If the D'Agapeyeff Cipher is indeed a numerical version of the ADFGX cipher, then I now possess the ability to check the entire likely keyspace . . . and in under a lifetime . . . with help from a fellow crypto-enthusiast.
As I outlined in my last post, the beauty of a fractionated cipher like the ADFGX is that each permutation of the columns will produce a unique phi value (or IC if you'd rather talk in terms of a ratio).
As I pointed out in my last post, for any even length transposition key there are only:
N! / ((N/2)! + (N/2)!) unique ways to create a phi value.
14! / (7! + 7!) = 8.7 million ways to create a phi value for the D'Agapeyeff cipher when treated as a fractionated transposition. I refer to each of these 8.7 million permutations as "key variants". From each key variant I can obtain all 7! = 5040 permutations of the column pairings.
For instance, let's say this is the correct key: 4, 5, 9, 11, 0, 1, 7, 13, 8, 3, 2, 12, 6, 10 and generates a phi value of 574.
I don't need to find this exact key to find the solution, I just need to find one of the 5,040 variations of the same column pairings, say: 9, 11, 4, 5, 7, 13, 0, 1, 6, 10, 2, 12, 8, 3. This also generates the same phi value because all the columns are pairing up the same to create the same "letters". If I were to check all 5,040 permutations of this key, I'd eventually come across 4, 5, 9, 11, 0, 1, 7, 13, 8, 3, 2, 12, 6, 10 and through log tetragraph scoring arrive at the correct plaintext after hill-climbing the substitution key.
Of course, I don't know the phi value of the plaintext in question, but I do know that it should fall into some acceptable range for English. While 634 is expected for English, the range actually spans from 480 to 780. I took a random sample of many different books from Project Gutenberg and randomly pulled different 98 characters of text and tested their phi values. As you can imagine, 75% of all the phi values fall between 540 and 660.
For the D'Agapeyeff cipher, only 0.001% of the 8.7 million key variants fall within the 540 to 660 range. This represents 74 key variants. If the D'Agapeyeff Cipher has a solution in this keyspace, I'll need only to check 74 * 5040 = 372,960 total keys to find the solution. This is much better than 87,178,291,200.
And what if the D'Agapeyeff cipher has a phi value closer to 500? There are 3,744 key variants with a phi value greater than or equal to 500. This represents 18,869,760 total keys to be checked. While this would take approximately 100 days to check each key in the solver, that's much better than 800 years.
As I outlined in my last post, the beauty of a fractionated cipher like the ADFGX is that each permutation of the columns will produce a unique phi value (or IC if you'd rather talk in terms of a ratio).
As I pointed out in my last post, for any even length transposition key there are only:
N! / ((N/2)! + (N/2)!) unique ways to create a phi value.
14! / (7! + 7!) = 8.7 million ways to create a phi value for the D'Agapeyeff cipher when treated as a fractionated transposition. I refer to each of these 8.7 million permutations as "key variants". From each key variant I can obtain all 7! = 5040 permutations of the column pairings.
For instance, let's say this is the correct key: 4, 5, 9, 11, 0, 1, 7, 13, 8, 3, 2, 12, 6, 10 and generates a phi value of 574.
I don't need to find this exact key to find the solution, I just need to find one of the 5,040 variations of the same column pairings, say: 9, 11, 4, 5, 7, 13, 0, 1, 6, 10, 2, 12, 8, 3. This also generates the same phi value because all the columns are pairing up the same to create the same "letters". If I were to check all 5,040 permutations of this key, I'd eventually come across 4, 5, 9, 11, 0, 1, 7, 13, 8, 3, 2, 12, 6, 10 and through log tetragraph scoring arrive at the correct plaintext after hill-climbing the substitution key.
Of course, I don't know the phi value of the plaintext in question, but I do know that it should fall into some acceptable range for English. While 634 is expected for English, the range actually spans from 480 to 780. I took a random sample of many different books from Project Gutenberg and randomly pulled different 98 characters of text and tested their phi values. As you can imagine, 75% of all the phi values fall between 540 and 660.
For the D'Agapeyeff cipher, only 0.001% of the 8.7 million key variants fall within the 540 to 660 range. This represents 74 key variants. If the D'Agapeyeff Cipher has a solution in this keyspace, I'll need only to check 74 * 5040 = 372,960 total keys to find the solution. This is much better than 87,178,291,200.
And what if the D'Agapeyeff cipher has a phi value closer to 500? There are 3,744 key variants with a phi value greater than or equal to 500. This represents 18,869,760 total keys to be checked. While this would take approximately 100 days to check each key in the solver, that's much better than 800 years.
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