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| Elijah 3 Cipher; challenge included | |
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| Topic Started: Jun 25 2010, 08:36 PM (1,583 Views) | |
| Elijah Cross | Jun 25 2010, 08:36 PM Post #1 |
Advanced Member
  ![]](http://z2.ifrm.com/static/1/pip_r.png)
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Here's the method for a new cipher I've been using: We use a keyword in combination with a keyed alphabet - in this example, our keyword is "luigi" and the alphabet begins with "mario": MARIOBCDEFGHJKLNPQSTUVWXYZ Next, we choose our plaintext, and above it, we write our repeated key:
Here's the method of encryption: find the distance from the keyword letter to the plaintext letter in the keyed alphabet, "wrapping around" the alphabet should we get to the end. in our example, we must count 16 letters from L to O. We then take the 16th letter in our keyed alphabet, and use it as the ciphertext - in this case, that letter is N. The encrypted text in our example will read "NZY BYLURXLI EL SH JUMEDTD RPLOTO" To decrypt a message, simply write the keyword over the ciphertext, find the keyword letter in the keyed alphabet, and move forward the distance represented by the ciphertext letter. In our example, starting at L and moving forward an N (16 spaces) will give us O, U plus Z gives us U, etc. It saves time when doing this by hand to number the letters in the keyed alphabet. As a variant, we could encrypt the plaintext by writing the key under it instead of above. Of course, if we already have the ciphertext using the first method, we can easily determine the ciphertext for the variant - the N in our example would then be an F, as their corresponding numbers total 26. Decryption would begin with starting at the keyword letter and the moving backwards through the alphabet. As yet another variant, we can write the keyword above the plaintext, start in the alphabet at the key letter, then move forward the number represented by the plaintext letter - in our example, the first letter of ciphertext would then be T. Decryption would mean starting in the alphabet at the ciphertext letter, then moving backwards the distance represented by the keyword letter. I have used the main method to create the following ciphertext - the goal, of course, is to determine the keyword and keyed alphabet.
Edited by Elijah Cross, Jun 27 2010, 07:51 PM.
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| fiziwig | Jun 25 2010, 10:23 PM Post #2 |
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Elite member
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Instead of counting letters, why not use a "slide rule". Line up 'A' on the slide with the first letter, and then read your distance off the sliding (bottom) scale just below the second letter. Instead of using numbers like '16' the slide rule returns a letter (e.g. 16 -> P) which you then just encrypt using your keyed alphabet, which is now just a pt/ct alphabet. Same effect, just easier mechanically. Slide rule set to "compute" M-T = H (H is equal to 8, but you don't really need to know the number if you key your ciphertext alphabet with letters instead of numbers.) |
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| Elijah Cross | Jun 26 2010, 12:33 AM Post #3 |
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The way I'm doing it, the distance from M to T in your example wouldn't be H, but G. Of course, I could still use the slide, just moving it over one space.
Edited by Elijah Cross, Jun 26 2010, 12:56 AM.
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| fiziwig | Jun 26 2010, 01:43 AM Post #4 |
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Elite member
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Ah, I see. On some of my ciphers I went base 26 with A=0, B=1, ... and on some of them I used base 27 with some imaginary null character Ø = 0, A=1, B=2, ... You could use the slot to the left of the 'A' for a pointer (the imaginary null character). Either way works, as long as you're consistent. I hope to get some time tomorrow to look at your challenge and give it a try, provided the grass doesn't need mowing.  It looks like fun.  I look forward to tackling it.--gary |
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| james | Jun 26 2010, 10:21 AM Post #5 |
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Super member
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Hey Elijah! I reckon you have a typo in your example cipher. I think it should read: NZY BYLURXLI EL SH JUMEDTD RPLOTO and not: NZY BYLURXLI EL SH JUMEDBD RPLOTO The key letter is 'L' and the plain letter is 'E'. From key 'L' to plain 'E' is 20 places in the mixed alphabet, which means the cipher letter is at position 20: that letter is a 'T', not a 'B'. Do I get an extra point for this?
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| fiziwig | Jun 26 2010, 03:15 PM Post #6 |
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Elite member
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Good morning, Well, here goes. I'm going to try to crack your challenge. It is 8:12 AM (Pacific) To start with, coincidence testing tells me your key is 7 letters long, so I'm going to re-write you message in blocks of 7: ON EDIT: My coincidence testing program (That I wrote in BASIC in the 1980's) had a grievous bug in it, so this conclusion was based on faulty programming. See post below. Edited by fiziwig, Jun 26 2010, 11:16 PM.
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| Elijah Cross | Jun 26 2010, 07:09 PM Post #7 |
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Advanced Member
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Good eye, sir - fortunately, an error like that won't make it impossible to figure out the surrounding text, but I am a perfectionist. Thanks! |
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| fiziwig | Jun 26 2010, 11:14 PM Post #8 |
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Elite member
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Scratch that previous post. I decided to take my old BASIC program from the 80's for coincidence testing and update it to PHP so I could put it on line. In the process I found a grievous bug in the old BASIC code, so my conclusion that your key word length is 7 was based on a faulty program. I'm back to square one while I fix my programming bug. |
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| fiziwig | Jun 27 2010, 04:37 AM Post #9 |
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After spending a number of hours on the challenge I believe the cipher can be cracked, but I don't think the message is long enough to provide enough data to crack it. The message is 232 characters, so if the key is 6 characters long, for example, that means there are only 38 examples of each ciphertext alphabet, which means we see each pt/ct combo an average of 38/26 = 1.5 times in the whole message. That means there is just not enough material to get a grip on. For example, suppose that if, in my feverd imagination, I imagined that I could fit a phrase to the ciphertext, say "has only been done using". I then copy the substitutions from that phrase into the rest of the cipher to see it it reveals anything else. Only the problem is the letters that lets me plug in are so sparse, and so few that there's no way to know if that phrase is actually part of the message or just something I made up because I imagined I saw a pattern that wasn't really there. I think it would either take quite a bit more ciphertext to work with before I could make anything out of it, OR maybe leaving the proper word divisions in the cipher like you did in your example. That might, with some luck, provide enough leverage to pry it open with so little text. Or maybe not. |
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| Elijah Cross | Jun 27 2010, 05:11 AM Post #10 |
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I'll create a much longer ciphertext and throw it up in place of the original challenge, perhaps tonight. I'll use the same keys. EDIT - The new ciphertext is in the first post. Edited by Elijah Cross, Jun 27 2010, 08:46 AM.
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| james | Jun 27 2010, 01:19 PM Post #11 |
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Super member
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Elijah, Thanks, with the longer text I've got the period no sweat. BUT THERE'S AN ERROR IN YOUR ENCIPHERING SYSTEM  You say that with the keyed alphabet: MARIOBCDEFGHJKLNPQSTUVWXYZ, numbering the letters from 1 to 26
But if the key letter and the plain letter are the same, then the distance apart is zero, and there is no zero letter. Edited by james, Jun 27 2010, 05:24 PM.
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| Elijah Cross | Jun 27 2010, 05:31 PM Post #12 |
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If the key and plain letters are the same, the distance is 26 - that's the number of spaces we'd move forward to get to the letter. EDIT - In my sleep deprived state last night, I made an error in moving the new challenge text to the thread. The first part was omitted, and one chunk was just wrong - I may have used the wrong key in my book for that section. My apologies - I have fixed the challenge text, and hopefully, you haven't gone so far in analysis that starting anew is a pain. |
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| fiziwig | Jun 27 2010, 06:57 PM Post #13 |
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Elite member
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I was going to ask if the repeated sequence: AWXW OQWC MACE PYFT XBMX SZTK SDAD SKTU GVAN AVZQ WKDA YOQD UAPL PTGY OTEU ENNJ ABRY AWXW OQWC MACE PYFT XBMX SZTK SDAD SKTU GVAN AVZQ WKDA YOQD UAPL PTGY OTEU ENNJ ABRY was an error, but I see it is not present in the corrected text. Thanks for fixing that. |
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| Elijah Cross | Jun 27 2010, 07:53 PM Post #14 |
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Fixed another minor error, one I caught before but forgot to change - The first block of four in the eleventh row now reads TOJC. OK, it's perfect now! Edited by Elijah Cross, Jun 27 2010, 07:53 PM.
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| james | Jun 29 2010, 05:19 PM Post #15 |
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Super member
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A few thoughts, as a cock-shy for others to critique. There's enough ciphertext to find the period. IC suggests two possibles, Kasiski shows period = 17. Now to find a 17 letter key, and a mixed alphabet with a keyword of unknown length. If the alphabet were normal, solving would be a doddle as in a Vigenere. Take each period, measure the cipher letter frequencies, cross-correlation with normal English to get key letter. No dice with unknown mixed alphabet. Every position in the period has a different mixed alphabet, so cross correlation is of no use. Trying to guess 'e' and 't' from the most frequent cipher letter in each period gets nowhere. In a plaintext of similar length to Elijah's ciphertext, chosen at random, and divided into periods of 17 letters, 'e' was the most frequent plaintext letter in just half of the 17 divisions. Also 'e' and 't' were the top two most frequent in just 15% of divisions. No basis for beginning to fill in the plaintext. A dictionary attack: try every 17-letter word in the dictionary and in a phrase list as the key, and every dictionary word as the basis of the mixed alphabet. Success depends on the 17-letter key used by Elijah being in the dictionary or the phrase list. If he has used an uncommon phrase, like "GETHOTCIPHERMUSIC", then there's no hope for this method. A list of 568 17-letter words as the key and a (short) 60,000 word dictionary for the mixed alphabet failed. Using 10,000 popular phrases did no better. Outlook: doubtful. Maybe there's a stochastic method? Perhaps a genetic algorithm to search for the key and hillclimbing every putative key to find the mixed alphabet. A long, drawn-out affair requiring weeks of computer time. Elijah's music is no doubt excellent, and 5% of his profits will buy a few drinks, but personally doubt that solving this cipher is worth the sweat required. |
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