Foundations of Network and Computer Security J J ohn Black Lecture #7 Sep 11 th 2009 CSCI 6268/TLEN 5550, Fall 2009.

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Presentation transcript:

Foundations of Network and Computer Security J J ohn Black Lecture #7 Sep 11 th 2009 CSCI 6268/TLEN 5550, Fall 2009

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)

Symmetric Authentication: The Intuitive Model Here’s the intuition underlying the authentication model: –Alice and Bob have some shared, random string K –They wish to communicate over some insecure channel –An active adversary is able to eavesdrop and arbitrarily insert packets into the channel Adversary Alice Key K Bob

Authentication: The Goal Alice and Bob’s Goal: –Alice wishes to send packets to Bob in such a way that Bob can be certain (with overwhelming probability) that Alice was the true originator Adversary’s Goal: –The adversary will listen to the traffic and then (after some time) attempt to impersonate Alice to Bob –If there is a significant probability that Bob will accept the forgery, the adversary has succeeded

The Solution: MACs The cryptographic solution to this problem is called a Message Authentication Code (MAC) –A MAC is an algorithm which accepts a message M, a key K, and possibly some state (like a nonce N), and outputs a short string called a “tag” MAC M K N tag = MAC K (M, N)

MACs (cont) Alice computes tag = MAC K (M, N) and sends Bob the message (M, N, tag) Bob receives (M’, N’, tag’) and checks if MAC K (M’, N’) == tag’ –If YES, he accepts M’ as authentic –If NO, he rejects M’ as an attempted forgery Note: We said nothing about privacy here! M might not be encrypted (M’, N’, tag’) MAC K (M’, N’) == tag’ ?? Y N ACCEPT REJECT Bob

Security for MACs The normal model is the ACMA model –Adaptive Chosen-Message Attack Adversary gets a black-box called an “oracle” –Oracle contains the MAC algorithm and the key K –Adversary submits messages of his choice and the oracle returns the MAC tag –After some “reasonable” number of queries, the adversary must “forge” To forge, the adversary must produce a new message M * along with a valid MAC tag for M * –If no adversary can efficiently forge, we say the MAC is secure in the ACMA model

Building a MAC with a Blockcipher Let’s use AES to build a MAC –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 AES K M1M1 tag M2M2 MmMm

CBC MAC notes Just like CBC mode encryption except: –No IV (or equivalently, IV is 0 n ) –We output only the last value Not parallelizable Insecure if message lengths vary

Breaking CBC MAC If we allow msg lengths to vary, the MAC breaks –To “forge” we need to do some (reasonable) number of queries, then submit a new message and a valid tag Ask M 1 = 0 n we get t = AES K (0 n ) back We’re done! –We announce that M * = 0 n || t has tag t as well –(Note that A || B denotes the concatenation of strings A and B)

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

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

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

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

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

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 %

T  A << 5 + g t (B, C, D) + E + K t + W t SHA-1... M1M1 M2M2 MmMm for i = 1 to m do W t = { 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

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

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

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 

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

MiMi T  A << 5 + g t (B, C, D) + E + K t + W t... M1M1 M2M2 MmMm for i = 1 to m do W t = { 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 bits

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: 1x x ½ + 3 x ½ = 2.5 What about 2 bits? –Expectation: 1 x x ¼ + 3 x ¾ ½ + 4 x ¾ ½ ¾ + 5 x ¾ ½ ¼ ≈ 3.22 This is too hard…

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… In class size is probably too small

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 x 364/365 ≈.997 For n=3, Pr[NBM] = 1 x 364/365 x 363/365 ≈.991 … For n=22, Pr[NBM] = 1 x … x 344/365 ≈.524 For n=23, Pr[NBM] = 1 x … x 343/365 ≈.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%

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

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

The Birthday Bound 1.0 Probability n2n Number of Hash Inputs 2 n/2

More Recently 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 lowered time-to-collision to 3 mins (July 2005) –SHA-0 isn’t used anymore (but see next slide)

Collisions in SHA-0 T  A << 5 + g t (B, C, D) + E + K t + W t W t = { 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!

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

A Provably-Secure Blockcipher-Based Compression Function E MiMi h i-1 hihi n bits

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)