CS346 Cryptography, Fall 2009
Homework 1, SOLUTIONS
1. (5 points) Problem 1 (Chapter 2.13)
Solution:
Antony knows that Caesar used a shift cipher, but he does not know the
key. He will not be able to ﬁgure out where to meet Caesar. When he tries to decrypt
the ciphertext EVIRE with all possible shifts, he will ﬁnd both the word ARENA
(shifting by 4) and the word RIVER (shifting by 13).
2. (5 points) Problem 5 (Chapter 2.13)
You should use the following table for the standard numeric values associated with
letters of the English alphabet.
A
0
N
13
B
1
O
14
C
2
P
15
D
3
Q
16
E
4
R
17
F
5
S
18
G
6
T
19
H
7
U
20
I
8
V
21
J
9
W
22
K
10
X
23
L
11
Y
24
M
12
Z
25
Solution:
The encryption function is of the form
y
=
ax
+
b
mod 26. We need to ﬁnd the value
of a and b. Knowing that the plaintext starts with “ha”, we get the following system
of two congruences (using the table for the numeric values for the letters):
2
≡
a
·
7 +
b
mod 26
17
≡
b
mod 26
This gives away the value of the key: a=9, b=17. Thus, the encryption function is
y
= 9
x
+ 17 mod 26, and the corresponding decryption function is
x
= 3
y
+ 1 mod 26.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '10
 GAL
 Cryptography, Greatest common divisor, Euclidean algorithm, Extended Euclidean algorithm, Euclidean domain

Click to edit the document details