1. Encoding and Ciphering
Tomáš Vaníček
Faculty of Civil Engeneering CTU
Thákurova 7, Praha-Dejvice, B407

2. Ciphering (symetrical)
Eva
(Enemy)
Alice
ciphering
deciphering
Bob
key

3. Encoding Message disturbing Alice encoding Bob
decoding
Bob
Automatical repair or at least alert of transmission mistake

4. Alphabet Finite set of symbols
A={AÁBCČDEÉĚFGHIÍJKLMNŇOÓPQRŘSŠTŤUÚVWXYÝZŽ}
A={ABCDEFGHIJKLMNOPQRSTUVWXYZ} (26 symbols)
A+ - Set of all words (sequences of symbols from A).
A* - Set of all sequences from A with the empty word.

5. Cipher Cryptografical transformation (cipher) is the injection mapping
Φ: A*x K  B*,
K is the key set

6. Caesar cipher f(x)=x+k mod N Key K = 3B C D E F G H I J K L M N O P Q R S T U V W X Y Z

7. Caesar cipher f(x)=x+k mod N Key K = 3
Tomorrow morning we will cross the Rubicon.
Wrpruurz pruqlqj zh zloo furvv wkh Uxelfrq.
ABCDEFGHIJKLMNOPQRSTUVWXYZ
DEFGHIJKLMNOPQRSTUVWXYZABC

8. Multiplicative cipher f(x)=x*k mod N Key K = 3B C D E I J S Z 1 2 3 4 8 9 18 25 6 12 24 27 54 75 23 G M Y X

9. Multiplicative cipher f(x)=x*k mod N KLÍČ K = 3
ABCDEFGHIJKLMNOPQRSTUVWXYZ
ADGJMPSVYBEHKNQTWZCFILORUX

10. Multiplicative „cipher“ key K=2
A 0 → 0 A
B 1 → 2 C …
N 13 → 26 → 0 A
O 14 → 28 → 2 C
Is not a bijection

11. Multiplicative cipher
If d(K,N)-1 then there exist only one L such that K*L = 1 mod N.
For K=3 and N=26 it is L=9.
K is ciphering key and L is deciphering key.
For example w=22 is ciphered to *3 mod 26 = 14 = O
a deciphered: 14*9 mod 26 = 126 mod 26 = 22 = w

12. General Affine Cipher f(x) = K*x + P mod N, d(K,N)=1
Ciphering key is a pair K,P
Deciphering key is a pair L,Q, where L is the only number such that K * L = 1 mod N and Q= 26-P mod N.

13. General Monoalphabetical Cipher
Šciphering key is the complete function (table) of the special letter images:
ABCDEFGHIJKLMNOPQRSTUVWXYZ
VMAIVLDRHQCSYKBXGOTZPEUVFN

14. General Monoalphabetical Cipher
Tommorrow morning we will cross the Rubicon.
Tbssbiibq sbikxkz qw qxuu mibcc oew Irgxmbk. A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

15. Frequence analysis (in %)
Letter EN FR GE CZ SK
A ,96 7,68 5,52 8,99 9,49
B ,60 0,80 1,56 1,86 1,90
C ,84 3,32 2,94 3,04 3,45
D ,01 3,60 4,91 4,14 4,09
E ,86 17,76 19,18 10,13 9,16
F ,62 1,06 1,96 0,33 0,31
G ,99 1,10 3,60 0,48 0,40
H ,39 0,64 5,02 2,06 2,35
I ,77 7,23 8,21 6,92 6,81
J ,16 0,19 0,16 2,10 2,12
K ,41 0,00 1,33 3,44 3,80
L ,51 5,89 3,48 4,20 4,56

16. Frequence analysis (in %)
Letter EN FR GE CZ SK
M ,43 2,72 1,69 2,99 2,97
N ,51 7,61 10,20 6,64 6,34
O ,62 5,34 2,14 8,39 9,34
P ,81 3,24 0,54 3,54 2,87
Q ,17 1,34 0,01 0,00 0,00
R ,83 6,81 7,01 5,33 5,12
S ,62 8,23 7,07 5,74 5,94
T ,72 7,30 5,86 4,98 5,06
U ,48 6,05 4,22 3,94 3,70
V ,15 1,27 0,84 4,50 4,85
W ,80 0,00 1,38 0,06 0,06
X ,17 0,54 0,00 0,04 0,03
Y ,52 0,21 0,00 2,72 2,57
Z ,05 0,07 1,17 3,44 2,72

17. E.A. Poe: The Golden Bug
53‡‡†305))6*;4826)4‡.)4‡);806*;48†8π60))85;1‡(;:‡*8†83(88)5*†;46(;88*96*?;8)*‡(;485);5*†2:*‡(;4956*2(5*-4)8 π8*; );)6†8)4‡‡;1(‡9;48081;8:8‡1;48†85;4)485†528806*81(‡9;48;(88;4(‡?34;48)4‡;161;:188;‡?;

18. Frequence Analysis
53‡‡†305))6*;4826)4‡.)4‡);806*;48†8π60))85;1‡(;:‡*8†83(88)5*†;46(;88*96*?;8)*‡(;485);5*†2:*‡(;4956*2(5*4)8 π8*; );)6†8)4‡‡;1(‡9;48081;8:8‡1;48†85;4)485†528806*81(‡9;48;(88;4(‡?34;48)4‡;161;:188;‡?;
5 12x x x ? 3x
3 4x * 13x π 2x x
‡ 16x ; 26x x
† 8x x ( 10x
0 6x x : 4x
) 16x x x -

19. The Side Channel
By the signature (the little goat = kid = Capitain Kid) there is an information that the language of the text is English

20. Most common english letters
Most common letters in the text
E 12,86%
T 9,72%
A 7,96%
I ,77%
N 7,51%
O 6,56%
S ,56%
33x
; 26x
19x
‡ 16x
The hypothesis: 8 = E

21. Confirmation
In english there is very common bigram EE. V the text there is the bigram 88 5 times.
53‡‡†305))6*;4826)4‡.)4‡);806*;48†8π60))85;1‡(;:‡*8†83(88)5*†;46(;88*96*?;8)*‡(;485);5*†2:*‡(;4956*2(5*-4)8 π8*; );)6†8)4‡‡;1(‡9;48081;8:8‡1;48†85;4)485†528806*81(‡9;48;(88;4(‡?34;48)4‡;161;:188;‡?;

22. Try to substitute e for 8
53‡‡†305))6*;4e26)4‡.)4‡);e06*;4e†eπ60))e5;1‡(;:‡*e†e3(ee)5*†;46(;ee*96*?;e)*‡(;4e5);5*†2:*‡(;4956*2(5*-4)e πe*;40692e5);)6†e)4‡‡;1(‡9;4e0e1;e:e‡1;4e†e5;4)4e5†52ee06*e1(‡9;4e;(ee;4(‡?34;4e)4‡;161;:1ee;‡?;

23. Continue In English there is common trigram THE
In the text there is 7x the trigram ;48, so ;4e
More ; is the second most common symbol, corresponding to the letter T.
Try ; = t, 4 = h
53‡‡†305))6*;4e26)4‡.)4‡);e06*;4e†eπ60))e5;1‡(;:‡*e†e3(ee)5*†;46(;ee*96*?;e)*‡(;4e5);5*†2:*‡(;4956*2(5*-4)e πe*;40692e5);)6†e)4‡‡;1(‡9;4e0e1;e:e‡1;4e†e5;4)4e5†52ee06*e1(‡9;4e;(ee;4(‡?34;4e)4‡;161;:1ee;‡?;

24. So we have
53‡‡†305))6*the26)h‡.)h‡)te06*the†eπ60))e5t1‡(t:‡*e†e3(ee)5*†th6(tee*96*?te)*‡(the5)t5*†2:*‡(th956*2(5*-h)e πe*th0692e5)t)6†e)h‡‡t1(‡9the0e1te:e‡1the†e5th)he5†52ee06*e1(‡9thet(eeth(‡?3hthe)h‡t161t:1eet‡?t

25. Continue
53‡‡†305))6*the26)h‡.)h‡)te06*the†eπ60))e5t1‡(t:‡*e†e3(ee)5*†th6(tee*96*?te)*‡(the5)t5*†2:*‡(th956*2(5*-h)e πe*th0692e5)t)6†e)h‡‡t1(‡9the0e1te:e‡1the†e5th)he5†52ee06*e1(‡9thet(eeth(‡?3hthe)h‡t161t:1eet‡?t
Another common symbol is ‡. In the text there is 2 times the bigram ‡‡. Seems to be a letter O
53oo†305))6*the26)ho.)ho)te06*the†eπ60))e5t1o(t:o*e†e3(ee)5*†th6(tee*96*?te)*o(the5)t5*†2:*o(th956*2(5*-h)e πe*th0692e5)t)6†e)hoot1(o9the0e1te:eo1the†e5th)he5†52ee06*e1(o9thet(eeth(o?3hthe)hot161t:1eeto?t

26. Continue
53oo†305))6*the26)ho.)ho)te06*the†eπ60))e5t1o(t:o*e†e3(ee)5*†th6(tee*96*?te)*o(the5)t5*†2:*o(th956*2(5*-h)e πe*th0692e5)t)6†e)hoot1(o9the0e1te:eo1the†e5th)he5†52ee06*e1(o9thet(eeth(o?3hthe)hot161t:1eeto?t
We can guess words thirteen and the tree, so 6 = i and ( = r
53oo†305))i*the2i)ho.)ho)te0i*the†eπi0))e5t1ort:o*e†e3ree)5*†thirtee*9i*?te)*orthe5)t5*†2:*orth95i*2r5*-h)e πe*th0i92e5)t)i†e)hoot1ro9the0e1te:eo1the†e5th)he5†52ee0i*e1ro9thetreethro?3hthe)hot1i1t:1eeto?t

27. Continue
53oo†305))i*the2i)ho.)ho)te0i*the†eπi0))e5t1ort:o*e†e3ree)5*†thirtee*9i*?te)*orthe5)t5*†2:*orth95i*2r5*-h)e πe*th0i92e5)t)i†e)hoot1ro9the0e1te:eo1the†e5th)he5†52ee0i*e1ro9thetreethro?3hthe)hot1i1t:1eeto?t
Antoher common bigram is )). Corresponding to the letter S and bigram SS. We can also guess the word through, so ? = u, 3 = g
5goo†g05ssi*the2isho.shoste0i*the†eπi0sse5t1ort:o*e†egrees5*†thirtee*9i*utes*orthe5st5*†2:*orth95i*2r5*-hse πe*th0i92e5stsi†eshoot1ro9the0e1te:eo1the†e5thshe5†52ee0i*e1ro9thetreethroughtheshot1i1t:1eetout

28. Continue
5goo†g05ssi*the2isho.shoste0i*the†eπi0sse5t1ort:o*e†egrees5*†thirtee*9i*utes*orthe5st5*†2:*orth95i*2r5*-hse πe*th0i92e5stsi†eshoot1ro9the0e1te:eo1the†e5thshe5†52ee0i*e1ro9thetreethroughtheshot1i1t:1eetout
5 is common symbol but never in bigram 55, corresponding to A
In the text begining 5 = a, † = d, 0 = l, (a good glass) could be guessed
agoodglassi*the2isho.shosteli*thedeπilsseat1ort:o*edegreesa*dthirtee*9i*utes*ortheasta*d2:*orth9ai*2ra*-hse πe*thli92eastsideshoot1ro9thele1te:eo1thedeathsheada2eeli*e1ro9thetreethroughtheshot1i1t:1eetout

29. Continue
agoodglassi*the2isho.shosteli*thedeπilsseat1ort:o*edegreesa*dthirtee*9i*utes*ortheasta*d2:*orth9ai*2ra*-hse πe*thli92eastsideshoot1ro9thele1te:eo1thedeathsheada2eeli*e1ro9thetreethroughtheshot1i1t:1eetout
Other guessing
* = n, 2 = b, . = p, π = v, 1 = f
agoodglassinthebishopshostelinthedevilsseatfort:onedegreesandthirteen9inutesnortheastandb:north9ainbran-hse venthli9beastsideshootfro9thelefte:eofthedeathsheadabeelinefro9thetreethroughtheshotfif1t:feetout

30. Finishing And the treasure cold be found
a good glass in the bishops hostel in the devils seat fort: one degrees and thirteen 9inutes north east and b: north 9ain bran-h se venth li9b east side shoot fro9 the left e:e of the deaths head a bee line fro9 the tree through the shot fif1t: feet out
: = y, 9 = m, - = c
a good glass in the bishops hostel in the devils seat forty one degrees and thirteen minutes north east and by north main branch seventh limb east side shoot from the left eye of the deaths head a bee line from the tree through the shot fif1ty feet out
And the treasure cold be found
Frequence analysis (in %)

31. Coincidence Index
The method to determine wheather the text was ciphered by a monoalphabetical cipher and even to determine the original language of the text without enciphering it.

32. The frequence analysis (in czech)

33. Caesar Cipher

34. Po použití monoalfabetické šifry

35. Graf vypadá pořád stejně
Jen sloupce jsou přeházené
Jak to vyjádřit číselně?
Nabízí se rozptyl veličiny, tedy průměrná odchylka od střední hodnoty

36. Variance
Var (X) = E (X - E(X))2

37. For frequency Analysis
n*Var (p) = ∑(p(i)-1/n)2 =
= ∑p(i)2 - ∑2*p(i)/n + ∑1/n2 =
= ∑p(i)2 - 2/n + 1/n =
= ∑p(i)2 - 1/n -=

38. Coincidence Index IC(T) = ∑p(i)2 = n*var(T)+1/n
Greater or equal 1/n = 1/26 = 0,03846.
Close to a value 0,03846 for random text.
Invariant for monoalphabetial cipher.

39. Coincidence Indeces for Languages
CZ 0,0577
SK 0,0581
EN 0,0676
FR 0,0801
GE 0,0824
IT 0,0754
ES 0,0769
RU 0,0470
Random text 0,0385