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Has anyone had any luck decoding the message on the Spirit Mars
Exploration Rover DVD? Have a look at http://planetary.org/news/2004/crack-the-code.html for more. P. |
#2
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#3
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Yup, I've been trying, but no luck so far. Actually thought I was
making some headway until I got their 1-21 clue, which actually confused me more than helped...but it is great fun. I'm sure some more experience cryptographic types could break it readily, but its an interesting project if nothing else... Good luck! -David (Pierre) wrote in message . com... Has anyone had any luck decoding the message on the Spirit Mars Exploration Rover DVD? Have a look at http://planetary.org/news/2004/crack-the-code.html for more. P. |
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(David Whitney) wrote in message . com...
Yup, I've been trying, but no luck so far. Actually thought I was making some headway until I got their 1-21 clue, which actually confused me more than helped...but it is great fun. I'm sure some more experience cryptographic types could break it readily, but its an interesting project if nothing else... Good luck! -David (Pierre) wrote in message . com... Has anyone had any luck decoding the message on the Spirit Mars Exploration Rover DVD? Have a look at http://planetary.org/news/2004/crack-the-code.html for more. P. Let me see if I can help a little? take a line from the example ---ll---lll--l-----llll-llll------lll-lll-lllll-ll-ll-llll--ll---lll----l-ll--, now break it into 3's (it's 3 bit binary) --- ll- --l ll- -l- --- -ll ll- lll l-- --- -ll l-l ll- lll ll- ll- ll- lll l-- ll- --l ll- --- l-l l--, the - = 0 and l = 1 so the breakdown into numbers is --- ll- --l ll- -l- --- -ll 000 110 001 110 010 ect.... now.. we have to turn those into real numbers we can understand.. it's simple math now.. the places in the 3 bit binary are 4 2 1 if a 1 is in the 4 place it = 4 if a 1 is in the 2 place it = 2 if a 1 is in the 1 place... you get the idea! if a 0 is in a place it = 0 no matter what. so now you take a group and add the values up.. for example 010 would really be 0+2+0 which = 2 110 would be 4+2+0 which = 6 000 would be 0+0+0 = 0 and so on... when you finish that step you get a long line of numbers the trick is figuring out what numbers = what letters... I hope this helps somewhat without taking the fun out! Raht |
#6
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Hi, Raht...
Thanks for the tip, but I was already to the point of decoding the bits that way. I was modeling my own theory of the code as some sort of Huffman tree against a particular frequency table, with the trick being to find the right table. Their clues were certainly indicative of some sort of frequency-based substitution, so I threw together some code to build a Huffman tree, then threw some frequency tables at it, but the results never came close to anything intelligible. I finally broke it the other day, realizing I had come closer with a separate theory I had abandoned. And, yes, I certainly had the "ovaltine" effect when I realized the answer...but it was a great mental exercise to see if the stuff you *think* know about encryption had sunk anywhere into the gray matter in *practice*.. -dew (raht) wrote in message om... (David Whitney) wrote in message . com... Yup, I've been trying, but no luck so far. Actually thought I was making some headway until I got their 1-21 clue, which actually confused me more than helped...but it is great fun. I'm sure some more experience cryptographic types could break it readily, but its an interesting project if nothing else... Good luck! -David (Pierre) wrote in message . com... Has anyone had any luck decoding the message on the Spirit Mars Exploration Rover DVD? Have a look at http://planetary.org/news/2004/crack-the-code.html for more. P. Let me see if I can help a little? take a line from the example ---ll---lll--l-----llll-llll------lll-lll-lllll-ll-ll-llll--ll---lll----l-ll--, now break it into 3's (it's 3 bit binary) --- ll- --l ll- -l- --- -ll ll- lll l-- --- -ll l-l ll- lll ll- ll- ll- lll l-- ll- --l ll- --- l-l l--, the - = 0 and l = 1 so the breakdown into numbers is --- ll- --l ll- -l- --- -ll 000 110 001 110 010 ect.... now.. we have to turn those into real numbers we can understand.. it's simple math now.. the places in the 3 bit binary are 4 2 1 if a 1 is in the 4 place it = 4 if a 1 is in the 2 place it = 2 if a 1 is in the 1 place... you get the idea! if a 0 is in a place it = 0 no matter what. so now you take a group and add the values up.. for example 010 would really be 0+2+0 which = 2 110 would be 4+2+0 which = 6 000 would be 0+0+0 = 0 and so on... when you finish that step you get a long line of numbers the trick is figuring out what numbers = what letters... I hope this helps somewhat without taking the fun out! Raht |
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#8
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I don't know if Raht is following this thread any more,but I might be
able to shed some light on this inference. First, any time you deal with bits, and can start grouping them in a fashion that lets you represent an octet (0-7), its fairly natural (at least to me) to at least give it a try and see if can make inferences about the result. Furthering that theory is the fact that the message was 75 bits in length, which is evenly divisible by three. Its not to say there's a hard rule that says it "has" to be in three bits, but it was a logical place to start looking. Yes, five bits can provide a symbol for every letter, but in applications where bandwidth is at a premium, there's almost always an effort to compress the data being transmitted. If every symbols is 5 bits, your average symbol length is 5 bits; if you use shorter codes to represent frequent letters, you can decrease that average symbol length and save bandwidth. That means use the fewest number of bits to represent the most frequent letters, with frequency determined by some sort of table. You can represent 6 symbols in three bits, reserving two as prefixes for longer codes. Once the clues started matching up with this idea, interpreting the data as Raht did started revealing some interesting patterns that were confirmed by later clues. No need to go further into that here, as some may still be working on the code, but perhaps that explains a bit why you wouldn't necessarily assume a 5-bit code... -David (Pierre) wrote in message . com... (raht) wrote in message om... Let me see if I can help a little? take a line from the example ---ll---lll--l-----llll-llll------lll-lll-lllll-ll-ll-llll--ll---lll----l-ll--, now break it into 3's (it's 3 bit binary) --- ll- --l ll- -l- --- -ll ll- lll l-- --- -ll l-l ll- lll ll- ll- ll- lll l-- ll- --l ll- --- l-l l--, the - = 0 and l = 1 Interesting approach, raht. But just how did you figure out it was 3-bit binary in the first place? At first glance, since 3 bits allows only for 8 possibilities, you'd need at least 5 bits to represent all 26 letters of the alphabet. So, at this point, why not 6 bits instead of 3? P. |
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