[gwml] [Quiz] Reversing Simple Password 2

[gwml] [Quiz] Reversing Simple Password 2

[Ammar Faizi]

Hi everyone,

We have another reversing challenge to play.

#quiz #endianess #xor
Find the password
https://www.gnuweeb.org/quiz/002

002: ELF 64-bit LSB executable, x86-64, version 1 (SYSV), statically 
linked, BuildID[sha1]=cb1dcfe619ded0d6051c712739204dfbd02da347, not stripped

   Ammar Faizi

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Re: [gwml] [Quiz] Reversing Simple Password 2

[Ammar Faizi]

Don't forget to post your solution on Telegram group when you solve it.

   Ammar Faizi


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Re: [gwml] [Quiz] Reversing Simple Password 2

[Ammar Faizi]

Hi everyone,

This is the answer of latest GNU/Weeb quiz.

Lookie here.

   ...
   ...
   ...
   .Ldont_cut:
           movq    $0x0062656557554e47, %rax
           xorq    %rax, (%rsp)            # xor the input with %rax
           movq    $0x00565c5c66000000, %rax
           cmpq    %rax, (%rsp)            # compare the xor'ed input
           jne     .Lwrong_password
           leaq    str3(%rip), %rsi
           movl    $end_str - str3, %edx
           jmp     .Lprint_last_str
   .Lwrong_password:
           leaq    str2(%rip), %rsi
           movl    $str3 - str2, %edx
   .Lprint_last_str:
           movl    $1, %eax
           movl    $1, %edi
           syscall
   .Lexit:
           movl    $60, %eax
           xorl    %edi, %edi
           syscall


The input is located at (%rsp). And then it gets xor'ed with
0x0062656557554e47. It is actually a C string "GNUWeeb\0" in little
endian representation.

Note:
This is just a trick to make `strings -o 002` yields GNUWeeb, so it
looks like the password is found by static analysis. While this is
actually not. The password is not "GNUWeeb".

Next...

And then after the input gets xor'ed with that number, it compares
the result with 0x00565c5c66000000. If it equals, then the password is
correct, otherwise it is wrong password.

Now, consider the nature of XOR operation:
   A ^ B = C
   B ^ C = A
   A ^ C = B

Hence, we have:
   0x0062656557554e47 ^ ??? = 0x00565c5c66000000
   0x0062656557554e47 ^ 0x00565c5c66000000 = ???

For the easy, use GDB to calculate it:
   gef➤  p 0x0062656557554e47 ^ 0x00565c5c66000000
   $1 = 0x34393931554e47

Okay, so we know that we have to make the QWORD (%rsp) be
0x34393931554e47. So what kind of input corresponds to that little
endian representation?

We can use PHP hex2bin function for the easy.

   ammarfaizi2@integral:/tmp$ php -a
   Interactive shell

   php > echo hex2bin('34393931554e47');
   4991UNG
   php >

Oops, we have to make it yields little endian, so we reverse the
string.

   php > echo strrev(hex2bin('34393931554e47'));
   GNU1994
   php >

Finally, we got the password "GNU1994".

   ammarfaizi2@integral:/tmp$ ./002
   Enter the password: GNU1994
   Congratulation, you have solved the quiz!
   ammarfaizi2@integral:/tmp$

If you have any question, kindly to reply to this email. Or drop your
question on GNU/Weeb Telegram group.

Happy hacking!

   Ammar Faizi

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