CS 608 Quiz 1 Spring 2019: Cryptography - Ciphers and Equations

This document provides a detailed solution to a CS 608 quiz, covering various aspects of cryptography. The solution includes encryption of the plaintext "Peter Bayiokos" using a substitution cipher (Caesar cipher), permutation cipher with the keyword “Newark”, and PlayFair cipher with the keyword “MilitarPark”. It also involves decryption of the ciphertext in reverse order. Furthermore, the solution includes solving a modular equation (27*a*x+131221) mod 9743=310 to find the unknown x and computing x ≡ 862318691a mod 1997 using the smallest number of operations. Finally, the document performs cryptoanalysis on a given ciphertext to determine the cipher used, which involves row and column shifts.

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Table of Contents
QUESTION 1...................................................................................................................................................... 3
QUESTION 2...................................................................................................................................................... 7
QUESTION 3...................................................................................................................................................... 7
QUESTION 4...................................................................................................................................................... 8
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Question 1
a)
a = num (F2) + 3* num (L1) +7
Where num(x) is the Latin alphabets (0-25)
F2 > second character of first name
L1> First character of last name
Encrypt plain text “Peter Bayiokos”
Substitution Cipher
A>0, B>1,C>2, D>3,E>4,F>5, G>6, H>7,I>8 J>9,K>10, L>11, M>12, N>13, O>14, P>15,
Q>16, R>17,S>18,T>19,U>20,V>21,W>22,X>23,Y>24,Z>25.
Caesar cipher,
P=Peter Bayiokos
K=3
C=(p+k) mod 26
C= 18 mod 26
C=18 = S
C=S
C=(p+K) mod 26
C=(4+3) mod 26
C=7 mod 26
C=7=H
C=(19+3) mod 26
C=22 mod 26
C=22=W
P=(4+3) mod 26
= 7 mod 26
= 7 = H
R =(17+3) mod 26
=20 mod 26
= 20 = U
B=(1+3) mod 26
= 4 mod 26
= 4=E
3

A=(0+1) mod 26
= 1 mod 26
= 1= B
Y=(24+3) mod 26
=27 mod 26
=1 = B
I=(8+3) mod 26
=21 mod 26
=5 = F
O=(14+3) mod 26
=17 mod 26
=17 = R
K=(10+3) mod 26
=13 mod 26
=2 = C
O=(14+3) mod 26
=17 mod 26
=17 = R
S=(18+3) mod 26
=21 mod 26
=5 = F
Description P = (c-k) mod 26
=(18-3) mod 26
=15 mod 26
=15
=P
The encrypt cipher text value is SHWHU EBBFRCRF
Inverse Multiplication
a=13*mod 26
=13 mod 26
=13-1 mod 26
=13 mod 26
=13*2 mod 26
4

13 mod 26
q r1 r2 r t1 t2 t
3 73 13 13 0 1 -3
1 20 13 7 1 -3 4
1 13 7 6 -3 4 -7
1 7 6 1 4 -7 11
6 6 1 0 -7 11 -73
1 0 13 -73
k-1 of 13 domain 26 = 13
A= 13* mod 26
14 mod 26
q r1 r2 r t1 t2 t
3 14 3 2 0 1 -2
1 3 2 7 7 1 5
1 2 7 2 2 1 3
1 7 2 1 5 6 1
5 2 1 0 7 2 14
1 0 14 14
k-1 of 14 domain 26 = 14
A= 14* mod 26
15 mod 26
q r1 r2 r t1 t2 t
3 15 3 11 0 1 -3
1 3 11 4 1 3 2
2 11 4 6 5 6 11
2 4 6 4 6 2 2
2 6 4 6 1 5 2
2 4 6 1 5 6 2
1 6 1 0 15 26 2
1 0 15 15
k-1 of 15 domain 26 = 15
A= 15* mod 26
5

b)
Permutation of the plain text is plain text : Peter Bayiokos
Keyword is Newark
N E W A R K
P E T E R B
A Y I O K O
S * * * * *
PAS EY TI EO RK BO
The cipher text value is EO EY BO PAS KTI
c)
Play fair cipher text
Plain text is Peter bayiokos
Keyword is militar parlr
M I L I T
A R P A R
K B C D E
F G H J N
O Q S U V
Eliminating PS EU TI EO RK BO
M I L I T
A R P A R
K B P D E
F G S J N
O Q H U V
The new text value is CH DV KV AB KQ
6

Question 2
(27*a*x+131221) mod 9743 = 310
(27*a*x+131221) mod 9743
27 v≡1(mod 9743)
X=310 (mod 9743)
27 v≡1- 9743 w
27 v+1- 9743 w = 1
=27=1*9744+5
=9744=1*5+3
=5=2*3+2
=3=2*1+1
=5=9744-27
=3=27-5
=3=5-2*1
1=1-2*1
27x≡310 mod 9743
=310*27 mod 9743
X=16 mod 9743
X =16
Question 3
X=8623 186919 mod 1997
=8623 186919 mod 1996 -1
=8623 186919 mod 1996
=1996826
mod 1996
=96 mod 1996
X=20 is the smallest mod value of multiplication
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Question 4
A mod 52+1
A B C D E
F G H I K
L M N O P
Q R S T U
V W X Y X
Row shift
F G H I K
L M N O P
Q R S T U
V W X Y X
A B C D E
Column shift
B C D E A
G H I K F
M N O P L
W X Y Z V
We can take the plain text value is AT TA CK AT DA WN
Cipher text: DQ QD EH DQ EB XM
Pairs value
AT- DQ
TA- QD
CK- EH
AT- DQ
DA- EB
WN- XM
A mod 52+1
8 mod 52+1
8 mod 53 = 5
Cipher Text value is 5
Cipher Text:
8

AFLQVBGMRWCHNSXDIOTYEKPUXFLQVAGMRWBHNSXCIOTXDKPOXEBG
MWCHNXDIOYEKPZAFLNATTACKATDAWNDQQDEHDQEBXM
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