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| jdege | May 20 2009, 01:19 PM |
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You have a homophonic alphabet - a single alphabet with variants. 1 maps to A, B, or C, and 0 maps to D, E, or F. Sometimes 1 will map to A, sometimes to B, but 1 will never map to D. The substitution is one-to-many, but it's always the same substitution. For it to be polyalphabetic, you'd need to have a number of bit to letter mappings, and to alternate between them in some pattern. As for your calculation of the number of possibilities, I'm not challenging that. I'm just pointing out that the number of possible keys isn't an indication of strength. You have two stages, the first is a simple substitution, the second is a binary homophonic substitution. The first stage on its own is trivial to crack, the second adds security only to the degree that it cannot be stripped off separately. And that depends upon whether we have a reliable test to indicate that a trial decryption is correct. In this case, we do, the Index of Coincidence. Do you understand what a hillclimber is, and how it works? Do you understand what the Index of Coincidence is, and how it can be used as a reliable test to indicate that the second stage of encryption has been successfully stripped? It'd take me half a day to write a hillclimber to attack this. I don't have time for that, and when it worked, it'd not give you greater understanding of where the weaknesses in your cipher are. Why don't you try to write a hillclimber to attack it? I'm sure you'd learn a lot from it. |
| When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl. | |
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12:43 AM Nov 28