Exercises (7, 8) p. 292:
Short Use Case
Press controller button
Garage door open
Garage door close
Laser beam detects an object
Obstacle sensor detects an obstacle
Garage door stops functioning
Exercise (1, 2) page 292:
1. The scenario in the case of if the customer enters three digits from his/her PIN, the last
digit is remaining, and then leaves. The user is now on screen 2 (Please enter your PIN).
The system should detect that when there is n
Chapter 9: Exercise 3 p. (183)
3. According to the book we can use the following definition and using nodes
DEF (commission, 33), DEF (commission, 37), DEF (commission, 39) and USE (commission,
P1= <33, 34, 35, 36, 37, 38, 39, 40, 41, 42>
P2= <38, 39,
The Lowest Score Drop Program:
public class LowestScoreDrop cfw_
static int s1,s2,s3,s4,s5, lowest;
public static void getScore()cfw_
Scanner k= new Scanner(System.in);
System.out.print("Please enter the first
Exercise (1, 7) page 157:
1. The cyclomatic complexity of graph 8.3 given n= 9, e= 13, p= 1
V (G )=en+ p=139+ 1=5
7. No, it will not cause a problem. Since the transitive property of equality might ignore the
possibility that a triangle can be formed from
Exercises (1,2,4,5,6) page 75:
1. The length of a path in a graph can be defined as the number of edges in the path.
For example, in the following graph the length of the path from node a to node c is 2,
which is the number of edges in the path.
2. If the
Exercises p.130 #(1, 2, 3):
1. Since we have 7 conditions there must be 27= 128 independent rules
c1: a< b + c?
c2: b< a + c?
c3: c< a + b?
c4: a<= b <= c?
c5: a = b?
c6: a = c?
c7: b = c?
a1: not a triangle
Exercises p. 114: (1, 3, 5)
1. By revising the day classes in the NextDate function we can have five day classes instead
of four day classes, that will result in the following equivalence classes:
M1= cfw_month: month has 30 days
M2= cfw_month: month has
Exercise (1,2,3) page 96:
1. There are seven test cases in the Robustness testing, which are (min-, min, min+,
nom, max-, max, max+). Therefore, each variable now has to assume 6 different
values each whilst the other values are assuming their nominal val
Chapter 3 Exercises:
a. The symmetric difference of A and B.
b. The relative complement of (A intersection B) with respect to (A union B).
c. For example we have two sets A=cfw_1,3,6,7 and B=cfw_3,6,7,9
A B=cfw_ 1,3,6,7,9 , A B=cfw_ 3, 6,7 (