Melden

Teilen

•0 gefällt mir•514 views

•0 gefällt mir•514 views

Melden

Teilen

Downloaden Sie, um offline zu lesen

Security Goals, What is Cryptography

- 2. Security Goals Consider the following security risks that could face two communicating entities in an unprotected environment: 2 A B C could view the secret message by eavesdropping on the communication. Loss of privacy / confidentiality C m(1)
- 3. 3 C could alter/corrupt the message, or the message could change while in transit. If B does not detect this, then we have Loss of Integrity C A Bm A B C m it could send a massage to B pretending to be A If B cannot verify the source entity of the information then we lack authentication (2) (3)
- 4. 4 A Bm A might repudiate having sent m to B Hence, some possible goals for communication: • Privacy/confidentiality - information not disclosed to unauthorized entities • Integrity - information not altered deliberately or accidentally • Authentication - validation of identity of source of information • Non-repudiation – Sender should not be able to deny sending a message (4)
- 5. What is Cryptography Cryptography is the study of mathematical techniques related to aspects of information security such as confidentiality, data integrity, authentication, and non-repudiation. 5
- 6. What is a cryptographic system composed of? 6 (encryption) (encryption key) C PP (decryption) Sender Receiver (decryption key) • Plaintext : original message or data (also called cleartext) • Encryption : transforming the plaintext, under the control of the key • Ciphertext : encrypted plaintext • Decryption : transforming the ciphertext back to the original plaintext • Cryptographic key: used with an algorithm to determine the transformation from plaintext to ciphertext, and v.v.
- 7. Attack classification 7 (encryption) (key) CP Ciphertext Alone attack: The attacker has available only the intercepted cryptogram C. From C , try to find P or (even better) the key.
- 8. Attack classification 8 Known Plaintext attack: The attacker knows a small amount of plaintext (Pi) and its ciphertext Equivalent (Ci). (encryption) (key) Ci Pi Ci+1 Pi+1 Attacker tries to find key or to infer Pi+1 (next plaintext)
- 9. Attack classification 9 Chosen Plaintext attack: The attacker can choose plaintext (Pi) and obtain its ciphertext (Ci). A careful selection of (Pi) would give a pair of (Pi, Ci) good for analyzing Enc. Alg. + key and in finding Pi+1 (next plaintext of sender) (encryption) (key) Ci Pi Ci+1 Pi+1
- 10. Forms of Cryptosystems 10 • Private Key (symmetric) : • A single key (K) is used for both encryption and decryption and must be kept secret. • Key distribution problem a secure channel is needed to transmit the key before secure communication can take place over an unsecure channel. (encryption) (K) C MM (decryption) Sender Receiver (K) EK(M) = C DK(C) = M
- 11. Forms of Cryptosystems • Public Key (asymmetric): • The encryption procedure (key) is public while the decryption procedure (key) is private. • Each participant has a public key and a private key. • May allow for both encryption of messages and creation of digital signatures.
- 12. Forms of Cryptosystems 12 Public Key (asymmetric): Requirements: 1. For every message M, encrypting with public key and then decrypting resulting ciphertext with matching private key results in M. 2. Encryption and Decryption can be efficiently applied to M 3. It is impractical to derive decryption key from encryption key. (encryption) (public key of Receiver) C MM (decryption) Sender Receiver (private key of Receiver)
- 13. Combining Public/Private Key Systems 13 • Public key encryption is more expensive than symmetric key encryption • For efficiency, combine the two approaches 1. Use public key encryption for authentication; once authenticated, transfer a shared secret symmetric key 2. Use symmetric key for encrypting subsequent data transmissions (1) (2)A B
- 14. Rivest Shamir Adelman (RSA) Method Named after the designers: Rivest, Shamir, and Adleman Public-key cryptosystem and digital signature scheme. Based on difficulty of factoring large integers For large primes p & q, n = pq Public key e and private key d calculated 14
- 15. RSA Key Generation 15 Every participant must generate a Public and Private key: 1. Let p and q be large prime numbers, randomly chosen from the set of all large prime numbers. 2. Compute n = pq. 3. Choose any large integer, d, so that: GCD( d, ϕ(n)) = 1 (where ϕ(n) = (p1)(q1) ) 4. Compute e = d-1 (mod ϕ(n)). 5. Publish n and e. Keep p, q and d secret. Note: • Step 4 can be written as: • Find e so that: e x d = 1 (modulo ϕ(n)) • If we can obtain p and q, and we have (n, e), we can find d
- 16. Rivest Shamir Adelman (RSA) Method 16 Assume A wants to send something confidentially to B: • A takes M, computes C = Me mod n, where (e, n) is B’s public key. Sends C to B • B takes C, finds M = Cd mod n, where (d, n) is B’s private key A Me mod n Cd mod n Encryption Key for user B (B’s Public Key) Decryption Key for user B (B’s PrivateKey) C (e, n) (d, n) B M M + Confidentiality
- 17. RSA Method 17 Example: 1. p = 5, q = 11 and n = 55. (p1)x(q1) = 4 x 10 = 40 2. A valid d is 23 since GCD(40, 23) = 1 3. Then e = 7 since: 23 x 7 = 161 modulo 40 = 1 in other words e = 23-1 (mod 40) = 7
- 18. Digital Signatures Based on RSA 18 • In RSA algorithm the encryption and decryption operations are commutative: ( me ) d = ( md ) e = m • We can use this property to create a digital signature with RSA.
- 19. Digital Signatures (Public Key) 19 • Public Key System: Sender, A: (EA : public, DA : private) Receiver, B: (EB : public, DB : private) • A signs the message m using its private key, the result is then encrypted with B’s public key, and the resulting ciphertext is sent to B: • C= EB (DA (M)) • B receives ciphertext C decrypts it using its private key, the result is then encrypted with the senders public key (A’s public key) and the message m is retrieved. • M = EA (DB (C))
- 20. Hashing 20 A one-way hash function h is a public function h (which should be simple and fast to compute) that satisfies three properties: • A message m of arbitrary length must be able to be converted into a message digest h(m) of fixed length. • It must be one-way, that is given y = h(m) it must be computationally infeasible to find m. • It must be collision free, that is it should be computationally infeasible to find m1 and m2 such that h(m1) = h(m2). Examples: MD5 , SHA-1
- 21. Hash Function 21 …M… H (M) Hash Function H Message of arbitrary length Fixed length output
- 22. Producing Digital Signatures 22 Step 1: A produces a one-way hash of the message. Step 2: A encrypts the hash value with its private key, forming the signature. Step 3: A sends the message and the signature to B. Hash Function Encryption Algorithm Digital Signature A’s private key message digestMessage H(M) Sig A M
- 23. Verifying Digital Signature 23 Step 4: B forms a one-way hash of the message. Step 5: B uses A’s public key to decrypt the signature and obtain the sent hash. Step 6: compare the computed and sent hashes Hash Function Decryption Algorithm Digital Signature received sender’s (A’s) public key message digest H(M’) H(M) CompareSig A M’ H(M’) Message received
- 24. Security of Digital Signatures 24 • If the hashes match then we have guaranteed the following: Integrity: if m changed then the hashes would be different Authenticity & Non-repudiation: A is who sent the hash, as we used A’s public key to reveal the contents of the signature A cannot deny signing this, nobody else has the private key. Satisfies the requirements of a Digital Signature • If we wanted to further add confidentiality, then we would encrypt the sent m + signature such that only B could reveal the contents (encrypt with B’s public key) Possible problem: If signing modulus > encrypting modulus -> Reblocking Problem
- 25. Secure Communication (Public Key) 25 BA Handshaking If B sees the same nonce at a later time, then it should suspect a replay attack. EPKA (IA, IB) EPKB, (IA, A) EPKB (IB) • IA, IB are “nonces” nonces can be included in each subsequent message • PKB: public key of B; PKA: public key of A; C EPKB (IB)
- 26. Data Encryption Standard US standard 64 bit plain text blocks 56 bit key
- 29. Strength of DES Declared insecure in 1998 Electronic Frontier Foundation DES Cracker machine DES now worthless Alternatives include TDEA
- 30. Introduction to Elliptic Curves
- 31. Lets start with a puzzle… What is the number of balls that may be piled as a square pyramid and also rearranged into a square array? Soln: Let x be the height of the pyramid… Thus, We also want this to be a square: Hence, 2 2 2 2 ( 1)(2 1) 1 2 3 ... 6 x x x x 2 ( 1)(2 1) 6 x x x y
- 32. Graphical Representation X axis Y axis Curves of this nature are called ELLIPTIC CURVES
- 33. Method of Diophantus Uses a set of known points to produce new points (0,0) and (1,1) are two trivial solutions Equation of line through these points is y=x. Intersecting with the curve and rearranging terms: We know that 1 + 0 + x = 3/2 => x = ½ and y = ½ Using symmetry of the curve we also have (1/2,-1/2) as another solution 3 23 1 0 2 2 x x x
- 34. Diophantus’ Method Consider the line through (1/2,-1/2) and (1,1) => y=3x-2 Intersecting with the curve we have: Thus ½ + 1 + x = 51/2 or x = 24 and y=70 Thus if we have 4900 balls we may arrange them in either way 3 251 ... 0 2 x x
- 35. Elliptic curves in Cryptography Elliptic Curve (EC) systems as applied to cryptography were first proposed in 1985 independently by Neal Koblitz and Victor Miller. The discrete logarithm problem on elliptic curve groups is believed to be more difficult than the corresponding problem in (the multiplicative group of nonzero elements of) the underlying finite field.
- 36. Discrete Logarithms in Finite Fields Alice Bob Pick secret, random X from F Pick secret, random Y from F gy mod p gx mod p Compute k=(gy)x=gxy mod p Compute k=(gx)y=gxy mod p Eve has to compute gxy from gx and gy without knowing x and y… She faces the Discrete Logarithm Problem in finite fields F={1,2,3,…,p-1}
- 37. Elliptic Curve on a finite set of Integers Consider y2 = x3 + 2x + 3 (mod 5) x = 0 y2 = 3 no solution (mod 5) x = 1 y2 = 6 = 1 y = 1,4 (mod 5) x = 2 y2 = 15 = 0 y = 0 (mod 5) x = 3 y2 = 36 = 1 y = 1,4 (mod 5) x = 4 y2 = 75 = 0 y = 0 (mod 5) Then points on the elliptic curve are (1,1) (1,4) (2,0) (3,1) (3,4) (4,0) and the point at infinity: Using the finite fields we can form an Elliptic Curve Group where we also have a DLP problem which is harder to solve…
- 38. Definition of Elliptic curves An elliptic curve over a field K is a nonsingular cubic curve in two variables, f(x,y) =0 with a rational point (which may be a point at infinity). The field K is usually taken to be the complex numbers, reals, rationals, algebraic extensions of rationals, p-adic numbers, or a finite field. Elliptic curves groups for cryptography are examined with the underlying fields of Fp (where p>3 is a prime) and F2 m (a binary representation with 2m elements).
- 39. General form of a EC An elliptic curve is a plane curve defined by an equation of the form baxxy 32 Examples
- 40. Weierstrass Equation A two variable equation F(x,y)=0, forms a curve in the plane. We are seeking geometric arithmetic methods to find solutions Generalized Weierstrass Equation of elliptic curves: 2 2 2 1 3 2 4 6y a xy a y x a x a x a Here, A, B, x and y all belong to a field of say rational numbers, complex numbers, finite fields (Fp) or Galois Fields (GF(2n)).
- 41. If Characteristic field is not 2: If Characteristics of field is neither 2 nor 3: 22 2 3 23 31 1 2 4 6 2 3 ' 2 ' ' 1 2 4 6 ( ) ( ) ( ) 2 2 4 4 a aa x a y x a x a x a y x a x a x a ' 1 2 2 3 1 1 1 /3x x a y x Ax B
- 42. Points on the Elliptic Curve (EC) Elliptic Curve over field L It is useful to add the point at infinity. The point is sitting at the top of the y-axis and any line is said to pass through the point when it is vertical. It is both the top and at the bottom of the y-axis. 2 3 ( ) { } {( , ) | ... ...}E L x y L L y x
- 43. The Abelian Group P + Q = Q + P (commutativity) (P + Q) + R = P + (Q + R) (associativity) P + O = O + P = P (existence of an identity element) there exists ( − P) such that − P + P = P + ( − P) = O (existence of inverses) Given two points P,Q in E(Fp), there is a third point, denoted by P+Q on E(Fp), and the following relations hold for all P,Q,R in E(Fp).
- 44. Elliptic Curve Picture Consider elliptic curve E: y2 = x3 - x + 1 If P1 and P2 are on E, we can define P3 = P1 + P2 as shown in picture Addition is all we need P1 P2 P3 x y
- 45. Addition in Affine Co-ordinates x y 1 1 2 2 3 3 ( , ), ( , ) ( ) ( , ) P x y Q x y R P Q x y y=m(x-x1)+y1 Let, P≠Q, y2=x3+Ax+B
- 46. Doubling of a point Let, P=Q What happens when P2=∞? 2 2 1 1 1 1 2 3 2 2 2 3 1 3 1 3 1 2 3 3 2 , 0 (since then P +P = ): 0 ... 2 , ( ) dy y x A dx dy x A m dx y If y x m x x m x y m x x y
- 47. Why do we need the reflection? P2=O=∞ P1 y P1=P1+ O=P1
- 48. Sum of two points 21 1 2 1 21 12 12 _ 2 3 _ xxfor y ax xxfor xx yy Define for two points P (x1,y1) and Q (x2,y2) in the Elliptic curve Then P+Q is given by R(x3,y3) : 1133 213 )( yxxy xxx
- 49. P+P = 2P Point at infinity O • As a result of the above case P=O+P • O is called the additive identity of the elliptic curve group. • Hence all elliptic curves have an additive identity O.
- 50. Projective Co-ordinates Two-dimensional projective space over K is given by the equivalence classes of triples (x,y,z) with x,y z in K and at least one of x, y, z nonzero. Two triples (x1,y1,z1) and (x2,y2,z2) are said to be equivalent if there exists a non-zero element λ in K, st: (x1,y1,z1) = (λx2, λy2, λz2) The equivalence class depends only the ratios and hence is denoted by (x:y:z) 2 KP
- 51. Projective Co-ordinates If z≠0, (x:y:z)=(x/z:y/z:1) What is z=0? We obtain the point at infinity. The two dimensional affine plane over K: 2 2 2 {( , ) } Hence using, ( , ) ( : :1) K K K A x y K K x y X Y A P There are advantages with projective co-ordinates from the implementation point of view
- 52. Singularity For an elliptic curve y2=f(x), define F(x,y)=y2-F(x). A singularity of the EC is a pt (x0,y0) such that: 0 0 0 0 0 0 0 0 ( , ) ( , ) 0 ,2 '( ) 0 , ( ) '( ) f has a double root F F x y x y x y or y f x or f x f x It is usual to assume the EC has no singular points
- 53. 1. Hence condition for no singularity is 4A3+27B2≠0 2. Generally, EC curves have no singularity 0 0 0 0 0 0 0 0 2 3 3 2 2 4 2 2 2 2 2 2 3 2 ( , ) ( , ) 0 , 2 '( ) 0 , ( ) '( ) f has a double root For double roots, 3 0 / 3. Also, +Bx=0, 0 9 3 2 9 2 3( ) 0 9 4 27 0 F F x y x y x y or y f x or f x f x y x Ax B x Ax B x A x A x Ax A A Bx A x B A A B A B 2 3 ( )y f x x Ax B If Characteristics of field is not 3:
- 54. Elliptic Curves in Characteristic 2 Generalized Equation: If a1 is not 0, this reduces to the form: If a1 is 0, the reduced form is: Note that the form cannot be: 2 3 2 y xy x Ax B 2 3 2 1 3 2 4 6y a xy a y x a x a x a 2 3 y Ay x Bx C 2 3 y x Ax B
- 55. Outline of the Talk… Introduction to Elliptic Curves Elliptic Curve Cryptosystems Implementation of ECC in Binary Fields
- 57. Public-Key Cryptosystems Secrecy: Only B can Decrypt the message Authentication: Only A can generate the encrypted message
- 60. What Is Elliptic Curve Cryptography (ECC)? Elliptic curve cryptography [ECC] is a public-key cryptosystem just like RSA, Rabin, and El Gamal. Every user has a public and a private key. Public key is used for encryption/signature verification. Private key is used for decryption/signature generation. Elliptic curves are used as an extension to other current cryptosystems. Elliptic Curve Diffie-Hellman Key Exchange Elliptic Curve Digital Signature Algorithm
- 61. Using Elliptic Curves In Cryptography The central part of any cryptosystem involving elliptic curves is the elliptic group. All public-key cryptosystems have some underlying mathematical operation. RSA has exponentiation (raising the message or ciphertext to the public or private values) ECC has point multiplication (repeated addition of two points).
- 62. Generic Procedures of ECC Both parties agree to some publicly-known data items The elliptic curve equation values of a and b prime, p The elliptic group computed from the elliptic curve equation A base point, B, taken from the elliptic group Similar to the generator used in current cryptosystems Each user generates their public/private key pair Private Key = an integer, x, selected from the interval [1, p-1] Public Key = product, Q, of private key and base point (Q = x*B)
- 63. Example – Elliptic Curve Cryptosystem Analog to El Gamal Suppose Alice wants to send to Bob an encrypted message. Both agree on a base point, B. Alice and Bob create public/private keys. Alice Private Key = a Public Key = PA = a * B Bob Private Key = b Public Key = PB = b * B Alice takes plaintext message, M, and encodes it onto a point, PM, from the elliptic group
- 64. Example – Elliptic Curve Cryptosystem Analog to El Gamal Alice chooses another random integer, k from the interval [1, p-1] The ciphertext is a pair of points PC = [ (kB), (PM + kPB) ] To decrypt, Bob computes the product of the first point from PC and his private key, b b * (kB) Bob then takes this product and subtracts it from the second point from PC (PM + kPB) – [b(kB)] = PM + k(bB) – b(kB) = PM Bob then decodes PM to get the message, M.
- 65. Example – Compare to El Gamal The ciphertext is a pair of points PC = [ (kB), (PM + kPB) ] The ciphertext in El Gamal is also a pair. C = (gk mod p, mPB k mod p) -------------------------------------------------------------------------- Bob then takes this product and subtracts it from the second point from PC (PM + kPB) – [b(kB)] = PM + k(bB) – b(kB) = PM In El Gamal, Bob takes the quotient of the second value and the first value raised to Bob’s private value m = mPB k / (gk)b = mgk*b / gk*b = m
- 66. Diffie-Hellman (DH) Key Exchange
- 67. ECC Diffie-Hellman Public: Elliptic curve and point B=(x,y) on curve Secret: Alice’s a and Bob’s b Alice, A Bob, B a(x,y) b(x,y) • Alice computes a(b(x,y)) • Bob computes b(a(x,y)) • These are the same since ab = ba
- 68. Example – Elliptic Curve Diffie-Hellman Exchange Alice and Bob want to agree on a shared key. Alice and Bob compute their public and private keys. Alice Private Key = a Public Key = PA = a * B Bob Private Key = b Public Key = PB = b * B Alice and Bob send each other their public keys. Both take the product of their private key and the other user’s public key. Alice KAB = a(bB) Bob KAB = b(aB) Shared Secret Key = KAB = abB
- 69. Why use ECC? How do we analyze Cryptosystems? How difficult is the underlying problem that it is based upon RSA – Integer Factorization DH – Discrete Logarithms ECC - Elliptic Curve Discrete Logarithm problem How do we measure difficulty? We examine the algorithms used to solve these problems
- 70. Security of ECC To protect a 128 bit AES key it would take a: RSA Key Size: 3072 bits ECC Key Size: 256 bits How do we strengthen RSA? Increase the key length Impractical?
- 71. Applications of ECC Many devices are small and have limited storage and computational power Where can we apply ECC? Wireless communication devices Smart cards Web servers that need to handle many encryption sessions Any application where security is needed but lacks the power, storage and computational power that is necessary for our current cryptosystems
- 72. Benefits of ECC Same benefits of the other cryptosystems: confidentiality, integrity, authentication and non-repudiation but… Shorter key lengths Encryption, Decryption and Signature Verification speed up Storage and bandwidth savings
- 73. Summary of ECC “Hard problem” analogous to discrete log Q=kP, where Q,P belong to a prime curve given k,P “easy” to compute Q given Q,P “hard” to find k known as the elliptic curve logarithm problem k must be large enough ECC security relies on elliptic curve logarithm problem compared to factoring, can use much smaller key sizes than with RSA etc for similar security ECC offers significant computational advantages
- 74. Public-Key Cryptosystems Secrecy: Only B can Decrypt the message Authentication: Only A can generate the encrypted message