Bounds on the Quantum Satisfiability Threshold Cristopher Moore - PowerPoint PPT Presentation

Bounds on the Quantum Satisfiability Threshold Cristopher Moore Center for Quantum Information and Control Computer Science / Physics and Astronomy, UNM Santa Fe Institute joint work with Sergey Bravyi (IBM) and Alexander Russell (Connecticut) Thursday, October 1, 2009

A phase transition • Random 3-SAT formulas with n variables and α n clauses 1.0 0.8 n = 10 n = 15 n = 20 n = 30 n = 50 0.6 n = 100 P 0.4 0.2 0.0 7 3 4 5 6 ! Thursday, October 1, 2009

A phase transition • Search times appear to peak at the transition 6 10 5 10 DPLL calls 4 10 3 10 2 10 1 2 3 4 7 8 5 6 ! Thursday, October 1, 2009

Quantum k-SAT [Bravyi] • Classical SAT: each clause forbids one out of 8 truth values. Think of this as forbidding a basis vector: ( x 1 ∨ x 2 ∨ x 3 ) ⇔ � 010 | x � = 0 • Quantum SAT: forbid an arbitrary vector in , C 2 ⊗ C 2 ⊗ C 2 � v | x � = 0 Π c | ψ � = | ψ � • For each clause c , we have where Π c = (1 − | v �� v | ) ⊗ 1 n − 3 Thursday, October 1, 2009

A local Hamiltonian • Alternately, ask whether there is a zero-energy state of a local, disordered | ψ � Hamiltonian: � | v �� v | ⊗ 1 H = c • What is its ground state energy? QMA 1 -complete [Bravyi] • When are its ground states entangled? Thursday, October 1, 2009

Forbidden and satisfying subspaces | v � ⊗ | 00 �       | v � ⊗ | 01 �     | v � | w �       | v � ⊗ | 10 �           | v � ⊗ | 11 �     V forbidden = span | 00 � ⊗ | w �       | 01 � ⊗ | w �           | 10 � ⊗ | w �           | 11 � ⊗ | w �   V sat = V ⊥ • The satisfying subspace is forbidden • With probability 1, , so rank V forbidden = 8 rank V sat = 32 − 8 = 24 Thursday, October 1, 2009

Generic clause vectors | v � ⊗ | 00 �       | v � ⊗ | 01 �     | v � | w �       | v � ⊗ | 10 �           | v � ⊗ | 11 �     V forbidden = span | 00 � ⊗ | w �       | 01 � ⊗ | w �           | 10 � ⊗ | w �         • These ranks take generic values with probability 1   | 11 � ⊗ | w �   • Coincidences can only decrease , and increase rank V forbidden rank V sat • For a given hypergraph, if any choice of clause vectors make it unsatisfiable, it is generically unsatisfiable [Laumann et al.] Thursday, October 1, 2009

Random quantum k-SAT formulas • Two sources of randomness: • A random hypergraph with n vertices and m hyperedges (clauses), where m = α n C ⊗ k • Random clause vectors, chosen uniformly from unit-length vectors in 2 • Threshold conjecture: � 1 α < α c n →∞ Pr[ H ( n, m = α n ) is generically satisfiable] = lim 0 α > α c Thursday, October 1, 2009

A classical upper bound • Compute the expected number of satisfying assignments. For k-SAT, 1 − 2 − k � m = 2(1 − 2 − k ) α � n E [ X ] = 2 n � � • This is an upper bound on the probability of satisfiability: Pr[ X > 0] ≤ E [ X ] • This becomes exponentially small when α is large enough: α c ≤ log 1 / (1 − 2 − k ) 2 ≈ 2 k ln 2 • This is asymptotically tight [Achlioptas&Moore, Achlioptas&Peres] Thursday, October 1, 2009

A simple quantum upper bound • Number of solutions is analogous to rank V sat • Expectation of a clause projector: E v Π c = (1 − E v | v �� v | ) ⊗ 1 = (1 − 2 − k ) 1 � • Since the clauses are independent, if then Π φ = Π c c rank V sat ≤ E { v } tr Π † φ Π φ = 2 n (1 − 2 − k ) m α q • So, the quantum bound is at most the classical one: c ≤ α c Thursday, October 1, 2009

Quantum SAT is more restrictive • 2-SAT problem on a star of degree d 2 ⌊ d/ 2 ⌋ + 2 ⌈ d/ 2 ⌉ • Classical: at least solutions • Quantum: only n + 1 = d + 2 Thursday, October 1, 2009

Quantum SAT is more restrictive • Remember that any choice of forbidden vectors gives an upper bound 1 • Forbid singlets: | v � = √ 2 ( | 01 � − | 10 � ) • if and only if is symmetric under transpositions � v | ψ � = 0 | ψ � • If the graph is connected, must be symmetric under all permutations | ψ � Thursday, October 1, 2009

Entangled states Thursday, October 1, 2009

Entangled states • This 2-SAT formula is satisfiable: Thursday, October 1, 2009

Entangled states • This 2-SAT formula is satisfiable: Thursday, October 1, 2009

Entangled states • This 2-SAT formula is satisfiable: • Is this one? Thursday, October 1, 2009

Entangled states • This 2-SAT formula is satisfiable: • Is this one? Thursday, October 1, 2009

Entangled states • This 2-SAT formula is satisfiable: • Is this one? • Classical: of course! Use the new variable to satisfy the new clause. Thursday, October 1, 2009

Entangled states • This 2-SAT formula is satisfiable: • Is this one? • Classical: of course! Use the new variable to satisfy the new clause. • Quantum: no! In entangled states, single variables don’t have values. Similarly, single variables can’t satisfy entangled clauses. Thursday, October 1, 2009

Better upper bounds • For any gadget H on t vertices , E [ Π H ] = rank V sat ( H ) 1 2 t • Any time we add a gadget, we reduce the generic rank. With probability 1, rank V sat ( G ∪ H ) ≤ rank V sat ( H ) rank V sat ( G ) 2 t • Partition a random hypergraph into gadgets: rank V sat ( H i ) rank V sat ≤ 2 n � 2 t i Thursday, October 1, 2009

The Sunflower • Partition the hypergraph into n d sunflowers of degree d : • This gives �� n d �� 3 � d � d ∞ � rank V sat ≤ 2 n 6 + 1 4 d =1 Thursday, October 1, 2009

Sunflower partitions • Naive: at each step, choose a random vertex, declare it and its clauses to be a sunflower, and remove them • Continuous time: give each vertex an index , and remove in t ∈ [0 , 1] decreasing order • The degree of a sunflower of index t is the number of clauses whose variables k α t k − 1 all have index < t . Poisson distribution with mean ln rank V sat • Setting gives α q c ≤ 3 . 894 = 0 n • Greedier partition: taking high-degree vertices first gives α q c ≤ 3 . 689 (analyze with system of differential equations) Thursday, October 1, 2009

nosegay |ˈnōzˌgā| noun The Nosegay a small bunch of flowers, typically one that is sweet-scented. • Bigger gadgets: more conflict, smaller rank • At each step, choose a random clause, and take it and its neighbors • Gives , far below the classical α q c ≤ 3 . 594 α c ≈ 4 . 267 Thursday, October 1, 2009

When k is large • Asymptotically, we have α c ≤ 2 k b • where is the root of ln 2 − 2 b + ln( b + 1) = 0 b ≈ 0 . 573 < ln 2 • Classically, α c = (1 − o (1)) ≤ 2 k ln 2 so the quantum threshold is a constant smaller. Thursday, October 1, 2009

Open questions • Classical: counting satisfying assignments of a 3-SAT formula is #P-complete. Quantum analog: computing . What it its complexity? rank V sat Might not be in #P: entanglement again. • Similarly, is generic satisfiability of a hypergraph in NP? Is it NP-hard? • Is there a satisfiable-but-entangled phase, in which random formulas are satisfiable, but all satisfying states are highly entangled? 2 k • Assuming there is a transition, does grow as ? Does it even grow α q c without bound as k increases? Best lower bounds so far are less than 1! • What is the adversarial classical threshold, where the hypergraph is random, but the adversary chooses which literals to negate? Thursday, October 1, 2009

Graph Isomorphism Factoring appears to be outside P, but not NP-complete. (Indeed, we believe that BQP does not contain all of NP.) Another candidate problem like this: ? Thursday, October 1, 2009

The Hidden Subgroup Problem We have a function f : G → X We want to know its symmetries H ⊆ G Essentially all quantum algorithms that are exponentially faster than classical are of this form: = factoring Z ∗ n = Graph Isomorphism S n = some cryptographic lattice problems D n Thursday, October 1, 2009

The Story So Far... It turns out that this naïve generalization of Shor’s algorithm doesn’t work: the permutation group is “too non-Abelian.” S n Tantalizingly, we know a measurement exists, but we don’t know if we can do it efficiently. How much can quantum computing really do? How “special” is factoring? Thursday, October 1, 2009

Scattering Algorithms false false false true true true true true false true Thursday, October 1, 2009

Scattering Algorithms false false false true true true true true false true Thursday, October 1, 2009