Week 3: Class Exercise solutions¶
In-class exercise: multiple conditions for if-else¶
Code:
# In-class exercise: multiple conditions for if-else (Instructor Solution)
# Author: Safwan Zulfazli and Tan Nian Wei
woke_up_late = True # or False
weather = "rainy" # or "sunny"
if woke_up_late:
if weather == "rainy":
print("Eat Maggi TomYam")
else:
print("Eat Maggi Asam Laksa")
else:
print("Eat At Dewan Makan")
Output:
Eat Maggi TomYam
elif¶
Code:
# In-class exercise: multiple conditions for if-else (Instructor Solution)
# Author: Safwan Zulfazli and Tan Nian Wei
item = "food waste" # Can be substituted with other items as above
# Did you write the following?
if item == "glass":
print("Put glass in brown bin")
else:
if item == "paper":
print("Put paper in blue bin")
else:
if item == "aluminium":
print("Put aluninium cans in orange bin")
else:
if item == "plastic":
print("Put plastic in orange bin")
else:
print("Food waste is not for recycling!")
# Use elifs makes it much cleaner!
if item == "glass":
print("Put glass in brown bin")
elif item == "paper":
print("Put paper in blue bin")
elif item == "aluminium":
print("Put aluninium cans in orange bin")
elif item == "plastic":
print("Put plastic in orange bin")
elif item == "food waste":
print("Food waste is not for recycling!")
Output:
Food waste is not for recycling!
In-class exercise: functions¶
Code:
# In-class exercise: functions (instructor solution)
# Author: Safwan Zulfazli and Tan Nian Wei
def what_int(x):
if(x < 0):
print(x, " is negative")
elif(x > 0):
print(x, " is positive")
else:
print(x , " is not positive or negative!")
what_int(1)
what_int(-3)
what_int(0)
Output:
1 is positive
-3 is negative
0 is not positive or negative!
Exercise: Fizz Buzz¶
Code:
for i in range(1, 101):
if (i % 3 == 0) and (i % 5 == 0):
print("FizzBuzz")
elif (i % 3 == 0):
print("Fizz")
elif (i % 5 == 0):
print("Buzz")
else:
print(i)
Output:
1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz
16
17
...
...
...
89
FizzBuzz
91
92
Fizz
94
Buzz
Fizz
97
98
Fizz
Buzz
Exercise: Quadratic solver¶
Code:
def solve_root(a, b, c);
discr = b ** 2 - 4 * a * c
if discr == 0:
return (-b + discr ** 0.5) / (2 * a)
else:
x1 = (-b + discr ** 0.5) / (2 * a)
x2 = (-b - discr ** 0.5) / (2 * a)
return x1, x2