Hellman’s TMTO 1 Hellman’s TMTO Attack

Hellman’s TMTO 2 Popcnt  Before we consider Hellman’s attack, consider simpler Time-Memory Trade-Off  “Population count” or popcnt o Let x be a 32-bit integer o Define popcnt(x) = number of 1 ’s in binary expansion of x  How to compute popcnt(x) efficiently?

Hellman’s TMTO 3 Simple Popcnt  Most obvious thing to do is popcnt(x) // assuming x is 32-bit value t = 0 for i = 0 to 31 t = t + ((x >> i) & 1) next i return t end popcnt  Is this the most efficient method?

Hellman’s TMTO 4 More Efficient Popcnt  Pre-compute popcnt for all 256 bytes o Store pre-computed values in a table  Given x, lookup its bytes in this table o Sum these values to find popcnt(x)  See next slide for pseudo-code…

Hellman’s TMTO 5 More Efficient Popcnt Initialize: table[i] = popcnt(i) for i = 0,1,…,255 popcnt(x) // assuming x is 32-bit word p = table[ x & 0xff ] + table[ (x >> 8) & 0xff ] + table[ (x >> 16) & 0xff ] + table[ (x >> 24) & 0xff ] return p end popcnt

Popcnt TMTO  Advantage(s)? o Each popcnt now requires 4 steps, not 32  Disadvantage? o Pre-computation (one time work) o Additional storage  Balance “time” vs “memory” where… o Time == popcnt computation(s) o Memory == storage/pre-computation Hellman’s TMTO 6

Hellman’s TMTO 7 TMTO Basics  Pre-computation o One-time work o Results stored in a table  Pre-computation results used to make each subsequent computation faster  In general, larger pre-computation requires more initial work and larger “memory” o But then each computation takes less “time”  Try to balance “memory” and “time”

Hellman’s TMTO 8 Block Cipher Notation  Consider a block cipher C = E(P, K) where P is plaintext block of size n C is ciphertext block of size n K is key of size k

Hellman’s TMTO 9 Block Cipher as Black Box  For TMTO, treat block cipher as black box  Details of crypto algorithm not relevant

Hellman’s TMTO Attack  This is a chosen plaintext attack  We choose P  Then we are given corresponding C o That is, C = E(P, K)  We want to find the key K Hellman’s TMTO 10

TMTO Attack  Given: P and C = E(P, K)  Find: key K  Two “obvious” approaches 1.Exhaustive key search “Memory” is 0, but “time” of 2 k-1 for each attack 2.Pre-compute C = E(P, K) for all keys K Then given any C, simply look up key K in table “Memory” of 2 k but “time” of 0 per attack  TMTO lies between 1. and 2. Hellman’s TMTO 11

Hellman’s TMTO 12 Chain of Encryptions  For simplicity, we assume block length n and key length k are equal, that is, n = k o Attack still works if this does not hold  Define a chain of encryptions as SP = K 0 = Starting Point K 1 = E(P, SP) K 2 = E(P, K 1 ) : EP = K t = E(P, K t  1 ) = End Point

Hellman’s TMTO 13 Encryption Chain  Note that ciphertext from one iteration is used as key at next iteration  Note also that the same (chosen) plaintext P used at each iteration

Hellman’s TMTO 14 Pre-computation  Pre-compute m encryption chains, each of length t +1  Save only the start and end points (SP 0, EP 0 ) (SP 1, EP 1 ) : (SP m-1, EP m-1 ) EP 0 SP 0 SP 1 SP m-1 EP 1 EP m-1

Hellman’s TMTO 15 TMTO Attack  Memory: Pre-compute encryption chains and save (SP i, EP i ) for i = 0,1,…,m  1 o This is one-time work o Sort based on EP i  For each attack for unknown key K do… o We have P and C = E(P, K) where P is same plaintext used to compute chains o Time: Compute a chain (max of t steps) X 0 = C, X 1 = E(P, X 0 ), X 2 = E(P, X 1 ),…

Hellman’s TMTO 16 TMTO Attack  Consider the computed chain X 0 = C, X 1 = E(P, X 0 ), X 2 = E(P, X 1 ), …  Suppose for some i we find X i = EP j SP j EP j C K  Since C = E(P, K) key K appears to lie just before ciphertext C in this chain!

Hellman’s TMTO 17 TMTO Attack  Summary of attack phase: compute X 0 = C, X 1 = E(P, X 0 ), X 2 = E(P, X 1 ),…  If for some i we find X i = EP j then reconstruct chain from SP j Y 0 = SP j, Y 1 = E(P,Y 0 ), Y 2 = E(P,Y 1 ),…  We find C = Y t  i = E(P, Y t  i  1 ) (always?)  So that K = Y t  i  1 (always?)

Hellman’s TMTO 18 Trudy’s Perfect World  Suppose block cipher has k = 56 o That is, the key length is 56 bits  Suppose we can find m = 2 28 chains each of length t = 2 28 with no overlap o Is this realistic?  Then all 2 56 possible keys covered by exactly one element of a chain  This looks promising…

Hellman’s TMTO 19 Trudy’s Perfect World  If we can find m = 2 28 chains each of length t = 2 28 and no intersection, then…  Memory: 2 28 pairs (SP j, EP i )  Time: about 2 28 (per attack) o Start at C, find some EP j in about 2 27 steps o Find K with about 2 27 more steps o So, work per attack is about 2 28  Trivial work and attack never fails! o For Trudy, life doesn’t get any better…

Hellman’s TMTO 20 Trudy’s Perfect World  No chains intersect  Every ciphertext C is in one chain EP 0 SP 0 C SP 1 SP 2 EP 1 EP 2 K

Hellman’s TMTO 21 The Real World  Real chains are not so well-behaved  Chains can cycle and merge  Chain beginning at C goes to EP  But chain from SP to EP does not give K  This looks like Trudy’s nightmare… EP SP CK

Hellman’s TMTO 22 The Real World  Chains can cycle and merge  This adversely affects attacks, as mentioned on previous slide  It also, adversely affects pre- computation and probability of success  Why? EP SP CK

Hellman’s TMTO 23 Real-World TMTO Issues  Merging chains, cycles, false alarms, …  Pre-computation is lots of work o Must attack many times to amortize cost  Success is not assured o Depends on initial work, merging, cycles, …  What if block size not equal key length? o This is easy to deal with  What is the probability of success? o This is not so easy to compute…

Hellman’s TMTO 24 To Reduce Merging  Compute chain as F(E(P, K i  1 )) where F permutes the bits o Then we have F(E(P, K)) = F(C)  Chains computed using different functions can intersect, but will not merge EP 1 SP 0 SP 1 EP 0 F 0 chain F 1 chain

Hellman’s TMTO 25 Hellman’s TMTO in Practice  Let m = random starting points for each F t = encryptions in each chain r = number of “tables”, i.e., functions F  Then mtr pre-computed chain elements o Pre-computation is about mtr work  Each TMTO attack requires o About mr “memory” (storage), and tr “time”  If we choose m = t = r = 2 k/3 then probability of success is at least 0.55

Hellman’s TMTO 26 Success Probability?  Throw n balls into m urns  Expected number of urns that have at least one ball?  This is classic “occupancy problem” o See Feller, Intro. to Probability Theory  Why is this relevant to TMTO ? o “Urns” correspond to keys o “Balls” correspond to constructing chains

Hellman’s TMTO 27 Success Probability  Using occupancy problem approach…  Assuming k -bit key and m,t,r as defined on previous slide o Can choose m = t = r but not necessary  Then, approximately, P( success ) = 1  e  mtr/k  An upper bound can be given that is slightly “better”

Hellman’s TMTO 28 Success Probability  Success probability P( success ) = 1  e  mtr/k

Hellman’s TMTO 29 Distributed TMTO  Employ “distinguished points”  Do not use fixed-length chains  Instead, compute chain until some distinguished point is found  Example of distinguished point:

Hellman’s TMTO 30 Distributed TMTO  Similar pre-computation as before, except we have triples: (SP i, EP i, l i ) for i = 0,1,…,rm o Where l i is the length of the chain o And r is number of tables o And m is number of random starting points  Let M i be the maximum l j for the i th table  Each table has a fixed random function F

Hellman’s TMTO 31 Distributed TMTO  Suppose r computers are available  Each computer deals with one table o So, only one random function F per computer  “Master” gives computer i values F i, M i, C o And definition of distinguished point  Computer i computes chain beginning with C and using F i of (at most) length M i

Hellman’s TMTO 32 Distributed TMTO  If computer i finds a distinguished point within M i steps o Returns result to “master” for secondary test o Master searches for K on corresponding chain, using same approach as in non-distributed TMTO o False alarms can occur (e.g., distinguished points)  If no distinguished point found in M i steps o Computer i gives up, since key not on any F i chains  Note that computer i does not need to know any SP or EP or l  Master knows (SP i, EP i, l i ) for i = 0,1,…,rm

Hellman’s TMTO 33 TMTO for DES  Attack is feasible against DES  Recall that for DES, we have k = 56  Assume that we use m = t = r = 2 k/3 ≈  Then pre-computation is about 2 56 work o And we store about 2 37 starting/end point pairs  So, each time attack is conducted… o …use about 2 37 “memory” and requires 2 37 “time”

Hellman’s TMTO 34 TMTO: The Bottom Line  Attack is not specific to DES o Applies to any block cipher  Inner working of cipher is not relevant o But key length is…  No fancy math required o Just a clever algorithm  Lesson: Clever algorithms can break crypto!