10. Exercises¶

What do these expressions evaluate to?
1. 3 == 3
2. 3 != 3
3. 3 >= 4
4. not (3 < 4)
1. True
2. False
3. False
4. False
Write a function which is given an exam score, and it returns the corresponding letter grade as a string according to this scheme:

score Grade

>= 90 A

[80-90) B

[70-80) C

[60-70) D

< 60 F

The square and round brackets denote closed and open intervals, respectively. A closed interval includes the number, an open interval excludes it. So 79.99999 gets grade C , but 80 gets grade B.

Test your function by printing the score and the grade for a number of different scores.

def grade(score): if score >= 90: return "A" else: if score >= 80: return "B" else: if score >= 70: return "C" else: if score >= 60: return "D" else: return "F" def main(): score = 83 print( "Score:", str(score), "Grade:", grade(score)) if __name__ == "__main__": main()
Modify the turtle bar chart program from the previous chapter so that the bar for any value of 200 or more is filled with red, values between [100 and 200) are filled yellow, and bars representing values less than 100 are filled green.
In the Turtle bar chart program, what do you expect to happen if one or more of the data values in the list is negative? Go back and try it out. Then change the program so that when it prints the text value for the negative bars, it puts the text above the base of the bar (on the 0 axis).

import turtle def draw_bar(t, height): """ Get turtle t to draw one bar, of height. """ t.begin_fill() # start filling this shape if height < 0: t.write(str(height)) t.left(90) t.forward(height) if height >= 0: t.write(str(height)) t.right(90) t.forward(40) t.right(90) t.forward(height) t.left(90) t.end_fill() # stop filling this shape def main(): data = [48, -50, 200, 240, 160, 260, 220] max_height = max(data) min_height = min(data) num_bars = len(data) border = 10 tess = turtle.Turtle() # create tess and set some attributes tess.color("blue") tess.fillcolor("red") tess.pensize(3) wn = turtle.Screen() # Set up the window and its attributes wn.bgcolor("lightgreen") if min_height > 0: bottom = 0 else: bottom = min_height - border wn.setworldcoordinates(0-border, bottom, 40 * num_bars + border, max_height + border) for x in data: draw_bar(tess, x) wn.exitonclick() if __name__ == "__main__": main()
Write a function called is_even(n) that takes an integer as an argument and returns True if the argument is an even number and False if it is odd. Note that instead of printing out the results we are using test statements. The goal is to pass all the tests that are listed underneath the function you will write. You do not need to add a main function to this code to run it.
from test import testEqual def is_even(n): # your code here testEqual(is_even(10), True) testEqual(is_even(5), False) testEqual(is_even(1), False) testEqual(is_even(0), True)
from test import testEqual def is_even(n): if n % 2 == 0: return True else: return False testEqual(is_even(10), True) testEqual(is_even(5), False) testEqual(is_even(1), False) testEqual(is_even(0), True)
Now write the function is_odd(n) that returns True when n is odd and False otherwise.
from test import testEqual def is_odd(n): # Your code here testEqual(is_odd(10), False) testEqual(is_odd(5), True) testEqual(is_odd(1), True) testEqual(is_odd(0), False)
Modify is_odd so that it uses a call to is_even to determine if its argument is an odd integer.
from test import testEqual def is_odd(n): # your code here testEqual(is_odd(10), False) testEqual(is_odd(5), True) testEqual(is_odd(1), True) testEqual(is_odd(0), False)
from test import testEqual def is_even(n): if n % 2 == 0: return True else: return False def is_odd(n): if is_even(n): return False else: return True testEqual(is_odd(10), False) testEqual(is_odd(5), True) testEqual(is_odd(1), True) testEqual(is_odd(0), False)
Write a fruitful function called pick_activity to help you pick an activity to engage in based on the current weather. It has two parameters, one for how hot it is and one for how wet it is. If it is hot and wet, it should tell you to watch Netflix. If it hot and dry, it should tell you to go swimming. If it is cold and wet, it should tell you to paint. If it is cold and dry, it should tell you to go to a cafe and read. Use the elif construct.

def pick_activity(temp, damp): if temp == "hot" and damp == "wet": message = "Watch Netflix" elif temp == "hot" and damp == "dry": message = "Go swimming" elif temp == "cold" and damp == "wet": message = "Paint!" elif temp == "cold" and damp == "dry": message = "Got to a cafe and read" else: message = "Invalid input. Enter hot or cold, wet or dry." return message def main(): print(pick_activity("hot", "wet")) if __name__ == "__main__": main()
Write a function is_rightangled which, given the length of three sides of a triangle, will determine whether the triangle is right-angled. Assume that the third argument to the function is always the longest side. It will return True if the triangle is right-angled, or False otherwise.

Hint: floating point arithmetic is not always exactly accurate, so it is not safe to test floating point numbers for equality. If a good programmer wants to know whether x is equal or close enough to y, they would probably code it up using the abs() function like so:
```
if  abs(x - y) < 0.001:      # if x is approximately equal to y
    ...
```
from test import testEqual def is_rightangled(a, b, c): # your code here testEqual(is_rightangled(1.5, 2.0, 2.5), True) testEqual(is_rightangled(4.0, 8.0, 16.0), False) testEqual(is_rightangled(4.1, 8.2, 9.1678787077), True) testEqual(is_rightangled(4.1, 8.2, 9.16787), True) testEqual(is_rightangled(4.1, 8.2, 9.168), False) testEqual(is_rightangled(0.5, 0.4, 0.64031), True)
Extend the above program so that the sides can be given to the function in any order.
from test import testEqual def is_rightangled(a, b, c): # your code here testEqual(is_rightangled(1.5, 2.0, 2.5), True) testEqual(is_rightangled(16.0, 4.0, 8.0), False) testEqual(is_rightangled(4.1, 9.1678787077, 8.2), True) testEqual(is_rightangled(9.16787, 4.1, 8.2), True) testEqual(is_rightangled(4.1, 8.2, 9.168), False) testEqual(is_rightangled(0.5, 0.64031, 0.4), True)
from test import testEqual def is_rightangled(a, b, c): rightangled = False if a > b and a > c: rightangled = abs(b**2 + c**2 - a**2) < 0.001 elif b > a and b > c: rightangled = abs(a**2 + c**2 - b**2) < 0.001 else: rightangled = abs(a**2 + b**2 - c**2) < 0.001 return rightangled testEqual(is_rightangled(1.5, 2.0, 2.5), True) testEqual(is_rightangled(4.0, 8.0, 16.0), False) testEqual(is_rightangled(4.1, 8.2, 9.1678787077), True) testEqual(is_rightangled(4.1, 8.2, 9.16787), True) testEqual(is_rightangled(4.1, 8.2, 9.168), False) testEqual(is_rightangled(0.5, 0.4, 0.64031), True)
Implement the calculator for the date of Easter.

The following algorithm computes the date for Easter Sunday for any year between 1900 to 2099 inclusive.

Ask the user to enter a year. Compute the following:
1. a = year % 19
2. b = year % 4
3. c = year % 7
4. d = (19 * a + 24) % 30
5. e = (2 * b + 4 * c + 6 * d + 5) % 7
6. date = 22 + d + e
Special note: The algorithm can give a date greater than 31 (the number of days in March). When this happens, it signifies a date in April. Thus, 32 is April 1, 35 is April 4, and so on. Also, if the year is one of four special years (1954, 1981, 2049, or 2076) then subtract 7 from the date.

Your program should print an error message if the user provides a date that is out of range.
def date_of_easter(year): # Your code here def main(): # Your code here if __name__ == "__main__": main()
def date_of_easter(year): if year >= 1900 and year <= 2099: a = year % 19 b = year % 4 c = year % 7 d = (19*a + 24) % 30 e = (2*b + 4*c + 6*d + 5) % 7 date = 22 + d + e if year == 1954 or year == 1981 or year == 2049 or year == 2076: date = date - 7 if date > 31: return("April " + str(date - 31)) else: return("March " + str(date)) else: return("ERROR...year out of range") def main(): year = int(input("Please enter a year")) print(date_of_easter(year)) if __name__ == "__main__": main()

score	Grade
>= 90	A
[80-90)	B
[70-80)	C
[60-70)	D
< 60	F

Graded Lesson Assignment¶

A year is a leap year if it is divisible by 4, unless it is the first year of a century and it is not divisible by 400.

For example:

1956 is a leap year because 1956 is divisible by 4.
1957 is not a leap year, because it is not divisible by 4.
1900 is not a leap year, despite the fact that it is divisible by 4, because 1900 is the first year of a century and 1900 is not divisible by 400.
1600 is a leap year, because 1600 is divisible by 4 and 1600 (although it is the first year of a century) is divisible by 400

Write a function is_leap that takes a year as a parameter and returns True if the year is a leap year, False otherwise.

Ready to submit your code? Time to fork, and submit a Pull Request to: this Github Repository.