Jarlve wrote:@Quicktrader,
The way I see it is that the solver uses another n-gram list for the n-grams you wish to mark. You supply this list yourself so you can do whatever you want. For instance I make a secondary 5-gram list with entries: ZODIAC, OFTHE, ABCDE. The solver will select a 5-gram from that list randomly to the 5-grams you marked in the cipher everytime it makes a change. So eventually it'll end up with the best fitting 5-gram for that piece of ciphertext. Is that okay?
Hi Jarlve,
that's absolutely perfect..
I currently do the logical way via Excel - and it's simply too slow. The procedure, however, is quite simple and efficient, imo:
- Setting the IoFBc 5-gram plus a + letter (e.g. 'L') delivers some letters
- Adding the w symbol plus the circle with the horizontal line adds some more (each A-Z)
- Choosing finally the reverse L symbol gives another one
This is what I can come up with, so far (line 17 & 18, here together in one line):
The colored ones I got from the 5-gram plus the + symbol. All others are not colored. Nevertheless, by defining the 5-grams (plus +) as well as only three (!) other symbols, I get a variety of potential cleartext results of 12 letters (orange line). It should be considered that the circle with the horizontal line shall, in case of + being an 'L' - be no other alphabetical letter than AEIOURWY, thus most likely is a vowel (for + being an 'L').
Even if the IoFBc section is not a double frequent trigram, it may well be expected that it is amongst the 20 if not 100 most frequent 5-grams (consisting of frequent trigrams). I got some solid statistics about that and will post it below.
This leads me to approximately half a billion varieties. The procedure can be repeated for a + being a 'T' or 'E'.
Currently I do get loads of results, those I select should contain either longer words (e.g. >5 letters) or two words (e.g. both >4 letters). This is guessing. The results are astonishing: '..AGENTCALL..' and similar results are not rare at all. Actually there are plenty of them, however it's about finding the correct one somehow.. Changing the final letter currently takes me 15 minutes of Excel's calculation time (each).
Still collecting results, so nothing in detail so far. But I write this because a 5-gram function indeed would support your superquick tool..thanks for your efforts.
QT
