1 / 29

Electronic Payment Systems 20-763 Lecture 6 Epayment Security II

Electronic Payment Systems 20-763 Lecture 6 Epayment Security II. Public-Key (Asymmetric) Encryption. 3. SITE USES ITS PRIVATE KEY FOR DECRYPTION. 2. SENDERS USE SITE’S PUBLIC KEY FOR ENCRYPTION. 4. ONLY WEBSITE CAN DECRYPT THE CIPHERTEXT. NO ONE ELSE KNOWS HOW.

alena
Download Presentation

Electronic Payment Systems 20-763 Lecture 6 Epayment Security II

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Electronic Payment Systems20-763Lecture 6Epayment Security II

  2. Public-Key (Asymmetric) Encryption 3. SITE USES ITS PRIVATE KEY FOR DECRYPTION 2. SENDERS USE SITE’S PUBLIC KEY FOR ENCRYPTION 4. ONLY WEBSITE CAN DECRYPT THE CIPHERTEXT. NO ONE ELSE KNOWS HOW 1. USERS WANT TO SEND PLAINTEXT TO RECIPIENT WEBSITE SOURCE: STEIN, WEB SECURITY

  3. Public-Key Encryption • Alice wants to send Bob a secure message M. • Alice uses Bob’s public key to encrypt M. • Bob uses his private key to decrypt M. • Bob is the ONLY ONE who can do this,so M is secure. • Problem: Anyone could have sent it. Was it really Alice? BOB’SPUBLIC KEY BOB’SPRIVATE KEY BOB DECRYPTS WITH HIS PRIVATE KEY ALICE ENCRYPTS WITH BOB’S PUBLIC KEY ALICE’SCLEARTEXT ALICE’SCODEDTEXT ALICE’SCODEDTEXT ALICE’SCLEARTEXT TRANSM ISSION

  4. Digital Authentication • Alice wants to send Bob a message M so that Bob is sure Alice is the sender. • Alice uses her own private key to encrypt M. • Bob uses Alice’s public key to decrypt M. • Alice is the ONLY ONE who could have sent it. • Problem 1: Anyone can read it! Problem 2: Replay attack! ALICE’S PRIVATE KEY ALICE’S PUBLIC KEY BOB DECRYPTS WITH ALICE’S PUBLIC KEY ALICE ENCRYPTS WITH HER PRIVATE KEY ALICE’SCLEARTEXT ALICE’SCODEDTEXT ALICE’SCODEDTEXT ALICE’SCLEARTEXT TRANSM ISSION

  5. BOB’S PUBLIC ALICE’S PRIVATE ALICE ENCRYPTS WITH HER PRIVATE KEY ALICE ENCRYPTS WITH BOB’S PUBLIC KEY ALICE’S PUBLIC BOB’S PRIVATE ALICE’SCODED ANDSIGNED TEXT ALICE’SCODED ANDSIGNED TEXT T R A N S M I BOB DECRYPTS WITH ALICE’S PUBLIC KEY BOB DECRYPTS WITH HIS PRIVATE KEY T ALICE’SCLEAR TEXT (DECRYPTED AND AUTHENTICATED) ALICE’SCODEDTEXT (AUTHENTICATED) ALICE’SCLEARTEXT ALICE’SCODEDTEXT Secure Authenticated Messages • Alice must send Bob a secret & authenticated message M so Bob is sure it was sent by Alice. Use both encryption and signature. 4 KEYS NEEDED:

  6. One-Way Trapdoor Function • A function that is easy to compute • Computationally difficult to invert without knowing the secret (the “trapdoor”) • Example: f(x, y) = x•y • Given f(x, y), it is difficult to find either x or y • Given f(x, y) and x (the secret), it is easy to find y • Any one-way trapdoor function can be used in public-key cryptography.

  7. Rivest-Shamir-Adelman (RSA) • It is easy to multiply two numbers but apparently hard to factor a number into a product of two others. • Given p, q, it is easy to compute n = p • q • Example: p = 5453089; q = 3918067 • Easy to find n = 21365568058963 • Given n, hard to find two numbers p, q with p • q = n • Now suppose n = 7859112349338149 What are p and q such that p • q = n ? • Multiplication is a one-way function • RSA exploits this fact in public-key encryption

  8. RSA Encryption • Select two large prime numbers p, q (e.g. 1024 bits) • Let n = p • q • Choose a small odd integer e that does not dividem = (p - 1)(q - 1). Then x(p-1)(q-1) = 1 (mod n) • Compute d = e-1(mod m) • That is, d • e gives remainder 1 when divided by m • Then xe•d = x (mod n) (by Fermat’s “Little” Theorem) • Public key is the pair (e, n) • Private key is the pair (d, n) • Knowing (e, n) is of no help in finding d. Still need p and q, which involves factoring n

  9. 6 • 2 = 12 WHEN DIVIDED BY 7 GIVES REMAINDER 5 MULTIPLICATION MOD 7 INVERSE OF 5 IS 3 EACH ROW EXCEPT THE ZERO ROW HAS EXACTLY ONE 1 EACH ELEMENT HAS A UNIQUE INVERSE Multiplicative InversesOver Finite Fields • The inverse e-1 of a number e satisfies e-1 • e = 1 • The inverse of 5 is 1/5 • If we only allow numbers from 0 to n-1 (mod n), then for special values of n, each e has a unique inverse

  10. RSA Encryption • Message M is a number • To encrypt message M using key (e, n): • Compute C(M) = Me (mod n) • To decrypt message C using key (d, n): • Compute P(C) = Cd (mod n) • Note that P(C(M)) = C(P(M)) = (Me)d (mod n) = Me•d (mod n) = Mbecause e • d = 1 and m = (p-1)(q-1) • DEMO

  11. RSA Example p = 61; q = 53 n = pq = 3233 (modulus, can be given to others) e = 17 (public exponent, can be given to others) d = 2753 (private exponent, kept secret!) PUBLIC KEY = (3233, 17) PRIVATE KEY = (3233, 2753) To encrypt 123, compute 12317 (mod 3233) =337587917446653715596592958817679803 mod 3233 = 855 To decrypt 855, compute 8552753 (mod 3233) = 123 (intermediate value has 8072 digits) 37 digits SOURCE: FRANCIS LITTERIO

  12. Trapdoor Functions for Cryptogrpahy • Any one-way trapdoor function f(x) can be used for public-key cryptography • Alice wants to send message m to Bob • Bob’s public key e is a parameter to the trapdoor function fe(x) (the inverse fe-1(x) is easy to compute knowing Bob’s private key d but difficult without d) • Alice computes fe(m), sends it to Bob • Bob computes fe-1(fe(m)) = m (easy if d is known) • Eavesdropper Eve can’t compute m = fe-1(fe(m)) without the trapdoor d to find the inverse fe-1

  13. Digital Signatures • A handwritten signature is a function of the signer only, not the message • Handwritten signatures can be copied and forged • The digital equivalent of a handwritten signature would be useless in eCommerce • Must be able to • Compare it with the “real” signature; AND • Must be sure it isn’t copied or forged • How can A prove his identity over the Internet?

  14. Digital Signatures • A digital signature is a function of both the signer and the message • A digital signature is a digest of the message encrypted with the signer’s private key MESSAGE M (LONG) USE SECURE HASH ALGORITHM (SHA) TO PRODUCE HASH (MESSAGE DIGEST) HASH ENCRYPT HASH USING SIGNER’S PRIVATE KEY PRIVATE KEY OF MR. A SIG THIS IS THE DIGITAL SIGNATURE OF MR. A ON MESSAGE M

  15. SIG MESSAGE (LONG) Authentication by Digital Signature RECIPIENT RECEIVES SIG + MESSAGE MESSAGE (LONG) RECIPIENT USES SHA TO COMPUTE HASH RECIPIENT DECRYPTS SIG WITH SIGNER’S PUBLIC KEY =? HASH HASH IF HASHES ARE EQUAL, MESSAGE IS AUTHENTIC. WHY? IF ANY BIT OF M OR SIG IS ALTERED, HASH CHANGES.

  16. Digital Signature • Message digest encrypted with signer’s private key MESSAGE (LONG) USE SHA TO PRODUCE HASH (MESSAGE DIGEST) HASH ENCRYPT HASH WITH SIGNER’S PRIVATE KEY PRIVATE KEY SIG APPEND SIGNATURE TO MESSAGE; SEND BOTH SIG MESSAGE (LONG) Recipient decrypts SIG with signer’s public key. Recipient computes the message digest. If it matches the SIG, the SIG is genuine AND the message has not been altered!

  17. Discrete Logarithms • If ab = c, we say that logac = b • Example: 232 = 4294927296 so log2(4294927296) = 32 • Computing ab and logac are both easy for real numbers • In a finite field, it is easy to calculate c = ab mod p but given c, a and p it is very difficult to find b • This is the “discrete logarithm” problem • Analogy: Given x it is easy to find two real numbers y, z such that x = y•z • Given an integer n it is hard to find two integers p, q such that n = p•q

  18. Diffie-Hellman Key Exchange • Object: allow Alice and Bob to exchange a secret key • Protocol has two public parameters: a prime p and a number g < p such that given 0 < n < p there is some k such that gk = n (g is called a generator) • Alice and Bob generate random private values a, b between 1 and p-2 • Alice’s public value is ga (mod p); Bob’s is gb (mod p) • Alice and Bob share their public values • Alice computes (gb)a (mod p) = gba • Bob computes (ga)b (mod p) = gab =gba • Let key = gab. Now both Alice and Bob have it. • No one else can compute it -- they don’t know a or b

  19. El Gamal Encryption • Based on the discrete logarithm • Bob’s public key is (p, q, r) • Bob’s private key is s such that r = qs mod p • Alice sends Bob the message m by picking a random secret number k and sending (a, b) = (qk mod p, mrk mod p) • Bob computes b (as )-1 mod p = mrk (qks)-1 = mqks (qks)-1 = m • (Bob knows s; nobody else can do this)

  20. x y Elliptic Curve Cryptography (ECC) • An elliptic curve is the set of points (x, y) satisfyingy2 + axy + by = x3 + cx2 + dx + e An elliptic curve has the property that aline drawn between two points of the curveintersects the curve at a single point.(Warning: need to include the point at infinity.) This allows us to define P + Q so that the sum is always another point on the curve. If the sum P + Q is always on the curve, so are the points P, P + P, P + P + P, . . . = P, 2P, 3P, 4P, . . . ONLINE TUTORIAL

  21. Elliptic Curve Operations The point at infinity Ois an identity element for addition SOURCE: INTEGRITY SCIENCES

  22. Elliptic Curves Over Finite Fields • Select a large prime number p • Choose two non-negative integers a and b with4a2 + 27b2 0 (mod p) • The pairs (x, y) with x, y < p that satisfyy2 = x3 + ax + b (mod p) are the elliptic group mod p • addition is closed and associative (x + y) + z = x + (y + z) • there is an identity element O such that x + O = x • every element x has an inverse x-1 such that x + x-1 = O • If y = k x (mod p), then given k and x it is easy to find y but given x and y it is computationally hard to find k • So elliptic curves can be used for cryptography

  23. Elliptic Curves for El Gamal • Multiplication in the elliptic group corresponds to exponentiation of real numbers • Solving y = k x (mod p) for k in the elliptic group is similar to solving c = ab (mod p) for b in El Gamal (discrete logarithm) • Choose a special point g of the group (called a generator) • Bob’s private key is s; Bob’s public key is (g, sg) • A plaintext message m is transformed to a point x in the group • Alice encrypts x by picking a random value k and sending(kg, x + ksg) • Bob decrypts by computing (x + ksg) - (kg)s = x Can’t find sfrom g and sg g and sg are public; Alice knows x and k Alice sent him these Bob knows s (his private key)

  24. Security of ECC versus RSA ECC Advantages 1. The elliptic curve logarithm problem is harder than the discrete logarithm problem. 2. Key size in ECC is much smaller for a given security level. 3. ECC is complicated; fewer people understand it. 4. ECC is not patented. GRAPHIC: RICHARD SOUTHERN

  25. Birthday Attacks • Dave’s birthday is Jan. 29. How many people must be in a room for the probability to be > 1/2 that someone else was born on Jan. 29? • Probability that 1 person was not born on Jan. 29 =364/365. • Probability that n people were not born on Jan. 29 is p(n) = (364/365)n. Now choose n so that p(n) < 0.5 • log p(n) < n log (364/635) • n > log(1/2)/log(364/365)  253 • If n = 183 (half of 366), p(n) = 0.6053. Less then 40% chance that someone else has same birthday

  26. Birthday Probabilities • Suppose a year has d days. How many people must be in a room for the probability to be > 1/2 that some pair of people have the same birthday? • Label the people 1 … n • Probability that person i has no birthday in common with people 1 … i -1 is (d - i + 1)/d, so • If d = 365 and n = 23, p(n)  0.4927 • If d = 365 and n = 50, p(n)  0.0296 • For large d, taking n  1.17 gives p(n) > 1/2

  27. Attacking Hash Algorithms • If two strings M and M* can be found such thatH(M) = H(M*) then a hash algorithm can be compromised • Let M = PO for $100; M* = PO for $100,000 • John digitally signs H(M), so it can’t be altered! • If H(M*) = H(M) then we can “prove” in court that John signed the $100,000 PO • Birthday attack: If the hash length is b bits, then d = 2b ;= 2b/2 • Try about 2b/2 small variations of the message. Prob.~ 50% we will find one that hashes to the same value • If the digest is 64 bits, try 232 variations. Possible!

  28. Major Ideas • Digital signature = message digest encrypted with signer’s private key • Dual signature: two people sign a document without being able to read the other person’s content • Blind signature: one person signs a document without being able to read it • Any trapdoor function can be used for public-key cryptography • Great care must be used with public-key systems to avoid protocol failure (allowing cracking through mistakes) • Elliptic-curve cryptography (ECC) is replacing RSA • Shorter keys for the same level of security

  29. Q A &

More Related