Homecooked was in the Cryptography category, but it was actually an optimization challenge.
The file attached with the challenge is a python file known as decrypt.py.After running the decrypt file, it outputs the following character by character and then executes indefinitely```flag{pR1m3s_4```which is part of the flag.This indicates that this file itself is the decrypt file and it will output our flag, but in indefinite amount of time.
The code of the decrypt.py file was```import base64num = 0count = 0cipher_b64 = b"MTAwLDExMSwxMDAsOTYsMTEyLDIxLDIwOSwxNjYsMjE2LDE0MCwzMzAsMzE4LDMyMSw3MDIyMSw3MDQxNCw3MDU0NCw3MTQxNCw3MTgxMCw3MjIxMSw3MjgyNyw3MzAwMCw3MzMxOSw3MzcyMiw3NDA4OCw3NDY0Myw3NTU0MiwxMDAyOTAzLDEwMDgwOTQsMTAyMjA4OSwxMDI4MTA0LDEwMzUzMzcsMTA0MzQ0OCwxMDU1NTg3LDEwNjI1NDEsMTA2NTcxNSwxMDc0NzQ5LDEwODI4NDQsMTA4NTY5NiwxMDkyOTY2LDEwOTQwMDA="
def a(num):    if (num > 1):        for i in range(2,num):            if (num % i) == 0:                return False                break        return True    else:        return False def b(num):    my_str = str(num)    rev_str = reversed(my_str)    if list(my_str) == list(rev_str):       return True    else:       return False
cipher = base64.b64decode(cipher_b64).decode().split(",")
while(count < len(cipher)):    if (a(num)):        if (b(num)):            print(chr(int(cipher[count]) ^ num), end='', flush=True)            count += 1            if (count == 13):                num = 50000            if (count == 26):                num = 500000    else:        pass    num+=1
print()```
Reverse Engineering the challenge and reimplementing the program in an optimized way was, I guess, was the intended solution.
The way i solved the challenge is by looking at the code a bit and optimizing that code itself.
When we open the file in a text editor, we see that it is not much code just 40 lines of code.
Lets start to understand the code:
It first base64 decodes the cipher_b64 and creates the list out of it which is```['100', '111', '100', '96', '112', '21', '209', '166', '216', '140', '330', '318', '321', '70221', '70414', '70544', '71414', '71810', '72211', '72827', '73000', '73319', '73722', '74088', '74643', '75542', '1002903', '1008094', '1022089', '1028104', '1035337', '1043448', '1055587', '1062541', '1065715', '1074749', '1082844', '1085696', '1092966', '1094000']```
Starting from the bottom, we see a while loop which runs for 40 times so we can assume that the length of the flag is 40 characters.
```while(count < len(cipher)):    if (a(num)):        if (b(num)):```
Inside the while loop there is are two nested if statements which calls function **a** and **b** and expects a true value to execute it, if any of the value returned, either from function a or b, is false then it will not go inside if and jump to else which is just pass and then increments the num variable```    else:        pass    num+=1```The function **a** takes a number as an argument and checks if it is a prime```        for i in range(2,num):            if (num % i) == 0: # checks for prime numbers numbers returns false if not a prime number                return False                break        return True
```
Function **b** takes a number (which is prime) and checks if its a palindrome (i.e. if the number is same if read in reverse, example 111, 10101).```    my_str = str(num)    rev_str = reversed(my_str)    if list(my_str) == list(rev_str):  # checks if its a palindrome       return True```
The *slowness* of the program is due to the increment of the **num** variable on 13th and 26th iteration.
So, in a nutshell, the program takes a number, first checks if its prime and then checks if its a palindrome.
The weakness lies here that it first checks for prime and then for palindrome.
To optimize it we can first check for palindrome and then for prime because it has to check for palindromic primes, and the prime checking function **a** takes a lot of time.
**The Solution** is just to swap the functions in the if clause, i.e. change a to b and b to a in the while loop's if clause.
After changing it, it takes around 5 seconds to output the flag.
The solution script is ```import base64num = 0count = 0cipher_b64 = b"MTAwLDExMSwxMDAsOTYsMTEyLDIxLDIwOSwxNjYsMjE2LDE0MCwzMzAsMzE4LDMyMSw3MDIyMSw3MDQxNCw3MDU0NCw3MTQxNCw3MTgxMCw3MjIxMSw3MjgyNyw3MzAwMCw3MzMxOSw3MzcyMiw3NDA4OCw3NDY0Myw3NTU0MiwxMDAyOTAzLDEwMDgwOTQsMTAyMjA4OSwxMDI4MTA0LDEwMzUzMzcsMTA0MzQ0OCwxMDU1NTg3LDEwNjI1NDEsMTA2NTcxNSwxMDc0NzQ5LDEwODI4NDQsMTA4NTY5NiwxMDkyOTY2LDEwOTQwMDA="
def a(num):    if (num > 1):        for i in range(2,num):            if (num % i) == 0: # checks for prime numbers nos returns false if not a prime number                return False                break        return True    else:        return False def b(num):    my_str = str(num)    rev_str = reversed(my_str)    if list(my_str) == list(rev_str):  # checks if its a palindrome       return True    else:       return False
cipher = base64.b64decode(cipher_b64).decode().split(",")
while(count < len(cipher)):    if (b(num)):        # if number is palindrome, the only change from the original script        if (a(num)):    # if number is prime             print(chr(int(cipher[count]) ^ num), end='', flush=True)            count += 1            if (count == 13):                num = 50000            if (count == 26):                num = 500000    else:        pass    num+=1
print()```And it happily gives us the flag```flag{pR1m3s_4re_co0ler_Wh3n_pal1nDr0miC}```