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Foundations of Network and Computer Security J J ohn Black Lecture #7 Sep 13 th 2005 CSCI 6268/TLEN 5831, Fall 2005
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CBC MAC (again) Review: –A common method is the CBC MAC: CBC MAC is stateless (no nonce N is used) Proven security in the ACMA model provided messages are all of once fixed length Resistance to forgery quadratic in the aggregate length of adversarial queries plus any insecurity of AES Widely used: ANSI X9.19, FIPS 113, ISO 9797-1 AES K M1M1 tag M2M2 MmMm
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Varying Message Lengths: XCBC There are several well-known ways to overcome this limitation of CBC MAC XCBC, is the most efficient one known, and is provably- secure (when the underlying block cipher is computationally indistinguishable from random) –Uses blockcipher key K1 and needs two additional n-bit keys K2 and K3 which are XORed in just before the last encipherment A proposed NIST standard (as “CMAC”) AES K1 M1M1 tag M2M2 MmMm K2 if n divides |M| K3 otherwise
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UMAC: MACing Faster In many contexts, cryptography needs to be as fast as possible –High-end routers process > 1Gbps –High-end web servers process > 1000 requests/sec But AES (a very fast block cipher) is already more than 15 cycles-per-byte on a PPro –Block ciphers are relatively expensive; it’s possible to build faster MACs UMAC is roughly ten times as fast as current practice
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UMAC follows the Wegman-Carter Paradigm Since AES is (relatively) slow, let’s avoid using it unless we have to –Wegman-Carter MACs provide a way to process M first with a non-cryptographic hash function to reduce its size, and then encrypt the result Message M hash function hash key encrypt encryption key hash(M) tag
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The Ubiquitous HMAC The most widely-used MAC (IPSec, SSL, many VPNs) Doesn’t use a blockcipher or any universal hash family –Instead uses something called a “collision resistant hash function” H Sometimes called “cryptographic hash functions” Keyless object – more in a moment HMAC K (M) = H(K © opad || H(K © ipad || M)) opad is 0x36 repeated as needed ipad is 0x5C repeated as needed
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Notes on HMAC Fast –Faster than CBC MAC or XCBC Because these crypto hash functions are fast Slow –Slower than UMAC and other universal-hash-family MACs Proven security –But these crypto hash functions have recently been attacked and may show further weaknesses soon
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What are cryptographic hash functions? Output Message e.g., MD5,SHA-1 Hash Function A cryptographic hash function takes a message from {0,1} * and produces a fixed size output Output is called “hash” or “digest” or “fingerprint” There is no key The most well-known are MD5 and SHA-1 but there are other options MD5 outputs 128 bits SHA-1 outputs 160 bits % md5 Hello There ^D A82fadb196cba39eb884736dcca303a6 %
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T A << 5 + g t (B, C, D) + E + K t + W t SHA-1... M1M1 M2M2 MmMm for i = 1 to m do Wt ={Wt ={ t-th word of M i 0 t 15 ( W t-3 © W t-8 © W t-14 © W t-16 ) << 1 16 t 79 A H 0 i-1 ; B H 1 i-1 ; C H 2 i-1 ; D H 3 i-1 ; E H 4 i-1 for t = 1 to 80 do E D; D C; C B >> 2; B A; A T H 0 i A H 0 i-1 ; H 1 i B + H 1 i-1 ; H 2 i C+ H 2 i-1 ; H 3 i D + H 3 i-1 ; H 4 i E + H 4 i-1 end return H 0 m H 1 m H 2 m H 3 m H 4 m 512 bits 160 bits
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Real-world applications Message authentication codes (HMAC) Digital signatures (hash-and-sign) File comparison (compare-by-hash, eg, RSYNC) Micropayment schemes Commitment protocols Timestamping Key exchange... Hash functions are pervasive
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A cryptographic property BAD: H(M) = M mod 701 (quite informal) 1. Collision resistance given a hash function it is hard to find two colliding inputs H M {0,1} n H M’M’ Strings
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More cryptographic properties 1. Collision resistance given a hash function it is hard to find two colliding inputs 3. Preimage resistance given a hash function and given an hash output it is hard to invert that output 2. Second-preimage given a hash function and resistance given a first input, it is hard to find a second input that collides with the first
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Merkle-Damgard construction IV M1M1 M2M2 M3M3 h1h1 h2h2 h 3 = H (M) n k Fixed initial value Chaining value Compression function fff k MD Theorem: if f is CR, then so is H
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MiMi T A << 5 + g t (B, C, D) + E + K t + W t... M1M1 M2M2 MmMm for i = 1 to m do Wt ={Wt ={ t-th word of M i 0 t 15 ( W t-3 W t-8 W t-14 W t-16 ) << 1 16 t 79 A H 0 i-1 ; B H 1 i-1 ; C H 2 i-1 ; D H 3 i-1 ; E H 4 i-1 for t = 1 to 80 do E D; D C; C B >> 2; B A; A T H 0 i A H 0 i-1 ; H 1 i B + H 1 i-1 ; H 2 i C+ H 2 i-1 ; H 3 i D + H 3 i-1 ; H 4 i E + H 4 i-1 end return H 0 m H 1 m H 2 m H 3 m H 4 m 512 bits 160 bits H 0..4 i- 1 160 bits
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Hash Function Security Consider best-case scenario (random outputs) If a hash function output only 1 bit, how long would we expect to avoid collisions? –Expectation: 1 £ 0 + 2 £ ½ + 3 £ ½ = 2.5 What about 2 bits? –Expectation: 1 £ 0 + 2 £ ¼ + 3 £ ¾ ½ + 4 £ ¾ ½ ¾ + 5 £ ¾ ½ ¼ ¼ 3.22 This is too hard…
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Birthday Paradox Need another method –Birthday paradox: if we have 23 people in a room, the probability is > 50% that two will share the same birthday Assumes uniformity of birthdays –Untrue, but this only increases chance of birthday match Ignores leap years (probably doesn’t matter much) –Try an experiment with the class…
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Birthday Paradox (cont) Let’s do the math –Let n equal number of people in the class –Start with n = 1 and count upward Let NBM be the event that there are No-Birthday-Matches For n=1, Pr[NBM] = 1 For n=2, Pr[NBM] = 1 £ 364/365 ¼.997 For n=3, Pr[NBM] = 1 £ 364/365 £ 363/365 ¼.991 … For n=22, Pr[NBM] = 1 £ … £ 344/365 ¼.524 For n=23, Pr[NBM] = 1 £ … £ 343/365 ¼.493 –Since the probability of a match is 1 – Pr[NBM] we see that n=23 is the smallest number where the probability exceeds 50%
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Occupancy Problems What does this have to do with hashing? –Suppose each hash output is uniform and random on {0,1} n –Then it’s as if we’re throwing a ball into one of 2 n bins at random and asking when a bin contains at least 2 balls This is a well-studied area in probability theory called “occupancy problems” –It’s well-known that the probability of a collision occurs around the square-root of the number of bins If we have 2 n bins, the square-root is 2 n/2
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Birthday Bounds This means that even a perfect n-bit hash function will start to exhibit collisions when the number of inputs nears 2 n/2 –This is known as the “birthday bound” –It’s impossible to do better, but quite easy to do worse It is therefore hoped that it takes (2 64 ) work to find collisions in MD5 and (2 80 ) work to find collisions in SHA-1
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The Birthday Bound 1.0 Probability 0.0 0.5 2n2n Number of Hash Inputs 2 n/2
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Latest News At CRYPTO 2004 (August) –Collisions found in HAVAL, RIPEMD, MD4, MD5, and SHA-0 (2 40 operations) Wang, Feng, Lai, Yu Only Lai is well-known –HAVAL was known to be bad –Dobbertin found collisions in MD4 years ago –MD5 news is big! CU team has lowered time-to-collision to 3 mins (July 2005) –SHA-0 isn’t used anymore (but see next slide)
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Collisions in SHA-0 T A << 5 + g t (B, C, D) + E + K t + W t Wt ={Wt ={ t-th word of M i 0 t 15 ( W t-3 W t-8 W t-14 W t-16 ) << 1 16 t 79 A H 0 i-1 ; B H 1 i-1 ; C H 2 i-1 ; D H 3 i-1 ; E H 4 i-1 for t = 1 to 80 do E D; D C; C B >> 2; B A; A T H 0 i H 0 i-1 ; H 1 i A + H 1 i-1 ; H 2 i C+ H 2 i-1 ; H 3 i D + H 3 i-1 ; H 4 i E + H 4 i-1 end H 0..4 i- 1 65 not in SHA-0 M1,M1, M1’M1’ Collision!
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What Does this Mean? Who knows –Methods are not yet understood –Will undoubtedly be extended to more attacks –Maybe nothing much more will happen –But maybe everything will come tumbling down?! But we have OTHER ways to build hash functions
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A Provably-Secure Blockcipher-Based Compression Function E MiMi h i-1 hihi n bits
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The Big (Partial) Picture Primitives Block Ciphers Hash Functions Hard Problems Stream Ciphers First-Level Protocols Symmetric Encryption Digital Signatures MAC Schemes Asymmetric Encryption Second-Level Protocols SSH, SSL/TLS, IPSec Electronic Cash, Electronic Voting (Can do proofs) (No one knows how to prove security; make assumptions)
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Symmetric vs. Asymmetric Thus far we have been in the symmetric key model –We have assumed that Alice and Bob share some random secret string –In practice, this is a big limitation Bootstrap problem Forces Alice and Bob to meet in person or use some mechanism outside our protocol Not practical when you want to buy books at Amazon We need the Asymmetric Key model!
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Asymmetric Cryptography In this model, we no longer require an initial shared key –First envisioned by Diffie in the late 70’s –Some thought it was impossible –MI6 purportedly already knew a method –Diffie-Hellman key exchange was first public system Later turned into El Gamal public-key system –RSA system announced shortly thereafter
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But first, a little math… A group is a nonempty set G along with an operation # : G £ G ! G such that for all a, b, c 2 G –(a # b) # c = a # (b # c) (associativity) – 9 e 2 G such that e # a = a # e = a (identity) – 9 a -1 2 G such that a # a -1 = e (inverses) If 8 a,b 2 G, a # b = b # a we say the group is “commutative” or “abelian” –All groups in this course will be abelian
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Notation We’ll get tired of writing the # sign and just use juxtaposition instead –In other words, a # b will be written ab –If some other symbol is conventional, we’ll use it instead (examples to follow) We’ll use power-notation in the usual way –a b means aaaa a repeated b times –a -b means a -1 a -1 a -1 a -1 repeated b times –Here a 2 G, b 2 Z Instead of e we’ll use a more conventional identity name like 0 or 1 Often we write G to mean the group (along with its operation) and the associated set of elements interchangeably
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Examples of Groups Z (the integers) under + ? Q, R, C, under + ? N under + ? Q under £ ? Z under £ ? 2 £ 2 matrices with real entries under £ ? Invertible 2 £ 2 matrices with real entries under £ ? Note all these groups are infinite –Meaning there are an infinite number of elements in them Can we have finite groups?
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Finite Groups Simplest example is G = {0} under + –Called the “trivial group” Almost as simple is G = {0, 1} under addition mod 2 Let’s generalize –Z m is the group of integers modulo m –Z m = {0, 1, …, m-1} –Operation is addition modulo m –Identity is 0 –Inverse of any a 2 Z m is m-a –Also abelian
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The Group Z m An example –Let m = 6 –Z 6 = {0,1,2,3,4,5} –2+5 = 1 –3+5+1 = 3 + 0 = 3 –Inverse of 2 is 4 2+4 = 0 We can always pair an element with its inverse a : 0 1 2 3 4 5 a -1 : 0 5 4 3 2 1 Inverses are always unique An element can be its own inverse –Above, 0 and 0, 3 and 3
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Another Finite Group Let G = {0,1} n and operation is © –A group? –What is the identity? –What is the inverse of a 2 G? We can put some familiar concepts into group-theoretic notation: –Caesar cipher was just P + K = C in Z 26 –One-time pad was just P © K = C in the group just mentioned above
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Multiplicative Groups Is {0, 1, …, m-1} a group under multiplication mod m? –No, 0 has no inverse Ok, toss out 0; is {1, …, m-1} a group under multiplication mod m? –Hmm, try some examples… m = 2, so G = {1} X m = 3, so G = {1,2} X m = 4, so G = {1,2,3} oops! m = 5, so G = {1,2,3,4} X
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Multiplicative Groups (cont) What was the problem? –2,3,5 all prime –4 is composite (meaning “not prime”) Theorem: G = {1, 2, …, m-1} is a group under multiplication mod m iff m is prime Proof: Ã : suppose m is composite, then m = ab where a,b 2 G and a, b 1. Then ab = m = 0 and G is not closed ! : follows from a more general theorem we state in a moment
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The Group Z m * a,b 2 N are relatively prime iff gcd(a,b) = 1 –Often we’ll write (a,b) instead of gcd(a,b) Theorem: G = {a : 1 · a · m-1, (a,m) = 1} and operation is multiplication mod m yields a group –We name this group Z m * –We won’t prove this (though not too hard) –If m is prime, we recover our first theorem
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Examples of Z m * Let m = 15 –What elements are in Z 15 * ? {1,2,4,7,8,11,13,14} –What is 2 -1 in Z 15 * ? First you should check that 2 2 Z 15 * It is since (2,15) = 1 –Trial and error: 1, 2, 4, 7, 8 X –There is a more efficient way to do this called “Euclid’s Extended Algorithm” Trust me
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Euler’s Phi Function Definition: The number of elements of a group G is called the order of G and is written |G| –For infinite groups we say |G| = 1 –All groups we deal with in cryptography are finite Definition: The number of integers i < m such that (i,m) = 1 is denoted (m) and is called the “Euler Phi Function” –Note that |Z m * | = (m) –This follows immediately from the definition of ()
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Evaluating the Phi Function What is (p) if p is prime? –p-1 What is (pq) if p and q are distinct primes? –If p, q distinct primes, (pq) = (p) (q) –Not true if p=q –We won’t prove this, though it’s not hard
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Examples What is (3)? –|Z 3 * | = |{1,2}| = 2 What is (5)? What is (15)? – (15) = (3) (5) = 2 £ 4 = 8 –Recall, Z 15 * = {1,2,4,7,8,11,13,14}
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