Squeezing a key through a carry bit @ 34c3

Squeezing a key through a carry bit @ 34c3

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Filippo Valsorda

December 27, 2017
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  1. 2.
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  3. 5.

    The code a = a - b mod p a

    = a - b x = a a = a + p
  4. 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
  5. 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
  6. 8.

    a = a - b x = a a =

    a + p The bug a = a - b t = a t += p a ?= t
  7. 9.
  8. 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
  9. 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
  10. 14.

    Double and add Q2 = dQ1 d is BIG. Like,

    256 bit. Can't add Q to itself 2256 times.
  11. 15.

    Double and add Q2 = dQ1 1 0 1 0

    1 1 1 0 1 0 1 1 0 1 +Q1 Z +Q
  12. 16.

    Double and add 1 0 1 0 1 1 1

    0 1 0 1 1 0 1 x2 Z +Q x2 Q2 = dQ1
  13. 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
  14. 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
  15. 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
  16. 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
  17. 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
  18. 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
  19. 24.

    secret = ScalarMult(point, scalar) ← Q2 = dQ └─ p256PointAddAffineAsm

    └─ p256SubInternal attacker supplied secret key session key
  20. 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
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    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
  23. 28.
  24. 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 ...
  25. 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} ...
  26. 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 ...
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  30. 39.

    The first limb 3 3 x2 x2 x2 x2 x2

    → 3 x25 Precomp Doubling Limb
  31. 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
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    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
  33. 43.

    Kangaroo jumps depend from the terrain at the start point.

    Let a tracked kangaroo loose. Place a trap at the end.
  34. 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.
  35. 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!
  36. 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
  37. 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
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