Help with A.I password cracker

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1
import random target_password = str(input("47983848:/n")) password_length = len(target_password) # Characters to create random passwords characters = ["1", "2", "3", "4", "5", "6", "7", "8", "9", "0",\ "a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m",\ "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z"] # Flag to stop guessing when attempt is sucessfull status = "ongoing" def crack(): global target_password, status, password_length # Number of guesses count = 0 while status == "ongoing": guess = "" # Create a string from randomly chosen characters while len(guess) < password_length: # Add a random index from the characters list guess += random.choice(characters) # Compare guess to the real password if guess == target_password: print("Password cracked!: " + str(guess)) status = "finished" count += 1 print(str(count) + " guesses") else: count+= 1 while password != "": status = "ongoing" crack() print("________") target_password = str(input("Enter another password you wish to test:\n")) password_length = len(target_password)

it says that the line with else: count += 1 has invalid syntax can anyone help with the issue. this is for my stem fair

2

Hello @raisbello0516962928, welcome to the forums! As it’s currently indented, line 21 ( global … ) throws an indentation error, but I assume in the environment you’re coding in, it doesn’t? Can you please repost your code with indentation preserved?

3

@codeneutrino i’ve now edited the message with indentation. now the error should be the else statement

4

The error is because you have code between the if statement and the else statement that isn’t within the if block. An else block should follow directly from an if block, meaning this works:

if condition:
    code
else:
    code

But this will throw an error:

if condition:
    code
print("This line is not within the if block")
else:
    code

Similarly, you cannot have an empty else block, so this throws an error:

if condition:
   code
else:
print("This line is not indented within the else block")
3 Likes
5

thank you for the assistance :+1:

2 Likes
6

but I just changed the code and removed the print, but it still says invalid syntax

7

Can you repost your code with all indentation preserved, and say what error message you’re getting?

8

I just reposted the code with indent preserved but its still providing the same error message

9

Did you edit your original message, because that’s throwing the same errors as before. Could you please post your new code in a new message?

10
import random target_password = str(input("47983848:/n")) password_length = len(target_password) # Characters to create random passwords characters = ["1", "2", "3", "4", "5", "6", "7", "8", "9", "0",\ "a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m",\ "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z"] # Flag to stop guessing when attempt is sucessful status = "ongoing" def crack(): global target_password, status, password_length # Number of guesses count = 0 while status == "ongoing": guess = "" # Create a string from randomly chosen characters while len(guess) < password_length: # Add a random index from the characters list guess += random.choice(characters) # Compare guess to the real password if guess == target_password: print("Password cracked!: " + str(guess)) status = "finished" count += 1 else: count+= 1 while password != "": status = "ongoing" crack() print("________") target_password = str(input("Enter another password you wish to test:\n")) password_length = len(target_password)

still says “SyntaxError: invalid syntax” on line 26

11

That’s because you have code between your if block and your else block. This throws an error:

if condition:
    pass
print("This line is not within the if block")
else:
    pass

Notice how your if block ends on line 22, and your else block doesn’t start until line 25.

3 Likes
12

thank you. its working now.

1 Like