Python Cyber Security - Function to suggest character substitutions for stronger passwords
Python Cyber Security: Exercise-4 with Solution
Write a Python function that takes a password as input and returns a list of common character substitutions that could be used to create a stronger password.
For example, if the password is "myspass", the function should return
['my$pass', 'my5pass', 'mySpass', 'mysp@ss', 'mysp4ss', 'myspAss', 'myspa$s', 'myspa5s', 'myspaSs', 'myspas$', 'myspas5', 'myspasS', 'myspass!', 'myspass123', 'myspass@', 'myspass#', 'myspass$', 'myspass%', 'myspass&', 'myspass*', 'myspass-', 'myspass_', 'myspass=', 'myspass+']
Sample Solution:
Python Code:
def get_password_variants(password):
pass_variants = []
substitutions = {
'a': ['@', '4', 'A'],
'e': ['3', 'E'],
'i': ['1', '!', 'I'],
'o': ['0', 'O'],
's': ['$', '5', 'S'],
't': ['7', 'T'],
'z': ['2', 'Z']
}
for i in range(len(password)):
if password[i] in substitutions:
for sub in substitutions[password[i]]:
pass_variant = password[:i] + sub + password[i+1:]
pass_variants.append(pass_variant)
pass_variants.append(password + '!')
pass_variants.append(password + '123')
pass_variants.append(password + '@')
pass_variants.append(password + '#')
pass_variants.append(password + '$')
pass_variants.append(password + '%')
pass_variants.append(password + '&')
pass_variants.append(password + '*')
pass_variants.append(password + '-')
pass_variants.append(password + '_')
pass_variants.append(password + '=')
pass_variants.append(password + '+')
return pass_variants
password = input("Input your password: ")
result_variants = get_password_variants(password)
print(result_variants)
Sample Output:
Input your password: pqre3sd ['pqr33sd', 'pqrE3sd', 'pqre3$d', 'pqre35d', 'pqre3Sd', 'pqre3sd!', 'pqre3sd123', 'pqre3sd@', 'pqre3sd#', 'pqre3sd$', 'pqre3sd%', 'pqre3sd&', 'pqre3sd*', 'pqre3sd-', 'pqre3sd_', 'pqre3sd=', 'pqre3sd+']
Flowchart:
Python Code Editor:
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