Lecture 4 Overview. Data Encryption Standard Combination of substitution and transposition – Repeated for 16 cycles – Provides confusion and diffusion

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

CS 450/650 – Lecture 4: DES 2

A High Level Description of DES

Input - P

16 Cycles

Output - C

Key

IP

Inverse IP

3CS 450/650 – Lecture 4: DES

A Cycle in DES


ERn-1 E(Rn-1 )

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.


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))


Types of Permutations

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

Lecture 5 DES & Rivest-Shamir-Adelman

CS 450/650

Fundamentals of Integrated Computer Security

Slides are modified from Hesham El-Rewini

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


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


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


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

• 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


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.


Algorithm Background

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 16

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)

17CS 450/650 Lecture 5: Algorithm Background

Growth Rate• We Compare functions by comparing their

relative rates of growth.

1000n vs. n2


Definitions 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)


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 ProblemGiven 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


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


Bin Packing ProblemExample (Optimal; Solution) for 7 items of sizes:

0.2, 0.5, 0.4, 0.7, 0.1, 0.3, 0.8.

0.8

0.2

0.3

0.7

0.50.10.4

Bin 1 Bin 2 Bin 325CS 450/650 Lecture 5: Algorithm Background

Rivest-Shamir-Adelman

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

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

• Choose e relatively prime to (p1)(q1)– How?

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

• Public key is (N,e)• Private key is d

Key Choice

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

• To decrypt C compute– M = Cd mod N

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 (p1)(q1)

• Factoring the modulus breaks RSA

• It is not known whether factoring is the only way to break 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) = 1 mod n

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 = MMed-1 = MMk(N) = M(M(N)) k mod N = M1 k mod N = M mod N

Documents

Lecture 4 Overview. Data Encryption Standard Combination of substitution and transposition – Repeated for 16 cycles – Provides confusion and diffusion