MAT 260 WEEK 2
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MAT 260: Cryptology Dr. Brian Stout
9 July 2023
2 - 2 Assignment: Chapter 2
SECTION 2.1 1.
By using Caesar Cipher decipher the phrases. By using If we use the way Caesar did in the Gallic Wars: A=D, B=E, C=F, … ETC. to translate we can see the messages translation.
(a)
HOYLV ZDV VLJKWHG DW PDLQ DQG XQLRQ
ELVIS WAS SIGHTED AT MAIN AND UNION
(b) SUHVL GHQWD QGFRQ JUHVV UHDFK EXGJH WDJUH HPHQW
PRESI DENTA NDCON GRESS REACH BUDGE TAGRE EMENT
With proper spacing the message reads. PRESIDENT AND CONGRESS REACH
BUDGET AGREEMENT
3.
In each of following find q
and r such the b = qm + r
. To find q you need to divide b
by m
. r would be the remainder between 0
r
m.
a.
b = 127, m = 7
127 MOD 7 = 1
q = 18
r = 1
127 = (18)7 + 1
b.
b = 473, m = 26
473 MOD 26 = 1
q = 18
r = 1
473 = (18)26 + 1
c.
b = 1024, m = 16
473 MOD 26 = 0
q = 64
r = 0
1024 = (64)16 + 0
4.
Find the smallest nonnegative integer x that satisfies the congruence. If we use the
formula x = y (mod m).
We take
y divided by m to see what it is equal too. Then we take the number rounded down to the nearest whole number (
c
). Multiply c
by m to get z..
Take y
minus z to find x.
x
y (
mod m)
y
m
c
c
m = z
y – z = x
a.
x
14 (
mod 3)
14
3
4.6666…
4
3 = 12
14 – 12 = 2
b.
x
130 (
mod 26)
130
26
5
5
26 = 130
130 – 130 = 0
c.
x
-1 (
mod 5)
-1
5
-.2
1
5 = 5 -1 – 5 = 4
d.
x
-258 (
mod 16)
-258
16
-16.125
-16
16 = -256
-258 – -256 = -2
SECTION 2.2
3. Solve for the following congruences for x.
ax + b
y (mod m)
-- minus b from both sides
ax + b - b
y - b (mod m)
-- (y - b = k)
ax
k (mod m)
Create a table with
k and add or subtract
m to find a number divisible by
a closest to 0.
k – (2m)
k - m
k
k + m
k + (2m)
(a, m) = d --- Find (d) being the GCD, ax
d (mod m) --- replace k with d
ax
k (mod m) --- multiply by inverse of a
ax
k (mod m) ---substitute k with appropriate answer of (
k
m)
we will call this new variable u
ax
u (mod m) --- divide both sides by a
to get x
w (mod m)
x
w (mod m)
(a) 2x + 1
4 (mod 5)
2x + 1
4 (mod 5)
2x
3 (mod 5)
3 – (2
5) = -7
3 - 5 = -2
3
3 + 5 = 8
3 + (2
5) = 13
(2x)
2
(8)
2 (mod 5) x
4 (mod 5)
4
(b) 9x + 5
1 (mod 10)
9x + 5
1 (mod 10)
9x
-4 (mod 10)
-4 - 10 = -14
-4
-4 + 10 = 6
-4 + (2
10) = 16
-4 + (3
10) = 26
-4 + (4
10) = 36
(9x)
9
(36)
9 (mod 10) x
4 (mod 10)
4
(c) 4x - 3
5 (mod 7)
4x - 3
5 (mod 7)
4x
8 (mod 7)
--- No need for table as 8 is divisible by 4.
(4x)
4
(8)
4 (mod 7) x
2 (mod 7)
2
SECTION 2.3
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Write the message in numerical equivalent form. A=0, B=1, C=2,… ETC. To encrypt the message, we will sub in each numerical equivalent into the given equation.
Then use the ciphertext and translate it to the corresponding letter.
7. Encipher the following message using affine function y = (3x+1) (mod 26)
IMITATION IS THE SINCEREST FORM OF FLATTERY
8, 12, 8, 19, 0, 19, 8, 14, 13 8, 18 19, 7, 4 18, 8, 13, 2, 4, 17, 4, 18, 19 5, 14, 17, 12 14, 5 5, 11, 0, 19, 19, 4, 17, 24
0 becomes…
y = (3x+1) (mod 26)
y = (3(0)+1) (mod 26)
y = (1) (mod 26)
1
26
0.038
0
26 = 0
1 – 0 = 1
2 …
y = (3x+1) (mod 26)
y = (3(2)+1) (mod 26)
y = (7) (mod 26)
7
26
0.270
0
26 = 0
7 – 0 = 7
4 …
y = (3x+1) (mod 26)
y = (3(4)+1) (mod 26)
y = (13) (mod 26)
13
26
0.5
0
26 = 0
13 – 0 = 13
5 …
y = (3x+1) (mod 26)
y = (3(5)+1) (mod 26)
y = (16) (mod 26)
16
26
0.615
0
26 = 0
16 – 0 = 16
7 …
y = (3x+1) (mod 26)
y = (3(7)+1) (mod 26)
y = (22) (mod 26)
22
26
0.846
0
26 = 0
22 – 0 = 22
8 … y = (3x+1) (mod 26)
y = (3(8)+1) (mod 26)
y = (25) (mod 26)
25
26
0.962
0
26 = 0
25 – 0 = 25
11 …
y = (3x+1) (mod 26)
y = (3(11)+1) (mod 26)
y = (34) (mod 26)
40
26
1.308
1
26 = 26
34 – 26 = 8
12 …
y = (3x+1) (mod 26)
y = (3(12)+1) (mod 26)
y = (37) (mod 26)
37
26
1.423
1
26 = 26
37 – 26 = 11
13 …
y = (3x+1) (mod 26)
y = (3(13)+1) (mod 26)
y = (40) (mod 26)
40
26
1.538
1
26 = 26
40 – 26 = 14
14 …
y = (3x+1) (mod 26)
y = (3(14)+1) (mod 26)
y = (43) (mod 26)
43
26
1.654
1
26 = 26
43 – 26 = 17
17 …
y = (3x+1) (mod 26)
y = (3(17)+1) (mod 26)
y = (52) (mod 26)
52
26
2
2
26 = 52
52 – 26 = 26
y = (26) (mod 26)
26
26
1
1
26 = 26
26 – 26 = 0
18 …
y = (3x+1) (mod 26)
y = (3(18)+1) (mod 26)
y = (55) (mod 26)
55
26
2.115
2
26 = 52
55 – 52 = 3
19 … y = (3x+1) (mod 26)
y = (3(19)+1) (mod 26)
y = (58) (mod 26)
58
26
2.231
2
26 = 52
58 – 52 = 6
24 …
y = (3x+1) (mod 26)
y = (3(24)+1) (mod 26)
y = (73) (mod 26)
73
26
2.808
2
52 = 52
73 – 52 = 21
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ORIGINAL
NUMERICAL EQUIVALENT CIPHER TEXT
8, 12, 8, 19, 0, 19, 8, 14, 13 8, 18 19, 7, 4 18, 8, 13, 2, 4, 17, 4, 18, 19 5, 14, 17, 12 14, 5 5, 11, 0, 19, 19, 4, 17, 24
25, 11, 25, 6, 1, 6, 25, 17, 13
25, 3
6, 22, 13
3, 25, 14, 7, 13, 0, 13, 3, 6
16, 17, 0, 11
17, 16
15, 8, 1, 6, 6, 13, 0, 21
We then translate the new numerical equivalent to letters and get:
ZLZGBGZRO ZD GWN DZOHNANDG QRAL RQ QIBGGNAV
SECTION 2.3
2.
Keyword is PRIME MINISTER construct a mixed cipher alphabet by columnar transposition. Spell out PRIME MINISTER in the first row with out repeating any letters. PRIME MINISTER becomes PRIME NST. P
R
I
M
E
N
S
T
A
B
C
D
F
G
H
J
K
L
O
Q
U
V
W
X
Y
Z
(a)
We then can write the cipher alphabet by columns going down, A=P, B=A, C=K… ETC.
PLAIN
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
CIPHE
R
P
A
K
Y
R
B
L
Z
I
C
O
M
D
Q
E
F
U
N
G
V
S
H
W
T
J
X
(b) Encipher the phrase – IT IS MUCH EASIER TO BE CRITICAL THAN TO BE CORRECT
IV IG DSKZ RPGIRN VE AR KNIVIKPM VZPQ VE AR KENNRKV
(c)
Decipher the phrase – PFNRK RYRQV RDAPM DGPFN IQKIF MR
APREC EDENT EMBAL MSAPR INCIP LE
With proper spacing reads – A PRECENDENT EMBALMS PRINCIPLE
SECTION 2.4
3.
Encipher a simple columnar transposition with 5 columns. Pad any partial lines with T. I DO NOT KNOW OF WHICH MAKES A MAN MORE CONSERVATIVE TO KNOW NOTHING OF THE PRESENT OR NOTHING OF THE PAST.
You start with writing the first 5 letters across, then the next five right below it, and continue this pattern till there are no letters left and if the last is not completely filled we will use the letter T to
fill in the last row.
IDONO TKNOW
WHICH
MAKES
AMANM
ORECO
NSERV
ATIVE
TOKNO
WNOTH
INGOF
THEPR
ESENT
ORNOT
HINGO
FTHEP
ASTTT
We then write the first column going down spacing after every 5 letters. We then get the following:
ITWMA ONATW ITEOH FADKH AMRST ONNHS RITSO NIKAE EIKOG EENNH TNOCE NCRVN TOPNO GETOW HSMOV EOHFR TTOPT
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