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Teaching Inverse Functions with Cryptography
An Interactive Approach
Lisa Harden1
Leandro Junes2
1 St.
Louis CC - Meramec
Kirkwood, MO.
[email protected]
2 University
of South Carolina-Sumter
Sumter, SC.
[email protected]
2011 AMATYC Conference
November 11
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Caesar Cipher
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Ciphers from Bijective Functions
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1
Caesar Cipher
2
Ciphers
3
Ciphers from Bijective Functions
Dictionary
An Example
4
Preview of the software
5
Group Activity
6
References
7
Questions
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Codes in Ancient Roman Battle
If Julius Caesar wished to call his
troops home, he may have sent a
message such as
-Uhwxuq wr Urph.
Can be deciphered if person knows the
key.
Secrecy maintained if a person does
not know the key. Picture taken from [2].
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Codes in Ancient Roman Battle
To obtain the encoded message “Uhwxuq wr Urph" each letter
of the original message was simply replaced with the letter
three places to the right.
- “Caesar Cipher”
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Codes in Ancient Roman Battle
“Uhwxuq wr Urph”
decodes to
“Return to Rome”
Picture taken from [2]
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How does this relate to functions?
Instead of thinking “Shift each letter 3 places to the right” we
can think of functions with input and output.
− Input: a number that represents a letter.
− Output: the number plus 3.
Input: x.
Output: x + 3
f (x) = x + 3
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Ciphers
A cipher is a method for creating secret messages.
The purpose of using a cipher is to exchange information
securely.
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An Example
Associate each Letter to a Number
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
A
26
B
27
C
28
D
29
E
30
F
31
G
32
H
33
I
34
J
35
K
36
L
37
M
38
N
39
O
40
P
41
Q
42
R
43
S
44
T
45
U
46
V
47
W
48
X
49
Y
50
Z
51
?
52
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Each Integer Correspond to a Letter
If n is any integer, then the remainder r of n ÷ 53 is a number
in {0, 1, . . . , 52}.
This correspondence is not injective.
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Each Integer Correspond to a Letter
a
0
53
..
.
K
36
89
..
.
b
1
54
..
.
c
2
55
..
.
d
3
56
..
.
e
4
57
..
.
f
5
58
..
.
g
6
59
..
.
h
7
60
..
.
i
8
61
..
.
j
9
62
..
.
k
10
63
..
.
l
11
64
..
.
m
12
65
..
.
n
13
66
..
.
o
14
67
..
.
p
15
68
..
.
q
16
69
..
.
r
17
70
..
.
t
19
72
..
.
u
20
73
..
.
v
21
74
..
.
w
22
75
..
.
x
23
76
..
.
y
24
77
..
.
z
25
78
..
.
A
26
79
..
.
B
27
80
..
.
C
28
81
..
.
D
29
82
..
.
E
30
83
..
.
F
31
84
..
.
G
32
85
..
.
H
33
86
..
.
I
34
87
..
.
J
35
88
..
.
X
49
102
..
.
Y
50
103
..
.
L
37
90
..
.
M
38
91
..
.
N
39
92
..
.
O
40
93
..
.
P
41
94
..
.
Q
42
95
..
.
R
43
96
..
.
S
44
97
..
.
L. Harden and L. Junes
T
45
98
..
.
U
46
99
..
.
V
47
100
..
.
W
48
101
..
.
Z
51
104
..
.
s
18
71
..
.
?
52
105
..
.
Caesar Cipher
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An Example
Each Integer Modulo 53 Corresponds to a Letter
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
A
26
B
27
C
28
D
29
E
30
F
31
G
32
H
33
I
34
J
35
K
36
L
37
M
38
N
39
O
40
P
41
Q
42
R
43
S
44
T
45
U
46
V
47
W
48
X
49
Y
50
Z
51
?
52
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An Example
All non-zero Integers Modulo 53 have Multiplicative Inverses
Examples
1
= 17.
25
1
Since 4 × 40 = 160 and 160 ÷ 53 leaves remainder 1, = 40.
4
Since 25 × 17 = 425 and 425 ÷ 53 leaves remainder 1,
Theorem
If p is a prime number, then every non-zero integer module p has a
multiplicative inverse.a
a For
a proof see [1, Fraleigh]
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All Integers Modulo 53 have “Exact” Cubic Roots
Definition
We say that
√
3
x = y , if y 3 = x.
Examples
Since 13 = 1,
√
3
1 = 1.
Since 183 = 5832 and 5832 ÷ 53 leaves remainder 2,
√
3
2 = 18.
Theorem
If p is a prime number of the form 3k + 2, then every integer
√
module p has a unique cubic root. That is, 3 x is a unique integer
module p.
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An Example
An Example
In 47 BC, Julius Caesar conquered Pharnaces II of Pontus in the
city of Zela in present day Turkey. He claimed to have done so in 4
hours. Caesar decides to send an encrypted message back to the
Senate in Rome.
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Before Caesar leaves, they decide to use the linear function
f (x) = 3x + 1 to encrypt/decrypt all comunications.
Caesar
Senate
f (x) = 3x + 1
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Caesar is now gone.
Caesar
Senate
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Caesar wants to send the Senate a message.
Caesar
Senate
Veni, vidi, vici
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f (x) = 3 x + 1
Text
x
V
47
e
4
n
13
i
8
v
21
L. Harden and L. Junes
Veni, vidi, vici
i
8
d
3
i
8
v
21
i
8
c
2
i
8
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f (x) = 3 x + 1
Text V e n i
x 47 4 13 8
f (x) = 3 x + 1 142 13 40 25
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Veni, vidi, vici
v i d i
21 8 3 8
64 25 10 25
v i c i
21 8 2 8
64 25 7 25
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f (x) = 3 x + 1
Text
x
f (x) = 3 x + 1
Modulo 53
V
47
142
36
e
4
13
13
n
13
40
40
i
8
25
25
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Veni, vidi, vici
v
21
64
11
i
8
25
25
d
3
10
10
i
8
25
25
v
21
64
11
i
8
25
25
c
2
7
7
i
8
25
25
Caesar Cipher
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Text
x
f (x) = 3 x + 1
Modulo 53
Cipher Text
f (x) = 3 x + 1 Veni, vidi, vici
V e n i
v i d i
47 4 13 8 21 8 3 8
142 13 40 25 64 25 10 25
36 13 40 25 11 25 10 25
K n O z
l z k z
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v
21
64
11
l
i
8
25
25
z
c
2
7
7
h
i
8
25
25
z
Caesar Cipher
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Caesar
Veni, vidi, vici
KnOz, lzkz, lzhz
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Caesar
KnOz, lzkz, lzhz
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Senate
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Caesar
Senate
Enemy
KnOz, lzkz, lzhz
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−→
Caesar Cipher
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Caesar
Senate
Enemy
KnOz, lzkz, lzhz
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An Example
An Example
Senate
f (x) = 3 x + 1
Cipher Text K n O z
x 36 13 40 25
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KnOz, lzkz, lzhz
l z k z
11 25 10 25
l z h z
11 25 7 25
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Senate
f −1 (x) =
1
1
x−
3
3
Cipher Text K n O z
x 36 13 40 25
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KnOz, lzkz, lzhz
l z k z
11 25 10 25
l z h z
11 25 7 25
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An Example
1
= 18 modulo 53 because 18 × 3 = 54 that leaves
3
residue 1 when divided by 53. Thus,
We know that
Senate
f −1 (x) = 18 x − 18
Cipher Text K n O z
x 36 13 40 25
L. Harden and L. Junes
KnOz, lzkz, lzhz
l z k z
11 25 10 25
l z h z
11 25 7 25
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An Example
Senate
f −1 (x) = 18 x − 18
KnOz, lzkz, lzhz
Cipher Text K n O z
l z k z
l z h z
x 36 13 40 25 11 25 10 25 11 25 7 25
f −1 (x) = 18 x − 18 630 216 702 432 180 432 162 432 180 432 108 432
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Senate
f −1 (x) = 18 x − 18
Cipher Text
x
−1
f (x) = 18 x − 18
Module 53
K
36
630
47
n
13
216
4
O
40
702
13
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KnOz, lzkz, lzhz
z
l z k
25 11 25 10
432 180 432 162
8 21 8 3
z
l z h
25 11 25 7
432 180 432 108
8 21 8 2
z
25
432
8
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Senate
f −1 (x) = 18 x − 18
Cipher Text
x
f −1 (x) = 18 x − 18
Module 53
Text
K
36
630
47
V
n
13
216
4
e
O
40
702
13
n
L. Harden and L. Junes
KnOz, lzkz, lzhz
z
l z k
25 11 25 10
432 180 432 162
8 21 8 3
i
v i d
z
l z h
25 11 25 7
432 180 432 108
8 21 8 2
i
v i c
z
25
432
8
i
Caesar Cipher
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isn’t this too much for students to do?
Our software simplifies the process by handling all minor detail
computations (software computes everything modulo 53).
Students only need to know how to compute the inverse of a
function. They do NOT need to know arithmetic modulo 53.
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isn’t this too much for students to do?
Our software simplifies the process by handling all minor detail
computations (software computes everything modulo 53).
Students only need to know how to compute the inverse of a
function. They do NOT need to know arithmetic modulo 53.
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Function Name
Linear Function
Rational Function
Cubic Function
Cubic Root Function
Algebraic Form
a
c
f (x) = x +
b
d
f (x) =
ax +b
53c x + d
a 3 c
x +
b
d
r
a
c
f (x) = 3 x +
b
d
f (x) =
Restrictions
a, b and d cannot
be multiples of 53.
a and d cannot be
multiples of 53.
a, b and d cannot
be multiples of 53.
a, b and d cannot
be multiples of 53.
All numbers a, b, c and d must be integers in the interval [−2147483648, 2147483647]
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The function f is given by f (x) = 3 x + 1 =
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3
1
x+
1
1
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User needs to know the inverse function at this point. In our case
1
1
f −1 (x) = x −
3
3
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Let’s Decrypt Some Messages!
1
In groups, find all four inverse functions.
2
After all inverse functions have been found, send a pair from
your group to the computer to enter one message and its
inverse function.
3
When you receive your decrypted message, copy it on your
paper, copy it on the board for the class to see, and return to
your group.
4
As each pair returns, send out a new pair to decrypt and
record.
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Class Results: Excerpt # 1
What is the name of this poem?
Who is author?
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Class Results: Excerpt # 1
What is the name of this poem?
The Raven.
Who is author?
Edgar Allan Poe.
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Class Results: Excerpt # 2
What is the title of this story?
Who is author?
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Class Results: Excerpt # 2
What is the title of this story?
The Masque of the Red Death.
Who is author?
Edgar Allan Poe.
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Class Results: Excerpt # 3
What is the name of this poem?
Who is author?
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Class Results: Excerpt # 3
What is the name of this poem?
Annabele Lee.
Who is author?
Edgar Allan Poe.
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Class Results: Excerpt # 4
What is the name of this story?
Who is author?
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Class Results: Excerpt # 4
What is the name of this story?
The Gold Bug.
Who is author?
Edgar Allan Poe.
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Poe’s Challenge
See [4].
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Poe’s Challenge
He correctly solve all submissions from December 1839 to May
1840.
- Approximately 34 submissions. See [3].
The Gold Bug.
-Read Poe’s famous story
- Free copy from The Oxford Text Archive at
http://ota.ahds.ac.uk/
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References
[1]J.B. Fraleigh, A First Course in Abstract Algebra, 7th ed.,
Addison Wesley, NY, USA, 2002, page 181.
[2]http://questgarden.com/62/15/1/080310122438/process.htm, July 2011.
[3]http://oerrecommender.org/visits/13032, July 2011.
[4]C.S. Brigham, Edgar Allan Poe’s Contributions to
Alexander’s Weekly Messenger, American Antiquarian Society,
1943.
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Any suggestions or comments, please let us know.
Thank you!
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