Manual for the Solution of Military CiphersHitt, Parker
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Manual for the Solution of Military Ciphers
Hitt, Parker
Ciphers; Cryptography
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
A A A
R E N _ O R _ E _ E N T _ _ _ _ _ N T O N _ N _
I I I
This is clearly the word REINFORCEMENTS and, using the letters thus
found, the rest of the line becomes AMMUNITIONAND. We have then the
following letters determined:
Arbitrary letters A B C D E F G H I J K L M
Plain Text R E I N F O C M T S A U D
If these be substituted we have for the message:
REINFORCEMENTS AMMUNITION AND RATIONS MUST ARRI_E _EFORE T_E
FIFTEENT_ OR _E CANNOT _O_D OUT_.
From this the remainder of the letters are determined:
Arbitrary letters N O P Q R S
Plain text V B H W L X
Now let us substitute the two-letter groups for the arbitrary letters:
Arbitrary letters K O G M B E P C R H D F A J I L N Q S
Two-letter groups GG RG AG NG TG GR AR NR GA RA AA NA RN AN NN TN GT RT AT
Plain text A B C D E F H I L M N O R S T U V W X
It is evident that the cipher was prepared with the letters of the
word GRANT chosen by means of a square of this kind:
G R A N T
G A B C D E
R F G H I K
A L M N O P
N Q R S T U
T V W X Y Z
Thus TG=E, AN=S, etc., as we have already found.
Case 9-b
Message
1950492958 3123252815 4418452815 2048115041
2252115345 5849134124 5028552526 5933195222
5245113215 6215584143 2861361265 2945565015
2342455850 6345542019 1550185311 2115415828
1124174553 4554205950 2552454132 1533492048
5018152364
An examination of the groups of two numerals each which make up this
message, shows that we have 11 to 36 and 41 to 65 with eleven groups
missing. Now the 11 to 36 combination is a very familiar one in numeral
substitution ciphers (See Case 6-c) and it will be noted that 41 to
66 would give us a similar alphabet. Let us make a frequency table
in this form:
Group Frequency Group Frequency
Public-domain text, read in full here on John Shaqi.
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