ADFGX cipher

[Shootinfish]

@Nieno69 

The total of the three squares equal the top left square

4+1+5+2+1+3+4+4+2+8+15+5 = 13+15+6+20

Also if you look at the columns the totals are reversed one side and doubled the other

13+6+4+1 = 24
15+20+5+2 = 42
2+8+1+3 = 14
15+5+4+4 = 28

Maybe nothing and just coincidence but i thought i would mention it

[Nieno69]

19 hours ago, Shootinfish said:

@Nieno69 

The total of the three squares equal the top left square

4+1+5+2+1+3+4+4+2+8+15+5 = 13+15+6+20

Also if you look at the columns the totals are reversed one side and doubled the other

13+6+4+1 = 24
15+20+5+2 = 42
2+8+1+3 = 14
15+5+4+4 = 28

Maybe nothing and just coincidence but i thought i would mention it

Ok now I get it ..

 

About the cipher

We have the first two lines... it's Eiter 6-5 or 5-6

 

So if you write it like this: 

Gd ** ** gf ** ** gf ** ** ff ** ** fg ** ** fg ** ** ff ** ** ** aa ** ** dx ** ** gf ** ** ff ** ** af 

You get 12 letters... maybe it's easier to guess the  missing letters than get confused about the wrong order of the others...what do you think?

 

Combined with e.g. Excel at the PC you can easy do a Table board and 

E.g. 1 2 3 4 5 6 ....34

put in the letters you get from your keyed alphabet. if you want to you can concentrate at the double letters (ff, aa) because this is either way correct 

Sry for Bad english 

Edit2:

"Streetinpanic" a big coincidence that it Starts and Ends like the cipher(first line of cipher ="S" from street) ? 

[Jami San]

there was a comment from 3arch (allegedly)  where they say , I think we made that one too hard (adfgx cipher)

I am starting to put more weight behind the theory that the keyword (transposition key) uses repeat letters,

perhaps they did this specifically so we would not be able to use the existing adfgx decrypters, but would have to write it out old school, OR adjust a decrypter program accordingly .

And bringing back up an old question, why is the final AF on the right hand side and not the left, I get that there is no problem having the final AF on its own, but why is it positioned where it is, doesnt this suggest something?

[Nieno69]

9 minutes ago, Jami San said:

there was a comment from 3arch (allegedly)  where they say , I think we made that one too hard (adfgx cipher)

I am starting to put more weight behind the theory that the keyword (transposition key) uses repeat letters,

perhaps they did this specifically so we would not be able to use the existing adfgx decrypters, but would have to write it out old school, OR adjust a decrypter program accordingly .

And bringing back up an old question, why is the final AF on the right hand side and not the left, I get that there is no problem having the final AF on its own, but why is it positioned where it is, doesnt this suggest something?

Its because this:

transposition key e.g. yzabcd 561234)

Under that you write the cipher so you have under 5+6 the AF

Then you rearrange the Order in alphabetical order = 123456 / abcdyz

So under 5 and 6 there is still the AF but on the right side of the cipher because the transposition...

Hope you understand what I want to say

[Jami San]

I do get that, thank you, but i am not completely convinced. So those two columns contained one more letter each, Got that, but... When you finalize the ciphertext and write it out , it is not written in column form, so those two columns with (extra) A and F , would have just been written out in continuation with the rest of the ciphertext. The point in the decipher method when you have those 2 columns with more letter, is NOT the final part , and NOT what gets written out as the final ciphertext...

I hope this makes sense, I Truly do understand what you said, and it is correct,  but now that we have those columns and the last two have extra letter...we are only half way thru the process, From the way I am seeing it , if the A and F are where they are because of what you say, then the process had not been completed.

Going to go re-check the whole cipher process to make sure what I am sayng is even correct...cuz there is the whole horizontal/vertical switch

[Jami San]

K, double checked , I believe I am right, After we have the columns arranged alphabetically , we then write out the final version of the ciphertext going DOWN each column , and the text becomes a horizontal string , with breaks in the text between each column group, If the motd cipher follows the adfgx rules then the AF would have been in thier own column...and when written out , likely would have been written left to right, as all the others were.

I am starting to understand why some people said it was not "in its final phase"

[Nieno69]

51 minutes ago, Jami San said:

K, double checked , I believe I am right, After we have the columns arranged alphabetically , we then write out the final version of the ciphertext going DOWN each column , and the text becomes a horizontal string , with breaks in the text between each column group, If the motd cipher follows the adfgx rules then the AF would have been in thier own column...and when written out , likely would have been written left to right, as all the others were.

I am starting to understand why some people said it was not "in its final phase"

Yeah that's correct its npt the final form.. but if they had write it that way it cannot be deciphered....they had to give both Keys...so the "only" hint they gave was the Arrangement of the column 

[Nieno69]

So back to the cipher: I tried Yesterday to play around with the cipher instead of playing zombies....

So what have i found:

There are not 24/48 possible combinations - you can break it down so there are only 6 (12)

Lets try - hope you understand - it's because the double letters in the third row and in the end of the cipher:

-deleted in edit- will correct the Post if I have time - save you the scrolling

[Shootinfish]

@Nieno69

Well spotted and I think its worth testing these combinations first

The only problem I can see is its a 25 letter alphabet and two letters are shared and we don't know if there the double diagraphs in the third row

 

[certainpersonio]

@Nieno69  So your logic is not sound for this one. The reason is that you are independently moving around the middle portion of the cipher text without considering the rest of the text in the same columns. Note that if this were true, the total number of possibilities would be greater than 48(24). This may not seem like an obvious thing right now, but it's easily seen in the total, raw data of all the possibilities: http://pastebin.com/hvW1HV97 Here you can see that there are actually 48 unique combinations of the text.

It has been noted that there are possibly on 24 options. This is true when you convert the text from digrams to monograms. The inverses of each other become the same monogram-text (Ex: 123456 - 214365). I haven't seen anyone mention the consequence of this: the polybeius squares are different. As you invert the sequences, you're basically "flipping" the encryption/decryption square. This doesn't actually have a real effect decryption, but I just wanted to point it out so that when people inevitably do crack this cipher, there is less confusion.

Here is the evidence: http://pastebin.com/PkpTgLC3

[Nieno69]

12 minutes ago, Shootinfish said:

@Nieno69

Well spotted and I think its worth testing these combinations first

The only problem I can see is its a 25 letter alphabet and two letters are shared and we don't know if there the double diagraphs in the third row

 

Sorry i dont get the point...

It's always 5x5 grid... ?

...know I get it about the third row... sorry made a mistake 

Will correct it and Edit it in this Post when I have time

But you get 6 letters with this - That's correct?

[Nieno69]

3 hours ago, certainpersonio said:

@Nieno69  So your logic is not sound for this one. The reason is that you are independently moving around the middle portion of the cipher text without considering the rest of the text in the same columns. Note that if this were true, the total number of possibilities would be greater than 48(24). This may not seem like an obvious thing right now, but it's easily seen in the total, raw data of all the possibilities: http://pastebin.com/hvW1HV97 Here you can see that there are actually 48 unique combinations of the text.

It has been noted that there are possibly on 24 options. This is true when you convert the text from digrams to monograms. The inverses of each other become the same monogram-text (Ex: 123456 - 214365). I haven't seen anyone mention the consequence of this: the polybeius squares are different. As you invert the sequences, you're basically "flipping" the encryption/decryption square. This doesn't actually have a real effect decryption, but I just wanted to point it out so that when people inevitably do crack this cipher, there is less confusion.

Here is the evidence: http://pastebin.com/PkpTgLC3

Yeah yeah i get the point 

Because english is Not my First language i will try another attempt and correct what I have done wrong i forget to write down the columns in order but in my head they were in the correct one *facepalm*

 

I will try this which I made:

The cipher is written like this: 123456 (in alphabetical order abcdef)

Iwe know because of the end of the ciphertext that it will Start with 5 or 6 

-56"1234" or 65 "1234"

Then you get "1234"- that are 24 possibilities 

I will Post the Permutations of "fg fg"

 

1ABCD 
=FGFG 
2ABDC 
=FGGF 
3ACBD 
=FFGG  
4ACDB   
=FFGG 
5ADBC   
=FGGF 
6ADCB
=FGFG 

7BACD   
=GFFG
8BADC   
=GFGF
9BCAD   
=GFFG
10BCDA   
=GFGF
11BDAC   
=GGFF
12BDCA
=GGFF

13CABD
=FFGG   
14CADB 
=FFGG 
15CBAD  
=FGFG
16CBDA 
=FGGF  
17CDAB  
=FGFG 
18CDBA
=FGGF

19DABC   
=GFGF
20DACB
=GFFG
21DBAC  
=GGFF
22DBCA
=GGFF
23DCAB  
=GFFG
24=DCBA
GFGF

So in the end decide if it 56**** or 65****

Delete the double ones: position 3+4 both are "FFGG" 

There are 6 (12)different possibilities....

Soooo.... let me know again: is this wrong? 

The same permutation will be with FXFX- there are the same 6 possibilities which lead to at least 6 letters ...

EDIT:

the original cipher is:

FDFDAA

FXFXDX

XFXDGF

The correct order is 

AA ** ** 

DX ** **

GF ** **

Or

AA ** **

XD ** **

FG ** **

Edit2:

Take this as key:

3EFACBD 
4EFACDB   

Its AA FF GG DX FF XX GF

Either way you take the order 3 or 4....Know what I mean? 

Edit: all 24 combinations:

AA FGFG *1* FXFX *2*= 1,6,15,17

AA FGGF *1* FXFX *2*= 2,5,16,18

AA FFGG *1* FF XX *2*= 3,4,13,14

AA GFFG *1* FXFX *2*= 7,9,20,23

 

AA GFGF ** XFXF *2*= 8,10,19,24

AA GGFF *1* XXFF *2*= 11,12,21,22

*1*= XD or DX

*2*= GF or FG

So i must correct the letter numbers - you definitely get 7 letters in a row which should give something...

...I will Edit this further but hope for you thoughts if this is correct (now)

The key for the alphabet is missing but with the correct alphabet you get 6 letters ...

 

 

 

[Jami San]

thank you nieno, the idea that the cipher is mid form, and that THAT is the big clue , is refreshing and very possible. good luck on the reasoned brute force, i hope it works. I am not fully understanding it yet tho, if u guess\try the correct permutation of 1234 to go with your 56 or 65, would that only give us a new column arragement for the origal ciphertext, wouldnt we then still need a polybius ?

great work btw

p.s. thoughts on the motd codex (polybius?) where are we at with this, would have been obvious if it were 5x5 instead of 4x4, cuz 5x5 would be nice neat fit with 5x5 alphbet polybius, but it 4x4...why? is it the key to the missing poly, or key to tranposing 123456 ?

kinda feel THIS is huge clue yet to be sussed out.

[Nieno69]

On 2.3.2016 at 8:15 PM, Jami San said:

thank you nieno, the idea that the cipher is mid form, and that THAT is the big clue , is refreshing and very possible. good luck on the reasoned brute force, i hope it works. I am not fully understanding it yet tho, if u guess\try the correct permutation of 1234 to go with your 56 or 65, would that only give us a new column arragement for the origal ciphertext, wouldnt we then still need a polybius ?

great work btw

p.s. thoughts on the motd codex (polybius?) where are we at with this, would have been obvious if it were 5x5 instead of 4x4, cuz 5x5 would be nice neat fit with 5x5 alphbet polybius, but it 4x4...why? is it the key to the missing poly, or key to tranposing 123456 ?

kinda feel THIS is huge clue yet to be sussed out.

That's the point... the polybius alphabet is the thing we should concentrate now...you could use a normal adfgx decoder:

Write it e.g. like this: 

Copy & paste in your decoder..

I use this: (link) 

In passwort 1 you make the alphabet key

Dont enter anything in Passwort 2(it's for columns)

Dont forget to change to adfgx...(sry for the Black Background i'm on the phone)

https://www.kryptographiespielplatz.de/index.php?aG=902b2f942e08be3c7e2e78cf89d2850c6c2286ba&action=yes

AA FG FG  XD FX FX FG

AA FG GF XD FX XF FG

AA FFGG XD FF XX FG

AA GFFG XD xffx FG

 

AA GF GF XD XF XF FG

AA GG FF XD XX FF FG

[certainpersonio]

I agree with you guys, and that the polybius square is a good potential point of attack for this cipher. It's been on my mind ever since it was discovered that you can't get a legible solution from any of the 48 possible arrangements of the cipher with a  standard alphabet. To me, this signals that there is a unique alphabet that they're using.

If we can assume it's a unique alphabet, I think there are 2 ways they could have made it:

1. Totally random: if this was the case then we're pretty screwed because it leaves brute force as the main avenue for attack.
2. Keyword generated: they took a keyword and used it to create an alphabet. In my research I've found 2 mains ways to set-up an alphabet.
   1. Priming: starting the alphabet with the keyword
      • Ex: (TIGER) TIGERABCDFHJKLMNOPQSUVWXYZ
   2. Mixed: using the keyword to perform a transposition of the alphabet
      • Ex: (TIGER) TAHOVIBKPWGCLQXEDMSYRFNUZ

I personally think the keyword is more likely.

Also, there are multiple ways that the polybius square could be created depending on the route the alphabet uses. However, in all of their other ciphers that use a polybius square (on SoE and DE), Treyarch has used a L-->R, top-->bottom route (like you would read a page of English text).  I think we should assume that this route was used in the ADFGX cipher too. 

I think this ties in nicely with the link @Nieno69 posted to a ADFGX solver. We can use these programs to test out alphabets and keywords more efficiently. 

[Nieno69]

Further:

Row 5 and 6 of the cipher

FD GF FG

FD GF FG

give each transpositions the same three letters ...e.g bye bye

Dont know if it helps someone 

 

[Shootinfish]

I'm unsure if it means anything or not but I was struggling to find keywords that are suitable for both a column transpose and put in a polybuis and the only thing I can think of is trough the only other link I have to this is the loadscreen newspaper with the market crash as trough has an economic meaning https://en.wikipedia.org/wiki/Trough_(economics)

[Shootinfish]

A few of us are working on computing and attacking the 48 keys but were also working on a side project where we have computed every transposition key in the dictionary and have a digest of 41238 cipher texts that the diagraphs can be manually edited to swap for letters to hopefully reveal the cipher incase we find nothing with the 48 keys

This would involve search and replace manually in a text editor and then checking the whole document to see if any patterns emerge and documenting the alphabets you have discounted 

This is far from ideal and we are working on systemising this but its are side project so I wondered if anyone is interested in the data or helping with ideas in how to best sort the data ? 

[Nieno69]

I take a break from that cipher....

I dont know how you try to decipher this cipher - but the only thing really needed is the used alphabet - I will love to try every alphabet that you give me - i tried already all combinations that came in my mind ....if you put all 48 combinations in a row  (without newlines) you can just put the alphabet in a decoder and look at the "plain text" and just add a newline after the 34th letter....

 

[WaterKH]

Alright, so I may have figured out the best way of attacking this monster. I found that for every 48 combinations that we are assuming for this cipher lines 5-6 are identical to each other. Now, I don't mean that we get GF FG DF GF FG DG in every single combination, but rather we get the sequence "A B C A B C" in each combination. I don't believe anyone has pointed this out, so just wanted to make a reply to this and get everyone on board with this. I may have made a mistake in transposing the columns, so if I did, let me know! But I think this is correct, and if we can figure out what letters go here, the rest should unravel from here.

The hard part is not knowing where the word breaks. We could easily have "ringing" be it since I N G I N G follow the pattern above, so it could be one word, 2, 3 or 4.

Anyways, hope this helps!

[Nieno69]

On 23.3.2016 at 9:24 PM, vigiliisgaming said:

Alright, so I may have figured out the best way of attacking this monster. I found that for every 48 combinations that we are assuming for this cipher lines 5-6 are identical to each other. Now, I don't mean that we get GF FG DF GF FG DG in every single combination, but rather we get the sequence "A B C A B C" in each combination. I don't believe anyone has pointed this out, so just wanted to make a reply to this and get everyone on board with this. I may have made a mistake in transposing the columns, so if I did, let me know! But I think this is correct, and if we can figure out what letters go here, the rest should unravel from here.

The hard part is not knowing where the word breaks. We could easily have "ringing" be it since I N G I N G follow the pattern above, so it could be one word, 2, 3 or 4.

Anyways, hope this helps!

I pointed this out already but thank you for trying Figure this cipher out - the columns are not the Problem - the only thing we need to know is the used alphabet... 

I thought about "bye bye" in my post but your Part about" Ing Ing " is more likely 

You could build the alphabet around this and look if it has an algorythm in the poly...

I will try to explain: you have your square with 25 Slots in it

You get the permutations of 

Gf 

Fg 

Df   

And put "i" "n"  "g" in the Slots and build the alphabet around it...

Maybe it's Worth a try

I will look up for it if I have time...

 

 

[WaterKH]

Oh, apologies then for not seeing it! This is the most probably attack for yielding results I believe. We actually are doing what you are saying - Finding a letter for the pair

[Nieno69]

It could look like this:

That are the permutations of dffg with "starting" "FG"

So if the word should truly be "Ing ing" you get "rid off" some permutations.. (No double letter - so DF FG vanishes)

If you Start with GF its vice versa e.g. GF DF doesnt work 

 

 

I tried to find words like singing ringing bringing etc in Google but english is Not my native language so I dont know how to search exactly 

Can someone help out? 

Edit2: could be words with double letters too... if you know some...

[WaterKH]

Currently working on a program with someone that will find and map letters to these based on words. Should be done in the coming week or so. 

[Koslans]

The Giant is in France...... 

EDIT: the loading screen said: Find the PORTALS. that's the word for it.