cover image for post ' – Matteo KeygenMe by Matteo' – Matteo KeygenMe by Matteo

The crackme Matteo KeygenMe by Matteo has been published February 24, 2015. It is rated at 4 - Needs special knowledge. The crackme is written in Assembler and runs on Windows.

The crackme has two major parts. The first part is all about trying to stop you from getting to the relevant code by throwing a handful of anti-debugging and anti-disassembly techniques at you. The second part then validates the key file that you provide.

I used IDA Pro and WinDBG to solve the crackme. No anti-debugging scripts have been used — all tricks have been manually defused.

Part 1: Anti-Debugging Measures

Trick 1: TLS Callbacks

The crackme uses TLS callbacks. Those are invoked before the main entry point:

.data:0040400C TlsCallbacks_ptr dd offset TlsCallbacks
.data:00404010 TlsSizeOfZeroFill dd 0
.data:00404014 TlsCharacteristics dd 0
.data:00404018 TlsIndex        db    0                 
.data:00404019                 db    0
.data:0040401A                 db    0
.data:0040401B                 db    0
.data:0040401C TlsCallbacks    dd offset TlsCallback_0 
.data:00404020 dword_404020    dd 0                  
.data:00404024 dword_404024    dd 0                    

Make sure you have set a breakpoint at the TlsCallback_0 offset, or that the debugger is set to stop at TLS callbacks.

Trick 2: Dynamically Added TLS Callbacks

All the first TLS callback does is create another TLS callback:

00401EC7 TlsCallback_0 proc near               
00401EC7 mov     dword_404020, offset loc_401E76
00401ED1 retn
00401ED1 TlsCallback_0 endp

Add a breakpoint at loc_401E76 and run there.

Trick 3: Anti-Disassembly with Fake Jumps and Garbage Bytes

The second TLS callback routine at first probably looks like that:

00401E76 push    (offset loc_401E7F+1)
00401E7B stc
00401E7C jnb     short loc_401E7F
00401E7E retn
00401E7F ; -------------------------------------------------------------------------
00401E7F loc_401E7F:                             
00401E7F jmp     dword ptr [esi-74h]
00401E82 ; -------------------------------------------------------------------------
00401E82 shl     byte ptr [esi-72h], 1
00401E85 shl     byte ptr [esi-64h], 1

The above code uses a fake conditional jump (the jump is always taken), a PUSH/POP-jump (the RETN instruction actually jumps to loc_401E7F+1), and garbage bytes. This causes the wrong disassembly above. Once you step into the code with debugger, IDA should show the correct disassembly. You can also manually fix the disassembly:

00401E76 push    offset loc_401E80
00401E7B stc
00401E7C jnb     short near ptr byte_401E7F
00401E7E retn
00401E7E ; -------------------------------------------------------------------------
00401E7F byte_401E7F db 0FFh                     
00401E80 ; -------------------------------------------------------------------------
00401E80 loc_401E80:                             
00401E80 mov     ax, ss
00401E83 mov     ss, ax

Trick 4: Trap Flag

Next come theses instructions:

00401E80 mov     ax, ss
00401E83 mov     ss, ax
00401E86 pushfw
00401E88 test    byte ptr [esp+1], 1
00401E8D jnz     short loc_401E9C
00401E8F popfw

The second MOV-instruction writes SS, this causes the processor to lock all interrupts until after the following instruction. The PUSHFW instruction then pushes the flags on the stack, and [esp+1] accesses the trap flag. Since interrupts, including INT 1, are locked until after PUSHFW, the debugger has no chance to unset the flag.

Either don’t single step over the pushfw instruction, or manually set the zero flag at the jump. The routine sets a third callback routine.

Trick 5: Push/Ret-Jumps

Because of the anti-disassembly techniques before, the third callback routine is not yet marked as code:

00401DBB db  68h ; h
00401DBC db 0C5h ; +
00401DBD db  1Dh
00401DBE db  40h ; @
00401DBF db    0
00401DC0 db 0F8h ; °
00401DC1 db  72h ; r
00401DC2 db    1

Hitting C in IDA produces the following disassembly:

00401DBB ; -----------------------------
00401DBB push    offset off_401DC5
00401DC0 clc
00401DC1 jb      short near ptr unk_401DC4
00401DC3 retn
00401DC3 ; -----------------------------

This snippet uses the PUSH/RET-jump. Pushing the offset and then returning is almost equivalent to a regular jump, but they might prevent the disassembler from properly detecting the function boundaries.

Trick 6: Self-Modifying Code

The following instructions follow:

00401DCF loc_401DCF:                             
00401DCF mov     decrypted, 1
00401DD6 mov     eax, 771881844                  ; key
00401DDB mov     ecx, offset start_of_ciphertext
00401DE0 mov     edx, offset end_of_ciphertext
00401DE5 decryption_loop:                        
00401DE5 xor     [ecx], al
00401DE7 ror     eax, 1
00401DE9 inc     ecx
00401DEA cmp     ecx, edx
00401DEC jnz     short decryption_loop
00401DEC ; ------------------------------------------------------------------
00401DEE start_of_ciphertext db 23h   
00401DEF db 0DEh ; ¦
00401DF0 db  56h ; V
00401DF1 db  7Bh ; {
00401DF2 db  87h ; ç
00401DF3 db 0DBh ; ¦

After the decryption the cipertext turns into meaningful code:

00401DEE start_of_ciphertext
00401DEE push    edi
00401DEF mov     edx, large fs:30h

The crackme uses similar snippets in many places. Make sure to not set memory breakpoint inside the modified section. Memory breakpoints affect the memory, and those changes will get encrypted or decrypted too, leading to corrupt code.

Trick 7: PEB->BeingDebugged

The code then checks the BeingDebugged-flag of the PEB (equivalent to calling IsDebuggerPresent, but more sneaky):

00401DEE push    edi
00401DEF mov     edx, large fs:30h
00401DF6 mov     al, [edx+2]
00401DF9 test    al, al
00401DFB jnz     short being_debugged

Manually patch the flag, or set the zero flag at the jump.

Trick 8: NtGlobalFlag

With edx still pointing to the PEB, the crackme also checks the NtGlobalFlag field. The field has three flags that indicate the presence of a debugger:


The proper way to check all three flags would therefore be with bitmask 0x70. The crackme is cruder and just checks if the entire NtGlobalFlag field is zero:

00401DFD cmp     dword ptr [edx+68h], 0
00401E01 jnz     short being_debugged

Again patch the field or manually set the zero flag.

Trick 9: Software Breakpoint Detection

The entry point of any executable is an obvious place to put software breakpoints (opcode 0xCC). The next lines search 0x26 bytes of the memory starting at the entry point for 0xCC:

00401E04 mov     edi, offset start
00401E09 mov     ecx, 26h
00401E0E mov     al, 0CCh
00401E10 repne scasb
00401E12 jz      short being_debugged

Remove software breakpoints before the check, or manually unset the zero flag.

Trick 10: SEH - Triggered with Single Step Exception

After three more self modifying loops, we end here:

00401D01 push    8E4C9A90h
00401D06 push    838042730
00401D13 mov     edx, large fs:18h
00401D1A mov     ecx, [edx]
00401D1C xchg    ebx, ecx
00401D1E pop     eax
00401D1F add     [esp], eax
00401D22 push    ebx
00401D23 mov     ebx, esp
00401D25 xchg    ebx, ecx
00401D27 mov     [edx], ecx
00401D29 pushfw
00401D2B or      byte ptr [esp+1], 1
00401D30 popfw
00401D32 mov     eax, offset off_404041
00401D37 mov     byte ptr [eax], 0D4h

The offset fs:18h points to the Structured Exception Handler (SEH) on the stack. The code adds another handler with address 8E4C9A90h + 31F3846Ah = 0x401efa (mod 2**32). The exception is triggered by setting the trap flag (pushfw, or, popfw). Set a breakpoint at the new SEH handler. Then make sure you pass the single step exception 0x80000004 to the application. Debuggers often consume this exception (it is intended for debuggers after all).

Trick 11: Exception’s Context Structure - Safe Place

The exception handler at 0x401EFA first decrypts the following block:

00401F12 start_of_ciphertext_0:                  
00401F12 mov     eax, [esp+0Ch]
00401F16 mov     dword ptr [eax+0B8h], offset loc_401D3A

The third argument to the exception handler, passed in [eax+0Ch], is the context structure of the exception:

struct _CONTEXT
    +B8     EIP;  # register 

The field 0xB8 is set to 0x401D37, or the EIP we would return to after the exception handler. By overwriting the register to 0x401D3A, the crackme changes the flow to a different address. Take note of the address to later set a breakpoint there when returning from the SEH.

Trick 11: Exception’s Context Structure - Hardware Registers

The crackme then also checks four other offsets into the context structure:

00401F23 test    [eax+4], edx
00401F26 jnz     being_debugged_0
00401F2C test    [eax+8], edx
00401F2F jnz     being_debugged_0
00401F35 test    [eax+0Ch], edx
00401F38 jnz     being_debugged_0
00401F3E test    [eax+10h], edx
00401F41 jnz     being_debugged_0
00401F47 push    esi
00401F48 xor     edx, edx

Those four values are the debug registers:

struct _CONTEXT
    +04     Dr0;
    +08     Dr1;
    +0C     Dr2;
    +10     Dr3;

They indicate if the hardware breakpoints are set. Disable all hardware breakpoints, or manually unset the zero flag at each jump.

Trick 12: Checksum

The following disassembly calculates a checksum of 0x332 bytes from the entry point of the crackme:

00401F4A mov     esi, offset start
00401F4F mov     ecx, 332h
00401F54 cld
00401F55 loc_401F55:                             
00401F55 lodsd
00401F56 add     edx, eax
00401F58 rol     edx, 1
00401F5A dec     ecx
00401F5B jnz     short loc_401F55
00401F5D pop     esi
00401F5E cmp     edx, checksum
00401F64 jz      short checksum_is_good
00401F66 mov     eax, [esp+0Ch]
00401F6A mov     dword ptr [eax+0B8h], offset loc_401D32
00401F74 jmp     being_debugged_0
00401F79 ; ------------------------------------------------------------------

If the checksum does not match a hardcoded value, we are busted. The checksum method detects software breakpoints as well any kind of patches. Remove breakpoints and patches, or manually set the zero flag at 401f64.

Trick 13: Correct Exception Record -> C3

The first parameter of the exception handler points to an EXCEPTION_RECORD:

    +00:    ExceptionCode DWORD
    +04:    ExceptionFlags DWORD

The next instructions of the crackme read the exception code:

00401F79 checksum_is_good:                       
00401F79 mov     edx, [esp+4]
00401F7D mov     edx, [edx]

We know it was a single step exception that triggered the exception, see Trick 10. This exception has the code 0x80000004:

#define STATUS_SINGLE_STEP               ((DWORD) 0x80000004)

By shifting that right 5 bytes and adding 0x33h, we get the value 0xC3

00401F7F rol     edx, 5       ; ---> 80
00401F82 add     edx, 33h     ; ---> C3

The value 0xC3 is stored at 2 bytes into 0x404041

00401F85 mov     eax, offset byte_404041
00401F8A mov     [eax+2], dl

So the data becomes:

.data:00404041 dword_404041 dd 16C381B9h   

The effect of the change will be relevant at the start of the entry point. Make sure that C3 is written, and manually correct the byte if your debugger changed the exception code.

The crackme then checks if there is a software breakpoint at the jump table entry for GetProcessAddress:

00401F8D mov     edx, offset get_process_address
00401F92 cmp     byte ptr [edx], 0CCh
00401F95 jz      being_debugged_1

The get_process_address address disassembles to:

00402162 get_process_address:                    
00402162                                         ; 004020C5p ...
00402162 jmp     ds:GetProcAddress

The crackme also checks the actual offset of GetProcAddress by extracting the offset from the jmp instruction:

00401F9B mov     edx, [edx+2]
00401F9E mov     edx, [edx]
00401FA0 cmp     byte ptr [edx], 0CCh
00401FA3 jz      being_debugged_1

The crackme repeats the same two checks for GetModuleHandleA and GetModuleHandleW. Disable plugins that add breakpoints at these API, or just unset the zero flag.

Trick 15: Thread Hiding

The crackme dynamically loads the NtSetInformationThread function:

0040207A apis_clean:                             
0040207A push    offset aNtdll_dll               ; "NtDll.dll"
0040207F call    get_module_handle_w
00402084 test    eax, eax
00402086 jz      loc_40211A
0040208C push    offset aNtsetinformationth      ; "NtSetInformationThread"
00402091 push    eax
00402092 call    get_process_address
00402097 test    eax, eax
00402099 jz      short loc_40211A

The crackme also checks if there is a breakpoint at the beginning of the routine – hinting a potential anti-anti measure:

0040209B mov     edx, eax
0040209D cmp     byte ptr [edx], 0CCh
004020A0 jz      short being_debugged_1

Next, NtSetInformationThread is called with 0x11 as the second parameter:

004020A2 push    0
004020A4 push    0
004020A6 push    11h
004020A8 push    0FFFFFFFEh
004020AA call    eax

0x11 stands for HideThreadFromDebugger, which will causes the debugger to no longer receive any events. To skip the call in WinDbg you can use the following commands:

>r @eip = @eip + 2
>r @esp = @esp + 0x10

Trick 16: Patch Vectored Exception Handlers

The following disassembly follows. It patches the beginning of the API function AddVectoredExceptionHandler with the 5 bytes at return_null:

004020AC push    offset aKernel32_dll_0          ; "Kernel32.dll"
004020B1 call    get_module_handle_w
004020B6 push    ebx
004020B7 push    esi
004020B8 push    edi
004020B9 mov     ebx, eax
004020BB test    eax, eax
004020BD jz      short loc_402117
004020BF push    offset aWriteprocessmemory      ; "WriteProcessMemory"
004020C4 push    ebx
004020C5 call    get_process_address
004020CA test    eax, eax
004020CC jz      short loc_402117
004020CE mov     edi, eax
004020D0 cmp     byte ptr [edi], 0CCh
004020D3 jz      short being_debugged_1
004020D5 push    offset aAddvectoredexcepti      ; "AddVectoredExceptionHandler"
004020DA push    ebx
004020DB call    get_process_address
004020E0 test    eax, eax
004020E2 jz      short loc_402117
004020E4 push    0
004020E6 push    5
004020E8 push    offset return_null
004020ED push    eax
004020EE push    0FFFFFFFFh
004020F0 call    edi

The stub return_null is:

00402075            return_null:                           
00402075 33 C0      xor     eax, eax
00402077 C2 08 00   retn    8

So crackme replaces AddVectoredExceptionHandler with return 0. It does the same with AddVectoredContinueHandler. I had to ask the crackme author what the purpose of patching the VEH is. According to Matteo, they prevent VEH-based debugger from working, for example Cheat Engine.

After Trick 16, we leave the exception handler and return to the address that was set in Trick 11. First, the handler is removed:

00401D3A                         loc_401D3A:  
00401D3A 64 8B 15 18 00 00 00    mov     edx, large fs:18h
00401D41 8F 02                   pop     dword ptr [edx]
00401D43 44                      inc     esp
00401D44 83 C4 08                add     esp, 8
00401D47 83 EC 05                sub     esp, 5

Next, we again calculate a checksum for the first bytes of the entry point:

00401D4A 33 D2                   xor     edx, edx
00401D4C 33 C0                   xor     eax, eax
00401D4E 56                      push    esi
00401D4F 57                      push    edi
00401D50 BE 00 10 40 00          mov     esi, offset start
00401D55 B9 26 00 00 00          mov     ecx, 26h
00401D5A FC                      cld
00401D5B                         loc_401D5B:                             
00401D5B AC                      lodsb
00401D5C 03 D0                   add     edx, eax
00401D5E D1 C2                   rol     edx, 1
00401D60 49                      dec     ecx
00401D61 75 F8                   jnz     short loc_401D5B
00401D63 81 FA 2B C6 16 5D       cmp     edx, offset checksum2
00401D69 74 17                   jz      short checksum_is_good2

This does the same as Trick 12. After that, we finally are done with the TlsCallbacks and enter the entry point of the crackme.

Trick 17: Timing Check

The first step of the entry point is to decrypt the content at 401026. The decryption key is stored at 0x404041, this is where Trick 13 made the modification based on the exception record:

00401000 public start
00401000 start:
00401000 mov     esi, offset loc_404041
00401005 mov     ecx, 0CA5h
0040100A mov     edx, offset loc_401026
0040100F cld
00401010 loc_401010:                             
00401010 lodsb
00401011 cmp     esi, offset loc_404047
00401017 jl      short loc_40101E
00401019 mov     esi, offset loc_404041
0040101E loc_40101E:                             
0040101E dec     ecx
0040101F xor     [ecx+edx], al
00401022 test    ecx, ecx
00401024 jnz     short loc_401010

After some irrelevant code we get to these lines:

004010B4                 add     esp, 4
004010B7                 mov     time_in_secs, eax
004010BC                 call    ds:GetTickCount
004010C2                 xor     edx, edx
004010C4                 mov     ecx, 3E8h
004010C9                 div     ecx
004010CB                 mov     tickcount_in_secs, eax

Here we store the current time in seconds (retrieved by an earlier call to time, not shown), and the tick count in seconds. These values become relevant further down in the crackme, here:

0040153D                 cmp     tickcount_in_secs, 0
00401544                 jl      short loc_40154F
00401546                 cmp     tickcount_in_secs, 4
0040154D                 jle     short good      ; At most 4 seconds passed

and here:

00401567                 cmp     time_in_secs, 0
0040156E                 jl      short bad 
00401570                 cmp     time_in_secs, 4
00401577                 jg      short bad 

These two blocks check if at most 4 seconds passed according to time or the tick count. If either one is true, we are fine. If on the other hand a debugger causes a greater delay than four seconds, then later the crackme jumps over setting a flag at offset 004015BC:

004015B9 loc_4015B9:                             
004015B9                 jmp     short loc_4015BD
004015B9 ; ----------------------------------------------------------------
004015BB unk_4015BB      db 0EBh ; d             
004015BC ; ----------------------------------------------------------------
004015BC timing_ok:                              
004015BC                                         ; 004015A9j
004015BC                 inc     ecx
004015BD loc_4015BD:                             
004015BD                                         ; loc_4015B9j
004015BD                 pop     edx
004015BE                 and     ecx, edx

These are all the anti-debugging checks that I found. Part 2 shows how the key validation works.

Part 2: Key Validation

The Keyfile

The registration information is stored in a file called TheKey.k in the same directory as the crackme:

004010E0                 push    0
004010E2                 push    80h
004010E7                 push    3
004010E9                 push    0
004010EB                 push    0
004010ED                 push    80000000h
004010F2                 push    offset aThekey_k ; "TheKey.k"
004010F7                 call    ds:CreateFile
004010FD                 xor     dword ptr ds:aThekey_k, offset unk_218F6F18 
00401107                 xor     dword ptr ds:aThekey_k+4, offset unk_218F6F18
00401111                 mov     fileHandle, eax
00401116                 test    eax, eax
00401118                 jz      fail
0040111E                 inc     eax
0040111F                 test    eax, eax
00401121                 jz      fail
00401127                 push    0
00401129                 push    offset ContentLength
0040112E                 push    40h
00401130                 push    offset keyContent
00401135                 push    fileHandle
0040113B                 call    ds:ReadFile
00401141                 push    fileHandle
00401147                 call    ds:CloseHandle

The keyfile needs to have 3 lines. The first two lines need to be terminated with 0xD (carriage return). The last line must not have a line terminator. The following disassembly determines the length of the three lines, and store pointer to each line:

0040115A                 mov     pKeyContent, offset keyContent
00401164                 push    0Dh
00401166                 push    15h
00401168                 push    offset keyContent
0040116D                 call    line_length
00401172                 cmp     eax, 0FFFFFFFFh
00401175                 jz      fail
0040117B                 mov     line1_len, eax
00401180                 mov     edx, offset keyContent
00401185                 add     edx, eax
00401187                 mov     byte ptr [edx], 0
0040118A                 inc     edx
0040118B                 inc     edx
0040118C                 mov     dword ptr pLine2, edx
00401192                 sub     ecx, eax
00401194                 push    0Dh
00401196                 push    15h
00401198                 push    edx
00401199                 call    line_length
0040119E                 cmp     eax, 0FFFFFFFFh
004011A1                 jz      fail
004011A7                 mov     line2_len, eax
004011AC                 mov     edx, offset keyContent
004011B1                 add     edx, eax
004011B3                 add     edx, line1_len
004011B9                 add     edx, 2
004011BC                 mov     byte ptr [edx], 0
004011BF                 add     edx, 2
004011C2                 mov     pLine3, edx
004011C8                 push    15h
004011CA                 push    edx
004011CB                 call    line_length_f
004011D0                 cmp     eax, 0FFFFFFFFh
004011D3                 jz      fail
004011D9                 mov     line3_len, eax

The content of the keyfile needs to be alphanumeric, i.e., only contain letters and digits:

004011EC                 mov     esi, pKeyContent
004011F2 loc_4011F2:                             
004011F2                 cmp     ebx, 0
004011F5                 jz      short loc_401205
004011F7                 mov     al, [esi]
004011F9                 push    eax
004011FA                 call    is_alpha_numeric
004011FF                 and     edi, eax
00401201                 inc     esi
00401202                 dec     ebx
00401203                 jmp     short loc_4011F2
00401205 ; ---------------------------------------------------------------
00401205 loc_401205:                             
00401205                 mov     ebx, line2_len
0040120B                 mov     esi, dword ptr pLine2
00401211 loc_401211:                             
00401211                 cmp     ebx, 0
00401214                 jz      short loc_401224
00401216                 mov     al, [esi]
00401218                 push    eax
00401219                 call    is_alpha_numeric
0040121E                 and     edi, eax
00401220                 inc     esi
00401221                 dec     ebx
00401222                 jmp     short loc_401211
00401224 ; ---------------------------------------------------------------
00401224 loc_401224:                             
00401224                 mov     ebx, line3_len
0040122A                 mov     esi, pLine3
00401230 loc_401230:                             
00401230                 cmp     ebx, 0
00401233                 jz      short loc_401243
00401235                 mov     al, [esi]
00401237                 push    eax
00401238                 call    is_alpha_numeric
0040123D                 and     edi, eax
0040123F                 inc     esi
00401240                 dec     ebx
00401241                 jmp     short loc_401230
00401243 ; ---------------------------------------------------------------
00401243 loc_401243:                             
00401243                 mov     eax, edi

Finally, there is an obfuscated check to see if the first and second line of the key have the same length:

004012EF                 mov     ecx, line2_len
004012F5                 mov     unpredictable, eax
004012FA                 add     eax, 93E8h
004012FF                 sub     eax, unpredictable
00401305                 add     ecx, eax
00401307                 mov     edx, line1_len
0040130D                 sub     ecx, edx
0040130F                 xor     ecx, 75382
00401315                 cmp     ecx, 112030
0040131B                 jnz     fail

The above check boils down to:

(line2len - line1len + 0x93e8) XOR 75382 = 112030
(line2len - line1len + 0x93e8)           = 0x93e8
line1len                                 = line2len

Valid Second Line

The crackme applies a series of transformations to the first line:

00401267                 mov     ecx, line1_len
0040126D                 cmp     ecx, 3
00401270                 jb      fail
00401276                 xor     ecx, 5Ch

In pseudo-code:

line1_len = len(line1)
line1[0] ^= line1_len ^ 0x5c

Then we XOR characters from the end with characters at the start:

00401283                 mov     ecx, line1_len
00401289                 add     edx, ecx
0040128B                 dec     edx
0040128C loc_40128C:                             
0040128C                 mov     cl, [edx]
0040128E                 xor     [eax], cl
00401290                 inc     eax
00401291                 dec     edx
00401292                 cmp     eax, edx
00401294                 jl      short loc_40128C

In pseudo-code:

i = 1
j = line1_len-1
while i < j:
    line1[i] ^= line1[j]
    i += 1
    j -= 1

A similar routine follows:

00401353                 mov     ecx, line1_len
00401359                 dec     ecx
0040135A                 shr     ecx, 1
0040135C                 mov     eax, pLine1
00401361                 mov     edx, eax
00401363                 add     edx, ecx
00401365                 inc     edx
00401366                 mov     al, [eax]
00401368 loc_401368:                             
00401368                 xor     [edx], al
0040136A                 inc     al
0040136C                 inc     edx
0040136D                 cmp     byte ptr [edx], 0
00401370                 jnz     short loc_401368

It does:

i = (line1_len-1)//2 + 1
c = line1[0]
for i in range(i, line1_len):
    line1[i] ^= c
    c += 1

Finally, the bytes in the line are made alphanumeric by calling:

00401378                 push    pLine1
0040137E                 call    make_alphanumeric

The routine make_alphanumeric is:

0040166A ; =============== S U B R O U T I N E =============================
0040166A ; Attributes: bp-based frame
0040166A make_alphanumeric proc near             
0040166A data            = dword ptr  8
0040166A length          = dword ptr  0Ch
0040166A                 push    ebp
0040166B                 mov     ebp, esp
0040166D                 push    edi
0040166E                 mov     edi, [ebp+data]
00401671 loc_401671:                             
00401671                 mov     cl, 25h
00401673                 cmp     [ebp+length], 0
00401677                 jz      short loc_4016A9
00401679 loc_401679:                             
00401679                 cmp     byte ptr [edi], '9'
0040167C                 jg      short loc_401685
0040167E                 cmp     byte ptr [edi], '0'
00401681                 jl      short loc_401685
00401683                 jmp     short loc_4016A3
00401685 ; -----------------------------------------------------------------
00401685 loc_401685:                             
00401685                                         ; make_alphanumeric+17j
00401685                 cmp     byte ptr [edi], 'Z'
00401688                 jg      short loc_401691
0040168A                 cmp     byte ptr [edi], 'A'
0040168D                 jl      short loc_401691
0040168F                 jmp     short loc_4016A3
00401691 ; -----------------------------------------------------------------
00401691 loc_401691:                             
00401691                                         ; make_alphanumeric+23j
00401691                 cmp     byte ptr [edi], 'z'
00401694                 jg      short loc_40169D
00401696                 cmp     byte ptr [edi], 'a'
00401699                 jl      short loc_40169D
0040169B                 jmp     short loc_4016A3
0040169D ; -----------------------------------------------------------------
0040169D loc_40169D:                             
0040169D                                         ; make_alphanumeric+2Fj
0040169D                 add     [edi], cl
0040169F                 inc     cl
004016A1                 jmp     short loc_401679
004016A3 ; -----------------------------------------------------------------
004016A3 loc_4016A3:                             
004016A3                                         ; make_alphanumeric+25j ...
004016A3                 inc     edi
004016A4                 dec     [ebp+length]
004016A7                 jmp     short loc_401671
004016A9 ; -----------------------------------------------------------------
004016A9 loc_4016A9:                             
004016A9                 pop     edi
004016AA                 leave
004016AB                 retn    8
004016AB make_alphanumeric endp

This routine decompiles to:

def make_alphanumeric(chars):
    for i in range(len(chars)):
        c = 37 
        while not chr(chars[i]).isalnum():
            chars[i] += c
            c += 1
            chars[i] &= 0xFF
            c &= 0xFF
    return chars

The crackme then XORs the result with unpredictable data:

00401384                 mov     edi, pLine1
0040138A                 mov     edx, unpredictable
00401390                 mov     ecx, line1_len
00401396                 add     ecx, edi
00401398 loc_401398:                             
00401398                 xor     [edi], dl
0040139A                 inc     edi
0040139B                 rol     edx, 8
0040139E                 cmp     edi, ecx
004013A0                 jnz     short loc_401398

The crackme also XORs the second line with the same key. It then compares the transformed first and second line:

00401398 loc_401398:                             
00401398                 xor     [edi], dl
0040139A                 inc     edi
0040139B                 rol     edx, 8
0040139E                 cmp     edi, ecx
004013A0                 jnz     short loc_401398
004013A2                 push    esi
004013A3                 mov     esi, pLine1
004013A9                 mov     edi, dword ptr pLine2
004013AF                 mov     ecx, line1_len
004013B5                 cld
004013B6                 repe cmpsb
004013B8                 pop     esi
004013B9                 pop     edi
004013BA                 jnz     fail

The XOR encryption of the first and second line with the same key can be omitted, such that we have the following relationship between first and second line:

line1 = "phildunphy"
line1_len = len(line1)
line1_codes = [ord(c) for c in line1]
line1_codes[0] ^= line1_len ^ 0x5c
i = 1
j = line1_len-1

while i < j:
    line1_codes[i] ^= line1_codes[j]
    i += 1
    j -= 1

i = (line1_len-1)//2 + 1
c = line1_codes[0]
for i in range(i, line1_len):
    line1_codes[i] ^= c
    c += 1

x = make_alphanumeric(line1_codes)
line2 = ''.join([chr(xx) for xx in x])

# >>> second line is: K6LAUSIXAS

The Third Line

The third line is the trickiest. The crackme first generates two seeds based on the third and first line. The first seed is determined as follows:

004013C4                 mov     edx, pLine3
004013CA                 mov     eax, [edx]
004013CC                 add     eax, [edx+4]
004013CF                 mov     ecx, [edx+8]
004013D2                 rol     eax, cl
004013D4                 mov     ecx, [edx+0Ch]
004013D7                 ror     eax, cl
004013D9                 xor     eax, [edx+10h]
004013DC                 mov     edx, dword ptr line1_copy
004013E2                 add     edx, dword ptr line1_copy+4
004013E8                 mov     ecx, dword ptr line1_copy+8
004013EE                 rol     edx, cl
004013F0                 mov     ecx, dword ptr line1_copy+0Ch
004013F6                 ror     edx, cl
004013F8                 xor     edx, dword ptr line1_copy+10h
004013FE                 xor     edx, eax
00401400                 mov     seed1, edx

This boils down to the following pseudo-code:

def hash1(line):
    eax = get_int_from_string(line[:4])
    eax += get_int_from_string(line[4:8])
    ecx = get_int_from_string(line[8:12])
    eax = rol(eax, ecx & 0xFF)
    ecx = get_int_from_string(line[12:16])
    eax = ror(eax, ecx & 0xFF)
    ecx = get_int_from_string(line[16:20]) 
    eax ^= ecx
    return eax

eax = hash1(line3)
edx = hash1(line1)
eax ^= edx
seed1 = eax

With get_int_from_string converting a string into the little endian integer:

def get_str_from_int(val):
    s = ""
    for i in range(4):
        s += chr(val & 0xFF)
        val >>= 8

    return s 

The second seed is calculated as follows:

00401406                 mov     eax, pLine3
0040140B                 add     eax, 9
0040140E                 mov     eax, [eax]
00401410                 rol     eax, 8
00401413                 xor     al, al
00401415                 ror     eax, 8
00401418                 mov     seed2, eax

Which is:

s = get_int_from_str(line3[9:13])

The seeds are then used to build a 16 bytes hash:

00401422                 mov     eax, seed1
00401427                 push    eax
00401428                 call    ds:srand
0040142E                 add     esp, 4
00401431                 jmp     short loc_401442
00401433 ; ----------------------------------------------------------------
00401433 loc_401433:                             
00401433                 mov     eax, seed2
00401438                 push    eax
00401439                 call    ds:srand
0040143F                 add     esp, 4
00401442 loc_401442:                             
00401442                 mov     esi, 4
00401447 loc_401447:                             
00401447                                         ; 00401473j
00401447                 call    ds:rand
0040144D                 push    4
0040144F                 push    eax
00401450                 call    modulus
00401455                 mov     edi, eax
00401457                 cmp     word ptr hash[ebx+edi*4], 0
00401460                 jnz     short loc_401447
00401462                 call    ds:rand
00401468                 add     word ptr hash[ebx+edi*4], ax
00401470                 dec     esi
00401471                 test    esi, esi
00401473                 jnz     short loc_401447
00401475                 call    ds:rand
0040147B                 xor     ecx, ecx
0040147D loc_40147D:                             
0040147D                 xor     word ptr hash[ecx*4], ax
00401485                 inc     ecx
00401486                 cmp     ecx, 4
00401489                 jnz     short loc_40147D
0040148B                 test    ebx, ebx
0040148D                 jnz     short loc_40149C
0040148F                 mov     ebx, 2
00401494                 xor     seed2, eax
0040149A                 jmp     short loc_401433
0040149C ; ----------------------------------------------------------------
0040149C loc_40149C:                             
0040149C                 call    ds:rand
004014A2                 mov     ecx, eax
004014A4                 rol     eax, 10h
004014A7                 mov     ax, cx
004014AA                 xor     ecx, ecx
004014AC loc_4014AC:                             
004014AC                 xor     dword ptr hash[ecx*4], eax
004014B3                 inc     ecx
004014B4                 cmp     ecx, 4
004014B7                 jnz     short loc_4014AC

The following C-code shows how the hash is calculated for given two seeds:

#include <stdio.h>

inline int rand(int *seed) {
    *seed = *seed*0x343fd + 0x269EC3;
    return ((*seed) >> 0x10) & 0x7FFF;

long int main (long int argc, char *argv[]) {
    unsigned char hash[16];
    unsigned int i;
    unsigned int base = 0;
    unsigned int seed1 = 0x0110469A;
    unsigned int seed2 = 0x006C7972;
    unsigned int seed = seed1;
    unsigned int offset;
    unsigned int tmp;

    for(i = 0; i < 16; i++)
        hash[i] = 0;

    for(base = 0; base <= 2; base += 2) {
        for(i = 0; i < 4; i++)
                offset = rand(&seed) % 4;
            } while ( hash[base + offset*4] != 0 || hash[base + offset*4 + 1] != 0 );
            tmp = rand(&seed);
            hash[base + offset*4 + 1] += (tmp >> 8); 
            hash[base + offset*4] += (tmp & 0xFF); 

        tmp = rand(&seed);
        for(i = 0; i < 4; i++) {
            hash[i*4 + 1] ^= (tmp >> 8);
            hash[i*4] ^= tmp & 0xFF;

        if(base == 0) {
            seed2 ^= tmp;
            seed = seed2;

    tmp = rand(&seed);
    tmp = (tmp<<16) + tmp;
    for(base = 0; base < 4; base++) {
        for(i = 0; i < 4; i++) {
            hash[base*4 + i] ^= (tmp & 0xFF);
            tmp = (tmp >> 8) | ((tmp & 0xFF) << 24);

    printf("hash is:\n");
    for(i = 0; i < 16; i++)
        printf("%x ", hash[i]);

The calculated hash is finally compared to a hardcoded value:

004014BB                 cmp     dword ptr hash, 3C0E7DEBh
004014C5                 jnz     short loc_4014EC
004014C7                 cmp     dword ptr hash+4, 3AD11611h
004014D1                 jnz     short loc_4014EC
004014D3                 cmp     dword ptr hash+8, 0B070195h
004014DD                 jnz     short loc_4014EC
004014DF                 cmp     dword ptr hash+0Ch, 36263E26h
004014E9                 jnz     short loc_4014EC
004014EB                 inc     ecx             ; set correct flag

Determining Valid Seeds

The first step to crack the algorithm is two find seeds that lead to the correct hash. I did this by first brute forcing the seeds for the second round, see program brute_force2.c. It should produce the following four seeds four the second round:

  • 006f445a
  • 406f445a
  • 806f445a
  • c06f445a

The starting seed of the second round is actually calulated like this:

seed2 ^= tmp;
seed = seed2;

where tmp is the last random number. All random numbers have at most 2 bytes. Also the seed2 is at most 0xFFFFFF. Therefore, only the first of the four seeds can actually be produced by the code. The value of the last random number call is also output by the program, the number is 0x210e and we need to XOR the seed with this value to get the actual calculated value in seed2.

The seed for the first round seed1 can be found similarly, I used the program brute_force1.c:

  • 01a01234
  • 41a01234
  • 81a01234
  • c1a01234

The seed 81a01234 did not work with the crackme, I don’t know why but all we need is one working seed anyways.

For the keygen we need a way to find lines that produce the desired hash. I used the following properties of the hashing to quickly find good lines:

  1. The seed1 is calculated with a hash routine, that uses bytes 16 to 19 of the second line in an XOR operation. We get the correct seed by simply adjusting these four bytes. Some of the adjustments will lead to strings that are not alpha numeric. I therefore simply loop over random strings until one corrects to an alphanumeric string.

  2. The seed2 is equivalent to the three bytes 9, 10, 11 of the second line. The seed2 is known to be 0x6f445a XOR 0x210e = 6F6554, or “Teo” as a little endian ASCII string.

The keygen first generates a random string of 20 characters. It then sets bytes 9, 10, and 11 to “Tor”. After that, it patches the last four bytes to match the first seed. If the result is not alphanumeric, it simply tries again.

The Keygen

The following Python code expects a name as the first and only argument. The name must be alphanumeric (so no spaces allowed). It then calculates the second line and third line. The result is writen to TheKey.k, which you need to copy to the crackme directory. The Crackme should show a message box with PERFECT as the text.

import random
import sys

def keygen(name):

    def get_int_from_string(s):
        val = 0
        for x in s[::-1]:
            val <<= 8
            val += ord(x)
        return val

    def get_str_from_int(val):
        s = ""
        for i in range(4):
            s += chr(val & 0xFF)
            val >>= 8

        return s 

    def rol(val, places):
        shift = places % 32;
        val = (val << shift)  + (val >> (32-shift))
        val &= 0xFFFFFFFF
        return val

    def ror(val, places):
        shift = places % 32;
        val = (val >> shift)  + (val << (32-shift))
        val &= 0xFFFFFFFF
        return val

    def make_alphanumeric(txt_codes):
        for i in range(len(txt_codes)):
            c = 37 
            while not chr(txt_codes[i]).isalnum():
                txt_codes[i] += c
                c += 1
                txt_codes[i] &= 0xFF
                c &= 0xFF
        return txt_codes

    def random_alphanumeric(l):
        return ''.join(random.sample(map(chr, range(48, 57) + range(65, 90) +\
            range(97, 122)), l))

    def hash1(line):
        eax = get_int_from_string(line[:4])
        eax += get_int_from_string(line[4:8])
        ecx = get_int_from_string(line[8:12])
        eax = rol(eax, ecx & 0xFF)
        ecx = get_int_from_string(line[12:16])
        eax = ror(eax, ecx & 0xFF)
        ecx = get_int_from_string(line[16:20]) 
        eax ^= ecx
        return eax

    def find_line3_seed1(line1, line3):
        #valid_seed1 = [0x1a01234, 0x41a01234, 0x81a01234, 0xc1a01234]
        valid_seed1 = [0x1a01234, 0x41a01234, 0xc1a01234]
        eax = hash1(line3)
        edx = hash1(line1)
        eax ^= edx
        for s in valid_seed1:
            diff = eax ^ s
            ecx = get_int_from_string(line3[16:20])
            ecx ^= diff 
            new_str = get_str_from_int(ecx)
            if new_str.isalnum():
                return line3[0:16] + new_str
        return None

    def find_line3_seed2():
        line3 = random_alphanumeric(20)
        valid_seed2 = 0x6f445a ^ 0x210e
        s = get_str_from_int(valid_seed2)[:3]
        line3 = line3[0:9] + s + line3[12:]
        return line3

        check if name is alphanumeric, crackme won't accept other names
    if not name.isalnum():
        print("The name must be alphanumeric")

    if not 3 <= len(name) <= 20:
        print("Name too long or too short")

        generate second line from name
    line1 = name 
    line1_len = len(line1)
    line1_codes = [ord(c) for c in line1]
    line1_codes[0] ^= line1_len ^ 0x5c

    i = 1
    j = line1_len-1
    while i < j:
        line1_codes[i] ^= line1_codes[j]
        i += 1
        j -= 1

    i = (line1_len-1)//2 + 1
    c = line1_codes[0]
    for i in range(i, line1_len):
        line1_codes[i] ^= c
        c += 1

    x = make_alphanumeric(line1_codes)
    line2 = ''.join([chr(xx) for xx in x])

        semi brute force valid third line
    while True:
        line3 = find_line3_seed2()
        line3 = find_line3_seed1(line1, line3)
        if line3:

    with open("TheKey.k", "wb") as w:


Example Keyfile

The following example for name “phildunphy” should give the PERFECT message: