Frequency Table

ABCDEFGHIJKLMNOPQRSTUVWXYZ
231176247326160188153632814171113032

Superficially, this looks like a normal frequency table, but O is the dominant letter, followed by H, E, A, T, I, N, S, in the order named. It is certainly Case 6 if it is a substitution cipher at all.

Let us see what can be done by assuming O=E; the triplet ENO, occurring six times might well be THE and E=T and N=H. A glance at the frequency table shows this to be reasonable. Now substitute these letters in some likely groups. FNOHOENO becomes _HE_ETHE; FTEN becomes _TH; ENOENHO becomes THETH_E; ENOHO becomes THE_E. A bit of study will show that F=W, T=I and H=R and the frequency table bears this out except that H(=R) seems to occur too frequently. The recurring groups containing DAC (see above) occur in such a way that we may be sure DAC is one word, FTRR is another and FTEN(=WITH) is a third. Now FTRR becomes WI__, which can only be completed by a double letter. LL fills the bill and we may say R=L. As DAC starts the message and is followed by FTRR (=WILL) it is reasonable to try DAC=YOU. Looking up DAC in the frequency table it is evident that we strain nothing by this assumption. We now have:

Letters of cipherONTAHECFD
Letters of messageEHIORTUWY

Now take the group ENOUTHOMEAH which occurs twice. This becomes THE_IRE_TOR and if we substitute U=D and M=C we have THE DIRECTOR. Next the group (FTRR)BHAMOOUEA becomes (WILL) _ROCEEDTO and the context gives word with missing letter as PROCEED, from which B=P. Next the group (ENO) IZTIETASMOSEOHIEYOCK(FNOHO) becomes (THE)__I_TIO_CE_TER_T_EU_(WHERE) and the group (FTEN)EFAPHOSMNIZTIEAHL becomes (WITH)TWO_RE_CH__I_TOR_. The substitution of A for I, V for Z, N for S and F for P makes the latter group read (WITH TWO FRENCH AVIATORS and the former read (THE)AVIATION CENTER AT _EU_(WHERE).

Now the word YOCK = (_EU_) is the name of a place, evidently. We find another group containing Y, viz: ENOSTSMAYBISD which becomes THENINCO_PANY so that evidently we should substitute M for Y. The other occurrence of Y (=M) is in the group EAYOEQISU which becomes TOMET_AND. A reasonable knowledge of geography gives us the words MEUX and METZ so that X should be substituted for K and Z for Q.

We now have sufficient letters for a complete deciphering of the message.

Letters of cipherABCDEFGHIKLMNOPQRSTUVWYZ
Letters of message OPUYTW_RAXSCHEFZLNID__MV

The message deciphers:

YOU WILL PROCEED TO THE AVIATION CENTER AT MEUX WHERE THE DIRECTOR HAS _EEN ORDERED TO FURNISH YOU WITH A HI_H POWER _LERIOT AEROPLANE. YOU WILL THEN IN COMPANY WITH TWO FRENCH AVIATORS ASSI_NED _Y THE DIRECTOR PROCEED TO METZ AND DESTROY THE THREE ZEPPELINS REPORTED PREPARIN_ THERE FOR A RAID ON PARIS.

The substitution of B for G, G for W and K for V completes the cipher. This cipher is difficult only because the cipher alphabet is made up, not haphazard, but scientifically with proper consideration for the natural frequency of occurrence of the letters. In cipher work it is dangerous to neglect proper analysis and jump at conclusions.

In the study of Mexican substitution ciphers, several alphabets have been found which are made up in a general way, like the one discussed in this case.

Case 6-c.—It is a convenience in dealing with ciphers made up of numbers or conventional signs to substitute arbitrary letters for the numbers and signs. Suppose we have the message:

”??2&45x15)“8&#&&1x4%&4&%
6x?&”8&*x46°*°&%“4&”

By arbitrary substitution of letters this is made

ABBCDEFGHFIJKDLDDHGEMDEDM
NGBDAKDOGENPOPDMAEDA

This message is now in convenient shape to handle as Case 6-a and on solution is found to read:

ALL PERSONS HAVE BEEN ORDERED TO LEAVE FORTIFIED AREA.

In the same way the message

172332232825182836302336142328272324312023173123
303621202415302915122831172127152811271519233030
1215113021283623

is found to be made up entirely of numbers between 11 and 36 with the numbers 23, 28 and 30 occurring most frequently. This immediately suggests an alphabet made up of the numbers from 11 to 36 inclusive and each cipher group of figures represents two letters. By arbitrary substitution of letters for groups of two numbers we obtain:

ABCBDEFDGHBGIBDJBKLMBALB
HGNMOPHQPRDLANJPDSJPTBHH
RPSHNDGB

and this message is also in shape to handle as Case 6-a. It reads, on solution,

SEVEN HUNDRED MEN LEFT YESTERDAY FOR POINTS ON LOWER RIO GRANDE.