Manual for the Solution of Military Ciphers - Parker Hitt

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

A brief study of this table and the distribution in the cipher leads to the conclusion that B, F and C are certainly vowels and are, if the normal frequency holds, equal to E, O, and A or I. Similarly D and I are consonants and we may take them as N and T. I is taken as T because of the combination IP (=possibly TH) occurring three times. The next letter in order of frequency is A; it is certainly a consonant and may be taken as R on the basis of its frequency. Let us now try these assumptions on the first two lines of the message. We have