Clarification

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Clarification
Topic Started: Jul 4 2009, 02:08 AM (191 Views)
Gerry StPierre	Jul 4 2009, 02:08 AM Post #1
Just registered Posts: 14 Group: Members Member #3,269 Joined: June 5, 2009	From the Chaocipher Clearing House, Progress Note #11, Mike Cowan to Moshe Rubin (21 April 2009) Quote: I can see that with 2 cipher wheels the machine will use 626 different alphabets during enciphering. I don't understand this. With 2 cipher wheels, wouldn't there only be 26 different alphabets?



jdege	Jul 5 2009, 02:30 PM Post #2
Elite member Posts: 262 Group: Admin Member #246 Joined: December 7, 2006	With one cipher wheel, there are 26 different positions, so 26 different alphabets. With two wheels that can rotate independently of each other, there are 26^2 positions, so 676 different alphabets.
	When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl.


osric	Jul 5 2009, 06:33 PM Post #3
Member Posts: 46 Group: Members Member #3,254 Joined: March 15, 2009	A single enciphering disk enables 26 different alphabets. If you imagine a disk with a mixed alphabet written around its perimeter and positioned in an annular ring with the normal alphabet inscribed, you have such a disk. It can be represented like this: ring: abcdefghijklmnopqrstuvwxyz disk: FROGSWALTZDUCKJIBVEXNYMPHQ This arrangement is the ‘F’ alphabet. Now if you rotate the disk one place anticlockwise you get: ring: abcdefghijklmnopqrstuvwxyz disk: ROGSWALTZDUCKJIBVEXNYMPHQF which is the ‘R’ alphabet. An enciphering machine with 2 disks can be set up in a number of ways. For example, each disk could have a different mixed alphabet, and be positioned within a ring marked with the normal alphabet: ring 1: abcdefghijklmnopqrstuvwxyz disk 1: FROGSWALTZDUCKJIBVEXNYMPHQ ring 2: abcdefghijklmnopqrstuvwxyz disk 2: WHYJUMPFOLDINGBRACKETSQVXZ To encipher ‘a’, you first look at ‘a’ in the ring of disk 1 and find ‘F’ on the disk. Then you look at ‘f’ on the ring of disk 2 and find ‘M’ on the disk. ‘M’ is the cipher equivalent. You can encipher every letter in this way and get the equivalent alphabet: abcdefghijklmnopqrstuvwxyz MCBPKQWLEZJTYDLOHSUVGXNRFA By moving disk 1 anticlockwise one step at a time, you will get 26 different alphabets. Then if you move disk 2 one step anticlockwise, and again move disk 1 a/c a step at a time you will get 26 more different alphabets. All told you will find 2626=676 different alphabets. Edited by osric*, Jul 5 2009, 06:40 PM.



Gerry StPierre	Jul 7 2009, 02:27 AM Post #4
Just registered Posts: 14 Group: Members Member #3,269 Joined: June 5, 2009	OK my confusion is coming from this statement in Progress Report #7 under the Experimenting with Cipher Disks heading about half-way down the page Quote: From descriptions by John F. Byrne and Henry E. Langen the Chaocipher involves two concentric (?) cipher disks with mixed alphabet components on the rims. And also the How to Construct a Cipher Disk page. Looking at the example completed cipher disk on the 'How To' page, you can see these two concentric alphabets outer: EINZVSPJFLUORXQTAKBHWMDGYC inner: WLCZFREDOHXMQYTVNJKSBPUIAG I had assumed that one of these two disks would be the plaintext, but that is not the case? Instead, a 'ring' of plaintext is also used? abcdefghijklmnopqrstuvwxyz EINZVSPJFLUORXQTAKBHWMDGYC WLCZFREDOHXMQYTVNJKSBPUIAG So in this example the current setting would be the 'EW' alphabet? The word 'too' would be encrypted as 'DNN'. If the outer disk is moved one position clockwise and the inner disk is moved one position anticlockwise after each letter then 'too' would encrypt as 'DAS'? Starting position 'EW' alphabet, t>H>D abcdefghijklmnopqrstuvwxyz EINZVSPJFLUORXQTAKBHWMDGYC abcdefghijklmnopqrstuvwxyz WLCZFREDOHXMQYTVNJKSBPUIAG Position after one move, 'CL' alphabet, o>X>A abcdefghijklmnopqrstuvwxyz CEINZVSPJFLUORXQTAKBHWMDGY abcdefghijklmnopqrstuvwxyz LCZFREDOHXMQYTVNJKSBPUIAGW Position after two moves, 'YC' alphabet, o>R>S abcdefghijklmnopqrstuvwxyz YCEINZVSPJFLUORXQTAKBHWMDG abcdefghijklmnopqrstuvwxyz CZFREDOHXMQYTVNJKSBPUIAGWL I realize that the whole mystery is what type of device the chaocipher is. But is the working assumption that there is a third fixed alphabet that would be the starting plaintext? By the way, I love the term 'anticlockwise'



osric	Jul 7 2009, 09:02 AM Post #5
Member Posts: 46 Group: Members Member #3,254 Joined: March 15, 2009	Quote: I realize that the whole mystery is what type of device the chaocipher is. But is the working assumption that there is a third fixed alphabet that would be the starting plaintext? I don't think there is any 'working assumption' ! I gave you one possibility to demonstrate how 676 enciphering alphabets can be formed using 2 disks but, as I pointed out, this is just one of many configurations -- including concentric disks. Jeff Hill describes other possible machines in his papers at The Chaocipher Clearing House, and these are well worth reading. Regarding working assumptions, I assumed that there are only two disks because of what Langan reported, but Jeff has pointed out that Langan may not have fully understood the blueprints shown him by Byrne, and thus there may be a third disk. The only thing we know for certain is that nobody has yet found the correct machine. So you are absolutely right that the whole thing is a mystery, and I would add that we are all waiting for someone to have a flash of inspiration. Perhaps this Forum will stimulate that moment, and maybe you will be the man of the moment. Good Luck!



jdege	Jul 7 2009, 01:02 PM Post #6
Elite member Posts: 262 Group: Admin Member #246 Joined: December 7, 2006	Gerry StPierre Jul 7 2009, 02:27 AM I had assumed that one of these two disks would be the plaintext, but that is not the case? Instead, a 'ring' of plaintext is also used? No, you're right. One ring of the disk is plaintext, the other is ciphertext. But the disk has 26 different alignments between the plaintext and the ciphertext, resulting in 26 different mappings between plaintext and ciphertext. In cryptography, an "alphabet" isn't a sequence of letters, it's a mapping of plaintext letters to ciphertext letters. {a->B, b->c, c->d, ... z->A} is an alphabet, and {a->C, b->D, c->E, ... z->B} is another, distinct alphabet.
	When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl.


mosher	Jul 8 2009, 07:30 AM Post #7
Member Posts: 38 Group: Moderators Member #3,268 Joined: May 26, 2009	Gerry, Thanks for pointing out the "626' typo. I've corrected it on the Chaocipher Clearing House web site. Moshe



Gerry StPierre	Jul 9 2009, 03:29 AM Post #8
Just registered Posts: 14 Group: Members Member #3,269 Joined: June 5, 2009	I have read through all the info that mosher has gathered. Several times now. And for the most part it is understandable. At least when I read it, which is why I keep reading it! Hmmm, I guess what I am trying to get at - if there are only two rings of text then there are 26 alphabets. However, since no one can be sure that the device has been described correctly there could in fact be a third ring. If there was a third ring then there could be 676 alphabets. So what I was looking to clarify - at this point there is no certainty as to whether there are 26 or 676 alphabets. While the analysis that has been done on the available texts indicates certain attributes, it does not indicate whether in fact there are two or three rings. Correct? Finally, is the challenge that seems more solvable (since it came from Crpytologia editors) is deciphering exhibit 5?



osric	Jul 9 2009, 08:20 AM Post #9
Member Posts: 46 Group: Members Member #3,254 Joined: March 15, 2009	Quote: Finally, is the challenge that seems more solvable (since it came from Crpytologia editors) is deciphering exhibit 5? Yes, I think so. As Moshe has explained elsewhere in this Forum, the fact that these 3 messages are 'in depth' should help us. If we could marry up the 3 ciphertexts to their correct plaintexts, we have a chance of working out the keys and how the machine works. By the 'keys' I mean those alphabets (presumably they are mixed alphabets) and the disk movements after each encipherment. Some of us (Greg Mellen, Jeff Hill, Moshe and myself) have tried to build a model of the Byrne machine that will replicate Exhibit 1 ciphertext from the plaintext but these attempts have not succeeded. Jeff and Moshe especially have also spent time trying to find the plaintext for the 3 ciphertexts from Deavours and Kruh. So far that has also been a dead end, but -- provided there are no mistakes -- this is probably a more promising attack because there is the extra 'information' from arising from the depth. Your intuitive gut-feel that the plaintexts may be quotations is the sort of new approach that is very useful -- and prompts me to have another look! Well done.



aloos	Oct 28 2009, 02:12 PM Post #10
Just registered Posts: 2 Group: Members Member #3,291 Joined: October 12, 2009	Hello all, I find Osrics idea about how two disks are moved by some pattern of steps along with the position in text intriguing. Mathematically, one could say that the idea is the following: We have two (different?) permuations of the alphabet, realized by matrices P_1 and P_2, and a matrix S[ i ] shifting the alphabet cyclic by i steps. Moreover, we have one function f(i) : i -> i mod 26 and one function g(i): i -> something in 1...26, following some cycling pattern. The encryption is then done by the permutation corresponding to the matrix product S[ f(i) ] * P_2 * S[ g(i) ] * P_1 So, the final permutation depends only on the actual position i of the letter that has to be encrypted. As far as I know, this is somewhat similar to the Enigma-encryption. However, Osrics idea does not take into account something Deavours and Kruh are heavily emphasizing in their article: ``He [John Byrne Jr.] said that perfect accuracy was essential when enciphering a message because one error would distort the rest of the message.'' (Byrne, Deavours, Kruh, ``Chaocipher enters the coumputer age when its method is disclosed to cryptologia editors.'') ``And, an important shortcoming of the cipher -- make one error during encipherment and the message is garbled beyond repair -- is eliminated by its implementation on a computer.'' (Ibid.) Therefore, I think that the encryption depends not only on the position as Osric thought, but (also?) on preceding (cipher or plain) letters. The problem is: Simply replacing one of the functions f(i) or g(i) by a function depending on -- say -- the preceding cipher letter would mean that cycling would emerge pretty fast (as long as we don't assume to have a long instruction about how to move the other disk forward or backward, which would contradict to the fact that the system is easy). Nevertheless, I tested this idea assuming that the x_{i-1} (the preceding plaintext letter) or y_{i-1} (the preceding ciphertext letter) play a role when encoding letter x_i. No surprise: I could not find any cycle. Any ideas how we could modify the model? andreas



osric	Oct 31 2009, 07:12 PM Post #11
Member Posts: 46 Group: Members Member #3,254 Joined: March 15, 2009	Quote: Therefore, I think that the encryption depends not only on the position as Osric thought, but (also?) on preceding (cipher or plain) letters. Hi there aloos (& anyone else following this thread), I am sorry not to have responded earlier to your comment quoted above. You may be absolutely correct that preceding plain or ciphertext is used in some way to determine the movement of the enciphering disks. This 'autokey' principle, as many have called it, is quite popular in many quarters -- including with mosher However it is also possible that disk movement is controlled by some other means. Perhaps a textual key, perhaps a pseudo-Fibonacci series... Even in these cases the comments of Byrne jr and of Deavoure and Kruh, concerning the need for absolute accuracy, are highly relevant. For example say a textual key calls for moves of v,w,x,y,z... to encipher a certain 5 letters, and the operator mistakenly moves v,w+1,x,y,z then the decipherer -- following the correct sequence v,w,x,y,z -- will get a garbled decrypt from the 2nd letter onwards. The need for perfect accuracy is present, I believe, independant of how the movement stream is generated, so long as the disk movement is continuous -- and the disk (or disks) are not returned to an initial position after each letter is enciphered, which I have not read that anyone is proposing. There is a problem with using plaintext as a determinant of the movement key in that it can cause repeats in ciphertext for the first 5500 letters of Exhibit 1, which comprise a 55-letter plaintext message repeated 100 times, that are not present in Chaocipher. In this regard the latest paper from Jeff Hill is interesting (progress report #14 at TCCH). He proposes a 2-commutator switching system, activated by the plaintext, and claims it has such a long cycle that it will not cause cipher repeats in Exhibit 1. On my list of things to do is to have a good look at Jeff's system, by building a computer model, enciphering Exhibit 1 and examining the output. I am sure there will be many other people who also will be very interested in Jeff's latest ideas.



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