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| A New Challenge | |
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| Topic Started: Jun 23 2007, 11:49 PM (385 Views) | |
| tidmiste | Jun 23 2007, 11:49 PM Post #1 |
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Just registered
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I'm kind of new to this particular forum, but not to cryptography. And I believe this is the best forum for cryptography, so I'm glad I found it. About this code: It was made by myself, so there's nothing bad about it. Also, it's a passage from a book, but no hint toward which book. Others have attempted to solve it on different forums, email accounts, and even my teachers... So I hope someone here can... Other hints will be put at the end as spoilers. 10748 24478 12412 63110 92980 25125 27425 24319 10225 21441 06111 05262 24710 34522 42295 96045 41341 06424 36740 56873 72242 85262 63612 15610 85449 36823 78525 48118 94914 05847 34842 31071 68218 14146 12252 49262 05610 76662 34692 76675 85451 26318 52810 42685 55426 24357 60468 32412 42110 54740 66415 56811 85190 SPOILER 1: (Not very helpful) It's substitution, but not direct simple substitution. SPOILER 2: (Sort of helpful) It is paper-and-pencil encrypted. (Well, not really paper and pencil... but in the head.) SPOILER 3: (Helpful) It uses basic algebra or pre-algebra. Those are the only ones I could think of as being good hints without giving it away. Ok. Good luck on this! |
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Ancient chinese wise man once told me... SPOILER:GO AWAY!!!!! | |
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| jdege | Jun 24 2007, 01:51 AM Post #2 |
Super member
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265 characters factors to 5*53. Which means that either characters don't encrypt to a constant number of digits, or each character or group of characters encrypts to five digits. But there are zero repetitions of the five-digit groups. Which would unlikely if each five-dgit group represented either a single character or a bigram. (And a five-digit group can't represent more than a bigram.) Which sends us back to characters not encrypting to constant numbers of digits. Which raises the question, how does the decipherer know how many digits make up a character? And which ends my patience for exploring this. Describe the cipher, provide plaintext/ciphertext pairs, so I can test to ensure that I understand how it works. And then encrypt ten plain text messages, each of >200 characters, with the same key, and I'll see if I can break it. |
| When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl. | |
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| tidmiste | Jun 24 2007, 04:08 AM Post #3 |
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Just registered
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Ok. I'll do that, since it's only fair. And if you haven't seen the hints, they have information that will really help, but if you want to try without it, it's up to you. They don't help that much... but they help enough. 1) 10929 60572 54628 21635 84615 10317 Translates to This code is hard 2) 45368 21567 37142 53874 64047 83246 25312 73666 38614 54654 87591 02750 Translates to Finally someone is willing to try And for some shorter challenges: 1) 49266 63847 2) 49481 04151 03241 42431 13000 3) 89484 06721 56222 34347 60651 05000 I'll send some more up later, because the process takes a while.... Hopefully those help a little... OH! And the information! Well, I was very interested in codes, and I made like a dozen different varieties, all of which are pretty original, but not sure if entirely not-made-up-before-by-someone-else. One day, I wanted a cipher that was entirely my own, and incredibly simple to remember, and very hard to solve without the key. So this came out. The way it's encrypted is in one of the hints (though it's not entirely explained, for the reason that it's supposed to be solved). I showed this to two of my teachers (a computer teacher and an Algebra II teacher), and asked if they could attempt to solve it. The closest my teachers got was finding two sets of the five that were alike, and once I saw that, I quickly fixed it to be a lot more secure, and still as simple as the original. Now, I have spread this out all over various forums, and no one thus far has gotten any answers to any of my previous challenges. I figured that a longer message would be easier for them (sort of), and even more easier with a passage, so I researched a passage, and the message at the beginning of the topic came out. And that's all of the information I can give for now. Once you begin to understand, I will PM you some more hints, ok? PS: PM me the last of the mini-challenges' plaintext so I can verify that you understand how it works. 
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Ancient chinese wise man once told me... SPOILER:GO AWAY!!!!! | |
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| Revelation | Jun 24 2007, 09:58 AM Post #4 |
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Welcome to the forum! This looks like a hard cipher. Some of the things I noticed: - it is possible to have an odd amount of numbers, so it can't be bigrams only - the last zeroes look like they are used to fill things up: 05000. Meaning they have no value. "This code is hard" has got 14 characters, excluding spaces. Inc it has got 17. This should be encrypted to 30 characters. Which thou canst not do with bigrams  Then we've got the second with 28 characters, ex spaces. This should be encryped to 60 characters. So if the plain text is twice as long, the ciphertext is too. So I think we can jump to the conclusion that spaces are removed. Now we need to find the right connection between 14 and 30 characters Or there are just 2 spam characters with a plaintext with length 14, and with 28 characters 4 spam characters and we do have a bigram
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RRRREJMEEEEEPVKLWENFNVJKEEEEEAOLKAFKLXCFZAASDJXZTTTTTTTLSIOWJXMOKLAFJNNKFNXN RAGRBAQEMHIGDJVDSEOXVIYCELFHWLELJFIENXLRATALSJFSLCYTKLASJDKMHGOVOKAJDNMNUITN RRRRLJVEEEEECLYVYHNVPFTAEEEEEMWLMEIRNGLARWJAKJDFLWNTIERJMIPQWOTZEOCXKNUBNXCN RJIRPOWEANFUSNCZVDVZNMSFEKLOEPZLDKDJWSAAAAAAAOERHJCTNCKFRIMVKSOFOMKMANREWNBN RZUDRGXEEEEENFQIDVLQNCKNEEEEEDGLLLLLLAWIOSNCDARLODMTOEJXMILDFJROTKJSDNLVCZNN | |
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| jdege | Jun 24 2007, 11:31 PM Post #5 |
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I have no interest whatsoever in guessing how it works. I'm willing to try to break the encryption, provided that you explain how it works. If you're not confident enough in the system to provide a challenge in which all is known except the key, I see no reason to be bothered with it. |
| When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl. | |
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| tidmiste | Jun 25 2007, 02:10 AM Post #6 |
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Just registered
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Well, if you must know... There aren't any nulls, any fake digits, except as Revelation pointed out, the zeros are just fillers. And this applies for ALL of the ciphertext. |
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Ancient chinese wise man once told me... SPOILER:GO AWAY!!!!! | |
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| tidmiste | Aug 15 2008, 02:33 AM Post #7 |
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Just registered
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Ok. I'm going to release (just about) everything related to this cipher now, to make it a fair challenge, as you've insisted. Sorry that it's about a year late, but... I'd love to see someone solve it. Let me start with how it's encrypted. As I've said, it uses basic arithmetic, and to be more exact, it uses it in the form Ax (+ or -) B, where x is the numerical value of a given letter. The resulting number would be the number that appears in the code. Also, after encryption using one equation on one number, the equation shifts for the second. There could be between four and ten (even though that number can still be higher) equations that get shifted, and the order never changes. After you reach the last of the equations, you start back over on the first equation. The letters are numbered using a special system (which is also part of the key), but an example would like... 101=A, 102=B, 103=C, 104=D, 201=E, etc. That isn't what it is, but it's sort of like it. After I convert the numerical values of the letters, I combine them into 5-number groups. If there isn't enough numbers in the last group, I'll add as many zeroes as necessary to create a five number group. In addition, though I said it wasn't in the text I had originally posted, I could add nulls and fake digits, and the recipient of the message would know what they are, despite the arrangement of the digits. The system of encryption I created also creates a system that makes decryption in the prescence of unknown nulls still easy. Also, there could be billions of different variations because of it's math-based encryption process. The numbers created are definitely not just bigrams, but some are trigrams too. As I said above, because of the keys (the varying equations and numerical values), decrypting the values despite the varying bi- and trigrams is very easy. Hopefully, this helps a bit, jdege and Revelation. |
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Ancient chinese wise man once told me... SPOILER:GO AWAY!!!!! | |
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| jdege | Aug 15 2008, 02:40 PM Post #8 |
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In your example, you're using three-digit numbers to represent your alphabet. Are your equations done mod-1000? If not, how do you know which digits make up a single letter? If so, are you aware that A must be relatively prime to 1000 in order for your equation to be reversible? That means you have 400 choices for A. You said you pad with zeros - if we strip the one zero off the end, we have 264 digits, which is 88*3, which doesn't rule out mod-1000 arithmetic. |
| When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl. | |
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| tidmiste | Aug 15 2008, 10:20 PM Post #9 |
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Just registered
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I was just using three digit numbers for examples, but they could be two digit numbers or even one digit numbers. I will use an example here. I will encrypt the phrase "We are at the inn." Let's let the letters equal:
That will be Key 1. Now the encryption method would represent Key 2.
Now, to encrypt, I would send each letter through the equations, first letter going through the first equation, the second through the second, the third in the third, the fourth in the first, and so on.
70644 06164 40631 28643 18286 43000 As you can see, this wasn't a very good example because both the first and second E and first and second A were in the same position in the equation list, therefore having the same numbers. It is easily untraceable though with the five number groups. It can still be spotted and analyzed, but without the key, it won't do much to help, except maybe give the cryptanalyst the number of equations. However, as shown with the T's back to back, the numbers aren't always the same, and it's possible that two letters may equal the same value. Now for decryption. It's just as simple as long as you have both of the keys. You could separate it first by bigrams. So it would end up like:
Here's the kicker. The recipient would know immediately that this isn't the correct message, because there is no way any of the numbers would equal 12, so he could easily find out that it's a trigram in that instance, and then make the rest bigrams, changing the message to this:
Then, after this separation, all the numbers are sent through the equations, but in the y value instead of the x. Once you're done, it is easily readable. For nulls in this example, you could use "18", because, even if they used the trigram rule above, it would rule out the number as a whole because no number going through those equations would be over 142. Hopefully that clears things up. (PS: I'm not sure if I said it, but it is not mod-1000. Also, it isn't strictly limited to using prime numbers, as you can see with the list of values for letters. I kind of dodged it, it seemed. Sorry.) |
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Ancient chinese wise man once told me... SPOILER:GO AWAY!!!!! | |
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| jdege | Aug 17 2008, 02:28 AM Post #10 |
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Sure does clear things up. It's not a reversible cipher. There isn't a unique decryption, you're expecting the recipient to try out multiple decryptions, and to try to guess which one is correct. That renders this impractical. There's nothing wrong at all with having your cipher switch between digrams and trigrams. But you have to have a mechanism by which your recipient knows what he's dealing with. You can't expect him to guess. |
| When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl. | |
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| Donald | Aug 29 2008, 04:00 AM Post #11 |
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And THAT is one reason cryptographers insist on enforcing Kerckhoff's Principle. The METHOD of encryption should never be secret. ESPECIALLY when you are trying to analyze a cipher. The attackers should approach the METHOD first, the crypt text second. Don't feel too bad though, it's a mistake we have ALL made. Not just ignoring Kerckhoff's Principle, but designing irreversable ciphers. I designed a REALLY cool cipher once based on a very clever (I thought) system that used a single playfair square to encrypt trigrams. I've forgotten the exact details, but I haven't forgotten the result. I had been playing with my genius discovery for quite some time before I realized that the process was irreversible without the original plain text. Oops! Might make an interesting hash, but was certainly of no use as a cipher. 
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| jdege | Aug 29 2008, 11:14 PM Post #12 |
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Yep. What, though, would it take to turn this into a reversible cipher? Not a lot. It has a number of alphabets, each of which is a multiplication+addition. If each was a multiplication+addition done modulo some third number, and was expressed in a constant number of digits, it'd be reversible. That is, if the number of digits in each alphabet was part of the key, we'd have a reversible cipher. Of course, once we had that, we'd have a cipher that was pretty easy to break. |
| When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl. | |
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