# Root-Me - Xor Madness - Walkthrough

### Denis Nuțiu ・3 min read

Hello,

In this article I will describe how I solved the **PE x86 - Xor Madness** challenge from Root-Me

This challenge will ask you for a password and the password is also used to validate the flag. What makes this challenge interesting is that it only uses `xor`

, `sub`

, `call`

and `ret`

.

Here's how I approached the challenge:

Since the binary had a few function and some strings were in plain text, I tried to figure out the big picture and labeled all the function accordingly:

`win`

,`lose`

,`check`

and so on.I've figured out that there are 6 stages, each stage tries to process a part of the password. The start function is also a stage and other stages are similar.

Next, I've spend a few hours debugging this challenge and documenting everything it does in order to see what it wants.

To solve this challenge is used IDA Pro and it's debugger, this challenge can also be solved with the help of x64dbg.

To solve this challenge you need to know how the XOR operation works, please check that you know this before moving forward.

The behaviour of the challenge is simple, it computes an addr, let's call this address SOLADDR, and it makes it point to the value 1. `[SOLADDR] = 1`

.

It takes a byte of the password you've gave and XORs it two times. The resulting value will be an address, we can call it PASSADDR. The trick is to look at how your password is XORed and make it equal to SOLADDR.

The function that will let you proceed to the next stage only does so when the location pointed to by PASSADDR has the value 1 assigned to it. This is stored into edx. `edx = [PASSADDR]`

.

The following function lets you proceed to the next stage if and only if EDX is 1:

```
weird_ret_addr_change proc near
arg2_next_stage_address= dword ptr 4
xor eax, eax
xor eax, [esp+arg2_next_stage_address]
sub eax, [esp+0] ; subtract return address from EAX
mul dl ; multiply eax with DL. Lower part of EDX
add eax, [esp+0] ; add back the return address
xor ecx, ecx ; reset ecx
xor ecx, [esp+0] ; set ecx to [esp]
xor [esp+0], ecx ; reset [esp]; [esp] = 0
xor [esp+0], eax ; set [esp] to eax
retn 4
```

We already know that EDX is set to [PASSADDR], which is EDX = *PASSADDR. Each stages performs this step.

In the main function, scanf scans for a maximum 63 characters (don't trust anything you read, verify!) and stores the result in EDI, then it moved it into ESI and gets the first byte of the password_input into EBX.

```
call ds:scanf
xor esp, esp
xor esp, edi
xor esi, esi
xor esi, esp
xor edi, edi
xor edi, esi
sub esp, 100h
xor ebx, ebx
xor bl, [esi] ; ebx contains the first byte of the password
```

EBX is then xored with a value, the first 3 bytes are just for show, we only care about the resulting last byte.

```
xor ebx, 59307554h ; ebx contains first byte
.text:004010C0 ;
.text:004010C0 ; Example:
.text:004010C0 ;
; in xor res
.text:004010C0 ; 0x30 ^ 0x54 -> 0x64 test case
.text:004010C0 ; 0x6C ^ 0x54 -> 0x38 attempt1
.text:004010C0 ; 0x24 ^ 0x54 -> 0x70 attempt2 - success
.text:004010C0 ;
.text:004010C0 ; The First Xor
```

The next operations loads ESP into EBX, XORs them with the *last resulting byte* from the previous operation and saves it somewhere.

```
.text:004010D0
xor eax, esp ; eax = 0x0019FD74
xor eax, ebx ; eax is now xored with the our byte 0x64 (res)
;
; 0x0019FD74 ^ 0x64 = 0x0019FD10
; the resulting address is used when setting EDX
; in this case it is invalid, the correct address is
; === 0x0019FD04
xor edx, edx
xor edx, eax ; edx = 0x0019FD10 (Addr resulting our xored byte)
xor ecx, ecx ; ecx = 0
xor ecx, [eax] ; try to get value from [eax]
xor [eax], ecx ; [EAX] = 0xFFFFFFFE
; eax = 0
```

Note that EDX now contains the address that we control 19FD10, which we can call PASSADDR. We obtained 19FD10 after running the program and giving it `0`

as the input.

The next piece of code computes the SOLADDR by XORing ESP with two hardcoded values and makes it point to the value of 1.

```
.text:004010E0
xor eax, esp ; esp = 0x0019FD74
xor eax, 0F2h ; 0x0019FD74 ^ 0xF2 = 0x0019FD86
xor eax, 82h ; 0x0019FD86 ^ 0x82 = 0x0019FD04
xor ecx, ecx ; ecx = 0
xor ecx, [eax] ; ecx = *(0x0019FD04)
xor [eax], ecx ; [eax] = [eax] ; eax = 0
xor byte ptr [eax], 1 [eax] = 1
```

Now, the code assigns the value pointed by PASSADDR to EDX. `EDX = [PASSADDR]`

.

```
xor eax, eax ; eax = 0;
xor eax, edx ; eax = 0x0019FD10; edx is set above; PASSADDR
xor edx, edx ; edx = 0
xor edx, [eax] ; edx = [0x0019FD10]; edx = [PASSADDR]
```

This is the only information you need to know to solve this challenge. You can now find the password by statically analyzing the next stages.

Thanks for reading!