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Public Key Cryptography Presentation Transcript

Slide 1 - Public Key Cryptography Part 1  Cryptography 1
Slide 2 - Public Key Cryptography Two keys Sender uses recipient’s public key to encrypt Receiver uses his private key to decrypt Based on trap door, one way function Easy to compute in one direction Hard to compute in other direction “Trap door” used to create keys Example: Given p and q, product N=pq is easy to compute, but given N, it is hard to find p and q Part 1  Cryptography 2
Slide 3 - Public Key Cryptography Encryption Suppose we encrypt M with Bob’s public key Only Bob’s private key can decrypt to find M Digital Signature Sign by “encrypting” with private key Anyone can verify signature by “decrypting” with public key But only private key holder could have signed Like a handwritten signature (and then some) Part 1  Cryptography 3
Slide 4 - Knapsack Part 1  Cryptography 4
Slide 5 - Knapsack Given a set of n weights W0,W1,...,Wn-1 and a sum S, is it possible to find ai  {0,1} so that S = a0W0+a1W1 +...+ an-1Wn-1 (technically, this is subset sum problem) Example Weights (62,93,26,52,166,48,91,141) Problem: Find subset that sums to S=302 Answer: 62+26+166+48=302 The (general) knapsack is NP-complete Part 1  Cryptography 5
Slide 6 - Knapsack General knapsack (GK) is hard to solve But superincreasing knapsack (SIK) is easy SIK each weight greater than the sum of all previous weights Example Weights (2,3,7,14,30,57,120,251) Problem: Find subset that sums to S=186 Work from largest to smallest weight Answer: 120+57+7+2=186 Part 1  Cryptography 6
Slide 7 - Knapsack Cryptosystem Generate superincreasing knapsack (SIK) Convert SIK into “general” knapsack (GK) Public Key: GK Private Key: SIK plus conversion factors Part 1  Cryptography 7 Easy to encrypt with GK With private key, easy to decrypt (convert ciphertext to SIK) Without private key, must solve GK (???)
Slide 8 - Knapsack Example Let (2,3,7,14,30,57,120,251) be the SIK Choose m = 41 and n = 491 with m, n rel. prime and n greater than sum of elements of SIK General knapsack 2  41 mod 491 = 82 3  41 mod 491 = 123 7  41 mod 491 = 287 14  41 mod 491 = 83 30  41 mod 491 = 248 57  41 mod 491 = 373 120  41 mod 491 = 10 251  41 mod 491 = 471 General knapsack: (82,123,287,83,248,373,10,471) Part 1  Cryptography 8
Slide 9 - Knapsack Example Private key: (2,3,7,14,30,57,120,251) m1 mod n = 411 mod 491 = 12 Public key: (82,123,287,83,248,373,10,471), n=491 Example: Encrypt 10010110 82 + 83 + 373 + 10 = 548 To decrypt, 548 · 12 = 193 mod 491 Solve (easy) SIK with S = 193 Obtain plaintext 10010110 Part 1  Cryptography 9
Slide 10 - Knapsack Weakness Trapdoor: Convert SIK into “general” knapsack using modular arithmetic One-way: General knapsack easy to encrypt, hard to solve; SIK easy to solve This knapsack cryptosystem is insecure Broken in 1983 with Apple II computer The attack uses lattice reduction “General knapsack” is not general enough! This special knapsack is easy to solve! Part 1  Cryptography 10
Slide 11 - RSA Part 1  Cryptography 11
Slide 12 - RSA Invented by Cocks (GCHQ), independently, by Rivest, Shamir and Adleman (MIT) Let p and q be two large prime numbers Let N = pq be the modulus Choose e relatively prime to (p-1)(q-1) Find d s.t. ed = 1 mod (p-1)(q-1) Public key is (N,e) Private key is d Part 1  Cryptography 12
Slide 13 - RSA To encrypt message M compute C = Me mod N To decrypt C compute M = Cd mod N 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 Part 1  Cryptography 13
Slide 14 - 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 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1k mod N = M mod N Part 1  Cryptography 14
Slide 15 - Simple RSA Example Example of RSA Select “large” primes p = 11, q = 3 Then N = pq = 33 and (p-1)(q-1) = 20 Choose e = 3 (relatively prime to 20) Find d such that ed = 1 mod 20, we find that d = 7 works Public key: (N, e) = (33, 3) Private key: d = 7 Part 1  Cryptography 15
Slide 16 - Simple RSA Example Public key: (N, e) = (33, 3) Private key: d = 7 Suppose message M = 8 Ciphertext C is computed as C = Me mod N = 83 = 512 = 17 mod 33 Decrypt C to recover the message M by M = Cd mod N = 177 = 410,338,673 = 12,434,505  33 + 8 = 8 mod 33 Part 1  Cryptography 16
Slide 17 - More Efficient RSA (1) Modular exponentiation example 520 = 95367431640625 = 25 mod 35 A better way: repeated squaring 20 = 10100 base 2 (1, 10, 101, 1010, 10100) = (1, 2, 5, 10, 20) Note that 2 = 1 2, 5 = 2  2 + 1, 10 = 2  5, 20 = 2  10 51= 5 mod 35 52= (51)2 = 52 = 25 mod 35 55= (52)2  51 = 252  5 = 3125 = 10 mod 35 510 = (55)2 = 102 = 100 = 30 mod 35 520 = (510)2 = 302 = 900 = 25 mod 35 Never have to deal with huge numbers! Part 1  Cryptography 17
Slide 18 - More Efficient RSA (2) Let e = 3 for all users (but not same N or d) Public key operations only require 2 multiplies Private key operations remain “expensive” If M < N1/3 then C = Me = M3 and cube root attack For any M, if C1, C2, C3 sent to 3 users, cube root attack works (uses Chinese Remainder Theorem) Can prevent cube root attack by padding message with random bits Note: e = 216 + 1 also used Part 1  Cryptography 18
Slide 19 - Diffie-Hellman Part 1  Cryptography 19
Slide 20 - Diffie-Hellman Invented by Williamson (GCHQ) and, independently, by D and H (Stanford) A “key exchange” algorithm Used to establish a shared symmetric key Not for encrypting or signing Security rests on difficulty of discrete log problem: given g, p and gk mod p find k Part 1  Cryptography 20
Slide 21 - Diffie-Hellman Let p be prime, let g be a generator For any x  {1,2,…,p-1} there is n s.t. x = gn mod p Alice generates secret value a Bob generates secret value b Alice sends ga mod p to Bob Bob sends gb mod p to Alice Both compute shared secret gab mod p Shared secret can be used as symmetric key Part 1  Cryptography 21
Slide 22 - Diffie-Hellman Bob & Alice use gab mod p as symmetric key Attacker can see ga mod p and gb mod p Note ga gb mod p = ga+b mod p  gab mod p If Trudy can find a or b, system is broken If Trudy can solve discrete log problem, then she can find a or b Part 1  Cryptography 22
Slide 23 - Diffie-Hellman Public: g and p Secret: Alice’s exponent a, Bob’s exponent b Part 1  Cryptography 23 Alice, a Bob, b ga mod p gb mod p Alice computes (gb)a = gba = gab mod p Bob computes (ga)b = gab mod p Could use K = gab mod p as symmetric key
Slide 24 - Diffie-Hellman Subject to man-in-the-middle (MiM) attack Part 1  Cryptography 24 Alice, a Bob, b ga mod p gb mod p Trudy, t gt mod p gt mod p Trudy shares secret gat mod p with Alice Trudy shares secret gbt mod p with Bob Alice and Bob don’t know Trudy exists!
Slide 25 - Diffie-Hellman How to prevent MiM attack? Encrypt DH exchange with symmetric key Encrypt DH exchange with public key Sign DH values with private key Other? You MUST be aware of MiM attack on Diffie-Hellman Part 1  Cryptography 25
 

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