pubic-key cryptography
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Public-Key Cryptography
The convergence of prime numbers, the history of math, inverse functions, and a
contemporary application
Introduction to Cryptography
» Cryptography is the study of ways of writing a message that hides its meaning from everyone except the intended recipient.
» Encryption is a method of changing plaintext, the message to be hidden, to ciphertext, the message in its hidden form.
» Decryption is the procedure that changes ciphertext back to plaintext.
Basic Example
Plaintext - MATH RULES
Encryption rule - Write the plaintext backwards
Ciphertext - SELUR HTAM
Function Example - Encrypt
» Choose a function that has an inverse.
» Rewrite the message as blocks of numbers. 2210 2917 3627 3021 1428
M A T H R U L E S » Evaluate the function at each block. This
becomes the encrypted message.
f (x) =2x+1
(2210) 2(2210) 1 4421f = + =
Function Example -Decrypt
» Find the inverse of the encryption function. This is the decryption function.
» Evaluate the decryption function at each received block.
1 1 1( )
2 2f x x =
f −1(4421) =
12(4421)−
12=2210
Path to a Public Function
» Create a function whose inverse is extremely difficult to determine without precise details of how the function was created.
» Publish this function in a data base of public functions.
» Use the inverse function only you can determine to decrypt messages intended for you.
Egyptian Multiplication
1 26
2 52
4 104
8 208
16 416
32 > 23
Find 23 26
Egyptian Multiplication
1 26
2 52
4 104
8 208
16 416
23⋅26 =(1+ 2 + 4 +16)26=26 +52 +104 + 416=598
Find 23⋅26
New Egyptian Multiplication
23 26
1 2L {1,1,1,0,1} L {26,52,104,208,416}= =
( )1 2
1 2
L L {26,52,104,0,416}
Sum L L 598
=
=
10 223 10111=
Egyptian Exponentiation
2326
L1⋅L2 = 26,262 ,264 ,0,2616{ }
L1 = 1,1,1,0,1{ } L2 = 26,262 ,264 ,268 ,2616{ }
Prod L1⋅L2( )=2623 Nonzero elem ents
Modular Arithmetic
( ) ( ) ( )mod mod moda n b n a b n+ = +
( ) ( ) ( )mod mod moda n b n a b n =
( )mod modp pa n a n=
Modular Exponentiation
23325
mod 537
1 233 mod 537
2 2332 mod 537 = 52 mod 537
4 2334 mod 537 = 52
2 mod 537 = 19 mod 537
8 2338 mod 537 = 19
2 mod 537 = 361 mod 537
16 23316
mod 537 = 3612 mod 537 = 367 mod 537
Modular Exponentiation
23325
mod 537
1 233 mod 537
2 2332 mod 537 = 52 mod 537
4 2334 mod 537 = 52
2 mod 537 = 19 mod 537
8 2338 mod 537 = 19
2 mod 537 = 361 mod 537
16 23316
mod 537 = 3612 mod 537 = 367 mod 537
Modular Exponentiation
Because the exponent 25 = 1 + 8 + 16, the product of the nonzero elements is
23325 mod537 =(233m od537)(361m od537)(367m od537) =(341m od537)(367m od537) =26m od537
Public-Key Cryptography
» Choose two large prime numbers, p and q, and form their product n = pq.
» Calculate » Randomly choose e such that
and » The values of e and n are the public key.» The ciphertext, c, is c = me mod n, where m
is the message being encrypted.
ϕ(n)=(p−1)(q−1).
2<e<ϕ(n)
gcd(e,ϕ(n))=1.
An Encryption Example
» Let p = 83 and q = 89. Then n = 7387.
= (83 – 1)(89 – 1) = 7216» Randomly choose e = 23. Verify
» The encryption function is c = m23 mod 7387, where m is a plaintext message block and c is a cipher block.
gcd(e,ϕ(n))=1.
ϕ (n)
An Encryption Example
Encrypt M A T H R U L E S 2210 2917 3627 3021 1428
23
23
23
23
23
2210 mod7387 6117
2917 mod7387 1088
3627 mod7387 6030
3021 mod7387 1874
1428 mod7387 5878
=
=
=
=
=
Decryption Function
The decryption function is m = c1255 mod 7387.1255
1255
1255
1255
1255
6117 mod7387 2210
1088 mod7387 2917
6030 mod7387 3627
1874 mod7387 3021
5878 mod7387 1428
=
=
=
=
=
Public-Key Cryptography
» Theorem: The decryption function is given by m = cd mod n, where d is the solution of
» Basically, we have to prove that
cd = (me mod n)d = med mod n = m.
de≡1mod ϕ(n).
Other Applications
» Digital signatures• Olivia encrypts a message using her private
key. Henry decrypts the message using her public key.
• Better: Olivia first encrypts her message using Henry’s public key. Then uses her private key to encrypt that message.
» HTTPS
Contact Information
» galoisgroup@mac.com» http://public.me.com/galoisgroup
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