# Isabelle Stanton Chalermpong Worawannotai

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Isabelle Stanton Chalermpong Worawannotai
MD5 Collisions Isabelle Stanton Chalermpong Worawannotai

Description of MD5 Takes any message and outputs an 128-bit hash.
A message is padded so the length is a multiple of 512 by concatenating a 1 then 0’s and it’s length as a 64 bit number. Each 512 bit block is compressed individually

Continued Description
The 512-bit block is divided into bit words There are 4 32-bit registers a, b, c and d. These are initially loaded with IV0 and carry the hash values from one 512-bit block to the next It works in an iterative (chaining) process: Hi+1 = f(Hi,Mi) IV0=H0 where Mi is a 512 bit block.

… Hash Chaining M1 M2 Mn f f f H0=IV0 fixed Hn = H H1 H2 Mi 512 bits
Hi 128 bits

One small step For each f there are 4 rounds and each round has 16 steps Ti and Si are fixed constant and depend only on the steps. Courtesy of

The Rounds Mi=(w0,…,w15) For fixed i, 4 consecutive steps will yield
ai+4 =bi +((ai +Fi (bi,ci,di)+wi+ti)<<<si) di+4=ai+((di+Fi+1 (ai,bi,ci)+wi+1+ti+1)<<<si+1) ci+4=di+((ci+Fi+2 (di,ai,bi)+wi+2+ti+2)<<<si+2) bi+4=ci+((bi+Fi+3 (ci,di,ai)+wi+3+ti+3)<<<si+3) ti and si are predefined step dependant constants

The Non-Linear Functions
Fi changes every 16 steps Fi(X,Y,Z)=(X^Y)ν(~X^Z) 0≤i ≤15 Fi(X,Y,Z)=(X^Z) ν(Y^~Z) 16 ≤i ≤31 Fi(X,Y,Z)=X  Y  Z 32 ≤i ≤47 Fi(X,Y,Z)=Y  (X ν ~Z) 48 ≤i ≤63 This provides non-linearity so you can not extract the message from the hash

Finding Collisions MD5 has a 128 bit hash so a brute force attack to find a collision requires at most 2128 applications of MD5 and 264 by the birthday paradox Xiaoyun Wang and Hongbo Yu have an attack that requires 239 operations This attack takes at most an hour and 5 minutes on a IBM P690 (supercomputer)

Recall: Differential Cryptanalysis
Find a particular ∆M such that a particular ∆H occurs with high probability In collision case, want ∆H = 0.

Differentials The attack uses two types of differentials
XOR differential: ΔX=X  X’ Modular differential: ΔX=X-X’ mod 232 For M=(m0,…,mn-1) and M’=(m’0,…m’n-1) the full hash differential is for a message of length 512n bits ΔH0 -> ΔH1 ->…-> ΔHn= ΔH If M and M’ are a collision pair ΔH=0

Round differentials ΔHi -> ΔHi+1 can be split into round differentials as well ΔHi ΔR ΔR ΔR ΔR3=ΔHi+1 P3 P0 P1 P2

Probability Each of these differentials has a probabilistic relationship with the next. Ideally, we’d like to be able to set up 2 messages where we can guarantee with probability 1 that ΔH=0 This can be assured by modifying M so the first round differential will be what you want More modifications will improve the probability for the second, third and fourth round differentials ΔM0 has been picked to improve this as well

The Attack Find M=(M0,M1 ) and M’=(M’0,M’1)
i.e. M0 and messages that does this is not a collision ΔM0 has been picked to improve the probability that the round differentials will hold M’0 differ in the 5th, 12th and 15th words only Same for M1 and M’1. Every set of messages that does this is not a collision ΔM0 has been picked to improve this as well

Message Modification It is easy to modify a message word so that the first non-zero step differential (after the 5th step) is anything you want with probability 1 Modify multiple words to guarantee the round differentials with high probability Each modification to make one condition hold may make another not hold

Sufficient Conditions
Δw5 is first non-zero differential At the 8th step Δw5 has affected a, d and c so (Δc2, Δd2, Δa2, Δb1 )-> Δb2 since Δb1=0 There are 13 conditions on a2, c2 and d2 that will guarantee Δb2 to be whatever you like with high probability Each characteristic has between 1 and 28 conditions for 30 characteristics for M0 and 29 characteristics with between 2 and 25 conditions for M1 for well over 200 conditions

Conditions for bi b1,7 = 0 b1,8 = c1,8 b1,9 = c1,9
b1,10 = c1,10 b1,11 = c1,11 b1,12 = 1 b1,13 = c1,13 b1,14 = c1,14 b1,15 = c1,15 b1,16 = c1,16 b1,17 = c1,17 b1,18 = c1,18 b1,19 = c1,19 b1,20 = 1 b1,21 = c1,21 b1,22 = c1,22 b1,23 = c1,23 b1,24 = 0 b1,32 = 1

Technique for M0 Select random M0
Modify M0 so as many of the conditions hold as possible Create M0’=M0+ ΔM0 This will result in ΔH1 with probability 2-37 Test this works This doesn’t require more then 239 MD5 operations

Technique for M1 Select a random message M1
Modify M1 so it meets the conditions M1’ =M1+ ΔM0 Starting with ΔH1 as IV the probability that H(M1)=H(M1’) is 2-30 Test the pair of messages for collisions

Creating More Collisions
There are many M1s that will collide with any properly crafted M0 You can also change the last two words of M0 and maintain the conditions This reduces the amount of work needed

Actual Collisions M0 = 2dd31d1 c4eee6c5 69a3d69 5cf9af98 87b5ca2f ab7e4612 3e ffbb8 634ad55 2b3f e483 5a e fc9cdf7 f2bd1dd9 5b3c3780 M1=d11d0b96 9c7b41dc f497d8e4 d555655a c79a7335 cfdebf0 66f fb109d1 797f2775 eb5cd530 baade822 5c15cc79 ddcb74ed 6dd3c55f d80a9bb1 e3a7cc35 M0’=2dd31d1 c4eee6c5 69a3d69 5cf9af98 7b5ca2f ab7e4612 3e ffbb8 634ad55 2b3f e483 5a41f125 e fc9cdf7 72bd1dd9 5b3c3780 M1’=d11d0b96 9c7b41dc f497d8e4 d555655a 479a7335 cfdebf0 66f fb109d1 797f2775 eb5cd530 baade822 5c154c79 ddcb74ed 6dd3c55f 580a9bb1 e3a7cc35 Hash: f a30f9dbf 9f65ffbc f41fc7ef

References How To Break MD5 and Other Hash Functions – Xiaoyun Wang and Hongbo Yu (they did the SHA-1 break as well) Guide to Hash Functions Cryptographic Hash Lounge (lists what functions have been broken and links to how) Questions?