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Lecture 3.1: Public Key Cryptography I. CS 436/636/736 Spring 2015 Nitesh Saxena. Today’s Informative/Fun Bit – Acoustic Emanations. http://www.google.com/search?source=ig&hl=en&rlz=&q=keyboard+acoustic+emanations&btnG=Google+Search http://tau.ac.il/~tromer/acoustic/.
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Lecture 3.1: Public Key Cryptography I CS 436/636/736 Spring 2015 Nitesh Saxena
Today’s Informative/Fun Bit – Acoustic Emanations • http://www.google.com/search?source=ig&hl=en&rlz=&q=keyboard+acoustic+emanations&btnG=Google+Search • http://tau.ac.il/~tromer/acoustic/ Public Key Cryptography -- I
Course Administration • HW1 posted – due at 11am on Feb 2 (Mon) • Any questions? • Regarding programming portion of the homework • Submit the whole modified code that you used to measure timings • Comment the portions in the code where you modified the code • Include a small “readme” for us to understand this Public Key Cryptography -- I
Outline of Today’s Lecture • Public Key Crypto Overview • Number Theory • Modular Arithmetic Public Key Cryptography -- I
Recall: Private Key/Public Key Cryptography • Private Key: Sender and receiver share a common (private) key • Encryption and Decryption is done using the private key • Also called conventional/shared-key/single-key/ symmetric-key cryptography • Public Key: Every user has a private key and a public key • Encryption is done using the public key and Decryption using private key • Also called two-key/asymmetric-key cryptography Public Key Cryptography -- I
Private key cryptography revisited. • Good: Quite efficient (as you’ll see from the HW#1 programming exercise on AES) • Bad: Key distribution and management is a serious problem – for N users O(N2) keys are needed Public Key Cryptography -- I
Public key cryptography model • Good: Key management problem potentially simpler • Bad: Much slower than private key crypto (we’ll see later!) Public Key Cryptography -- I
Public Key Encryption • Two keys: • public encryption key e • private decryption key d • Encryption easy when e is known • Decryption easy when d is known • Decryption hard when d is not known • We’ll study such public key encryption schemes; first we need some number theory. Public Key Cryptography -- I
Public Key Encryption: Security Notions • Very similar to what we studied for private key encryption • What’s the difference? Public Key Cryptography -- I
Group: Definition (G,.) (where G is a set and . : GxGG) is said to be a group if following properties are satisfied: • Closure : for any a, b G, a.b G • Associativity : for any a, b, c G, a.(b.c)=(a.b).c • Identity : there is an identity element such that a.e = e.a = a, for any a G • Inverse : there exists an element a-1 for every a in G, such that a.a-1 = a-1.a = e Abelian Group: Group which also satisfies commutativity , i.e., a.b = b.a
Groups: Examples • Set of all integers with respect to addition --(Z,+) • Set of all integers with respect to multiplication (Z,*) – not a group • Set of all real numbers with respect to multiplication (R,*) • Set of all integers modulo m with respect to modulo addition (Zm, “modular addition”) Public Key Cryptography -- I
Divisors • xdividesy (written x | y) if the remainder is 0 when y is divided by x • 1|8, 2|8, 4|8, 8|8 • The divisors of y are the numbers that divide y • divisors of 8: {1,2,4,8} • For every number y • 1|y • y|y Public Key Cryptography -- I
Prime numbers • A number is prime if its only divisors are 1 and itself: • 2,3,5,7,11,13,17,19, … • Fundamental theorem of arithmetic: • For every number x, there is a unique set of primes {p1, … ,pn} and a unique set of positive exponents {e1, … ,en} such that Public Key Cryptography -- I
Common divisors • The common divisors of two numbers x,y are the numbers z such that z|x and z|y • common divisors of 8 and 12: • intersection of {1,2,4,8} and {1,2,3,4,6,12} • = {1,2,4} • greatest common divisor: gcd(x,y) is the number z such that • z is a common divisor of x and y • no common divisor of x and y is larger than z • gcd(8,12) = 4 Public Key Cryptography -- I
Euclidean Algorithm: gcd(r0,r1) Main idea: If y = ax + b then gcd(x,y) = gcd(x,b) Public Key Cryptography -- I
Example – gcd(15,37) • 37 = 2 * 15 + 7 • 15 = 2 * 7 + 1 • 7 = 7 * 1 + 0 • gcd(15,37) = 1 Public Key Cryptography -- I
Relative primes • x and y are relatively prime if they have no common divisors, other than 1 • Equivalently, x and y are relatively prime if gcd(x,y) = 1 • 9 and 14 are relatively prime • 9 and 15 are not relatively prime Public Key Cryptography -- I
Modular Arithmetic • Definition: x is congruent to y mod m, if m divides (x-y). Equivalently, x and y have the same remainder when divided by m. Notation: Example: • We work in Zm = {0, 1, 2, …, m-1}, the group of integers modulo m • Example: Z9 ={0,1,2,3,4,5,6,7,8} • We abuse notation and often write = instead of Public Key Cryptography -- I
Addition in Zm : • Addition is well-defined: • 3 + 4 = 7 mod 9. • 3 + 8 = 2 mod 9. Public Key Cryptography -- I
Additive inverses in Zm • 0 is the additive identity in Zm • Additive inverse of a is -a mod m = (m-a) • Every element has unique additive inverse. • 4 + 5= 0 mod 9. • 4 is additive inverse of 5. Public Key Cryptography -- I
Multiplication in Zm : • Multiplication is well-defined: • 3 * 4 = 3 mod 9. • 3 * 8 = 6 mod 9. • 3 * 3 = 0 mod 9. Public Key Cryptography -- I
Multiplicative inverses in Zm • 1 is the multiplicative identity in Zm • Multiplicative inverse (x*x-1=1 mod m) • SOME, but not ALL elements have unique multiplicative inverse. • In Z9 : 3*0=0, 3*1=3, 3*2=6, 3*3=0, 3*4=3, 3*5=6, …, so 3 does not have a multiplicative inverse (mod 9) • On the other hand, 4*2=8, 4*3=3, 4*4=7, 4*5=2, 4*6=6, 4*7=1, so 4-1=7, (mod 9) Public Key Cryptography -- I
Which numbers have inverses? • In Zm, x has a multiplicative inverse if and only if x and m are relatively prime or gcd(x,m)=1 • E.g., 4 in Z9 Public Key Cryptography -- I
Extended Euclidian: a-1 mod n • Main Idea: Looking for inverse of a mod n means looking for x such that x*a – y*n = 1. • To compute inverse of a mod n, do the following: • Compute gcd(a, n) using Euclidean algorithm. • Since a is relatively prime to m (else there will be no inverse) gcd(a, n) = 1. • So you can obtain linear combination of rm and rm-1 that yields 1. • Work backwards getting linear combination of ri and ri-1 that yields 1. • When you get to linear combination of r0 and r1 you are done as r0=n and r1= a. Public Key Cryptography -- I
Example – 15-1 mod 37 • 37 = 2 * 15 + 7 • 15 = 2 * 7 + 1 • 7 = 7 * 1 + 0 Now, • 15 – 2 * 7 = 1 • 15 – 2 (37 – 2 * 15) = 1 • 5 * 15 – 2 * 37 = 1 So, 15-1 mod 37 is 5. Public Key Cryptography -- I
Modular Exponentiation:Square and Multiply method • Usual approach to computing xc mod n is inefficient when c is large. • Instead, represent c as bit string bk-1 … b0 and use the following algorithm: z = 1 For i = k-1 downto 0 do z = z2 mod n if bi = 1 then z = z* x mod n Public Key Cryptography -- I
Example: 3037 mod 77 z = z2 mod n if bi = 1 then z = z* x mod n Public Key Cryptography -- I
Other Definitions • An element g in G is said to be a generator of a group if a = gi for every a in G, for a certain integer i • A group which has a generator is called a cyclic group • The number of elements in a group is called the order of the group • Order of an element a is the lowest i (>0) such that ai = e (identity) • A subgroup is a subset of a group that itself is a group Public Key Cryptography -- I
Lagrange’s Theorem • Order of an element in a group divides the order of the group Public Key Cryptography -- I
Euler’s totient function • Given positive integer n, Euler’s totient function is the number of positive numbers less than n that are relatively prime to n • Fact: If p is prime then • {1,2,3,…,p-1} are relatively prime to p. Public Key Cryptography -- I
Euler’s totient function • Fact: If p and q are prime and n=pq then • Each number that is not divisible by p or by q is relatively prime to pq. • E.g. p=5, q=7: {1,2,3,4,-,6,-,8,9,-,11,12,13,-,-,16,17,18,19,-,-,22,23,24,-,26,27,-,29,-,31,32,33,34,-} • pq-p-(q-1) = (p-1)(q-1) Public Key Cryptography -- I
Euler’s Theorem and Fermat’s Theorem • If a is relatively prime to n then • If a is relatively prime to p then ap-1 = 1 mod p Proof : follows from Lagrange’s Theorem Public Key Cryptography -- I
Euler’s Theorem and Fermat’s Theorem EG: Compute 9100 mod 17: p =17, so p-1 = 16. 100 = 6·16+4. Therefore, 9100=96·16+4=(916)6(9)4 . So mod 17 we have 9100 (916)6(9)4 (mod 17) (1)6(9)4 (mod 17) (81)2 (mod 17) 16 Public Key Cryptography -- I
Some questions • 2-1 mod 4 =? • What is the complexity of • (a+b) mod m • (a*b) mod m • xc mod (n) • Order of a group is 5. What can be the order of an element in this group? Public Key Cryptography -- I
Further Reading • Chapter 4 of Stallings • Chapter 2.4 of HAC Public Key Cryptography -- I