Exercises¶

1. • Question
   • Answer

   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

2. 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
3. • Question
   • Answer

   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.

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   (ex_6_3)

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   def grade(mark): if mark >= 90: return "A" else: if mark >= 80: return "B" else: if mark >= 70: return "C" else: if mark >= 60: return "D" else: return "F" mark = 83 print( "Mark:", str(mark), "Grade:", grade(mark))

   (q3_question)

4. 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)

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   from test import testEqual def findHypot(a,b): # your code here testEqual(findHypot(12.0, 5.0), 13.0) testEqual(findHypot(14.0, 48.0), 50.0) testEqual(findHypot(21.0, 72.0), 75.0) testEqual(findHypot(1, 1.73205), 1.999999)

   (ex_6_6)

5. • Question
   • Answer

   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.

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   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)

   (ex_6_7)

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   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)

   (q7_answer)

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

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   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)

   (ex_6_8)

7. • Question
   • Answer

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

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   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)

   (ex_6_9)

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   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)

   (q9_answer)

8. 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
       ...
   

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   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)

   (ex_6_10)

9. • Question
   • Answer

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

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   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)

   (ex_6_11)

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   from test import testEqual def is_rightangled(a, b, c): is_rightangled = False if a > b and a > c: is_rightangled = abs(b**2 + c**2 - a**2) < 0.001 elif b > a and b > c: is_rightangled = abs(a**2 + c**2 - b**2) < 0.001 else: is_rightangled = abs(a**2 + b**2 - c**2) < 0.001 return is_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)

   (q11_answer)

10. Write the function is_prime(n) that returns True when n is prime and False otherwise.

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    (ex_6_11b)

11. A year is a leap year if it is divisible by 4 unless it is a century that is not divisible by 400. Write a function that takes a year as a parameter and returns True if the year is a leap year, False otherwise.

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    from test import testEqual def isLeap(year): # your code here testEqual(isLeap(1944), True) testEqual(isLeap(2011), False) testEqual(isLeap(1986), False) testEqual(isLeap(1800), False) testEqual(isLeap(1900), False) testEqual(isLeap(2000), True) testEqual(isLeap(2056), True)

    (ex_6_12)

12. • Question
    • Answer

    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.

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    (ex_6_13)

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    year = int(input("Please enter a 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 dateofeaster = 22 + d + e if year == 1954 or year == 2981 or year == 2049 or year == 2076: dateofeaster = dateofeaster - 7 if dateofeaster > 31: print("April", dateofeaster - 31) else: print("March", dateofeaster) else: print("ERROR...year out of range")

    (answer_ex_6_13)