Squeezing a key through a carry bit @ 34c3

https://events.ccc.de/congress/2017/Fahrplan/events/9021.html

Recording: https://media.ccc.de/v/34c3-9021-squeezing_a_key_through_a_carry_bit

Demo: https://youtu.be/aIFiaCcKfW8

1. Squeezing a key through a carry bit Sean Devlin, Filippo Valsorda
4. One month later
5. The code a = a - b mod p a = a - b x = a a = a + p
6. The code a = a - b mod p a = a - b x = a a = a + p a = a - b t = a t += p a ?= t
7. The code a = a - b mod p a = a - b x = a a = a + p a < b a = a - b t = a t += p a ?= t
8. a = a - b x = a a = a + p The bug a = a - b t = a t += p a ?= t
9. The bug
10. The bug Wrong result with probability 2-32
11. A carry propagation bug
12. ECCCCCCC Elliptic Curve Cryptography Crash Course for CCC • Field: numbers modulo p • Points: like (3, 7); fitting an equation • Group: a generator point and addition • Multiplication: repeated addition
13. ECCCCCCCC Elliptic Curve Cryptography Crash Course for CCC (cont.) • Multiplication: 5Q = Q + Q + Q + Q + Q • ECDH private key: a big integer d • ECDH public key: Q = dG (think y = ga) • ECDH shared secret: Q2 = dQ1
14. Double and add Q2 = dQ1 d is BIG. Like, 256 bit. Can't add Q to itself 2256 times.
15. Double and add Q2 = dQ1 1 0 1 0 1 1 1 0 1 0 1 1 0 1 +Q1 Z +Q
16. Double and add 1 0 1 0 1 1 1 0 1 0 1 1 0 1 x2 Z +Q x2 Q2 = dQ1
17. Double and add 1 0 1 0 1 1 1 0 1 0 1 1 0 1 x2 Z +Q x2 x2 Q2 = dQ1
18. Double and add 1 0 1 0 1 1 1 0 1 0 1 1 0 1 +Q1 Z +Q x2 x2 +Q Q2 = dQ1
19. Double and add 1 0 1 0 1 1 1 0 1 0 1 1 0 1 Z +Q x2 x2 +Q x2 x2 Q2 = dQ1
20. Double and add 1 0 1 0 1 1 1 0 1 0 1 1 0 1 Z +Q x2 x2 +Q x2 +Q +Q1 Q2 = dQ1
21. Double and add 1 0 1 0 1 1 1 0 1 0 1 1 0 1 Z +Q x2 x2 +Q x2 +Q x2 x2 Q2 = dQ1
22. Double and add 1 0 1 0 1 1 1 0 1 0 1 1 0 1 Z +Q x2 x2 +Q x2 +Q x2 x2 ... x2 Q2 = dQ1
23. Back to the carry bug
24. secret = ScalarMult(point, scalar) ← Q2 = dQ └─ p256PointAddAffineAsm └─ p256SubInternal attacker supplied secret key session key
25. Q1 → ScalarMult(Q1, ) Q2 → ScalarMult(Q2, ) 1 1 1 0 1 Z +Q1 x2 x2 +Q1 x2 +Q1 x2 +Q1 0 1 1 0 1 Z +Q2 x2 x2 +Q2 x2 +Q2 x2 x2
26. Q1 → ScalarMult(Q1, ) → Q2 → ScalarMult(Q2, ) → ✅ ? 1 1 0 1 ? 1 1 0 1 1 1 1 0 1
27. Q1 → Q2 → 0 1 1 0 1 1 1 1 0 1 Q1 → Q2 → 0 0 1 1 0 1 1 0 1 1 0 1 Q1 → Q2 → 0 1 0 1 1 0 1 1 1 0 1 1 0 1
29. Go implementation of ScalarMult Booth's multiplication in 5-bit windows. Precomputed table of 1Q to 16Q. Add, double 5 times. 01 00010 01110 01010 01010 10010 00001 01111 10011 01101 ...
30. Precomp table
31. Multiplication loop
32. Go implementation of ScalarMult Booth's multiplication in 5-bit windows. Precomputed table of 1Q to 16Q. Add, double 5 times. Limbs representation: less overlap and aliasing problems. 01 00010 01110 01010 01010 10010 00001 01111 10011 01101 ... {1 0} {15 1} {7 0} {5 0} {5 0} {9 0} {1 0} {8 1} {6 1} {9 1} ...
33. Go implementation of ScalarMult Booth's multiplication in 5-bit windows. Precomputed table of 1Q to 16Q. Add, double 5 times. Attack one limb at a time, instead of one bit. 34 limb values → 17 points / 5 key bits on average. 01 00010 01110 01010 01010 10010 00001 01111 10011 01101 ...
34. Multiplication loop
35. Assembly hook
39. The first limb 3 3 x2 x2 x2 x2 x2 → 3 x25 Precomp Doubling Limb
40. The first limb 3 3 x2 x2 x2 x2 x2 → 3 x25 3 x2 6 x2 x2 x2 x2 x2 → 3 x26 3 x2 x2 12 x2 x2 x2 x2 x2 → 3 x27 Precomp Doubling Limb
41. The first limb 3 3 x2 x2 x2 x2 x2 → 3 x25 3 x2 6 x2 x2 x2 x2 x2 → 3 x26 3 x2 x2 12 x2 x2 x2 x2 x2 → 3 x27 Precomp Doubling Limb
42. The last bits
43. Kangaroo jumps depend from the terrain at the start point. Let a tracked kangaroo loose. Place a trap at the end.
44. Kangaroo jumps depend from the terrain at the start point. If the wild kangaroo intersects the path at any point,
 it ends up in the trap.
45. Back to elliptic curves. A jump is QN+1 = QN + H(QN) where H is a hash. Same starting point, same jump. You run from a known starting point, then from dG.
 If you collide, you traceback to d!
46. A target • JSON Object Signing and Encryption, JOSE (JWT) • ECDH-ES public key algorithm • go-jose and Go 1.8.1 • Check if the service successfully decrypts payload
47. Spot instance infrastructure Sage dispatcher /work /result
48. Figures! • Each key: ~52 limbs, modulo the kangaroo • Each limb: ~16 points on average • Each point: ~226 candidate points • (226 * 16) candidate points: ~85 CPU hours • 85 CPU hours: $1.26 EC2 spot instances • Total: 4,400 CPU hours / $65 on EC2
49. Demo
50. Demo
51. Demo
52. Filippo Valsorda @FiloSottile Sean Devlin @spdevlin Thank you! No bug is small enough.