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Nir Dahan Inner circle Munich, Germany 1390 Posts |
Guys I just saw that in the above post the different widths assigned to different characters make the columns misaligned - sorry for that, but I don't know any other way to write it up.
Nir |
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TomasB Inner circle Sweden 1144 Posts |
Beautiful, Nir. It _is_ the same as I wrote earlier but never found a compact way to describe the 2048 sequences going with each of the 16 symbols.
In your decoder it is obvious that you can flip all bits and the result will be the same. You can also flip groups of three bits (for example the bits below (1), (1,2) and (2). There are six such groups) or groups of four bits (such as those below (0), (1), (2), (0,1,2). There are three such groups) or you can flip the bits below (0), (1), (2), (3) and (0,1,2,3). Any of these changes in any combination will result in an even number of changes within each "p" so the decoded word will be the same and the number of different 15-bit sequences mapping to each of the 16 symbols then becomes 2^(1+6+3+1)=2048 as predicted. Seeing the decoder written Nir's way it gets all so much clearer and the coding and decoding of a message is a breeze. Perfect work! /Tomas |
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Nir Dahan Inner circle Munich, Germany 1390 Posts |
Tomas,
I am not a CS guy but do you know if that is related to CRC coding? seems too close in concept to CRC, that it smells suspicious... Nir |
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TomasB Inner circle Sweden 1144 Posts |
Nir, I don't think it is related in any other way than that the coding and decoding can probably be expressed in polynomial divisions. In CRC and such codes (Reed Solomon for example) a single bit error (or more) during transfer should mean that you still decode the correct symbol, but in the case here we assume a perfect channel with zero probability of error AND a single bit error can lead to ANY symbol. Not quite a code you'd want to use over a real channel.
I'm already talking over my head since I haven't studied encoding since 1997 and really would need to refresh my knowledge in the subject. /Tomas |
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Nir Dahan Inner circle Munich, Germany 1390 Posts |
Ok, just gave it a try...
another interesting question is whether this can be turned into some sort of a (not boring) magical effect... Nir |
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TomasB Inner circle Sweden 1144 Posts |
The first idea for a trick that comes to mind would be to have a spectator shuffle the deck Triumph style and remove any 15 cards. The packet is spread on the table. He then decides to remember a face up card or peek a face down card. He should indicate his decision to you. Explain that he is to turn his card over and replace it exactly where it is. You demo this and it is at this moment you turn a card in a specific position over, if needed. Rare times no card at all needs to be turned over.
The spectator carries out the instructions, then squares the pile and carries it to another room where a medium is sitting. She analyzes the cards and eventually gives him one of the cards face down to bring it back to you. You still have one codeword left to send so maybe you can have it mean that no card at all was selected or to mean that the center card was selected AND the spectators middle name is Robert. So what happens if the spectator happens to (or is given a choice to do so) turns the packet over before giving it to the medium? Would it be possible to map the symbols to positions in a way to make a turnover of the packet mean that the new symbol maps to 16 minus the previous position? /Tomas |
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TomasB Inner circle Sweden 1144 Posts |
The reason I think that the last suggestion is possible is that if you analyze the case of just three cards and change the parity functions to
11- 0-0 xxx Turning the packet over would mean that a decoded 1 becomes 3, 2 becomes 2 and 3 becomes 1, which is exactly what we are looking for. 0 becomes 0 but we could use that to mean that no card was selected. The improtant thing is that 0 still is 0 when the packet is turned over. So the question is how the parity functions need to be altered to have this same property on 15 bits. Turning a packet over is the same flipping all bits and reversing the order of the bits. /Tomas |
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