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| Sometimes Things Are Easier Than They Look | |
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| Topic Started: May 24 2008, 01:11 AM (510 Views) | |
| jdege | May 24 2008, 01:11 AM Post #1 |
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In the absence of any new challenges here, I've been looking at the problems in The Cryptogram that I've never tried before. The ACA has dozens of cipher types, and there are only a few of them I'm really familiar with. One of the low-numbered ones, (which means easy), was a checkerboard. Didn't have a clue what a checkerboard was, so I looked it up. There are two forms. In each, you use a keyword to fill a 5x5 Polybius square. In the simple form, you use two five letter keywords as coordinates, replacing each letter with its coordinate values: In the more complicated, two keywords are used for each coordinate, and letters from either can be used for coordinates, giving you four alternatives for each plaintext letter: This seems to be done both preserving and not preserving word separations. Because this was a low-numbered problem, it was likely the simpler form. And, in this case, word separations were preserved. So, what do we have, here? A simple substitution cipher. Replace each pair of letters with any single arbitrary letter, and you have an Aristocrat. Pattern word search cracked it with ease. In the more complicated form, you have a homophonic cipher. with four ciphertext alternatives for each plaintext letter. The general techniques for solving homophones are statistical, and these puzzles are way too short for them to be of any use. So the best attack I can envision is trying to reconstruct the square. But for the simple form, I'd hardly expect it to be necessary. |
| When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl. | |
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| jdege | May 25 2008, 02:27 PM Post #2 |
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Another one that turned out to be not very difficult - monome/dinome. In this, you need a permutation of the alphabet and a permutation of the digits. Use a keyword to shuffle the alphabet. Write out the keyword, followed by the rest of the alphabet, leaving out J and Z, in three rows of eight columns each. Use a keyword to shuffle the digits. Write the first two digits to the left of the second and third rows the text, write the other eight over the top of the columns. Plaintext letters in the first row are replaced by the digit above, letters in the second and third row are replaced by the digit to the left followed by the digit above. This is clearly in the same family as Donald's Spread Cipher. The trick is to determine which are the two row digits. Once these are identified, you can convert this into a Pat. In the above case, replace 70 with A, 71 with B, 72 with C, 74 with D, etc. Skip 73, because 3 is your other row digit. Then replace 30 with J, 31 with K, etc. And when you're done with that, replace each remaining digit with a letter. Now you have a simple substitution cipher. The question is how to identify the row digits. Two things we know - they should be fairly frequent, and they should never appear in pairs or together. Frequent because they are each involved in a third of the letters in the alphabet, never in pairs because they don't appear as column digits. So start by identifying the paired digits. 11, 33, 66. Eliminate these. They cannot be the row digits. Then do a frequency count of the digits. Look at the high-frequency digits that never appear paired. Check each possible pair to see if they ever appear adjacent to each other. If they do, they can't be the row digits. The highest frequency digits that never appear doubled and never appear adjacent to each other are very likely the row digits. Use them to convert the ciphertext to a Pat, and see if the result looks like a monoalphabetic substitution. Worst case there are only a hundred possibilities. In the problem I was trying to solve, there were four digits that appeared doubled, which cuts it down to 36 possibles. Look at them in order of descending frequency. In my case, the highest-frequency pair of remaining letters never appeared adjacent to each other, so I did a test conversion to letters. The result contained a match for the crib, so I attacked it, and found the solution. (If you don't have a crib, you should be able to judge whether a test conversion is correct using the IC.) What's fascinating to me is how many of these ciphers can be solved by converting them to Pats. |
| When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl. | |
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| Revelation | May 29 2008, 07:35 PM Post #3 |
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I've become a member of the ACA too 
This is indeed a major weakness! |
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RRRREJMEEEEEPVKLWENFNVJKEEEEEAOLKAFKLXCFZAASDJXZTTTTTTTLSIOWJXMOKLAFJNNKFNXN RAGRBAQEMHIGDJVDSEOXVIYCELFHWLELJFIENXLRATALSJFSLCYTKLASJDKMHGOVOKAJDNMNUITN RRRRLJVEEEEECLYVYHNVPFTAEEEEEMWLMEIRNGLARWJAKJDFLWNTIERJMIPQWOTZEOCXKNUBNXCN RJIRPOWEANFUSNCZVDVZNMSFEKLOEPZLDKDJWSAAAAAAAOERHJCTNCKFRIMVKSOFOMKMANREWNBN RZUDRGXEEEEENFQIDVLQNCKNEEEEEDGLLLLLLAWIOSNCDARLODMTOEJXMILDFJROTKJSDNLVCZNN | |
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| jdege | May 29 2008, 10:40 PM Post #4 |
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The weakness is that the row digits never appear as column digits. The classical Straddling Checkerboard does not have this weakness. I've just started reading through Friedman & Callimahos' "Military Cryptanalytics." I noticed it had a chapter on monome-dinome ciphers, and discussed quite a few, including the Straddling Checkerboard. None of the monome-dinome variants discussed had this weakness. Given the general level of expertise of the folks in the ACA who standardized on these ciphers, I can only presume that this weakness was intentional - to give novices who are just moving beyond the Aristos a place to start. Much like the ACA's homophonic cipher - far easier to crack than a generic homophone.
Welcome. |
| When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl. | |
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| jdege | Jun 9 2008, 06:27 PM Post #5 |
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I was trying to explain this to someone, and noticed something else, when trying to pick out the row digits. We know that the row digits will never appear doubled, and that they'll never appear adjacent to each other. There are 10*9 = 90 possibilities. In the cryptogram I was trying to solve, there were four digits that appeared doubled, leaving 6 * 5 = 30 possibilities. If you create a grid: , then for every digit that appears doubled, put an X in all the cells for the row and the column for that digit (1,3,4,5): , then for every pair of digits, fill in the cell for that pair (ignoring order, either 4+5 or 5+4 in the cipertext will X the cell for 4+5: . In my case, there were only five unchecked cells left. Odds are that of these cells, it'll be the pair with the highest frequency that is correct, as it was in my case. But when it's not, there won't be many others you need to look at. |
| When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl. | |
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| jdege | Jun 24 2008, 04:17 PM Post #6 |
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Yet again, I've gotten my statistics wrong. We're talking sets - order doesn't matter. 10 choose 2 = 10!/(2!*8!) = (10*9)/(2*1) = 45 6 choose 2 = 6!/(2!*4!) = (6*5)/(2*1) = 15 |
| When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl. | |
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6:18 AM Nov 8
