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Grow & Cut - Using Representations in Cryptanalysis Alexander Meurer , Ruhr-Universitt Bochum ECRYPT Summer School on Tools, Mykonos, 2012 Motivation Motivation Generic technique for combinatorial problems Simple (no complicated

  1. Knapsack Representation Technique Knapsack Representation Technique Expand H into larger stack H'

  2. Expanding the Knapsack Haystack Expanding the Knapsack Haystack ● Recall: Classical MitM haystack H = {0,1} n/2 x {0} n/2 of size 2 n/2

  3. Expanding the Knapsack Haystack Expanding the Knapsack Haystack ● Recall: Classical MitM haystack H = {0,1} n/2 x {0} n/2 of size 2 n/2 Contains unique Needle ( x 1 , … , x n/2 )

  4. Expanding the Knapsack Haystack Expanding the Knapsack Haystack ● Recall: Classical MitM haystack H = {0,1} n/2 x {0} n/2 of size 2 n/2 ● New expanded haystack H' = { x 2 {0,1} n : wt( x ) = n/4 }

  5. Expanding the Knapsack Haystack Expanding the Knapsack Haystack ● Recall: Classical MitM haystack H = {0,1} n/2 x {0} n/2 of size 2 n/2 ● New expanded haystack H' = { x 2 {0,1} n : wt( x ) = n/4 }

  6. Knapsack Representation Technique Knapsack Representation Technique Expanding H' introduces r many representations N 1 , … , N r

  7. Number of Knapsack Representations Number of Knapsack Representations ● Example (n=8) ● Denote x * = ( 1 0 1 1 0 0 1 0) the solution ● x * can be represented as ( 1 0 1 0 0 0 0 0 ) + ( 0 0 0 1 0 0 1 0 ) ( 1 0 0 1 0 0 0 0 ) + ( 0 0 1 0 0 0 1 0 ) ( 1 0 0 0 0 0 1 0 ) + ( 0 0 1 1 0 0 0 0 ) ( 0 0 1 1 0 0 0 0 ) + ( 1 0 0 0 0 0 1 0 ) ( 0 0 1 0 0 0 1 0 ) + ( 1 0 0 1 0 0 0 0 ) ( 0 0 0 1 0 0 1 0 ) + ( 1 0 1 0 0 0 0 0 )

  8. Number of Knapsack Representations Number of Knapsack Representations ● Example (n=8) ● Denote x * = ( 1 0 1 1 0 0 1 0) the solution ● x * can be represented as ( 1 0 1 0 0 0 0 0 ) + ( 0 0 0 1 0 0 1 0 ) ( 1 0 0 1 0 0 0 0 ) + ( 0 0 1 0 0 0 1 0 ) ( 1 0 0 0 0 0 1 0 ) + ( 0 0 1 1 0 0 0 0 ) ( 0 0 1 1 0 0 0 0 ) + ( 1 0 0 0 0 0 1 0 ) ( 0 0 1 0 0 0 1 0 ) + ( 1 0 0 1 0 0 0 0 ) ( 0 0 0 1 0 0 1 0 ) + ( 1 0 1 0 0 0 0 0 )

  9. Number of Knapsack Representations Number of Knapsack Representations ● Example (n=8) ● Denote x * = ( 1 0 1 1 0 0 1 0) the solution ● x * can be represented as ( 1 0 1 0 0 0 0 0 ) + ( 0 0 0 1 0 0 1 0 ) ( 1 0 0 1 0 0 0 0 ) + ( 0 0 1 0 0 0 1 0 ) ( 1 0 0 0 0 0 1 0 ) + ( 0 0 1 1 0 0 0 0 ) ( 0 0 1 1 0 0 0 0 ) + ( 1 0 0 0 0 0 1 0 ) ( 0 0 1 0 0 0 1 0 ) + ( 1 0 0 1 0 0 0 0 ) ( 0 0 0 1 0 0 1 0 ) + ( 1 0 1 0 0 0 0 0 )

  10. Number of Knapsack Representations Number of Knapsack Representations ● Example (n=8) ● Denote x * = ( 1 0 1 1 0 0 1 0) the solution ● x * can be represented as ( 1 0 1 0 0 0 0 0 ) + ( 0 0 0 1 0 0 1 0 ) ( 1 0 0 1 0 0 0 0 ) + ( 0 0 1 0 0 0 1 0 ) ( 1 0 0 0 0 0 1 0 ) + ( 0 0 1 1 0 0 0 0 ) ( 0 0 1 1 0 0 0 0 ) + ( 1 0 0 0 0 0 1 0 ) ( 0 0 1 0 0 0 1 0 ) + ( 1 0 0 1 0 0 0 0 ) ( 0 0 0 1 0 0 1 0 ) + ( 1 0 1 0 0 0 0 0 )

  11. Number of Knapsack Representations Number of Knapsack Representations ● Example (n=8) ● Denote x * = ( 1 0 1 1 0 0 1 0) the solution ● x * can be represented as ( 1 0 1 0 0 0 0 0 ) + ( 0 0 0 1 0 0 1 0 ) ( 1 0 0 1 0 0 0 0 ) + ( 0 0 1 0 0 0 1 0 ) ( 1 0 0 0 0 0 1 0 ) + ( 0 0 1 1 0 0 0 0 ) ( 0 0 1 1 0 0 0 0 ) + ( 1 0 0 0 0 0 1 0 ) ( 0 0 1 0 0 0 1 0 ) + ( 1 0 0 1 0 0 0 0 ) ( 0 0 0 1 0 0 1 0 ) + ( 1 0 1 0 0 0 0 0 )

  12. Number of Knapsack Representations Number of Knapsack Representations ● Example (n=8) ● Denote x * = ( 1 0 1 1 0 0 1 0) the solution ● x * can be represented as ( 1 0 1 0 0 0 0 0 ) + ( 0 0 0 1 0 0 1 0 ) ( 1 0 0 1 0 0 0 0 ) + ( 0 0 1 0 0 0 1 0 ) ( 1 0 0 0 0 0 1 0 ) + ( 0 0 1 1 0 0 0 0 ) ( 0 0 1 1 0 0 0 0 ) + ( 1 0 0 0 0 0 1 0 ) ( 0 0 1 0 0 0 1 0 ) + ( 1 0 0 1 0 0 0 0 ) ( 0 0 0 1 0 0 1 0 ) + ( 1 0 1 0 0 0 0 0 )

  13. Number of Knapsack Representations Number of Knapsack Representations ● Example (n=8) ● Denote x * = ( 1 0 1 1 0 0 1 0) the solution ● x * can be represented as ( 1 0 1 0 0 0 0 0 ) + ( 0 0 0 1 0 0 1 0 ) ( 1 0 0 1 0 0 0 0 ) + ( 0 0 1 0 0 0 1 0 ) ( 1 0 0 0 0 0 1 0 ) + ( 0 0 1 1 0 0 0 0 ) ( 0 0 1 1 0 0 0 0 ) + ( 1 0 0 0 0 0 1 0 ) ( 0 0 1 0 0 0 1 0 ) + ( 1 0 0 1 0 0 0 0 ) ( 0 0 0 1 0 0 1 0 ) + ( 1 0 1 0 0 0 0 0 )

  14. Number of Knapsack Representations Number of Knapsack Representations ● Example (n=8) ● Denote x * = ( 1 0 1 1 0 0 1 0) the solution ● x * can be represented as ( 1 0 1 0 0 0 0 0 ) + ( 0 0 0 1 0 0 1 0 ) We can choose 2 out of 4 of the 1-entries, so ( 1 0 0 1 0 0 0 0 ) + ( 0 0 1 0 0 0 1 0 ) many representations ( 1 0 0 0 0 0 1 0 ) + ( 0 0 1 1 0 0 0 0 ) ( 0 0 1 1 0 0 0 0 ) + ( 1 0 0 0 0 0 1 0 ) ( 0 0 1 0 0 0 1 0 ) + ( 1 0 0 1 0 0 0 0 ) ( 0 0 0 1 0 0 1 0 ) + ( 1 0 1 0 0 0 0 0 )

  15. Number of Knapsack Representations Number of Knapsack Representations ● Example (n=8) ● Denote x * = ( 1 0 1 1 0 0 1 0) the solution ● x * can be represented as ( 1 0 1 0 0 0 0 0 ) + ( 0 0 0 1 0 0 1 0 ) We can choose 2 out of 4 of the 1-entries, so ( 1 0 0 1 0 0 0 0 ) + ( 0 0 1 0 0 0 1 0 ) In general: many representations many representations ( 1 0 0 0 0 0 1 0 ) + ( 0 0 1 1 0 0 0 0 ) ( 0 0 1 1 0 0 0 0 ) + ( 1 0 0 0 0 0 1 0 ) ( 0 0 1 0 0 0 1 0 ) + ( 1 0 0 1 0 0 0 0 ) ( 0 0 0 1 0 0 1 0 ) + ( 1 0 1 0 0 0 0 0 )

  16. Knapsack Representation Technique Knapsack Representation Technique Examine a 1/r – fraction of H' to fnd one N i

  17. The Cutting Phase The Cutting Phase ● We need to flter out a 2 -n/2 fraction of H' = { x 2 {0,1} n : wt( x ) = n/4 } ● For compute two lists ● Find one (x,y) with over

  18. The Cutting Phase The Cutting Phase ● We need to flter out a 2 -n/2 fraction of H' = { x 2 {0,1} n : wt( x ) = n/4 } ● For compute two lists ● Find one (x,y) with over

  19. The Cutting Phase The Cutting Phase ● We need to flter out a 2 -n/2 fraction of H' = { x 2 {0,1} n : wt( x ) = n/4 } ● For compute two lists ● Find one (x,y) with over

  20. The Cutting Phase The Cutting Phase ● We need to flter out a 2 -n/2 fraction of Complexity H' = { x 2 {0,1} n : wt( x ) = n/4 } ● For compute two lists ● Find one (x,y) with over

  21. The Cutting Phase The Cutting Phase ● We need to flter out a 2 -n/2 fraction of Computing the is in fact Complexity H' = { x 2 {0,1} n : wt( x ) = n/4 } more expensive! ● For compute two lists ● Find one (x,y) with over

  22. - 1 st st Attempt Computing L Computing 1 - 1 Attempt L 1 ● Idea: Do it in a Meet-in-the-Middle way ● Choose a (random) partition ● Merge the lists

  23. - 1 st st Attempt Computing L Computing 1 - 1 Attempt L 1 ● Idea: Do it in a Meet-in-the-Middle way ● Choose a (random) partition ● Merge the lists

  24. - 1 st st Attempt Computing L Computing 1 - 1 Attempt L 1 Choose labels ● Idea: Do it in a Meet-in-the-Middle way ● Choose a (random) partition ● Merge the lists

  25. st Attempt Complexity Analysis: 1 st Complexity Analysis: 1 Attempt List sizes

  26. st Attempt Complexity Analysis: 1 st Complexity Analysis: 1 Attempt List sizes

  27. st Attempt Complexity Analysis: 1 st Complexity Analysis: 1 Attempt List sizes

  28. st Attempt Complexity Analysis: 1 st Complexity Analysis: 1 Attempt List sizes

  29. st Attempt Complexity Analysis: 1 st Complexity Analysis: 1 Attempt List sizes x

  30. st Attempt Complexity Analysis: 1 st Complexity Analysis: 1 Attempt List sizes Complexity* = Max x = 2 0.4057n

  31. st Attempt Complexity Analysis: 1 st Complexity Analysis: 1 Attempt List sizes If assign uniform labels Complexity* = Max x = 2 0.4057n

  32. and ¸ are almost always good L and R are almost always good ¸ ¸ L ¸ R ● Recall ● Example: Good and bad distributed labels (n=20) random a i a i = 0 for i=1,...,n/2

  33. and ¸ are almost always good L and R are almost always good ¸ ¸ L ¸ R ● Recall One can show: Fraction of bad knapsacks (a 1 , … , a n ) is exponentially small! ● Example: Good and bad distributed labels (n=20) random a i a i = 0 for i=1,...,n/2

  34. - 2 nd nd Attempt Computing Computing L 1 - 2 Attempt L 1 ● Aim: Compute 2 -n/2 fraction of H' = { x 2 2 {0,1} n : wt( x ) = n/4 } ● Idea: Use representations again! ● Decompose x 2 H' as x = y + z where y , z 2 2 H'' and 2 {0,1} n : wt( y ) = n/8 } H'' = { y 2 → Every x has representations ( y , z )

  35. - 2 nd nd Attempt Computing Computing L 1 - 2 Attempt L 1 ● Aim: Compute 2 -n/2 fraction of H' = { x 2 2 {0,1} n : wt( x ) = n/4 } ● Idea: Use representations again! ● Decompose x 2 H' as x = y + z where y , z 2 2 H'' and 2 {0,1} n : wt( y ) = n/8 } H'' = { y 2 → Every x has representations ( y , z )

  36. - 2 nd nd Attempt Computing Computing L 1 - 2 Attempt L 1 ● Choose coprime M (1) and M (2) of size 2 0.25n ● Choose random target values (of size 2 0.25n ) with with ● Compute lists ● Merge into

  37. - 2 nd nd Attempt Computing Computing L 1 - 2 Attempt L 1 ● Choose coprime M (1) and M (2) of size 2 0.25n ● Choose random target values (of size 2 0.25n ) with with ● Compute lists ● Merge into

  38. - 2 nd nd Attempt Computing Computing L 1 - 2 Attempt L 1 ● Choose coprime M (1) and M (2) of size 2 0.25n ● Choose random target values (of size 2 0.25n ) with with ● Compute lists ● Merge into

  39. - 2 nd nd Attempt Computing Computing L 1 - 2 Attempt L 1 ● Choose coprime M (1) and M (2) of size 2 0.25n ● Choose random target values (of size 2 0.25n ) with Can be computed as before by a MitM approach! with ● Compute lists ● Merge into

  40. - 2 nd nd Attempt Computing Computing L 1 - 2 Attempt L 1 ● Choose coprime M (1) and M (2) of size 2 0.25n ● Choose random target values (of size 2 0.25n ) with with ● Compute lists ● Merge into

  41. nd Attempt Complexity Analysis: 2 Complexity Analysis: 2 nd Attempt List sizes Randomly partioned base lists

  42. nd Attempt Complexity Analysis: 2 Complexity Analysis: 2 nd Attempt List sizes Randomly partioned base lists

  43. nd Attempt Complexity Analysis: 2 Complexity Analysis: 2 nd Attempt List sizes Randomly partioned base lists

  44. nd Attempt Complexity Analysis: 2 Complexity Analysis: 2 nd Attempt List sizes Randomly partioned base lists

  45. nd Attempt Complexity Analysis: 2 Complexity Analysis: 2 nd Attempt List sizes Randomly partioned base lists

  46. nd Attempt Complexity Analysis: 2 Complexity Analysis: 2 nd Attempt List sizes Randomly partioned base lists

  47. nd Attempt Complexity Analysis: 2 Complexity Analysis: 2 nd Attempt List sizes Randomly partioned base lists Two modular constraints:

  48. nd Attempt Complexity Analysis: 2 Complexity Analysis: 2 nd Attempt List sizes Randomly partioned base lists x

  49. nd Attempt Complexity Analysis: 2 Complexity Analysis: 2 nd Attempt List sizes Randomly partioned base lists Complexity = Max = 2 0.3113n Lower bound achieved? x

  50. nd Attempt Complexity Analysis: 2 Complexity Analysis: 2 nd Attempt List sizes Randomly partioned base lists No! Forgot merge costs! x

  51. nd Attempt Complexity Analysis: 2 Complexity Analysis: 2 nd Attempt List sizes Randomly partioned base lists x

  52. Summary (so far) Summary (so far) ● Represenation Technique gives fastest (randomized) algorithm for random hard knapsacks Time Space Brute-Force 2 n 1 MitM 2 n/2 2 n/2 Schroeppel-Shamir 2 n/2 2 n/4 Representations 2 0.3373n 2 0.2936n

  53. Generic Decoding of Random Generic Decoding of Random Binary Linear Codes Binary Linear Codes

  54. Recap Binary Linear Codes Recap Binary Linear Codes ● C = random binary [n,k,d] code ● n = length / k = dimension / d = minimum distance Bounded Distance Decoding (BDD) Bounded Distance Decoding (BDD) ● Given x = c + e with c C · · 2 4 5 d-1 4 5 and w := wt( e ) = d-1 2 2 x · c · · ● Find e and thus c = x + e

  55. Why do we care about BDD? Why do we care about BDD? ● BDD is connected to ● Message Recovery for classical McEliece c = m * G + e ● Key recovery for Stern's identifcation scheme id = H * s ● Hardness of the LPN problem (Learning Parities with noise)

  56. Comparing Running Times Comparing Running Times How to compare performance of decoding algorithms ● Running time T(n,k,d) ● Fixed code rate R = k/n ● For n→∞, k and d are related via Gilbert-Varshamov bound, thus T(n,k,d) = T(n,k) = T(n,R) ● Compare algorithms by complexity coeffcient F(R), i.e. T(n,R) = 2 F(R) • n + o(n)

  57. Comparing Running Times Comparing Running Times How to compare performance of decoding algorithms ● Running time T(n,k,d) Minimize F(R)! Minimize F(R)! ● Fixed code rate R = k/n ● For n→∞, k and d are related via Gilbert-Varshamov bound, thus T(n,k,d) = T(n,k) = T(n,R) ● Compare algorithms by complexity coeffcient F(R), i.e. T(n,R) = 2 F(R) • n + o(n)

  58. Syndrome Decoding Syndrome Decoding (BDD) Given x = c + e with c C and wt( e )=w, fnd e ! 2 ● H = parity check matrix ● Consider syndrome s := s( x ) = H · x = H ·( c + e ) = H · e → Find linear combination of w columns of H matching s weight w n H s n-k = + + =

  59. Syndrome Decoding Syndrome Decoding (BDD) Given x = c + e with c C and wt( e )=w, fnd e ! 2 ● H = parity check matrix ● Consider syndrome s := s( x ) = H · x = H ·( c + e ) = H · e → Find linear combination of w columns of H matching s weight w n Brute-Force complexity Brute-Force complexity H s n-k = + + = T(n,k,d) = T(n,k,d) =

  60. Complexity Coeffcients (BDD) Complexity Coeffcients (BDD) Brute-Force F(R) 0,05 0,06 0,3868

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