EBK DATABASE SYSTEM CONCEPTS
7th Edition
ISBN: 9781260049268
Author: SILBERSCHATZ
Publisher: MCG COURSE
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Expert Solution & Answer
Chapter 7, Problem 33E
a.
Explanation of Solution
Candidate keys
- A key is known as a candidate key when it has the ability to derive all the attributes from a relation...
b.
Explanation of Solution
Canonical cover for F
- The steps for canonical conversion is to break the functional dependencies so that right hand side will have only one single attribute.
- Then all the extra unnecessary functional dependencies are removed.
- Then all the functional dependencies are combined.
- Hence here the given functional dependencies include AB -> G, AB -> D ...
c.
Explanation of Solution
Remaining steps of
- The given relation is in 1NF as there is no multivalued attribute or complex attribute.
- The given relation is not in 2NF as there is partial functional dependency...
d.
Explanation of Solution
Final decomposition
- Given functional dependencies are AB -> CD, ADE -> GDE, B -> GC, G -> DE.
- Here ADE -> GDE have some trivial attributes.
- So those attributes can be removed and hence the dependency is ADE -> G...
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Exercise 1 Function and Structure [30 pts]
Please debug the following program and answer the following questions. There is a cycle in a linked
list if some node in the list can be reached again by continuously following the next pointer.
#include
typedef struct node {
int value;
struct node *next;
} node;
int 11_has_cycle (node *first)
if (first
==
node *head =
{
NULL) return 0;
first;
while (head->next != NULL) {
}
if (head
first) {
return 1; }
head = head->next;
return 0;
void test ll_has_cycle () {
int i;
node nodes [6];
for (i = 0; i < 6; i++) {
nodes [i] .next = NULL;
nodes [i].value = i;
}
nodes [0] .next
=
&nodes [1];
nodes [1] .next
=
&nodes [2];
nodes [2] .next
=
&nodes [3];
nodes [3] .next
nodes [4] .next
&nodes [4];
NULL;
nodes [5] .next = &nodes [0];
printf("1. Checking first list for cycles. \n Function 11_has_cycle says it
has s cycle\n\n", 11_has_cycle (&nodes [0])?"a":"no");
printf("2. Checking length-zero list for cycles. \n Function 11_has_cycle
says it has %s…
Chapter 7 Solutions
EBK DATABASE SYSTEM CONCEPTS
Ch. 7 - Prob. 1PECh. 7 - Prob. 2PECh. 7 - Explain how functional dependencies can be used to...Ch. 7 - Prob. 4PECh. 7 - Prob. 5PECh. 7 - Prob. 6PECh. 7 - Prob. 7PECh. 7 - Prob. 8PECh. 7 - Prob. 9PECh. 7 - Prob. 10PE
Ch. 7 - Prob. 11PECh. 7 - Prob. 12PECh. 7 - Prob. 13PECh. 7 - Prob. 14PECh. 7 - Prob. 15PECh. 7 - Prob. 16PECh. 7 - Prob. 17PECh. 7 - Prob. 18PECh. 7 - Prob. 19PECh. 7 - Prob. 20PECh. 7 - Prob. 21ECh. 7 - Prob. 22ECh. 7 -
Explain what is meant by repetition of...Ch. 7 -
Why are certain functional dependencies called...Ch. 7 - Prob. 25ECh. 7 - Prob. 26ECh. 7 - Prob. 27ECh. 7 - Prob. 28ECh. 7 - Prob. 29ECh. 7 - Prob. 30ECh. 7 - Prob. 32ECh. 7 - Prob. 33ECh. 7 - Prob. 35ECh. 7 - Prob. 36ECh. 7 - Prob. 37ECh. 7 - Prob. 38ECh. 7 - Prob. 39ECh. 7 - Prob. 40ECh. 7 - Prob. 41ECh. 7 - Prob. 42ECh. 7 - Prob. 43E
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