Post on 14-Dec-2015
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Intro to Crypto and Mod
Supplementary Notes
Prepared by Raymond WongPresented by Raymond Wong
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e.g.1 (Page 4)
E.g., m = 21 n = 9 21 can be expressed as 9 x 2 + 3 (i.e., nq + r)
q = 2r = 3
0 r < n
r is defined to be 21 mod 9
21 mod 9 is equal to 3
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e.g.2 (Page 9) Illustration of
[(-m) mod n] = n – [m mod n] E.g., m = 4
n = 5
E.g., m = 9 n = 5
-4 mod 5
-9 mod 5
4 mod 5 = 4
9 mod 5 = 4
= 5 – (4 mod 5)
= 5 – (9 mod 5)
= 5 – 4 = 1
= 5 – 4 = 1
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e.g.3 (Page 10)
2 x 8 mod 9 = 16 mod 9 = 7
(2 mod 9) (8 mod 9) = 2 x 8 = 16
(2 + 8) mod 9 = 1
(2 mod 9) + (8 mod 9) = 2 + 8 = 10
Conclusion: 2 x 8 mod 9 (2 mod 9) (8 mod 9)
Conclusion: (2 + 8) mod 9 (2 mod 9) + (8 mod 9)
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e.g.4 (Page 11)
Illustration of Lemma 2.2 E.g.,
i = 1 n = 5
1 mod 5 = 1
(1 + 5) mod 5 = 6 mod 5 = 1
(1 + 2x5) mod 5 = 11 mod 5 = 1
(1 + 3x5) mod 5 = 16 mod 5 = 1
(1 + k.5) mod 5 = 1
(1 + (-1)x5) mod 5 = -4 mod 5 = 1
(1 + (-2)x5) mod 5 = -9 mod 5 = 1
(1 + (-3)x5) mod 5 = -14 mod 5 = 1
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e.g.5 (Page 11)
Illustration of Lemma 2.2 E.g.,
i = 11 n = 5
11 mod 5 = 1
(11 + 5) mod 5 = 16 mod 5 = 1
(11 + 2x5) mod 5 = 21 mod 5 = 1
(11 + 3x5) mod 5 = 26 mod 5 = 1
(11 + k.5) mod 5 = 1
(11 + (-1)x5) mod 5 = 6 mod 5 = 1
(11 + (-2)x5) mod 5 = 1 mod 5 = 1
(11 + (-3)x5) mod 5 = -4 mod 5 = 1
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e.g.6 (Page 11)
Prove that 11 mod 5 = (11 + k.5) mod 5for all integers k
Let r = 11 mod 5
By Euclid’s Division Theorem, we can write 11 = 5q + rwhere q and r are two unique integers and 0 r < 5
Consider 11 + k.5= (5q + r) + k.5
= 5q + r + k.5
= 5q + k.5 + r
= 5(q + k) + r
By the definition of Euclid’s division theorem, (11 + k.5) mod 5 = r
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e.g.7 (Page 12)
Illustration of Lemma 2.3 E.g.,
(2 + 8) mod 9 = (2 + (8 mod 9)) mod 9
= ((2 mod 9) + 8) mod 9
= ((2 mod 9) + (8 mod 9)) mod 9
(2.8) mod 9 = (2 . (8 mod 9)) mod 9
= ((2 mod 9) . 8) mod 9
= ((2 mod 9) . (8 mod 9)) mod 9
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e.g.8 (Page 12)
2 x 8 mod 9 = 16 mod 9 = 7
((2 mod 9) (8 mod 9)) mod 9 = 2 x 8 mod 9 = 16 mod 9 = 7
(2 + 8) mod 9 = 1
((2 mod 9) + (8 mod 9)) mod 9 = (2 + 8) mod 9 = 10 mod 9 = 1
Conclusion: 2 x 8 mod 9 = ((2 mod 9) (8 mod 9)) mod 9
Conclusion: (2 + 8) mod 9 = ((2 mod 9) + (8 mod 9)) mod 9
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e.g.8
(2 + 8) mod 9 = 1
((2 mod 9) + (8 mod 9)) mod 9 = (2 + 8) mod 9 = 10 mod 9 = 1
Conclusion: (2 + 8) mod 9 = ((2 mod 9) + (8 mod 9)) mod 9
Claim: (20 + 17) mod 9 = ((20 mod 9) + (17 mod 9)) mod 9
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e.g.8Claim: (20 + 17) mod 9 = ((20 mod 9) + (17 mod 9)) mod 9
Why is it correct?By Euclid’s Division Theorem, we can write 20 as follows.
20 = 9q1 + r1 where q1 and r1 are some unique integers.= 9q1 + (20 mod 9)
By Euclid’s Division Theorem, we can write 17 as follows. 17 = 9q2 + r2 where q2 and r2 are some unique integers.
= 9q2 + (17 mod 9)
Consider (20 + 17) mod 9
= {[9q1 + (20 mod 9)] + [9q2 + (17 mod 9)]} mod 9= {9q1 + (20 mod 9) + 9q2 + (17 mod 9)} mod 9= [ (20 mod 9) + (17 mod 9) + 9q1 + 9q2] mod 9
= [ (20 mod 9) + (17 mod 9) + 9(q1 + q2)] mod 9
= [ (20 mod 9) + (17 mod 9)] mod 9
Lemma 2.2 (Y + 9k) mod 9 = Y mod 9
(by Lemma 2.2)
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e.g.9 (Page 14)
E.g. 0 +5 2 = 2
E.g., 1.52 = 2
(0 + 2) mod 5 = 2 mod 5 = 2
(1.2) mod 5 = 2 mod 5 = 2
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e.g.10 (Page 15) Illustration of Theorem 2.4
Commutative law
Associative law
Distributive law
2 x9 8 = 8 x9 2
2 +9 8 = 8 +9 2
2 x9 (8 x9 1) = (2 x9 8) x9 1
2 +9 (8 +9 1) = (2 +9 8) +9 1
2 x9 (8 +9 1) = (2 x9 8) +9 (2 x9 1)
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e.g.11 (Page 24)
I love UST. kfjEfklje$3 I love UST.
encryptiondecryption
password password
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e.g.11
I love UST. kfjEfklje$3 I love UST.
encryptiondecryption
keroro keroro
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e.g.11
I love UST. kfjEfklje$3 I love UST.
encryptiondecryption
keroro keroro
sender receiverkfjEfklje$3
attacker
Undecipherable(cannot be decrypted easily)
keroro keroro
17
e.g.11
I love UST. I love UST.
encryption
0
18
I love UST. J mpwf VTU.
encryption
1
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
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 a
19
I love UST. K nqxg WUV.
encryption
2
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
c d e f g h i j k
l m n o p q r s t
u v w x y z a b
20
I love UST. L oryh XVW.
encryption
3
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
d e f g h i j k l
m n o p q r s t u
v w x y z a b c
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e.g.12 (Page 32)
7 11
encryption
a = 5n = 12
Encrypted value = 5 .
12 7= 5 . 7 mod 12= 35 mod 12= 11
Encryption function= 5 .
12 x Multiplication modn
Is there any division modn ?
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e.g.13 (Page 32)
Examples that an inverse function exists S T
T S
f
f-1
123
4
a
bc
d
a
bc
d
123
4
A function!
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e.g.14 (Page 32)
Examples that an inverse function does not existS T
T S
f
No such f-1
123
4
a
bc
a
bc
123
4
Not a function!
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e.g.15 (Page 34)S
f4, 12
012
3
4
5
6
7
8
9
10
11
T
012
3
4
5
6
7
8
9
10
11
The inverse of f4, 12 does not exist
sender receiver3
Encrypted
0
sender receiver6
Encrypted
0
The receiver cannot determine the original number.
Case (a): a = 4, n = 12
f4, 12 (x) = 4.x mod 12
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e.g.16 (Page 35)S
f3, 12
012
3
4
5
6
7
8
9
10
11
T
012
3
4
5
6
7
8
9
10
11
The inverse of f3, 12 does not exist
sender receiver2
Encrypted
6
sender receiver6
Encrypted
6
The receiver cannot determine the original number.
Case (b): a = 3, n = 12
f3, 12 (x) = 3.x mod 12
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e.g.17 (Page 36)S
f5, 12
012
3
4
5
6
7
8
9
10
11
T
012
3
4
5
6
7
8
9
10
11
The inverse of f5, 12 exists
sender receiver7
Encrypted
11
sender receiver1
Encrypted
5
The receiver can uniquely determine the original number.
Case (c): a = 5, n = 12
f5, 12 (x) = 5.x mod 12
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e.g.18 (Page 41)
Private-key cryptosystems
sender receiver1
Encrypted
5
private-key (e.g., 2) private-key (e.g., 2)
Encrypt this message with this private-key
1
Decrypt this message with the same private-key
Encrypted
5
Encrypted
5
It should be kept privately at the sender’s side.
It should be kept privately at the receiver’s side.
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e.g.19 (Page 41)
Public-key cryptosystems
sender receiver1
Encrypted
5
public key (e.g., 2) secret-key (e.g., 4)
Encrypt this message with a public-key
1
Decrypt this message with a secret-key
Encrypted
5
Encrypted
5
It can be kept publicly.
It should be kept privately at the receiver’s side.
It has some relationships with the public key.
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e.g.19 (Page 41)
Public-key cryptosystems
sender receiver1
Encrypted
5
public key (e.g., 2) secret-key (e.g., 4)
Encrypt this message with a public-key
1
Decrypt this message with a secret-key
Encrypted
5
Encrypted
5
It should be kept privately at the receiver’s side.
Raymond
Public Key Directory
Raymond 2
Peter
Peter 7
This directory is accessible to the public.
It can be kept publicly.
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e.g.20 (Page 44)
An ideal key pair (public key and secret key) Given a public key,
it is difficult for the adversary to deduce the secret key
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e.g.20
Public-key cryptosystems
sender receiver167
Encrypted
338
public key secret-key
rev (1000 – M) = rev (1000 – 167) = rev (833) = 338
167
1000 – rev(C) = 1000 – rev (338) = 1000 – 833 = 167
Encrypted
338
Encrypted
338
It is not secure. Why?