Xorathrust

Overview

• Overall difficulty for me (From 1-10 stars): ★☆☆☆☆☆☆☆☆☆

• Challenge static score: 25

Backgrond

In the challenge’s description, we can download an attachment. Let’s download it:

┌[root♥siunam]-(~/ctf/KnightCTF-2023/Cryptography/Xorathrust)-[2023.01.21|18:21:12(HKT)]
└> mv /home/nam/Downloads/Xorathrust-20230121T102216Z-001.zip .
┌[root♥siunam]-(~/ctf/KnightCTF-2023/Cryptography/Xorathrust)-[2023.01.21|18:22:29(HKT)]
└> unzip Xorathrust-20230121T102216Z-001.zip 
Archive:  Xorathrust-20230121T102216Z-001.zip
  inflating: Xorathrust/flag.enc.txt  
  inflating: Xorathrust/encrypt.py
┌[root♥siunam]-(~/ctf/KnightCTF-2023/Cryptography/Xorathrust)-[2023.01.21|18:22:40(HKT)]
└> cd Xorathrust

Find the flag

flag.enc.txt:

┌[root♥siunam]-(~/ctf/KnightCTF-2023/Cryptography/Xorathrust/Xorathrust)-[2023.01.21|18:24:46(HKT)]
└> xxd flag.enc.txt 
00000000: 2d25 3220 1d0c 1353 1239 5239 0452 530f  -%2 ...S.9R9.RS.
00000010: 0539 1e56 141b                           .9.V..

encrypt.py:

def main():

    flag_enc = ""

    with open("flag.txt", "r") as infile:
        flag = infile.read()
        flag = list(flag)

        for each in flag:
            each = chr(ord(each) ^ 0x66)
            flag_enc += each


    with open("flag.enc", "w") as outfile:
        outfile.write(flag_enc)



if __name__ == "__main__":
    main()

As you can see, the original flag.txt is being XOR’ed, and every characters of flag.txt is XOR’ed by hex 66.

Armed with above information, we can reverse it by XOR’ing hex 66:

#!/usr/bin/env python3

XORedflag = ''

with open('flag.enc.txt', 'rb') as file:
    flag = file.read()
    flag = list(flag)
    
    for character in flag:
        XORedflag += chr(character ^ 0x66)

print(XORedflag)

┌[root♥siunam]-(~/ctf/KnightCTF-2023/Cryptography/Xorathrust/Xorathrust)-[2023.01.21|18:30:30(HKT)]
└> python3 solve.py
KCTF{ju5t_4_b45ic_x0r}

Found the flag!

• Flag: KCTF{ju5t_4_b45ic_x0r}

Conclusion

What we’ve learned:

1. Decrypting XOR’ed File