Span-program-based G ( k ) G ( 1 ) k 1 quantum algorithm for . - PowerPoint PPT Presentation

G ( ρ ) ρ NOT Span-program-based G ( ρ k ) G ( ρ 1 ) ρ k ρ 1 quantum algorithm for . . . . . . OR formula evaluation G ( ρ 1 ) G ( ρ 2 ) G ( ρ 3 ) ρ 1 ρ 2 ρ 3 MAJ Ben Reichardt Robert Š palek G ( ρ k ) G ( ¬ ρ k ) G ( ρ 1 ) G ( ¬ ρ 1 ) ρ 1 ρ k Caltech Google . . . . . . . . . EQUAL [quant-ph/0710.2630] | � t e u r t A f o f o 0 n a s n p m s u l o c ) ) ) ) ρ ρ

Farhi, Goldstone, Gutmann ‘07 algorithm • Theorem ([FGG ‘07, CCJY ‘07]) : A balanced binary NAND formula can be evaluated in time N ½ +o(1) . x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 AND AND AND AND OR OR AND ϕ ( x )

Farhi, Goldstone, Gutmann ‘07 algorithm • Theorem ([FGG ‘07, CCJY ‘07]) : A balanced binary NAND formula can be evaluated in time N ½ +o(1) . x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 NAND NAND NAND NAND NAND NAND NAND ϕ ( x )

Farhi, Goldstone, Gutmann ‘07 algorithm • Theorem ([FGG ‘07, CCJY ‘07]) : A balanced binary NAND formula can be evaluated in time N ½ +o(1) . • Convert formula to a tree: NAND • Attach an infinite line to the root x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8

Farhi, Goldstone, Gutmann ‘07 algorithm • Theorem ([FGG ‘07, CCJY ‘07]) : A balanced binary NAND formula can be evaluated in time N ½ +o(1) . • Convert formula to a tree: NAND • Attach an infinite line to the root • Add edges above leaf nodes evaluating to one… =0 =1

Continuous-time quantum walk [FGG ‘07] x 11 = 1 x 11 = 0

FGG quantum walk | ψ t � = e iA G t | ψ 0 �

FGG quantum walk | ψ t � = e iA G t | ψ 0 �

FGG quantum walk | ψ t � = e iA G t | ψ 0 � ϕ ( x ) = 0 ϕ ( x ) = 1 Wave transmits! Wave reflects!

Farhi, Goldstone, Gutmann ‘07 algorithm • Theorem ([FGG ‘07, CCJY ‘07]) : A balanced binary NAND formula can be evaluated in time N ½ +o(1) . Questions: 1. Why does it work? 2. How does it connect to what we know already? 3. How does it generalize? 4. What kinds of problems can we hope to solve with this technique?

Farhi, Goldstone, Gutmann ‘07 algorithm • Theorem ([FGG ‘07, CCJY ‘07]) : A balanced binary NAND formula can be evaluated in time N ½ +o(1) . Questions: Answers: 1. Why does it work? 2. How does it connect to what we know already? “span programs” [Karchwer/Wig. ‘93] 3. How does it generalize? formula evaluation problem over 4. What kinds of problems can we extended gate sets hope to solve with this technique?

Farhi, Goldstone, Gutmann ‘07 algorithm • Theorem ([FGG ‘07, CCJY ‘07]) : A balanced binary NAND formula can be evaluated in time N ½ +o(1) . Questions: Answers: 1. Why does it work? 2. How does it connect to what we know already? “span programs” [Karchwer/Wig. ‘93] 3. How does it generalize? formula evaluation problem over 4. What kinds of problems can we extended gate sets hope to solve with this technique? ??? 5. No, really, WHY does it work?

Def: Read-once formula ' on gate set S = Tree of nested gates from S , with x 7 x 8 each input appearing once x 1 x 2 x 3 x 4 x 6 x 9 x 10 x 11 x 12 OR x 1 x 5 AND AND AND OR Ex: S = {AND, OR}: OR AND ϕ ( x ) Problem: Evaluate ' (x).

Def: Read-once formula ' on gate set S = Tree of nested gates from S , with x 7 x 8 each input appearing once x 1 x 2 x 3 x 4 x 6 x 9 x 10 x 11 x 12 OR x 1 x 5 AND AND AND OR Ex: S = {AND, OR}: OR Gates cannot have fan-out! AND (unlike in a circuit ) ϕ ( x ) Problem: Evaluate ' (x).

x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 [FGG ‘07] algorithm NAND NAND NAND NAND • NAND Theorem ([FGG ‘07, CCJY ‘07]) : A balanced NAND binary AND-OR formula can be evaluated NAND in time N ½ +o(1) . balanced, Analysis by scattering theory. more gates unbalanced [R Š ‘07] algorithm AND-OR • Theorem: A balanced (“adversary- [ACR Š Z ‘07] algorithm bound-balanced”) formula φ over a • Theorem : gate set including all three-bit gates • An “approximately balanced” AND-OR (and more…) can be evaluated in O(ADV( φ )) queries (optimal!) . formula can be evaluated with O( √ N) queries (optimal for read-once!) . (Some gates, e.g., AND, OR, PARITY, can • A general AND-OR formula can be be unbalanced —but not most! ) evaluated with N ½ +o(1) queries. Running time is N ½ +o(1) in each case, after preprocessing.

Recursive 3-bit majority tree 3 d . . . • Best quantum lower bound is x 1 x 1 x 1 [LLS‘05] ADV( ϕ ) = 2 d � � Ω MAJ MAJ MAJ MAJ MAJ MAJ MAJ MAJ MAJ • Expand majority into {AND, OR} gates: d MAJ MAJ MAJ MAJ 3 ( x 1 , x 2 , x 3 ) = ( x 1 ∧ x 2 ) ∨ ( x 3 ∧ ( x 1 ∨ x 2 )) MAJ ϕ ( x ) ∴ {AND, OR} formula size is ≤ 5 d ∴ O( √ 5 d ) = O(2.24 d )-query algorithm [FGG, ACRSZ ‘07] [R Š ‘07] algorithm • Theorem: A balanced (“adversary- • New: O(2 d )-query quantum algorithm bound-balanced”) formula φ over a gate set including all three-bit gates (and more…) can be evaluated in O(ADV( φ )) queries (optimal!) . (Some gates, e.g., AND, OR, PARITY, can be unbalanced —but not most! )

Classical complexity of formula evaluation x N x 1 x 2 · · · Θ (N) OR Balanced AND-OR x 1 x 3 x 4 x 5 x 6 x 7 x 8 x 2 AND AND AND AND Θ (N 0.753… ) [Snir ‘85, Saks & Wigderson ‘86, Santha ‘95] (fan-in two, tight bounds also for “well- OR OR balanced” formulas) AND R(f) = Ω (N 0.51 ) [Heiman, W ‘91] General read-once AND-OR Conjecture: R(f) = Ω (D(f) 0.753… ) [SW ‘86] . . . x 1 x 1 x 1 Balanced MAJ 3 Ω ((7/3) depth ) = R 2 (f) = O((2.6537…) depth ) MAJ MAJ MAJ MAJ MAJ MAJ MAJ MAJ MAJ [Jayram, Kumar, Sivakumar ’03] MAJ MAJ MAJ MAJ ϕ ( x )

Classical Quantum x N x 1 x 2 · · · Θ (N) Θ ( √ N) [Grover ‘96] OR This morning: Balanced AND-OR x 1 x 3 x 4 x 5 x 6 x 7 x 8 x 2 AND AND AND AND Θ (N 0.753… ) Θ ( √ N) (fan-in two) OR OR [S‘85, SW‘86, S‘95] [FGG, ACR Š Z ‘07] AND Ω (N 0.51 ) [HW‘91] Ω ( √ N), √ N ⋅ 2 O( √ (log N)) General read-once AND-OR Conj.: Ω (D(f) 0.753… ) [SW ‘86] [BS ‘04] [ACR Š Z ‘07] . . . x 1 x 1 x 1 This afternoon: Balanced MAJ 3 MAJ MAJ MAJ MAJ MAJ MAJ MAJ MAJ MAJ Ω ((7/3) d ), O((2.6537…) d ) Θ ( 2 d =N log 3 2 ) MAJ MAJ MAJ [JKS ’03] . and much more… . . MAJ ϕ ( x )

More substitution rules G ( ρ 1 ) G ( ρ 2 ) G ( ρ 3 ) ρ 1 ρ 2 ρ 3 MAJ 3 NAND G ( ρ 1 ) G ( ¬ ρ 2 ) G ( ¬ ρ 1 ) G ( ρ 2 ) ρ 2 ρ 1 PARITY G ( ρ k ) G ( ¬ ρ k ) G ( ρ 1 ) G ( ¬ ρ 1 ) ρ 1 ρ k . . . . . . . . . EQUAL . . (with appropriate edge weights) .

⇒ ⇒ • Main Theorem: • φ (x)=1 A G has λ =0 eigenstate with Ω (1) support on the root. • φ (x)=0 A G has no eigenvectors overlapping the root with | λ |<1/ √ N. Converges to steady- state inside tree

⇒ ⇒ • Main Theorem: • φ (x)=1 A G has λ =0 eigenstate with Ω (1) support on the root. • φ (x)=0 A G has no eigenvectors overlapping the root with | λ |<2/ √ N. Faster Algorithm: ( times faster than FGG scattering) √ log N 2 • Start at the root • Apply phase estimation to the quantum walk with precision 1/ √ N • If measured phase is 0, output “ φ (x)=1.” Otherwise, output “ φ (x)=0.” Running time is √ N Recall: Precision- δ phase estimation on a unitary U, starting at an corr. e-state, returns the e-value | eigenvector � | λ � λ ± δ eigenvalue ± δ to precision δ , except w/ prob. 1/4. It uses O(1/ δ ) calls to c-U.

• Theorem: φ (x)=1 ∃ a λ =0 eigenstate of A G( φ , x) supported on root r. ⇔ x= 1 0 0 0 1 1 0 0 0 0 1 11 1 0 1 Proof: λ =0 eigenstate: � Want | α � = α v | v � v such that ∀ v, � � v | A G | α � = α w = 0 w ∼ v G( φ , x): r

• Theorem: φ (x)=1 ∃ a λ =0 eigenstate of A G( φ , x) supported on root r. ⇔ Induction Claim: Each edge x= 1 0 0 0 1 1 0 0 0 0 1 11 1 0 1 gives a “dual-rail” encoding for the evaluation of the subformula above that edge… subformula φ v The λ =0 eigenstate v “output edge” of G( φ v ,x) is: Supported here ⇔ φ v (x)=false Supported here r ⇔ φ v (x)=true

• Proof of the Induction Claim: Base case: v an input x i =0: x i =1: c c v v λ =0 eigenvector constraint α v is not constrained since v at c is α v =0. ✓ and c are not connected. ✓

Induction Claim: Each edge (p,v) gives a “dual-rail” encoding… c 1 c 2 The λ =0 eigenstate v of G( φ v ,x) is: Supported on v p ⇔ φ v (x)=false Supported on p ⇔ φ v (x)=true Proof (induction step): • α v ≠ 0 ⇔ both input subtrees φ c 1 and φ c 2 are true ⇔ NAND( φ c 1 , φ c 2 ) = 0 • Since α p + α c 1 + α c 2 = 0, α p can be ≠ 0 ⇔ either α c i ≠ 0 ⇔ NAND( φ c 1 , φ c 2 ) = 1 ☑

⇒ ⇒ Induction Claim: Each edge (p,v) gives a “dual-rail” encoding… c 1 c 2 The λ =0 eigenstate v of G( φ v ,x) is: Supported on v p ⇔ φ v (x)=false Supported on p ⇔ φ v (x)=true ☐ • Theorem: φ (x)=1 ∃ a λ =0 eigenstate of A G( φ , x) supported on root r. ⇔ ☐ • Main Theorem: • φ (x)=1 A G has eigenvalue-0 e.v. with Ω (1) support on the root. • φ (x)=0 A G has no eigenvectors overlapping the root with | λ |<1/ √ N.

3-Majority gate gadget | α T 1 � | α T 2 � | α T 3 � T 1 T 2 T 3 Input subformulas with λ =0 FALSE w 3 w 1 w 2 eigenstates constructed by induction TRUE TRUE v 1 v 2 v 3 ω 2 ω 1 s c MAJ 3 r