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CS 445 / 645 Internet Security Mon & Wed, 11:30 AM ~ 12:45 PM @ SEB 1240 Spring, 2012

School of Computer Science. CS 445 / 645 Internet Security Mon & Wed, 11:30 AM ~ 12:45 PM @ SEB 1240 Spring, 2012. Wednesday, Feb. 1, 2012. Cyber war between Israeli and Saudi hackers (Jan 21, 2012).

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CS 445 / 645 Internet Security Mon & Wed, 11:30 AM ~ 12:45 PM @ SEB 1240 Spring, 2012

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  1. School of Computer Science CS 445 / 645 Internet Security Mon & Wed, 11:30 AM ~ 12:45 PM @ SEB 1240 Spring, 2012 Wednesday, Feb. 1, 2012 CS 448/648 – Computer Security

  2. Cyber war between Israeli and Saudi hackers (Jan 21, 2012) • http://tech2.in.com/news/general/israeli-and-saudi-hackers-wage-cyber-war-against-each-other/274832 • Israeli credit card company was hacked with 20,000 cards compromised - by a Saudi hacker, 0xOmar. • Arab hackers hacked into the servers of Israel’s Anti Drug Authority • Israeli hackers retaliated by hacking the websites of two major Arab banks. (Central Bank of the United Arab Emirates, Arab Bank) • Pro-Palestinian hackers hacked the websites of the Tel Aviv Stock Exchange, El Al Airlines, three other Israeli banks • Pro-Israel hackers hack banks in Saudi Arabia. CS 448/648 – Computer Security

  3. The war is still going on… • Pro-Israel hackers published the log-in details of 20,000 Arab Facebook users • Pro-Palestinian hackers attacked the websites of two Israeli hospitals (Jan 25, 2012.) CS 448/648 – Computer Security

  4. Review Quiz at the end of class CS 448/648 – Computer Security

  5. 1. Electronic Code Book (ECB) • Each block is encrypted independently of the other • Easy to use in parallel , higher performance. • But identical blocks of plaintext  same cipher text • Repetitions in message may appear in the ciphertext • Easier cryptanalysis • Uses: secure transmission of a few block of data CS 448/648 – Computer Security

  6. Block1 Block2 Block3 Block4 IV + + + + DES DES DES DES Cipher1 Cipher2 Cipher3 Cipher4 2. Cipher Block Chaining (CBC) • Message blocks are linked together • Each previous cipher blocks is chained with current plaintext block • A change in the message affects all ciphertext blocks after the change (good!) • But difficult to parallelize • Uses Initialization Vector (IV) to start process • IV: a random number generated by sender and sent together with the ciphertext • Uses: bulk data encryption CS 448/648 – Computer Security

  7. One-way Hash Function • Secure hash or cryptographic hash • One-way • easy to generate a code, but computationally infeasible to find data mapping to specific hash value • Collision Free • computationally infeasible to find two data to same hash value • Note • in h = f (M), M is called preimage • M can be multiple CS 448/648 – Computer Security

  8. One way hash function requirements CS 448/648 – Computer Security

  9. One-way (Secure) hash functions • SHA • Developed by NIST in 1993 and SHA-1 in 1995 • 160 bit Message digest • Every bit of hash code is a function of every bit of input • SHA-2 (SHA-224, 256, 384, and 612) in 2002 • MD5 (RFC 1321) • Developed by Ron Rivest in 1991 • Generates 128-bit MD • RIPEMD-160 • Developed under EUROPEAN RACE Integrity Primitives Evaluation project (1996) • Generates 160-bit MD CS 448/648 – Computer Security

  10. Level of Safety • For n-bit hash, the possible outcomes are 2n (= H). • Since it would take roughly trials to find a collision with 50% chance, (or 2n/2) messages should be checked before finding two messages with the same hash value (level of safety) • It requires O(2n/2) evaluations of H to find two messages m and m’ that have a collision, H(m)=H(m’). • Level of safety for SHA-1 : 280 (half of 160 bits) • Level of safety for MD5 : 264 (half of 128 bits) • N = number of bits in digest CS 448/648 – Computer Security

  11. Message Authentication Code (MAC) • Why don’t we use one-way hash as MAC? • Hash-based MAC (HMAC) • Faster than block cipher algorithms • But hash functions do not take keys  need to add key • Keyed Hash • Using a shared secret key, generate a Message Authentication Code (MAC) • HMAC (RFC 2104, in 1996) • HMAC (Key, message) • Any hash function can be used • Used for IP security, TLS, SSL CS 448/648 – Computer Security

  12. Public Key Algorithm • Proposed by Diffie and Hellman in 1976 • Each use has a public key (kPUB) and a private key (kPRIV) • One key is kept secret, the other is public • Now only two keys are needed per user • P = D(kPRIV, E(kPUB, P)and • P = D(kPUB, E(kPRIV, P)with RSA Plaintext Plaintext Internet Encrypt with public key Decrypt with private key Ciphertext Asymmetric key cryptography CS 448/648 – Computer Security

  13. Public Key Algorithms • RSA (1978) • By Rivest, Shamir,Adleman • Flexible key length (512-bits, 1024-bits common) • Diffie-Hellman (1976) • Does not transmit symmetric session keys • Instead, it combines the receiver’s private key and the sender’s public key to generate symmetric session key • Elliptic Curve (1985) • Neal Koblitz (UW) and Victor Miller (IBM Watson) • Since then many were developed • To this date, usages are • 1st: RSA • 2nd: DH • 3rd: Elliptic curves CS 448/648 – Computer Security

  14. Public Key Algorithms RSA

  15. RSA • Best known & widely used public-key scheme • Based on exponentiation in a finite (Galois) field over integers modulo a prime (Modulo exponentiation) • 35 = 243, 243 % 7 = 5 • Easy • Security due to cost of factoring large numbers • Hard CS 448/648 – Computer Security

  16. RSA Encryption and Decryption • Encryption • Plaintext M < n • Ciphertext C= Me (mod n) • Decryption • Ciphertext C • Calculate Cdmod n = (Me)dmod n = Medmod n = M CS 445 – Internet Security

  17. Background in RSA - I • How many numbers (less than n) are relatively prime to n? • E.g, 5 is relatively prime (or coprime) to 12 • The answers are, • (n-1), if n is prime • (p-1)(q-1), if n = pq, and p,q are prime numbers • Number of relatively prime = n – (# not prime) • Numbers not relatively prime: p, 2p,… qp, q, 2q, ,,,, pq  total of q + p - 1 • Then, n – (# not relatively prime) = pq - (p+q-1) = (p-1)(q-1) • Euler totient function: (n) • (n) = (p-1) * (q-1) • E.g., (10) = 4, because 10 = 2*5, and (n) = (2-1)*(5-1) : i.e., {1,3,7,9} CS 445 – Internet Security

  18. Background in RSA - II • Modular exponentiation • 46 = 4096 = 6 (mod 10) • Magic formula: xy % n = x(y mod (n)) % n • e.g. (recall (10) = 4 ) • 37 mod 10 = 2187 mod 10 = 7 mod 10 • 3(7 mod 4) mod 10 = 27 mod 10 = 7 mod 10 • 34 mod 10 = 81 mod 10 = 1 mod 10 • 3(4 mod 4) mod 10 = 1 mod 10 = 1 mod 10 • 35 mod 10 = 243 mod 10 = 3 mod 10 • 3(5 mod 4) mod 10 = 3 mod 10 = 3 mod 10 same Our particular interest. Why? CS 445 – Internet Security

  19. Background in RSA - III • Magic formula: xy % n = x(y mod (n)) % n • A special case of this where y = 1 mod (n) • Then, xy % n = x(y mod (n)) % n = x( ((n) * k + 1) mod (n)) % n • For any number x, xy = xmod n e.g. (take any y,1 greater than 4 * k, i.e., 5, 9, 13, 17…) y = 5, then 35 = 243 = 3 mod 10 y = 5, then 75 = 16807 = 7 mod 10 y = 5, then 95 = 59049 = 9 mod 10 y = 9, then 89 = 134217728 = 8 mod 10 y = 13, then 813 = 549755813888 = 8 mod 10 =1 CS 445 – Internet Security

  20. RSA Principle • A special case of this where y = 1 mod (n) • Then, for any number x, xy = xmod n • RSA principle • Let y =e * d • Choose e, d such that e*d = 1 mod (n), then xed = x mod n • e: encryption key, d: decryption key, • x=plaintext (<n), xe % n = ciphertext, xed % n = decrypted text (=x) • Note: xde= xed • This gives the interchangeability of public and private keys • Why secure? (N and e are known to public!) • It takes long time to find out d given e, if you don’t know (n). • Easy, if you know (n) = (p-1)(q-1). • But to get (n), you need to know p and q. • So, it depends on the difficulty of factoring n into p and q CS 445 – Internet Security

  21. RSA factoring challenge • It is now closed (1991 – 2007) • RSA-768 is factored in Dec. 2009 • http://www.rsa.com/rsalabs/node.asp?id=2094 • http://en.wikipedia.org/wiki/RSA_Factoring_Challenge CS 448/648 – Computer Security

  22. RSA Algorithm: Key Generation • Select p, q (p,q are both primes) (secret) • Calculate n = p * q (public) • Calculate (n) = (p-1)(q-1) (secret) Euler totient of n (number of positive integers less than and relatively prime to n) • Chose e (public key) and d(private key) • Select integer e • e is usually a prime larger than (p-1) or (q-1) • Relatively prime to (n) and 1 < e < (n • Calculate d = e-1mod (n) (from de = 1 mod (n) ) • Public Key KS = {e, n} • Private Key K-1S = {d, n} CS 445 – Internet Security

  23. RSA Encryption and Decryption • Encryption • Plaintext M < n • Ciphertext C= Me (mod n) • Decryption • Ciphertext C • Calculate Cd mod n = (Me)dmod n = Medmod n = M • Why? • xde = x mod n, if (d * e) = 1 mod (n) CS 445 – Internet Security

  24. RSA Example • Select p=7, q=17 • Calculate n = pq = 7*17 = 119 • Calculate (n) = (p-1)(q-1) = 96 • Select e, relatively prime to (n) and less than (n) • e = 5 • Determine d such that de = 1 mod 96 • d = 77 because 77*5 = 385 = 4*96+1 • Created Keys: • KA = {5,119}, K-1A = (77,119} • Encryption: Message = 19 • 195 = 2476099 = 66 mod 119 • Ciphertext = 66 • Decryption • 6677 = 19 mod 119 CS 445 – Internet Security

  25. Finding the Inverse – example • Given e = 5, (n) = 96, find smallest d such that de = 1 mod 96 • 96 * i = 96, 192, 288, 384, 480, … • 96 * i + 1 = 97, 193, 289, 385, 481,… • Check them if they are divisible by e (=5) • 97, 193, 289: obviously no • 385 = 77 * 5 • So, d = 77 Candidates for d * e CS 445 – Internet Security

  26. Cryptool • Educational tool for cryptograph, originally developed by German Universities • https://www.cryptool.org/en/ CS 448/648 – Computer Security

  27. Tool to generate RSA Keys • Use PuTTygen • http://the.earth.li/~sgtatham/putty/latest/x86/puttygen.exe • Usage: http://the.earth.li/~sgtatham/putty/0.55/htmldoc/Chapter8.html CS 448/648 – Computer Security

  28. RSA discussions • N is known to be product of two primes, p, q • N is large > 100 digits (or even 600 digits at 2048 bits) • It is even impractical to verify whether p and q are prime: 1050 possible factors • How do you choose p and q? • Heuristic algorithm by Solovay and Strassen • Fermat’s primality test (used in PGP) • And many other (Miller-Rabin primality test, deterministic tests, number theoretic methods…) CS 445 – Internet Security

  29. Public Key Algorithm Diffie Hellman

  30. Diffie-Hellman Key Exchange • Cannot be used to exchange an arbitrary message • Rather it can establish a common key • Based on exponentiation in a finite (Galois) field (modulo a prime or a polynomial) – easy • Security relies on the difficulty of computing discrete logarithms (similar to factoring) – hard CS 448/648 – Computer Security

  31. Diffie-Hellman Key Exchange • Depends on the difficulty of computing discrete logarithms • First, what is exponentiation and logarithm? • 24 = 16, log216 = log2 24 = 4 • Then, what is discrete exponentiation? • Find k-th power as an integer and find the remainder of modulo p. E.g., • 35 = 243, 243 % 7 = 5 • Discrete logarithm is the inverse operation • E.g., find the smallest k for 3k 5 (mod 7) CS 448/648 – Computer Security

  32. DH Key Exchange Algorithm - I • Alice and Bob can determine same secret key • Start by agreeing on a large prime number N (public) • Then choose a generator G, where G<N (public) • Each follows these steps • Each chooses a large random number X < N • (X is the private key) • Each calculates the value Y = Gx % N. • (Y is the public key and sent to the other party) • Y’ = GX’ % N • Each computes the K = Y’X % N = GX’X % N • (Y’ is the other party’s public key) • K is the shared secret key • X = own private key, X’ is the other’s private key • Y = own public key, Y’ is the other’s public key CS 448/648 – Computer Security

  33. DH Key Exchange Algorithm - II • Note K = Y’X mod N = (GX’ mod N)X mod N = GXX’ mod N • GXX’ and GX’X are equal !!! • Alice and Bob know it • But intermediaries only know N, G, Y (=Gx % N) and Y’ (=Gx’ % N) • Nobody can calculate GXX’ in a reasonable amount of time • Why? Because the intermediaries must know X or X’. • But it is difficult to compute discrete logarithm, i.e., figuring out X from GX % N (or X’ from GX’ % N) CS 448/648 – Computer Security

  34. Diffie-Hellman Example • N=11, G=2 • Alice picks 4 for K-1A • Bob picks 3 for K-1B • Alice computes • (24 mod 11) = (16 mod 11)= 5 • Sends this to Bob • Bob computes • (23 mod 11) = (8 mod 11) = 8 • Sends this to Alice • Bob computes • (53 mod 11) = (125 mod 11) = 4 • Alice computes • (84 mod 11) = (4096 mod 11)= 4 CS 448/648 – Computer Security

  35. Speed comparison RSA Enc ~ 1024 bit/8 * 1000/0.18 = 711 KB/sec RSA Dec ~ 1024 bit/8 * 1000/4.77 = 27 KB/sec http://www.cryptopp.com/benchmarks.html CS 448/648 – Computer Security

  36. Other public key algorithms • NTRU • Short for N-th degree truncated polynomial ring, R[x] / (xN − 1) • Created by four math professors at Brown University • http://securityinnovation.com/cryptolab/index.shtml • Speed • NTRUEncrypt-263 vs. RSA-1024: 27.6 times faster • NTRUEncrypt key pair generation takes about 3.0 milliseconds, while Crypto++ takes over 1 second to generate an RSA key pair (on 300 MHz machines)  300 times faster CS 448/648 – Computer Security

  37. Cryptographic Algorithms Digital signature

  38. Digital signature • This? • No, but this. CS 448/648 – Computer Security

  39. Digital Signature • Digital signatures provide the ability to: • verify author, date & time of signature • authenticate message contents • be verified by third parties to resolve disputes • No need for confidentiality • How? Use public key algorithm • Under RSA, public key and private keys are commutative D ( E (M, Kpub), Kpriv) = M = D ( E(M, Kpriv), Kpub) • Encrypt with sender’s private key, receiver decrypt it with sender’s public key • In practice, the message could be big, so sign the hash value of the message instead CS 448/648 – Computer Security

  40. Confirming Digital Signature • Receiver verifies that: • Sender signed M ANDno one else signed M • Non-repudiation: • Receiver can take the message and the signature to court and prove that Sender signed M • The purported maker of a statement will not be able to successfully challenge the validity of the statement CS 448/648 – Computer Security

  41. Digital Signature Usage • US Government Printing Office Deploys Digital Signatures for FY2009 Budget • http://blogs.adobe.com/security/2008/02/us_government_printing_office.html CS 448/648 – Computer Security

  42. Digital Signature Usage • Microsoft Word • http://www.ehow.com/how_4925786_install-digital-signature-microsoft-word.html • PowerPoint • http://www.brighthub.com/computing/windows-platform/articles/8398.aspx CS 448/648 – Computer Security

  43. Digital Signature Usage • View code signature with Process Explorer • Signed software is more trustworthy • http://technet.microsoft.com/en-us/sysinternals/bb896653.aspx CS 448/648 – Computer Security

  44. Wrapping up • Public key algorithms • RSA • DH • Digital signature • Next class • Digital certificate • Computer Networks review • Quiz now! CS 448/648 – Computer Security

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