Here is my next solution for HITB CTF 2009 Daemon1. Similar to daemon 6, the flag is the content of errorcode.txt file located in the same directory with daemon’s binary.

home suto # netstat -tulpan
Active Internet connections (servers and established)
Proto Recv-Q Send-Q Local Address           Foreign Address         State       PID/Program name
tcp        0      0 0.0.0.0:4444            0.0.0.0:*               LISTEN   6174/daemon1

So you can see it listens on port 4444. Next I tried to find where the daemon processes my input.

.text:080494F1                 push    eax
.text:080494F2                 call    _recv
.text:080494F7                 add     esp, 10h
.text:080494FA                 cmp     eax, 0
.text:080494FD                 jle     loc_80495D2
.text:08049503                 push    esi
.text:08049504                 push    eax
.text:08049505                 lea     esi, [ebp-538h]
.text:0804950B                 push    esi
.text:0804950C                 mov     ecx, [ebp-548h]
.text:08049512                 push    ecx
.text:08049513                 call    sub_804A2B0
.text:08049518                 mov     eax, offset aIcvykbmukcrwdp ; "iCvYkBMuKcrwDPkAqmCFgOKVeV34"
.text:0804951D                 mov     ecx, 1Ch
.text:08049522                 cld
.text:08049523                 mov     esi, [ebp-560h]
.text:08049529                 mov     edi, eax
.text:0804952B                 repe cmpsb
.text:0804952D                 setnbe  dl
.text:08049530                 setb    al
.text:08049533                 add     esp, 10h
.text:08049536                 cmp     dl, al
.text:08049538                 jnz     loc_80495FF
.text:0804953E                 call    sub_8048F10
.text:08049543                 push    0
.text:08049545                 sub     esp, 8
.text:08049548                 push    offset s
.text:0804954D                 call    _strlen
.text:08049552                 add     esp, 0Ch
.text:08049555                 push    eax
.text:08049556                 push    offset s
.text:0804955B
.text:0804955B loc_804955B:                            ; CODE XREF: .text:08049608j
.text:0804955B                 mov     edx, [ebp-548h]
.text:08049561                 push    edx
.text:08049562                 call    _send

And here is what sub_8048F10 does:

lea     edi, [ebp+var_40]
mov     esi, offset unk_80553D2
mov     ecx, edx
rep movsd
mov     ax, ds:word_80553EA
mov     [edi], ax
push    (offset aSocketError+0Bh) ; modes
push    offset filename ; "/home/d1/errorcode.txt"
call    _fopen
<snip>

The code compares “iCvYkBMuKcrwDPkAqmCFgOKVeV34” with the input string. If it’s matched, the encrypted content of errorcode.txt will be returned.

home suto #nc localhost 4444

iCvYkBMuKcrwDPkAqmCFgOKVeV34

ddddddddddPfddddfdssqpfdddddddddhfh

“ddddddddddPfddddfdssqpfdddddddddhfh” is the return data. It’s the encrypted content of errorcode.txt (which is “1″ in this case).

After few hours trying to reverse the binary, I got stuck with the encoding algorithm so I tried to analysis the output data instead.

Input: 1
Ouput: ddddddddddPfddddfdssqpfdddddddddhfh

Input: 2
Output: ddddddddddPfdddddfdssqpfhfh

Input: 3
Output: ddddddddddPfdddddfdssqpfdhfh

Input: 4
Output: ddddddddddPfdddddfdssqpfddhfh

==>Output string begins with ddddddddddPfdddddfdssqpf and ends with hfh, number 1 is the special case.

9
ddddddddddPfdddddfdssqpfdddddddhfh

Next, we test with 2 numbers:

24
ddddddddddPfdddddfdssqpfhddhfh

3 numbers:

247
ddddddddddPfdddddfdssqpfhddhdddhfh

We can see that the string with red color is the same as the output for 24, and the green part is addition part for 7, so I guess h is character to begin a new number, let’s see with 6 numbers:

247398
ddddddddddPfdddddfdssqpfhddhdddhqqqqhddddddhqhfh

Now the algorithm is more clear :), the length of input number is the number of ‘h’ in the encoded data + 1 (we don’t count the last ‘hfh’). But how about q and d?

From 247398:
ddddddddddPfdddddfdssqpfhddhdddhqqqqhddddddhqhfh
4 is hdd
7 is hddd
3 is hqqqq
9 is hdddddd
8 is hq

Yeah! when the next number is increased, it uses a d for +1 (7 = 4 + 3 = hddd).
q is used for decrease (-1).

35896742
ddddddddddPfdddddfdssqp fd[3] hdd[5] hddd[8] hd[9] hqqq[6] hd[7] hqqq[4] hqq[2]hfh

Why 3? You answer yourself !

Now we come back to special cases for number 1 and 0

358967421
ddddddddddPfdddddfddddfdsssqpfdhddhdddhdhqqqhdhqqqhqqhfdddddddddhfh

Here is output for 35896742
ddddddddddPfdddddfdssqpfdhddhdddhdhqqqhdhqqqhqqhfh

The different parts are marked with Red color.

Put 1 in the middle:
3589617421
ddddddddddPfdddddfddddfdsssqpfdhddhdddhdhqqqhfdddddddddhsdhqqqhqqhfhfh

358967421
ddddddddddPfdddddfddddfdsssqpfdhddhdddhdhqqqhdhqqqhqqhfdddddddddhfh

35896742

ddddddddddPfdddddfdssqpfdhddhdddhdhqqqhdhqqqhqqhfh

So the output will be fdddddddddh for number 1. If 1 is in the middle, it will be dddfds.
And another notes is hsd , one “d” character because it is calculated from the number before “1″ – 6- and increases it to -7-.

Another test:

4668981445134
ddddddddddPfdddddfddddfdsssqpfdd(4)hdd(6)h(6)hdd(8)hd(9)hq(8)hfddddddddd(1)hsqqqq(4)h(4)

hd(5) hf(1)hs qq(3) hd(4) hffh

Now replace the number 1 with 0 from previous input:

ddddddddddPfdddddfddddfdsssqpfdd(4)hdd(6)h(6)hdd(8)hd(9)hq(8)hfdddddddd(0)hsqqqq(4)h(4)hd(5)

hf(0)hsqq(3) hd(4)hffh

We see 0 is quite similar to 1 with one ‘d’ less.

Now it’s just a simple task to decode the return content of errorcode.txt (flag) from the daemon.

And it’s all about daemon1 in HITB CTF 2009!

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