A pragmatic starting point for the d’Agapeyeff cipher is to sequentially replace its digit pairs with letters, i.e.
** .1 .2 .3 .4 .5
6. _0 17 12 16 11 --> A B C D E
7. _1 _9 _0 14 17 --> F G H I J
8. 20 17 15 11 17 --> K L M N O
9. 12 _3 _2 _1 _0 --> P Q R S T
0. _0 _0 _0 _1 _0 --> U V W X Y
If you then “re-flow” those letters into a 14×14 grid, many of its oddities are to be found in the final right hand column:-
[ 0] J B L O P B P D K D P I O N
[ 1] D I I L N M K C K K I I L B
[ 2] D J M L N P J I E M J J J R
[ 3] C E E K C K J O J J D B L Q
[ 4] O I C L J I M K E K N O D O
[ 5] D O O C L G B M B K K G K D
[ 6] C J L K D M C L O K C C C X
[ 7] I K P P N C O N E D O E B S
[ 8] B B O P O P I P G J D E J F
[ 9] E M B D I K L N B L D P K R
[10] E B D N N P M O I P K E G I
[11] M M O L M D B G B E B M J Q
[12] G C L L G G M L O N J L K M
[13] G N B L M J K D J I O K B Q
The ‘X’ (’04’) right at the end of row #6 is highly suspicious: at least one person before me has suspected that this might somehow be a padding ‘X’ appended to the end of the (pre-transposition-stage) plaintext to bring it up to a 14×14 multiple.
However, I think that the three ‘Q’ (’92’) symbols in the same rightmost column are even more suspicious: this symbol occurs exactly three times in the cryptogram, and only ever in this column. I think these are even more likely than the ‘X’ to be the final three letters of the plaintext, appended to pad it up to 14×14 = 196 characters in length.
In fact, I’m now almost certain that the correct starting point for cryptanalysis should be the diagonal transposition of the 14×14 grid, which transformation would flip all these oddities across onto the bottom (final) row of the transposed grid, leaving (presumably) a 14-column transposition cipher to solve:-
[ 0'] J D D C O D C I B E E M G G
[ 1'] B I J E I O J K B M B M C N
[ 2'] L I M E C O L P O B D O L B
[ 3'] O L L K L C K P P D N L L L
[ 4'] P N N C J L D N O I N M G M
[ 5'] B M P K I G M C P K P D G J
[ 6'] P K J J M B C O I L M B M K
[ 7'] D C I O K M L N P N O G L D
[ 8'] K K E J E B O E G B I B O J
[ 9'] D K M J K K K D J L P E N I
[10'] P I J D N K C O D D K B J O
[11'] I I J B O G C E E P E M L K
[12'] O L J L D K C B J K G J K B
[13'] N B R Q O D X S F R I Q M Q
Here I’ve highlighted the two tripled letters (“LLL” on row #3′, and “KKK” on row #9′): here LLL is on a row with 6 L’s (so it’s hardly surprising that it ended up as a tripled letter post-transposition), while KKK is on a row with 4 K’s. Here are the overall letter instance counts for the cryptogram:-
.K .B .J .L .O .D .M .I .C .P .E .N .G .Q .R .F .S .X
20 17 17 17 17 16 15 14 12 12 11 11 9 3 2 1 1 1
It’s interesting to compare this set with the letter frequency table of the text mini-corpus taken from d’Agapeyeff’s “Codes and Ciphers” (which I also generated recently). If you normalize that to 196 characters, here’s what you would expect to see in the cryptogram:-
.E .T .A .I .O .S .N .R .H .D .L .C .U .M .F .P .W .G .Y .B .V .K
25 18 15 14 14 14 13 12 11 .8 .7 .6 .5 .4 .4 .4 .4 .4 .3 .3 .2 .1
From this, it looks as though K probably –> E, while B/J/L/O/D seem likely to go to T/A/I/O/S. What I’m thinking here is that if this is right, all we need to solve it is to generate a moderate number of best-guess substitution values and feed those into a transposition cipher solver, i.e.:-
(a) guess that K –> E
(b) generate the 5! = 120 permutations of B/J/L/O/D –> T/A/I/O/S
(c) assign plausible values to the remainder of the used letters (in matching descending frequency order)
(d) feed the 120 versions of the transposed 14×14 grid into a reliable columnar transposition solver
My prediction is that even though this will still be wrong, getting the 6 most popular letters right (i.e. 20 + 17 + 17 + 17 + 17 + 16 = 104 characters, ~53% of the cryptogram) and possibly some of the others (by chance) will allow the transposition solver to get us close enough to the answer, that we can tell from its output what the correct transposition order is. Does that sound reasonable?
PS: if the final row is partly artificial, it may be a good idea not to feed that into the transposition solver, i.e. only try to solve a 14×13. Incidentally, a very good freeware cipher solver Windows application is CryptoCrack, but more about that another day… 🙂