My solution for 1.4 was:
n = 11
acc = 0
for j in range(n):
acc = acc + 2 ** j
print(acc)
If we use while we have to implement the couting which is done by for automatically. So we introduce the variable j which starts with 0, we iterate up to n - 1 and must not forget to increase this counter at the end of the code block after while:
n = 11
acc = 0
j = 0
while j < n:
acc = acc + 2 ** j
j = j + 1
print(acc)
The rewritten solution of 2.3.B is:
n = 5
odd_numbers_added = 0
even_numbers_added = 0
i = 1
while i <= n:
if i % 2 == 1:
odd_numbers_added = odd_numbers_added + i
else:
even_numbers_added = even_numbers_added + i
i = i + 1
print(even_numbers_added, odd_numbers_added)
In case the user enters "x" at the beginning the count is zero and the computation of the average is not defined. We handle this case explicitly.
If you test your programs do not forget to test corner cases !!!
summed = 0
count = 0
while True:
number = input("please input a number, use 'x' or 'X' to finish: ")
if number == 'x' or number == 'X':
break
summed = summed + float(number)
count = count + 1
if count > 0:
average = summed / count
else:
average = "not computable"
print("you provided", count, "numbers. their sum is", summed, "and the average is", average)
The integier division // produces an integer. If you use regular division / we get "ugly" floating point values with trailing .0 in the output:
n = int(input("n = ? "))
length = 1
while n != 1:
print(n, end=" ")
if n % 2 == 0:
n = n // 2
else:
n = 3 * n + 1
length += 1
print(1)
print("collatz sequence has length", length)
a = int(input("a ? "))
b = int(input("b ? "))
while a != b:
if a > b:
a = a - b
else:
b = b - a
print("gcd is", a)
summed = 0
count = 0
number = input("please input a number, use 'x' or 'X' to finish: ")
while number != "x" and number != "X":
summed = summed + float(number)
count = count + 1
number = input("please input a number, use 'x' or 'X' to finish: ")
if count > 0:
average = summed / count
else:
average = "not computable"
print("you provided", count, "numbers. their sum is", summed, "and the average is", average)
The following snippet tries to solve the computation of the largest of the given input values (I stripped the other computations so we can focus on the actual problem).
If you test it you will see that it fails if the user enters x at the beginning, or if the user provides only numbers small than -1000:
max_so_far = -1000
while True:
number = input("please input a number, use 'x' or 'X' to finish: ")
if number == "x" or number == "X":
break
number = float(number)
if number > max_so_far:
max_so_far = number
print("the maximum of the provided numbers is", max_so_far)
Even if you change the initial value -1000 the problem is not solved, you always find user inputs which break the computation.
So the actual problem is to find a proper starting value for max_so_far.
For this we must handle the first input differently to the other inputs. Either we use a separate input for the first value, or we need an extra variable to track if the current input is the first input or another one.
So try to understand the following solutions and then read the preceeding sentences again !
"""
here we use an extra input for the first value
"""
summed = 0
count = 0
number = input("please input a number, use 'x' or 'X' to finish: ")
if number != "x" and number != "X":
max_so_far = float(number)
while number != "x" and number != "X":
summed = summed + float(number)
count = count + 1
number = input("please input a number, use 'x' or 'X' to finish: ")
if number != "x" and number != "X":
number = float(number)
if number > max_so_far:
max_so_far = number
if count > 0:
average = summed / count
else:
average = "not computable"
max_so_far = "unknown"
print("you provided", count, "numbers. their sum is", summed, "and the average is", average)
print("further the maximum of the provided numbers is")
If you use an infinite while loop instead you have to follow a different strategy to handle the input of the first value:
"""
here we use an start value with a different type to distinguish the first from the other inputs
"""
summed = 0
count = 0
max_so_far = "unknown"
while True:
number = input("please input a number, use 'x' or 'X' to finish: ")
if number == "x" or number == "X":
break
number = float(number)
summed = summed + number
count = count + 1
if max_so_far == "unknown":
max_so_far = number
elif number > max_so_far:
max_so_far = number
if count > 0:
average = summed / count
else:
average = "not computable"
print("you provided", count, "numbers. their sum is", summed, "and the average is", average)
print("further the maximum of the provided numbers is", max_so_far)
The solution below uses an extra indicator variable to handle the first input aproppriately:
"""
here we use an extra indictator variable (also called 'flag') to distinguish the first from the other inputs
"""
summed = 0
count = 0
max_so_far = 0
first_value_handled = False
while True:
number = input("please input a number, use 'x' or 'X' to finish: ")
if number == "x" or number == "X":
break
number = float(number)
summed = summed + number
count = count + 1
if not first_value_handled:
max_so_far = number
first_value_handled = True
elif number > max_so_far:
max_so_far = number
if count > 0:
average = summed / count
else:
average = "not computable"
print("you provided", count, "numbers. their sum is", summed, "and the average is", average)
if first_value_handled:
print("further the maximum of the provided numbers is", max_so_far)
else:
print("further the maximum of the provided numbers is unknown")
This exercise was very tricky ! We have to use two variable to track the longest collatz sequence and the according valueof n. As the collatz loop changes n we have to use anothe variable for the outer loop:
n_max = 100
max_length = 0
optimal_n = 0
for n_start in range(1, n_max + 1):
length = 1
n = n_start
while n != 1:
if n % 2 == 0:
n = n // 2
else:
n = 3 * n + 1
length += 1
if length > max_length:
max_length = length
optimal_n = n_start
print("n =", optimal_n, "produces longest collatz sequence of length", max_length, "for starting values 2 ..", n_max)