#python : I want the correction of this exercise

Question

Loop
Write a program that asks the user for a 4-digit code and leaves him 3 attempts
to find the correct password (the password to find is password = “2021"

Answer 1

isnt an attempt required to be corrected?

Answer 2

attempts = 0
start = True
while start:
attempts += 1
in = int(input())
if in == 2021:
print("correct")
break
if attempts == 3:
start = False

Answer 3

Mamadou Drame I saw this same question with a code attempt. Is this a duplicate?

EDIT: 👍 maybe it was someone else that has a similar similar class. In addition to ∆BH∆Y comment a counter variable to test and branch elsewhere after 3 attempts.

Answer 4

Mamadou Drame
I'm not writing the code to improve yourself. I'm just hinting.😉

1. Use loop
2. Ask Password
3. Compare input with Password
4. Before loop, define a variable that holds the number of attempts
5. Increase this variable by 1 each time and compare with 3 attempts
6. If the variable is 3, print a warning and end the loop.

I wish you success😊

Answer 5

where is the code to correct?

Answer 6

I have an exam that our teacher gave us

Answer 7

what did u tried so far?

Answer 8

Hello ! well I have tried beautiful and there is no rush

Answer 9

There is a small modification

print ('password combo to continue')
count = 0
while count <3:
password = input ('Enter password:')
if password == '2021'
print ('Access granted')
break

Answer 10

#this is an advanced code, hopefully you will understand it and learn smthg from it

import re

trials = 0

pwd = 2021

pat = "\d{4}"

while trials <= 2:
try:
t = int(input("Enter password: "))
if t == pwd:
print("access granted")
break
elif re.search(pat, str(t)) == None:
print("only 4 digits allowed!")
trials += 1
print("Trial N°", trials)
else:
print("Access denied")
trials += 1
print("Trial N°", trials)

except ValueError:
print("Invalid input! only digits allowed!")
trials += 1
print("Trial N°", trials)

Answer 11

for i in "21":
if input() == "2021":
print("Access Granted!")
exit()
else: print("Attempt Failed!\n{} attempt{} left".format(i, "s" * (i > "1")))
raise ValueError("INTRUDER ALERT!")

# Hope this helps