Case 9 can be recognized by some or all of the following points; the number of characters in the cipher is always an even number; often only a few, say five to ten, of the letters of the alphabet appear; either a frequency table for pairs of the cipher text resembling the normal single letter frequency table can be made, or groups of four letters will show a regular recurrence, from which the cipher can be solved as in Case 7. Case 9a.-- Message RNTGN RAAGR NARNA GTGRA TGAAN NANGG RARAT NAANR NNNRN AAAGG AANGR NGGNN NRNAA AANRA TNANN NGGRN RNNRG TTGRG TGGRN ARNTG NNART GGRNR GRNNT GTGAA NNARN ARNRT TGAGG GAAAA NANNA RNAGA NGNAT NNNAT This message contains 160 letters and it will be noted that the only letters used are A, G, N, R and T. We may expect a simple two-letter substitution cipher at once. It will simplify the work if we divide the cipher into groups of two letters and then, if we find there are 26 or less recurring groups, to assign an arbitrary letter to each group and work out the cipher by the method of Case 6. RN TG NR AA GR NA RN AG TG RA TG AA NN AN GG RA RA TN AA NR NN NR NA AA GG AA NG RN GG NN NR NA AA AN RA TN AN NN GG RN RN NR GT TG RG TG GR NA RN TG NN AR TG GR NR GR NN TG TG AA NN AR NA RN RT TG AG GG AA AA NA NN AR NA GA NG NA TN NN AT With arbitrary letters substituted, we have A B C D E F A G B H B D I J K H H L D C I C F D K D M A K I C F D J H L J I K A A C N B O B E F A B I P B E C E B B D I P F A Q B G K D D F I P F R M F L I S Now, preparing a frequency table, with note of prefixes and suffixes we have: Frequency Prefix Suffix A 7 1111111 FMKAFF BGKACBQ B 10 1111111111 AGHNOAPIBQ CHDOEIEBDG C 6 111111 BDIIAE DIFFNE D 9 111111111 CBLFKFBKD EICKMJIDF E 4 1111 DBBC FFCI F 8 11111111 ECCEPDPM ADDAAIRL G 2 11 AB BK H 4 1111 BKJH BHLL I 9 111111111 DCKJBEDFL JCCKPBPP J 3 111 IDL KHI K 5 11111 JDAIG HDIAD L 3 111 HHF DJI M 2 11 DR AF N 1 1 C B O 1 1 B B P 3 111 III BFF Q 1 1 A B R 1 1 F M S 1 1 I