Frequency Tables

Colum 1Column 2Column 3Column 4Column 5
A11A111111111A1A1A11
BB111B111BB1111111
C1111111C1C111C1111C
D11D11D1DD111
E1111EE11E1111111E
F111FF111111111F111F11111
G111111111GG111G11G11
H111H11111H111H111H11
I11I11I1111111I11111111111111111I11
J11111J1J111111JJ111111111
K111111K11111KK1K1
LL1111111111111111111L11L11111L1
MMM1111111M1111M111
N1111111N111N1111NN11111
O11111OO111111111O1O
P1111111P1111111P11111111P1111P
Q11111QQQ11Q111111
RR1R1R111111R1
SS11111111S111111S111111111111S1111111
T1111111T111T11111T1T11111111111111
U1111111U111U111111UU1
V11111VV11V11111V
W111W1111WW11111W1111111
X11XX1111X11111111X111111
Y1111Y11111YY111Y1111111
ZZ11111Z111ZZ111

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).

In the fourth table, I is clearly E and A=E. The fifth table shows T=14 and J=9. If we take T=E we find that we would have many letters which should not occur. On the other hand, if we take J=E then T=O and in view of the many E’s already accounted for in the other columns, this may be all right. It checks as correct if we apply the last three alphabets to the second word of our message, OSB, which deciphers NOW. Using these alphabets to decipher the whole message, we find it to read:

“M. B. Am now safe on board a barge moored below Tower Bridge where no one will think of looking for me. Have good friends but little money owing to action of police. Trust, little girl, you still believe in my innocence although things seem against me. There are reasons why I should not be questioned. Shall try to embark before the mast in some outward bound vessel. Crews will not be scrutinized so sharply as passengers. There are those who will let you know my movements. Fear the police may tamper with your correspondence but later on when hue and cry have died down will let you know all.”

The key to this message is CHBEF which is not intelligible as a word but if put into figures indicating that the 2d, 7th, 1st, 4th, and 5th letter beyond the corresponding letter of the message has been used the key becomes 27145 and we may connect it with the “personal” which appeared in the same paper the day before reading:

“M. B. Will deposit £27 14s 5d tomorrow.”

Case 7-b.

Message

DDLRMERGLMUJTLLCHERSLSOEESMEJU
ZJIMUDAEESDUTDBGUGPNRCHOBEQEIE
OOACDEIOOGCOLJLPDUVMIGIYXQQTOT
DJCPJOISLYDUASIUPFNEAECOBOESHO
BETNDQXUCYLUQOYEHYDULXPEQFIXZE
PDCNZENELQMJTSQECFIEARNDNETSCF
IFQSETDDNPUUZHQCDTXQIRMERGLXBE
IQRXJFBSQDLDSVIXUMTBAEQEBYLECO
IYCUDQTPYSVOQBLULYROYHEFMOYMUY
ROYMUEQBLVUBREYGHYTQCMUBREQTOF
VSDDUDAFFSCEBSVTIOYETCLQXDVNLQ
XYTSIMZULXBAXQRECVTDETGOBCCUYF
TTNXLUNEFSIVIJRZHSBYLLTSI

On the preliminary determination, we have the following count of letters out of a total of 385:

A8L23J9
E38N11Q22
I19R14V9
O21S20X13
U24T21Z6
Total110Total89Total59
28%23%15%

Every letter except K and W occurs at least six times. We may say then that it is a substitution cipher, Spanish text, and certainly not Case 4, 5 or 6. We will now analyze it for recurring pairs or groups to determine, if it be Case 7, how many alphabets were used. The following is a complete list of such recurring groups and pairs with the number of letters intervening and the factors thereof. In work of this kind, the groups of three or more letters are always much more valuable than single pairs. For example, the groups, HOBE, OYMU, RMERGL and UBRE show, without question, that six alphabets were used. It is not necessary, as a rule, to make a complete list like the following:

AE74=2×37IE110=2×5×11RE50=2×5×5
AE120=2×2×2×3×5IM302=2×151RMERGL198=2×3×3×11
BE88=2×2×2×11IO250=2×5×5×5SC132=2×2×3×11
CD132=2×2×3×11IX78=2×3×13SD262=2×131
CFI12=2×2×3LY158=2×79SI230=2×5×23
CH36=2×2×3×3JT150=2×3×5×5SI34=2×17
CO42=2×3×7LL367 No factorsSI264=2×2×2×3×11
CO126=2×3×3×7LQ164=2×2×41SI12=2×2×3
CU114=2×3×19LQX6=2×3SL78=2×3×13
DD186=2×3×31LU124=2×2×31SQ54=2×3×3×3
DD116=2×2×29LU110=2×5×11SV27=3×3×3
DE285=5×57LU234=2×3×3×13SV63=3×3×7
DL218=2×109LX66=2×3×11SV90=2×3×3×5
DN14=2×7LXB132=2×2×3×11TD47 No factors
DQ120=2×2×2×3×5LY158=2×79TD165=3×5×11
DU36=2×2×3×3ME22=2×11TD96=2×2×2×2×2×3
DU24=2×2×2×3MU24=2×2×2×3TN239 No factors
DU38=2×19MU240=2×2×2×2×3×5TS14=2×7
DU165=3×5×11MU18=2×3×3TS156=2×2×3×13
EA30=2×3×5ND47 No factorsTSI50=2×5×5
EB78=2×3×13NE48=2×2×2×2×3UBRE12=2×2×3
EC180=2×2×3×3×5NE18=2×3×3UD60=2×2×3×5
ECO126=2×3×3×7NE192=2×2×2×2×2×2×3UDA270=2×3×3×3×5
EES14=2×7OB6=2×3UL114=2×3×19
EF105=3×5×7OB234=2×3×3×13ULX198=2×3×3×11
EI8=2×2×2OE93=3×31UY89 No factors
EI152=2×2×2×19OI144=2×2×2×2×3×3UZ162=2×3×3×3×3
EQ88=2×2×2×11OO7 No factorsVI148=2×2×37
EQ264=2×2×2×3×11OY6=2×3VT33=3×11
EQ44=2×2×11OY46=2×23XQ114=2×3×19
EQE176=2×2×2×2×11OYMU6=2×3XQ 144=2×2×2×2×3×3
ER12=2×2×3PD75=3×5×5XU99=3×3×11
ES78=2×3×13QBL24=2×2×2×3YE184=2×2×2×23
ET135=3×3×5QC95=5×19YL106=2×53
ET9=3×3QE108=2×2×3×3×3YL 144=2×2×2×2×3×3
ET54=2×3×3×3QE68=2×2×17YM6=2×3
ET31 No factorsQR132=2×2×3×11YRO12=2×2×3
HE245=5×7×7QTO210=2×3×5×7ZE6=2×3
HOBE66=2×3×11QX198=2×3×3×11ZH183=3×61

Out of one hundred and one recurring pairs we have fifty with the factors 2×3=6; out of twelve recurring triplets, nine have these factors; and the four recurring groups of four or more letters all have these factors. The percentages are respectively 49.5%, 75% and 100% and we may be certain from this that six alphabets were used. But, before the six frequency tables are made up, there is one more point to be considered; why are there so many recurring groups which do not have six as a factor? The answer is that one or more of the alphabets is repeated in each cycle; that is, a key word of the form HAVANA has been used. If this were the key word, the second, fourth and sixth alphabets would be the same. We will see later that in this example the second and sixth alphabets are the same and this introduces the great number of recurring groups without the factor 6.

We will now proceed to make a frequency table for each alphabet. As the message is written in thirty columns, we take the first, seventh, thirteenth, etc., as constituting the first alphabet; the second, eighth, fourteenth, etc., as constituting the second alphabet and so on. The prefix and suffix letter is noted for each occurrence of each letter. The importance of this will be appreciated when the form of the frequency tables is examined. None bears any resemblance to the normal frequency table except that each is evidently a mixed up alphabet. The numbers after “Prefix” and “Suffix” refer to the alphabet to which these belong, for convenience in future reference.