1. Once Upon a Time-Memory Tradeoff Mark Stamp Department of Computer Science San Jose State University

2. TMTO2 This talk…  Non-cryptanalytic TMTOs  Crypto background  Hellman’s cryptanalytic TMTO  Distributed TMTO  Conclusions

3. TMTO3 Non-crypto TMTOs  Popcnt  Shank’s algorithm

4. TMTO4 Popcnt  Let x be a 32-bit integer  Define popcnt(x) = number of 1’s in binary expansion of x  How to compute popcnt(x) ?

5. TMTO5 Simple popcnt popcnt(x) t = 0 for i = 0 to 31 t = t + (x >> i) & 1 next i return t end popcnt

6. TMTO6 Efficient popcnt Initialize: table[i] = popcnt(i) for i = 0,1,…,255 popcnt(x) t = table[ x & 0xff ] + table[ (x >> 8) & 0xff ] + table[ (x >> 16) & 0xff ] + table[ (x >> 24) & 0xff ] return t end popcnt

7. TMTO7 Discrete Log  Let p be prime, g  {1,2,…, p  1} s.t. for any n there is a k with n = g k mod p  Discrete log: given m  {1,2,…, p  1} find e s.t. m = g e mod p  Notation: e = log g (m)  Could try each value in {1,2,…, p  1} to find e that works

8. TMTO8 Shank’s algorithm  Shank’s is a TMTO for discrete log  Given m, we want e = log g (m) A.Compute list L r as follows 1. Let r =  sqrt(p  1)  and compute g rj mod p for j = 0,1,…,r  1 2. Let L r be the list of ( j, g rj mod p) sorted on second coordinate

9. TMTO9 Shank’s alg (continued) B.Compute list L m as follows 1. Compute mg  i mod p for i = 0,1,…,r  1 2. Let L m be the list of (i, mg  i mod p) sorted on second coordinate

10. TMTO10 Shank’s alg. (cont. again) C.Then e = log g (m) is found by 1. Find elements of L r and L m that agree in 2nd coordinates, say, ( j, x)  L r and (i, x)  L m 2. Then e = log g (m) = rj + i mod (p  1) since g rj = mg  i mod p  Shank’s: baby step, giant step

11. TMTO11 Shank’s algorithm (example)  Suppose p = 257, g = 3. Then r = 16 and L r is (0,1) (3,2) (6,4) (9,8) (12,16) (15,32) (2,64) (5,128) (13,129) (10,193) (7,225) (4,241) (1,249) (14,253) (11,255) (8,256)  Suppose m = 132. Then L m is (9,23) (1,44) (3,62) (5,64) (8,69) (12,77) (15,79) (6,107) (0,132) (10,179) (2,186) (4,192) (13,197) (7,207) (11,231) (14,237)  From L r and L m we find (2,64 ) and (5,64). Then log 3 (132) = 2  16 + 5 = 37 and easy to verify 3 37 = 132 mod 257

12. TMTO12 Block cipher Consider a block cipher C = E(P, K) where P is plaintext of length n C is ciphertext of length n K is key of length k

13. TMTO13 Block Cipher

14. TMTO14 Chosen plaintext attack  We choose P and obtain C, where C = E(P, K)  Want to find the key K 1. Exhaustive key search 2. Table pre-computation  TMTO lies between 1. and 2.

15. TMTO15 Chain of encryptions Assume n = k. Then a chain is 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

16. TMTO16 Chain (another view)

17. TMTO17 Pre-computation  Compute m encryption chains, each of length t +1  Save only start and end points (SP 0, EP 0 ) (SP 1, EP 1 ) : (SP m-1, EP m-1 )

18. TMTO18 TMTO Attack  Memory: Given (SP i, EP i ), i = 0,1,…,m  1  For chosen P compute C = E(P, K)  The key K is unknown  Time: Compute chain (max of t steps) X 0 = C, X 1 = E(P, X 0 ), X 2 = E(P, X 1 ),…

19. TMTO19 Attack (continued)  Given the chain X 0 = C, X 1 = E(P, X 0 ), X 2 = E(P, X 1 ),…  Suppose we find X i = EP j  Then C might be in chain (SP j, EP j )  Assume C is in chain (SP j, EP j )

20. TMTO20 Attack (continued again)  Given C is in the chain (SP j, EP j ) and X i = EP j  Re-compute chain (SP j, EP j ) Y 0 = SP j, Y 1 = E(P,Y 0 ), Y 2 = E(P,Y 1 ),…  Then C = Y t  i = E(P, Y t  i  1 )  And Y t  i  1 = K (always?)

21. TMTO21 In a perfect world  Suppose the block cipher has 56 bit key  Suppose we find m = 2 28 chains, each of length t = 2 28 and no chains overlap  Memory: 2 28 pairs (SP j, EP j )  Time: about 2 28 (for attack) 1. Find C in about 2 27 tries 2. Find K with about 2 27 more tries

22. TMTO22 In a perfect world  All chains distinct  Ciphertext C lies within a chain EP 0 SP 0 C SP 1 SP 2 EP 1 EP 2

23. TMTO23 In the real world  Chains are not so well-behaved  Chains cycle and merge EP SP C

24. TMTO24 To reduce merging  Compute chain as K i = F(E(P, K i  1 )) where F is a permutation  Choose r different functions F  For each F choose m random SP  Each chain of length t

25. TMTO25 Notation  m = number of random starting points for each function F  t = length of each chain  r = number of “random” functions F  Note: mtr = total number of computed chain elements

26. TMTO26 Real-world issues  False alarms, avoid cycles, reduce merging, etc.  Pre-computation is lots of work (must be amortized over many attacks)  Success is not assured  What if block size not equal key length?  What is the probability of success?

27. TMTO27 Probability of success  Occupancy problem: b balls distributed with uniform probability to c cells  Let p l (b,c) be probability of l empty cells. Feller [3] shows

28. TMTO28 Success probability (continued)  Poisson approx to p l (b,c) is p l ( ) = e  l /l! where = ce  b/c  So expected number of empty cells is

29. TMTO29 Success probability (still more)  Expected number of occupied cells is c  = c(1  e  b/c )  Therefore P(cell i is occupied) = 1  e  b/c  TMTO attack succeeds if and only if the “cell” with key K is “occupied”

30. TMTO30 Success prob (last word, almost) mtr P(success) = 1  e  mtr/2 k ---- ------------- 2 k  2 0.22 2 k  1 0.39 2 k 0.63 2 k+1 0.86 2 k+2 0.98

31. TMTO31 The bottom line  Choose m = t = r = 2 k/3 and probability of success is about 0.63 (at least 0.55 by a more careful analysis)  Pre-computation is O(mtr) work  Each TMTO attack requires O(mr) “memory” and O(tr) “time”

32. TMTO32 Distinguished points  Let a distinguished point be of the form (x 0,x 1,…,x s  1,0,0,…,0)  Construct chain until distinguished point is found  If no distinguished point is found within max steps, don’t save chain  Then every EP is distinguished

33. TMTO33 Distinguished points +/-  Disadvantages  Chains are variable length  Some extra work to find chains  Triples (SP, EP, length)  Advantage  Distributed attack is very nice  Why? One client for each F then client only needs (P, C) and F and max chain length  no data!

34. TMTO34 References [1] M. Hellman, A cryptanalytic time-memory tradeoff, IEEE Trans on Info Thy, Vol. 26, No. 4, July 1980, pp. 401-406 [2] J.Borst, et al., On the time-memory tradeoff between exhaustive key search and table precomputation, http://www.esat.kuleuven.ac.be/~borst/downloadable/tm.ps.gz http://www.esat.kuleuven.ac.be/~borst/downloadable/tm.ps.gz [3] W. Feller, An Introduction to Probability Theory and Its Applications, volume 1, Wiley (1968) [4] M. Stamp, Once upon a time-memory tradeoff, http://www.cs.sjsu.edu/faculty/stamp/articles/tmto.pdf http://www.cs.sjsu.edu/faculty/stamp/articles/tmto.pdf