Classic Cryptology. Modified version by Dr M. Sakalli Marmara University.
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Classic Cryptology
Modified version by Dr M. Sakalli
Marmara University
Cipher message:
53++!305))6*;4826)4+.)4+);806*;48!8`60))85;]8*:+*8!83(88)5*!;
46(;88*96*?;8)*+(;485);5*!2:*+(;4956*2(5*-4)8`8*; 4069285);)6
!8)4++;1(+9;48081;8:8+1;48!85;4)485!528806*81(+9;48;(88;4(+?3
4;48)4+;161;:188;+?;
e t a o i n s r h l d c u m f p g w y b v k x j q z
sfiilfcsoorntaeuroaikoaiotecrntaeleyrcooestvenpinelefheeosnlt
arhteenmrnwteonihtaesotsnlupnihtamsrnuhsnbaoeyentacrmuesotorl
eoaiitdhimtaecedtepeidtaelestaoaeslsueecrnedhimtaetheetahiwfa
taeoaitdrdtpdeetiwt
53++!305))6*the26)h+.)h+)te06*the!e`60))e5t]e*:+*e!e3(ee)5*!t
h6(tee*96*?te)*+(the5)t5*!2:*+(th956*2(5*h)e`e*th0692e5)t)6!e
)h++t1(+9the0e1te:e+1the!e5th)he5!52ee06*e1(+9thet(eeth(+?3hthe)h+t161t:1eet+?t
AGOODGLASSINTHEBISHOPSHOSTELINTHEDEVILSSEATWENYONEDEGRE
ESANDTHIRTEENMINUTESNORTHEASTANDBYNORTHMAINBRANCHSEVENT HLIMBEASTSIDESHOOTFROMTHELEFTEYEOFTHEDEATHSHEADABEELINE
FROMTHETREETHROUGHTHESHOTFIFTYFEETOUT
A GOOD GLASS IN THE BISHOP’S HOSTEL IN THE DEVIL’S SEA,
TWENY ONE DEGREES AND THIRTEEN MINUTES NORTHEAST AND BY NORTH,
MAIN BRANCH SEVENTH LIMB, EAST SIDE, SHOOT FROM THE LEFT EYE OF
THE DEATH’S HEAD A BEE LINE FROM THE TREE THROUGH THE SHOT,
FIFTY FEET OUT.
cctgatagacgctatctggctatccacgtacgtaggtcctctgtgcgaatctatgcgtttccaaccatagtactggtgtacatttgatacgtacgtacaccggcaacctgaaacaaacgctcagaaccagaagtgcaaacgtacgtgcaccctctttcttcgtggctctggccaacgagggctgatgtataagacgaaaattttagcctccgatgtaagtcatagctgtaactattacctgccacccctattacatcttacgtacgtatacactgttatacaacgcgtcatggcggggtatgcgttttggtcgtcgtacgctcgatcgttaacgtacgtc
Y is the cipher text, where K = [k1, k2, ..., kj] are the set of keys, and each y is the member of finite cipher set.
X= D(E(X, K), K)
PrmDm kK{m=g|E(m,K)=c} = PrmDM{m=g}..
meet me after the toga party
PHHW PH DIWHU WKH WRJD SDUWB
"GCUA VQ DTGCM"
Y = Ea,b(x) = (a* xi + b) %m
X=D(y) = (a-1 *(yi-b)) %m
If (a-1%m) does exist which means some numbers cannot be included.. .
then a-1= 21..
E(p1+p2) = a(p1+p2)+b modm, not linear
E(p1) + E(p2) = a(p1+p2)+2b modm
K(m1, b1), K(m2, b2) K(m1m2, m2b1+b2),
Give a proof!!.
xy == 1 mod m
EXTENDED EUCLID(m, b)
(B1, B2, B3)=(0, 1, b)
2. if B3 = 0
return A3 = gcd(m, b); no inverse
3. if B3 = 1
return B3 = gcd(m, b); B2 = b–1 mod m
4. Q = A3 div B3
5. (T1, T2, T3)=(A1 – Q B1, A2 – Q B2, A3 – Q B3)
6. (A1, A2, A3)=(B1, B2, B3)
7. (B1, B2, B3)=(T1, T2, T3)
8. goto 2
VUEPHZHMDZSHZOWSFPAPPDTSVPQUZWYMXUZUHSX
EPYEPOPDZSZUFPOMBZWPFUPZHMDJUDTMOHMQ
it was disclosed yesterday that several informal but
direct contacts have been made with political
representatives of the vietcong in moscow
MONAR
CHYBD
EFGIK
LPQST
UVWXZ
C1 k11 k12k13 P1
C3 k31k32k33 P3
Φr = 0.0385 N (N – 1). N is the length of the text.
Φo = ΦA + ΦB + …+ ΦZ = Σf(f-1)
And the index of coincidences for phi test ΔIC= Φo/Φr
Φ2r = 0.0015 N (N – 1),
Φ2p= 0.0069 N (N – 1),
Φ 2o = Σf(f-1)
And the index of coincidences for digraphic phi test
ΔIC2p= Φo/Φr
Remember xor is lossless..
xi =D(yi) = (E(xi) - ki)mod(m).
4 (E) 16 (Q) 13 (N) 21 (V) 25 (Z) ciphertext
- 23 (X) 12 (M) 2 (C) 10 (K) 11 (L) key
= -19 4 11 11 14 ciphertext - key
= 7 (H) 4 (E) 11 (L) 11 (L) 14 (O) (ciphertext - key) (mod 26)
>> plaintext
http://en.wikipedia.org/wiki/One-time_pad
yi =E(xi) = (xi + ki)mod(m) where y is a random sequence.
7 (H) 4 (E) 11 (L) 11 (L) 14 (O) message
+ 23 (X) 12 (M) 2 (C) 10 (K) 11 (L) key
= 30 16 13 21 25 message + key
= 4 (E) 16 (Q) 13 (N) 21 (V) 25 (Z) (message + key) (mod (26)
>> ciphertext
http://en.wikipedia.org/wiki/One-time_pad
m e m a t r h t g p r y
e t e f e t e o a a t
MEMATRHTGPRYETEFETEOAAT
o s t p o n e
d u n t i l t
w o a m x y z
t t n a a p t
m t s u o a o
d w c o i x k
n l y p e t z
01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28
03 10 17 24 04 11 18 25 02 09 16 23 01 08 15 22 05 12 19 26 06 13 20 27 07 14 21 28
17 09 05 27 24 16 12 07 10 02 22 20 03 25 15 13 04 23 19 14 11 01 26 21 18 08 06 28
46
Types of Attacks