lecture 5 overview

Lecture 5 Overview

Does DES Work?• Differential Cryptanalysis Idea– Use two plaintext that barely differ– Study the difference in the corresponding cipher

text– Collect the keys that could accomplish the change– Repeat

2CS 450/650 – Lecture 5: DES

Cracking DES• Diffie and Hellman then outlined a "brute

force" attack on DES– By "brute force" is meant that you try as many of

the 256 possible keys as you have to before decrypting the ciphertext into a sensible plaintext message

– They proposed a special purpose "parallel computer using one million chips to try one million keys each" per second


Cracking DES (cont.)• In 1998, Electronic Frontier Foundation spent

$220K and built a machine that could go through the entire 56-bit DES key space in an average of 4.5 days– On July 17, 1998, they announced they had

cracked a 56-bit key in 56 hours– The computer, called Deep Crack• used 27 boards each containing 64 chips• was capable of testing 90 billion keys a second


Cracking DES (cont.)• In early 1999, Distributed. Net used the DES Cracker

and a worldwide network of nearly 100K PCs to break DES in 22 hours– combined they were testing 245 billion keys per second

• This just serves to illustrate that any organization with moderate resources can break through DES with very little effort these days


Double DES• E(k1, E(k2, M) )

– As strong as 57-bit key !

– Given message M and ciphertext c– Encrypt M with all possible keys• 256 steps

– Decrypt c with all possible keys and match Ms

• 256 steps

CS 450/650 Fundamentals of Integrated Computer Security 6

Triple DES – Two keys• E(k1, D(k2, E(k1, M) ) )

• The first key is used to DES-encrypt the message• The second key is used to DES-decrypt the encrypted

message– Since the second key is not the right key, this decryption just

scrambles the data further

• The twice-scrambled message is then encrypted again with the first key to yield the final ciphertext

• As strong as 80-bit key !


Triple DES – Three keys• E(k3, D(k2, E(k1, M) ) )

• The first key is used to DES-encrypt the message• The second key is used to DES-decrypt the encrypted

message– Since the second key is not the right key, this decryption just

scrambles the data further

• The twice-scrambled message is then encrypted with the third key to yield the final ciphertext

• As strong as 112-bit key !


Analysis of Algorithms• Algorithms– Time Complexity– Space Complexity

• An algorithm whose time complexity is bounded by a polynomial is called a polynomial-time algorithm– An algorithm is considered to be efficient if it runs

in polynomial time.

CS 450/650 Lecture 5: Algorithm Background 9

Growth Rate T(n) = O(f(n)): T is bounded above by fThe growth rate of T(n) <= growth rate of f(n)

T(n) = (g(n)): T is bounded below by gThe growth rate of T(n) >= growth rate of g(n)

T(n) = (h(n)): T is bounded both above and below by hThe growth rate of T(n) = growth rate of h(n)

T(n) = o(p(n)): T is dominated by pThe growth rate of T(n) < growth rate of p(n)

10CS 450/650 Lecture 5: Algorithm Background

Time Complexity C O(n) O(log n) O(nlogn) O(n2) … O(nk)

O(2n) O(kn) O(nn)


Polynomial

Exponential

P, NP, NP-hard, NP-complete• A problem belongs to the class P if the problem can be

solved by a polynomial-time algorithm• A problem belongs to the class NP if the correctness of the

problem’s solution can be verified by a polynomial-time algorithm

• A problem is NP-hard if it is as hard as any problem in NP– Existence of a polynomial-time algorithm for an NP-hard problem

implies the existence of polynomial solutions for every problem in NP

• NP-complete problems are the NP-hard problems that are also in NP


Relationships between different classes

NP

P

NP-complete

NP-hard


Partitioning Problem• Given a set of n integers, partition the integers

into two subsets such that the difference between the sum of the elements in the two subsets is minimum– NP-complete

13, 37, 42, 59, 86, 100


Sum 165 17286 10042 5937 13

Bin Packing Problem• Suppose you are given n items of sizes

s1, s2,..., sn

• All sizes satisfy 0 si 1

• The problem is to pack these items in the fewest number of bins, – given that each bin has unit capacity– NP-hard


Lecture 6 RSA

CS 450/650

Fundamentals of Integrated Computer Security

Slides are modified from Hesham El-Rewini

RSA• Invented by Cocks (GCHQ), independently, by

Rivest, Shamir and Adleman (MIT)• Two keys e and d used for Encryption and

Decryption– The keys are interchangeable • M = D(d, E(e, M) ) = D(e, E(d, M) )

– Public key encryption• Based on problem of factoring large numbers– Not in NP-complete– Best known algorithm is exponential

17CS 450/650 Lecture 6: RSA

RSA• To encrypt message M compute– c = Me mod N

• To decrypt ciphertext c compute– M = cd mod N


• Let p and q be two large prime numbers• Let N = pq

• Choose e relatively prime to (p1)(q1)– a prime number larger than p-1 and q-1

• Find d such that ed mod (p1)(q1) = 1

Key Choice


RSA• Recall that e and N are public

• If attacker can factor N, he can use e to easily find d – since ed mod (p1)(q1) = 1

• Factoring the modulus breaks RSA• It is not known whether factoring is the only

way to break RSA20CS 450/650 Lecture 6: RSA

Does RSA Really Work?

• Given c = Me mod N we must show – M = cd mod N = Med mod N

• We’ll use Euler’s Theorem– If x is relatively prime to N then x(N) mod N =1• (n): number of positive integers less than n that are

relatively prime to n.• If p is prime then, (p) = p-1


Does RSA Really Work?• Facts: – ed mod (p 1)(q 1) = 1– ed = k(p 1)(q 1) + 1 by definition of mod– (N) = (p 1)(q 1)– Then ed 1 = k(p 1)(q 1) = k(N)

• Med = M(ed-1)+1 = MMed-1 = MMk(N) = M(M(N)) k mod N = M1 k mod N = M mod N


Example• Select primes p=11, q=3.• N = p* q = 11*3 = 33

• Choose e = 3

• check gcd(e, p-1) = gcd(3, 10) = 1 – i.e. 3 and 10 have no common factors except 1

• check gcd(e, q-1) = gcd(3, 2) = 1• therefore gcd(e, (p-1)(q-1)) = gcd(3, 20) = 1


Example (cont.)• p-1 * q-1 = 10 * 2 = 20 • Compute d such that

e * d mod (p-1)*(q-1) = 13 * d mod 20 = 1

d = 7

Public key = (N, e) = (33, 3)Private key = (N, d) = (33, 7)


Example (cont.)• Now say we want to encrypt message m = 7

• c = Me mod N = 73 mod 33 = 343 mod 33 = 13– Hence the ciphertext c = 13

• To check decryption, we computeM' = cd mod N = 137 mod 33 = 7


More Efficient RSA• Modular exponentiation example– 520 = 95367431640625 = 25 mod 35

• A better way: repeated squaring – Note that 20 = 2 10, 10 = 2 5, 5 = 2 2 + 1, 2 = 1 2– 51= 5 mod 35– 52= (51) 2 = 52 = 25 mod 35– 55= (52) 2 51 = 252 5 = 3125 = 10 mod 35– 510 = (55) 2 = 102 = 100 = 30 mod 35– 520 = (510) 2 = 302 = 900 = 25 mod 35

• No huge numbers and it’s efficient!

CS 450/650 Lecture 6: RSA 26

RSA key-length strength• RSA has challenges for different key-lengths– RSA-140

• Factored in 1 month using 200 machines in 1999

– RSA-155 (512-bit)• Factored in 3.7 months using 300 machines in 1999

– RSA-160• Factored in 20 days in 2003

– RSA-200• Factored in 18 month in 2005

– RSA-210, RSA-220, RSA-232, … RSA-2048


Group Work

Find keys d and e for the RSA cryptosystem with p = 7 and q = 11

Solution– p*q = 77– (p-1) * (q-1) = 60– e = 37– d = 13– n = 13 * 37 = 481 = 1 mod 60


lecture 5 overview

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