Example 4: Decrypting Substitution Cipher
Suppose the following message was enciphered using a substitution cipher.
WE WHCVE VZE IEVVEHK PB D QEMWPHY WCVZPRV HEJEVCVCPSK CS
PHYEH PB DJJEDHDSNE
The single letter count is:
Letter Count: E = 11; V = 8; H = 7; P = 6; C = 5; D = 4; W = 4; J = 3; S = 3; K = 2; B =
2; Y = 2; Z = 2; N = 1; Q = 1; R = 1; I = 1; M = 1; O = 0; T = 0; U = 0; A = 0; G = 0; X =
0; L = 0; F = 0;
The two letter count is:
Letter Count: CV = 3; VE = 2; VZ = 2; EV = 2; EH = 2; PB = 2; PH = 2; HY = 2; JE = 2;
VC = 2; WH = 1; HK = 1; HC = 1; QE = 1; EM = 1; MW = 1; WP = 1; ZE = 1; IE = 1;
WC = 1; ZP = 1; PR = 1; RV = 1; HE = 1; EJ = 1; WE = 1; VV = 1; CP = 1; PS = 1; SK
= 1; CS = 1; YE = 1; DJ = 1; JJ = 1; ED = 1; DH = 1; HD = 1; DS = 1;
and the three letter count is:
Letter Count: PHY = 2; HCV = 1; CVE = 1; VZE = 1; IEV = 1; EVV = 1; VVE = 1;
VEH = 1; EHK = 1; QEM = 1; EMW = 1; MWP = 1; WPH = 1; WHC = 1; WCV = 1;
CVZ = 1; VZP = 1; ZPR = 1; PRV = 1; HEJ = 1; EJE = 1; JEV = 1; EVC = 1; VCV = 1;
CVC = 1; VCP = 1; CPS = 1; PSK = 1; HYE = 1; YEH = 1; DJJ = 1; JJE = 1; JED = 1;
EDH = 1; DHD = 1; HDS = 1; DSN = 1; Solution:
1. From the one letter count it appears E is itself.
2. -E ----E --E -E--E-- -- - -E----- ------- -E-E------- -- ---E- -- ---E-----E
3. Let’s assume V is T: -E ---TE T-E -ETTE-- -- - -E----- --T---T -E-ET-T---- -- --E- -- ---E-----E
4. A two letter word that ends in E, BE, HE, WE. (To save time) Set W to W.
5. For BE would imply W is B: WE W---E --E -E--E-- -- - -E-W--- W------ -E-E------ -- ---E- -- ---E-----E
6. T-E suggests that Z is H: -E ---TE THE -ETTE-- -- - -E----- --TH--T -E-ET-T----- ---E- -- ---E-----E
7. After T in frequency should be A, Perhaps H is A: -E -A-TE THE -ETTEA- -- - E---A- --TH--T AE-ET-T---- -- -A-EA -- ---E-A---E. (After some work we come
to the conclusion that this is bad)
8. How about H is R. -E -R-TE THE -ETTER- -- - -E---R- --TH--T RE-ET-T---- -- R-ER -- ---E-R---E
9. –ETTER- suggests that K is S: -E -R-TE THE -ETTERS -- - -E---R- --TH--T REET-T---S -- -R-ER -- ---E-R---E
10. -ETTERS suggest I is L. -E -R-TE THE LETTERS -- - -E---R- --TH--T RE-ETT---S -- -R-ER -- ---E-R---E
11. –R-TE suggests Write: So W is W and C is I: WE WRITE THE LETTERS -- - E-W-R- WITH--T RE-ETITI--S I- -R-ER -- ---E-R---E
12. WITH—T suggests OU. Therefore, PR is OU: WE WRITE THE LETTERS O- -E-WOR- WITHOUT RE-ETITIO-S I- OR-ER O- ---E-R---E
13. O- is either OF or ON. Try F. Which means B is F.
14. The only singleton word is A. So D is A: WE WRITE THE LETTERS OF A -EWOR- WITHOUT RE-ETITIO-S I- OR-ER OF A--EARA—E
15. OR-ER suggests Y is D: WE WRITE THE LETTERS OF A -E-WORD
WITHOUT RE-ETITIO-S I- ORDER OF A--EARA—E
16. Two letters that appear often together are LL and PP. We have already used L so
lets try P. J is P: WE WRITE THE LETTERS OF A -E-WORD WITHOUT
REPETITIO-S I- ORDER OF APPEARA—E
17. Solution: WE WRITE THE LETTERS OF A KEYWORD WITHOUT
REPETITIONS IN ORDER OF APPEARANCE
Answer: The keyword is: decipher it
D E C I P H R T
A B F G J K L M
N O Q S U V W X
Y Z
A
D
B
A
C
N
D
Y
E
E
F
B
G
O
H
Z
I
C
J
F
K
Q
L
I
M
G
N
S
O
P
P
J
Q
U
R
H
S
K
T
V
U
R
V
L
W
W
X
T
Y
M
WE WRITE THE LETTERS OF A KEYWORD WITHOUT REPETITIONS IN ORDER
OF APPEARANCE
Z
X