SRL Archive - Mathematics

xxx

Yeah, well that is a hard one. You would need the keyword to simply decrypt it. Where have you got this?

Could this help? Sandstorm made a Encrytper/decrypter. And a challenge…

http://www.villavu.com/forum/showthread.php?t=51400&highlight=encrypter

Since you have the cipher it should be easy, I'll give it a shot.

end 175.333333333

Interesting…

You're on your own for this one :-P

175.333333333 ??

How did you get that number ?.

Actually I think you're right :X.. I've got an Idea already, but the solution should be somewhat simpler than a bruteforce attempt. But i know that the key-word is "short".. so i may give it a shot…..

You got any idea what it should output?

From: Frement
You got any idea what it should output?

It should be some german text.

We would need the key-word, else it would be almost impossible to decode.

From: Harry
We would need the key-word, else it would be almost impossible to decode.

no its not ;).

ADFGVX was cryptanalysed by French Army Lieutenant Georges Painvin. The work was exceptionally difficult by the standards of classical cryptography, and Painvin became physically ill during it. His method of solution relied on finding messages with stereotyped beginnings, which would fractionate the same, then form similar patterns in the positions in the ciphertext that had corresponded to column headings in the transposition table. (Considerable statistical analysis was required after this step had been reached — all done by hand.)

This meant it was only effective during times of very high traffic — but, fortunately for the cryptanalysts, that was also when the most important messages were sent.

Painvin broke the ADFGX cipher in April 1918, a few weeks after the Germans launched their Spring Offensive. As a direct result, the French army discovered where Ludendorff intended to attack. The French concentrated their forces at that point and stopped the Spring Offensive.

The ADFGX and ADFGVX ciphers are now regarded as insecure for any purpose.

Note: the claim that Painvin's breaking of the ADFGX cipher stopped the German Spring Offensive of 1918, while frequently made,[1] is disputed by some. In his 2002 review of Sophie de Lastours' book on this subject, La France gagne la guerre des codes secrets 1914-1918, in the Journal of Intelligence History, (Journal of Intelligence History: volume 2, Number 2, Winter 2002) Hilmar-Detlef Brückner states:

Regrettably, Sophie de Lastours subscribes to the traditional French view that the solving of a German ADFGVX-telegram by Painvin at the beginning of June 1918 was decisive for the Allied victory in the First World War because it gave timely warning of a forthcoming German offensive meant to reach Paris and to inflict a critical defeat on the Allies. However, it has been known for many years, that the German Gneisenau attack of 11 June was staged to induce the French High Command to rush in reserves from the area up north, where the Germans intended to attack later on. To achieve this, its aim had to be grossly exaggerated. This the German High Command did by spreading rumors that the attack was heading for Paris and beyond; disinformation proved effective then - and apparently still does. But the German offensive was not successful because the French had a sufficient number of reserves at hand to stop the assault and did not need to bring in additional reinforcements. Moreover, it is usually overlooked that the basic version of the ADFGVX cipher had been particularly created for the German spring offensive in 1918, meant to deal the Allies a devastating blow. It was hoped that the cipher ADFGX would protect German communications against Allied cryptographers during the assault and this is what it indeed did. Telegrams in ADFGX appeared for the first time on 5 March, the German attack started on 21 March. When Painvin presented his first solution of the code on 5 April, the German offensive had already petered out.

Well i could make a program that gous through the whole shit, but for XXXXXX keyword it would take like 3 hours to run.

You would need a programme that could "guess" a Polybios-Square.

I'm currently working on one. Anyone else wann give it a shot :) ?!

Could you tell where you have got this and what you get from decrypting it?

just decrypt and see! ;)

Umm…does this have anything to do with maths? And someone want to explain what 'decrypt' means?

From: Santa_Clause
Umm…does this have anything to do with maths? And someone want to explain what 'decrypt' means?

dycrypt is pretty much a puzzel with words.

the random letters are all patterns, and if you crack the pattern, you can find out what the message says :)

lets say:

c = g
a = z
t = k

if i had an encrypted message called gzk, and you knew what the pattern was, then you could easily say that the message says cat, although anyone else wouldnt know what it means

Encrypted text: SGHIHDGJLHRJHGJDS (example)
Clear text: Buy me apples.

To turn the encrypted text into text that you can read, you have to decrypt the text.

Really this takes ages to decrypt.

From: Zyt3x
Encrypted text: SGHIHDGJLHRJHGJDS (example)
Clear text: Buy me apples.

To turn the encrypted text into text that you can read, you have to decrypt the text.

From: Zyt3x

To turn the encrypted text into text that you can read, you have to decrypt the text.

No shit?!

Back to topic please :)..

I'm working on a programme to Brute-Force ADFGX To gain valid Polybios-Squares. When it's done you'll get the sauce : D. I really want to decipher this :).

~caused

Based on the wiki article, I've broken it down to what should be its original string, now only requiring you to bruteforce the alphabet it uses and compare the output to common German words. It will still take you forever though, so I would suggest looking at the number of different pairings - it should have 36, but if there's only 26 then it's likely the numbers have been omitted, which should cut down on possibilities a lot. Logic would also state that the vowels would probably be the most common, so if you analyzed it for the most common pairings and assumed those were vowels, you could make it more easy to detect common words.

XG XA DX GF AD VX GF VA XX VA VV VF FG DG DF XD FF AF FD AG FD VF AA AF FG VF VD AA FD DX DX
FG AX DX DF VA FF FD VX XG GG XD DF GX XG VD AA XG XD GX VF XD DD XG XX GD VV DX VX DA VG VG
AA VD AD AX DD VD VX VF XX AF AX AA DA AG VX XV DF DD AG VX DA DG DX AX FG XV XX DF GX FD FD
AA DD AX FD XF DF GX AX XG AX XD VD AD FD DG VA VF AX DX DD FD DX VD AA DG FD GA AD DG VD VD
DV VX DF DX VD AX AG GG XF DD VD DD FX GD AD GV AF VF GF VX AF XA DG GF FD DV FF DA DD AF AF
DV AD FX FA AD AA GX XG VX DX GG GX XD VX DD XA DA AA XF AV DD AF DD DA FD FA XA XX VF VF VF
AF AG DV AX DA FD DA AF AD XA DA DG XA XD AX AX XA DV DA XA AX XV AG VD DX FA GX DF FX DV DV
GG DD XG AA AV DD AX XD FA FA FA GV VD DX GF GF FF DX XX AV FG FF XA DX AV AG FF AA VA AX AX
AD AF VA GF XX FG VF VX XG AA AD VG GX DD DG XF DX DD DD DF GX GV DG DX FA GV DV DX FD XV XV
AD FD DX AG XV XX AX FV DD VD FA FX VF AX DX XA AG VA FX AD AX VX AD XA AX AA XD XD AV XD XD
GX GX DG AV DF FD VX DD XX AX VD DA GX FD AV GA AD XF FA AD AD DX VA XD FX FF DX FD GG XA XA
FV XF AX GF AF XF DF DG XA XA DX XD XD DV FF FX FD FV AD DF VA AV AD FF FG FV DX XA FA AG AG
XF FA VA XV DD XX GA VX GD FA AA VF DA XG VF DD DF DA AD AF XG AD GA FG XF AD AF GA FX AX AX
VD XX XD GF DV GX XX AA VD DV AX AA DA DA XF VD DX GX XF GG DV FF GD DD XD XG GA GG AF XA XA
DF VD DD GX DF VX AF GV AX GF AF AX XD VF AD FF XD DX DA DG XF VG AD AD DX DA DX AD XG AX AX
XD DG AX VD DD XD DA VX AG DD AD AX GG DV DG FX VV XG XD VD DD AD XX GF XV AA GF VA VX GV GV
XF XX FA FD XD DA GA GD AG VX GG FA AD DX AG XD XD VA AV XV GA DA DA AV FD VG VV XF XD AA AA

Oh, you got the Columnar key ?.. What is it ?

And more imporatnly.. How did you get it :O ?

Well, as you said you had "short" as the key, I assumed that was the transposition key. Using that and a bit of voodoo magic, I worked backwards to get the original, encrypted key. I've got a Scar script that does it, but it's just a bit of backwards working based on what the wiki article says - you still have to do the hard bit and produce the encryption alphabet.

From: mixster
Well, as you said you had "short" as the key, I assumed that was the transposition key. Using that and a bit of voodoo magic, I worked backwards to get the original, encrypted key. I've got a Scar script that does it, but it's just a bit of backwards working based on what the wiki article says - you still have to do the hard bit and produce the encryption alphabet.

No, you got me wrong, i dont have the transposition key.

I do only know that the length is "short". Which would probably mean, under 10 letters.

It's a really tough one. But i know that its doable. And i just found an interesting forum topic about it (http://s13.zetaboards.com/Crypto/topic/123878/1/).

~caused

If you don't know how long the transposition key is, then not only do you have to work out which of the possible squares it could be, you also have to go through all the possible transposition keys. The square itself is a 6x6 grid, so that's just about an unfeasible number as it is and so your only bet would be quite advanced compared to bruteforcing by combining the bruteforce with word recognition to find combinations that produce large amounts of legible words - cryptanalysis of some sorts.

From: mixster
If you don't know how long the transposition key is, then not only do you have to work out which of the possible squares it could be, you also have to go through all the possible transposition keys. The square itself is a 6x6 grid, so that's just about an unfeasible number as it is and so your only bet would be quite advanced compared to bruteforcing by combining the bruteforce with word recognition to find combinations that produce large amounts of legible words - cryptanalysis of some sorts.

And the program takes forever to run. (Unless it finds the key fast. :))

Well, there isn't actually any key. It's rearranged, but not by a key - by a pattern. I'm guessing they just use a key because it's easier to remember "cargo" than "21453". The square itself is also not a key since it's more an encryption using an unknown method. A bit like converting integers to hexadecimal, but changing what every character is worth, ie A is 4, 9 is 7, F is 0.

Thanks for your efforts mixster. I think I found a solution. But its late over here, gotta read thorugh it tommorow. It's here: http://dsns.csie.nctu.edu.tw/research/crypto/HTML/PDF/C84/339.PDF