Find the External Key of a message with help of a "mot probable" =============================================================== {Excerpt of TICOM I-175 document, but the example is mine.} Position of key wheels pins Wheel 1 (26): AB_D___HI_K_MN____ST_VW___ Wheel 2 (25): A__DE_G__JKL__O__RS_U_X__ Wheel 3 (23): AB____GH_J_LMN___RSTU_X Wheel 4 (21): __C_EF_HI___MN_P__STU Wheel 5 (19): _B_DEF_HI___LN_P__S Wheel 6 (17): AB_D___H__K__NO_Q Lugs settings: - Number of lugs front of a wheel: Wheel 26 = 2, Wheel 23 = 1, Wheel 19 = 10 Wheel 25 = 12, Wheel 21 = 5, Wheel 17 = 3 - Number of Overlaps beetween two wheels: Wheels 23-17 = 1, Wheels 21-19 = 1, Wheels 25-19 = 2 Wheels 26-19 = 1, Wheels 26-17 = 1 Cipher formula P = K -1 -C P: Plain letter, K: Key, C: Cipher letter Remarks: 1) In the German text, the "0" and "1" are represented by "-" and "+". 2) In the German text, K (Key) is named "Space". Spaces 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 26 01 01001 1 001 11 0 0 1 001 11 0 01 01 00011 01 00011 25 01 01111 0 000 00 0 0 0 000 00 0 00 10 00100 01 01101 23 01 11010 0 011 01 0 1 0 011 01 0 10 01 01101 00 10101 21 01 01111 0 000 00 1 1 1 111 11 0 00 00 00000 10 10010 19 01 01111 0 000 00 0 0 0 000 00 1 11 01 11011 10 10010 17 00 01111 0 110 11 0 0 0 110 11 0 00 00 11011 00 01100 Spaces 16 17 18 19 20 21 22 23 24 25 26 111 00011 0 1 0001 0111 1 0011 0 01 25 011 00100 1 1 1111 1111 1 1111 1 11 23 101 01001 1 0 0011 1001 1 0101 0 10 21 100 11111 1 1 0111 0011 0 0000 1 11 19 100 11011 0 0 1000 1100 1 1111 1 11 17 011 11011 0 0 0110 0011 0 1111 0 00 {begining of the exercpt:} A "mot probable" is now tried at the beginning of the message. The spaces that must have been made, if the supposition is correct, are derived from the given cipher, and the possible clear, text. A table with the possible spaces and pin positions is drawn acoordingly. E.g. Cipher Text: W U H D U A J R J Plain text: A T T A C K Z A T Spaces: 23 14 1 4 23 11 9 18 3 Wheel 26 0011 01 01001 11 0011 01 11 0 001 Wheel 25 1111 01 01111 00 1111 00 00 1 000 Wheel 23 0101 00 11010 01 0101 10 01 1 011 Wheel 21 0000 10 01111 00 0000 00 11 1 000 Wheel 19 1111 10 01111 00 1111 11 00 0 000 Wheel 17 1111 00 01111 11 1111 00 11 0 110 {Solution added by author} ? ? ? ? 1 ? ? 1 0 ? S 1 ? ? 0 1 0 0 1 0 ? ? 0 ? ? ? ? ? 1 ? G 0 ? ? 0 0 0 1 1 0 D 1 ? ? 0 1 1 0 0 0 O 1 0 ? 1 1 0 1 0 ? {Then:} ? ? ? ? 1 ? ? 1 0 ? S 1 0 1 0 1 0 0 1 0 ? ? 0 ? ? ? ? ? 1 ? G 0 1 1 0 0 0 1 1 0 D 1 1 1 0 1 1 0 0 0 O 1 0 1 1 1 0 1 0 0 {Then (with a correct vertical)} E ? 0 ? 1 ? ? 1 0 1 S 1 0 1 0 1 0 0 1 0 E ? 0 ? ? ? ? ? 1 1 G 0 1 1 0 0 0 1 1 0 D 1 1 1 0 1 1 0 0 0 O 1 0 1 1 1 0 1 0 0 {We have found the active pins at the begining: ESEGDO then, the external key is PEOPLE} When more than one pin position arrangement is possible for a space, so many tables have to be drawn, as are possible. Then the horizontal lines are compared with the sequels of pin positions for every wheel. If they are possible, i.e. if they are sections of these sequels, the table is correct, and the supposed word really stands in this part of the clear text. The sequels of pin positions are then completed sideways (or to both sides) through the whole of the message. From the vertical columns the spaces are derived. The cypher letters and the spaces belonging to them give the clear letters. The message can be read throughout. If the horizontal lines in the table are not part of the sequels of pin positions on the wheels, the table is wrong and the supposed word is not there. Then the "mot probable" has to be tried in the same way right through the message. If one "mot probable" has been tried throughout without success, others have to be tried, until the table of "0" and "1" is correct.