EXAMINATION OF SUBSTITUTION CIPHERS
When an unknown cipher has been put into the substitution class by
the methods already described we may proceed to decide on the variety
of substitution cipher which has been used.
There are a few purely mechanical ways of solving some of the simple
cases of substitution ciphers but as a general rule some or all of
the following determinations must be made:
1. By preparation of a frequency table for the message we determine
whether one or more substitution alphabets have been used and, if
one only has been used, this table leads to the solution.
2. By certain rules we determine how many alphabets have been used,
if there are more than one, and then isolate and analyze each alphabet
by means of a frequency table.
3. If the two preceding steps give no results we have to deal with a
cipher with a running key, a cipher of the Playfair type, or a cipher
where two or more characters are substituted for each letter of the
text. Some special cases under this third head will be given but,
in general, military ciphers of the substitution class will usually
be found to come under the first two heads, on account of the time
and care required in the preparation and deciphering of messages by
the last named methods and the necessity, in many cases, of using
complicated machines for these processes.
Case 4-a.
Message
OBQFO BPBRP QBAML OBHIF PILFQ FJBOX OFLNR BIXOZ EL
From the recurrence of B, F and O, we may conclude that a single
substitution alphabet was used for this message. If so and if the
alphabet runs in the same order and direction as the regular alphabet,
the simplest way to discover the meaning of the message is to take
the first two words and write alphabets under each letter as follows,
until some line makes sense:
O B Q F O B P B R P
P C R G P C Q C S Q
Q D S H Q D R D T R
R E T I R E S E U S
The word RETIRESE occurs in the fourth line, and, if the whole message
be handled in this way we find the rest of the fourth line to read
USTED POR EL MISMO ITINERARIO QUE MARCHO. The message was enciphered
using an alphabet where A = X, B = Y, C = Z, D = A, etc. noting that
as this message is in Spanish the letters K and W do not appear in
the alphabet.
Case 4-b.
Message
HUJZH UIUPN OZYTS VQXMI SMOMX MQHUD UMREI SESJU AG
This is a message in Spanish. We will handle it as in case 4-a,
setting down the whole message.
HUJZHUIU PNOZY TSV QX MISMO MXMQHUDUMR EIS ESJUAG
IVLAIVJV QOPAZ UTX RY A=A NYNRIVEVNS FJT FTLVBH
JXMBJXLX RPQBA VUY SZ OZOSJXFXOT GLU GUMXCI
LYNCLYMY SQRCB XVZ TA PAPTLYGYPU HMV HVNYDJ
MZODMZNZ TRSDC YXA UB QBQUMZHZQV INX IXOZEL
NAPENAOA USTED ZYB VC RCRVNAIARX JOY JYPAFM
OBQFOBPB A=U AZC XD SDSXOBJBSY LPZ LZQBGN
PCRGPCQC BAD YE TETYPCLCTZ MQA MARCHO
QDSHQDRD CBE ZF UFUZQDMDUA NRB A=S
RETIRESE DCF AG VGVARENEVB OSC
A=Q EDG BH XHXBSFOFXC PTD
FEH CI YIYCTGPGYD QUE
GFI DJ ZJZDUHQHZE A=O
HGJ EL ALAEVIRIAF
IHL A=M BMBFXJSJBG
JIM CNCGYLTLCH
LJN DODHZMUMDI
MLO EPEIANVNEJ
NMP FQFJBOXOFL
ONQ GRGLCPYPGM
POR HSHMDQZQHN
A=E ITINERARIO
A=D
Here each word of the message comes out on a different line, and
noting in each case the letter corresponding to A, we have the word
QUEMADOS which is the key. The cipher alphabet changed with each word
of the message.
A variation of this case is where the cipher alphabet changes according
to a key word but the change comes every five letters or every ten
letters of the message instead of every word. The text of the message
can be picked up in this case with a little study.
Note in using case 4 that if we are deciphering a Spanish message
we use the alphabet without K or W as a rule, altho if the letters
K or W appear in the cipher it is evidence that the regular English
alphabet is used.
Case 5-a.
Message
DNWLW MXYQJ ANRSA RLPTE CABCQ RLNEC LMIWL XZQTT QIWRY ZWNSM
BKNWR YMAPL ASDAN
This message contains K and W and therefore we expect the English
alphabet to be used. The frequency of occurrence of A, L, N, R and
W has lead us to examine it under case 4 but without result. Let us
set down the first two words and decipher them with a cipher disk
set A to A and then proceed as in case 4.
Cipher message DNWLWMXYQJ
Deciphered A to A XNEPEODCKR
B YOFQFPEDLS
C ZPGRGQFEMT
D AQHSHRGFNU
E BRITISHGOV
The message is thus found to be enciphered with a cipher disk set
A to E and the text is: BRITISH GOVERNMENT PLACED CONTRACTS WITH
FOLLOWING FIRMS DURING SEPTEMBER.
Case 5-b.
Same as case 4-b except that the cipher message must be deciphered
by means of a cipher disk set A to A before proceeding to make up
the columns of alphabets. The words of the deciphered message will
be found on separate lines, the lines being indicated as a rule by
a key word which can be determined as in case 4-b.
The question of alphabetic frequency has already been discussed in
considering the mechanism of language. It is a convenient thing to put
the frequency tables in a graphic form and to use a similar graphic
form in comparing unknown alphabets with the standard frequency
tables. For instance the standard Spanish frequency table put in
graphic form is here presented in order to compare with it the
frequency table for the message discussed in case 4-a.
Standard Spanish Table for Message
frequency table Case 4-a
A 111111111111111111111111111 27 A 1 1
B 11 2 B 1111111 7
C 111111111 9 C
D 1111111111 10 D
E 1111111111111111111111111111 28 E 1 1
F 11 2 F 11111 5
G 111 3 G
H 11 2 H 1 1
I 111111111111 12 I 111 3
J 1 1 J 1 1
L 1111111111 10 L 111 3
M 111111 6 M 1 1
N 111111111111 12 N 1 1
O 1111111111111111 16 O 111111 6
P 11111 5 P 111 3
Q 11 2 Q 111 3
R 111111111111111 15 R 11 2
S 11111111111111 14 S
T 11111111 8 T
U 1111111 7 U
V 11 2 V
X X 11 2
Y 11 2 Y
Z 1 1 Z 1 1
Our first assumption might be that B = A and F = E but it is evident
at once that in that case, S, T, U and V (equal to R, S, T and U)
do not occur and a message even this short without R, S, T or U is
practically impossible. By trying B = E we find that the two tables
agree in a general way very well and this is all that can be expected
with such a short message. The longer the message the nearer would
its frequency table agree with the standard table. Note that if a
cipher disk has been used, the alphabet runs the other way and we
must count upward in working with a graphic table. Note also that if,
in a fairly long message, it is impossible to coördinate the graphic
table, reading either up or down, with the standard table and yet some
letters occur much more frequently than others and some do not occur
at all, we have a mixed alphabet to deal with. The example chosen for
case 6-a is of this character. An examination of the frequency table
given under that case shows that it bears no graphic resemblance to the
standard table. However, as will be seen in case 7-b, the preparation
of graphic tables enables us to state definitely that the same order
of letters is followed in each of a number of mixed alphabets.
General Remarks
Any substitution cipher, enciphered by a single alphabet composed of
letters, figures or conventional signs, can be handled by the methods
of case 6. For example, the messages under case 4-a and 5-a are easily
solved by these methods. But note that the messages under case 4-b
and 5-b cannot so be solved because several alphabets are used. We
will see later that there are methods of segregating the different
alphabets in some cases where several are used and then each of the
alphabets is to be handled as below.
Case 6-a.
Message
QDBYP BXHYS OXPCP YSHCS EDRBS ZPTPB BSCSB PSHSZ AJHCD OSEXV HPODA
PBPSZ BSVXY XSHCD
This message was received from a source which makes us sure it is in
Spanish. The occurrence of B, H, P and S has tempted us to try the
first two words as in case 4 and 5 but without result. We now prepare
a frequency table, noting at the same time the preceding and following
letter. This latter proceeding takes little longer than the preparation
of an ordinary frequency table and gives most valuable information.
Frequency Table
Prefix Suffix
A 11 2 ZD JP
B 11111111 8 DPRPBSPZ YXSBSPPS
C 11111 5 PHSHH PSSDD
D 11111 5 QECOC BROA
E 11 2 SS DX
F
G
H 111111 6 XSSJVS YCSCPC
I
J 1 1 A H
L
M
N
O 111 3 SDP XSD
P 111111111 9 YXCZTBHAB BCYTBSOBS
Q 1 1 D
R 1 1 D B
S 111111111111 12 YYCBBCPHOPBX OHEZCBHZEZVH
T 1 1 P P
U
V 11 2 XS HX
X 11111 5 BOEVY HPVYS
Y 1111 4 BHPX PSSX
Z 111 3 SSS PAB
It is clear from an examination of this table that we have to deal
with a single alphabet but one in which the letters do not occur in
their regular order.
We may assume that P and S are probably A and E, both on account of
the frequency with which they occur and the variety of their prefixes
and suffixes. If this is so, then B and H, are probably consonants
and may represent R and N respectively. D and X are then vowels by
the same method of analysis. Noting that HC occurs three times and
taking H as N we conclude that C is probably T. Substitute these
values in the last three words of the message because the letters
assumed occur rather frequently there.
PBPSZBSVXYXSHCD
I I I
ARAE_RE_ _ ENT
O O O
Now Z is always prefixed by S and may be L. Taking X=I and D=O,
(they are certainly vowels), V=G and Y=M, we have
ARA EL REGIMIENTO
Substituting these values in the rest of the message we have
Q DBYPBXH YSOXPCP YSHCSED RBSZPTPB
_ ORMARIN ME_IATA MENTE_O _RELA_AR
BSCSB PSHSZ AJHCD OSEXVHPODA
RETER AENEL __NTO _E_IGNA_O_
We may now take Q=F, O=D, E=S, R=B, T=C, A=P and J=U and the message
is complete. We are assisted in our last assumption by noting that
S=E and E=S, etc., and we may on that basis reconstruct the entire
alphabet. The letters in parenthesis do not occur in the message but
may be safely assumed to be correct.
Ordinary A B C D E F G H I J L M N O P Q R S T U V X Y Z
Cipher P R T O S (Q) (V) N (X) (U) (Z) (Y) (H) D A F B E C J G I M L
It is always well to attempt the reconstruction of the entire
alphabet for use in case any more cipher messages written in it are
received.----
Case 6-b.
Message
Lt. J. B. Smith, Royal Flying Corps, Calais, France.
DACFT RRBHA MOOUE AENOI ZTIET
ASMOS EOHIE YOCKF NOHOE NOUTH
OMEAH NILGO OSAHU OHOUE APCHS
TLNDA CFTEN INTWN BAFOH GROHT
AEIOH ABRIS ODACF TRREN OSTSM
AYBIS DFTEN EFAPH OSMNI ZTIEA
HLILL TWSOU GDENO UTHOM EAHBH
AMOOU EAYOE QISUU OLEHA DENOE
NHOOQ OBBOR TSLHO BAHEO UBHOB
IHTSW ENOHO PAHIH ITUAS BIHTL
Graham-White.
The address and signature indicate that this message is in English.
There are 250 letters in the cipher; the vowels AEIOU occur 109 times
or 43.6%, the letters LNRST occur 62 times or 24.8%, and the letters
KQVXZ occur 5 times or 2%. The proportion in the case of the vowels
is somewhat too large and, in the case of the letters LRNST, it is
too small. It is then questionable whether this is a transposition
cipher altho, at first glance it might appear to be one.
On examination for parts of possible words we are at once struck by
the occurrence at irregular intervals of recurring groups, viz:
DACFTRR ENO BHAMOOUEA
DACFTEN ENOUTHOMEAH BHAMOOUEA
ENO
DACFTRR DENOUTHOMEAH IZTIE
FTEN DENO IZTIE
ENO
This is a strong indication that the cipher is a substitution cipher,
so, to make an examination a frequency table will be constructed.
Frequency Table
A B C D E F G H I J K L M
23 11 7 6 24 7 3 26 16 0 1 8 8
N O P Q R S T U V W X Y Z
15 36 3 2 8 14 17 11 1 3 0 3 2
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 cipher ONTAHECFD
Letters of message EHIORTUWY
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 cipher ABCDEFGHIKLMNOPQRSTUVWYZ
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&*x4 6°*°& %"4&"
By arbitrary substitution of letters this is made
ABBCD EFGHF IJKDL DDHGE MDEDM
NGBDA KDOGE NPOPD MAEDA
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
1723 3223 2825 1828 3630 2336 1423 2827 2324 3120 2317 3123
3036 2120 2415 3029 1512 2831 1721 2715 2811 2715 1923 3030
1215 1130 2128 3623
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:
AB CB DE FD GH BG IB DJ BK LM BA LB
HG NM OP HQ PR DL AN JP DS JP TB HH
RP SH ND GB
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.