A new Cipher... or not?

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A new Cipher... or not?
Topic Started: Jun 23 2009, 07:01 PM (201 Views)
Kestrix	Jun 23 2009, 07:01 PM Post #1
Just registered Posts: 9 Group: Members Member #3,271 Joined: June 14, 2009	Okay this is what I'v come up with, not sure if I'v re-invented someone esle's work I hope this makes sence it's the third time I'v tried to explain on paper, the first two attempts left my friends scratching thier heads. Kestrix Attached to this post: My_cipher_2.11.doc (105.5 KB)



jdege	Jun 24 2009, 04:58 PM Post #2
Elite member Posts: 263 Group: Admin Member #246 Joined: December 7, 2006	I'm having a hard time following your explanation. Quote: How it all works: We'll take the first letter W. As it's the first character, it's at box 1. Travel down the plane text characters directly below it until you reach W. Then follow that row to the right until you reach the cipher text which is F. W has now been converted into F. Now we apply the key to F. The first character of the key is F (6) so we count six boxes up to get Q. That's the finale enciphered letter. I run down the first column until I find the plaintext letter 'W'. Fine, so far. Then I run right along that row until I find the ciphertext letter 'F'. Do you mean the key letter 'F'? You don't have any ciphertext, yet, so I don't see how we could use it in the process. In any case, we've found the row in column 1 that contains a 'W', and then found the cell in that row that contains an 'F'. Then we "count six boxes up to get Q". When I count six boxes up from that cell I get a '0'. The 'Q' in that column is 11 cells down, the Q in that row is 11 cells to the right. Now there is a 'Q' six cells up from the 'W' in the first column, but if that had been what you meant, why this talk about running along the row?
	When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl.


Kestrix	Jun 24 2009, 05:32 PM Post #3
Just registered Posts: 9 Group: Members Member #3,271 Joined: June 14, 2009	I'm having trouble getting my idea across I've made some diagrams hope this helps. Kestrix Attached to this post: Explanations.doc (248 KB)



jdege	Jun 24 2009, 07:05 PM Post #4
Elite member Posts: 263 Group: Admin Member #246 Joined: December 7, 2006	Maybe part of the problem is terminology. In an encryption process you start with the plaintext and the key, and the result is the ciphertext. But in your description, you have the ciphertext at the beginning, before you've encrypted a single character. How? Where did it come from? It seems as if what you are referring to as the "ciphertext" is simply a set of mixed alphabets that you are using as part of the key. In normal usage, the ciphertext is "KMY" - the output of the cipher. How are these four alphabets generated? You mention the four sides of the square having different directions of travel, but your example always works along the right side of the square. Under what circumstances are the other sides used?
	When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl.


Kestrix	Jun 24 2009, 07:30 PM Post #5
Just registered Posts: 9 Group: Members Member #3,271 Joined: June 14, 2009	you'll notice the numbers go all the way round, from number 1 to 144. The cipher encrypts in blocks of 144 letters. the table is generated before the cipher is started as long as both the sender and reciver have the same table everythings ok. You start with the key and enter it on the table going all the way round until you get back to the beginning again. The four mixed alphabets are just randomly picked. You can as an aide enter each letter of the plain text into the (yellow) number boxes. The key and plain text now run all the way around the square ( if the message is that long) if it is longer carry on going round the square again. To decrypt start from the last letter apply the oppisit ceaser shift then follow the table back to the plaintext letter... Does this make it clearer? Kestrix Edited by Kestrix, Jun 24 2009, 07:33 PM.



jdege	Jun 24 2009, 08:17 PM Post #6
Elite member Posts: 263 Group: Admin Member #246 Joined: December 7, 2006	So, after I've encrypted the first 36 letters, I start working down the side of the grid, moving left on the rows to find the plaintext letter, moving to the bottom to find its replacement, and then counting to the left by the value of the keyword letter? Also, suppose I'm on the 12th letter, and I'm encrypting the letter 'I'. That puts me at the letter 'H' on the right-size orange column. If my keyword letter is 'M', I need to count forward by 13 places. Do I continue around to the bottom orange row, counting clockwise?
	When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl.


Kestrix	Jun 25 2009, 04:01 PM Post #7
Just registered Posts: 9 Group: Members Member #3,271 Joined: June 14, 2009	"So, after I've encrypted the first 36 letters, I start working down the side of the grid, moving left on the rows to find the plaintext letter, moving to the bottom to find its replacement, and then counting to the left by the value of the keyword letter?" Almost you travel to the right by the value of the key word letter "Also, suppose I'm on the 12th letter, and I'm encrypting the letter 'I'. That puts me at the letter 'H' on the right-size orange column. If my keyword letter is 'M', I need to count forward by 13 places. Do I continue around to the bottom orange row, counting clockwise?" No, when you reach to top of the mixed alphabet go to the bottom and keep counting up.



jdege	Jun 25 2009, 06:22 PM Post #8
Elite member Posts: 263 Group: Admin Member #246 Joined: December 7, 2006	Can you give me an example of how you'd encrypt the 37th letter of the cipher?
	When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl.


Kestrix	Jun 25 2009, 08:48 PM Post #9
Just registered Posts: 9 Group: Members Member #3,271 Joined: June 14, 2009	Assuming you understand now how letters 1 to 36 are encrypted, it my be simpler for you to understand letters 37 to 72 if you envision the table rotating anti-clockwise. Each time you reach the end of the row rotate the table until it has returned to it's original position. That will be the first block of 144 letters. Edited by Kestrix, Jun 25 2009, 08:51 PM.



jdege	Jun 26 2009, 02:00 AM Post #10
Elite member Posts: 263 Group: Admin Member #246 Joined: December 7, 2006	I'd envisioned running the letters around the square, rather than rotating the square and continuing across the top. That's not a real difference, the output is the same. OK, I think I understand it now.
	When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl.


Kestrix	Jun 26 2009, 05:08 AM Post #11
Just registered Posts: 9 Group: Members Member #3,271 Joined: June 14, 2009	I had a friend back when I woirked for Time Computers who wrote this into a program for me but I'v lost it now (was on a floppy disk) and I've lost contact with him. Edited by Kestrix, Jun 26 2009, 05:14 AM.



jdege	Jun 27 2009, 12:26 AM Post #12
Elite member Posts: 263 Group: Admin Member #246 Joined: December 7, 2006	I've always thought that having software to be essential. For one thing, it allows you to create ciphertexts automatically.. As you said, your cipher is a variant on the Vig. In fact, it's a variant on the mixed-alphabet Vig, unusual in that it switches between four different mixed alphabets every 36 letters. Are you aware of the usual attacks on the Vig?
	When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl.


Kestrix	Jun 27 2009, 06:23 AM Post #13
Just registered Posts: 9 Group: Members Member #3,271 Joined: June 14, 2009	I've seen a few tools created by others on the net. I've played with them but they have left me scratching my head. http://cs.colgate.edu/faculty/nevison/Core139Web/tools/index.html But would the usual attacks on the Vig be effective against this variant of the Vig?



jdege	Jun 27 2009, 04:11 PM Post #14
Elite member Posts: 263 Group: Admin Member #246 Joined: December 7, 2006	Yes. It's a matter of how long the ciphertext has to be in order for the statistics to show through clearly enough. For a standard Vig, using ordinary alphabets, the longer the keyword the greater the security. Using a mixed alphabet adds more security, but truthfully not all that much. I remember seeing an estimate that it was equivalent to adding five more letters to the key, so that a mixed-alphabet Vig using a key of seven letters would be about as Secure as an ordinary-alphabet Vig using a key of 12 letters. In your case, you could take the ciphertext encrypted with the first alphabet, attack it as an ordinary mixed-alphabet Vig, and recover the keyword. Then you could subtract the key from the ciphertext encrypted with each of the other three alphabets, and have three ordinary substitution ciphers to solve. The trick is that it is quite hard to crack a mixed-alphabet Vig with only 36 characters unless the key is very short, so your cipher would probably be safe against classical methods for messages of less than 144 characters. Theoretically, there should be a way to pull information from the parts of the text encrypted with the other alphabets, but I can't see one right now. OTOH, dictionary attacks always work, if you use a keyword that's in the dictionary, and there's nothing in your system that would hinder a hill-climber. So it's vulnerable to the usual computer attacks. One thing you might ponder. What happens if the enemy gets his hands on the plaintext of one of your messages? He'd very easily be able to retrieve the keyword and all four alphabets. If you've sent other messages using the same keyword and alphabets, they'd all be compromised. If you'd been changing the keyword daily, but keeping the same alphabets for a longer period, knowing the alphabets allows the enemy to convert messages into an ordinary Vig encrypted by just the keyword, which is easily attacked. The standard, these days, is for a cipher from which the enemy cannot recover the key, even if he has both plaintext and ciphertext. Few of the classical ciphers could offer that protection.
	When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl.


Kestrix	Jun 27 2009, 04:55 PM Post #15
Just registered Posts: 9 Group: Members Member #3,271 Joined: June 14, 2009	So I would have to, as a rule, use a key that is not a word in any dictionary and is longer than 36 characters long. Keys can be a max of 144 characters. Then if the message finishes before the end of the key I just add random characters untill I reach the 144th charater.



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