An excerpt from TM11-485 War Department Technical Manual published in 1944 about complex route transposition, which shows a matrix and describes a complicated route transposition in terms of chess moves:
We briefly discussed "Knight’s Tour" here: viewtopic.php?f=81&t=2617&hilit=tour&start=940 and here: viewtopic.php?f=81&t=2617&hilit=tour&start=950.
However, the Technical Manual discusses other similar options, similar to Knight’s Tour.
Interesting observations, smokie. The herringbone pattern reminds me of interference patterns, such as https://en.wikipedia.org/wiki/Moir%C3%A9_pattern
I wonder if we can determine whether this arises naturally or as a consequence of a transposition scheme.
Simple Test for Playfair
I did a simple test for Playfair, thinking that the message could have been drafted vertically into 15 or 19 columns, then encoded vertically with Playfair, then re-drafted into 17 columns. Or, the plaintext were encoded with a period of 15 or 19 using Playfair, something that the War Department Technical Manual makes brief discussion of. With Playfair, an odd even pair of symbols will never be the same, provided of course that Playfair rules are adhered to. I drafted the message into 19 columns LRTB and RLBT and 15 columns RLTB and LRBT. I found repeating ciphertext in odd even pairs in every example.
What do you think smokie, is playfair a possibility or not? Could you share the ciphers?
What do you think smokie, is playfair a possibility or not? Could you share the ciphers?
That is the 340 two posts above. Because I was able to find repeating symbols in odd / even rows, this is likely not a Playfair cipher that follows the rules. With Playfair, if I was going to encode the letters "LL", then I would replace the second "L" with an "X", and encode "LX" instead. "LX" would become two different, new symbols which would not be the same. Just a very simple test, but if he had drafted the message vertically into 15 or 19 columns, then encoded two letters at a time, then redrafted it into 17 columns, I would not be able to re-draft into 15 or 19 columns and find repeating symbols that occupy odd / even rows.
In other words, it isn’t Playfair encoded starting from one of the four corners.
There are other types of ciphers that encode two letters at a time, and I am still considering those at this time. I think that I may still make a Playfair suite and make some Playfair messages just for fun and to study them a little bit, before and after homophonic encoding.
EDIT: I want to add that I have been still working on it. I have been making messages with route transposition. Even with a bigram repeat count spike, there is not necessarily a spike in probability scores. If you score each bigram repeat for a period, and then take the average of those scores for that period. And then do that for all periods and compare the average probability score for each period, then there is not necessarily a spike at the same period. Sometimes there is, and with the 340 there is. This is what made me think that perhaps two letters were encoded at a time.
The 340 LRTB:
I just found an even cooler cryptanalysis manual.
FIELD MANUAL
NO 34-40-2 HEADQUARTERS
DEPARTMENT OF THE ARMY
Washington, DC, 13 September 1990
http://www.umich.edu/~umich/fm-34-40-2/#pdf
It explains identification and cryptanalysis worksheets for all different types of ciphers. I will post the link on my TOC as well.
Cool manual smokie, there is a program in there.
I am about to release an AZdecrypt update with some new features you people might like. For the moment I am testing and retesting everything to work properly.
Smokie, it will include a true polyphone solver, in such that you can state that on a per symbol basis, for example, the "+" symbol has X many polyphones resulting in a more minimal increase of multiplicity over the symbol expansion method which we have used previously. If you state that for instance, the "+" symbol has 2 polyphones then the solver will be able to assign 2 letters to the "+" symbol, it will figure out the distribution and letters by itself. It is great solver for aperiodical polyalpabetism where multiplicity is not a problem. I would like to know if it can solve the smokie33 if the right amount of polyphones per symbol are set.
Mr lowe, I have added a built transposition matrix creater which should make it much easier for you to create these so that you don’t have to use the pattern drawer anymore.
Also support for 4, 7 and 8-grams have been added though I am unable to test 8-grams, AZdecrypt is very memory hungry by design.
When rushing through the manual I again spotted the pivot affinity of a vigenere square, that is a sequential string of unique letters with an incremental offset going through the rows:
abcdefghijklmnopqrstuvwxyz bcdefghijklmnopqrstuvwxyza cdefghijklmnopqrstuvwxyzab defghijklmnopqrstuvwxyzabc efghijklmnopqrstuvwxyzabcd fghijklmnopqrstuvwxyzabcde ghijklmnopqrstuvwxyzabcdef hijklmnopqrstuvwxyzabcdefg ijklmnopqrstuvwxyzabcdefgh jklmnopqrstuvwxyzabcdefghi klmnopqrstuvwxyzabcdefghij lmnopqrstuvwxyzabcdefghijk mnopqrstuvwxyzabcdefghijkl nopqrstuvwxyzabcdefghijklm opqrstuvwxyzabcdefghijklmn pqrstuvwxyzabcdefghijklmno qrstuvwxyzabcdefghijklmnop rstuvwxyzabcdefghijklmnopq stuvwxyzabcdefghijklmnopqr tuvwxyzabcdefghijklmnopqrs uvwxyzabcdefghijklmnopqrst vwxyzabcdefghijklmnopqrstu wxyzabcdefghijklmnopqrstuv xyzabcdefghijklmnopqrstuvw yzabcdefghijklmnopqrstuvwx zabcdefghijklmnopqrstuvwxy
And here is another one in a 17 by 20 grid but with sequence length of 19:
abcdefghijklmopqr stabcdefghijklmop qrstabcdefghijklm opqrstabcdefghijk lmopqrstabcdefghi jklmopqrstabcdefg hijklmopqrstabcde fghijklmopqrstabc defghijklmopqrsta bcdefghijklmopqrs tabcdefghijklmopq rstabcdefghijklmo pqrstabcdefghijkl mopqrstabcdefghij klmopqrstabcdefgh ijklmopqrstabcdef ghijklmopqrstabcd efghijklmopqrstab cdefghijklmopqrst abcdefghijklmopqr
It has interesting some properties after encoding:
– Bigram peak period 19, same as the sequence length.
– Would create allot of 340 like pivots if the sequence length is 20.
– Is ultra cyclic because the plaintext already cycles on its own. Random homophonic substitution could be used and the cipher would appear cyclic.
But in general creates far to many bigrams to even consider in the case of the 340 and a keyword length of 19 is very strongly detected. Though it gets me to think that the pivots may be period 20 shift areas. I also tried to mix it with horizontal fragments of text but it didn’t help much to diffuse the many bigrams.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 20 21 22 23 24 25 26 27 28 29 30 31 32 15 16 33 34 35 1 2 36 37 5 38 39 8 40 41 42 12 13 43 15 16 44 45 46 1 20 47 48 23 49 50 26 51 52 53 30 31 54 15 16 55 56 57 1 2 3 58 5 6 7 8 9 59 60 12 13 61 15 16 17 18 62 1 20 21 4 23 24 25 26 27 63 11 30 31 14 15 16 33 34 19 1 2 36 22 5 38 39 8 40 10 29 12 13 32 15 16 44 45 35 1 20 47 37 23 49 50 26 51 28 42 30 31 43 15 16 55 56 46 1 2 3 48 5 6 7 8 9 41 53 12 13 54 15 16 17 18 57 1 20 21 58 23 24 25 26 27 52 60 30 31 61 15 16 33 34 62 1 2 36 4 5 38 39 8 40 59 11 12 13 14 15 16 44 45 19 1 20 47 22 23 49 50 26 51 63 29 30 31 32 15 16 55 56 35 1 2 3 37 5 6 7 8 9 10 42 12 13 43 15 16 17 18 46 1 20 21 48 23 24 25 26 27 28 53 30 31 54 15 16 33 34 57 1 2 36 58 5 38 39 8 40 41 60 12 13 61 15 16 44 45 62 1 20 47 4 23 49 50 26 51 52 11 30 31 14 15 16 55 56 19 1 2 3 22 5 6 7 8 9 59 29 12 13 32 15 16 17 18 35 1 20 21 37 23 24 25 26 27 63 42 30 31 43 15 16 33
Cool manual smokie, there is a program in there.
I am about to release an AZdecrypt update with some new features you people might like. For the moment I am testing and retesting everything to work properly.
Smokie, it will include a true polyphone solver, in such that you can state that on a per symbol basis, for example, the "+" symbol has X many polyphones resulting in a more minimal increase of multiplicity over the symbol expansion method which we have used previously. If you state that for instance, the "+" symbol has 2 polyphones then the solver will be able to assign 2 letters to the "+" symbol, it will figure out the distribution and letters by itself. It is great solver for aperiodical polyalpabetism where multiplicity is not a problem. I would like to know if it can solve the smokie33 if the right amount of polyphones per symbol are set.
Mr lowe, I have added a built transposition matrix creater which should make it much easier for you to create these so that you don’t have to use the pattern drawer anymore.
Also support for 4, 7 and 8-grams have been added though I am unable to test 8-grams, AZdecrypt is very memory hungry by design.
That’s awesome!
Smokie33 info:
Symbol 6 maps to three plaintext
Symbol 14 maps to four plaintext
Symbol 21 maps to two plaintext
Symbol 23 maps to three plaintext
Symbol 33 maps to two plaintext
Symbol 54 maps to four plaintext
Symbol 57 maps to two plaintext
I am working on a digraph encoding experiment now. The results and smokie57 should be interesting.
Cool manual smokie, there is a program in there.
I am about to release an AZdecrypt update with some new features you people might like. For the moment I am testing and retesting everything to work properly.
Mr lowe, I have added a built transposition matrix creater which should make it much easier for you to create these so that you don’t have to use the pattern drawer anymore.
Also support for 4, 7 and 8-grams have been added though I am unable to test 8-grams, AZdecrypt is very memory hungry by design.
Thanks Jarlve looking forward to your matrix creator i will give it my best.. i have lots of patterns to try . looking forward to the updated AZ program as well.
i have been busy trying different patterns. one with an unusually high bigram repeat.. no good solve but interesting all the same .. will post it up soon for every ones thoughts.
cheers all
Smokie, can you double check the polyphones for smokie33? Or share me the original cipher or plaintext.
Thanks
Box corner & pivot visual clue; period 1 digraph encoded + transposition + homophonic OR period 15 / 19 digraph encoded + homophonic hypothesis
Question: Is it possible to digraph encode a message, then transpose it, then encode it with homophones and still get a high count of period 15 / 19 repeats? Or, is it possible to digraph encode two symbols at a time at period 15 / 19, then encode with homophones and still get a high count of period 15 / 19 repeats?
The short answer is yes, and it not particularly difficult.
I have been wanting to do this for quite some time, and found some inspiration from the War Department Technical Manual TM 11-484, which discusses digraph encoding in great detail, and has one short paragraph about encoding at some period other than period 1. For some reason I cannot access the website right now, so will continue on.
The new spreadsheet suite doesn’t fool around with the complicated Playfair rules, or makeup a four corner style grid of letters and symbols. It just randomly generates a digraph encoding chart, with 26 As, Bs, Cs etc. on the left side of the ciphertext digraphs, and 26 As, Bs, and Cs etc. on the right side of the ciphertext digraphs. It is 26 x 26, changes every time I generate a message, and simulates a four corner – opposite corner style cipher, of which there are different variations. Discussions of digraph encoding in the book begins with pictures of charts like this one. You encode two plaintext at a time, the plaintext on the left side of the pair find in the left column on the chart, the plaintext on the right side of the pair find on the top row. Then find the intersection and those are your two symbols.
The idea is to change the frequency distribution of the symbols, and also the count of bigram repeats to work with. Only about half of the high count bigram repeats, like TH in the English language, will be encoded into AB.
I almost did not even try this because I assumed that the cipher would decrease the count of period 1 bigram by one half, and then after homophonic substitution, there would not be enough period 1 bigram repeats to make a spike like period 15 / 19. But I was wrong.
Here is "I LIKE KILLING", encoded with a chart.
The frequency distribution gets flatter, but there are more symbols at low count. In other words, the plaintext graph has a higher slope, and the ciphertext graph has a lower slope and all 26 possible symbols. This part of it is similar to bifid with an odd period.
Here is a distribution of the plaintext bigrams, top, and ciphertext bigrams, bottom. See red stars, where there is a difference in high count bigrams, but as you move to the right, the distributions get more and more like each other, green stars and brown stars. They are really not substantially different, and that is what makes it possible to get spikes like at period 15 / 19.
And the same distributions, only zoomed out so that you can see all of them. After encoding, there are more unique bigrams, but still plenty of bigram repeats to work with.
I am going to make 100 messages, encoded with the digraph chart, then transposed with this matrix:
Then encode with homophones with a key like this ( with 26 to work with, it is easier to pack +/- 63 symbols into the key ) and map one symbol to 4 of the digraph ciphertext to simulate the +. About 75% cyclic, 25% randomness in cycles.
It works pretty well to make symbol count distributions like this one. High count symbols in the 340 red star, are not particularly easy to duplicate with an inefficient key and similar settings with a standard + / – 23 plaintext key. But it is a little easier when encoding digraph ciphertext. It is pretty easy to generate messages that look like the 340 here.
Then, check for spikes with period 19, and the average of the probability scores of all period 19 repeats. See if there is a trend.
