Description

Practical Examination 2
Time allowed: 2 hours
Instructions (please read carefully):

1. This is an open-book exam. You are allowed to bring in any course or reference materials in printed form. No electronic media or storage devices are allowed.
2. This practical exam consists of two questions. The time allowed for solving this test is 2 hours.
3. The maximum score of this test is 20 marks. Note that the number of marks awarded for each question IS NOT correlated with the difficulty of the question.
4. You are advised to attempt all questions. Even if you cannot solve a question correctly, you are likely to get some partial credit for a credible attempt.
5. While you are also provided with the templates q1-template.c and q2-template.c to work with, your answers should be submitted on Coursemology. Note that you can only run the test cases on Coursemology for a limited number of tries because they are only for checking that your code is submitted correctly. You are expected to test your own code for correctness locally on VS Code and not depend only on the provided test cases. Do ensure that you pasted your answers correctly by running the test cases at least once.
6. In case there are problems with Coursemology.org and we are not able to upload the answers to Coursemology.org, you will be required to name your file q1-<student no>.c and q2-<student no>.c where <student no> is your student number and leave the file in the tic1001-pe1 folder on the Desktop. If your file is not named correctly, we will choose any .c file found at random.
7. Please note that it shall be your responsibility to ensure that your solution is submitted correctly to Coursemology and correctly left in the desktop folder at the end of the examination. Failure to do so will render you liable to receiving a grade of ZERO for the Practical Exam, or the parts that are not uploaded correctly.
8. Please note that while sample executions are given, it is not sufficient to write programs that simply satisfy the given examples. Your programs will be tested on other inputs and they should exhibit the required behaviours as specified by the problems to get full credit. There is no need for you to submit test cases.
ALL THE BEST!
Question 1: [15 marks]
Temasek Railway Metro System (TRMS) has recently upgraded the signalling system on their subway lines.
In the new system, each rail line is divided into blocks of a unit length. Each station on the line will occupy a particular block. For example, in the figure below, Station A is on block 44 and Station B is on block 89.

Station A Station B
Âa˘
Fixed speed trains. A train travelling at speed s covers exactly s blocks in each time unit.
Suppose a time unit is the smallest divisible unit, that means the train cannot exactly cover a distance smaller than s.
For example, if the train in the figure above is travelling at speed s = 2, it will never be able to stop exactly at Station B if it leaves from Station A. Because from block 44, it will go to block 46 in the next time unit, then block 48, and so on, until block 88. But in the time unit it will overshoot to block 90.
However, at speed s = 3, the train will move from block 44 to 47, then 47 to 50, and eventually reach block 86 and will be able to stop exactly at block 89.
A. Write a function int best_speed(int start, int end, int max_speed) which takes in the block number of the starting and ending station, and the maximum speed which the train can travel. The function returns the maximum integer speed that the train can travel at and yet be able to exactly reach the destination station. [5 marks]
Sample Execution:
Function call Return value

best_speed(44, 89, 4) 3 best_speed(44, 89, 5) 5 best_speed(44, 89, 10) 9

Variable Speed trains Suppose now the trains have to accelerate up to their max speed from 0 and decelerate back to 0 to stop. The acceleration and deceleration rate for a train is 1 unit speed per unit time, i.e., in one time unit, the speed and increase or decrease by 1.
Using the same example in the figure, if the maximum speed of the train is s = 5, then the following table shows the location and speed of the train over time:
Time Block Speed
0 44 0
1 44 1
2 45 2
3 47 3
4 50 4
5 54 5
6 59 5
7 64 5
8 69 5
9 74 5
10 79 4
11 83 3
12 86 2
13 88 1
14 89 0
At t = 0 the train’s speed is s = 0. Thus, at t = 1 it is still at the same place, but its speed has increased to s = 1. So in t = 2 it has moved to the next block. Its speed then keep increasing until it accelerates up to its maximum speed.
Notice then that it has to start decelerating at t = 10 in order to be able to stop at the ending block.
B. Write a function int stopping_distance(int speed) that takes as input the current speed of the train, and returns the minimum number of blocks required for the train to come to a complete stop at the given speed. [5 marks]
Sample Executions:
Function call Return value

stopping_distance(4) 10 stopping_distance(5) 15

C. Write a function travel(int start, int end, int max_speed) which takes in the block number of the starting and ending station, and the maximum speed which the train can travel. The function then prints the location and speed of the train for each time unit like in the table above. Each time unit is on a new line and the numbers are separated by a single space.
[5 marks]
Sample Executions (on next page):
travel(44, 89, 5); travel(44, 89, 10); travel(44, 89, 4);
0 44 0 0 44 0 0 44 0
1 44 1 1 44 1 1 44 1
2 45 2 2 45 2 2 45 2
3 47 3 3 47 3 3 47 3
4 50 4 4 50 4 4 50 4
5 54 5 5 54 5 5 54 4
6 59 5 6 59 6 6 58 4
7 64 5 7 65 6 7 62 4
8 69 5 8 71 5 8 66 4
9 74 5 9 76 4 9 70 4
10 79 4 10 80 3 10 74 4
11 83 3 11 83 3 11 78 4
12 86 2 12 86 2 12 82 3
13 88 1 13 88 1 13 85 2
14 89 0 14 89 0 14 87 1
15 88 1
16 89 0
Question 2: Hello World! from space! [5 marks]
Each deep space message is represented as a character string. Write a function void filter(const char *original, const char threshold, char *passed, char *failed) which separates (filters) the original string into two output strings:
1. passed string: Contains all characters in the original that is larger or equal to the threshold character in terms of ASCII value.
2. failed string: Contains all characters in the original that is lesser than the threshold character.
Sample runs

Original Threshold Passed Failed

“ABC 1234!@#”
“ABCDEF abcdef 1234!@#” ‘D’ “DEFabcdef”
“PE is FUN!” ” “PE is FUN!” “”

Note that:
• The order of the characters in passed and failed strings should be the same as in the original.
• All characters of the original string should be in either passed or failed.
• It is possible to have empty string as an answer.
[5 marks]
Appendix: ASCII Table
ASCII Control Characters
Num Sym Description
0 NUL Null char
1 SOH Start of Heading
2 STX Start of Text
3 ETX End of Text
4 EOT End of Transmission
5 ENQ Enquiry
6 ACK Acknowledgment
7 BEL Bell
8 BS Backspace
9 HT Horizontal Tab
10 LF Line Feed
11 VT Vertical Tab
12 FF Form Feed
13 CR Carriage Return
14 SO Shift Out / X-On
15 SI Shift In / X-Off
16 DLE Data Line Escape
17 DC1 Device Control 1
18 DC2 Device Control 2
19 DC3 Device Control 3
20 DC4 Device Control 4
21 NAK Negative Acknowl
22 SYN Synchronous Idle
23 ETB End of Trans Block
24 CAN Cancel
25 EM End of Medium
26 SUB Substitute
27 ESC Escape
28 FS File Separator
29 GS Group Separator
30 RS Record Separator
31 US Unit Separator
— E N D

ASCII Printable Characters

Num
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63 Sym
!
”
#
$
%
&
‘
(
)
*
+
,
.
/
0
1
2
3
4
5
6
7
8
9 :
;
<
=
> ? Num
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95 Sym
@
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
Q
R
S
T
U
V
W
X
Y
Z
[

]
^
_
O F P A P E R —

Num Sym
96 `
97 a
98 b
99 c
100 d
101 e
102 f
103 g
104 h
105 i
106 j
107 k
108 l
109 m
110 n
111 o
112 p
113 q
114 r
115 s
116 t
117 u
118 v
119 w
120 x
121 y
122 z
123 {
124 |
125 }
126 ~
127 <DEL>