge (cellphone);
745 \node[multi attribute] (app) [above of=cellphone] {App\_Installed} edge (cellphone);
746 \node[attribute] (plan) [right of=cellphone] {Plan} edge (cellphone);
747 \node[attribute] (carrier) [above right of=plan] {Carrier} edge (plan);
748 \node[attribute] (price) [right of=plan] {Price} edge (plan);
749 \end{tikzpicture}
750 \end{solution}
751
752
753 \begin{exercise}[tags={hw,qz}]
754 Name one difference between a primary key in the relational model, and a key attribute in the ER model.
755 \end{exercise}
756
757 \begin{solution}
758 There can be more than one key in the ER model.
759 \end{solution}
760
761 \begin{exercise}[tags={hw}]
762 What is the difference between an entity type and a weak entity type?
763 \end{exercise}
764
765 \begin{solution}
766 The weak entity type doesn't have a primary key.
767 \end{solution}
768
769 \begin{exercise}[tags={hw}]
770 What is the degree of a relationship type?
771 \end{exercise}
772
773 \begin{solution}
774 The number of participating entity types.
775 \end{solution}
776
777 \begin{exercise}[tags={hw}]
778 What is a self-referencing, or recursive, relationship type? Give two examples.
779 \end{exercise}
780
781 \begin{solution}
782 A relationship type where the same entity type participates more than once.
783 On rooms, \enquote{is to the left of}, on persons, \enquote{is married to}.
784 \end{solution}
785
786 \begin{exercise}[tags={hw,qz}]
787 What does it mean for a binary relationship type \enquote{Owner} between entity types \enquote{Person} and \enquote{Computer} to have a cardinality ration $M:N$?
788 \end{exercise}
789
790 \begin{solution}
791 That a person can own multiple computers, and that a computer can have multiple owners.
792 \end{solution}
793
794 \begin{exercise}[tags={hw}]
795 What are the two possible structural constraints on a relationship type?
796 \end{exercise}
797
798 \begin{solution}
799 Cardinality ration and participation constraints.
800 \end{solution}
801
802 \begin{exercise}[tags={hw}]
803 Under what condition(s) can an attribute of a binary relationship type be migrated to become an attribute of one of the participating entity type?
804 \end{exercise}
805
806 \begin{solution}
807 When the cardinality is $1 : N$, $1 : 1$ or $N : 1$.
808 \end{solution}
809
810 \begin{exercise}[tags={hw}]
811 What is a partial key?
812 \end{exercise}
813
814 \begin{solution}
815 For a weak entity attribute, it is the attribute that can uniquely identify weak entites that are related to the same owner entity.
816 \end{solution}
817
818 \begin{exercise}[tags={hw}]
819 For the following binary relationships, suggest cardinality ratios based on the common-sense meaning of the entity types.
820
821 {\centering
822
823 \begin{tabular}{l c l}
824 \textbf{Entity 1} & \textbf{Cardinality Ratio} & \textbf{Entity 2}\\
825 STUDENT & & MAJOR\\ % 1 : N
826 CAR & & TAG \\ % 1 : 1
827 INSTRUCTOR & & LECTURE \\ % N :1
828 INSTRUCTOR & & OFFICE \\ % 1 : N
829 COMPUTER & & OPERATING\_SYSTEM % N : M
830 \end{tabular}
831
832 }
833 \end{exercise}
834
835 \begin{solution}
836 N:1 , 1 : 1, 1 : N , 1:N, M : N
837 \end{solution}
838
839 \begin{exercise}[tags={hw}]
840 Give an example of a binary relationship type of cardinality $1 : N$.
841 \end{exercise}
842
843 \begin{solution}
844 SUPERVISION, a recursive relationship on EMPLOYEE.
845 \end{solution}
846
847 \begin{exercise}[tags={hw}]
848 Give an example of a binary relationship type of cardinality $N:1$, and draw the corresponding diagram (you don't have to include details on the participating entity types).
849 \end{exercise}
850
851 \begin{solution}
852 \begin{tikzpicture}[node distance=7em]
853 \node[relationship] (contains){BELONGS};
854 \node[entity] (burger) [left = 1cm of contains] {HAND} edge node[above] {$N$} (contains);
855 \node[entity] (ingredient) [right = 1cm of contains] {PERSON} edge node[above]{$1$} (contains);
856 \end{tikzpicture}
857
858 \end{solution}
859
860 \begin{exercise}[tags={hw}]
861 Draw an ER diagram with a single entity type, with two stored attributes and one derived attribute. In your answer, it should be clear that the value for the derived attribute will always be obtained from the value(s) for the other attribute(s).
862 \end{exercise}
863
864 \begin{solution}
865
866 \begin{tikzpicture}[node distance=7em]
867 \node[entity] (person) {PERSON};
868 \node[attribute] (ssn) [left of=person] {\key{SSN}} edge (person);
869 \node[attribute] (dob) [above of=person] {Date\_Of\_Birth} edge (person);
870 \node[derived attribute] (age) [right of=person] {Age} edge (person);
871 \end{tikzpicture}
872
873 \end{solution}
874
875 \begin{exercise}[tags={hw}]
876 Draw an ER diagram expressing the total participation of an entity type \enquote{BURGER} in a binary relation \enquote{CONTAINS} between \enquote{BURGER} and \enquote{INGREDIENT}.
877 What would be the ratio of such a relation?
878 \end{exercise}
879
880 \begin{solution}
881 \begin{tikzpicture}[node distance=7em]
882 \node[relationship] (contains){CONTAINS};
883 \node[entity] (burger) [left = 1cm of contains] {BURGER} edge[total] node[above] {$N$} (contains);
884 \node[entity] (ingredient) [right = 1cm of contains] {INGREDIENT} edge node[above]{$M$} (contains);
885 \end{tikzpicture}
886 \end{solution}
887
888 \begin{exercise}[tags={qz}]
889 Convert the following ER diagram into a relational model:
890
891 \begin{tikzpicture}[node distance=7em]
892 \node[relationship] (contains){STAYS\_AT};
893 \node[entity] (person) [above = 1cm of contains] {PERSON} edge node[right] {$N$} (contains);;
894 \node[attribute] (ssn) [left of=person] {\key{SSN}} edge (person);
895 \node[attribute] (dob) [above = .5cm of person] {Date\_Of\_Birth} edge (person);
896 \node[entity] (burger) [below of= contains] {PLACE} edge node[right] {$1$} (contains);
897 \node[attribute] (dob) [below = .5cm of burger] {\key{Address}} edge (burger);
898 \node[attribute] (ssn) [left of=burger] {Rooms} edge (burger);
899 \end{tikzpicture}
900 \end{exercise}
901
902 \begin{solution}
903 \begin{tikzpicture}[node distance=7em]
904 \node[relationship] (contains){CONTAINS};
905 \node[entity] (burger) [left = 1cm of contains] {BURGER} edge[total] node[above] {$N$} (contains);
906 \node[entity] (ingredient) [right = 1cm of contains] {INGREDIENT} edge node[above]{$M$} (contains);
907 \end{tikzpicture}
908 \end{solution}
909
910 \begin{exercise}[tags={hw}]
911 Why do weak entity type have a total participation constraint?
912 \end{exercise}
913
914 \begin{solution}
915 Otherwise, we couldn't identify entities in it without owner entity.
916 \end{solution}
917
918 \IfTagSetF{hw}{
919 \end{document}
920 }
921
922 \section{Part II --- Problems}
923
924 The three exercises from this homework requires some time to be completed.
925 The first one will help you to become better at reading ER diagrams, the second will ask you to design one, and the last one will ask you to apply the algorithm to your ER diagram to obtain a relational model.
926 These three exercises are similar to what you will be asked to do during the second exam.
927
928 I'll assume that you will have successfully completed them by the time Homework \#\the\numexpr\hwNum +1\relax{} is released (\hwDateNext), so don't wait and let me know if you had difficulties solving them.
929
930 \begin{problem}[tags={hw}]
931
932 Consider the ER schema for the MOVIES database in Figure~\ref{fig:movies}.
933
934 \begin{figure}
935 \begin{tikzpicture}
936 \node (a) at (0,0) {\includegraphics[page=1,width=1\textwidth]{img/te.pdf}};
937 \fill[white] (-8.3, 6.5) rectangle (-3.5, 4);
938 \end{tikzpicture}
939 \caption{An ER diagram for a MOVIES database schema.} \label{fig:movies}
940 \end{figure}
941
942 Assume that MOVIES is a populated database.
943 ACTOR is used as a gender-netral term.
944 Given the constraints shown in the ER schema, respond to the following statements with \emph{True} or \emph{False}.
945 Justify each answer.
946
947 \begin{enumerate}
948 \item There are no actors in this database that have been in no movies.
949 \item There might be actors who have acted in more than ten movies.
950 \item Some actors could have done a lead role in multiple movies.
951 \item A movie can have only a maximum of two lead actors.
952 \item Every director have to have been an actor in some movie.
953 \item No producer has ever been an actor.
954 \item A producer cannot be an actor in some other movie.
955 \item There could be movies with more than a dozen actors.
956 \item Producers can be directors as well.
957 \item A movie can have one director and one producer.
958 \item A movie can have one director and several producers.
959 \item There could be some actors who have done a lead role, directed a movie, and produced some movie.
960 \item It is impossible for a director to play in the movie (s)he directed.
961 \end{enumerate}
962
963 \end{problem}
964
965 \begin{answer}[print = true]
966 \begin{enumerate}
967 \item true
968 \item true
969 \item true
970 \item true
971 \item false
972 \item false
973 \item false
974 \item true
975 \item true
976 \item true
977 \item true
978 \item true
979 \item false
980 \end{enumerate}
981 \end{answer}
982
983 \begin{problem}[tags={hw}]
984 Draw the ER diagram for the following situation:
985 A car-insurance company wants to have a database of accidents.
986 An accident involves cars, drivers, and it has several aspects: the moment and place where it took place, the amount of damages, and a (unique) report number.
987 A car has a license, a model, a year, and an owner.
988 A driver has an id, an age, a name, and an address.
989
990 One of the interesting choice is: should \enquote{accident} be an entity type or a relationship type?
991
992 \begin{answer}
993
994 \includegraphics[page=1,width=1\textwidth]{img/p.pdf}
995
996 OR
997
998
999
1000 \begin{tikzpicture}[node distance=7em]
1001 \node[entity] (person) {Person};
1002 \node[attribute] (pid) [left of=person] {\key{id}} edge (person);
1003 \node[attribute] (name) [above left of=person] {Name} edge (person);
1004 \node[attribute] (address) [below left of=person] {address} edge (person);
1005
1006 \node[relationship] (owns) [above right of=person] {Owns} edge node[above, sloped]{$M$} (person);
1007 \node[entity] (car) [above right of=owns] {Car} edge node[above, sloped]{$N$} (owns);
1008 \node[attribute] (licence) [above left of=car] {\key{licence}} edge (car);
1009 \node[attribute] (model) [above of =car] {model} edge (car);
1010 \node[attribute] (year) [above right of =car] {year} edge (car);
1011
1012 \node[relationship] (accident) [right = 5cm of person] {Accident} edge[total] node[above, sloped]{$M$} (person);
1013 \node[attribute] (report_number) [below of = accident] {\key{Report Number}} edge (accident);
1014 \node[attribute] (time) [right of = accident] {Time} edge (accident);
1015 \node[attribute] (place) [below right of = accident] {Place} edge (accident);
1016 \node[attribute] (damage_amount) [above right of = accident] {Damage\_Amount} edge (accident);
1017
1018 \draw (car) edge[total] node[above, sloped]{$N$} (accident);
1019 \end{tikzpicture}
1020 \end{answer}
1021 \end{problem}
1022
1023 \begin{problem}[tags={hw}]
1024 Apply the ER-to-Relation mapping to your ER diagram from the previous problem.
1025 \end{problem}
1026
1027 %%% 05
1028
1029
1030 \begin{exercise}[tags={hw}]
1031 What is insertion anomaly? Give an example.
1032 \end{exercise}
1033
1034 \begin{solution}
1035 When you have to invent a primary key or add a lot of \sqli{NULL} value to be able to add a tuple.
1036 I want to add a room in my DB, but the only place where rooms are listed are as an attribute on a Instructor table, so I have to "fake" an instructor to add a room.
1037 \end{solution}
1038
1039 \begin{exercise}[tags={hw}]
1040 Why should we avoid attributes whose value will often be \sqli{NULL}?
1041 Can the usage of \sqli{NULL} be completely avoided?
1042 \end{exercise}
1043
1044 \begin{solution}
1045 Because they waste space, and because they are ambiguous (N/A, or unknown, or not communicated?).
1046 No, it is necessary sometimes.
1047 \end{solution}
1048
1049 \begin{exercise}[tags={qz}]
1050 Consider the following relation:
1051
1052 \begin{tabular}{p{3em} l}
1053 & PROF(\underline{SSN}, Name, Department, Bike\_brand)
1054 \end{tabular}
1055
1056 Why is it a poor design to have a \enquote{Bike\_brand} attribute in such a relation?
1057 How should we store this information?
1058 \end{exercise}
1059
1060 \begin{solution}
1061 Because it will be \sqli{NULL} most of the time.
1062 In a separate table.
1063 \end{solution}
1064
1065 \begin{exercise}[tags={hw}]
1066 Consider the following relation:
1067
1068 \begin{tabular}{p{3em} l}
1069 & STUDENT(\underline{SSN}, Name, \(\hdots\), Sibling\_On\_Campus)
1070 \end{tabular}
1071
1072 Why is it a poor design to have a \enquote{Sibling\_On\_Campus} attribute in such a relation?
1073 How should we store this information?
1074 \end{exercise}
1075
1076 \begin{solution}
1077 Because it will be \sqli{NULL} most of the time, and because students could have more than one sibling on campus.
1078 In a separate table.
1079 \end{solution}
1080
1081 \begin{exercise}[tags={hw}]
1082 Consider the following relational database schema:
1083
1084 \begin{tabular}{p{3em} l}
1085 & STUDENT(\underline{Login}, Name, \(\hdots\), Major, Major\_Head)\\
1086 & DEPARTMENT(\underline{Code}, Name, Major\_Head)
1087 \end{tabular}
1088
1089 Assuming that \enquote{Major} is a foreign key referencing \enquote{DEPARTMENT.Code}, what is the problem with that schema? How could you address it?
1090 \end{exercise}
1091
1092 \begin{solution}
1093 Major\_Head will give update anomalies.
1094 By putting the Head of the department in the DEPARTMENT relation only, i.e., removing it from STUDENT.
1095 \end{solution}
1096
1097 \begin{exercise}[tags={hw}]
1098 Consider the relation \(R(A , B, C, D, E, F)\) and the following functional dependencies:
1099 \begin{enumerate}
1100 \item \(F \to \{D, C\}, D \to \{B, E\}, \{B, E\} \to A\)
1101 \item \(\{A, B\} \to \{C, D\}, \{B, E\} \to F\)
1102 \item \(A \to \{C, D\}, E \to F, D \to B\)
1103 \end{enumerate}
1104 For each set of functional dependency, give a key for \(R\). We want a key, so it has to be minimal.
1105 \end{exercise}
1106
1107 \begin{solution}
1108 \begin{enumerate}
1109 \item \(F\)
1110 \item \(\{A, B, E\}\)
1111 \item \(\{A, E\}\)
1112 \end{enumerate}
1113 \end{solution}
1114
1115 \begin{exercise}[tags={none}]
1116 Consider the relation \(R(A, B, C, D, E, F)\) and the following functional dependencies:
1117 \[ A \to \{D, E\}, D \to \{B, F\}, \{B, E\} \to A, \{A,C\} \to \{B, D, F\}, A \to F\]
1118 Answer the following:
1119 \begin{enumerate}
1120 \item How many candidate keys is there? List them.
1121 \item How many transitive dependencies can you find? Give them and justify them.
1122 \end{enumerate}
1123 \end{exercise}
1124
1125 \begin{solution}
1126 \begin{enumerate}
1127 \item \(\{A, C\}\),
1128 \item \(A \to F\) by \(A \to D, D\to F\).
1129 \end{enumerate}
1130 \end{solution}
1131
1132 \begin{exercise}[tags={hw}]
1133 Consider the relation \(R(A, B, C, D)\) and answer the following:
1134 \begin{enumerate}
1135 \item If \(\{A, B\}\) is the only key, is \(\{A, B\} \to \{C,D\}, \{B, C\} \to D\) a 2NF? List the nonprime attributes and justify.
1136 \item If \(\{A, B, C\}\) is the only key, is \(A \to \{B, D\}, \{A, B, C\} \to D\) a 2NF? List the nonprime attributes and justify.
1137 \end{enumerate}
1138 \end{exercise}
1139
1140 \begin{solution}
1141 \(1.\) Yes. \(C\) and \(D\) are non prime, and they fully depend on \(\{A,B\}\).
1142
1143 \(2.\) No. \(D\) is the only non prime, and it depends only on \(A\).
1144 \end{solution}
1145
1146 \begin{exercise}[tags={hw}]
1147 Consider the relation \(R(A, B, C, D, E, F)\) with candidate keys \(\{A, B\}\) and \(C\). Answer the following:
1148 \begin{enumerate}
1149 \item What are the prime attributes in \(R\)?
1150 \item Is \(\{C,D\} \to E\) a fully functional dependency?
1151 \item Write a set of functional dependencies containing at least one transitive depency, and justify your answer.
1152 \end{enumerate}
1153 \end{exercise}
1154
1155 \begin{solution}
1156 \begin{enumerate}
1157 \item\(A,B,C\)
1158 \item No, because we can remove \(D\),
1159 \item \(A \to D\), \(D \to E\) and \(A \to E\)
1160 \end{enumerate}
1161 \end{solution}
1162
1163 \begin{exercise}[tags={qz}]
1164 Consider the relation \(R(A , B, C, D, E)\) and the following functional dependencies:
1165 \begin{enumerate}
1166 \item \(C \to D, \{C, B\} \to A, A \to \{B, C, D\}, B \to E\)
1167 \item \(\ A \to \{C, D\}, C \to B, D \to E, \{E, C\} \to A\)
1168 % \item \(\{A, B\} \to D, D \to \{B, C\}, E \to C\)
1169 \end{enumerate}
1170 For each one, give one candidate key for \(R\).%two candidate keys for \(R\).
1171 \end{exercise}
1172
1173 \begin{solution}
1174 \begin{enumerate}
1175 \item \(\{B, C\}\), \(A\)
1176 \item \(A\), \(\{C,E\}\),
1177 \item \(\{A, D, E\}\), \(\{A, B, E\}\)
1178 \end{enumerate}
1179 \end{solution}
1180
1181 \begin{exercise}[tags={qz}]
1182 Consider the relation \(R(A, B, C, D, E)\) and answer the following:
1183 \begin{enumerate}[topsep=0pt,itemsep=10em,partopsep=1ex,parsep=1ex]
1184 \item If \(\{A, B\}\) is the primary key, is \(B \to E, C \to D\) a 2NF?% List the nonprime attributes and justify.
1185 \item If \(\{A\}\) is the primary key, is \(B \to C, B \to D\) a 2NF?% List the nonprime attributes and justify.
1186 \end{enumerate}
1187 \end{exercise}
1188
1189 \begin{solution}
1190 \(1.\) No. \(C, D, E\), and \(E\) has a partial relation to \(B\)
1191
1192 \(2.\) Yes. Since the primary key is a singleton, it is obvious.
1193 \end{solution}
1194
1195 \begin{exercise}[tags={qz}]
1196 Consider the relation \(R(A, B, C, D, E, F)\), and let \(\{B, D\}\) be the primary key, and have additionnaly the functional dependencies \(\{A, D\} \to E, C \to F\).
1197 This relation is not in 3NF, can you tell why?% List the nonprime attributes and justify.
1198 % \item If \(A\) is the primary key, is \(\{A, B\} \to C, \{A, B, C\} \to E\) a 3NF?% List the nonprime attributes and justify.
1199 \end{enumerate}
1200 \end{exercise}
1201
1202 \begin{solution}
1203 \(1.\) No. \(A, C, E, F\) are non-prime, \(\{B, D\} \to C \to F\) breaks the 3NF.
1204
1205 \(2.\) Yes. \(B, C, E, F\), \(\{A,B\}\) and \(\{A, B, C\}\) are not strict subset of the primary key.
1206 \end{solution}
1207
1208 \begin{exercise}[tags={hw}]
1209 Consider the relation \(R(A, B, C, D)\) and answer the following:
1210 \begin{enumerate}
1211 \item If \(A\) is the only key, is \(A \to \{B,C,D\}, \{A, B\} \to C, \{B, C\} \to D\) a 3NF? List the nonprime attributes and justify.
1212 \item If \(B\) is the only key, is \(B \to \{A, C, D\}, A \to \{C, D\}, \{A, C\} \to D\) a 3NF? List the nonprime attributes and justify.
1213 \end{enumerate}
1214 \end{exercise}
1215
1216 \begin{solution}
1217 \(1.\) No. \(B\), \(C\) and \(D\) are non prime, \(A \to \{B,C\} \to D\) breaks the 3NF.
1218
1219 \(2.\) No. \(A\), \(B\) and \(D\) are non prime, \(B \to \{A,C\} \to D\) breaks the 3NF.
1220 \end{solution}
1221
1222
1223 \IfTagSetF{hw}{
1224 \end{document}
1225 }
1226
1227 \section{Part II --- Problems}
1228
1229 This part will help you in assessing your level of understanding of this lecture, and give you an idea of the kind of problem you will be asked to solve during the exams.
1230 I'll assume that you will have successfully completed those two problems by the time Homework \#\the\numexpr\hwNum +1\relax{} is released (\hwDateNext), so don't wait and let me know if you had difficulties solving them.
1231
1232 \begin{problem}[tags={none}]
1233 Consider the following relation, and its functional dependencies:
1234
1235 \begin{tabular}{l}
1236 CAR\_SALE(\underline{Car\_no}, Date\_sold, \underline{Salesman\_no}, Commission, Discount\_amt)
1237 \end{tabular}
1238
1239 \begin{tabular}{rcl}
1240 \{Car\_no, Salesman\_no\} & \(\to\)& \{Date\_sold, Commission, Discount\_amt\}\\
1241 Date\_sold & \(\to\)& Discount\_amt\\
1242 Salesman\_no & \(\to\)& Commission
1243 \end{tabular}
1244
1245 Based on the given primary key, is this relation in 1NF, 2NF, or 3NF? Why or why not? Normalize it completely. Then draw an ER diagram for the resulting schema.
1246 \end{problem}
1247
1248 \begin{answer}
1249 This relation satisfies 1NF but not 2NF (Car\_no \(\to\) Date\_sold and Salesman\_no \(\to\) Commission so these two attributes are not fully functional dependent on the primary key) and not 3NF.
1250
1251 To normalize,
1252
1253 2NF:
1254 \begin{tabular}{l}
1255 Car\_Sale1(Car\_no, Date\_sold, Discount\_amt)\\
1256 Car\_Sale2(Car\_no, Salesman\_no)\\
1257 Car\_Sale3(Salesman\_no,Commission)
1258 \end{tabular}
1259 3NF:
1260 \begin{tabular}{c}
1261 Car\_Sale1-1(Car\_no, Date\_sold)\\
1262 Car\_Sale1-2(Date\_sold, Discount\_amt)\\
1263 Car\_Sale2(Car\_no, Salesman\_no)\\
1264 Car\_Sale3(Salesman\_no,Commission)
1265 \end{tabular}
1266 \end{answer}
1267
1268 \begin{problem}[tags={hw}]
1269 This problem asks you to convert business statements into dependencies.
1270 Consider the following relation:
1271 \begin{center}
1272 BIKE(Serial\_no, Manufacturer, Model, Batch, Wheel\_size, Retailer)
1273 \end{center}
1274
1275 Each tuple in the relation BIKE contains information about a bike with a serial number, made by a manufacturer, with a particular model number, released in a certain batch, which has a certain wheel size, and is sold by a certain retailer.
1276 \begin{itemize}
1277 \item Write each of the following dependencies as a functional dependency (I give you the first one as an example):
1278
1279 \begin{enumerate}
1280 \item A retailer can't have two bikes of the same model from different batches.
1281
1282 \emph{solution:} \{Retailer, Model\} \(\to\) Batch
1283 \item The manufacturer and serial number uniquely identifies the bike and where it is sold.
1284 \item A model number is registered by a manufacturer and therefore can't be
1285 used by another manufacturer.
1286 \item All bikes in a particular batch are of the same model.
1287 \item All bikes of a certain model have
1288 the same wheel size.
1289 \end{enumerate}
1290 \item Based on those statements, what could be a key for this relation?
1291 \item Assuming all those functional dependencies hold, and taking the primary key you identified at the previous step, what is the degree of normality of this relation? Justify your answer.
1292 \end{itemize}
1293 \end{problem}
1294
1295 \begin{answer}
1296 \begin{itemize}
1297 \item
1298 \begin{enumerate}
1299 \item \{ Manufacturer, Serial\_no \} \(\to\) \{ Model, Batch, Wheel\_size, Retailer\}
1300 \item Model \(\to\) Manufacturer
1301 \item Batch \(\to\) Model
1302 \item \{Model, Manufacturer\} \(\to\) Wheel\_size
1303 \end{enumerate}
1304 \item \{Manufacturer, Serial\_no \}
1305 \item If every attribute is atomic, it is in second nf.
1306 \{ Manufacturer, Serial\_no \} \(\to\) Batch \(\to\)Model breaks the 3NF.
1307
1308 \end{itemize}
1309 \end{answer}
1310
1311
1312 \begin{problem}[tags={hw}]
1313 Consider the relations \(R\) and \(T\) below, and their functional dependencies (on top of the one induced by the primary keys):
1314
1315 \begin{tabular}{l}
1316 R(\underline{EventId}, \underline{Email}, Time, Date, Location, Status)\\
1317 T(\underline{Invno}, Subtotal, Tax, Total, Email, Lname, Fname, Phone)
1318 \end{tabular}
1319
1320 \begin{tabular}{r c l}
1321 \{EventId, Email\} & \(\to\) & Status\\
1322 EventId & \(\to\) & \{Time, Date, Location\}\\
1323 Invno & \(\to\) & \{Subtotal, Tax, Total, Email\}\\
1324 Email & \(\to\) & \{Fname, Lname, Phone\}
1325 \end{tabular}
1326
1327 Normalize the relations to 2NF and 3NF. Show all relations at each stage (2NF and 3NF) of the normalization process.
1328 \end{problem}
1329
1330 \begin{answer}
1331 TO BE WRITTEN
1332 \end{answer}
1333
1334
1335 \begin{problem}[tags={hw}]
1336 Consider the following relation for published books:
1337 \begin{center}
1338 BOOK (Book\_title, Book\_type, Author\_name, List\_price, Author\_affil, Publisher)
1339 \end{center}
1340
1341 Suppose we have the following dependencies:
1342
1343 \begin{center}
1344 \begin{tabular}{r c l}
1345 Book\_title & \(\to\) & \{ Publisher, Book\_type \}\\
1346 Book\_type & \(\to\) & List\_price\\
1347 Author\_name & \(\to\) & Author\_affil
1348 \end{tabular}
1349 \end{center}
1350
1351 \begin{itemize}
1352 \item What would be a suitable key for this relation?
1353 \item How could this relation not be in first normal form? Explain your answer.
1354 \item This relation is not in second normal form: explain why and normalize it.
1355 \item Is the relations you obtained at the previous step in third normal form? Explain why, and normalize them if needed.
1356 \end{itemize}
1357 \end{problem}
1358
1359
1360 \begin{answer}
1361 \begin{itemize}
1362 \item \{Book Title, Author Name\}
1363 \item If an attribute is composite or multi-valued.
1364 \item Because of \{ Book\_title \}\(\to\) \{ Publisher, Book\_type \}.
1365 We can normalize it as
1366 (Book Title, Publisher, Book Type, List Price), (Author Name, Author Affiliation), (Author Name, Book Title).
1367 \item Because of \{Book title\} \(\to\) \{ Book\_type\} \(\to\) \{ List\_price\}\\
1368 (Book Title, Publisher, Book Type) and (Book Type, List Price), (Author Name, Author Affiliation), (Author Name, Book Title).
1369
1370 \end{itemize}
1371 \end{answer}
1372
1373 %%% 06
1374
1375
1376 \begin{exercise}[tags={hw}]
1377 Consider the relation \(R(A, B, C, D, E)\) and the functional dependencies \( \{A, B\} \to C, B \to D, C \to E\).
1378 Answer the following:
1379
1380 \begin{enumerate}
1381 \item \(A\) by itself is not a primary key, but what is the only key that contains \(A\)?
1382 \item List the non-prime attributes.
1383 \item This relation is not in 2NF: what transformation can you operate to obtain a 2NF?
1384 \item One of the relation you obtained at the previous step is likely not to be in 3NF.
1385 Can you normalize it?
1386 If yes, how?
1387 \end{enumerate}
1388 \end{exercise}
1389
1390 \begin{solution}
1391 \(\{A, B\}\),
1392
1393 \(C, D, E\),
1394
1395 \(R_1(A, B, C, E)\) and \(R_2(B, D)\)
1396
1397 \(R_1(A, B, C)\), \(R_2(C, E)\) and \(R_3(B, D)\)
1398 \end{solution}
1399
1400
1401 \begin{exercise}[tags={hw,qz}]
1402 What are the two different categories of U.M.L. diagram?
1403 \end{exercise}
1404
1405 \begin{solution}
1406 Behaviour and structure
1407 \end{solution}
1408
1409 \begin{exercise}[tags={hw,qz}]
1410 Can a \texttt{C++} developer working on Linux and a \texttt{Java} developer working on MacOS use the same class diagram as a basis to write their programs?
1411 Justify your answer.
1412 \end{exercise}
1413
1414 \begin{solution}
1415 Yes, U.M.L. diagram is language-independent and platform-independent.
1416 \end{solution}
1417
1418 \begin{exercise}[tags={hw}]
1419 What kind of diagram should we use if we want to \(\hdots\)
1420 \begin{enumerate}
1421 \item describe the functional behavior of the system as seen by the user?
1422 \item capture the flow of messages in a software?
1423 \item represent the workflow of actions of an user?
1424 \end{enumerate}
1425 \end{exercise}
1426
1427 \begin{solution}
1428 \begin{enumerate}
1429 \item Use-case
1430 \item Sequence diagram
1431 \item Activity diagram
1432 \end{enumerate}
1433 \end{solution}
1434
1435
1436 \begin{exercise}[tags={hw,qz}]
1437 Name two reasons why one would want to use a U.M.L. class diagram over an E.-R. diagram to represent a conceptual schema.
1438 \end{exercise}
1439
1440 \begin{solution}
1441 To use direction for association, to have a common language with someone less knowledgeable of other diagrammatic notations.
1442 For the concept of integration.
1443 \end{solution}
1444
1445
1446
1447 \begin{exercise}[tags={hw}]
1448 Consider the following diagram:
1449
1450 \begin{tikzpicture}[scale = 0.8]
1451 \begin{class}[text width=7cm]{Flight}{0,0}
1452 \attribute{flightNumber : Integer}
1453 \attribute{departureTime : Date}
1454 \attribute{flightDuration : Minutes}
1455 \attribute{departingAirport : String}
1456 \attribute{arrivingAirport : String}
1457 \operation{delayFlight(numberOfMinutes : Minutes)}
1458 \operation{getArrivalTime( ) : Date}
1459 \end{class}
1460 \begin{class}{Plane}{11, -0.8}
1461 \attribute{airPlaneType : String}
1462 \attribute{maximumSpeed : MPH}
1463 \attribute{maximumDistance : Miles}
1464 \attribute{tailID : String}
1465 \end{class}
1466 \association{Plane}{}{0..1}{Flight}{0..*}{}
1467 \end{tikzpicture}
1468
1469 Give the number of attributes for both classes, and suggest two operations for the class that doesn't have any.
1470 Discuss the multiplicities: why did the designer picked those values?
1471 \end{exercise}
1472
1473 \begin{solution}
1474 \(5\) and \(4\).
1475
1476 getLastFlightNumber() : Integer
1477
1478 stMaximumSpeed(MPH) : void
1479
1480 A flight could be assigned to no plane, and a plane could not be assigned to a flight.
1481
1482 A plane can be assigned to multiple (or no) flights, but a flight must have at most one plane (and could have none).
1483 \end{solution}
1484
1485 \begin{exercise}[tags={hw}]
1486 Briefly explain the difference between an aggregation and a composition association.
1487 \end{exercise}
1488
1489 \begin{solution}
1490 Aggregation: associated class can have an existence of its own.
1491
1492 Composition association: class doesn't exist without the association.
1493 \end{solution}
1494
1495 \begin{exercise}[tags={hw}]
1496 How is generalization (or inheritance) represented in a U.M.L. class diagram?
1497 Why is such a concept useful?
1498 \end{exercise}
1499
1500 \begin{solution}
1501 \tikz{\draw[>=open triangle 60,->] (0,0) to (1,0);}
1502 Because it avoids redundancy.
1503 \end{solution}
1504
1505 \begin{exercise}[tags={qz}]
1506 Convert the following E.R. diagram into a U.M.L. class diagram:
1507
1508 \begin{tikzpicture}[node distance=8em]
1509 \node[entity] (person) {PILOT};
1510 \node[attribute] (pid) [left of=person] {\key{ID}} edge (person);
1511 \node[attribute] (name) [above left of=person] {Name} edge (person);
1512 \node[attribute] (phone) [above of=person] {Experience} edge (person);
1513
1514 \node[relationship] (drives) [below of=person] {ASSIGNED\_TO} edge node[right, pos=0.1] {$N$} (person);
1515 \node[entity] (plane) [below of=drives] {PLANE} edge [total] node[right, pos=0.7] {$1$} (drives);
1516 \node[attribute] (make) [left of=plane] {\key{TailID}} edge (plane);
1517 \node[attribute] (maxS) [below left of =plane] {MaxSpeed} edge (plane);
1518 \node[attribute] (airPlaneType) [below of = plane] {AirPlaneType} edge (plane);
1519 \end{tikzpicture}
1520 \end{exercise}
1521
1522 \IfTagSetF{hw}{
1523 \vspace{-8em} % ugly hack…
1524 \end{document}
1525 }
1526
1527 \section{Part II --- Problems}
1528
1529 This part will help you in assessing your level of understanding of this lecture, prepare you for the rest of your cursus, and give you an idea of the kind of problem you will be asked to solve during the exams.
1530 I'll assume that you will have successfully completed those problems by the time Homework \#\the\numexpr\hwNum +1\relax{} is released (\hwDateNext), so don't wait and let me know if you had difficulties solving them.
1531
1532 \begin{problem}[tags={hw}]
1533 \label{probMYSQLW}
1534 Consider the following E.R. schema for the CAR\_INFO database:
1535
1536 \begin{tikzpicture}[node distance=8em]
1537 \node[entity] (person) {PERSON};
1538 \node[attribute] (pid) [left of=person] {\key{ID}} edge (person);
1539 \node[attribute] (name) [above left of=person] {Name} edge (person);
1540 \node[multi attribute] (phone) [above of=person] {Phone} edge (person);
1541 \node[attribute] (address) [above right of=person] {Address} edge (person);
1542 \node[attribute] (street) [above right of=address] {Street} edge (address);
1543 \node[attribute] (city) [right of=address] {City} edge (address);
1544 % \node[derived attribute] (age) [right of=person] {Age} edge (person);
1545
1546 \node[relationship] (drives) [below right of=person] {DRIVES} edge node[above, pos=0.1] {$1$} (person);
1547 \node[entity] (car) [below left of=drives] {CAR} edge node[above, pos=0.7] {$1$} (drives);
1548 \node[attribute] (make) [left of=car] {Make} edge (car);
1549 \node[attribute] (year) [below left of =car] {Year} edge (car);
1550 \node[attribute] (brand) [below of =car] {Brand} edge (car);
1551
1552 \node[relationship] (seats) [below left of=person] {SEATS\_IN} edge node[above, pos=0.1] {$N$} (person);
1553 \draw (seats) edge node[above, pos=0.3] {$1$} (car);
1554 \node[attribute] (position) [left of=seats] {Position} edge (seats);
1555
1556 \node[ident relationship] (insured) [right of=car] {INSURED} edge node[above, pos=0.3] {$1$} (car);
1557 \node[weak entity] (insurance) [right = 1cm of insured] {CAR\_INSURANCE} edge[total] node[above, pos=0.7] {$N$} (insured);
1558 \node[attribute] (amount) [above of =insurance] {Covered Amount} edge (insurance);
1559 \node[attribute] (policy) [above right of =insurance] {Policy Number} edge (insurance);
1560 \node[attribute] (company) [below right of =insurance] {Company Name} edge (insurance);
1561 \end{tikzpicture}
1562
1563 Note that a car can have at most one driver, \(N\) passengers, \(N\) insurances, and that car insurances exist only if they are \enquote{tied up} to a car (i.e., they are weak entities, and their identifying relationship is called \enquote{Insured}).
1564
1565 \begin{enumerate}
1566 \item Find the key attribute for \enquote{Car}, and the partial key for \enquote{Car Insurance}. If you can't think of any, add a dummy attribute and make it be the key.
1567 \item Convert that E.-R. diagram to a relational database schema. \label{ERtoREL}
1568 \item Convert the E.-R. diagram to a U.M.L. class diagram. Comparing Figure 7.16 with Figure 7.2 from your textbook should guide you. \label{ERtoUML}
1569 \end{enumerate}
1570 \end{problem}
1571
1572
1573 \begin{answer}
1574 For \enquote{Car}, we need to create an attribute, like \enquote{vin}.
1575 For \enquote{Car Insurance}, \enquote{Policy Number} is perfect.
1576
1577 \colorlet{lightgray}{gray!20}
1578
1579 \begin{tikzpicture}[relation/.style={rectangle split, rectangle split parts=#1, rectangle split part align=base, draw, anchor=center, align=center, text height=3mm, text centered}]\hspace*{-0.3cm}
1580
1581 % RELATIONS
1582
1583 \node (phonetitle) {\textbf{PHONE}};
1584
1585 \node [relation=2, rectangle split horizontal, rectangle split part fill={lightgray!50}, anchor=north west, below=0.6cm of phonetitle.west, anchor=west] (phone)
1586 {\underline{id}%
1587 \nodepart{two} \underline{number}
1588 };
1589
1590 \node [below=1.3cm of phone.west, anchor=west] (persontitle) {\textbf{PERSON}};
1591
1592 \node [relation=6, rectangle split horizontal, rectangle split part fill={lightgray!50}, below=0.6cm of persontitle.west, anchor=west] (person)
1593 {\underline{id}%
1594 \nodepart{two} Name
1595 \nodepart{three} Street
1596 \nodepart{four} City
1597 \nodepart{five} Seat
1598 \nodepart{six}{Position}
1599 };
1600
1601 \node [below=1.1cm of person.west, anchor=west] (cartitle) {\textbf{CAR}};
1602
1603 \node [relation=5, rectangle split horizontal, rectangle split part fill={lightgray!50}, anchor=north west, below=0.6cm of cartitle.west, anchor=west] (car)
1604 {\underline{Vin}%
1605 \nodepart{two} Make
1606 \nodepart{three} Year
1607 \nodepart{four} Brand
1608 \nodepart{five} Driver
1609 };
1610
1611 \node [below=1.4cm of car.west, anchor=west] (carinsurancetitle) {\textbf{CAR INSURANCE}};
1612
1613 \node [relation=4, rectangle split horizontal, rectangle split part fill={lightgray!50}, anchor=north west, below=0.6cm of carinsurancetitle.west, anchor=west] (carinsurance)
1614 {\underline{Policy Number}%
1615 \nodepart{two} Covered Amount
1616 \nodepart{three} Company Name
1617 \nodepart{four} Insured Car
1618 };
1619
1620 % FOREIGN KEYS
1621
1622 \draw[-latex] (phone.one south) -- ++(0,-0.2) --++(7,0) --++(0, -2) -| ($(person.one south) + (0.2,0)$);
1623 %\draw[-latex] (person.five south) -- ++(0,-0.2) --++(3,0) --++(0, -2) -| ($(car.one south) + (-0.2,0)$);
1624 \draw[-latex] (carinsurance.four south) -- ++(0,-0.2) --++(2,0) --++(0, 1.75) -| ($(car.one south) + (0.1,0)$);
1625 \draw[-latex] (car.five south) -- ++(0,-0.2) --++(2,0) --++(0, 1.5) -| ($(person.one south) + (-0.1,0)$);
1626 \end{tikzpicture}
1627
1628
1629 \begin{tikzpicture}[scale = 0.8]
1630 \node (person) at (0,0){\begin{tabular}{| r l |}
1631 \hline
1632 \multicolumn{2}{| c |}{\textbf{Person}}\\
1633 \hline
1634 id &: String\\
1635 name &: String\\
1636 address &: Street\\
1637 & City\\
1638 \hline
1639 getAge()&: Int\\
1640 \hline
1641 \end{tabular}
1642 };
1643 \node (phone) at (8, 1){\begin{tabular}{| r l |}
1644 \hline
1645 \multicolumn{2}{| c |}{\textbf{Phone}}\\
1646 \hline
1647 number &: String\\
1648 \hline
1649 \end{tabular}
1650 };
1651 \draw[open diamond-] (person) -- node[above, pos=0.1]{\(0..\star\)} node[above, pos=0.9]{\(1..1\)} (phone);
1652 \node (car) at (8, -3){\begin{tabular}{| r l |}
1653 \hline
1654 \multicolumn{2}{| c |}{\textbf{Car}}\\
1655 \hline
1656 vin &: String\\
1657 make &: String\\
1658 year &: Year\\
1659 brand &: String\\
1660 \hline
1661 \end{tabular}
1662 };
1663 \draw[-] ($(person)+(2.5,-0.5)$) -- node[above, pos=0.1]{\(0..1\)} node[above, pos=0.5]{drives} node[above, pos=0.9]{\(0..1\)} (car);
1664 %\draw[-] (person) -- node[below, pos=0.1]{\(0..1\)} node[below, pos=0.9]{\(0..4\)} ($(car)+(-2.5, 0.5)$);
1665 \node (seat) at (3.5, -5){\begin{tabular}{| r l |}
1666 \hline
1667 \multicolumn{2}{| c |}{\textbf{Seats In}}\\
1668 \hline
1669 position &: String\\
1670 \hline
1671 \end{tabular}
1672 };
1673 \draw[dashed] (seat) -- (3.5, -1.6);
1674
1675 \node (insu) at (14.3, -2){\begin{tabular}{| r l |}
1676 \hline
1677 \multicolumn{2}{| c |}{\textbf{Car Insurance}}\\
1678 \hline
1679 policy number &: String\\
1680 covered amount &: int\\
1681 compagny name &: String\\
1682 \hline
1683 \end{tabular}
1684 };
1685
1686 \draw[diamond -] (car) -- node[above, pos=0.1]{\(0..\star\)} node[above, pos=0.9]{\(1\)} (insu);
1687 \end{tikzpicture}
1688
1689 \end{answer}
1690
1691 \begin{problem}[tags={hw}]
1692 \label{probERtoREL}
1693 In this exercise, we will install and explore the basic functionalities of MySQL Workbench, which is a cross-platform, open-source, and free graphical interface for database design.
1694 \begin{enumerate}
1695 \item Install MySQL Workbench: use your package manager, or download the binaries from \url{https://dev.mysql.com/downloads/workbench/}.
1696 \item Once installed, execute the software. Under the panel \enquote{MySQL Connections}, you should see your local installation listed as \enquote{Local instance 3306}. Click on the top-right corner of that box, and then on \enquote{Edit Connections}. Alternatively, click on \enquote{Database}, on \enquote{Manage Connections}, and then on \enquote{Local instance 3306}.
1697 \item Check that all the parameters are correct. Normally, you only have to change the name of the user to \enquote{testuser}, and leave the rest as it is. Click on \enquote{Test the connection}, and enter your password (which should be \enquote{password}) when prompted. If you receive a warning about \enquote{Incompatible/nonstandard server version or connection protocol detected}, click on \enquote{Continue anyway}.
1698 \item Now, click on the box \enquote{Local instance 3306}, and enter your password. A new tab appears, you can see the list of schemas in the bottom part of the left panel.
1699 \item Click on \enquote{Database}, and then on \enquote{Reverse Engineering} (or hit \texttt{ctrl} + \texttt{r}), click on \enquote{next}, enter your password, and click on \enquote{next}. You should see the list of the schemas stored in your database. Select one (any one, we are just exploring the functionalities at that point), click on \enquote{next}, and then click on \enquote{execute}, \enquote{next}, and \enquote{close}.
1700 \item You're back on the previous view, but you should now see \enquote{E.E.R. diagram} on the top of the middle panel. Click on \enquote{E.E.R. diagram} twice, scroll down if needed, and you should see the E.E.R. diagram.
1701 \item This diagram isn't exaclty an E.-R. diagram, and it's not a U.M.L. diagram either. Yet, you should still be able to understand parts of it, and should try to modify it. Make some relations mandatory, change their name, add an attribute, change the name of another, insert a couple of elements in an entity, add a row in a table, etc. Make sure you understand the meaning of the lines between the entities.
1702 \item Once you're done, try to \enquote{Forward Engineer} by hitting \enquote{Ctrl} + \enquote{G}. Click on \enquote{next} twice, enter your password, click on lick on \enquote{next} once more, and you should see the \texttt{SQL} code needed to produce the table you just designed using the graphical tool.
1703 \end{enumerate}
1704 \end{problem}
1705
1706 \begin{problem}[tags={hw}]
1707 \emph{This problem requires you to have successfully completed Problem~\ref{probMYSQLW} and Problem~\ref{probERtoREL}.}
1708
1709 Using the relational database schema you obtained in question~\ref{ERtoREL} of Problem~\ref{probERtoREL}, write the \texttt{SQL} implementation of that database.
1710 Then, using MySQL Workbench, use the \enquote{Reverse Engineering} function to obtain a E.E.R. diagram of your database, and compare it with the U.M.L. diagram you draw in question~\ref{ERtoUML} of Problem~\ref{probERtoREL}.
1711 Apart from the difference inherent to the nature of the diagram (i.e., U.M.L. Vs E.E.R.), how are they the same? How do they differ? Is the automated tool as efficient and accurate as you are?
1712 \end{problem}
1713
1714 \begin{answer}
1715 \inputminted{sql}{include/sol.sql}
1716 \includegraphics{include/mysql_workbench_drawing-crop}
1717 \end{answer}
1718
1719 %%% 07
1720
1721
1722 \begin{exercise}[tags={hw}]
1723 What are the technologies that makes it possible for a Java application to communicate with a DBMS?
1724 \end{exercise}
1725
1726 \begin{solution}
1727 API + driver
1728 \end{solution}
1729
1730 \begin{exercise}[tags={hw}]
1731 What JDBC method do you call to get a connection to a database?
1732 \end{exercise}
1733 \begin{solution}
1734 DriverManager.getConnection()
1735 \end{solution}
1736
1737 \begin{exercise}[tags={hw}]
1738 Briefly explain what the \texttt{next()} method from the \texttt{ResultSet} class does, and what is its return type.
1739 \end{exercise}
1740 \begin{solution}
1741 It checks if there is data to read, and if move the cursor reads it.
1742 It returns a Boolean.
1743 \end{solution}
1744
1745 \begin{exercise}[tags={hw}]
1746 How do you submit a \mintinline{SQL}{SELECT} statement to the DBMS?
1747 \end{exercise}
1748
1749 \begin{solution}
1750 Using \mintinline{java}{.executeQuery(strSelect)}
1751 \end{solution}
1752
1753 \begin{exercise}[tags={hw}]
1754 Where is a ResultSet object's cursor initially pointing? How do you move the cursor forward in the result set?
1755 \end{exercise}
1756
1757 \begin{solution}
1758 Before the first line.
1759 Unsing the \mintinline{java}{next()} method.
1760 \end{solution}
1761
1762 \begin{exercise}[tags={hw}]
1763 Give three navigation methods provided by \mintinline{Java}{ResultSet}.
1764 \end{exercise}
1765
1766 \begin{solution}
1767 \mintinline{java}{first()}
1768 \mintinline{java}{last()}
1769 \mintinline{java}{next()}
1770 \mintinline{java}{previous()}
1771 \mintinline{java}{relative(rows)}
1772 \mintinline{java}{absolute(row} method.
1773 \end{solution}
1774
1775 \begin{exercise}[tags={hw}]
1776 Explain this JDBC URL format:
1777
1778 \mintinline[breaklines=true]{Java}{jdbc:mysql://localhost:3306/HW_NewDB?createDatabaseIfNotExist=true&useSSL=true}
1779 \end{exercise}
1780
1781 \begin{solution}
1782 Connect to localhost:3306 and create a new database if needed, and use secure connection.
1783 \end{solution}
1784
1785 \begin{exercise}[tags={hw}]
1786 In what class is the \mintinline{java}{getColumnName()} method?
1787 \end{exercise}
1788
1789 \begin{solution}
1790 ResultSetMetaData
1791 \end{solution}
1792
1793 \begin{exercise}[tags={hw}]
1794 What is a prepared statement?
1795 \end{exercise}
1796
1797 \begin{solution}
1798 A prepared statement is a feature used to execute the same (or similar) SQL statements repeatedly with high efficiency.
1799 \end{solution}
1800
1801 \begin{exercise}[tags={qz}]
1802 In the code below, there are \(5\) errors between line \(11\) and line \(30\). They are \emph{not} subtle Java errors (like misspelling a key word) and do not come from the DBMS (so you should assume that the password is correct, that the database exists, etc.). For each error, highlight it precisely and give a short explanation.
1803
1804 \begin{minted}[linenos, breaklines=true, baselinestretch=1.4]{Java}
1805 import java.sql.*;
1806
1807 public class MyProg{
1808 public static void main(String[] args) {
1809 try (
1810 Connection conn = DriverManager.getConnection("jdbc:mysql://localhost:3306/"
1811 +"HW_TestDB?user=testuser&password=password");
1812 Statement stmt = conn.createStatement();
1813 ) {
1814
1815 /* Errors after this point.*/
1816
1817 String strSelect = "SELECT title WHERE qty > 40 FROM disks;";
1818 ResultSet rset = stmt.executeUpdate(strSelect);
1819
1820 System.out.println("The records selected are: (listed last first):");
1821 rset.last();
1822
1823 while(rset.previous()) {
1824 String title = rset.getDouble("title");
1825 System.out.println(title + "\n");
1826 }
1827
1828 String sss = "SELECT title FROM disks WHERE Price <= ?";
1829 PreparedStatement ps = conn.prepareStatement(sss);
1830 ResultSet result = ps.executeQuery();
1831
1832 conn.close();
1833
1834 /* Errors before this point.*/
1835
1836 } catch(SQLException ex) {
1837 ex.printStackTrace();
1838 }
1839 }
1840 }
1841 \end{minted}
1842 \end{exercise}
1843
1844 %
1845 %METADATA
1846 %
1847 %ResultSet is array, other?
1848 %
1849 %Multiple queries
1850 %
1851 %qet datatype
1852 %
1853 %prepared statemernt.
1854 %
1855 %http://tutorials.jenkov.com/jdbc/resultset.html
1856
1857 \IfTagSetF{hw}{
1858 \vspace{-8em} % ugly hack…
1859 \end{document}
1860 }
1861
1862 \section{Part II --- Problems}
1863
1864
1865 This part will mainly hands on, and ask you to read code.
1866 It is important that you learn how to get a working environment for a new technology quickly, for your understanding of this lecture, for the exam, and as a CS major.
1867 I'll assume that you will have successfully completed those tasks by \hwDateNext, so don't wait and let me know if you had difficulties solving them.
1868
1869 \subsection*{References}\label{ref}
1870 \begin{itemize}
1871 \item This second part takes some inspiration from \url{https://www.ntu.edu.sg/home/ehchua/programming/java/JDBC_Basic.html}. If you experience troubles, \url{https://www.ntu.edu.sg/home/ehchua/programming/howto/ErrorMessages.html#JDBCErrors} might be a good read.
1872 \item Section 13.3.2 of your textbook is a condensed, but good read. Many textbook on Java includes a part on Databases, cf. for instance the Chapter 16 of \emph{Starting Out with Java: Early Objects} (5th Edition) by Tony Gaddis.
1873 \end{itemize}
1874
1875
1876 \begin{problem}[tags={hw}]
1877 \label{qu:batch}
1878 In the archive, navigate to \texttt{code/sql/}, open and read \texttt{HW\_ebookshop.sql}.
1879
1880 Then, open a terminal (or command-line interpreter), navigate to the folder where you stored that file (using \mintinline{bash}{cd}), and type
1881
1882 \begin{minted}{bash}
1883 mysql -u testuser -p < HW_ebookshop.sql
1884 \end{minted}
1885
1886 for linux, or (something like)
1887
1888 \begin{minted}[breaklines=true]{bash}
1889 "C:\Program Files\MySQL\MySQL Server 5.7\bin\mysql.exe" -u testuser -p < HW_ebookshop.sql
1890 \end{minted}
1891 for Windows.
1892
1893 You just discovered MySQL's batch mode, that perform \emph{series} of instructions from a file.
1894 You can easily make sure that the database and the table were indeed created, and the values inserted.
1895 \end{problem}
1896
1897 \begin{problem}[tags={hw}]
1898 This exercise supposes you successfully completed Problem~\ref{qu:batch}.
1899 We will compile and execute your first database application, using Java and MySQL.
1900 \begin{itemize}
1901 \item I will assume that you have MySQL installed and set-up as indicated in Homeworks \#1 and \#2.
1902 \item I will assume that you have Java installed. If not, please refer to \url{http://spots.augusta.edu/caubert/teaching/general/java/} for a simple program and the instructions to compile and execute it.
1903 \item We need to set up the \emph{driver} (or \emph{connector}) to make the java \texttt{sql} API and MySQL communicate. To do so,
1904 \begin{itemize}
1905 \item Go to \url{https://dev.mysql.com/downloads/connector/j/}
1906 \item Click on \enquote{Download} in front of \enquote{Platform Independent (Architecture Independent), ZIP Archive}
1907 \item Look for the (somewhat hidden) \enquote{No thanks, just start my download.}
1908 \item You will download a file named \enquote{mysql-connector-java-***.zip}, where \texttt{***} is the version number.
1909 \item Upon completion of the download, unzip the file, and locate the \enquote{mysql-connector-java-***-bin.jar} file.
1910 \item Copy that file in \texttt{code/java/}.
1911 \end{itemize}
1912 \item Open a terminal in that same folder, and compile \texttt{FirstProg.java}, using
1913 \begin{minted}{bash}
1914 javac FirstProg.java
1915 \end{minted}
1916 (or an equivalent command for windows).
1917 Normally, nothing will be printed, but a \texttt{FirstProg.class} file will be created.
1918 \item Now, execute that program, using
1919 \begin{minted}{bash}
1920 java -cp .:mysql-connector-java-***-bin.jar FirstProg
1921 \end{minted}
1922 in Linux, or
1923 \begin{minted}{bash}
1924 java -cp .;mysql-connector-java-***-bin.jar FirstProg
1925 \end{minted}
1926 in Windows.
1927 The \texttt{-cp} option lists the places where java should look for the class used in the program: we are explicitely asking java to use the \texttt{mysql-connector-java-***-bin.jar} executable to execute our \texttt{FirstProg} executable.
1928 Try to execute \texttt{FirstProg} without that flag, and see what happens.
1929 \end{itemize}
1930 \end{problem}
1931
1932 \begin{answer}
1933 \begin{minted}[breaklines=true]{bash}
1934 $ java FirstProg
1935 java.sql.SQLException: No suitable driver found for jdbc:mysql://localhost:3306/HW_ebookshop
1936 at java.sql.DriverManager.getConnection(DriverManager.java:689)
1937 at java.sql.DriverManager.getConnection(DriverManager.java:247)
1938 at FirstProg.main(FirstProg.java:9)
1939 \end{minted}
1940 \end{answer}
1941
1942 \begin{problem}[tags={hw}]
1943 There are three other programs in that archive:
1944 \begin{description}
1945 \item[Variation.java] is the program we looked at during class.
1946 \item[NullProg.java] is an attempt to answer the question whenever SQL's \sqli{NULL} is JAVA's \texttt{NULL} when it comes to strings.
1947 \item[AdvancedProg.java] is a long and commented program that introduces multiple new tools and ideas.
1948 \end{description}
1949
1950 Read, execute, break, edit, compile, patch, hack and (most importantly) understand those three programs, with, of course, a focus on the last one.
1951
1952 \end{problem}
1953
1954 %%% 08 -- Exam 1
1955
1956
1957 \begin{problem}[tags={ex}, points = {20}]
1958 %30
1959 A cinema company wants you to design a relational model for the following set-up:
1960 \begin{itemize}
1961 \item The company has movie stars. Each star has a name, gender, birthday, and unique id.
1962 \item The company has information about movies: title, year, length, genre, and studio. Each movie has a unique id, and features multiple stars.
1963 \item The company owns movie theaters as well. Each theater has a name, address, and a unique id.
1964 \item Furthermore, each theater has a set of auditoriums. Each auditorium has a unique number, and seating capacity.
1965 \item Each theater can schedule movies at show-times.
1966 Each show-time has a unique id, a start time, and is for a specific movie, at a theater auditorium.
1967 \item The company sells tickets for scheduled show-times. Each ticket has a unique ticket id, and a price.
1968 \end{itemize}
1969 %Your relations should be written as follow: relation names in uppercase, coma separated attributes enclosed in parenthesis, primary key underlined, and foreign key(s) detailed on the side (or drawn). That is, for instance,
1970 %\enquote{PROF(\underline{Login}, Name, Department) ; where Department references the Code attribute in the DEPARTMENT table.}
1971 Complete the following relational model.
1972 If you think there are missing relations, add them, and when the parenthesis isn't closed, close it or add attribute(s).
1973 Also, specify the primary as well as foreign keys.
1974
1975 {
1976 \renewcommand{\arraystretch}{3}%
1977 \large
1978 \hspace{3em}\begin{tabular}{l}
1979 STAR(Name, Gender, Birthday, Id\\
1980 MOVIE(Title\\
1981 FEATURE\_IN(StarId, MovieId)\\
1982 THEATER(Name\\
1983 AUDITORIUM(Id, Capacity, Theater\\
1984 SHOWTIME(
1985 \end{tabular}
1986 }
1987 \end{problem}
1988 \clearpage
1989
1990 \begin{problem}[tags={ex}, points ={10}]
1991 Your professor designed the following relational model at some point in his career, to help him organizing his exams and the students grades:
1992
1993 \begin{center}
1994 \begin{tabular}{l | l}
1995 Table Name and Attributes & Example of Value\\
1996 \hline
1997 EXAM(\underline{Number}, Date, Course) & \(\langle 1, \text{'2018-02-14'}, \text{'CSCI3410'}\rangle\)\\
1998 PROBLEM(\underline{Statement}, Points, Length, Exam) & \(\langle \text{'Your professor designed\dots'}, 10, \text{'00:10:00'}, 1\rangle\) \\
1999 STUDENT\_GRADE(\underline{Login}, \underline{Exam}, Grade) & \(\langle \text{'aalyx'}, 1, 83\rangle\)
2000 \end{tabular}
2001 \end{center}
2002
2003 The idea was to have
2004 \begin{itemize}
2005 \item The EXAM table storing information about exams,
2006 \item One entry per problem in the PROBLEM table, and to associate every problem to an exam (i.e., to have the Exam attribute referencing the Number attribute in the EXAM table),
2007 \item The grade of one student for one particular exam stored in the STUDENT\_GRADE table, where Exam is referencing the Number attribute in the EXAM table.
2008 \end{itemize}
2009 Unfortunately, this design turned out to be terrible.
2010 Describe at least one common and interesting situation where this model would fail to fulfill its purpose, and propose a way to correct the particular problem you identified.
2011 \end{problem}
2012
2013 \clearpage
2014
2015
2016 \begin{problem}[tags={ex}, points = {30}]
2017 Look at the SQL code below, and then answer the following questions.
2018
2019 \begin{minted}[linenos]{mysql}
2020 CREATE TABLE TRAIN(
2021 Id VARCHAR(30),
2022 Model VARCHAR(30),
2023 ConstructionYear YEAR(4)
2024 );
2025
2026 CREATE TABLE CONDUCTOR(
2027 Id VARCHAR(20),
2028 Name VARCHAR(20),
2029 ExperienceLevel VARCHAR(20)
2030 );
2031
2032 CREATE TABLE ASSIGNED_TO(
2033 TrainId VARCHAR(20),
2034 ConductorId VARCHAR(20),
2035 Day DATE,
2036 PRIMARY KEY(TrainId, ConductorId)
2037 );
2038 \end{minted}
2039
2040 \begin{enumerate}[topsep=0pt,itemsep=10em,partopsep=1ex,parsep=1ex]
2041 \item Modify the \sqli{CREATE} statement that creates the \sqli{TRAIN} table (l. 1--5), so that \sqli{Id} would be declared as the primary key.
2042 You can write only the line(s) that need to change.
2043 \item Write an \sqli{ALTER} statement that makes \sqli{Id} become the primary key of the \sqli{CONDUCTOR} table.
2044 \item Modify the \sqli{CREATE} statement that creates the \sqli{ASSIGNED_TO} table (l. 13--18), so that it has two foreign keys: \sqli{ConductorId} would be referencing the \sqli{Id} attribute in \sqli{CONDUCTOR} and \sqli{TrainId} would be referencing the \sqli{Id} attribute in \sqli{TRAIN}.
2045 You can write only the line(s) that need to change.
2046 \clearpage
2047 \item Write \sqli{INSERT} statements that insert one tuple of your invention in each relation.
2048 Your statements should respect all the constraints (including the ones we added at the previous questions) and result in actual insertions. (Remember that four digits is a valid value for an attribute with the \sqli{YEAR(4)} datatype.)
2049 \vspace{2em}
2050
2051 \item Write a statement that sets the value of the \sqli{ExperienceLevel} attribute to \enquote{Senior} in all the tuples where the \sqli{Id} attribute is \enquote{GP1029} in the \sqli{CONDUCTOR} relation.
2052 \vspace{-2em}
2053 \item For each of the following questions, write a \sqli{SELECT} statement that would answer it:
2054 \begin{enumerate}[topsep=0pt,itemsep=6em,partopsep=1ex,parsep=1ex]
2055 \item \enquote{What are the identification numbers of the trains?}
2056 \item \enquote{What are the names of the conductors with a \enquote{Senior} experience level?}
2057 \item \enquote{What are the construction years of the \enquote{Surfliner} and \enquote{Regina} models that we have?}
2058 \item \enquote{What is the id of the conductor that was responsible of the train referenced \enquote{K-13} on 2015/12/14?}
2059 \item \enquote{What are the models that were ever conducted by the conductor whose id is \enquote{GP1029}?}
2060 % \item \enquote{What is the name of the conductor that was responsible of the \enquote{Surfliner} model on October, 25th, 2015?}
2061 \end{enumerate}
2062 \end{enumerate}
2063 \end{problem}
2064
2065 \clearpage
2066
2067 \begin{problem}[tags={ex}, bonus-points = {4}, points={40}]
2068 Suppose we have the relational model depicted in \autoref{fig:coffee}, p.~\pageref{fig:coffee}, with the indicated data in it.
2069
2070 In the following, we will assume that this model was implemented in a DBMS (MySQL or MariaDB), the primary keys being the underlined attributes, and the foreign keys as follow:
2071
2072 \begin{tabular}{r c l}
2073 \textbf{FavCoffee} in the \textbf{CUSTOMER} relation & refers to & \textbf{Ref} in the \textbf{COFFEE} relation\\
2074 \textbf{Provider} in the \textbf{SUPPLY} & refers to & \textbf{Name} in the \textbf{PROVIDER} relation\\
2075 \textbf{Coffee} in the \textbf{SUPPLY} & refers to & \textbf{Ref} in the \textbf{COFFEE} relation
2076 \end{tabular}
2077
2078 You will be asked to read and write SQL commands.
2079 You should assume that
2080 \begin{enumerate}
2081 \item Datatype doesn't matter: we use only strings and appropriate numerical datatypes.
2082 \item Every statement respects SQL's syntax (there's no \enquote{a semi-colon is missing} trap).
2083 \item None of those commands are actually executed: the data is always in the state of \autoref{fig:coffee}, you are asked to answer \enquote{what if} questions.
2084 \end{enumerate}
2085
2086 In questions (\ref{ques:upd}) and (\ref{ques:del}), please, use \textbf{COFFEE}.\(1\) to denote the first tuple in \textbf{COFFEE}, and similarly for other relations and tuples (so that, for instance \textbf{SUPPLY}.\(4\) corresponds to \enquote{Johns \& Co., \(221\)}).
2087
2088 \begin{enumerate}
2089 \item Determine if the following insertion statements would violate the
2090 the \textbf{E}ntity integrity constraint, % (\enquote{primary key cannot be \texttt{NULL} and should be unique})
2091 the \textbf{R}eferential integrity constraint% (\enquote{the foreign key must refer to something that exists}),
2092 , if there would be some \textbf{O}ther kind of error% (ignoring the plausability / revelance of inserting that tuple)
2093 , or if it would result in \textbf{S}uccessful insertion.
2094
2095 { \renewcommand{\arraystretch}{2}
2096 \begin{tabular}{l | c | c | c | c |}
2097 & \textbf{E} & \textbf{R} & \textbf{O} & \textbf{S}\\
2098 \hline
2099 \sqli{INSERT INTO CUSTOMER VALUES(005, 'Bob Hill', NULL, 001);} & & & & \\ \hline
2100 \sqli{INSERT INTO COFFEE VALUES(002, "Peru", "Decaf", 3.00);} & & & & \\ \hline
2101 \sqli{INSERT INTO PROVIDER VALUES(NULL, "contact@localcof.com");} & & & & \\ \hline
2102 \sqli{INSERT INTO SUPPLY VALUES("Johns & Co.", 121);} & & & & \\ \hline
2103 \sqli{INSERT INTO SUPPLY VALUES("Coffee Unl.", 311, 221);} & & & & \\ \hline
2104 \end{tabular}
2105 }
2106
2107 \item Assuming that we cascade update statements, list the tuples modified by the following statements: \label{ques:upd}
2108 \begin{enumerate}[topsep=0pt,itemsep=4em,partopsep=1ex,parsep=1ex]
2109 \item \sqli{UPDATE CUSTOMER SET FavCoffee = 001 WHERE CardNo = 001;}
2110 \item \sqli{UPDATE COFFEE SET TypeOfRoast = 'Decaf' WHERE Origin = 'Brazil';}
2111 \item \sqli{UPDATE PROVIDER SET Name = 'Coffee Unlimited' WHERE Name = 'Coffee Unl.';}
2112 \item \sqli{UPDATE COFFEE SET PricePerPound = 10.00 WHERE PricePerPound > 10.00;}
2113 \end{enumerate}
2114 \clearpage
2115 \item Assuming that we cascade delete statements, list the tuples deleted by the following statements: \label{ques:del}
2116 \begin{enumerate}[topsep=0pt,itemsep=4em,partopsep=1ex,parsep=1ex]
2117 \item \mintinline{MYSQL}{DELETE FROM CUSTOMER WHERE Name LIKE '%S%';}
2118 \item \sqli{DELETE FROM COFFEE WHERE Ref = 001;}
2119 \item \sqli{DELETE FROM SUPPLY WHERE Provider = 'Coffee Unl.' AND Coffee = 001;}
2120 \item \sqli{DELETE FROM PROVIDER WHERE Name = 'Johns & Co.';}
2121 \vspace{4em}
2122 \end{enumerate}
2123
2124 \item %Starting here, assume that there is more data in our table than what was given at the beginning of the problem.
2125 Write queries that answer the following questions (the last two are bonus):
2126 \begin{enumerate}[topsep=0pt,itemsep=6em,partopsep=1ex,parsep=1ex]
2127 \item \enquote{What are the origins of your dark coffees?}
2128 \item \enquote{What is the reference of Bob's favorite coffee?} (nota: it doesn't matter if you return the favorite coffee of all the Bobs in the database.)
2129 \item \enquote{What are the names of the providers who didn't give their email?}
2130 \item \enquote{How many coffees does Johns \& co. provide us with?}
2131 \item \enquote{What are the names of the providers of my dark coffees?}
2132 \end{enumerate}
2133
2134 \end{enumerate}
2135
2136 \end{problem}
2137
2138
2139
2140 \begin{figure}
2141
2142 \begin{multicols}{2}
2143 \textbf{COFFEE}
2144
2145 \begin{tabular}{| c | c | c | c | }
2146 \hline
2147 \rowcolor{gray!50} \underline{\textbf{Ref}} & \textbf{Origin} & \textbf{TypeOfRoast} & \textbf{PricePerPound}\\ \hline
2148 \(001\) & Brazil & Light & 8.90\\
2149 \(121\) & Bolivia & Dark& 7.50\\
2150 \(311\) & Brazil & Medium & 9.00\\
2151 \(221\) & Sumatra & Dark & 10.25\\ \hline
2152 \end{tabular}
2153 \\[2em]
2154 \textbf{CUSTOMER}
2155
2156 \begin{tabular}{| c | c | c | c | }
2157 \hline
2158 \rowcolor{gray!50} \underline{\textbf{CardNo}} & \textbf{Name} & \textbf{Email} & \textbf{FavCoffee}\\ \hline
2159 \(001\) & Bob Hill & b.hill@isp.net & 221\\
2160 \(002\) & Ana Swamp & swampa@nca.edu & 311\\
2161 \(003\) & Mary Sea & brig@gsu.gov & 121\\
2162 \(004\) & Pat Mount & pmount@fai.fr & 121\\ \hline
2163 \end{tabular}
2164 %\\[2em]
2165 %\noindent
2166
2167 \hspace{5em}
2168 \textbf{SUPPLY}
2169
2170 \hspace{5em}
2171 \begin{tabular}{| c | c | }
2172 \hline
2173 \rowcolor{gray!50} \underline{\textbf{Provider}} & \underline{\textbf{Coffee}}\\ \hline
2174 Coffee Unl. & \(001\) \\
2175 Coffee Unl. & \(121\) \\
2176 Coffee Exp. & \(311\) \\
2177 Johns \& Co. & \(221\) \\ \hline
2178 \end{tabular}
2179 \\[2em]
2180
2181 \hspace{5em}
2182 \textbf{PROVIDER}
2183
2184 \hspace{5em}
2185 \begin{tabular}{| c | c | }
2186 \hline
2187 \rowcolor{gray!50} \underline{\textbf{Name}} & \textbf{Email}\\ \hline
2188 Coffee Unl. & bob@cofunl.com \\
2189 Coffee Exp. & pat@coffeex.dk \\
2190 Johns \& Co. & NULL \\ \hline
2191 \end{tabular}
2192 \end{multicols}
2193 \caption{Relational Model and Data for the DB\_COFFEE Database}
2194 \label{fig:coffee}
2195 \end{figure}
2196
2197 %% 09 -- Exam 2
2198
2199
2200 \begin{problem}[tags={ex}, points = {15}]
2201 Consider the relation
2202
2203 \begin{center}
2204 CONTACT(Phone, Call\_center, Email, Zip, Brand, Website)
2205 \end{center}
2206
2207 and the following functional dependencies:
2208
2209 \begin{center}
2210 \begin{tabular}{r c l}
2211 \{Zip, Brand\} & \(\to\) & \{Phone\}\\
2212 \{Brand\} & \(\to\) & \{Email\}\\
2213 \{ Brand\} & \(\to\) & \{Website\}\\
2214 \{Phone\} & \(\to\) & \{Call\_center\}
2215 \end{tabular}
2216 \end{center}
2217
2218 Assume that \{Zip, Brand\} is the primary key.
2219 Normalize this relation to the second normal form, and \emph{then} to the third normal form.
2220 Give the relations, their primary keys, and functional dependencies for both steps.
2221 \end{problem}
2222
2223 \clearpage
2224
2225 \begin{problem}[tags={ex}, points = {15}]
2226 \label{nf}
2227
2228 Consider the relation
2229
2230 \begin{center}
2231 CONSULTATION(Doctor\_no, Patient\_no, Date, Diagnosis, Treatment, Charge, Insurance)
2232 \end{center}
2233
2234 with the following functional dependencies:
2235
2236 \begin{center}
2237 \begin{tabular}{r c l}
2238 \{Doctor\_no, Patient\_no, Date\} & \(\to\) & \{Diagnosis\}\\
2239 \{Doctor\_no, Patient\_no, Date\} & \(\to\) & \{Treatment\}\\
2240 \{Treatment, Insurance\} & \(\to\) & \{Charge\}\\
2241 \{Patient\_no\} & \(\to\) & \{Insurance\}
2242 \end{tabular}
2243 \end{center}
2244
2245 \begin{enumerate}[topsep=0pt,itemsep=10em,partopsep=1ex,parsep=1ex]
2246 \item The designer decided not to add the functional dependency \{Diagnosis\} $\to$ \{Treatment\}.
2247 Explain what could be the designer's justification, at the level of the mini-world.
2248 \item Identify a primary key for this relation.
2249 \item What is the degree of normalization of this relation?
2250 Normalize it to the third normal form if necessary.
2251 \end{enumerate}
2252 \end{problem}
2253 \clearpage
2254
2255 \begin{solution}
2256 Answer:
2257 Because there are no partial dependencies, the given relation is in 2NF already. This however is not 3NF because the Charge is a nonkey attribute that is determined by another nonkey attribute, Treatment. We must decompose further:
2258
2259 R (Doctor\_no, Patient\_no, Date, Diagnosis, Treatment)
2260
2261 R1 (Treatment, Charge)
2262
2263 We could further infer that the treatment for a given diagnosis is functionally dependant, but we should be sure to allow the doctor to have some flexibility when prescribing cures.
2264 \end{solution}
2265
2266 \begin{problem}[tags={ex}, points = {20}]
2267 A friend of yours want you to review and improve the code for a role-playing game.
2268
2269 The original idea was that each character should have a name, a class (e.g., Bard, Assassin, Druid), a certain amount of experience, a level, one or more weapons (providing bonuses) and can complete quests.
2270 A quest have a name, and rewards the characters that completed it with a certain amount of experience, and sometimes (but rarely) with a special item.
2271
2272 Your friend came up with the code presented in \autoref{lst:RPG}, page~\pageref{lst:RPG}, but there are several problems:
2273
2274 \begin{itemize}
2275 \item As of now, a character can have only one weapon.
2276 All the attempts to \enquote{hack} the \sqli{CHARACTER} table to add an arbitrary number of weapons ended up creating horrible messes.
2277 \item Every time a character completes a quest, a copy of the quest must be created.
2278 Your friend is not so sure why, but nothing else works.
2279 Also it seems that a character can complete only one quest, but your friend is not so sure about that.
2280 \item It would be nice to be able to store features that are tied to the class, and not to the character, like the bonus they provide and the associated element (e.g., all bards use fire, all assassins use wind, etc.).
2281 But you friend simply can't figure out how to do that.
2282 \end{itemize}
2283
2284 Can you provide a \emph{relational database schema} (no need to write the \sqli{SQL} code, but remember to indicate the primary and foreign keys) that would solve all of your friend's troubles?
2285
2286 \end{problem}
2287
2288 \clearpage
2289
2290
2291 \begin{problem}[tags={ex}, points = {25}]
2292 This exercise asks you to convert business statements into dependencies.
2293 Consider the following relation:
2294 \begin{center}
2295 KEYBOARD(Manufacturer, Model, Layout, Retail\_Store, Price)
2296 \end{center}
2297
2298 A tuple in the KEYBOARD relation contains information about a computer keyboard: its manufacturer, its model, its layout (AZERTY, QWERTY, etc.), the place where it is sold, and its price.
2299
2300 \begin{enumerate}
2301 \item Write each of the following business statement as a functional dependency:
2302
2303 \begin{enumerate}[topsep=0pt,itemsep=7em,partopsep=1ex,parsep=1ex]
2304 \item A model has a fixed layout.
2305 \item A retail store can't have two different models produced by the same manufacturer.
2306 \item The price of one particular model can vary from one store to another, but all the keyboards of a particular model are sold at the same price at one particular store.\vspace{7em}
2307 \end{enumerate}
2308
2309 \item Based on those statements, what could be a key for this relation?
2310 \vspace{6em}
2311 \item Assuming all those functional dependencies hold, and taking the primary key you identified at the previous step, what is the degree of normality of this relation? Justify your answer.
2312 \end{enumerate}
2313 \end{problem}
2314
2315 \clearpage
2316
2317 \begin{problem}[tags={ex}, points={25}]
2318 Consider the E.R. schema of Figure~\ref{fig:erCOUNTRY}, p.~\pageref{fig:erCOUNTRY}.
2319 Map that E.-R. diagram to a relational database schema.
2320 \end{problem}
2321 \clearpage
2322
2323 \begin{figure}
2324 % \begin{listing}
2325 \begin{minted}[linenos, breaklines=true]{mysql}
2326 CREATE TABLE CHARACTER(
2327 Name VARCHAR(30) PRIMARY KEY,
2328 Class VARCHAR(30),
2329 XP INT,
2330 LVL INT,
2331 Weapon_Name VARCHAR(30),
2332 Weapon_Bonus INT,
2333 Quest_Completed VARCHAR(30)
2334 );
2335
2336 CREATE TABLE QUEST(
2337 Id VARCHAR(20) PRIMARY KEY,
2338 Completed_By VARCHAR(30),
2339 XP_Gained INT,
2340 Special_Item VARCHAR(20),
2341 FOREIGN KEY (Completed_By) REFERENCES CHARACTER(Name)
2342 );
2343
2344 ALTER TABLE CHARACTER ADD FOREIGN KEY (Quest_Completed) REFERENCES QUEST(Id);
2345 \end{minted}
2346 % \end{listing}
2347 \caption{A \sqli{SQL} Code for a Role-Playing Game}
2348 \label{lst:RPG}
2349 \end{figure}
2350
2351 \begin{figure}
2352 \begin{tikzpicture}[node distance=8em]
2353 \node[entity] (COUNTRY) {COUNTRY};
2354 \node[attribute] (name) [left of=COUNTRY] {\key{Name}} edge (COUNTRY);
2355 \node[attribute] (population) [below left of=COUNTRY] {Population} edge (COUNTRY);
2356
2357 \node[relationship] (SPEAKS) [below right of=COUNTRY] {SPEAKS} edge node[right, pos=0.4] {$M$} (COUNTRY);
2358 \node[entity] (LANGUAGE) [below right of=SPEAKS] {LANGUAGE} edge node[right, pos=0.5] {$N$} (SPEAKS);
2359 \node[attribute] (code) [left of = LANGUAGE] {\key{Code}} edge (LANGUAGE);
2360 \node[attribute] (symbol) [right of = LANGUAGE] {Name} edge (LANGUAGE);
2361
2362 \node[relationship] (BWF) [below of = LANGUAGE] {B\_W\_F};
2363 \draw (LANGUAGE) to node[left, pos=0.6] {$N$} (BWF.west);
2364 \draw (LANGUAGE) to node[right, pos=0.6] {$M$} (BWF.east);
2365
2366 %\node[relationship] (boundary) [above of = COUNTRY] {S\_B\_W};
2367 %\draw (COUNTRY) to node[left, pos=0.6] {$N$} (boundary.west);
2368 %\draw (COUNTRY) to node[right, pos=0.6] {$M$} (boundary.east);
2369
2370 %\node[relationship] (conversion) [below of =LANGUAGE] {CONVERSION};
2371 %\draw (LANGUAGE) to node[left, pos=0.6] {$N$} (conversion.west);
2372 %\draw (LANGUAGE) to node[right, pos=0.6] {$M$} (conversion.east);
2373
2374 %\node[attribute] (timestamp) [below left of=conversion] {\key{Timestamp}} edge (conversion);
2375 %\node[attribute] (rate) [below right of=conversion] {Exchange\_rate} edge (conversion);
2376
2377 \node[ident relationship] (hasfor) [right of=COUNTRY] {SINGS} edge node[above, pos=0.4] {$1$} (COUNTRY);
2378 \node[weak entity] (anthem) %[right of=hasfor]
2379 [right = 1cm of hasfor] {NATIONAL\_ANTHEM} edge[total] node[above, pos=0.6]{\(M\)} (hasfor);
2380 \node[multi attribute] (color) [right =1cm of anthem] {Creator} edge (anthem);
2381 \node[attribute] (name) [below right of = anthem] {\pkey{Name}} edge (anthem);
2382
2383 \node[relationship] (WIN) [above right of =LANGUAGE] {W\_IN};
2384 \draw (LANGUAGE) to node[left, pos=0.6] {$N$} (WIN);
2385 \draw (anthem) to node[right, pos=0.6] {$M$} (WIN);
2386 \end{tikzpicture}
2387
2388 %\enquote{S\_B\_W} stands for \enquote{SHARES\_A\_BORDER\_WITH}.
2389 \enquote{W\_IN} stands for \enquote{WRITTEN\_IN}, and
2390 \enquote{B\_W\_F} stands for \enquote{BORROWS\_WORDS\_FROM}.
2391 For this relationship, on the left-hand side is the language that borrows a word, and on the right-hand side is the language that provides the loanword.
2392 \caption{E.R. Schema for the COUNTRY\_INFO database}
2393 \label{fig:erCOUNTRY}
2394 \end{figure}
2395
2396 %%% 10 Final
2397
2398
2399 \begin{problem}[tags={ex}, points = {40}]
2400 Look at the relational model from Figure~\ref{fig:RelMod}, p.~\pageref{fig:RelMod}, and \enquote{reverse-engineer} it to obtain an E.-R. diagram.
2401 \end{problem}
2402
2403 \clearpage
2404
2405 \begin{problem}[tags={ex}, points = {30}]
2406 Answer the following in one or two sentences:
2407
2408 \begin{enumerate}[topsep=0pt,itemsep=12em,partopsep=1ex,parsep=1ex]
2409 \item What is deletion anomaly? Is it a desirable feature?
2410 \item What is polyglot persistence? Is it useful?
2411 \item What does it mean to be \enquote{schemaless}? What does it imply?
2412 \item What is denormalization?
2413 When could that be useful?
2414 \item What is the (object-relational) impedance mismatch? Is it an issue that can't be overcome?
2415 \end{enumerate}
2416 \end{problem}
2417
2418 \begin{solution}
2419 A Delete Anomaly exists when certain attributes are lost because of the deletion of other attributes.
2420
2421 When storing data, it is best to use multiple data storage technologies, chosen based upon the way data is being used by individual applications or components of a single application.
2422 \end{solution}
2423
2424 \clearpage
2425
2426
2427 \begin{problem}[tags={ex}, points = {35}]
2428 Consider the following relation:
2429 \vspace{1.5em}
2430 \begin{center}
2431 FLIGHT(From, \qquad To, \qquad Airline, \qquad Flight\#, \qquad DateHour, \qquad HeadQuarter, \qquad Pilot, \qquad TZDifference)
2432 \end{center}
2433 \vspace{1.5em}
2434
2435 A tuple in the FLIGHT relation contains information about an airplane flight: the airports of departure and arrival, the airline carrier, the number of the flight, its time of departure, the headquarter of the company chartering the flight, the name of the pilot(s), and the time zone difference between the departure and arrival airports.
2436
2437 The \enquote{Pilot} attribute is multi-valued (so that between \(1\) and \(4\) pilot's names can be stored in it).
2438 Given an airline and a flight number, one can determine the departure and arrival airports, as well as the date and hour and the pilot(s).
2439 Given the airline carrier, one can determine the headquarter.
2440 Finally, given the departure and arrival airports, one can determine their time zone difference.
2441
2442 Normalize the \enquote{FLIGHT} relation to its third normal form.
2443 You can indicate your steps, justify your reasoning, and indicate the foreign keys if you want to, but don't have to.
2444 \end{problem}
2445
2446 \clearpage
2447
2448 \begin{problem}[tags={ex}, points = {60}]
2449 Consider the set of requirements in Figure~\ref{fig:reqUNIV}, p.~\pageref{fig:reqUNIV}.
2450 Draw an E.-R. diagram for that schema.
2451 Specify key attributes of each entity type and structural constraints on each relationship type.
2452 Note any unspecified requirements, and make appropriate assumptions to make the specification complete.
2453 \end{problem}
2454 \clearpage
2455
2456 \begin{problem}[tags={ex}, points = {35}]
2457 Consider the UML diagram in Figure~\ref{fig:UML}, p.~\pageref{fig:UML}, and convert it to the relational model.
2458 Don't forget to indicate primary and foreign keys.
2459 \clearpage
2460 \end{problem}
2461
2462 \begin{figure}
2463 \colorlet{lightgray}{gray!20}
2464
2465 \begin{tikzpicture}[relation/.style={rectangle split, rectangle split parts=#1, rectangle split part align=base, draw, anchor=center, align=center, text height=3mm, text centered}]\hspace*{-0.3cm}
2466
2467 % RELATIONS
2468
2469 \node (actortitle) {\textbf{ACTOR}};
2470
2471 \node [relation=3, rectangle split horizontal, rectangle split part fill={lightgray!50}, anchor=north west, below=0.6cm of actortitle.west, anchor=west] (actor)
2472 {\underline{Id}%
2473 \nodepart{two} Name
2474 \nodepart{three} Birthdate
2475 };
2476
2477 \node [below=1.3cm of actor.west, anchor=west] (movietitle) {\textbf{MOVIE}};
2478
2479 \node [relation=4, rectangle split horizontal, rectangle split part fill={lightgray!50}, below=0.6cm of movietitle.west, anchor=west] (movie)
2480 {\underline{Title}%
2481 \nodepart{two} Year
2482 \nodepart{three} Length
2483 \nodepart{four} Studio
2484 };
2485
2486
2487 \node [right=8cm of actortitle.west, anchor=west] (actingtitle) {\textbf{ACTING}};
2488
2489 \node [relation=2, rectangle split horizontal, rectangle split part fill={lightgray!50}, below=0.6cm of actingtitle.west, anchor=west] (acting)
2490 {\underline{ActorID}%
2491 \nodepart{two} \underline{MovieTitle}%
2492 };
2493
2494 \node [below=1.3cm of movie.west, anchor=west] (theatretitle) {\textbf{THEATRE}};
2495
2496 \node [relation=5, rectangle split horizontal, rectangle split part fill={lightgray!50}, anchor=north west, below=0.6cm of theatretitle.west, anchor=west] (theatre)
2497 {\underline{Name}%
2498 \nodepart{two} Street
2499 \nodepart{three} City
2500 \nodepart{four} State
2501 \nodepart{five} Zip
2502 };
2503
2504 \node [below=1.3cm of acting.west, anchor=west] (showingtitle) {\textbf{SHOWING}};
2505
2506 \node [relation=4, rectangle split horizontal, rectangle split part fill={lightgray!50}, anchor=north west, below=0.6cm of showingtitle.west, anchor=west] (showing)
2507 {\underline{TheaterName}%
2508 \nodepart{two} \underline{MovieTitle}
2509 \nodepart{three} \underline{Day}
2510 \nodepart{four} \underline{Time}
2511 };
2512
2513
2514 % FOREIGN KEYS
2515
2516 %\draw[-latex] (acting) -- -| (actor.one south);%(actor.one south) -- ++(0,-0.2) --++(7,0) --++(0, -2) -| ($(movie.one south) + (0.2,0)$);
2517 \draw[-latex] ($(showing.two south) + (-0.3,0)$)-- ++(0,-0.3) -| (movie.one south);
2518 \draw[-latex] (showing.one south) -- ++ (0,-2.6) -| (theatre.one south);
2519 \draw[-latex] (acting.one south) -- ++ (0,-0.3) -| (actor.one south);
2520 \draw[-latex] (acting.two south) -- ++ (0,-0.5) -- ++(-3, 0) -- ++(0, -2.1)-| ($(movie.one south) + (-0.25,0)$);
2521 %\draw[-latex] (showing.four south) -- ++(0,-0.2) --++(2,0) --++(0, 1.75) -| ($(theatre.one south) + (0.1,0)$);
2522 %\draw[-latex] (theatre.five south) -- ++(0,-0.2) --++(2,0) --++(0, 1.5) -| ($(movie.one south) + (-0.1,0)$);
2523 \end{tikzpicture}
2524 \caption{Relational Model for a MOVIE Database}
2525 \label{fig:RelMod}
2526 \end{figure}
2527
2528 \begin{figure}
2529 Consider the following requirements for a UNIVERSITY database, used to keep track of students' transcripts.
2530 \begin{enumerate}
2531 \item The university keeps track of each student's name, student number, class (freshman, sophomore, \dots, graduate), major department, minor department (if any), and degree program (B.A., B.S., \dots, Ph.D.).
2532 Student number has unique values for each student.
2533 \item Each department is described by a name and has a (unique) department code.
2534 \item Each course has a course name, a course number, credit hours, and is offered by at least one department. The value of course number is unique for each course. A course has at least one section.
2535 \item Each section of a course has an instructor, a semester, a year, and a section number.
2536 The section number distinguishes different sections of the same course that are taught during the same semester/year; its values are 1, 2, 3, \dots, up to the number of sections taught during each semester. Students can enroll in sections and receive a letter grade, and grade point (0, 1, 2, 3, 4 for F, D, C, B, A, respectively).
2537 \end{enumerate}
2538 \caption{Requirements for a UNIVERSITY database}
2539 \label{fig:reqUNIV}
2540 \end{figure}
2541
2542 \begin{figure}
2543 \begin{tikzpicture}[scale = 0.8]
2544 \node (movie) at (0,0){\begin{tabular}{| r l |}
2545 \hline
2546 \multicolumn{2}{| c |}{\textbf{DRIVER}}\\
2547 \hline
2548 id &: String\\
2549 dob &: Date\\
2550 name &: String\\
2551 address &: Street\\
2552 & City\\
2553 \hline
2554 getAge()&: Int\\
2555 \hline
2556 \end{tabular}
2557 };
2558 \node (cdriver) at (8, 1){\begin{tabular}{| r l |}
2559 \hline
2560 \multicolumn{2}{| c |}{\textbf{COMMERCIAL\_DRIVER}}\\
2561 \hline
2562 Class &: String\\
2563 \hline
2564 \end{tabular}
2565 };
2566 %\draw[open diamond-] (movie) -- node[above, pos=0.1]{\(0..\star\)} node[above, pos=0.9]{\(1..1\)} (cdriver);
2567 \draw[{Triangle[open]}-] (movie) -- %node[above, pos=0.1]{\(0..\star\)} node[above, pos=0.9]{\(1..1\)}
2568 (cdriver);
2569 \node (car) at (9, -2){\begin{tabular}{| r l |}
2570 \hline
2571 \multicolumn{2}{| c |}{\textbf{CAR}}\\
2572 \hline
2573 vin &: String\\
2574 make &: String\\
2575 year &: Year\\
2576 brand &: String\\
2577 \hline
2578 \end{tabular}
2579 };
2580 \draw[-] ($(movie)+(2.5,-0.5)$) -- node[above, pos=0.1]{\(0..*\)} node[above=0.5em]{POSSESSES} node[above, pos=0.9]{\(0..*\)} (car);
2581 %\draw[-] (movie) -- node[below, pos=0.1]{\(0..1\)} node[below, pos=0.9]{\(0..4\)} ($(car)+(-2.5, 0.5)$);
2582 \node (seat) at (3.5, -5){\begin{tabular}{| r l |}
2583 \hline
2584 \multicolumn{2}{| c |}{\textbf{POSSESSION\_HISTORY}}\\
2585 \hline
2586 date of purchase &: Date\\
2587 \hline
2588 \end{tabular}
2589 };
2590 \draw[dashed] (seat) -- (3.5, -1.2);
2591
2592 \node (insu) at (15.5, 2){\begin{tabular}{| r l |}
2593 \hline
2594 \multicolumn{2}{| c |}{\textbf{CAR\_INSURANCE}}\\
2595 \hline
2596 policy number &: String\\
2597 covered amount &: int\\
2598 compagny name &: String\\
2599 \hline
2600 \end{tabular}
2601 };
2602
2603 \draw[diamond -] (car) -- node[below, pos=0.1, right=.2]{\(1..1\)} node[below, pos=0.9, right=0.2]{\(1..\star\)} node[right=0.2]{IS\_COVERED\_BY} (insu);
2604 \end{tikzpicture}
2605 \caption{UML Diagram for the CAR\_POSSESSION Database}
2606 \label{fig:UML}
2607 \end{figure}
2608
2609 \end{document}
2610
2611
2612
2613 \begin{problem}[tags={ex}, points = {40}]
2614 \label{nf}
2615
2616 Consider the relation
2617
2618 \begin{center}
2619 CONSULTATION(Doctor\_no, Patient\_no, Date, Diagnosis, Treatment, Charge, Insurance)
2620 \end{center}
2621
2622 with the following functional dependencies:
2623
2624 \begin{center}
2625 \begin{tabular}{r c l}
2626 \{Doctor\_no, Patient\_no, Date\} & \(\to\) & \{Diagnosis\}\\
2627 \{Doctor\_no, Patient\_no, Date\} & \(\to\) & \{Treatment\}\\
2628 \{Treatment, Insurance\} & \(\to\) & \{Charge\}\\
2629 \{Patient\_no\} & \(\to\) & \{Insurance\}
2630 \end{tabular}
2631 \end{center}
2632
2633 \begin{enumerate}[topsep=0pt,itemsep=10em,partopsep=1ex,parsep=1ex]
2634 \item The designer decided not to add the functional dependency \{Diagnosis\} $\to$ \{Treatment\}.
2635 Explain what could be the designer's justification, at the level of the mini-world.
2636 \item Identify a primary key for this relation.
2637 \item What is the degree of normalization of this relation?
2638 Normalize it to the third normal form if necessary.
2639 \end{enumerate}
2640 \end{problem}
2641 \clearpage
2642
2643 \begin{solution}
2644 Answer:
2645 Because there are no partial dependencies, the given relation is in 2NF already. This however is not 3NF because the Charge is a nonkey attribute that is determined by another nonkey attribute, Treatment. We must decompose further:
2646
2647 R (Doctor\_no, Patient\_no, Date, Diagnosis, Treatment)
2648
2649 R1 (Treatment, Charge)
2650
2651 We could further infer that the treatment for a given diagnosis is functionally dependant, but we should be sure to allow the doctor to have some flexibility when prescribing cures.
2652 \end{solution}
2653
2654 \begin{problem}[tags={ex}, points = {40}]
2655 A friend of yours want you to review and improve the code for a role-playing game.
2656
2657 The original idea was that each character should have a name, a class (e.g., Bard, Assassin, Druid), a certain amount of experience, a level, one or more weapons (providing bonuses) and can complete quests.
2658 A quest have a name, and rewards the characters that completed it with a certain amount of experience, and sometimes (but rarely) with a special item.
2659
2660 Your friend came up with the code presented in \autoref{lst:RPG}, page~\pageref{lst:RPG}, but there are several problems:
2661
2662 \begin{itemize}
2663 \item As of now, a character can have only one weapon.
2664 All the attempts to \enquote{hack} the \sqli{CHARACTER} table to add an arbitrary number of weapons ended up creating horrible messes.
2665 \item Every time a character completes a quest, a copy of the quest must be created.
2666 Your friend is not so sure why, but nothing else works.
2667 Also it seems that a character can complete only one quest, but your friend is not so sure about that.
2668 \item It would be nice to be able to store features that are tied to the class, and not to the character, like the bonus they provide and the associated element (e.g., all bards use fire, all assassins use wind, etc.).
2669 But you friend simply can't figure out how to do that.
2670 \end{itemize}
2671
2672 Can you provide a \emph{relational database schema} (no need to write the \sqli{SQL} code, but remember to indicate the primary and foreign keys) that would solve all of your friend's troubles?
2673
2674 \end{problem}
2675
2676 \clearpage
2677
2678
2679 \begin{problem}[tags={ex}, points = {40}]
2680 This exercise asks you to convert business statements into dependencies.
2681 Consider the following relation:
2682 \begin{center}
2683 KEYBOARD(Manufacturer, Model, Layout, Retail\_Store, Price)
2684 \end{center}
2685
2686 A tuple in the KEYBOARD relation contains information about a computer keyboard: its manufacturer, its model, its layout (AZERTY, QWERTY, etc.), the place where it is sold, and its price.
2687
2688 \begin{enumerate}
2689 \item Write each of the following business statement as a functional dependency:
2690
2691 \begin{enumerate}[topsep=0pt,itemsep=7em,partopsep=1ex,parsep=1ex]
2692 \item A model has a fixed layout.
2693 \item A retail store can't have two different models produced by the same manufacturer.
2694 \item The price of one particular model can vary from one store to another, but all the keyboards of a particular model are sold at the same price at one particular store.\vspace{7em}
2695 \end{enumerate}
2696
2697 \item Based on those statements, what could be a key for this relation?
2698 \vspace{6em}
2699 \item Assuming all those functional dependencies hold, and taking the primary key you identified at the previous step, what is the degree of normality of this relation? Justify your answer.
2700 \end{enumerate}
2701 \end{problem}
2702
2703 \clearpage
2704
2705 \begin{problem}[tags={ex}, points={40}]
2706 Consider the E.R. schema of Figure~\ref{fig:erCOUNTRY}, p.~\pageref{fig:erCOUNTRY}.
2707 Map that E.-R. diagram to a relational database schema.
2708 \end{problem}
2709 \clearpage
2710
2711 \begin{figure}
2712 % \begin{listing}
2713 \begin{minted}[linenos, breaklines=true]{mysql}
2714 CREATE TABLE CHARACTER(
2715 Name VARCHAR(30) PRIMARY KEY,
2716 Class VARCHAR(30),
2717 XP INT,
2718 LVL INT,
2719 Weapon_Name VARCHAR(30),
2720 Weapon_Bonus INT,
2721 Quest_Completed VARCHAR(30)
2722 );
2723
2724 CREATE TABLE QUEST(
2725 Id VARCHAR(20) PRIMARY KEY,
2726 Completed_By VARCHAR(30),
2727 XP_Gained INT,
2728 Special_Item VARCHAR(20),
2729 FOREIGN KEY (Completed_By) REFERENCES CHARACTER(Name)
2730 );
2731
2732 ALTER TABLE CHARACTER ADD FOREIGN KEY (Quest_Completed) REFERENCES QUEST(Id);
2733 \end{minted}
2734 % \end{listing}
2735 \caption{A \sqli{SQL} Code for a Role-Playing Game}
2736 \label{lst:RPG}
2737 \end{figure}
2738
2739 \begin{figure}
2740 \begin{tikzpicture}[node distance=8em]
2741 \node[entity] (COUNTRY) {COUNTRY};
2742 \node[attribute] (name) [left of=COUNTRY] {\key{Name}} edge (COUNTRY);
2743 \node[attribute] (population) [below left of=COUNTRY] {Population} edge (COUNTRY);
2744
2745 \node[relationship] (SPEAKS) [below right of=COUNTRY] {SPEAKS} edge node[right, pos=0.4] {$M$} (COUNTRY);
2746 \node[entity] (LANGUAGE) [below right of=SPEAKS] {LANGUAGE} edge node[right, pos=0.5] {$N$} (SPEAKS);
2747 \node[attribute] (code) [left of = LANGUAGE] {\key{Code}} edge (LANGUAGE);
2748 \node[attribute] (symbol) [right of = LANGUAGE] {Name} edge (LANGUAGE);
2749
2750 \node[relationship] (BWF) [below of = LANGUAGE] {B\_W\_F};
2751 \draw (LANGUAGE) to node[left, pos=0.6] {$N$} (BWF.west);
2752 \draw (LANGUAGE) to node[right, pos=0.6] {$M$} (BWF.east);
2753
2754 %\node[relationship] (boundary) [above of = COUNTRY] {S\_B\_W};
2755 %\draw (COUNTRY) to node[left, pos=0.6] {$N$} (boundary.west);
2756 %\draw (COUNTRY) to node[right, pos=0.6] {$M$} (boundary.east);
2757
2758 %\node[relationship] (conversion) [below of =LANGUAGE] {CONVERSION};
2759 %\draw (LANGUAGE) to node[left, pos=0.6] {$N$} (conversion.west);
2760 %\draw (LANGUAGE) to node[right, pos=0.6] {$M$} (conversion.east);
2761
2762 %\node[attribute] (timestamp) [below left of=conversion] {\key{Timestamp}} edge (conversion);
2763 %\node[attribute] (rate) [below right of=conversion] {Exchange\_rate} edge (conversion);
2764
2765 \node[ident relationship] (hasfor) [right of=COUNTRY] {SINGS} edge node[above, pos=0.4] {$1$} (COUNTRY);
2766 \node[weak entity] (anthem) %[right of=hasfor]
2767 [right = 1cm of hasfor] {NATIONAL\_ANTHEM} edge[total] node[above, pos=0.6]{\(M\)} (hasfor);
2768 \node[multi attribute] (color) [right =1cm of anthem] {Creator} edge (anthem);
2769 \node[attribute] (name) [below right of = anthem] {\pkey{Name}} edge (anthem);
2770
2771 \node[relationship] (WIN) [above right of =LANGUAGE] {W\_IN};
2772 \draw (LANGUAGE) to node[left, pos=0.6] {$N$} (WIN);
2773 \draw (anthem) to node[right, pos=0.6] {$M$} (WIN);
2774 \end{tikzpicture}
2775
2776 %\enquote{S\_B\_W} stands for \enquote{SHARES\_A\_BORDER\_WITH}.
2777 \enquote{W\_IN} stands for \enquote{WRITTEN\_IN}, and
2778 \enquote{B\_W\_F} stands for \enquote{BORROWS\_WORDS\_FROM}.
2779 For this relationship, on the left-hand side is the language that borrows a word, and on the right-hand side is the language that provides the loanword.
2780 \caption{E.R. Schema for the COUNTRY\_INFO database}
2781 \label{fig:erCOUNTRY}
2782 \end{figure}
2783
File notes/00_Questions_Notes.md changed (mode: 100644) (index 547f9e9..a9db505)
... ... I.e., a pet is in my DB only if it belongs to a friend, but a pet can belong to
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19 19 \\enquote\{([^}]+)?\} \\enquote\{([^}]+)?\}
20 20 "\1" "\1"
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22 \\mint\{([^}]*)\}\|([^|]*)\|
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24 ```\1
25 \2
26 ```
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34 \begin{minted}{mySQL}
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