Who's attacking you, and what with?

doronb, in the Puzzles and Riddles TA, with the cryptogram.

LOL ppl.. nice tries ;)

Ok, this is attacking as in a game... once you decrypt the text-block, it should be obvious someone is attacking you... Just tell me who (generally) and with what :)

It looks like someone is attacking me with a keyboard and that is the output on the screen.

>>> It looks like someone is attacking me with a keyboard and that is the output on the screen.

Not with a keyboard ;)

>>> From a decryption standpoint, there are way too many H's and T's for this to be a normal substitution cypher.

Quite true.

There are actually many repeated strings with just one or two character different. I think the key might be finding the differences between all the garbage and whatever is left over is the text.

e.g.

<.>/H#H4H$H5H%H6H^H7H_H=H+HqHQHwHWHeHiHIHoHOHpHPH[H{HSHdHDHfHFHgHGHhHjH

versus

<.>/H#H4H$H5H%H6H^H7H_H=H+HqHQHwHWHeHUHiHIHoHOHpHPH[H{HSHdHDHfHFHgHGHhHjH

Remove duplication and get only the letters 'HU' remaining in the second string.

I would do more but I have to go home now. Hopefully someone who has time can see if this is the correct way to break this code.

>>> I was thinking that there were maybe two different decodes here

True.

>>> I think the key might be finding the differences between all the garbage and whatever is left over is the text.

Always check the garbage before throwing it out ;)

>>> There are actually many repeated strings with just one or two character different.

Is that intentional, or perhaps a diversion? ;)

lots of string that are upper and lowercase in order on the key board separated  T or H

H6H^H7H_H=H+HqHQHwHWHeHUHiHIHoHOHpHPH[H{HSHdHDHfHFHgHGHhHjH'H"HzHZHxHXHcHCHvH,H<H.H>H/H?T`T~T1T5T%T6T^T7T&T8T*T9TqTQTwTWTeTETrTRTtTOTpTPT[T{T]T}T\T|TFTgTGThTjTJTkTKTlT

are these strings describing a path over the key board of 2 opponents, perhaps?

>>> are these strings describing a path over the key board of 2 opponents, perhaps?

Hmmm... no, they're not, but that's an interesting idea.

looks like ascci art

>>> looks like ascci art

You're close.. but its not Ascii :)

is this a .bmp file? too late here for me to test here going to bed.

Its not a .bmp file... but you're very close :)

Is it code for a virus? a worm?

No, its not code at all, but you probably have to write code to decrypt it :)

We're looking at a gif or jpeg?

Surely there would be non-printable characters if it were a standard image format (RLE, etc)...

you're right... it's not a standard image file...

still think we're looking at rows and columns though...

> still think we're looking at rows and columns though...

Yeah, maybe it's a dump of a map from Rogue!

The asterisk is hitting the equals-sign with an exclaimation mark ;)

>>> We're looking at a gif or jpeg?

Indeed.

>>> Surely there would be non-printable characters if it were a standard image format (RLE, etc)...

The file has been encoded to contain only printable characters.

>>> still think we're looking at rows and columns though
>>> Yeah, maybe it's a dump of a map from Rogue!
>>> The asterisk is hitting the equals-sign with an exclaimation mark ;)

Alas, but no, to all three comments :)

Would it help if I added another data-block containing the same attacker using the same weapon? :)

>  >>> We're looking at a gif or jpeg?
> Indeed.

Surely a GIF then?

>>> Surely a GIF then?

Actually, its a JPEG. :)

Are you being attacked with a sword by some creature possibly a man.

So much for copying to Notepad and saving it as a .jpg....

>>> Are you being attacked with a sword by some creature possibly a man.

If you have decrypted the image, please post it somewhere, or otherwise provide a proper description of the "creature" attacking you :)

>>> So much for copying to Notepad and saving it as a .jpg....

I wouldn't've posted it here if it were that simple ;)

It is UUEncoded or Base 64. Once that has been done, save it to a file, and try gif or jpg

>>> It is UUEncoded or Base 64. Once that has been done, save it to a file, and try gif or jpg

Have you tried it already?

Just like I promised... the same attacker, same weapon, different data-black...

o11"1R1"1I`*x'Z'`~~~`o@`o@``1"1y`L`$##4#3$44455$%^)7^66^q(08)eQrEeQWWRY]iRT[yWWIDO[\|sssriFHfah]Asa1"1y`L~555^6^+77+atWtaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa1"1$`9$`s`t2~T`!9~29~1"1^`r``~3~~~~~~````````~!2@3#4$5%61"1^`o>*`!~22!@233@@``~ot~!2`@93(t|K#0Bo34To-)aouoSo:$yloM15-n1q1fYAo#o_oU5%_=+qQuUiIOpsSdDfFgL;:'"zZxNmM,<.>/o4o$o5o%o6o^o7o&o=o+oqoQowoWoeoEoioIoOopoPo[o{o]odoDofoFogoGohoHojo'o"ozoZoxoXocoCovo,o<o.o>o/o?1`1~1!1%161^171&181*191(1Q1w1W1e1E1r1R1t1T1O1p1P1[1{1]1}1\1|1a1F1g1G1h1H1j1J1k1K1l1"1^`r~`2~~~~~~~~~``````~!2@3#4$5%61"1^`o>9`!~!@@2@43@@`~!ow`~!29@3t|#(KB4o3o-0Taou$)loSo:oM155yAn1f-o#o_1q%_Ys1Ou1F=+qQUiIOpSdDfFgL;:'"zZxNmM,<.>/o4o$o5o%o6o^o7o&o=o+oqoQowoWoeoEoUoioIoOopoPo[o{o]odoDofoFogoGohoHojo'o"ozoZoxoXocoCovo,o<o.o>o/o?1`1~1!1%161^171&181*191(1Q1w1W1e1E1r1R1t1T1p1P1[1{1]1}1\1|1a1g1G1h1H1j1J1k1K1l1"1T`^2~`!929`J`1j8+1koSoR{1O1to>`o}u15o3-1@o0oKoBt1H18o@~Ho<o\o$o%1@1yz1F1Do|o514op1d.o@ol1(1do2]+on-{1poSo}o-WndDJ@~ogHoWfonoG1Pi1;o]_oL1H1'oL1l]x1~1Io}G1`1;3o:1%{{ox1U1%o60XoYoMo.op:oSo)oB1~oYp1$o-1H@<oJo{oC<1T1g1512oeeof1WyXoHoz1yotv1Uo]`1$17-1\oBo>f1%o]Iox4ok1;n131!o<o8_o/1~oj#CGmoHcn-PT1To/1E+1'toDoi1Wo$1roI1kxYom1F}oi]oBA@1DoC1To"Zw-o%>+4@oao;oUKEoIoaop1#oK1F1(o<1;j1FToc[18x1!`~odVof1e1g1DoPUoj1p1A'oUw>1wnoU1]ro,opo\1P1_n!dz1yo\1P1Do-1A17o_,o\1Puf1uoV1iVXE1U<[H.Mo:,Bo\oko9=io5or1ioToP1W1#}olXo0^o~o`l1yo5o{1po+o*o6@zozEo8n1%1A1S1!oqo>1U1-oloRXU1Y1}o(1I&A1k1r3o$S1AoL81AdoL9o+1\o%1q:oV1=S16oh19Mb5o~oMo=1$wou2RoN4f18o;?K-oBob,1f:W\1^ouyoaS%oco;o@`4`~1&3o9voNGo>oj1H"oHo{161@H"VoHX1_]1A1Duo'o]PNok181Fo61;o{1T1}1:oi1}1"`1qo9o2aoWo#1-1i1&1'rbKoOS6,n1d>oFoI1$ol13o=1$19ho=o2C1ko<1';1So=1'151G1"`1\o/oNoJ1;|[o)oWo<"ojo7_fo"1#ouoK1Y1Om4@odVK?NXoF1Y1G|odo-oB6oy/oLmBco].1}oOUop:odO)agWoV21*Roao'oL.olohB:)~71uobD1Eo=1To`ox1kio|o'oSa'1{o3oUaFWQ1{1816r4oaojjJ1S1]o21']o]Oxpx1Q<3o"1]?oKEoHox>1\1To+,_*1KdoMo6o#o({o\odVYoD1Wojo?:)Yodo,wo912omo'o]Oo[1"1t

>aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

Obviously, whoever was attacking succeeded.

I noticed some smileys there too.  Cute.

>>> Obviously, whoever was attacking succeeded.

LOL

>>> I noticed some smileys there too.  Cute.

Thanks, but the smileys are a part of the image, they're not there just for good measure, or good mood ;)

u will only understand it if you took the red pill.

>>> u will only understand it if you took the red pill.

Umm... isn't the red pill for odd days and the blue pill for even days?

mmm... both data blocks have exactly 1891 characters... so, when decoded the "image" size is almost certainly the same.

Is the image size 42 x 42????

If so, did you encode it with the infite improbability drive?

still on the red pills here

Or 31*61?

Is it an ogre with a club?

>>> Is the image size 42 x 42????
>>> Or 31*61?

Just to stop the guessing game, the image is 33x52 in pixels.

>>> Is it an ogre with a club?

Nope, and once someone does decrypt the image, I'd require proof that you did, so please post the code you've used to decrypt the image so we can all see your code works :)

As far as I can tell it is a uuencoded file.  The problem I am having is tryin to decode it.  Tried using software but that didn't work.  I started trying to do a character exchange by changing it into binary, but that got to be tedious, although this might be the way to go cause it should come out into an Ascii art file

>>> it should come out into an Ascii art file

I thought I already said it was a JPG image that is 33x52 in pixels.

>>> As far as I can tell it is a uuencoded file.  The problem I am having is tryin to decode it.  Tried using software but that didn't work.

Well, that's the idea isn't it, if you could decode it with just any software, what'd be the point? :)

The encoding appears to be based on the layout of a Roman QWERTY keyboard.
I have not determined what particular national keyboard was used.

>>> The encoding appears to be based on the layout of a Roman QWERTY keyboard

True.

>>> I have not determined what particular national keyboard was used

Just a normal Microsoft Natural Keyboard Pro, made in Thailand ;)

doronb, If the data is a jpg file of any form then it does not have a header.

Is it a Microsoft GDI+ vulnerability? in a JPEG?

>>> If the data is a jpg file of any form then it does not have a header.

Its a JPEG file header and all, if you haven't been able to decrypt it yet, try try again :)

>>> Is it a Microsoft GDI+ vulnerability? in a JPEG?

Nope, its an image of a "creature" holding a weapon and ready to attack, in JPEG file-format, encrypted of course.

The image itself is intact, I assure you.  Before posting the text-blocks I check them to make sure they can be decrypted back into the original file, there's no tricks involved.

Are the two data-blocks the same JPEG file?

An encrypted .jpg, we could be here for years. You realise the only plain text would be in the header and that depends on the program that made it.

>>> Are the two data-blocks the same JPEG file?

Yes, they are.

>>> An encrypted .jpg, we could be here for years.

Well, I do want to test my idea for encoding a binary data-stream, and so far its holding tight :)

>>> You realise the only plain text would be in the header and that depends on the program that made it.

I didn't really understand what you meant to say... :o

......so this is not a reasonably solvable puzzle but an encryption test, that would have been nice to know ahead of time.

>>> so this is not a reasonably solvable puzzle but an encryption test, that would have been nice to know ahead of time.

I guess its reasonably solvable to someone who knows his encoding; I have been giving hints along the way...
and in my post [02/14/2005 08:42AM GMT-12:00] I said:

   "...once you decrypt the text-block..."

So I never kept that a secret and assumed anyone who'd post here would read the comments first.  I consider this reasonably solvable for a person experienced in encoding/decoding.  I can also tell you that there is no special key involved and there were no prime numbers involved in the encoding.  The only piece of the puzzle missing is the encoding/decoding process.

By the way, the graphics program used to create the image and save it as a JPEG was Microsoft Paint.

what do you mean by "data-black" ?

Sorry, "data-black" is a typo, I meant to say data-block as in a text-block containing the image.

Ok can you answer a few questions i have about your data block above.

The thing that puzzles me is that alot of these post almost seem to contradict or confuse the matter. It was mentioned that 2 different codes were used... Does this imply that part of your data block is encoded with one type of encryption, and the rest in another?
 OR does this imply that your image was encrypted completely with one type of encryption....and the newly encrypted file is then encrypted AGAIN in another/same encryption form. The result would be one file, encrypted twice!

The other thing i am puzzled with is what type of encryption you have used.... Most types of encryption require some type of header or footer, to tell the decrypter the origianl file name, or its size etc. Our problem is that we know the file size (as you have previously mentioned it) HOWEVER the headers and footers are missing...which puzzles me.
As there are no headers or footers, i can't see this code being UUencoded, Base64, XXencoded or yEncoded, unless of course you have removed key headers or used another form of encoding....

Am i running in circles, or am i remotely close?

>>> The thing that puzzles me is that alot of these post almost seem to contradict or confuse the matter. It was mentioned that 2 different codes were used... Does this
>>> imply that part of your data block is encoded with one type of encryption, and the rest in another?
>>> OR does this imply that your image was encrypted completely with one type of encryption....and the newly encrypted file is then encrypted AGAIN in another/same
>>> encryption form. The result would be one file, encrypted twice!

I apologize since I may have misdirected you here.  There are two steps to this encoding process, not two complete encoding processes.  I first start with an array of characters which I rotate around (and saying that in itself is a big hint).  Then the array is used to encode the contents of a stream of bytes.  However, after encoding the original bytes, the process is complete.  So yes, there are two phases of encoding here, which is also why the 2nd text-block is a bit different than the first but both text-blocks contain the same JPEG image.  There is no double-encryption here, just a random rotating of the base character-table and then encoding of the byte-stream.

>>> The other thing i am puzzled with is what type of encryption you have used.... Most types of encryption require some type of header or footer, to tell the decrypter
>>> the origianl file name, or its size etc. Our problem is that we know the file size (as you have previously mentioned it) HOWEVER the headers and footers are
>>> missing...which puzzles me. As there are no headers or footers, i can't see this code being UUencoded, Base64, XXencoded or yEncoded, unless of course you have
>>> removed key headers or used another form of encoding....

I have used my own encoding (of course) which gives the end result of a simple text-block.  I do not require a header or footer since my encoder/decoder use the same character-table and the file-size is evident from the length of the text-block itself (that is, if you understand the encoding process).  I assure you that I have not removed any headers or footers as the code I use is my own and not truly Base64 or any other encoder there were no extra info-blocks to remove.

>>> Am i running in circles, or am i remotely close?

I think you're closely running in circles ;)

Honestly, there's not a lot I can say now without giving you the full answer here (or giving a bigger hint than I already have).  What I did is to take a JPEG file, read its contents as a byte-stream, run the byte-stream through my encrypter/encoder, save the textual result in a text file.  I then re-read the text file's contents and decoded it back to a JPEG file which I then compared with the original image.  I did this process twice, once for each of the text-blocks I posted here.

Some of your comments have made me think of ways to improve my encoding algorithm:

@Arawn:            >>> From a decryption standpoint, there are way too many H's and T's for this to be a normal substitution cypher.

I've improved my encoding slightly thanks to your comment :)

@JohnK813:      >>> But the characters between those letters are basically the QWERTY sequence.
@ozo:            >>> The encoding appears to be based on the layout of a Roman QWERTY keyboard.

That's why I decided to call my encoding QwertyEncoding, thanks guys! :)

@ ric_m:            >>> ...both data blocks have exactly 1891 characters...

The same JPEG image, encoded with the new and improved QwertyEncoding algorithm is now 1659 bytes/characters long, 232 bytes smaller.

I'm still keeping the old code, but if anyone is interested, I can post a text-block from the improved QwertyEncoding algorithm.

So?  Everyone gave up already?! :o

I'm still trying =o(

Thank you :)

Well? anyone?

for the record, I'm out.  too specialist, and too much work.

sorry.

Decrypting is by no means something I'm good at.  Maybe if I have some leisure time over the next few days....

No problems muso and thanks for trying.
Thanks John813, I appriciate the attempt.

It's Harry Potter and he's holding his wand out at a creature.

*wink*

Nice try, but no :)

This question is interesting. You should keep it open just to see how good is your QwertyEncoding. However, decrypting a JPG file is too much of a bother for someone to answer your question. How about posting just a QwertyEncoded text pharse in plain english?

I took your advice diegoful, and here's the result:

mzt!,m&m8mqRtm=w6qQ8-)tRm8mqRtm8)Q6+6qQ8)7t]R?m-mwRtmq&-w(%tRtm966_tRm8mQRtm-_6)
tRtm*wqQtRmQm-Rmqm6m6Rm&m-meRtm7--%tRm8mqRtmr-w+tRntme6+Qrt:tm)5-%8)7t]RZtm-e6W6
+t[Rtm%65+r_Q8)7tRm4RXB"Rtm^8(6tRm8mqRmQm-m-Rtm0w5&tRm-m^Rm4Rtm$-Q&6+tRm^m-m+Rtm
q-06-)6tRmQm-Rtm4)qe6+tRtmr-w+tRtm=w6qQ8-)t]RZm-meRtm4$-wQtRtm_-qQ8)7tRtm*wqQtRm
4Rntme6+Qrt:tm)5-%6%tRtmQ6EQtRtm_&4+q6tRm8m)Rtm_(48)tRtm6)7(8q&tJ

The above text-block contains simple English text only.  The following text-block however contains the original JPEG image:

\*-`-*CUC{-~9czXz~111~\#~\#~~*C*o~:~$##4#3$44455$%^_7^66^w0)8_rWtRrWEETU}oTY{uEE
OFp{aAdddtoGJgsj}Sds*C*o~:1555^6^Q77QsyEy`32`s*C*^~($~d~y21Y~!(12(1*C*9~t~~13`6`
1`8`~1!2@3#4$5%6*C*9~*19~!122!@233@@~~1\u1!2~@(30yAL#)N\44Y\q_s\o\F\Z$u;\>*7=m*r
*jUS\$\Q\O5%+qQwWiIoOpPdDfFgGh:'"zZxXcM,<.>/?-\~5%6^7&89wWeErRtTpP[{]}|agGhHjJkK
lxXcCvVbBn/?*~1!2@3#&89(0)_=+RtTyYuUiI]}|aAsSdDfJkKlL;:'"zC9-~t1~2`9`1`6`~1!2@3#
4$5%6*C*9~*1(~!1!@@2@43@@~1!\r~1!2(@3yA#0LN4\4\q)Ys\o$_;\F\Z\>*75uSm*j=\$\Q*r%+U
d*]i*JqQwWIoOpPDfFgGh:'"zZxXcM,<.>/?-\~5%6^7&89wWeErRtTOpP[{]}|agGhHjJkKlxXcCvVb
Bn/?*~1!2@3#&89(0)_=+RtTyYuUiI}|aAsSdDfkKlL;:'"zCI-~^21~!(2(~K~*;8Q*'\F\Y]*]*i*1
~\Ai*7\4=*$\=\:\My*L*)\#1J-\?s%^-*$*ox*J*H\S\6*6\{*h/\#\'*+*h\3}Q\,=]*}\F\A\qEmf
FK@1\jJ\Rg\,\J*|o*x\a+\"*L*c\"*z}c*3*{\AH*@*x3\Z*&]]\v*P*&\7)C\I\>*~\{"\F\+\M*3\
IP*^\q*L@>\L\|\B>-*Ik74-\tr\h*TuC\K\c*o\ub*P\a~*^*(=*d\M*1g*&\aO\v4\;*xm*5*#\?\(
+*!*3\l#VH,\Kvm=[Y*I*!*YQ*cy\G\p*T\%*u\P*'cU\.*J|\p}\MS@*H\B*I\XXe=\^?Q4@\d\z\OL
R\P\d\{*%\:*J*+\?*xk*JY\b{*)c*#~1\gB\h*y*k*H\]I\l*}*Fz\Oe?*tm\O*st\/\{\s*|*Wm!fx
*o\s*|*H\q*F*(\Q.\s*|ig*p\N*[BCR*P>{J/<\Z.N\s\;\)qo\6\y*[\U\]*T*%|\'C\=^\2\!;*o\
6\|*}\W\0\7@x\cR\(m-*&FG#-\e*1*P*w\'\YCI*O*S\_*{&S*'*u3\%D*F\"8*Ff\"(\W*d\^*r"\N
*eD*8\k*=<n5\2\>\w*^e\o2T\<4g*)\z\1L=\M\m.*j"Ea*9\ou\dD%\b\z\#~4~1*03\)b\<H*1\l*
LZ\K\|*8*$JZB\KC*W}*F*Hi\x\a[M\;*)*J\7*x\|*I*S*X\p*S*C~*r\)\3s\R\$-*w[0c-tnL\[D6
.m*h?\H\P*^\'*5\w*^*=j\w\3V*'\?*c'*G\w-*c7KC-~*d*!\<\L*xA{\+\R\?Z\l\8+g\X*%\o\:*
O*],4@\gBL\1MC\H*O*KA\g\q\M6\i\~\",Nv\a/*S\[I\{"\gp_shE\N2*_T\d\x\"/\'\kN"_17*p\
mF*Y\w*I\!\v*'o\S\x\Fsz*A\4\OsGEW*A*)*8t4\d\lkK*G*s\3*c}\apcPc*R>3\X*s\1\:R\K\v?
*d*I\W.+9*"f-\>7$_-]\s\gBU\G*T\l*2"_U\g\/e\)*4\.\x\ap\}*C*i

I am sure that once you figure out what the English text that's contained in the first text-block, the JPEG image would be easier to figure out as well :)

Just a comment about the text-blocks, the QwertyEncoding algorithm is an improved version that created the text-blocks posted in the beginning of the question.  This means that although the 2nd text-block in my previous post does contain the same JPEG image I've used so far, the encoding is different!

Another comment, both text-blocks in my previous comment are formatted to have 80 characters per line (except the last line of text of course which usually would have less than 80 characters).  It would be easier to visualize this if the text-blocks would be copied and pasted into NOTEPAD for instance with a fixed-width font like Courier or Fixedsys.

Any luck diegoful yet?

it has something to do with the positioning of the *'s?

Got it!  Professor Plum, in the library, with the candle stick. ;)

No, its not Professor Plum in the library with the candlestick, nice try, but no ciggar ;)

And of course the positioning of the *'s is important, every character in the text block is important if you want to decode the JPEG file, or the encoded text message :)

The sorce code has kind of a pattern too... &quot &amp &gt &lt

source code? You mean the HTML source code to the post? If so thats because EE converts special characters to thier HTML entities.

There's no source code or HTML conversion involved since my QwertyEncoding makes sure all content-bytes is converted to text characters that appear on a normal QWERTY keyboard (hence the name, QwertyEncoding) :)

Ok, I've decided to make things really easy and post three words that define things most people would always like to have in life.

This is something almost everyone wants to find at least once in their lifetime [*^a/Cb<'] (4 letters)
This is a resource most people waste without realizing it [fjdq,XV"] (4 letters)
While this resource, most people just love to spend [xo;}VBb'?] (5 letters)

The square-brackets are not a part of the words and all three words are encoded with my QwertyEncoding.

Once you've figured out the three words and how to decode them, try the following block:

mzt!,m&m8mqRtm=w6qQ8-)tRm8mqRtm8)Q6+6qQ8)7t]R?m-mwRtmq&-w(%tRtm966_tRm8mQRtm-_6)
tRtm*wqQtRmQm-Rmqm6m6Rm&m-meRtm7--%tRm8mqRtmr-w+tRntme6+Qrt:tm)5-%8)7t]RZtm-e6W6
+t[Rtm%65+r_Q8)7tRm4RXB"Rtm^8(6tRm8mqRmQm-m-Rtm0w5&tRm-m^Rm4Rtm$-Q&6+tRm^m-m+Rtm
q-06-)6tRmQm-Rtm4)qe6+tRtmr-w+tRtm=w6qQ8-)t]RZm-meRtm4$-wQtRtm_-qQ8)7tRtm*wqQtRm
4Rntme6+Qrt:tm)5-%6%tRtmQ6EQtRtm_&4+q6tRm8m)Rtm_(48)tRtm6)7(8q&tJ

And if you've figured that text-block, try decoding the original JPEG image:

\*-`-*CUC{-~9czXz~111~\#~\#~~*C*o~:~$##4#3$44455$%^_7^66^w0)8_rWtRrWEETU}oTY{uEE
OFp{aAdddtoGJgsj}Sds*C*o~:1555^6^Q77QsyEy`32`s*C*^~($~d~y21Y~!(12(1*C*9~t~~13`6`
1`8`~1!2@3#4$5%6*C*9~*19~!122!@233@@~~1\u1!2~@(30yAL#)N\44Y\q_s\o\F\Z$u;\>*7=m*r
*jUS\$\Q\O5%+qQwWiIoOpPdDfFgGh:'"zZxXcM,<.>/?-\~5%6^7&89wWeErRtTpP[{]}|agGhHjJkK
lxXcCvVbBn/?*~1!2@3#&89(0)_=+RtTyYuUiI]}|aAsSdDfJkKlL;:'"zC9-~t1~2`9`1`6`~1!2@3#
4$5%6*C*9~*1(~!1!@@2@43@@~1!\r~1!2(@3yA#0LN4\4\q)Ys\o$_;\F\Z\>*75uSm*j=\$\Q*r%+U
d*]i*JqQwWIoOpPDfFgGh:'"zZxXcM,<.>/?-\~5%6^7&89wWeErRtTOpP[{]}|agGhHjJkKlxXcCvVb
Bn/?*~1!2@3#&89(0)_=+RtTyYuUiI}|aAsSdDfkKlL;:'"zCI-~^21~!(2(~K~*;8Q*'\F\Y]*]*i*1
~\Ai*7\4=*$\=\:\My*L*)\#1J-\?s%^-*$*ox*J*H\S\6*6\{*h/\#\'*+*h\3}Q\,=]*}\F\A\qEmf
FK@1\jJ\Rg\,\J*|o*x\a+\"*L*c\"*z}c*3*{\AH*@*x3\Z*&]]\v*P*&\7)C\I\>*~\{"\F\+\M*3\
IP*^\q*L@>\L\|\B>-*Ik74-\tr\h*TuC\K\c*o\ub*P\a~*^*(=*d\M*1g*&\aO\v4\;*xm*5*#\?\(
+*!*3\l#VH,\Kvm=[Y*I*!*YQ*cy\G\p*T\%*u\P*'cU\.*J|\p}\MS@*H\B*I\XXe=\^?Q4@\d\z\OL
R\P\d\{*%\:*J*+\?*xk*JY\b{*)c*#~1\gB\h*y*k*H\]I\l*}*Fz\Oe?*tm\O*st\/\{\s*|*Wm!fx
*o\s*|*H\q*F*(\Q.\s*|ig*p\N*[BCR*P>{J/<\Z.N\s\;\)qo\6\y*[\U\]*T*%|\'C\=^\2\!;*o\
6\|*}\W\0\7@x\cR\(m-*&FG#-\e*1*P*w\'\YCI*O*S\_*{&S*'*u3\%D*F\"8*Ff\"(\W*d\^*r"\N
*eD*8\k*=<n5\2\>\w*^e\o2T\<4g*)\z\1L=\M\m.*j"Ea*9\ou\dD%\b\z\#~4~1*03\)b\<H*1\l*
LZ\K\|*8*$JZB\KC*W}*F*Hi\x\a[M\;*)*J\7*x\|*I*S*X\p*S*C~*r\)\3s\R\$-*w[0c-tnL\[D6
.m*h?\H\P*^\'*5\w*^*=j\w\3V*'\?*c'*G\w-*c7KC-~*d*!\<\L*xA{\+\R\?Z\l\8+g\X*%\o\:*
O*],4@\gBL\1MC\H*O*KA\g\q\M6\i\~\",Nv\a/*S\[I\{"\gp_shE\N2*_T\d\x\"/\'\kN"_17*p\
mF*Y\w*I\!\v*'o\S\x\Fsz*A\4\OsGEW*A*)*8t4\d\lkK*G*s\3*c}\apcPc*R>3\X*s\1\:R\K\v?
*d*I\W.+9*"f-\>7$_-]\s\gBU\G*T\l*2"_U\g\/e\)*4\.\x\ap\}*C*i

If you've finally managed to decode the image, please post the code you've used to decode my text-blocks and also, what's in the JPEG image. :)

Arawn, you're in the right direction :)

The others could be "love" and "money" (not sure of the capitalization used)
which would mean they all have "e" in the fourth position
<'  V"  b'  all end in ' or " which go together on many qwerty keyboards
Patterns from other encodings suggest that each plaintext character maps to a fixed number of cypertext chracters,
To be decodeable, it would help it if could be determined from the first character whether a one or two character encoding is being used.
One possibility could be that ? always decodes as y
The 8 or 9 character samples don't seem to have the pairs of very common characters that seem to characterize the other samples

You're right about the other words, it is TIME, LOVE and MONEY.  Keep up the good work :)

Are you saying they should be capitalised?

Yes.

public class TestApplication {
      public static void main(String[] args) {
            String love = QwertyEncoding.encodeBytesAsString("LOVE".getBytes());
            String time = QwertyEncoding.encodeBytesAsString("TIME".getBytes());
            String money = QwertyEncoding.encodeBytesAsString("MONEY".getBytes());
            System.out.println(new String(QwertyEncoding.decodeBytesFromString(love)));
            System.out.println(new String(QwertyEncoding.decodeBytesFromString(time)));
            System.out.println(new String(QwertyEncoding.decodeBytesFromString(money)));
      }
}

The output is:
LOVE
TIME
MONEY

Now, figure out the rest ;)

So, any ideas yet?

>> Those riddles were easy enough without you giving away the answers.

True, but those riddles weren't the real riddle :)

>> If you're limiting yourself to the letters and numbers directly on a QWERTY keyboard

I am, but I'm also using the SHIFT key so I have 1 and ! and 'a' and 'A'.

>> You're converting the letters into XY-Qwerty Coordinates

You're quite close there.

>> The alternative is that you're adding on (possibly randomized) garbage that is placed in location where you can determine if it is garbage.

Actually, I'm not including any grabage on purpose.  There can be used characters however if the encrypted data is too short or not varied enough.

>> PS: I spy Java, how'd you learn that? I've been looking for ages for a decent guide and I've come up rather empty-handed.
>> PPS: Using the Altkey with the Numberpad (numlock enabled), the QWERTY or AZERTY keyboard can contain at least the whole ASCII set.

I'm not sure what you meant by those two comments.

I think it's an ANSI File, do you remember ansi??? i'll check it out right now..

Nope - no ansi there, ill check out RIP.

Sorry didn't follow the post from the begginning ..... I'll try to see if i can scramble some letters from the 3 words.

>> Nope - no ansi there, ill check out RIP.

What's RIP?

another kind of graphics like ansi

Ok, thanks, but no, its not ASCII graphics, the encoded graphic file is a JPEG file and the short messages are simple text messages really.

So... any ideas? :)

Are you doing any binary operation on the ASCII values to get them to fit into the printable range, or you just use the values and map them onto the keyboard?

>> I bet the array you use in the encoding is a multidimensional one and must somehow contain a representation of your keyboard, does it?

Multidimensional, wrong; contains a representation of an English QWERTY keyboard, correct.

>> Are you doing any binary operation on the ASCII values to get them to fit into the printable range, or you just use the values and map them onto the keyboard?

I'm actually mapping any byte-value (0-255) to a QWERTY sequence. :)

Any luck guys?

A friend of mine read this question-thread and suggested I put this question on a special PayPal site, so that
people pay something like $5 to participate in the decoding-contest and whoever decodes the JPEG file first
gets the accumulated cache... anyone think that's a good idea? >;)

> I'm thinking that perhaps there is a character to indicate "shift," so shifted chacarters are two characters long,
> and unshifted are one character long.

If that was true, the encoded output for 'MONEY' wouldn't have had seven characters. It's not that easy.

Also, there is no direct correspondence between characters. Letters O, E and M are repeated in all three words, but there's no similarity in the encoded strings. That seems to indicate either of:
- The encoding uses a random seed, which should somehow be embedded in or obtained from the encoded data.
- The encoding uses the length of the data as a seed. The problem with this alternative is that as we have seen, the data length isn't preserved during the encoding process, so...

uKER,

> Also, there is no direct correspondence between characters. Letters O, E and M are repeated
> in all three words, but there's no similarity in the encoded strings. That seems to indicate either of:
> - The encoding uses a random seed, which should somehow be embedded in or obtained from the
> encoded data.
> - The encoding uses the length of the data as a seed. The problem with this alternative is
> that as we have seen, the data length isn't preserved during the encoding process, so...

More or less what I was saying.  It could be seeded with the length of the input string, which IS known.

- qwaletee

> It could be seeded with the length of the input string, which IS known.

Is it? Really? How would you find out
xo;}VBb'?
translates into a five-character string?
Either I've missed something or...

Say the xo;} gives us a seed, and the VBb'? is straigh translation.  There could a more complex varuiant as well. xo could be the seed, and due to the nature of the hash, tghe last character could be filler.

> Say the xo;} gives us a seed, and the VBb'? is straigh translation.

I seem to like that idea. Will give it a deeper insight when I get home.

Bravo you two, you're on the right path for sure... now, all you need is to decide what characters are the direct translation, and what characters are the "header"... and then, to start translating the JPEG!!

/me laughs quietly as the traps within are laid out and yet unnoticed... <evil chuckle!> >;)

> Bravo you two, you're on the right path for sure...

I somewhat feel you're misguiding us here. There can't be a 'header' global to the file. There have to be per-character indicators for the decoding, as you are mapping the whole byte range into printable range, or what is worse, keyboard range. So there simply can't be a direct translation anywhere, in the sense of one data byte corresponding to just one encoded byte.

Unless there's some kind of 'block header' that you could recognize as not being data, and told you for example, that the following characters fall between a given ASCII range.

*Sigh!*...

Let me give another hint... each text-block is composed out of two "segments" of characters (in a keyboard range), the first part is a "header" of sorts, without which, the second part is un-decodable.  The second part is the actual encoded byte-stream.

I really can't give you more hints on this and if I would, I'd have to end up closing the question and giving the points to myself ;)

>> is your name Doron...

Yes it is :)

>> ...the first part of the cyphertext tells the decoder what version of direct translation is in use, and the rest is the direct translation...

Not exactly what version of direct translation, but very close.

>> The first part, more specifically, tell the decoder how to construct the mapping for the direct translation.
>> ...the vast majority of the block is direct translation (albeit, not necessarily 1:1 char count)...

Both statements are true.

>> It would then be a huge job just to figure out how the header creates that mapping, and then to generalize tha to all header classes.  On top of that, we have no large plain ASCII text /cryptotext to analyze for such frequencies.  Which means, it may be beyond me altogether, and is certainly beyond my ability to invest time in it.

Do remember that my encoder may produce different text-block results for the same byte-stream being encoded. >:)

Its been two months since I opened this question, is anyone closer to a solution yet?

You attacked me with a question.

You mossad abies get us every time, even with cyphers that are little more than plain text :(

I deny any ties to the Israeli Mossad! >;)

So...?  Does everyone give up?  Should I give away the solution just like that? >;)

Can someone confirm the US keyboard layout as I use a british layout and the layout is different, including characters such as £ and with various symbol keys in diferrent places. I have it as:

~!@#$%^&*()_+
`1234567890-=

QWERTYUIOP{}|
qwertyuiop[]\

ASDFGHJKL:"
asdfghjkl;'

ZXCVBNM<>?
zxcvbnm,./

(top line to bottom line, shifted line first then unshifted)

Damn. Give this guy a Nobel! :O

Just for a follow up - to say there is no header info appears to be a bit of a fib! The first two characters always seem to be the two key characters!

A few different ways to get rid of the extra 'Z's, I'm not sure I chose the correct method,
the same algorithm on the JPEGs don't find any JPEG headers.
Could you post some more examples.

Well, so far ozo's the closest, but even though those Z's can be removed from the English text, eliminating them from a JPEG byte-stream proves difficult.
Still, I'm pleased with the progress made. :)

paradoxism, great work on figuring out the basics, but you're totally wrong with the DUMMY characters, there are NO dummy characters at all, every character is required or else you get Z's in the wrong places just like ozo did.

Just to confirm  - so this isn't a basic cyclic 8-bit qwerty encryption? Shame, that one looked good paradoxism.

A small confession, those two character CAN actually be considered "dummies" IF (and only if) they are not used in the text-block itself (that is, if they appear only once in the whole text-block).

So.. how are things progressing? :)

Could you give a text example in which all the header characters appear in the text-block itself?
How about an encoding of
 !"#$%&'()*+,-./0123456789:;<=>?
 @ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_
 `abcdefghijklmnopqrstuvwxyz{|}~
 ¡¢£¤¥¦§¨©ª«¬­®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖ×ØÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿ

_iaxtTyYuUIoOpP[{]}\|AsSdDfFgGhHjJkKlL;:'"zZXcCvVbBnNmM,<.>/?a_`~1!2@3#4$5%6^7&8
*9(0)-=+qQwWeErRtTyYzZXcCvVbBnNmM,<.>/?i`~1!2@3#4$5%6^7&8*9(0)-=+qQwWeErRtTyYuUI
oOpP[{]}\|AsSdDfFgGhHjJkKlL;:'"zZXcCa

-<@`yYuUiIoOpP[{]}\|aAsSdDfFgGhHjJkKlL;:'"zZxXcCvVbBnNmM,.>/?@-~1!23#4$5%6^7&8*9
(0)_=+qQwWeErRtTyYuUzZxXcCvVbBnNmM,.>/?<~1!23#4$5%6^7&8*9(0)_=+qQwWeErRtTyYuUiIo
OpP[{]}\|aAsSdDfFgGhHjJkKlL;:'"zZxXc@

+(rjyYuUiIoOpP[{]}\|aAsSdDfFgGhHJkKlL;:'"zZxXcCvVbBnNmM,<.>/?r+`~1!2@3#4$5%6^7&8
*90)-_=qQwWeERtTyYuUZxXcCvVbBnNmM,<.>/?(`~1!2@3#4$5%6^7&8*90)-_=qQwWeERtTyYuUiIo
OpP[{]}\|aAsSdDfFgGhHJkKlL;:'"zZxXcCr

ta-+yYuUiIoOpP[{]}\|AsSdDfFgGhHjJkKlL;:'"zZxXcCvVbBnNmM,<.>/?-t`~1!2@3#4$5%6^7&8
*9(0)_=qQwWeErRTyYuUZxXcCvVbBnNmM,<.>/?a`~1!2@3#4$5%6^7&8*9(0)_=qQwWeErRTyYuUiIo
OpP[{]}\|AsSdDfFgGhHjJkKlL;:'"zZxXcC-

All of the above text blocks are encoded versions of your requested text ozo.  However, you will find that only the first three header-characters are in use.  An example using all four header-characters would have to include more variation in the target data. >:)

By the way, thank you for your participation in this because this is helping me see all the "holes" in my encoding :)

*** WARNING: Do not try to decode the text in this post! :WARNING ***

Like I said, thanks ozo, your previous question helped me see a potential hole in the encoding.  I think I've managed to improve the encoding chaos factor:

=/#?e7`Wq52+)@$0*~^8^E*6$Q)42-q!`9e#=Uy03Tr8%Ew6&Q_4(-(Y_9&Rw7%Wr53+y@1#Ck=~.Z;<
m:zNbKcVc=!bXz>m";M.LkB#=oL;|[Jk{\GhOsDf:fAsKh}\HkP[F;Io/G]ZxDa:'AdKl}gHjXjFg"lS
dL'|aJx{]W&Op+(Ui0_Ty8wErPrQwIy-_Yi9(Rp7&#
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Data Length: 189; Encoded Length: 206; Encoding Inflation: 9%

$;#(@w`7~r2`&052%=7Q=%9E0&_9r~Q_w@E#$8^t3)4u63-q86*e)Re*+Yq-W+u4RWt^Y#x"?Kvl$!'K
VNx'X<v/<Xn$1NV,n$!l/,?"#$O|]"{Spg|OdjS{aaFKjdH'gpKF"]'Hl;Z}DsCAG\kD}h:GAffJz:hL
ck\zJCscLZo8+-[)W*t+8euW)qqRIueYPt*IR[-PYo#
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Data Length: 189; Encoded Length: 207; Encoding Inflation: 10%

,&#>72`w5`2=@^$0~$^8E-9%Q9-3_eq1(qe#,u)%3t*3%e608q480-Yw=9R=w^Wyr$+ry2#C;k,`Zk;<
:cznKzcv,!mbx?bm'M,1/lB/#,olSo[j|s\g{\sdOhfa:fh]K;kpHk;iF&a]zG]a;DgdkAdgh}ljfXjl
s"x'\L'x[J08oW80u+w=t)=we*yrqPry-Ipi9Yip^R#
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Data Length: 189; Encoded Length: 207; Encoding Inflation: 10%

=#")E~~7Q@@~-55@*77w^99r$__92ww_`rr"=8Y440R664W886+00t(qqu&eeq%tte3uu"X=1llV/''l
,XX'nVV?vNN=!x<<N:??<K"=!pa:]]dKaapHdd]FgglSjj'|llg{''jO#\\fcsshzff\LhhsJkkZG;;C
DZZkACC;}88qP00eIqq8Yee0RttoWuu[+oot([[u&"
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Data Length: 189; Encoded Length: 206; Encoding Inflation: 9%

As you can see, the text-blocks are now even more hectic although you will not detect the last header-character being used.  I would give an example of the last header-characters being used in my next post.

The text-blocks contained in this post were created by a slightly modified QwertyEncoding so they are not fully compatible with the other text-blocks presented in this thread.  Decoding the text-blocks in this post should result in ozo's requested text-string:

!"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~¡¢£¤¥¦§¨©ª«¬­®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖ×ØÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿ

The QwertyEncoding used to encode the text-string is NOT compatible with the one used to encode the original JPEG image and this post serves only as a thank you note to ozo. :)

*** WARNING: Do not try to decode the text in this post! :WARNING ***

>> But I still need a fourth character example, so how about seeing if an encoding of the following has enough variation...

Well, since I know what the fourth character is for, I can tell you now that its not about enough variation so encoding your text example isn't going to help you there >:)

However, as a major hint to what the fourth character is responsible for, imagine an image that has many shapes of SOLID COLOURS.  Now, take this image and save its RAW data (i.e. pixel RGB values) to a file.  Encoding an image like that with the QwertyEncoder will produce a text-block where the fourth character is used, a LOT :)

Repetitions?

Do I even need to answer that? GET workin' already people, sheesh >;)

As you can see from the previous WARNING post I made four days ago, my QwertyEncoding just got more complex.  The last text-block sample contains duplicate characters where the original byte-stream does not.  This may prevent the occasional cracker thinking that my encoding is a direct-translation encoding and although this addition inflate the resulting text-block a bit, I think it is worth the extra confusion.

Encoding the JPEG with the new QwertyEncoding now gives a text-block that is about 540 bytes bigger than the JPEG file itself, but for the purpose of this question-thread I will use the older (easier) encoding.  Again, thank you ozo. :)

Now... has anyone made any further progress yet?

Hello...? ECHO... ECHo... ECho... Echo... echo... ... .. . *sigh!* :o

You are, in the Puzzles & Riddles section, with an encoded file that I won't get because I'm too lazy.

And your point is, tonsofpcs? :o

Come on ppl.. I wanna close this question and give points to ppl... :)

COme on, Doronb, don't you get it?  This is for fun and recreation.  At the point each of us decided it wasn't, well, that's how we get to a 3-month-long question.

Keep it open, keep it open!

qwaletee, what can i say.. sorry if it bugs you, but if this lasts 3 months with people who think about this sort of problem as fun and recreation, maybe it'd stand a chance of lasting a month with someone who thinks they can make a profit out of it :)

diegoful, at the moment, i can't really accept any answer but rather split points among the people who are the closest to getting it right, however, i do want to see this through to the end, so it'll stay open.

Diego, you want it kept open so you can work on it, or because you want the final answer?

Doron, I meant no disrespect.  If it buged me, I would have just unsubscribed.  I'm merely telling you that I don't think anyone is actively working on it, because those who had worked on it stopped, and it is rare for a new person to pick up an EE question if it has been open more than a day or two.

Tell you what... go to community support, request that this question be PAQ'ed with refund, then open a new 500 pointer. You might get new blood.  You can also give a few hints and lower the points to match.

qwaletee, all of your suggestions are valuable and interesting.  I posted this question here to sort of gauge the response-time with which a different encdoing scheme is met.  I don't know if this is realistic or not because as a friend of mine already told me, "points aren't a stimulating motivation", but until I evaluate your suggestions, I'll keep this open and see if others have anything more to say.

ok, show us some source code? :)

Ok synque, here's the newest encoded text-blocks.  The first one is the same as the first image I've posted:

qg{jqsg%{qO\_e{|Pft{qt***_{#qt5{q__kz8{a{q^+99W#Y4WWW00+3y+W6RR*4E47)e*E%e*tt~{b
q@lq~xq!'q^q^;zb>q`lMMMq5Vxq`:/VXxq2:qkqzq8a{q6===6R*$^^&{<q`qWnj32j<{qOn_W+8{"{
q^`&-{"{qt736Et-k{g%{q80^^R)j6j-j8j^R7=$Y9W)=3Rk{g%q8g8{qet7-55r0=44uu88-{Q{qR7=
^utQw`~{\q#q7h8q3?wq%<gzgZgWq)/Ig6qvq0fqzqJq@?()g"{q$t_9=7y2{v.Bml"mVq!Xq2LoSTJ[
F]|Fua[Jokt$)1e^+7*+@9^e$E~{gSL]XF"GJ']KGxaC]Hr8pwy6r&it0O8e=E0o@{qBg~qxg$qMg@q,
q>g2qxq?q,g4qvg%{qx.dpksGi;IkA\kIFs;P"IdsXtw-^Ej9j*j6j_6r&w4y(r0_#sXt{gp{q3^r*+$
$E0W4uu8-%{6{q8-%Etw4`~ew{GhqW35qrM<gzq4q%ugdg_g6qvq=X;Sq|qu()qnq3q^:Bq|,{qH9=7y
!{V?bNq@Zn:MCYK}DRAyhs}Ki\yAp2r&_`(!w0&r#8!(%{g|'Gl[x]'FS"|kFXSc[0E&O=TwE6&Rwp(W
&r_{qCg2q,g`qZg4qxg2qMqBg@qCq/qMg$qBg5{qZ]fo'aHSfUoGS"[doF|kZ8y5R8+WEt_{/{qt;Yuo
{gZ^cqlq?g8q^gHBq.3q0gTEgzgTCqOqzwq*vgRg:7_giqzTqUq]g;&gQg[qha*g"q>qO]Xq$g8q5cqK
gZgz-q^fZn'q0q-gJz9Xg)gfq}.qLgJqTgGqfqpgLq|X|grqDgflgEqAq4gUqbq1q1g^qCqb-qQ\{gd6
rX{}gP_gTgrgJVg#wqOq$PgJg'gQPq.qoqvg6(q9gKq<XGgcgpq>6oqCgPq_qVg5quqKg-gQmq?gFMg=
qWgZq[fg=gog0~q^geg7gHqeOlkgspoquq`"qZgIqVquqpCg]g\q<7qvg}qoR<gyq\"gkXgT?qwq]gEq
.gE]q3qu_kq$q(qwgSgigs\qr{gC\:ua{qLg#gRq[,qU"grzg1[g^q8q-gLsgcqcqhqGgF>gXqKqiDg|
q3AqxSg"qaq0gtgXgaqfqBoq7Z}qzgaqfq]5qig`)[gaqf;:g!gUg2stq)g@PzvK]g_idg:g;Wq-q1&5
g2@gFqCgu,glG4qeD[uqz&g'qp&3-q$AgQq)6oqbqlq\g^r{g8@1l{^Pq!g1qH$qLq5;q"qmq))mqdgL
qYqaVgLq3(qF6qzTgWqC'qcg}qXYDq=Dg60g#q_gHq5q~g*qWq@g4g-OUq0g(go|q|}q(>g%gDmgSbq3
gug(w{q8W^RmQ{5Sg*cgEg[q"dgcgkqBg+q`HugsPg~'qlq]q2g0gZvIg;g1qHWqHg'g$q;qJgk{qUO8
z{5Y/$@g1g~g!qKq5G\g]bq#iSqhA{g\}#l={0g#qXV0Yiq}gRqp]q;0{qpNIktI{g(g0gPqHlC%9g7R
g[!q^q@{gYRHz{qVq|uq3qug'kURFGgkqVqKlgL5g(qR=tgLudDgPoqUgcvg:TgS:q%<LqWgUq5q?cgC
g*g\aghgCO}q%q6qWg!gonqM+qNhgoqf.gdgugZ:tq'%gK<;qWq*qIg4qBq0q3gLgHB/qHqa]qK<gP/[
NRqX\qYg)qJHgaqQgsg^TqIqN([qTq#q'Vgi-@)ZgXgLsbg]q?g[gosq+q@g'g&qT-{ge^*J{:g"qkqx

This one however, is a bit longer and is a different image:

aiexea:>KJ#re}Jo}ea%rrr#e$a%=a#a#aHi5aETea&(YY6)3-66633(=(y7q88%T8=6_%twt%tQQe'a
1Bb'a`Vcararn.KXZnea!!!R@e/,a!ca!z.vMaHi5aETea7999q8%yQQr~e:a5Keax32x~Kei2ea#E(E
e<a&:ata3Cea%R-76&3HzEY&&5yx6x3x8x&5R4(3Y6#9=8HzEeiRear%R388^T499##EE3e0ea5R4&#&
5te:XRa)awO`a0z$aQa~i\idi6a#"Giwaba-pa.aklC!Qisea_*u=3E+eKa2"lvm<a1MB?na@hu[ykOR
}PdHs}g\kQ@6!r$`&%)w(&+8reiz|F}Cd[hDJXHGLjZ]&-^O(q+W_IT=RPE6YreaVi#aCamaXi%eaB"<
n/ei$a.a,i1a>i4eaZU|[LpdoD]OG\F;DSHfLo:M%^3&6x9xrx6x#7^t%9$u&3)+:M%ei+ea-&^r$((6
T69##E3*e9eaE3*6&%9e:Xa^atYOa640a*;a~i\a-aQPiKiuiwaba9a~mhaha(!Qi`a=aWVNak:eaJ=3
E+e;a@XKBV>a2a!la~ptjI]{|PG|{fJjth5~W#7^*%q*^_EW~QeifAZ\h]HS};flZkh'HCU7+6p9^=Er
UW+ypO^IeaNavi4eac<Z.bXei!aNi~i4a?a<i@a.i%eaKosU:[IAFgKfsj:;IlHei%eaE(85E$E6&#e.
a%aja@ba;i&saka\aS0iPiL&aQa`abi9oisa"&a*?a=icaBa&i;acaoafW:ata}2aUiAa\EicSiDdiGH
i#ltpNaM]1iFi@G`o?waIkiDi5]eaDYh2ocezah^eiHk^`eMidaUaZi:|LaK9u'hy[irirea91u%e2aN
aG&mi}$agi5iHl}*i]a6#iza]aLQKIa9a=iIipaji4i;aKaEi8dp&i7a.gaCIiYi@d7aQi;iUiga9_iY
ij.araMiIeaChL?eda;Qoi;saC'iCdi&iy/i;oa+1Pi*OiEi^i!a@yabiXaQFOki")Ra]>iqa{<Fa1]l
aAata(,eaX~BGreiZ1QigKaEa-aS|k.ikiPa'6hqi(a.a?i+i=ZigaEaOa8i@eapmJJueioaAiABi%i)
i8a1a]/fdi$aEiGi#a#a#4ikiYi[5igda6fi`a6iq?iS/i9awaba'vasaZi!iR3ioi#amw_vi`aC\zil
$a_LawijWa8?ea,5$0eQ4M1i9ayPiFa1Ai+iyLipi9eacG`v|e\LaQCi@a0iQa/i2Ui#ahan7iWisaEa
)aZi%i`acPiuacW$i]lif?a7a-a7@iU*eaE3"Ae[aIaJi7iX&i(ziQiQaDziSvagiW{alihaj;r!+a%a
rirR?qiwa[N<rYeat\?]i*9yze;<%alaLitna8i[aYYa&~i'aHisi3iPzi=eah<ZH%q]@eilYiGa5i$h
ri#aJiRitiPasa\RZ'i:i=iCa0iQa#4a*abIIa`ji1CzamaDqiW&i-i0R}$$wiKi-aU$wafOYRioaLEi
Oa~iEi!Da8itasCioiL.iTiYHikaRiE6iwX1Ka6ila4a&@iAd;aIaham+w=i&9>i5ag5=Esei-*@Ieab
_iril[aIak)Fa\DEa<va}7ik[T{)saX*i]i#a}i[ijavViQaoiXaSa1a@i]ifa@a&iXi@a4sa%aki2ya
c=5,9oBauihti$ciIi:ilgayarTli0li(i:a*li(i:a*a}afisi|a?a*i'aOa4aE(IiEa>iRB2i0aliT
i:'ida%E}iYa$6fiZea+B7"Veti@Zi$i2RaE-iXSja#aySja#-aKaz

If you can show some source-code that would decode both of these text-blocks back into their correct images, you'd get your share of the points :)

Open a new question if you want to know how good your improved algorithm is. The question title is "Who's attacking you, and what with?" and in your first comment you explain "Ok, this is attacking as in a game... once you decrypt the text-block, it should be obvious someone is attacking you... Just tell me who (generally) and with what :)". You already got once algorithm change free of charge :P

Just because you asked synque, here's the different question: http://www.experts-exchange.com/Miscellaneous/Puzzles_Riddles/Q_21425106.html

However, I'm sure you'd understand why I need you to show the source code you used in here, and also why it'd be a point-split here as well :)

I understand why you want to split points - it's your decision and a good choice. Whetever I want to post my source code is my decision though. You already got your decoded image.

Synque, the whole point of this excersize was for me to discover how "strong" this encoding is, for that, I need to know how long this actually took to crack, and taking a look into your code would also help in that.  However, as I have put that in more obvious terms in the other question I opened, I will split the points and close this question.  If you want to answer the other question and get any points for it, you will have to show me your code. :)

Okay, okay, your encryption took me most of yesterday to figure out. I got most of my clues by manually decrypting the English Text-Block.
After that, I decrypted the "old" text-blocks and figured out what the fourth header character does.

I was a bit hesitant to post the source code because I'm still new to C++ and hate showing ugly code. Oh well, here goes nothing:

#include <iostream>
#include <string>
#include <algorithm>
#include <sstream>
#include <fstream>
#include <ios>
using namespace std;

template <class Out>
void decode(const string& s, Out out) {
      string cset = "`~1!2@3#4$5%6^7&8*9(0)-_=+qQwWeErRtTyYuUiIoOpP[{]}\\|aAsSdDfFgGhHjJkKlL;:'\"zZxXcCvVbBnNmM,<.>/?";

      char sw_base1 = s[0],
             sw_base2 = s[1],
             sw_block = s[2],
             sw_rle   = s[3];

      cset.erase(cset.find(sw_base1), 1);
      cset.erase(cset.find(sw_base2), 1);
      cset.erase(cset.find(sw_block), 1);
      cset.erase(cset.find(sw_rle), 1);

      int base = 0;
      bool in_block = false;
      int i = 4;
      while (i != s.size())
      {
            if (sw_block == s[i]) {
                  base = 0;
                  in_block = in_block ? false : true;
            } else
            if (sw_base1 == s[i]) {
                  base = cset.size();
            } else
            if (sw_base2 == s[i]) {
                  base = cset.size()*2;
            } else
            if (sw_rle == s[i]) {
                  stringstream ss;
                  while (s[++i] != sw_rle)
                        ss << s[i];
                  int count;
                  ss >> hex >> count;
                  fill_n(out, count, cset.find(s[++i]));
            }
            else {
                  out++ = base + cset.find(s[i]);
                  if (!in_block) base = 0;
            }
            ++i;
      }
}

int main(int argc, char* argv[])
{
      if(argc != 3) {
            cout << "QwertyDec\nUsage: qwertydec infile outfile" << endl;
            return EXIT_FAILURE;
      }

      ifstream in(argv[1]);
      ofstream out(argv[2], ios::binary | ios::out);

      string data;
      string x;
      while (in >> x)
            data += x;

      decode(data, ostream_iterator<char>(out));
      return 0;
}

Binary, Source & Text-blocks: http://www.synque.de/qwertydec.zip

Oh I forgot: the above code only decrypts the new format. To make it work with the old cipher just remove the handling of the "sw_block" and "sw_rle" header/control characters.

The code above decodes the text-blocks appearing in my comment posted on "05/15/2005 09:25AM GMT-12:00"?

No, it decodes the English text-block starting with [mzt!,], the jpeg image starting with with [\*-`], and the single words you posted. If you remove the code dealing with
header character 3 and 4 it could also decode the first two text-blocks you posted - starting with [HTT"]  and [o11"].

I didn't even try to decrypt the most recent incarnation of your cipher. If I do, I'll post in your other thread.

It'll also decode the text-blocks ozo requested. Btw, are the text-blocks you posted on 04/22/2005 with the heading "WARNING: Do not try to decode the..." made using
the same encoder you used for the two newest text-blocks (starting with [qg{j] and [aiex])?

Yes.

Btw, forgot to ask synque, did you read the whole thread before you actually decode my text-blocks, or did you get get to it and did it all in a half-day's work? :)

Well, I found your question right after I joined EE (february) and tried decoding the text-block for half an hour. Back then I noticed that you use the
first two characters as a kind of "control characters". I forgot about your thread while I was trying to get my premium membership, however.

I read most of the thread before trying to decode it on sunday. Your hints were helpful, but maybe not in the way you intended. For example the
knowledge that you spelled the words LOVE, MONEY and TIME in capital letters was very helpful. I think the most helpful thing was the English
text-block. I dunno how long it'd have taken me to decode the JPG directly.