1. CSCE 790: Computer Network Security Chin-Tser Huang huangct@cse.sc.edu University of South Carolina

2. 9/9/20032 Points of Vulnerability Adversary can eavesdrop from a machine on the same LAN Adversary can eavesdrop by dialing into communication server Adversary can eavesdrop by gaining physical control of part of external links twisted pair, coaxial cable, or optical fiber radio or satellite links

3. 9/9/20033 Placement of Symmetric Encryption Two major placement alternatives Link encryption encryption occurs independently on every link implies must decrypt traffic between links requires many devices, but paired keys End-to-end encryption encryption occurs between original source and final destination need devices at each end with shared keys

4. 9/9/20034 Characteristics of Link and End-to-End Encryption

5. 9/9/20035 Placement of Encryption Can place encryption function at various layers in OSI Reference Model link encryption occurs at layers 1 or 2 end-to-end can occur at layers 3, 4, 6, 7 If move encryption toward higher layer less information is encrypted but is more secure application layer encryption is more complex, with more entities and need more keys

6. 9/9/20036 Scope of Encryption

7. 9/9/20037 Traffic Analysis When using end-to-end encryption, must leave headers in clear so network can correctly route information Hence although contents are protected, traffic patterns are not protected Ideally both are desired end-to-end protects data contents over entire path and provides authentication link protects traffic flows from monitoring

8. 9/9/20038 Key Distribution Symmetric schemes require both parties to share a common secret key Need to securely distribute this key If key is compromised during distribution, all communications between two parties are compromised

9. 9/9/20039 Key Distribution Schemes Various key distribution schemes for two parties  A can select key and physically deliver to B  third party C can select and deliver key to A and B  if A and B have shared a key previously, can use previous key to encrypt a new key  if A and B have secure communications with third party C, C can relay key between A and B

10. 9/9/200310 Key Distribution Scenario

11. 9/9/200311 Key Distribution Issues Hierarchies of KDC’s are required for large networks, but must trust each other Session key lifetimes should be limited for greater security Use of automatic key distribution on behalf of users, but must trust system Use of decentralized key distribution Controlling purposes keys are used for

12. 9/9/200312 Summary of Symmetric Encryption Traditional symmetric cryptography uses one key shared by both sender and receiver If this key is disclosed, communications are compromised Symmetric because parties are equal Provide confidentiality, but does not provide non-repudiation

13. 9/9/200313 Insufficiencies with Symmetric Encryption Symmetric encryption is not enough to address two key issues key distribution – how to have secure communications in general without having to trust a KDC with your key? digital signatures – how to verify that a received message really comes from the claimed sender?

14. 9/9/200314 Advent of Asymmetric Encryption Probably most significant advance in the 3000 year history of cryptography Use two keys: a public key and a private key Asymmetric since parties are not equal Clever application of number theoretic concepts instead of merely substitution and permutation

15. 9/9/200315 How Asymmetric Encryption Works Asymmetric encryption uses two keys that are related to each other a public key, which may be known to anybody, is used to encrypt messages, and verify signatures a private key, known only to the owner, is used to decrypt messages encrypted by the matching public key, and create signatures the key used to encrypt messages or verify signatures cannot decrypt messages or create signatures

16. 9/9/200316 Asymmetric Encryption for Confidentiality

17. 9/9/200317 Asymmetric Encryption for Authentication

18. 9/9/200318 Security of Asymmetric Encryption Like symmetric schemes brute-force exhaustive search attack is always theoretically possible, but keys used are too large (>512bits) Not more secure than symmetric encryption, dependent on size of key Security relies on a large enough difference in difficulty between easy (en/decrypt) and hard (cryptanalyse) problems Generally the hard problem is known, just made too hard to do in practise Require using very large numbers, so is slow compared to symmetric schemes

19. 9/9/200319 RSA Invented by Rivest, Shamir & Adleman of MIT in 1977 Best known and widely used public-key scheme Based on exponentiation in a finite (Galois) field over integers modulo a prime exponentiation takes O((log n) 3 ) operations (easy) Use large integers (eg. 1024 bits) Security due to cost of factoring large numbers factorization takes O(e log n log log n ) operations (hard)

20. 9/9/200320 RSA Key Setup Each user generates a public/private key pair by select two large primes at random: p, q compute their system modulus n=p·q note ø(n)=(p-1)(q-1) select at random the encryption key e where 1<e<ø(n), gcd(e,ø(n))=1 solve following equation to find decryption key d e·d=1 mod ø(n) and 0≤d≤n publish their public encryption key: KU= {e,n} keep secret private decryption key: KR= {d,n}

21. 9/9/200321 RSA Usage to encrypt a message M the sender: obtains public key of recipient KU={e,n} computes: C=M e mod n, where 0≤M<n to decrypt the ciphertext C the owner: uses their private key KR={d,n} computes: M=C d mod n note that the message M must be smaller than the modulus n (block if needed)

22. 9/9/200322 Why RSA Works Euler's Theorem: a ø(n) mod n = 1 where gcd(a,n)=1 In RSA, we have n=p·q ø(n)=(p-1)(q-1) carefully chosen e and d to be inverses mod ø(n) hence e·d=1+k·ø(n) for some k hence : C d = (M e ) d = M 1+k·ø(n) = M 1 ·(M ø(n) ) q = M 1 ·(1) q = M 1 = M mod n

23. 9/9/200323 RSA Example: Computing Keys 1. Select primes: p=17, q=11 2. Compute n=pq=17×11=187 3. Compute ø(n)=(p–1)(q-1)=16×10=160 4. Select e: gcd(e,160)=1 and e<160  choose e=7 5. Determine d: de=1 mod 160 and d<160  d=23 since 23×7=161=10×160+1 6. Publish public key KU={7,187} 7. Keep secret private key KR={23,187}

24. 9/9/200324 RSA Example: Encryption and Decryption Given message M = 88 ( 88<187 ) Encryption: C = 88 7 mod 187 = 11 Decryption: M = 11 23 mod 187 = 88

25. 9/9/200325 Exponentiation Use a property of modular arithmetic [(a mod n)  (b mod n)]mod n = (a  b)mod n Use the Square and Multiply Algorithm to multiply the ones that are needed to compute the result Look at binary representation of exponent Only take O(log 2 n) multiples for number n eg. 7 5 = 7 4 ·7 1 = 3·7 = 10 (mod 11) eg. 3 129 = 3 128 ·3 1 = 5·3 = 4 (mod 11)

26. 9/9/200326 RSA Key Generation Users of RSA must: determine two primes at random - p,q select either e or d and compute the other Primes p,q must not be easily derived from modulus n=p·q means p,q must be sufficiently large typically guess and use probabilistic test Exponents e, d are multiplicative inverses, so use Inverse algorithm to compute the other

27. 9/9/200327 Security of RSA Three approaches to attacking RSA brute force key search (infeasible given size of numbers) mathematical attacks (based on difficulty of computing ø(n), by factoring modulus n) timing attacks (on running of decryption)

28. 9/9/200328 Factoring Problem Mathematical approach takes 3 forms: factor n=p·q, hence find ø(n) and then d determine ø(n) directly and find d find d directly Currently believe all equivalent to factoring have seen slow improvements over the years as of Aug 99 best is 130 decimal digits (512) bit with GNFS biggest improvement comes from improved algorithm cf “Quadratic Sieve” to “Generalized Number Field Sieve” 1024+ bit RSA is secure barring dramatic breakthrough ensure p, q of similar size and matching other constraints

29. 9/9/200329 Timing Attacks Developed in mid-1990’s Exploit timing variations in operations eg. multiplying by small vs large number Infer operand size based on time taken RSA exploits time taken in exponentiation Countermeasures use constant exponentiation time add random delays blind values used in calculations

30. 9/9/200330 Next Class Key management with asymmetric encryption Diffie-Hellman key exchange Read Chapter 10