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Introduction to quantum algorithms and introduction to code-based cryptography Daniel J. Bernstein University of Illinois at Chicago & Technische Universiteit Eindhoven Data (state) stored in n bits: an element of { 0 ; 1 } n ,

  1. Reversible and clean: after finishing dirty computation, set non-outputs back to 0, by repeating same operations on non-outputs in reverse order. Original computation: (inputs) �→ (inputs ; dirt ; outputs). Dirty reversible computation: (inputs ; zeros ; zeros) �→ (inputs ; dirt ; outputs). Clean reversible computation: (inputs ; zeros ; zeros) �→ (inputs ; zeros ; outputs).

  2. Given fast circuit for p and fast circuit for p − 1 , build fast reversible circuit for ( x; zeros) �→ ( p ( x ) ; zeros).

  3. Given fast circuit for p and fast circuit for p − 1 , build fast reversible circuit for ( x; zeros) �→ ( p ( x ) ; zeros). Replace reversible bit operations with Toffoli gates etc. permuting C 2 n + z → C 2 n + z . Permutation on first 2 n entries is ( a 0 ; a 1 ; : : : ; a 2 n − 1 ) �→ ( a p − 1 (0) ; a p − 1 (1) ; : : : ; a p − 1 (2 n − 1) ). Typically prepare vectors supported on first 2 n entries so don’t care how permutation acts on last 2 n + z − 2 n entries.

  4. Warning: Number of qubits ≈ number of bit operations in original p; p − 1 circuits. This can be much larger than number of bits stored in the original circuits.

  5. Warning: Number of qubits ≈ number of bit operations in original p; p − 1 circuits. This can be much larger than number of bits stored in the original circuits. Many useful techniques to compress into fewer qubits, but often these lose time. Many subtle tradeoffs.

  6. Warning: Number of qubits ≈ number of bit operations in original p; p − 1 circuits. This can be much larger than number of bits stored in the original circuits. Many useful techniques to compress into fewer qubits, but often these lose time. Many subtle tradeoffs. Crude “poly-time” analyses don’t care about this, but serious cryptanalysis is much more precise.

  7. Fast quantum operations, part 2 “Hadamard”: ( a 0 ; a 1 ) �→ ( a 0 + a 1 ; a 0 − a 1 ).

  8. Fast quantum operations, part 2 “Hadamard”: ( a 0 ; a 1 ) �→ ( a 0 + a 1 ; a 0 − a 1 ). ( a 0 ; a 1 ; a 2 ; a 3 ) �→ ( a 0 + a 1 ; a 0 − a 1 ; a 2 + a 3 ; a 2 − a 3 ).

  9. Fast quantum operations, part 2 “Hadamard”: ( a 0 ; a 1 ) �→ ( a 0 + a 1 ; a 0 − a 1 ). ( a 0 ; a 1 ; a 2 ; a 3 ) �→ ( a 0 + a 1 ; a 0 − a 1 ; a 2 + a 3 ; a 2 − a 3 ). Same for qubit 1: ( a 0 ; a 1 ; a 2 ; a 3 ) �→ ( a 0 + a 2 ; a 1 + a 3 ; a 0 − a 2 ; a 1 − a 3 ).

  10. Fast quantum operations, part 2 “Hadamard”: ( a 0 ; a 1 ) �→ ( a 0 + a 1 ; a 0 − a 1 ). ( a 0 ; a 1 ; a 2 ; a 3 ) �→ ( a 0 + a 1 ; a 0 − a 1 ; a 2 + a 3 ; a 2 − a 3 ). Same for qubit 1: ( a 0 ; a 1 ; a 2 ; a 3 ) �→ ( a 0 + a 2 ; a 1 + a 3 ; a 0 − a 2 ; a 1 − a 3 ). Qubit 0 and then qubit 1: ( a 0 ; a 1 ; a 2 ; a 3 ) �→ ( a 0 + a 1 ; a 0 − a 1 ; a 2 + a 3 ; a 2 − a 3 ) �→ ( a 0 + a 1 + a 2 + a 3 ; a 0 − a 1 + a 2 − a 3 , a 0 + a 1 − a 2 − a 3 ; a 0 − a 1 − a 2 + a 3 ).

  11. Repeat n times: e.g., (1 ; 0 ; 0 ; : : : ; 0) �→ (1 ; 1 ; 1 ; : : : ; 1). Measuring (1 ; 0 ; 0 ; : : : ; 0) always produces 0. Measuring (1 ; 1 ; 1 ; : : : ; 1) can produce any output: Pr[output = q ] = 1 = 2 n .

  12. Repeat n times: e.g., (1 ; 0 ; 0 ; : : : ; 0) �→ (1 ; 1 ; 1 ; : : : ; 1). Measuring (1 ; 0 ; 0 ; : : : ; 0) always produces 0. Measuring (1 ; 1 ; 1 ; : : : ; 1) can produce any output: Pr[output = q ] = 1 = 2 n . Aside from “normalization” (irrelevant to measurement), have Hadamard = Hadamard − 1 , so easily work backwards from “uniform superposition” (1 ; 1 ; 1 ; : : : ; 1) to “pure state” (1 ; 0 ; 0 ; : : : ; 0).

  13. Simon’s algorithm Assume: nonzero s ∈ { 0 ; 1 } n satisfies f ( x ) = f ( x ⊕ s ) for every x ∈ { 0 ; 1 } n . Can we find this period s , given a fast circuit for f ?

  14. Simon’s algorithm Assume: nonzero s ∈ { 0 ; 1 } n satisfies f ( x ) = f ( x ⊕ s ) for every x ∈ { 0 ; 1 } n . Can we find this period s , given a fast circuit for f ? We don’t have enough data if f has many periods. Assume: { periods } = { 0 ; s } .

  15. Simon’s algorithm Assume: nonzero s ∈ { 0 ; 1 } n satisfies f ( x ) = f ( x ⊕ s ) for every x ∈ { 0 ; 1 } n . Can we find this period s , given a fast circuit for f ? We don’t have enough data if f has many periods. Assume: { periods } = { 0 ; s } . Traditional solution: Compute f for many inputs, sort, analyze collisions. Success probability is very low until #inputs approaches 2 n= 2 .

  16. Simon’s algorithm uses far fewer qubit operations if n is large and reversibility overhead is low.

  17. Simon’s algorithm uses far fewer qubit operations if n is large and reversibility overhead is low. Say f maps n bits to m bits using z “ancilla” bits for reversibility. Prepare n + m + z qubits in pure zero state: vector (1 ; 0 ; 0 ; : : : ).

  18. Simon’s algorithm uses far fewer qubit operations if n is large and reversibility overhead is low. Say f maps n bits to m bits using z “ancilla” bits for reversibility. Prepare n + m + z qubits in pure zero state: vector (1 ; 0 ; 0 ; : : : ). Use n -fold Hadamard to move first n qubits into uniform superposition: (1 ; 1 ; 1 ; : : : ; 1 ; 0 ; 0 ; : : : ) with 2 n entries 1, others 0.

  19. Apply fast vector permutation for reversible f computation: 1 in position ( q; 0 ; 0) moves to position ( q; f ( q ) ; 0). Note symmetry between 1 at ( q; f ( q ) ; 0) and 1 at ( q ⊕ s; f ( q ) ; 0).

  20. Apply fast vector permutation for reversible f computation: 1 in position ( q; 0 ; 0) moves to position ( q; f ( q ) ; 0). Note symmetry between 1 at ( q; f ( q ) ; 0) and 1 at ( q ⊕ s; f ( q ) ; 0). Apply n -fold Hadamard.

  21. Apply fast vector permutation for reversible f computation: 1 in position ( q; 0 ; 0) moves to position ( q; f ( q ) ; 0). Note symmetry between 1 at ( q; f ( q ) ; 0) and 1 at ( q ⊕ s; f ( q ) ; 0). Apply n -fold Hadamard. Measure. By symmetry, output is orthogonal to s .

  22. Apply fast vector permutation for reversible f computation: 1 in position ( q; 0 ; 0) moves to position ( q; f ( q ) ; 0). Note symmetry between 1 at ( q; f ( q ) ; 0) and 1 at ( q ⊕ s; f ( q ) ; 0). Apply n -fold Hadamard. Measure. By symmetry, output is orthogonal to s . Repeat n + 10 times. Use Gaussian elimination to (probably) find s .

  23. Example, 3 bits to 3 bits: f (0) = 4. f (1) = 7. f (2) = 2. f (3) = 3. f (4) = 7. f (5) = 4. f (6) = 3. f (7) = 2.

  24. Example, 3 bits to 3 bits: f (0) = 4. f (1) = 7. 4 ❁ 7 ❁ ❁ ❁ f (2) = 2. 2 3 f (3) = 3. 7 ❁ 4 ❁ f (4) = 7. ❁ ❁ f (5) = 4. 3 2 f (6) = 3. f (7) = 2.

  25. Example, 3 bits to 3 bits: f (0) = 4. f (1) = 7. 4 ❁ 7 ❁ ❁ ❁ f (2) = 2. 2 3 f (3) = 3. 7 ❁ 4 ❁ f (4) = 7. ❁ ❁ f (5) = 4. 3 2 f (6) = 3. f (7) = 2. Complete table shows that f ( x ) = f ( x ⊕ 5) for all x . Let’s watch Simon’s algorithm for f , using 6 qubits.

  26. Step 1. Set up pure zero state: 1 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 :

  27. Step 2. Hadamard on qubit 0: 1 ; 1 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 :

  28. Step 3. Hadamard on qubit 1: 1 ; 1 ; 1 ; 1 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 :

  29. Step 4. Hadamard on qubit 2: 1 ; 1 ; 1 ; 1 ; 1 ; 1 ; 1 ; 1 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 :

  30. Step 5. ( q; 0) �→ ( q; f ( q )): 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 1 ; 0 ; 0 ; 0 ; 0 ; 1 ; 0 ; 0 ; 0 ; 1 ; 0 ; 0 ; 1 ; 0 ; 1 ; 0 ; 0 ; 0 ; 0 ; 1 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 1 ; 0 ; 0 ; 1 ; 0 ; 0 ; 0 :

  31. Step 6. Hadamard on qubit 0: 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 1 ; 1 ; 0 ; 0 ; 1 ; 1 ; 0 ; 0 ; 1 ; 1 ; 0 ; 0 ; 1 ; 1 ; 1 ; 1 ; 0 ; 0 ; 1 ; 1 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 1 ; 1 ; 0 ; 0 ; 1 ; 1 ; 0 ; 0 : Notation: 1 = − 1.

  32. Step 7. Hadamard on qubit 1: 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 1 ; 1 ; 1 ; 1 ; 1 ; 1 ; 1 ; 1 ; 1 ; 1 ; 1 ; 1 ; 1 ; 1 ; 1 ; 1 ; 1 ; 1 ; 1 ; 1 ; 1 ; 1 ; 1 ; 1 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 1 ; 1 ; 1 ; 1 ; 1 ; 1 ; 1 ; 1 :

  33. Step 8. Hadamard on qubit 2: 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 2 ; 0 ; 2 ; 0 ; 0 ; 2 ; 0 ; 2 ; 2 ; 0 ; 2 ; 0 ; 0 ; 2 ; 0 ; 2 ; 2 ; 0 ; 2 ; 0 ; 0 ; 2 ; 0 ; 2 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 2 ; 0 ; 2 ; 0 ; 0 ; 2 ; 0 ; 2 :

  34. Step 8. Hadamard on qubit 2: 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 2 ; 0 ; 2 ; 0 ; 0 ; 2 ; 0 ; 2 ; 2 ; 0 ; 2 ; 0 ; 0 ; 2 ; 0 ; 2 ; 2 ; 0 ; 2 ; 0 ; 0 ; 2 ; 0 ; 2 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 2 ; 0 ; 2 ; 0 ; 0 ; 2 ; 0 ; 2 : Step 9. Measure. First 3 qubits are uniform random vector orthogonal to 101: i.e., 000, 010, 101, or 111.

  35. Grover’s algorithm Assume: unique s ∈ { 0 ; 1 } n has f ( s ) = 0. Traditional algorithm to find s : compute f for many inputs, hope to find output 0. Success probability is very low until #inputs approaches 2 n .

  36. Grover’s algorithm Assume: unique s ∈ { 0 ; 1 } n has f ( s ) = 0. Traditional algorithm to find s : compute f for many inputs, hope to find output 0. Success probability is very low until #inputs approaches 2 n . Grover’s algorithm takes only 2 n= 2 reversible computations of f . Typically: reversibility overhead is small enough that this easily beats traditional algorithm.

  37. Start from uniform superposition over all n -bit strings q . Step 1: Set a ← b where b q = − a q if f ( q ) = 0, b q = a q otherwise. This is fast. Step 2: “Grover diffusion”. Negate a around its average. This is also fast. Repeat Step 1 + Step 2 about 0 : 58 · 2 0 : 5 n times. Measure the n qubits. With high probability this finds s .

  38. Normalized graph of q �→ a q for an example with n = 12 after 0 steps: 1.0 0.5 0.0 −0.5 −1.0

  39. Normalized graph of q �→ a q for an example with n = 12 after Step 1: 1.0 0.5 0.0 −0.5 −1.0

  40. Normalized graph of q �→ a q for an example with n = 12 after Step 1 + Step 2: 1.0 0.5 0.0 −0.5 −1.0

  41. Normalized graph of q �→ a q for an example with n = 12 after Step 1 + Step 2 + Step 1: 1.0 0.5 0.0 −0.5 −1.0

  42. Normalized graph of q �→ a q for an example with n = 12 after 2 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0

  43. Normalized graph of q �→ a q for an example with n = 12 after 3 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0

  44. Normalized graph of q �→ a q for an example with n = 12 after 4 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0

  45. Normalized graph of q �→ a q for an example with n = 12 after 5 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0

  46. Normalized graph of q �→ a q for an example with n = 12 after 6 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0

  47. Normalized graph of q �→ a q for an example with n = 12 after 7 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0

  48. Normalized graph of q �→ a q for an example with n = 12 after 8 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0

  49. Normalized graph of q �→ a q for an example with n = 12 after 9 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0

  50. Normalized graph of q �→ a q for an example with n = 12 after 10 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0

  51. Normalized graph of q �→ a q for an example with n = 12 after 11 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0

  52. Normalized graph of q �→ a q for an example with n = 12 after 12 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0

  53. Normalized graph of q �→ a q for an example with n = 12 after 13 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0

  54. Normalized graph of q �→ a q for an example with n = 12 after 14 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0

  55. Normalized graph of q �→ a q for an example with n = 12 after 15 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0

  56. Normalized graph of q �→ a q for an example with n = 12 after 16 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0

  57. Normalized graph of q �→ a q for an example with n = 12 after 17 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0

  58. Normalized graph of q �→ a q for an example with n = 12 after 18 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0

  59. Normalized graph of q �→ a q for an example with n = 12 after 19 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0

  60. Normalized graph of q �→ a q for an example with n = 12 after 20 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0

  61. Normalized graph of q �→ a q for an example with n = 12 after 25 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0

  62. Normalized graph of q �→ a q for an example with n = 12 after 30 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0

  63. Normalized graph of q �→ a q for an example with n = 12 after 35 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0 Good moment to stop, measure.

  64. Normalized graph of q �→ a q for an example with n = 12 after 40 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0

  65. Normalized graph of q �→ a q for an example with n = 12 after 45 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0

  66. Normalized graph of q �→ a q for an example with n = 12 after 50 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0 Traditional stopping point.

  67. Normalized graph of q �→ a q for an example with n = 12 after 60 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0

  68. Normalized graph of q �→ a q for an example with n = 12 after 70 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0

  69. Normalized graph of q �→ a q for an example with n = 12 after 80 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0

  70. Normalized graph of q �→ a q for an example with n = 12 after 90 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0

  71. Normalized graph of q �→ a q for an example with n = 12 after 100 × (Step 1 + Step 2): 1.0 0.5 0.0 −0.5 −1.0 Very bad stopping point.

  72. q �→ a q is completely described by a vector of two numbers (with fixed multiplicities): (1) a q for roots q ; (2) a q for non-roots q . Step 1 + Step 2 act linearly on this vector. Easily compute eigenvalues and powers of this linear map to understand evolution of state of Grover’s algorithm. ⇒ Probability is ≈ 1 after ≈ ( ı= 4)2 0 : 5 n iterations.

  73. � � Notes on provability Textbook algorithm analysis: Proof of correctness New algorithm Proof of run time Mislead students into thinking that best algorithm = best proven algorithm.

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