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Data Encryption Standard (DES) Overview and Cracking Methods

Learn about DES, a product cipher with 16 cycles of substitution and transposition, providing confusion and diffusion for security. Explore DES cracking methods and the evolution of Triple-DES. Understand algorithm analysis with time and space complexity considerations.

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Data Encryption Standard (DES) Overview and Cracking Methods

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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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