7.11. Exercises

  1. What do these expressions evaluate to?

    1. 3 == 3

    2. 3 != 3

    3. 3 >= 4

    4. not (3 < 4)

    (ex_6_1)

    1. True

    2. False

    3. False

    4. False

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  1. Give the logical opposites of these conditions. You are not allowed to use the not operator.

    1. a > b

    2. a >= b

    3. a >= 18  and  day == 3

    4. a >= 18  or  day != 3

    (ex_6_2)

  1. Write a function which is given an exam mark, and it returns a string — the grade for that mark — according to this scheme:

    Mark

    Grade

    >= 90

    A

    [80-90)

    B

    [70-80)

    C

    [60-70)

    D

    < 60

    F

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

    Test your function by printing the mark and the grade for a number of different marks.

    (ex_6_3)

    18
     
    1
    def grade(mark):
    2
        if mark >= 90:
    3
            return "A"
    4
        else:
    5
            if mark >= 80:
    6
                return "B"
    7
            else:
    8
                if mark >= 70:
    9
                    return "C"
    10
                else:
    11
                    if mark >= 60:
    12
                        return "D"
    13
                    else:
    14
                        return "F"
    15
    ​
    16
    mark = 83
    17
    print( "Mark:", str(mark), "Grade:", grade(mark))
    18
    ​

    (q3_question)

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  1. 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.

    (ex_6_4)

  1. 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. 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).

    (ex_6_5)

    46
     
    1
    import turtle
    2
    ​
    3
    def drawBar(t, height):
    4
        """ Get turtle t to draw one bar, of height. """
    5
        t.begin_fill()               # start filling this shape
    6
        if height < 0:
    7
            t.write(str(height))
    8
        t.left(90)
    9
        t.forward(height)
    10
        if height >= 0:
    11
            t.write(str(height))
    12
        t.right(90)
    13
        t.forward(40)
    14
        t.right(90)
    15
        t.forward(height)
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        t.left(90)
    17
        t.end_fill()                 # stop filling this shape
    18
    ​
    19
    ​
    20
    ​
    21
    xs = [48, -50, 200, 240, 160, 260, 220]  # here is the data
    22
    maxheight = max(xs)
    23
    minheight = min(xs)
    24
    numbars = len(xs)
    25
    border = 10
    26
    ​
    27
    tess = turtle.Turtle()           # create tess and set some attributes
    28
    tess.color("blue")
    29
    tess.fillcolor("red")
    30
    tess.pensize(3)
    31
    ​
    32
    wn = turtle.Screen()             # Set up the window and its attributes
    33
    wn.bgcolor("lightgreen")
    34
    if minheight > 0:
    35
        lly = 0
    36
    else:
    37
        lly = minheight - border
    38
    ​
    39
    wn.setworldcoordinates(0-border, lly, 40*numbars+border, maxheight+border)
    40
    ​
    41
    ​
    42
    for a in xs:
    43
        drawBar(tess, a)
    44
    ​
    45
    wn.exitonclick()
    46
    ​

    (answer_ex_6_5)

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  1. Write a function findHypot. The function will be given the length of two sides of a right-angled triangle and it should return the length of the hypotenuse. (Hint: x ** 0.5 will return the square root, or use sqrt from the math module)

    (ex_6_6)

  1. 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.

    (ex_6_7)

    13
     
    1
    from test import testEqual
    2
    ​
    3
    def is_even(n):
    4
        if n % 2 == 0:
    5
            return True
    6
        else:
    7
            return False
    8
    ​
    9
    testEqual(is_even(10), True)
    10
    testEqual(is_even(5), False)
    11
    testEqual(is_even(1), False)
    12
    testEqual(is_even(0), True)
    13
    ​

    (q7_answer)

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  1. Now write the function is_odd(n) that returns True when n is odd and False otherwise.

    (ex_6_8)

  1. Modify is_odd so that it uses a call to is_even to determine if its argument is an odd integer.

    (ex_6_9)

    19
     
    1
    from test import testEqual
    2
    ​
    3
    def is_even(n):
    4
        if n % 2 == 0:
    5
            return True
    6
        else:
    7
            return False
    8
    ​
    9
    def is_odd(n):
    10
        if is_even(n):
    11
            return False
    12
        else:
    13
            return True
    14
    ​
    15
    testEqual(is_odd(10), False)
    16
    testEqual(is_odd(5), True)
    17
    testEqual(is_odd(1), True)
    18
    testEqual(is_odd(0), False)
    19
    ​

    (q9_answer)

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  1. 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 as

    if  abs(x - y) < 0.001:      # if x is approximately equal to y
    ...

    (ex_6_10)

  1. Extend the above program so that the sides can be given to the function in any order.

    (ex_6_11)

    20
     
    1
    from test import testEqual
    2
    ​
    3
    def is_rightangled(a, b, c):
    4
        is_rightangled = False
    5
    ​
    6
        if a > b and a > c:
    7
            is_rightangled = abs(b**2 + c**2 - a**2) < 0.001
    8
        elif b > a and b > c:
    9
            is_rightangled = abs(a**2 + c**2 - b**2) < 0.001
    10
        else:
    11
            is_rightangled = abs(a**2 + b**2 - c**2) < 0.001
    12
        return is_rightangled
    13
    ​
    14
    testEqual(is_rightangled(1.5, 2.0, 2.5), True)
    15
    testEqual(is_rightangled(4.0, 8.0, 16.0), False)
    16
    testEqual(is_rightangled(4.1, 8.2, 9.1678787077), True)
    17
    testEqual(is_rightangled(4.1, 8.2, 9.16787), True)
    18
    testEqual(is_rightangled(4.1, 8.2, 9.168), False)
    19
    testEqual(is_rightangled(0.5, 0.4, 0.64031), True)
    20
    ​

    (q11_answer)

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  1. 3 criteria must be taken into account to identify leap years:

    The year is evenly divisible by 4;

    If the year can be evenly divided by 100, it is NOT a leap year, unless;

    The year is also evenly divisible by 400. Then it is a leap year.

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

    (ex_6_12)

  1. Implement the calculator for the date of Easter.

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

    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. dateofeaster = 22 + d + e

    Special note: The algorithm can give a date in April. 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.

    (ex_6_13)

    19
     
    1
    year = int(input("Please enter a year"))
    2
    if year >= 1900 and year <= 2099:
    3
        a = year % 19
    4
        b = year % 4
    5
        c = year % 7
    6
        d = (19*a + 24) % 30
    7
        e = (2*b + 4*c + 6*d + 5) % 7
    8
        dateofeaster = 22 + d + e
    9
    ​
    10
        if year == 1954 or year == 2981 or year == 2049 or year == 2076:
    11
            dateofeaster = dateofeaster - 7
    12
    ​
    13
        if dateofeaster > 31:
    14
            print("April", dateofeaster - 31)
    15
        else:
    16
            print("March", dateofeaster)
    17
    else:
    18
        print("ERROR...year out of range")
    19
    ​

    (answer_ex_6_13)

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