We investigated a program which is probably used as one component of a bigger password breaking algorithm. We determined that th
1 answer:
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Answer:
<em>The program written in Python 3 is as follows;</em>
<em>The program does not make use of comments; however, see explanation section for detailed line by line explanation of the program</em>
num = int(input("Number of Inputs: "))
mylist = []
userinput = int(input("Enter a digit: "))
i = 1
while i < num:
mylist.append(userinput)
userinput = int(input("Enter a digit: "))
i += 1
try:
first7 = mylist.index(7)+1
print('The first position of 7 is ',first7)
last7 = num - 1 - mylist[::-1].index(7)
print('The last position of 7 is ',last7)
except:
print('7 is not on the list')
Explanation:
This line prompts user for the number of inputs
num = int(input("Number of Inputs: "))
This line declares an empty list
mylist = []
This line inputs the first number into the empty list
userinput = int(input("Enter a digit: "))
The italicized lines represent an iteration that allows user to input the numbers into the empty list.
<em>i = 1</em>
<em>while i < num:</em>
<em> mylist.append(userinput)</em>
<em> userinput = int(input("Enter a digit: "))</em>
<em> i += 1</em>
The try-except is used to check the presence of 7 in the list
try:
This checks the first occurrence of 7
first7 = mylist.index(7)+1
This prints the first occurrence of 7
print('The first position of 7 is ',first7)
This checks the last occurrence of 7
last7 = num - 1 - mylist[::-1].index(7)
This prints the last occurrence of 7
print('The last position of 7 is ',last7)
The following is executed if there's no occurrence of 7
except:
print('7 is not on the list')
See Attachments for sample runs
