Transcript Codes, Ciphers, and Cryptography-RSA Encryption Michael A. Karls Ball State University The RSA Encryption Scheme We now look at the public key cryptography scheme developed by Rivest, Shamir,

Codes, Ciphers, and
Cryptography-RSA
Encryption
Michael A. Karls
Ball State University
1
The RSA Encryption Scheme
We now look at the public key
cryptography scheme developed by
Rivest, Shamir, and Adleman (RSA) in
1977.
 In order to understand this scheme, we
need some definitions!
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Definition of Divisor
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Let a and b be integers, with b  0. We say that
b divides a or b is a divisor of a if a = b x c for
some integer c.
Notation: b|a
Example 1:
3|24 since 24 = 3 x 8.
Divisors of 12 are: -12, -6, -4, -3, -2, -1, 1, 2, 3,
4, 6, 12
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Definition of Greatest Common
Divisor (GCD)
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Let a and b be integers, not both zero. The
greatest common divisor (GCD) of a and b is the
largest integer that divides both a and b.
Notation: (a,b)
Example 2:
Divisors of 6: -6, -3, -2, -1, 1, 2, 3, 6
Divisors of 8: -8, -4, -2, -1, 1, 2, 4, 8
Thus, (6,8) = 2
Since the divisors of 7 are -7, -1, 1, 7, (7,8) = 1.
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Definition of Relatively Prime
Two integers whose GCD is 1 are said to
be relatively prime.
 Example 3:
Since (7,8) = 1, 7 and 8 are relatively
prime.
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Definition of Prime Number
A positive integer p is said to be prime if
p>1 and the only positive divisors of p are
1 and p.
 Example 4:
2, 3, and 7 are prime.
6, 8, 10, 100 are not prime (composite).
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RSA Scheme (with Alice and Bob!)
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Step 1:
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Example 5:
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Alice chooses two huge
prime numbers p and q.
Note: Alice keeps p and
q secret!
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p = 47 and q = 59.
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RSA Scheme (with Alice and Bob!)
(cont.)
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Step 2:
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Example 5 (cont.):
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Alice computes
N = p x q.
Then she computes k =
(p-1)(q-1).
Finally, she chooses an
integer e such that 1<e<N
and (e,k) =1.
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N = 47 x 59 = 2773.
k = 46 x 58 = 2668.
e = 17.
Choice of e is o.k., since
1<17<2773 and
(17,2668)= 1.
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RSA Scheme (with Alice and Bob!)
(cont.)
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Step 3:
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Example 5 (cont.):
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Alice computes
d = e-1 mod k.
Alice publishes her public
key: N, e.
Alice keeps secret her
private key: p, q, d, k.
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d = 17-1 mod 2668 = 157.
Alice’s public key:
N = 2773; e = 17.
Alice’s private key:
p = 47; q = 59; d = 157; k
= 2668.
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RSA Scheme (with Alice and Bob!)
(cont.)
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Step 4:
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Example 5 (cont.):
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Suppose Bob wants to
send a message to Alice.
To do so, he looks up
Alice’s public key,
converts the message
into numbers M<N.
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Plaintext is HELLO
HELLO  HE LL O_
Assign 00space;
01A; 02B, … , 26Z
(or use ASCII).
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Plaintext Plain #
¨
HE
0805
LL
1212
0_
1500
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RSA Scheme (with Alice and Bob!)
(cont.)
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Step 4 (cont.):
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Example 5 (cont.):
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Next Bob computes:
C = Me mod N (1)
for each plaintext number M to
get ciphertext number C.
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080517 mod 2773 = 542.
121217 mod 2773 = 2345.
150017 mod 2773 = 2417.
Encrypted message is
0542 2345 2417.
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RSA Scheme (with Alice and Bob!)
(cont.)
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Step 5:
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Example 5 (cont.):
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Bob emails Alice the encrypted
message.
To decrypt, Alice uses her
private key and computes:
M = Cd mod N (2)
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0542157 mod 2773 = 805.
2345157 mod 2773 =
1212.
2417157 mod 2773 =
1500.
Decrypted message is
HE LL 0_.
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