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Lecture 4 Overview

Lecture 4 Overview. Data Encryption Standard. Combination of substitution and transposition Repeated for 16 cycles Provides confusion and diffusion Product cipher Two weak but complementary ciphers can be made more secure by being applied together.

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Lecture 4 Overview

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  1. Lecture 4 Overview

  2. Data Encryption Standard • Combination of substitution and transposition • Repeated for 16 cycles • Provides confusion and diffusion • Product cipher • Two weak but complementary ciphers can be made more secure by being applied together CS 450/650 – Lecture 4: DES

  3. A High Level Description of DES Input - P IP 16 Cycles Key Inverse IP Output - C CS 450/650 – Lecture 4: DES

  4. A Cycle in DES CS 450/650 – Lecture 4: DES

  5. Expand each block Rn-1 • We'll call the use of this selection table the function E. • Thus E(Rn-1) has a 32 bit input block, and a 48 bit output block. E Rn-1 E(Rn-1 ) CS 450/650 – Lecture 4: DES

  6. The Calculation of the function f 1- Expand Rn-1 E(Rn-1 ) 2- XOR Kn + E(Rn-1) = B1B2B3B4B5B6B7B8 3- Substitution S-Boxes  S1(B1)S2(B2)S3(B3)S4(B4)S5(B5)S6(B6)S7(B7)S8(B8) 4- P permutation  f = P(S1(B1)S2(B2)...S8(B8)) CS 450/650 – Lecture 4: DES

  7. Types of Permutations Pattern of Expansion Permutation CS 450/650 Fundamentals of Integrated Computer Security

  8. Lecture 5 DES & Rivest-Shamir-Adelman CS 450/650 Fundamentals of Integrated Computer Security Slides are modified from Hesham El-Rewini

  9. 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 CS 450/650 – Lecture 5: DES

  10. Cracking DES • During the period NBS was soliciting comments on the proposed algorithm, the creators of public key cryptography registered some objections to the use of DES. • Hellman wrote: "Whit Diffie and I have become concerned that the proposed data encryption standard, while probably secure against commercial assault, may be extremely vulnerable to attack by an intelligence organization" • letter to NBS, October 22, 1975 CS 450/650 – Lecture 5: DES

  11. Cracking DES (cont.) • 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 CS 450/650 – Lecture 5: DES

  12. 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 CS 450/650 – Lecture 5: DES

  13. 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 • It has been shown that a dedicated hardware device with a cost of $1M (is much less in 2010) can search all possible DES keys in about 3.5 hours • This just serves to illustrate that any organization with moderate resources can break through DES with very little effort these days CS 450/650 – Lecture 5: DES

  14. Triple DES • Triple-DES is just DES with two 56-bit keys applied. • Given a plaintext message, 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. • This three-step procedure is called triple-DES. CS 450/650 – Lecture 5: DES

  15. Algorithm Background

  16. 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

  17. Time and Space • Should be calculated as function of problem size (n) • Sorting an array of size n, • Searching a list of size n, • Multiplication of two matrices of size n by n • T(n) = function of n (time) • S(n) = function of n (space) CS 450/650 Lecture 5: Algorithm Background

  18. Growth Rate • We Compare functions by comparing their relative rates of growth. 1000n vs. n2 CS 450/650 Lecture 5: Algorithm Background

  19. Definitions • T(n) = O(f(n)): T is bounded above by f The growth rate of T(n) <= growth rate of f(n) • T(n) = W (g(n)): T is bounded below by g The growth rate of T(n) >= growth rate of g(n) • T(n) = Q(h(n)): T is bounded both above and below by h The growth rate of T(n) = growth rate of h(n) • T(n) = o(p(n)): T is dominated by p The growth rate of T(n) < growth rate of p(n) CS 450/650 Lecture 5: Algorithm Background

  20. Time Complexity • C • O(n) • O(log n) • O(nlogn) • O(n2) • … • O(nk) • O(2n) • O(kn) • O(nn) Polynomial Exponential CS 450/650 Lecture 5: Algorithm Background

  21. 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 CS 450/650 Lecture 5: Algorithm Background

  22. Relationships between different classes NP-hard NP P NP-complete CS 450/650 Lecture 5: Algorithm Background

  23. 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 13, 37, 42, 59, 86, 100 CS 450/650 Lecture 5: Algorithm Background

  24. 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 CS 450/650 Lecture 5: Algorithm Background

  25. Bin Packing Problem Example (Optimal; Solution) for 7 items of sizes: 0.2, 0.5, 0.4, 0.7, 0.1, 0.3, 0.8. CS 450/650 Lecture 5: Algorithm Background

  26. Rivest-Shamir-Adelman

  27. RSA • Invented by Cocks (GCHQ), independently, by Rivest, Shamir and Adleman (MIT) • in 1978 • Two keys e and d are used for Encryption and Decryption • The keys are interchangeable • Based on the problem of factoring large numbers

  28. Key Choice • Let p and q be two large prime numbers • Let N = pq be the modulus • Choose e relatively prime to (p1)(q1) • How? • Find d such that ed = 1 mod (p1)(q1) • Public key is (N,e) • Private key is d

  29. RSA • To encrypt message M compute • C = Me mod N • To decrypt C compute • M = Cd mod N

  30. RSA • Recall that e and N are public • If attacker can factor N, he can use e to easily find d • since ed = 1 mod (p1)(q1) • Factoring the modulus breaks RSA • It is not known whether factoring is the only way to break RSA

  31. 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) = 1 mod n

  32. Does RSA Really Work? • Facts: • ed = 1 mod (p  1)(q  1) • By definition of “mod”, ed = k(p  1)(q  1) + 1 • (N) = (p  1)(q  1) • Then ed  1 = k(p  1)(q  1) = k(N) • Med = M(ed-1)+1 = MMed-1 = MMk(N) = M(M(N)) k mod N = M1 k mod N = M mod N

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