Manual for the Solution of Military Ciphers

by Parker Hitt

A	B	C	D	E	F	G	H	I	J	K	L	M	N	O	P	Q	R	S	T	U	V	W	X	Y	Z
15	13	15	8	13	20	16	16	30	21	13	27	14	19	15	26	13	9	33	30	17	12	19	20	19	11

This clearly eliminates Cases 4, 5 and 6.

Referring to the recurring words and groups above noted, we figure the number of letters between each.

AII	...	AII	45	=	3×3×5
BK	...	BK	345	=	23×3×5
CT	...	CT	403		No factors
CTW	...	CTW	60	=	2×2×3×5
DL	...	DL	75	=	3×5×5
ES	...	ES	14	=	2×7
FJ	...	FJ	187		No factors
NP	...	NP	14	=	2×7
OL	...	OL	120	=	2×2×2×3×5
OS	...	OS	220	=	11×2×2×5
OSB	...	OSB	465	=	31×3×5
PO	...	PO	105	=	7×3×5
SQ	...	SQ	250	=	2×5×5×5
TLF	...	TLF	80	=	2×2×2×2×5
TP	...	TP	405	=	3×3×3×3×5
UV	...	UV	115	=	23×5
XMKU	...	XMKU	120	=	2×2×2×3×5
UV	...	UV	73		No factors
YJ	...	YJ	85	=	17×5

The dominant factor is clearly 5, so we may consider that five alphabets were used, indicating a keyword of five letters. Writing the message in lines of five letters each and making a frequency table for each of the five columns so formed, we find the following:

Frequency Tables

Colum 1		Column 2		Column 3		Column 4		Column 5
A	11	A	111111111	A	1	A	1	A	11
B		B	111	B	111	B		B	1111111
C	1111111	C	1	C	111	C	1111	C
D	11	D	11	D	1	D		D	111
E	1111	E		E	11	E	1111111	E
F	111	F		F	111111111	F	111	F	11111
G	111111111	G		G	111	G	11	G	11
H	111	H	11111	H	111	H	111	H	11
I	11	I	11	I	1111111	I	11111111111111111	I	11
J	11111	J	1	J	111111	J		J	111111111
K	111111	K	11111	K		K	1	K	1
L		L	1111111111111111111	L	11	L	11111	L	1
M		M		M	1111111	M	1111	M	111
N	1111111	N	111	N	1111	N		N	11111
O	11111	O		O	111111111	O	1	O
P	1111111	P	1111111	P	11111111	P	1111	P
Q	11111	Q		Q		Q	11	Q	111111
R		R	1	R	1	R	111111	R	1
S		S	11111111	S	111111	S	111111111111	S	1111111
T	1111111	T	111	T	11111	T	1	T	11111111111111
U	1111111	U	111	U	111111	U		U	1
V	11111	V		V	11	V	11111	V
W	111	W	1111	W		W	11111	W	1111111
X	11	X		X	1111	X	11111111	X	111111
Y	1111	Y	11111	Y		Y	111	Y	1111111
Z		Z	11111	Z	111	Z		Z	111

In the table for Column 1, the letter G occurs 9 times. Let us consider it tentatively as E. Then if the cipher alphabet runs regularly and in the direction of the regular alphabet, C (7 times) = A and the cipher alphabet bears a close resemblance to the regular frequency table. Note TUV (= RST) occurring respectively 7, 7, and 5 times and the non-occurrence of B, L, M, R, S, Z, (= Z, J, K, P, Q, and X respectively.)

In the next table, L occurs 19 times and taking it for E with the alphabet running in the same way, A=H. The first word of our message, CT, thus becomes AM when deciphered with these two alphabets and the first two letters of the key are C H.

Similarly in the third table we may take either F or O for E, but a casual examination shows that the former is correct and A=B (even if we were looking for a vowel for the next letter of the keyword).