#                             - Challenge #1 -                                 #
#                                 by JaZZ                                      #

1) Localisation of the protection
2) Analysis
3) The keymaker


1) Localisation.

To work efficiently on this target, I've used Softice 2.80 for dos, and IDA 3.7,
which seems to be the only disassembler beeing (by far) able to produce a
meaningful listing that you can use in conjunction  with  softice. Well, there
are no anti-debuging tricks as soon as you start softice ! I suppose these
one were supposed to be the use of overlays but IDA treats them quite right as
We know that the babe needs a file to reg right so we just BPINT 21 ah=3D to
break whenever a file is opened. The second time Sice pops, ds:dx points to a
"terminat.key" string. Ok, this is the name. We then just create a bogus
terminat.key file, then BPINT 21 ah=3F to break on the reading. And we land
smack into it. In Ida it's referenced as ovr018.

2) Analysis of the  scheme.

I've provided a COMMENTED (with some vars renamed, please have a look...)listing
of the interesting part of overlay 18, to clarify my explanations, but i suppose
the reader has figured out how all this works :-) First the key file is divided
into 0B buffers of each 162h bytes long. The file must then be at least
162*B=F36 bytes long. There are several levels of
protection in this scheme. Let's examine the first.

a) A loop is reading all the buffers of the keyfile, and performs a global
checksum (through a dedicated function) on bytes [0,15D] of each one. The 2 last
words in each buffer (15E to 161) are either a checksum (local to this buffer or
global to the whole file) , or are not checked at all. It can be noted that the
checksum function actually uses a 1024 bytes area to perform its job (that's why
i've been forced to hardcode it in the keymaker). When the reading has been
achieved the resultant checksum is matched against the bytes 15E->161 of the
last 0B buffer. If the 2 words don't match, you are kicked off. During this
first loop, the author also tries to detect cracked terminat.key files. To
achieve this, he only checks constant patterns that always occur at the same
place in cracked files:

1)First buffer: cracked file if chars at positions B,1B,14 are respectively
equal to 'F','/',and '.'

2)Second buffer: cracked file if first five characters are equal.

3)0B buffer: cracked file if bytes 12C to 130 are equal OR if [15A,15D]= "_d+"
(ie DC 64 D9 E9)

If the program detects a cracked key, either it fills the general checksum with
bogus, or it displays a message saying you're using an anauthorized key file.
Anyway, it won't reg. To prevent the program from matching succesfully, i just
fill all the terminat.key with random at the beginning (the random generator is
the one used by terminat itself, seeded with the clock), and then put the
interesting data over it. It'll ensure that even for the SAME user input, a lot
of bytes will be different in the final keyfile. This will work as long as the
author doesn't check some "required" bytes !
b) the second part of the protection is dealing with the fourth buffer,
involving bytes from 5B to 161. Besides, the user input is encrypted in here.
The program will xor the whole area with FF at first. After it does the same
four times again, with a different xoring value each time. This value is
calculated by some kind of "random generator", seeded with defined DWORDS (not
really random then...). At the end of this process, the area is fully decrypted.
A new checksum then occurs onto this area, involving the same algo described in
a) : checksumming [5B,161] and comparing to the word in 15E. One more occasion
to be eventually rejected. If not the work is not done still: now you've got to
face the TERRIBLE integrity check call: though you may have gone through all the
previous doors, this call can enter a unfinite loop, or even worse crash the
PC while beeping continuously (apart from this, why this integrity check is
going backwards on the file ? a mystery). I summarize here quite a lot of work:
what is important are the parameters pushed before the call. Three of them are

0D67		    mov	    cx,	ss:[di+FEDE] ; this is computed with
0D6C		    mov	    si,	ss:[di+FEE0] ; the checksum of #5 buffer
0D71		    mov	    di,	ss:[di+FEE2]
0D76		    call    sub_1708_1707 ;-------> this time it MUST give ZERO
0D7B		    jnz	    set_a_fucking_mess
A detail I didn't notice at first was the calculation that was performed with 
the two last bytes of 5th buffer. Finally the result of this calculus is in 
FEDE FEE0 FEE2. If the call above doesn't return z on (it compares aldxbx with 
clsidi), this nasty prog messes every parameters that'll be pushed, leading to  
the unfinite loop. Ok, this is, I think, the real difficult part of the scheme: 
find the last two words of 5th buffer. The first idea that came to me was to 
put the SAME checksum in the 5th than the one calculated for 4th. buffer. When 
tracing with sice you see that both follow the same transformation algo, except 
for this:

[FEDE,FEE2] holds the algo result for #5 buf.
[FED8,FEDC]  ""    ""  ""    """  for #4 buf.

0D48		    mov	    ax,	ss:[di+FED8]
0D4D		    mov	    bx,	ss:[di+FEDA]
0D52		    mov	    dx,	ss:[di+FEDC]
0D57		    mov	    cx,	84h ; ''
0D5A		    xor	    si,	si
0D5C		    mov	    di,	2000h
0D5F		    call    sub_1708_16EB

and then it's compared (see above). when returning, bx is lowered by 28h & 
that's all. Putting then the same checksum for the two fails because of this 
difference. So we'll have to find a checksum for 5th that through calculation 
gives [FEDA]-28h. Now this seems quite a laborious task, but is not really, if 
you remark that lowering [15E] by 1 lowers the [FFE0] by 4. It's fully linear. 
To lower the final result by 40 (=28h), we'll only have to take the [15E] word 
of the 4th buffer, lower it by Oa, and put the result in [15E] in 5th buf.

3) the keymaker. 
One more thing we'll have to find is how the user's personal info is stored in 
the key file. To achieve this, i simply cracked terminate.exe to allow every 
bogus file (with the sufficient nbr of bytes) to be a valid one. I won't discuss 
that much as there's only 3 or 4 bytes to change, and we don't want to patch 
it! Pressing ALT O in the program displays all the user info. All the info is 
stored in 4th buffer, at these positions (with sice you go on the decrypting 
process until you find the bogus displayed by ALT O):
[7A]=name length
[AD]=adress length
[E0]=city length
[113]=country length

To realize the keymaker, we just follow the process the author uses in backwards 
direction (the listing is fully commented) 
- get user input with dos function. 
- initialize the area in memory with random. 
- copy the user data at the right 
- put a correct checksum in 4th buf and then deduct the one for 5th as explained 
- xoring the 4th buf succesively with 4 distinct values 
- performing the global checksum on all the buffers and putting the result in 
the right place of the last one. 

#------------------------------------------------------------------------------# # - Challenge #2 - # # by JaZZ # #------------------------------------------------------------------------------# I choose to code the patcher in C. The compiler i've used is Borland C++ 4.52. (Thanks fravia for the info of "quasi free" release from pc plus). The exe length is about 30K, which is not tiny, maybe this is the reason why almost patchers are DOS based ;-), although less functionnal. I choose to add two features to mine,that is: 1- The ability to patch at multiple locations. Well the experience proofs it's useful. 2- The possibility to make a backup of the original file. The exe will be renamed as ".bak". I think the listing is self-explanatory enough, the only things required are: long length= 593920; /* size of the file to be patched in bytes */ char *prgnam="bbrk32.exe"; /* name of the file to be patched */ int how_many=3; /* number of patches to be applied */ int sizes[]={1,1,3}; /* size of each one in bytes */ long offset[]= { 0x403B6,0x45226,0x6A07D }; /* offset of each patch */ unsigned char patch[]={0x33,0xEB,0x6A,0x01,0x5a}; /* the bytes of the patches, one after the other */ The program will put 33 at offet403B6 EB at 45226 6A,01,5A at 6A07D Before patching, i only check the size which in most of the cases is enough to differentiate between versions of the target. Backing up the file (if selected) is a quite easy using the CopyFile function. I provided an example for patching brainsbreaker v2.1. If you want to compile, don't forget to set the target to win32 and libraries to static. #------------------------------------------------------------------------------#
#------------------------------------------------------------------------------# # - Challenge #3 - # # by JaZZ # #------------------------------------------------------------------------------# The program has several levels of protection: 1- integrity check. 2- no trivial access to registration menu. 3- no trivial comparison between serials. 4- "vital" strings are encrypted. 5- "Hidden" protection features. This is, apart from the game itself which is imo very well designed (good windoze progs are not so frequent !), quite a good protection, for a high level language program. Hope that having included it in the strainer won't do any harm to that guy as he's capable. Back to the topic. The 5th category is quite unfair: for example you can't load a puzzle that's been created in another PC with the program unregistered on it. May be there are several tricks of that kind, i discovered this only one when showing the game a colleague at work, as he tried to send me a puzzle by the network. As I said, cracking it is all the more interesting as there ain't ANY "wide open doors" where you can get through. Disassembling it with wdasm is ok, but the strings references lead to nothing vital: the "Demo" references only the background of the image. You can disable it, it won't reg the program. Furthermore this nasty author made a great "pied de nez" to all crackers as he managed (i may be wrong, but i've never seen that before ;-) to associate his name & copyright to every 00000001 that appears in the listing. Well done Mr Trujillos, 1 to 0, we'll see your damned name at least a thousand times, but we're not dead though. Of course you can remove them with your editor, but dont try to use wdasm after as it'll be fooled. Anyway it's only a detail. Our tools will be Softice and Wdasm, as usual. ------------------- 1) At first let's access the registration process. It can be a box, or a file that you are emailed with for example. Checking the file accesses with filemon doesn't show anything suspicious, it must be a box triggered by a special keystroke. Let's call it zen or anything else (non academic approach...), I tried every function key combined with alt,ctrl etc...and CTRL F8 triggered it. Pack, name, id, key, nothing less than 4 fields to fill. At this point i recalled that the challenge was to explain the scheme, NOT to make a keygen so i didn't delve into "serial" manipulation and tracking at this moment, supposing this guy clever enough to have used a heavy encryption process. So where to start then? A very uncommon idea: let's type in some bogus stuff. We get a very common messagebox "Seems that the data etc....". Ok. With Sice you discover the main "registration box" routine (i dump just the end): :0046B455 E89BB6FEFF call 00456AF5 ---> check what you entered :0046B45A 8BF8 mov edi, eax :0046B45C 85FF test edi, edi :0046B45E 0F8585000000 jne 0046B4E9 ----> if eax!=0 you get out of the loop :0046B464 837DFC00 cmp dword ptr [ebp-04], 00000000 :0046B468 752D jne 0046B497 ---> it was bogus, go to messagebox :0046B46A C745FC01000000 mov [ebp-04], 00000001 :0046B471 6A02 push 00000002 :0046B473 8D45C4 lea eax, dword ptr [ebp-3C] :0046B476 50 push eax :0046B477 E88F9CFAFF call 0041510B :0046B47C 6A02 push 00000002 :0046B47E 8D55DC lea edx, dword ptr [ebp-24] :0046B481 52 push edx :0046B482 E8849CFAFF call 0041510B :0046B487 6A02 push 00000002 :0046B489 8D4DF0 lea ecx, dword ptr [ebp-10] :0046B48C 51 push ecx :0046B48D E8799CFAFF call 0041510B :0046B492 E968FDFFFF jmp 0046B1FF ---> give it a try again! ; error: messagebox :0046B497 83C4FC add esp, FFFFFFFC :0046B49A 66C704242E02 mov word ptr [esp], 022E -> index for "Seems that the data..." :0046B4A0 E803E8FCFF call 00439CA8 :0046B4A5 50 push eax :0046B4A6 FF35D4974800 push dword ptr [004897D4] :0046B4AC 6810200000 push 00002010 :0046B4B1 53 push ebx :0046B4B2 E8676CFBFF call 0042211E -> call messageboxa :0046B4B7 837DF800 cmp dword ptr [ebp-08], 00000000 :0046B4BB 7507 jne 0046B4C4 :0046B4BD BF7C150000 mov edi, 0000157C :0046B4C2 EB05 jmp 0046B4C9 ------------------- 2) Up to this point you can either enter the 456AF5 routine, i didn't do it at first as i was believing a lot of calls were involved. Infortunately it wasn't true, and sure it would have been easier to look closer at this snippet, BUT things always look easy once you've understand. That's why I tried another direction. I just wanted to follow this idea to proof that with just a little weakness the prog could be cracked. Look at the process that leads to the messagebox. We notice that the adress of the string is calculated, with some kind of "index". here it is 22E that through 439CA8 leads to the correct adress in memory. In return of this call, eax holds the adress of the string that'll be pushed for messagebox. for 22E the adress calculated is: 4A11F3. Then just look around this area with sice: it's simply the whole array of every interesting string used by bbrk, in addition they're fully decrypted. First error from the author, he should have dispatched them, or better, decrypted only if necessary. Sniffin' around only TWO strings appealed to me: "Thank you!" "Registration code detected" Lets choose the first. Its adress is 4A1301. My aim is to search the index that'll give this adress through the call 439CA8. A good old bruteforce attack: assemble this little routine "live" with sice: xor eax,eax adr: inc eax push eax add esp,-4 mov word ptr[esp],ax call 439CA8 cmp eax,4A1301 pop eax jnz adr ok: ; just bpx here! It'll break instantly with 0231 in eax (unique solution). So back to the dead listing. Let's search for the ", 0231" pattern in it: Only one match at 45CA36, a very stimulating result! The corresponding routine obviously starts in 45C9AD. Very disapointing that wdasm doesn't reference it though. Must be indirectly called (call [ebx+nn] or similar). It is neither hardcoded (with a push) so it's in a table and this will charge us with some extra work. With Sice again and bbrk loaded, let's search the above table with s 30:0 l ffffffff AD C9 45 00. Once again just one hit: 4852D4. Then same process again, search ", 004852D4" with the editor. Two hits this time: this is the first one: * Referenced by a CALL at Address: |:00456E04 -----> ****** THIS IS IMPORTANT ****** | :0045C755 55 push ebp :0045C756 8BEC mov ebp, esp :0045C758 83C488 add esp, FFFFFF88 :0045C75B 53 push ebx :0045C75C 56 push esi :0045C75D 57 push edi :0045C75E 8B5D08 mov ebx, dword ptr [ebp+08] :0045C761 6A00 push 00000000 :0045C763 53 push ebx :0045C764 E8DC2BFBFF call 0040F345 :0045C769 C703D4524800 mov dword ptr [ebx], 004852D4 --> HERE ! etc..... No matter what the second hit could be, Have you seen that it's referenced by a call in 456E04 ? This is deep into the 456AF5 "password checking" routine. So now it's time to have a look at it, and it'll be easier as we are SURE that to register, the processor must reach 456E04. Let's go then. At first sight all the: jmp 00457055 are EVIL, as they get (too quickly...) out of it with eax=0 (remember eax mustn't be 0 when returning). tracing with sice while entering bogus, you'll be kicked but not at the beginning though. Summary of the strategic jumps: :00456BA9 751B jne 00456BC6 ---> jump (ok) :00456BF8 7526 jne 00456C20 ---> jump (ok) :00456C4F 7531 jne 00456C82 ---> jump (ok) :00456D31 E8363D0100 call 0046AA6C :00456D36 85C0 test eax, eax :00456D38 741B je 00456D55 ---> jump (bad for us...) after this jmp, eax will be set to 0 and this will kick us out in 456DC9. So this 46AA6C routine is undoubtly the faulty one. Here's what it does: :0046AA6C 55 push ebp :0046AA6D 8BEC mov ebp, esp :0046AA6F 8B4508 mov eax, dword ptr [ebp+08] :0046AA72 8B00 mov eax, dword ptr [eax] :0046AA74 3B0590994800 cmp eax, dword ptr [00489990] :0046AA7A 0F94C2 sete dl :0046AA7D 83E201 and edx, 00000001 :0046AA80 8BC2 mov eax, edx :0046AA82 5D pop ebp :0046AA83 C20400 ret 0004 In short it returns 1 (the good_boy value) if [adress_pushed]=[489990]. Looking closer near this interesting call, we see that in fact there are two: :00456D23 8D55C0 lea edx, dword ptr [ebp-40] :00456D26 52 push edx :00456D27 53 push ebx :00456D28 8B0B mov ecx, dword ptr [ebx] :00456D2A FF5108 call [ecx+08] :00456D2D 8D45C0 lea eax, dword ptr [ebp-40] :00456D30 50 push eax :00456D31 E8363D0100 call 0046AA6C ----> FIRST NICE_BUYER TEST :00456D36 85C0 test eax, eax :00456D38 741B je 00456D55 ----> get out! :00456D3A 8D55BC lea edx, dword ptr [ebp-44] :00456D3D 52 push edx :00456D3E 6A00 push 00000000 :00456D40 6A00 push 00000000 :00456D42 53 push ebx :00456D43 8B0B mov ecx, dword ptr [ebx] :00456D45 FF5110 call [ecx+10] :00456D48 8D45BC lea eax, dword ptr [ebp-44] :00456D4B 50 push eax :00456D4C E81B3D0100 call 0046AA6C ----> SECOND ONE. :00456D51 85C0 test eax, eax :00456D53 7504 jne 00456D59 ----> nice buyer! So we just want eax to be non zero when returning 46AA6C. We'll change then: :0046AA7D 83E201 and edx, 00000001 to: 6A 01 push 01 5A pop edx At first with a "live" patch, while bpx-ing on 46AA6C. A nice box pops up with our "Thank you!" string in it (no field should be empty though) and it seems regged... Try a puzzle, the "demo" background has disapeared, no more reminder box anymore. Ok, now we try to hard patch, and we're stuck with a CRC check : "content seems to be altered...." This shouldn't be too hard, with the messagebox you'll easily find the dedicated routine: :00442CC2 E870DBFFFF call 00440837 ---> checksum routine :00442CC7 3D22334455 cmp eax, 55443322 ---> good boy value :00442CCC 7472 je 00442D40 ----> exe not altered ;messagebox("sucker!") Having a look at 440837, we quickly find that the good boy test is here: (yes we could have set the je to jmp in 442CCC, but who knows, may be it's called elsewhere, even if wdasm doesn't reference it) :00440DAE E8B504FCFF call 00401268 :00440DB3 83C40C add esp, 0000000C :00440DB6 85C0 test eax, eax :00440DB8 752E jne 00440DE8 ---> this is the faulty one! :00440DBA A1803B4800 mov eax, dword ptr [00483B80] :00440DBF 8B10 mov edx, dword ptr [eax] :00440DC1 8B0D243B4800 mov ecx, dword ptr [00483B24] :00440DC7 8911 mov dword ptr [ecx], edx :00440DC9 B822334455 mov eax, 55443322 ----> nice value... Just change a single byte: :00440DB6 85C0 test eax, eax to: 33C0 xor eax,eax When you try to register, the program will write an entry in the bbrk.ini file, where sections are defined according to the pack field. You can see then that the guy is always checking if you're a nice buyer or not. Just bpx 46AA6C... The release is now 2.3 (09/98), and the protection in this one is less simple. He doesn't check a simple flag anymore. This decided me to look deeper at the encryption scheme. But by now this is almost the deadline, only 2 days left, i couldn't manage to break it totally. Please have a look at braincrypt.txt for a little discussion of the encryption system. ------------------ 3) Finally let's enable the possibility to play every puzzle, even those created with an "unregged" computer. (when creating a puzzle, a header holding your "registration info" will be included in the file, program has just to check it when loading). With a little research we find the"loading" call: :00445CA4 E8B9E1FFFF call 00443E62 the routine begins at 445ABC. tracing through it with sice, comparing between a good and a bad puzzle, you'll see that the evil jump is: :00445C1E E8ED390200 call 00469610 :00445C23 3B45E8 cmp eax, dword ptr [ebp-18] :00445C26 743B je 00445C63 ---> regged user. :00445C28 68EE030000 push 000003EE :00445C2D 6A00 push 00000000 :00445C2F 6A00 push 00000000 :00445C31 8D4DFC lea ecx, dword ptr [ebp-04] :00445C34 51 push ecx :00445C35 E86243FDFF call 00419F9C :00445C3A 8D45FC lea eax, dword ptr [ebp-04] :00445C3D 50 push eax :00445C3E 56 push esi :00445C3F E8DD40FDFF call 00419D21 :00445C44 8BC6 mov eax, esi :00445C46 50 push eax :00445C47 6A02 push 00000002 :00445C49 8D55EC lea edx, dword ptr [ebp-14] :00445C4C 52 push edx :00445C4D E8B9F4FCFF call 0041510B :00445C52 6A02 push 00000002 :00445C54 8D4DFC lea ecx, dword ptr [ebp-04] :00445C57 51 push ecx :00445C58 E83441FDFF call 00419D91 :00445C5D 58 pop eax :00445C5E E98A000000 jmp 00445CED ---> exit so we'll change one byte: :00445C26 743B je 00445C63 to EB3B jmp 445C63 Frow now on, every puzzle will be loaded. The patcher has been designed to be tested on bbrk.(nb: whenever patched you must go through the CTRL F8 process and enter bogus (no empty fields). #------------------------------------------------------------------------------#
(please read challenge3.txt before) Just for reminding, the sensible area is: :00456D23 8D55C0 lea edx, dword ptr [ebp-40] :00456D26 52 push edx :00456D27 53 push ebx :00456D28 8B0B mov ecx, dword ptr [ebx] :00456D2A FF5108 call [ecx+08] ---> adr:459AC1 :00456D2D 8D45C0 lea eax, dword ptr [ebp-40] :00456D30 50 push eax :00456D31 E8363D0100 call 0046AA6C ----> FIRST NICE_BUYER TEST :00456D36 85C0 test eax, eax :00456D38 741B je 00456D55 ----> get out! :00456D3A 8D55BC lea edx, dword ptr [ebp-44] :00456D3D 52 push edx :00456D3E 6A00 push 00000000 :00456D40 6A00 push 00000000 :00456D42 53 push ebx :00456D43 8B0B mov ecx, dword ptr [ebx] :00456D45 FF5110 call [ecx+10] -----> 459C3A :00456D48 8D45BC lea eax, dword ptr [ebp-44] :00456D4B 50 push eax :00456D4C E81B3D0100 call 0046AA6C ----> SECOND ONE. :00456D51 85C0 test eax, eax :00456D53 7504 jne 00456D59 ----> nice buyer! I think huge asm dumps are just boring, so i'll try to sumarize what i've understand from the encryption: The important calls are 459AC1 and 459C3A. They finally set the flag that will be compared with 489990. If it's ok the value is 15A4. if not it'll be its opposite: FFFFEA5C. Seems that you've got two "chances" to trigger the ok box. The algos differ in that in each case the working area width vary. It can be 9 or D, and the "congruence" can be 1 or 3. For the moment i've just looked at the first case. A) First call: crashThe call 459AC1 checks the key. Here's how: the characters (uppercased) must be alphanumeric, values 1 and 0 are excluded, 1 is set to L (have you noticed that everything es when you put "1" in the key field ?). A rather "cryptic" routine (46ABE9) does the following: 1- give the position of each char in the string "ABCDEFHIKLMNOPQRSTUVWXYZ23456789". Only the 5 less significant bits of this position will be used. The program constructs a new key with these five bits by putting them aside. If the previous was L bytes long, the new one will be int(L*5/8)+1 bytes. I'll give an example: supposing your key is "23456", 2-->1A (the position in the string) --> 11010 3-->1B --> 11011 4-->1C --> 11100 5-->1D --> 11101 6-->1E --> 11110 2- Juxtaposing all this stuff beginning from the right gives the new pattern: 1 1110 1110 1111 0011 0111 1010 1 E E F 3 7 A In memory there will be: 7A F3 EE 01 + some constant bytes. the 9 bytes beginning with the new key are xored alltogether (459B50), and the result must give 0 (:459B76 call 401268 is the comparing routine). That triggers the first "nice buyer flag". This appears quite easy to bruteforce; i hadn't seen the rest.... B) Second call: The routine in 459C3A first uses the pack, name, id fields to construct some kind of 6 bytes long checksum: all bytes at the same position in each string are xored together and the result is put in this 6 bytes area beginning at adr.(the position from adr is evaluated modulo 6).Example for the following values : (adr is initialized to 00 00 00 00 00 00) pack: 212 name: abc def id : 12345678 [adr]= [adr] xor 2 xor a xor 1 [adr+1]=[adr+1] xor 1 xor b xor 2 ..... etc, then for the 7th: [adr]=[adr] xor f xor 7 (no more chars in pack) Finally you've got this 6 bytes large area which will be matched against another, triggering the second ok flag. And that's where i'm stuck, because the second one is not always the same even if you don't touch anything in the 4 fields. I came to the conclusion that one of the two calls: 46A4E0 or 46A77F obviously affects this area (its adress is pushed but not used) if some condition is met. Actually these 2 calls are doing something on the transformed key, but these routines are such a mess (recursivity involved...) Unfortunately that's all for now :-(
#------------------------------------------------------------------------------# # - Challenge #4 - # # by JaZZ # #------------------------------------------------------------------------------# All being well considered, the ultimate one may have been the tougher... Reproducing as close as possible the damned graphical effect is indeed a real challenge. I chose to code the program in C for conveniency. But before coding, a deep analysis of the process is needed. Setting a break on Ellipse or Bitblt after a puzzle has been achieved will get you in the heart of the routine, which starts at 42EAFB. The effect consists of 16 successives images, each of which is displayed by a Bitblt function at adress 429D5A. This adress is therefore a priveleged location to study the sparkling. A random location is chosen on the puzzle. Then there are two phases: - ascending (images 1 to 8) where the geometric pattern grows. - descending (im. 9 to 16) where it disappears (the 16th image beeing the original image of the screen), using the same images in reverse order. In addition there's a random amount of small sparks near the main figure. They last 6 images max, and can only be a pixel or a small cross. (only 2 colors for them) The main figure is made up of - a cross - one to three ellipses (depending on the progress of the process), each one having its own colour. There are 7 sets of colors. The program chooses one at the beginning (To retrieve them, just break on CreateSolidBrush). The author's method. -------------------- He uses several bitmaps (20x20). A first one is used to save the original screen:B0 He creates two bitmaps for the figures (B1,B2), and a working one, Bw, where he elaborates the final image that will be transfered to the screen at 429D5A. Then the process for building one image is the following, rather complicated (i chose a more simple way to achieve a similar result, you'll judge). 1) B0->Bw : working bmp is loaded with the "untouched" image (raster-op:SRCCPY) 2) fill B1 with white and draw the pattern in black on it. 3) fill B2 with black and draw the pattern with appropriate colors in it 4) B1->Bw (raster-op:SRCAND) the effect is to "black" the pattern in the original bitmap. 5) B2->Bw (raster-op:SRCINVERT) the effect is to insert the colored pattern in its "blacked" location in Bw. 6) Bw->Screen. And so on along the 16 steps. My method. ---------- I use only 2 bitmaps. With the first i save the original screen. In the second, the working one, i draw directly. 1) copy original image in it (SRCCPY) 2) draw pattern according to the drawing process.(SRCCPY) Drawing the correct figure... To do this, i traced every call to Ellipse and wrote down the coordinates. then i've hardcoded them in an array in C. Quite a laborious task indeed. For the LineTo and MoveToEx, I've used a more clever method: I just use the min and max of the coordinates and compute the relevant increments according to the number of steps. The possible sets of color are hardcoded too, one is chosen randomly at the beginning of the process. For the little sparks, the task is a little more comlicated, for they have a more random behavior: -random position "near" the ellipse. -random date of appearance. -random amount of these objects. To formalize this, i defined a structure for this object: -location -step of appearance -age of the object, to control its appearance: pixel or cross. At first, i decide how many sparks there will be. For every of them,i initialize the above structure with appropriate data. Especially, to find a good location i use a special random generator which exclude certain intervals, as the sparks must be "close" to the ellipse, ie. the center area and the edges shouldn't be used. Modifying sparkle.ini --------------------- Feel free to modify this initialisation file, which defines the behavior of the program. Here are the entries: fullscreen=1/0 ; The output will/will not take the entire screen. blackscreen=1/0; The background color will be black/unchanged. howmuch=20 ; The graphical effect will occur 20 times. waste=23 ; A temporisation in milliseconds. width=300 ; If fullscreen=0, these two parameters define the width & height=200 ; height of the rectangle (randomly located) where all ; the figures will be drawn. the "waste" param is quite sensible. It can affect dramatically the realism of the effect. This value gave a good result on my pc, for another machine i dont know... #------------------------------------------------------------------------------#