Database Systems: The Complete Book. Solutions to Selected Exercises. Solutions for Chapter 2 · Solutions for Chapter 3 · Solutions for Chapter 4 · Solutions. Database Systems: The Complete Book. Solutions for Chapter 2. Solutions for Section Solutions for Section Solutions for Section Solutions for. are instances of two relations that might constitute part of a banking database. The data related to one account holder represents a tuple of Accounts relation. View the primary ISBN for: Database Systems 2nd Edition Textbook Solutions.
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I have the book you are looking for >>> Database Systems: The Complete Book ( 2nd Edition) Database Systems: The Complete Book is ideal. Solutions Manual for Database Systems: The Complete Book, 2nd Edition. Hector Garcia-Molina. Jeffrey D. Ullman. Jennifer Widom. © | Pearson. 4/26/ Database Systems: The Complete Book: Solutions for Chapter 3. Database Systems: The Complete Book Solutions for Chapter 3. Solutions for Section.
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Make sure to check the newsgroup frequently enough to stay informed. Reading Read the textbook for the required reading before lectures, and study them more carefully after class.
Please note that all the required readings are fair materials for exams. These materials may not be fully covered in lectures.
Our lectures are intended to motivate as well as provide a road map for your reading-- with the limited lecture time we may not be able to cover everything in the readings. Assignments There will be four written assignments, spaced out over the course of the semester.
Projects There will be a semester-long project, which involves significant database application programming. The project will be structured with several milestones due in the course of the semester, leading to a demo and write-up near the end of the semester.
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Exercise 3. We do not have a relation for GivenBy. In part a , there is a relation for Courses and a relation for LabCourses that has only the key and the computer-allocation attribute.
It looks like: And for c , C o u r s e sand LabCourses are combined, as: However, we think that the attributes x, y, and z together form another key.
Solutions Manual for Database Systems: The Complete Book, 2nd Edition
The reason is that at no time can two molecules occupy the same point. A more systematic way is to consider the closures of all 15 nonempty sets of attributes.
Now consider pairs of attributes: The collection of 11 new dependencies mentioned above is: All other sets either do not have ABCD as the closure or contain one of these three sets. That is, a superkey that is not a key must contain B and more than one of A, C, and D. Since A1A For suppose A1A Then A1A An is not closed. Here are the calculations for the remaining six sets: These are: Thus, any dependency above that does not have one of these pairs on the left is a BCNF violation.
Since all attributes are in at least one key of ABCD, that relation is already in 3NF, and no decomposition is necessary. However, any other nontrivial, derived FD will have A and B on the left, and therefore will contain a key.
It is obtained starting with any of the four violations mentioned above. Thus, we know that R must have at least the nine tuples of the form a,b,c , where b is any of b1, b2, or b3, and c is any of c1, c2, or c3.
Solutions to Selected Exercises
That is, we can derive, using the definition of a multivalued dependency, that each of the tuples a,b1,c2 , a,b1,c3 , a,b2,c1 , a,b2,c3 , a,b3,c1 , and a,b3,c2 are also in R.
There are two multivalued dependencies that do not follow from these functional dependencies. First, the information about one child of a person is independent of other information about that person.
That is, if a person with social security number s has a tuple with cn,cs,cb, then if there is any other tuple t for the same person, there will also be another tuple that agrees with t except that it has cn,cs,cb in its components for the child name, Social Security number, and birthdate. Consider two tuples xyzw and xy'z'w' in the relation R in question.
Solutions to Selected Exercises
Similarly, we can take the pair xy'z'w' and xy'zw and swap Z's to get xy'zw' and xy'z'w. In conclusion, we started with tuples xyzw and xy'z'w' and showed that xyzw' and xy'z'w must also be in the relation. Note that the above statements all make sense even if there are attributes in common among X, Y, and Z.
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Search inside document. The Complete Book: Solutions for Chapter 3 Database Systems: Return to Top Solutions for Section 3.
Solutions for Chapter 3 Solutions for Section 3.
Solutions for Chapter 3 attributes A2 through An may independently be chosen in or out. Solutions for Chapter 3 Exercise 3.
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Zeb Aslam.To further illustrate, imagine a relational model with two different tables people and friend. In contrast, a graph database would search for all the users in "", then follow the back-links through the subscriber relationship to find the subscriber users.
View a full sample. Grading: Questions about the grading have to be raised with the TA within a week after the graded assignment was returned.
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