Introduction
For every fixed k ≥ 3 k \ge3 k ≥ 3 k k k [25 ] , and its exponential-time complexity underlies the Exponential Time Hypothesis (ETH) and the Strong Exponential Time Hypothesis (SETH) [38 , 39 ] . In this paper, we refute ETH by giving, for every fixed k ≥ 2 k \ge2 k ≥ 2 k k k
Write the input k k k n n n m m m
F ( x 1 , … , x n ) = ⋀ e = 1 m [ ( x i e , 1 ⊕ b e , 1 ) ∨ ⋯ ∨ ( x i e , k ⊕ b e , k ) ] . F(x_1,\ldots,x_n)=\bigwedge_{e=1}^{m}\left[(x_{i_{e,1}}\oplus b_{e,1})\vee\cdots\vee(x_{i_{e,k}}\oplus b_{e,k})\right]. F ( x 1 , … , x n ) = e = 1 ⋀ m [ ( x i e , 1 ⊕ b e , 1 ) ∨ ⋯ ∨ ( x i e , k ⊕ b e , k ) ] . where i e , j ∈ { 1 , … , n } i_{e,j}\in\{1,\ldots,n\} i e , j ∈ { 1 , … , n } b e , j ∈ { 0 , 1 } b_{e,j}\in\{0,1\} b e , j ∈ { 0 , 1 } j j j e e e ⊕ \oplus ⊕ k k k i e , j i_{e,j} i e , j b e , j b_{e,j} b e , j F F F x 1 , … , x n x_1,\ldots,x_n x 1 , … , x n
# F = ∣ { x ∈ { 0 , 1 } n : F ( x ) = 1 } ∣ . (1) \#F=\left|\left\{x\in\{0,1\}^{n}:F(x)=1\right\}\right|. \tag*{(1)} # F = ∣ { x ∈ { 0 , 1 } n : F ( x ) = 1 } ∣ . (1) Testing # F > 0 \#F>0 # F > 0 k k k n n n log \log log
Theorem 1.1. For each fixed k ≥ 2 k \ge2 k ≥ 2 k k k F F F n ≥ 20 n \ge20 n ≥ 20 m m m # F \#F # F (1) in
( n + m ) O k ( 1 ) 2 O k ( n log log n / log n ) (n+m)^{O_k(1)}2^{O_k(n\log\log n/\sqrt{\log n})} ( n + m ) O k ( 1 ) 2 O k ( n l o g l o g n / l o g n ) bit operations and ( n + m ) O k ( 1 ) (n+m)^{O_k(1)} ( n + m ) O k ( 1 ) D > 0 D>0 D > 0 m ≤ D n m \le Dn m ≤ D n
( n + m ) O k , D ( 1 ) 2 O k , D ( n / log n ) . (n+m)^{O_{k,D}(1)}2^{O_{k,D}(n/\sqrt{\log n})}. ( n + m ) O k , D ( 1 ) 2 O k , D ( n / l o g n ) . The rational interpolation argument in § 6.4 gives the weaker exponent O k ( ρ ( n ) ) O_k(\rho(n)) O k ( ρ ( n ))
ρ ( n ) = n ( log log n ) 2 log n = n ( log n ) 1 / 2 + o ( 1 ) . \rho(n)=\frac{n(\log\log n)^2}{\sqrt{\log n}}=\frac{n}{(\log n)^{1/2+o(1)}}. ρ ( n ) = log n n ( log log n ) 2 = ( log n ) 1/2 + o ( 1 ) n . Appendix § 10 proves both bounds in Theorem 1.1 by tensor contraction. When # F > 0 \#F>0 # F > 0 poly k ( n , m ) \operatorname{poly}_k(n,m) poly k ( n , m ) [41 Theorem 3.3(1)] .
The theorem also yields arbitrarily accurate subexponential-time approximation algorithms for Max-k k k [53 Section 8.1] implies NP-hardness bounds for Max-k k k [56 ] . Write OPT ( F ) \operatorname{OPT}(F) OPT ( F ) F F F
Corollary 1.2. For every fixed k ≥ 2 k \ge2 k ≥ 2 ε > 0 \varepsilon>0 ε > 0 k k k F F F n ≥ 20 n \ge20 n ≥ 20 m m m OPT ( F ) − ε m \operatorname{OPT}(F)-\varepsilon m OPT ( F ) − ε m 1 − 2 − Ω ( n ) 1-2^{-\Omega(n)} 1 − 2 − Ω ( n )
( n + m ) O k , ε ( 1 ) 2 O k , ε ( ρ ( n ) ) (n+m)^{O_{k,\varepsilon}(1)}2^{O_{k,\varepsilon}(\rho(n))} ( n + m ) O k , ε ( 1 ) 2 O k , ε ( ρ ( n )) time and ( n + m ) O k , ε ( 1 ) (n+m)^{O_{k,\varepsilon}(1)} ( n + m ) O k , ε ( 1 )
Proof. We may assume m > 0 m>0 m > 0 f ( x ) f(x) f ( x ) x x x f ^ ( x ) \widehat{f}(x) f ( x ) O ( n / ε 2 ) O(n/\varepsilon^2) O ( n / ε 2 ) 2 n 2^n 2 n 1 − 2 − Ω ( n ) 1-2^{-\Omega(n)} 1 − 2 − Ω ( n )
∣ f ^ ( x ) − f ( x ) ∣ ≤ ε / 2 for every x ∈ { 0 , 1 } n . \left|\widehat{f}(x)-f(x)\right|\le\varepsilon/2\qquad\text{for every }x\in\{0,1\}^n. f ( x ) − f ( x ) ≤ ε /2 for every x ∈ { 0 , 1 } n . Thus, if x ^ \widehat{x} x f ^ \widehat{f} f f ( x ^ ) ≥ max x f ( x ) − ε f(\widehat{x})\ge\max_x f(x)-\varepsilon f ( x ) ≥ max x f ( x ) − ε b b b O k , ε ( n ) O_{k,\varepsilon}(n) O k , ε ( n ) b b b [9 Lemmas 15.9–15.10] produces a formula with O k , ε ( n ) O_{k,\varepsilon}(n) O k , ε ( n ) Theorem 1.1 , binary search on the threshold, and self-reduction then find an exact sample optimizer within the stated bounds. Truncating rejection sampling after O ( n + log ( n / ε 2 ) ) O(n+\log(n/\varepsilon^2)) O ( n + log ( n / ε 2 )) poly ( n , m , ε − 1 ) \operatorname{poly}(n,m,\varepsilon^{-1}) poly ( n , m , ε − 1 ) 2 − Ω ( n ) 2^{-\Omega(n)} 2 − Ω ( n )
Note that runtime bounds of the form 2 o ( n ) 2^{o(n)} 2 o ( n ) 2 n o ( 1 ) 2^{n^{o(1)}} 2 n o ( 1 )
Corollary 1.3. On simple graphs with v v v e e e q ≥ 2 q\ge2 q ≥ 2 q q q 2 o ( v ) 2^{o(v)} 2 o ( v ) poly q ( v , e ) \operatorname{poly}_q(v,e) poly q ( v , e ) 2 o ( v + e ) 2^{o(v+e)} 2 o ( v + e ) poly ( v , e ) \operatorname{poly}(v,e) poly ( v , e )
Proof. For G = ( V , E ) G=(V,E) G = ( V , E ) x u x_u x u u ∈ V u \in V u ∈ V
⋀ { u , w } ∈ E ( ¬ x u ∨ ¬ x w ) \bigwedge_{\{u,w\}\in E}(\neg x_u \lor\neg x_w) { u , w } ∈ E ⋀ ( ¬ x u ∨ ¬ x w ) correspond bijectively to the independent sets { u ∈ V : x u = 1 } \{u \in V : x_u = 1\} { u ∈ V : x u = 1 } q q q q q q q v qv q v O q ( v + e ) O_q(v+e) O q ( v + e )
For perfect matchings, use one Boolean variable per edge and a linear-size circuit checking that exactly one selected edge meets each vertex. The standard circuit-to-3-CNF reduction [9 Lemmas 15.9–15.10] gives O ( v + e ) O(v+e) O ( v + e )
The fixed-density case of Theorem 1.1 gives a circuit lower bound for E N P \mathrm{E}^{\mathrm{NP}} E NP 2 O ( n ) 2^{O(n)} 2 O ( n )
Corollary 1.4. Let C L ( n ) C_L(n) C L ( n ) L L L { 0 , 1 } n \{0,1\}^n { 0 , 1 } n L ∈ E N P L \in\mathrm{E}^{\mathrm{NP}} L ∈ E NP c > 0 c > 0 c > 0
C L ( n ) > c n log n (2) C_L(n) > cn\sqrt{\log n} \tag*{(2)} C L ( n ) > c n log n (2) for infinitely many n n n
Appendix C.1 proves the corollary. Appendix C.2 gives algorithms for inversion and exact preimage counting of Boolean maps with constant output locality.
Related work
We first discuss earlier SAT and counting algorithms and the complexity hypotheses contradicted by Theorem 1.1. We then compare our approach with earlier algebraic algorithms and identify the existing constructions used in our proof, whose roles are developed in the roadmap in Section 2.
SAT algorithms and the exponential-time hypotheses
Schöning’s local-search algorithm and PPSZ are landmarks in the development of exponential-time algorithms for k k k [59 , 54 ] . Schöning’s randomized algorithm starts from a random assignment and repeatedly flips a uniformly chosen variable in an unsatisfied clause. Its bounded-error decision time is
( n + m ) O ( 1 ) ( 2 − 2 / k ) n (n+m)^{O(1)}(2-2/k)^n ( n + m ) O ( 1 ) ( 2 − 2/ k ) n for each fixed k ≥ 3 k \ge3 k ≥ 3 [59 Section 2] . In particular, this gives ( n + m ) O ( 1 ) ( 4 / 3 ) n (n+m)^{O(1)}(4/3)^n ( n + m ) O ( 1 ) ( 4/3 ) n
The PPSZ algorithm [54 ] uses a different mechanism: bounded resolution derives additional clauses, after which variables are assigned in a random order, taking forced values when available and guessing otherwise. For k = 3 , 4 k=3,4 k = 3 , 4 O ( 1.308 n ) O(1.308^n) O ( 1.30 8 n ) [35 ] . Hansen, Kaplan, Zamir, and Zwick improved the PPSZ bound for every fixed k ≥ 3 k \ge3 k ≥ 3 [34 ] . Scheder obtained an improvement for the same range of k k k [58 Theorem 1.5] . For general 3-SAT, a preprint by Jiang and Cai gives a randomized O ∗ ( 1.307031578 n ) O^*(1.307031578^n) O ∗ ( 1.30703157 8 n ) [42 Corollary 1.2] , where O ∗ O^* O ∗ O ( 1.32793 n ) O(1.32793^n) O ( 1.3279 3 n ) [48 Theorem 21] .
Randomized local search has also produced effective heuristics. The WalkSAT heuristic chooses an unsatisfied clause and flips one of its variables, combining random choices with greedy moves [61 ] . Experiments on formulas from circuit synthesis and diagnosis demonstrated the practical value of this approach [61 pp. 340–342, Tables 4–5] .
Among uniformly random k k k [68 Section 1] ; for fixed k ≥ 3 k \ge3 k ≥ 3 m / n ≤ 2 k / k C m/n \le2^k/k^C m / n ≤ 2 k / k C C C C [68 Theorem 3] give a PPZ-based algorithm that finds a solution in 2 n ( 1 − Ω ( log k / k ) ) 2^{n(1-\Omega(\log k/k))} 2 n ( 1 − Ω ( l o g k / k ))
The Exponential Time Hypothesis (ETH) asserted that 3-SAT requires 2 Ω ( n ) 2^{\Omega(n)} 2 Ω ( n ) k k k k ≥ 3 k \ge3 k ≥ 3 k k k 2 n ( 1 − o k ( 1 ) ) 2^{n(1-o_k(1))} 2 n ( 1 − o k ( 1 )) k k k ϵ ∈ ( 0 , 1 ) \epsilon\in(0, 1) ϵ ∈ ( 0 , 1 ) k k k k k k 2 ( 1 − ϵ ) n ( n + m ) O ( 1 ) 2^{(1-\epsilon)n}(n+m)^{O(1)} 2 ( 1 − ϵ ) n ( n + m ) O ( 1 ) k k k
The Gap Exponential Time Hypothesis (Gap-ETH) strengthens ETH by postulating constants δ , γ > 0 \delta, \gamma> 0 δ , γ > 0 O ( 2 γ n ) O(2^{\gamma n}) O ( 2 γ n ) δ \delta δ m = O ( n ) m=O(n) m = O ( n ) 2 / 3 2/3 2/3 [29 ] and dense constraint satisfaction problems [MR17]. It also rules out nontrivial fixed-parameter approximations for Clique and Dominating Set. For Clique, the excluded ratios are o ( O P T ) o(\mathrm{OPT}) o ( OPT ) O P T \mathrm{OPT} OPT
The Nondeterministic Strong Exponential Time Hypothesis (NSETH) asserts that, for every ϵ > 0 \epsilon> 0 ϵ > 0 k k k 2 ( 1 − ϵ ) n ( n + m ) O ( 1 ) 2^{(1-\epsilon)n}(n+m)^{O(1)} 2 ( 1 − ϵ ) n ( n + m ) O ( 1 ) k k k unsatisfiable , or equivalently whether a k k k
Theorem 1.1 refutes ETH, SETH, Gap-ETH, and NSETH. Indeed, testing whether the exact count is positive or zero decides both satisfiability and unsatisfiability in deterministic 2 o ( n ) ( n + m ) O k ( 1 ) 2^{o(n)}(n+m)^{O_k(1)} 2 o ( n ) ( n + m ) O k ( 1 ) k k k [28 Section 1] and SETH [27 Definition 2.1] , and of Gap-ETH. It does not conflict with unconditional exponential lower bounds for restricted proof systems such as regular resolution [BI13]. In 2016, Ryan Williams refuted the Merlin–Arthur version of SETH (MA-SETH) by giving a protocol that certifies k k k 2 n / 2 poly ( n , m ) 2^{n/2}\operatorname{poly}(n,m) 2 n /2 poly ( n , m )
Exact counting and counting complexity
In Valiant’s framework, #SAT and #3-SAT are #P-complete under parsimonious reductions [Val79, Section 3, Facts 1–2]. For fixed k ≥ 3 k \ge3 k ≥ 3 poly k ( n , m ) 2 n − Ω ( n / k ) \operatorname{poly}_k(n,m)2^{n-\Omega(n/k)} poly k ( n , m ) 2 n − Ω ( n / k ) [37 Lemma 3.5] and a deterministic bound in [CW21, Theorem 1.6]. For #2-SAT, Wahlström’s algorithm counts satisfying assignments in O ∗ ( 1.2377 n ) O^*(1.2377^n) O ∗ ( 1.237 7 n ) O ∗ ( 1.637 n ) O^*(1.637^n) O ∗ ( 1.63 7 n )
The counting Exponential Time Hypothesis (#ETH) asserted that exactly counting the satisfying assignments of a 3-CNF requires 2 Ω ( n ) 2^{\Omega(n)} 2 Ω ( n ) [26 ] rules out deterministic 2 o ( v ) 2^{o(v)} 2 o ( v ) v v v O ( v ) O(v) O ( v ) 2 o ( n ) poly ( n , m ) 2^{o(n)} \operatorname{poly}(n,m) 2 o ( n ) poly ( n , m )
Implications for fine-grained complexity
ETH and SETH became standard assumptions for parameterized and fine-grained complexity [39 , 67 ] . ETH-based reductions yielded lower bounds for the dependence on solution size in Clique [CHKX06], and on treewidth in constraint satisfaction and partitioned subgraph isomorphism [51 ] . SETH supported sharp exponential bounds for Hitting Set and Set Splitting [27 ] , and for Independent Set and Dominating Set parameterized by treewidth [49 ] . It also gave bounds on possible speedups of pseudopolynomial algorithms for Subset Sum and Bicriteria Path [1 ] .
For problems in P, SETH was used to rule out polynomial-factor speedups over quadratic running times for edit distance [8 ] , longest common subsequence and dynamic time warping [17 ] , Fréchet distance [16 ] , and regular-expression matching [7 ] . Other applications concerned approximating graph diameter [57 ] and maintaining reachability counts or the size of the largest strongly connected component under graph updates [2 ] . Together these results made SETH a common basis for explaining the apparent optimality of algorithms across quite different problems.
Our refutation of ETH and SETH removes one basis for the hardness claims above. Some of the same lower bounds follow from weaker hypotheses that may still hold. For example, the lower bounds for longest common subsequence and regular-expression matching follow from the Orthogonal Vectors Conjecture (OVC) [17 , 7 ] . OVC asserts that for every fixed ε > 0 \varepsilon> 0 ε > 0 O ( N 2 − ε poly ( d ) ) O(N^{2-\varepsilon}\operatorname{poly}(d)) O ( N 2 − ε poly ( d )) N N N d d d [70 , 67 ] . The lower bounds for Fréchet distance, graph diameter approximation, and the treewidth dependence of Independent Set and Dominating Set also follow from the hypothesis that unrestricted CNF-SAT admits no 2 ( 1 − δ ) n poly ( n , m ) 2^{(1-\delta)n}\operatorname{poly}(n,m) 2 ( 1 − δ ) n poly ( n , m ) δ > 0 \delta> 0 δ > 0 [16 , 57 , 49 ] . Theorem 1.1 gives a running-time guarantee for each fixed clause width k k k n n n
To obtain a faster algorithm for a target problem from our SAT algorithm, we need a reduction in the other direction, from that problem to fixed-width SAT, that controls the number of variables. Set Splitting with sets of size at most k k k [27 Observation 3.6] : use one Boolean variable per element to specify its side of the split. For each set, one clause requires at least one of its variables to be true, and a second requires at least one to be false. The resulting k k k k k k
Prior work on algebraic algorithms for combinatorial problems
Our algorithm is algebraic, based on polynomial interpolation of carefully chosen auxiliary variables. Below we discuss prior work on using algebraic methods to design algorithm for combinatorial problems. For broader background, see the surveys of Koutis and Williams [45 ] on algebraic fingerprints, Williams [72 ] on the polynomial method in algorithm design, and Husfeldt [36 ] on algorithmic inclusion–exclusion.
Williams’s split-and-list algorithm [70 ] partitions the variables of a 2-CSP into three groups, enumerates partial assignments, and combines them using fast matrix multiplication. Koivisto [43 ] gives a direct sum–product formulation. The algorithm of Sections 3–6 likewise enumerates assignments within groups. Fixing the shared measurements y ˉ \bar{y} y ˉ [60 Section 6] extend split-and-list to polynomial-valued weights and arbitrary rings. Our summation routine similarly uses truncated polynomial rings to retain several coefficients in one computation.
Björklund [14 Sections 2–3] uses inclusion–exclusion and polynomial interpolation to reduce permanent and directed Hamiltonian-cycle counting to a weighted sum over smaller instances. He adds the weights of identical instances in a table, then solves each distinct instance with nonzero total weight once. Björklund, Kaski, and Williams [BKW19] use low-degree curves and generalized Kakeya sets to recover polynomial values from evaluations on a small structured set, yielding faster algorithms for fermionants, including the permanent of integer matrices whose entries are bounded in absolute value by a fixed constant. Our polynomial replacements serve a similar purpose: they connect the desired count to structured evaluations from which it can be reconstructed. Here the useful structure is separation into groups after fixing a few shared measurements.
Bhargava, Ghosh, Kumar, and Mohapatra [13 Sections 3.1–3.2] evaluate multivariate polynomials over finite fields along low-degree curves, using Hasse derivatives and Hermite interpolation when values alone do not suffice. For a univariate restriction g g g j j j t t t z j z^j z j g ( t + z ) g(t+z) g ( t + z )
Woodruff and Yekhanin [74 ] use values and first derivatives along a line to recover a target value in private information retrieval. Kopparty, Saraf, and Yekhanin [KSY14] use higher-order Hasse derivatives for local correction of multiplicity codes, extending the line-based recovery of Reed–Muller codes. In our algorithm, the matrix construction supplies low-degree replacement forms that let us compute the Taylor coefficients at the interpolation points by fast summation after fixing a few shared measurements.
Antecedents of the matrix construction
Our counting algorithm evaluates sums of products of polynomials over Boolean assignments. The polynomials share variables, so these sums do not factor into independent smaller sums. We introduce linear substitutions, depending polynomially on an interpolation parameter, that make such a factorization possible after fixing a few integer linear measurements. An auxiliary square rational matrix M 0 M_0 M 0 rank ( I − M 0 ) \operatorname{rank}(I-M_0) rank ( I − M 0 ) M 0 M_0 M 0 § 2.7 explains this connection.
Related interpolation methods have been used to obtain low-rank approximations to structured matrices. In [5 Section 3] , a low-degree polynomial matches ( − 1 ) j (-1)^j ( − 1 ) j j j j [4 Lemma 3.1] . This gives low-rank matrices agreeing with Walsh–Hadamard matrices in most entries, a result in the study of matrix rigidity [65 ] . Related interpolation gives rigidity bounds for Kronecker powers in [3 Theorem 6.1] . Dvir and Liu [30 Claim 2.20 and Lemmas 3.3–3.4] use a spectral approach: sparse changes to a function invariant under permutations of its coordinates make many Fourier coefficients vanish, reducing the rank of its group-circulant matrix.
Our construction also interpolates ( − 1 ) j (-1)^j ( − 1 ) j I − M 0 I-M_0 I − M 0 M 0 M_0 M 0 § 4 .
Sparsification and parity tests
We use the counting-preserving version [28 Appendix A] of the sparsification lemma [39 ] , with quantitative bounds from [19 Section 6] and [18 Lemma 5.14 and equation (5.4)] . This partitions the satisfying assignments among sparse formulas on the same Boolean variables. Appendix § 7 records the disjoint branching argument and the bounds needed when its parameter varies with the input size.
The initial parity tests are the constraints of a Tanner code [63 Section II] , applied to the binary vector recording which clauses an assignment violates. We use the edge-based expander-code construction of [62 Sections IV and VI] : each graph edge carries one bit, and each vertex requires its incident bits to belong to a fixed binary linear code. The local code’s minimum distance, together with graph expansion, forces every nonzero vector satisfying all these constraints to have large support. This construction and distance argument are standard. Lemma 6.1 supplies the parameter choices needed here, including the bounds on the number and support size of the tests, and constructs the graph and local code by deterministic exhaustive searches. The graph is the bipartite double cover of a Cayley multigraph. The shifts defining that multigraph nearly balance every nonconstant parity on the base group: this is the small-bias condition [52 , 6 ] , stated explicitly in (63) . Fixing unused edge bits to zero gives a code on exactly m m m [55 ] , using exact counts to evaluate each candidate. Section 6 gives the constructions and quantitative bounds needed here.
Theorem 1.1 invites a renewed search for algorithms for problems whose known hardness results relied on ETH or SETH. The examples in § 1.1.3 suggest both immediate consequences and problems for which new algorithmic ideas are needed.
Notation
For an integer s ≥ 0 s \ge0 s ≥ 0 [ s ] = { 1 , … , s } [s] = \{1,\ldots,s\} [ s ] = { 1 , … , s } [ 0 ] = ∅ [0] = \varnothing [ 0 ] = ∅ E E E 1 { E } \mathbf{1}\{E\} 1 { E } E E E log 2 \log_2 log 2 poly ( t 1 , … , t s ) \operatorname{poly}(t_1,\ldots,t_s) poly ( t 1 , … , t s ) ( 1 + t 1 + ⋯ + t s ) O ( 1 ) (1+t_1+\cdots+t_s)^{O(1)} ( 1 + t 1 + ⋯ + t s ) O ( 1 ) poly k , c ( t 1 , … , t s ) \operatorname{poly}_{k,c}(t_1,\ldots,t_s) poly k , c ( t 1 , … , t s ) z z z O ( log ( ∣ z ∣ + 2 ) ) O(\log(|z|+2)) O ( log ( ∣ z ∣ + 2 ))
For a finite set S S S Q S \mathbb{Q}^{S} Q S S S S U ⊆ S U \subseteq S U ⊆ S 1 U ∈ { 0 , 1 } S \mathbf{1}_U \in\{0,1\}^{S} 1 U ∈ { 0 , 1 } S ( 1 U ) v = 1 { v ∈ U } (\mathbf{1}_U)_v = \mathbf{1}\{v\in U\} ( 1 U ) v = 1 { v ∈ U } v ∈ S v \in S v ∈ S 0 ⃗ \vec{0} 0 v ∈ S v \in S v ∈ S δ v ∈ Q 1 × S \delta_v \in\mathbb{Q}^{1\times S} δ v ∈ Q 1 × S v v v δ v x = x v \delta_v x = x_v δ v x = x v U ⊆ S U \subseteq S U ⊆ S S S S S ∖ U S \setminus U S ∖ U
We write X ⃗ = ( X 1 , … , X n ) \vec{X} = (X_1,\ldots,X_n) X = ( X 1 , … , X n ) P ⃗ f , v ( T ) \vec{P}_{f,v}(T) P f , v ( T ) p ⃗ = ( p i ) i ∈ S \vec{p} = (p_i)_{i\in S} p = ( p i ) i ∈ S z = ( z i ) i ∈ S z = (z_i)_{i\in S} z = ( z i ) i ∈ S ⟨ p ⃗ , z ⟩ = ∑ i ∈ S p i z i \langle\vec{p},z\rangle= \sum_{i\in S} p_i z_i ⟨ p , z ⟩ = ∑ i ∈ S p i z i ⟨ P ⃗ f , v ( T ) , X ⃗ ⟩ \langle\vec{P}_{f,v}(T),\vec{X}\rangle ⟨ P f , v ( T ) , X ⟩ X i X_i X i
For a polynomial P ( X ⃗ , z ) P(\vec{X},z) P ( X , z ) X ⃗ \vec{X} X z z z [ z j ] P [z^j]P [ z j ] P z j z^j z j X ⃗ \vec{X} X deg X ⃗ P \deg_{\vec{X}} P deg X P X ⃗ \vec{X} X Q \mathbb{Q} Q
Technique overview
We first count satisfying assignments of a 3-CNF in which each variable occurs in at most three clauses. The product of its clause polynomials has degree at most 3 n 3n 3 n § 2.8 then uses disjoint sparsification and variable copying to count the satisfying assignments of an arbitrary k k k ( n + m ) O k ( 1 ) 2 o ( n ) (n+m)^{O_k(1)}2^{o(n)} ( n + m ) O k ( 1 ) 2 o ( n ) k ≥ 2 k \ge2 k ≥ 2 k = 3 k=3 k = 3 § 2.9 explains how counting on the original variables gives the exponent O k ( ρ ( n ) ) O_k(\rho(n)) O k ( ρ ( n )) Proposition 6.4 .
The bounded-occurrence setting
Let F F F n n n F F F
We remove tautological clauses and repeated literals within a clause. An empty clause gives count zero immediately. These operations preserve the occurrence bound. For each remaining clause e e e I e ⊆ [ n ] I_e \subseteq[n] I e ⊆ [ n ] { v } \{v\} { v } F \mathcal{F} F
∣ I e ∣ ≤ 3 , # { e ∈ F : v ∈ I e } ≤ 3 , n ≤ M : = ∑ e ∈ F ∣ I e ∣ ≤ 3 n . (3) |I_e| \le3,\qquad\#\{e \in\mathcal{F}: v \in I_e\} \le3,\qquad n \le M := \sum_{e \in\mathcal{F}} |I_e| \le3n. \tag*{(3)} ∣ I e ∣ ≤ 3 , # { e ∈ F : v ∈ I e } ≤ 3 , n ≤ M := e ∈ F ∑ ∣ I e ∣ ≤ 3 n . (3) All sums below still range over the original n n n
A direct polynomial for the count
Write L e , v ( X v ) = X v L_{e,v}(X_v) = X_v L e , v ( X v ) = X v L e , v ( X v ) = 1 − X v L_{e,v}(X_v) = 1 - X_v L e , v ( X v ) = 1 − X v e e e
Φ e ( ( X v ) v ∈ I e ) = 1 − ∏ v ∈ I e ( 1 − L e , v ( X v ) ) . \Phi_e((X_v)_{v \in I_e}) = 1 - \prod_{v \in I_e}(1 - L_{e,v}(X_v)). Φ e (( X v ) v ∈ I e ) = 1 − v ∈ I e ∏ ( 1 − L e , v ( X v )) . For example, x a ∨ ¬ x b ∨ x c x_a \lor\lnot x_b \lor x_c x a ∨ ¬ x b ∨ x c
Φ e ( X a , X b , X c ) = 1 − ( 1 − X a ) X b ( 1 − X c ) . \Phi_e(X_a, X_b, X_c) = 1 - (1 - X_a)X_b(1 - X_c). Φ e ( X a , X b , X c ) = 1 − ( 1 − X a ) X b ( 1 − X c ) . On Boolean inputs this polynomial is one when the clause is satisfied and zero otherwise. For a dummy clause e e e Φ e = 1 \Phi_e = 1 Φ e = 1
# F = ∑ x ∈ { 0 , 1 } n ∏ e ∈ F Φ e ( ( x v ) v ∈ I e ) . (4) \#F = \sum_{x \in\{0,1\}^n} \prod_{e \in\mathcal{F}} \Phi_e((x_v)_{v \in I_e}). \tag*{(4)} # F = x ∈ { 0 , 1 } n ∑ e ∈ F ∏ Φ e (( x v ) v ∈ I e ) . (4) The product has total degree at most M ≤ 3 n M \le3n M ≤ 3 n
Equation (4) is an exact expression for the count, but evaluating its 2 n 2^n 2 n r r r S 1 , … , S r S_1,\ldots,S_r S 1 , … , S r
Each group contains at most 2 n / r 2n/r 2 n / r
For every clause, all of its variables belong to a single group.
This is a hypothetical, very restrictive assumption; we do not assume that the input formula admits such a partition.
Under this assumption, we could count satisfying assignments in each group independently. Let N g N_g N g S g S_g S g
# F = ∏ g = 1 r N g . \#F = \prod_{g=1}^{r} N_g. # F = g = 1 ∏ r N g . Computing each N g N_g N g
∑ g = 1 r 2 ∣ S g ∣ ≤ r 2 2 n / r \sum_{g=1}^{r} 2^{|S_g|} \le r2^{2n/r} g = 1 ∑ r 2 ∣ S g ∣ ≤ r 2 2 n / r local assignments, with polynomial work per assignment. For large r r r 2 n 2^n 2 n
We will replace the clause inputs so that different groups interact only through a small collection of shared linear summaries. Once the values of those summaries are fixed, we can again sum over each group separately.
A polynomial family with easier evaluation points
We introduce a parameter T T T T = 0 T = 0 T = 0 T = 1 , … , Q T = 1, \ldots, Q T = 1 , … , Q Q Q Q R R R
Write X ⃗ = ( X 1 , … , X n ) \vec{X} = (X_1, \ldots, X_n) X = ( X 1 , … , X n ) v ∈ I e v \in I_e v ∈ I e P ⃗ e , v ( T ) = ( p e , v , 1 ( T ) , … , p e , v , n ( T ) ) \vec{P}_{e,v}(T) = (p_{e,v,1}(T), \ldots, p_{e,v,n}(T)) P e , v ( T ) = ( p e , v , 1 ( T ) , … , p e , v , n ( T )) X v X_v X v e e e
⟨ P ⃗ e , v ( T ) , X ⃗ ⟩ = ∑ i = 1 n p e , v , i ( T ) X i . \langle\vec{P}_{e,v}(T), \vec{X}\rangle= \sum_{i=1}^{n} p_{e,v,i}(T)X_i. ⟨ P e , v ( T ) , X ⟩ = i = 1 ∑ n p e , v , i ( T ) X i . We call this expression the replacement linear form for the occurrence of variable v v v e e e X v X_v X v Φ e \Phi_e Φ e
⟨ P ⃗ e , v ( 0 ) , X ⃗ ⟩ = X v , deg P ⃗ e , v ≤ R . \langle\vec{P}_{e,v}(0), \vec{X}\rangle= X_v,\qquad\deg\vec{P}_{e,v} \le R. ⟨ P e , v ( 0 ) , X ⟩ = X v , deg P e , v ≤ R . Different clauses may use different replacement linear forms for the same variable. Define
H ( T ) = ∑ x ∈ { 0 , 1 } n ∏ e ∈ F Φ e ( ( ⟨ P ⃗ e , v ( T ) , x ⟩ ) v ∈ I e ) ∈ Q [ T ] . (5) H(T) = \sum_{x\in\{0,1\}^{n}} \prod_{e\in F} \Phi_e\left(\left(\left\langle\vec{P}_{e,v}(T),x\right\rangle\right)_{v\in I_e}\right) \in\mathbb{Q}[T]. \tag*{(5)} H ( T ) = x ∈ { 0 , 1 } n ∑ e ∈ F ∏ Φ e ( ( ⟨ P e , v ( T ) , x ⟩ ) v ∈ I e ) ∈ Q [ T ] . (5) At zero every replacement restores the original variable, while the degree of each clause polynomial grows by at most a factor of R R R
H ( 0 ) = # F , deg H ≤ M R ≤ 3 n R . (6) H(0) = \#F,\qquad\deg H \le MR \le3nR. \tag*{(6)} H ( 0 ) = # F , deg H ≤ M R ≤ 3 n R . (6) We will recover H ( 0 ) H(0) H ( 0 ) t = 1 , … , Q t = 1, \ldots, Q t = 1 , … , Q O ( 1 + n R / Q ) O(1 + nR/Q) O ( 1 + n R / Q ) § 2.5 explains. Our goal is therefore to make R R R Q Q Q
At a nonzero t t t ⟨ P ⃗ e , v ( t ) , x ⟩ \langle\vec{P}_{e,v}(t),x\rangle ⟨ P e , v ( t ) , x ⟩ { 0 , 1 } \{0,1\} { 0 , 1 } x x x
Separating the sum by fixing measurements
Fix t ∈ [ Q ] t \in[Q] t ∈ [ Q ] P ⃗ e , v ( t ) \vec{P}_{e,v}(t) P e , v ( t ) H ( t ) H(t) H ( t ) S g = S t , g S_g = S_{t,g} S g = S t , g C g = C t , g C_g = C_{t,g} C g = C t , g
Definition 2.1 (( r , τ , D ) (r,\tau,D) ( r , τ , D ) P ⃗ e , v ( t ) \vec{P}_{e,v}(t) P e , v ( t ) ( r , τ , D ) (r,\tau,D) ( r , τ , D ) t t t S g S_g S g σ e ( t ) \sigma_e(t) σ e ( t ) C g C_g C g
(1) Small groups. Partition the variables into disjoint groups [ n ] = S 1 ⊔ ⋯ ⊔ S r [n] = S_1 \sqcup\cdots\sqcup S_r [ n ] = S 1 ⊔ ⋯ ⊔ S r ∣ S g ∣ ≤ 2 n / r |S_g| \le2n/r ∣ S g ∣ ≤ 2 n / r
(2) One assigned group per clause. Assign each clause e e e σ e ( t ) ∈ [ r ] \sigma_e(t) \in[r] σ e ( t ) ∈ [ r ] e e e
(3) Few bounded measurements. Each group g g g C g C_g C g S g S_g S g c g c_g c g S g S_g S g 2 n 2n 2 n D D D x x x
x g = x ∣ S g , y g = C g x g , y ⃗ = ( y 1 , … , y r ) . x_g = x|_{S_g},\qquad y_g = C_gx_g,\qquad\vec{y} = (y_1,\ldots,y_r). x g = x ∣ S g , y g = C g x g , y = ( y 1 , … , y r ) . (4) Dependence on other groups only through measurements. Suppose clause e e e g = σ e ( t ) g = \sigma_e(t) g = σ e ( t ) S g S_g S g S h S_h S h
that group’s measurements y h y_h y h v ∈ I e v \in I_e v ∈ I e λ ⃗ e , v , h ∈ Q c h \vec{\lambda}_{e,v,h} \in\mathbb{Q}^{c_h} λ e , v , h ∈ Q c h h ≠ g h \ne g h = g x ∈ Q n x \in\mathbb{Q}^n x ∈ Q n
⟨ P ⃗ e , v ( t ) , x ⟩ = ⟨ P ⃗ e , v ( t ) ∣ S g , x g ⟩ + ∑ h ≠ g ⟨ λ ⃗ e , v , h , y h ⟩ . (7) \langle\vec{P}_{e,v}(t),x\rangle= \langle\vec{P}_{e,v}(t)|_{S_g},x_g\rangle+ \sum_{h\ne g}\langle\vec{\lambda}_{e,v,h},y_h\rangle. \tag*{(7)} ⟨ P e , v ( t ) , x ⟩ = ⟨ P e , v ( t ) ∣ S g , x g ⟩ + h = g ∑ ⟨ λ e , v , h , y h ⟩ . (7) Equivalently, P ⃗ e , v ( t ) ∣ S h \vec{P}_{e,v}(t)|_{S_h} P e , v ( t ) ∣ S h C h C_h C h h ≠ g h \ne g h = g
The matrices C h C_h C h y ⃗ \vec{y} y
The algorithm. Given an ( r , τ , D ) (r,\tau,D) ( r , τ , D ) t t t λ ⃗ e , v , h \vec{\lambda}_{e,v,h} λ e , v , h y ⃗ \vec{y} y X t , g ( y g ) \mathcal{X}_{t,g}(y_g) X t , g ( y g ) S g S_g S g y g y_g y g
X t , g ( y g ) = { u ∈ { 0 , 1 } S g : C g u = y g } . \mathcal{X}_{t,g}(y_g)=\{u\in\{0,1\}^{S_g}:C_g u=y_g\}. X t , g ( y g ) = { u ∈ { 0 , 1 } S g : C g u = y g } . For such an assignment u u u g g g e e e g g g (7) to compute the values of its replacement linear forms from u u u Φ e \Phi_e Φ e g g g A t , g ( u ; y ⃗ ) A_{t,g}(u;\vec{y}) A t , g ( u ; y )
Each measurement of a Boolean assignment is an integer in [ − D , D ] [-D,D] [ − D , D ] Z = 0 Z=0 Z = 0
Try every measurement tuple y ⃗ ∈ Y : = ∏ g = 1 r { − D , … , D } c g \vec{y}\in\mathcal{Y}:=\prod_{g=1}^{r}\{-D,\ldots,D\}^{c_g} y ∈ Y := ∏ g = 1 r { − D , … , D } c g
For each group g g g u u u C g u = y g C_g u=y_g C g u = y g A t , g ( u ; y ⃗ ) A_{t,g}(u;\vec{y}) A t , g ( u ; y )
B t , g ( y ⃗ ) = ∑ u ∈ X t , g ( y g ) A t , g ( u ; y ⃗ ) . B_{t,g}(\vec{y})=\sum_{u\in\mathcal{X}_{t,g}(y_g)}A_{t,g}(u;\vec{y}). B t , g ( y ) = u ∈ X t , g ( y g ) ∑ A t , g ( u ; y ) . If no assignment satisfies C g u = y g C_g u=y_g C g u = y g B t , g ( y ⃗ ) = 0 B_{t,g}(\vec{y})=0 B t , g ( y ) = 0
Multiply the group sums and add the result ∏ g = 1 r B t , g ( y ⃗ ) \prod_{g=1}^{r}B_{t,g}(\vec{y}) ∏ g = 1 r B t , g ( y ) Z Z Z
After all measurement tuples have been processed, return Z Z Z
To see why this returns H ( t ) H(t) H ( t ) y ⃗ \vec{y} y X t , g ( y g ) \mathcal{X}_{t,g}(y_g) X t , g ( y g ) y ⃗ \vec{y} y A t , g A_{t,g} A t , g y ⃗ \vec{y} y
H ( t ) = ∑ y ⃗ ∈ Y ∏ g = 1 r ( ∑ u ∈ X t , g ( y g ) A t , g ( u ; y ⃗ ) ) . (8) H(t)=\sum_{\vec{y}\in\mathcal{Y}}\prod_{g=1}^{r}\left(\sum_{u\in\mathcal{X}_{t,g}(y_g)}A_{t,g}(u;\vec{y})\right). \tag*{(8)} H ( t ) = y ∈ Y ∑ g = 1 ∏ r u ∈ X t , g ( y g ) ∑ A t , g ( u ; y ) . (8) Every Boolean assignment has exactly one measurement tuple, so this sum counts each assignment’s contribution exactly once.
There are at most ( 2 D + 1 ) 2 τ n (2D+1)^{2\tau n} ( 2 D + 1 ) 2 τ n ∑ g 2 ∣ S g ∣ ≤ r 2 n / r \sum_g 2^{|S_g|}\le r2^{n/r} ∑ g 2 ∣ S g ∣ ≤ r 2 n / r
( 2 D + 1 ) 2 τ n 2 2 n / r (2D+1)^{2\tau n}2^{2n/r} ( 2 D + 1 ) 2 τ n 2 2 n / r up to polynomial factors. The first factor bounds the number of measurement tuples; the second bounds the number of assignments within one group. The algorithm scans one group at a time and keeps only the current measurement tuple, the current assignment, and running sums and products; see Lemma 3.2 .
For the parameter choices in § 2.6 , we can construct replacement forms of degree at most R R R ( r , τ , D ) (r,\tau,D) ( r , τ , D ) t ∈ [ Q ] t \in[Q] t ∈ [ Q ] § 2.7 gives the construction. We first explain how these properties let us recover the count efficiently.
Recovering the count from Taylor coefficients
Our goal is to recover H ( 0 ) = # F H(0)=\# F H ( 0 ) = # F t = 1 , … , Q t=1,\ldots,Q t = 1 , … , Q deg H ≤ 3 n R \deg H \le3nR deg H ≤ 3 n R Q Q Q H ( t ) H(t) H ( t ) H H H
q = ⌊ 3 n R Q ⌋ , J = q + 1. q=\left\lfloor\frac{3nR}{Q}\right\rfloor,\qquad J=q+1. q = ⌊ Q 3 n R ⌋ , J = q + 1. Suppose we know the coefficients of z 0 , … , z q z^0,\ldots,z^q z 0 , … , z q H ( t + z ) H(t+z) H ( t + z ) t ∈ [ Q ] t\in[Q] t ∈ [ Q ] H H H 3 n R 3nR 3 n R ( T − t ) J (T-t)^J ( T − t ) J t ∈ [ Q ] t\in[Q] t ∈ [ Q ] ∏ t = 1 Q ( T − t ) J \prod_{t=1}^{Q}(T-t)^J ∏ t = 1 Q ( T − t ) J Q J > 3 n R QJ>3nR Q J > 3 n R H H H H H H H ( 0 ) H(0) H ( 0 )
It remains to compute these Taylor coefficients. Fix t ∈ [ Q ] t\in[Q] t ∈ [ Q ] H ( t + z ) H(t+z) H ( t + z ) q q q z z z t t t t t t t + z t+z t + z
Keeping the shifted inputs separate. Write
Δ P ⃗ e , v ( z ) = P ⃗ e , v ( t + z ) − P ⃗ e , v ( t ) . \Delta\vec{P}_{e,v}(z)=\vec{P}_{e,v}(t+z)-\vec{P}_{e,v}(t). Δ P e , v ( z ) = P e , v ( t + z ) − P e , v ( t ) . Keep x x x X ⃗ \vec{X} X
G x ( X ⃗ , z ) = ∏ e ∈ F Φ e ( ( ⟨ P ⃗ e , v ( t ) , x ⟩ + ⟨ Δ P ⃗ e , v ( z ) , X ⃗ ⟩ ) v ∈ I e ) . (9) G_x(\vec{X},z)=\prod_{e\in F}\Phi_e\left(\left(\langle\vec{P}_{e,v}(t),x\rangle+\langle\Delta\vec{P}_{e,v}(z),\vec{X}\rangle\right)_{v\in I_e}\right). \tag*{(9)} G x ( X , z ) = e ∈ F ∏ Φ e ( ( ⟨ P e , v ( t ) , x ⟩ + ⟨ Δ P e , v ( z ) , X ⟩ ) v ∈ I e ) . (9) Here x x x X ⃗ \vec{X} X z z z X ⃗ = x \vec{X}=x X = x ⟨ P ⃗ e , v ( t + z ) , x ⟩ \langle\vec{P}_{e,v}(t+z),x\rangle ⟨ P e , v ( t + z ) , x ⟩
H ( t + z ) = ∑ x ∈ { 0 , 1 } n G x ( x , z ) . H(t+z)=\sum_{x\in\{0,1\}^n}G_x(x,z). H ( t + z ) = x ∈ { 0 , 1 } n ∑ G x ( x , z ) . On the other hand, fixing X ⃗ = 1 U \vec{X}=\mathbf{1}_U X = 1 U U ⊆ [ n ] U\subseteq[n] U ⊆ [ n ] ⟨ Δ P ⃗ e , v ( z ) , 1 U ⟩ \langle\Delta\vec{P}_{e,v}(z),\mathbf{1}_U\rangle ⟨ Δ P e , v ( z ) , 1 U ⟩ x x x x x x ⟨ P ⃗ e , v ( t ) , x ⟩ \langle\vec{P}_{e,v}(t),x\rangle ⟨ P e , v ( t ) , x ⟩ X ⃗ = x \vec{X}=x X = x X ⃗ \vec{X} X
The degree bound makes this possible with few fixed substitutions. Every entry of Δ P ⃗ e , v ( z ) \Delta\vec{P}_{e,v}(z) Δ P e , v ( z ) z z z X i X_i X i z z z
deg X ⃗ ( [ z j ] G x ( X ⃗ , z ) ) ≤ j . \deg_{\vec{X}}\left([z^j]G_x(\vec{X},z)\right)\le j. deg X ( [ z j ] G x ( X , z ) ) ≤ j . Thus, after discarding powers of z z z q q q q q q X ⃗ \vec{X} X
Recovering the shifted sum by inclusion–exclusion. Work modulo z J z^J z J z q z^q z q X i X_i X i G x G_x G x X i X_i X i X ⃗ \vec{X} X
G x ( X ⃗ , z ) ≡ ∑ E ⊆ [ n ] ∣ E ∣ ≤ q a E ( x , z ) ∏ i ∈ E X i ( m o d z J ) . G_x(\vec{X},z)\equiv\sum_{\substack{E\subseteq[n]\\ |E|\le q}}a_E(x,z)\prod_{i\in E}X_i\pmod{z^J}. G x ( X , z ) ≡ E ⊆ [ n ] ∣ E ∣ ≤ q ∑ a E ( x , z ) i ∈ E ∏ X i ( mod z J ) . where a E ( x , z ) a_E(x,z) a E ( x , z ) 1 U 1_U 1 U U U U U U U
a E ( x , z ) ≡ ∑ U ⊆ E ( − 1 ) ∣ E ∣ − ∣ U ∣ G x ( 1 U , z ) ( m o d z J ) . a_E(x,z) \equiv\sum_{U\subseteq E}(-1)^{|E|-|U|}G_x(1_U,z) \pmod{z^J}. a E ( x , z ) ≡ U ⊆ E ∑ ( − 1 ) ∣ E ∣ − ∣ U ∣ G x ( 1 U , z ) ( mod z J ) . When we set X ⃗ = x \vec{X}=x X = x ∏ i ∈ E x i \prod_{i\in E}x_i ∏ i ∈ E x i E E E x x x ∑ x : x i = 1 ( i ∈ E ) a E ( x , z ) \sum_{x:x_i=1\ (i\in E)}a_E(x,z) ∑ x : x i = 1 ( i ∈ E ) a E ( x , z ) x x x
Computing the restricted sums. For U ⊆ E ⊆ [ n ] U\subseteq E\subseteq[n] U ⊆ E ⊆ [ n ] ∣ E ∣ ≤ q |E|\le q ∣ E ∣ ≤ q
Z t , E , U ( z ) = ∑ x ∈ { 0 , 1 } n x i = 1 ( i ∈ E ) G x ( 1 U , z ) in Q [ z ] / ( z J ) . (10) \mathcal{Z}_{t,E,U}(z)=\sum_{\substack{x\in\{0,1\}^n\\x_i=1\ (i\in E)}}G_x(1_U,z)\quad\text{in }\mathbb{Q}[z]/(z^J). \tag*{(10)} Z t , E , U ( z ) = x ∈ { 0 , 1 } n x i = 1 ( i ∈ E ) ∑ G x ( 1 U , z ) in Q [ z ] / ( z J ) . (10) Here U U U X ⃗ = 1 U \vec{X}=1_U X = 1 U E E E x x x
H ( t + z ) ≡ ∑ E ⊆ [ n ] ∣ E ∣ ≤ q ∑ U ⊆ E ( − 1 ) ∣ E ∣ − ∣ U ∣ Z t , E , U ( z ) ( m o d z J ) . (11) H(t+z)\equiv\sum_{\substack{E\subseteq[n]\\|E|\le q}}\sum_{U\subseteq E}(-1)^{|E|-|U|}\mathcal{Z}_{t,E,U}(z)\pmod{z^J}. \tag*{(11)} H ( t + z ) ≡ E ⊆ [ n ] ∣ E ∣ ≤ q ∑ U ⊆ E ∑ ( − 1 ) ∣ E ∣ − ∣ U ∣ Z t , E , U ( z ) ( mod z J ) . (11) This recovers all J = q + 1 J=q+1 J = q + 1 Lemma 3.3 states the general identity used here.
Each sum Z t , E , U \mathcal{Z}_{t,E,U} Z t , E , U § 2.4 . With X ⃗ = 1 U \vec{X}=1_U X = 1 U ⟨ Δ P ⃗ e , v ( z ) , 1 U ⟩ \langle\Delta\vec{P}_{e,v}(z),1_U\rangle ⟨ Δ P e , v ( z ) , 1 U ⟩ x x x x x x ⟨ P ⃗ e , v ( t ) , x ⟩ \langle\vec{P}_{e,v}(t),x\rangle ⟨ P e , v ( t ) , x ⟩ y ⃗ \vec{y} y x i = 1 x_i=1 x i = 1 X ⃗ = 1 U \vec{X}=1_U X = 1 U z z z J J J a E ( x , z ) a_E(x,z) a E ( x , z )
Repeating this calculation at t = 1 , … , Q t=1,\ldots,Q t = 1 , … , Q H ( 0 ) = # F H(0)=\#F H ( 0 ) = # F
The three costs and the running time
Jumping ahead, § 2.7 will construct replacement forms that restore the original variables at zero, have degree at most R R R ( r , τ , D ) (r,\tau,D) ( r , τ , D ) t ∈ [ Q ] t\in[Q] t ∈ [ Q ] n n n
r = Θ ( log n ( log log n ) 2 ) , Q = Θ ( log n log log n ) , r=\Theta\left(\frac{\sqrt{\log n}}{(\log\log n)^2}\right),\qquad Q=\Theta\left(\frac{\log n}{\log\log n}\right), r = Θ ( ( log log n ) 2 log n ) , Q = Θ ( log log n log n ) , R = Θ ( log n ) , τ = Θ ( 1 r log n ) , D = r 2 Q ≤ ( 2 n ) 1 / 128 . (12) R=\Theta(\sqrt{\log n}),\qquad\tau=\Theta\left(\frac{1}{r\log n}\right),\qquad D=r^{2Q}\le(2n)^{1/128}. \tag*{(12)} R = Θ ( log n ) , τ = Θ ( r log n 1 ) , D = r 2 Q ≤ ( 2 n ) 1/128 . (12) We first show how these bounds give a fast counting algorithm; the next subsection will justify that they can all be achieved simultaneously.
Recall that q = J − 1 = ⌊ 3 n R / Q ⌋ q=J-1=\lfloor3nR/Q\rfloor q = J − 1 = ⌊ 3 n R / Q ⌋ t t t § 2.5 computes one restricted sum for each pair U ⊆ E U\subseteq E U ⊆ E ∣ E ∣ ≤ q |E|\le q ∣ E ∣ ≤ q ( n j ) \binom{n}{j} ( j n ) E E E j j j 2 j 2^j 2 j U U U
Thus the three costs multiply to give, up to polynomial factors,
∑ j = 0 min ( q , n ) 2 j ( n j ) ⋅ ( 2 D + 1 ) 2 τ n ⋅ 2 2 n / r . (13) \sum_{j=0}^{\min(q,n)} 2^j \binom{n}{j} \cdot(2D+1)^{2\tau n} \cdot2^{2n/r}. \tag*{(13)} j = 0 ∑ m i n ( q , n ) 2 j ( j n ) ⋅ ( 2 D + 1 ) 2 τ n ⋅ 2 2 n / r . (13) To bound the first cost, note that 1 ≤ q ≤ n / 2 1 \le q \le n/2 1 ≤ q ≤ n /2 n n n R / Q → 0 R/Q \to0 R / Q → 0 ( n j ) ≤ ( e n / j ) j \binom{n}{j} \le(en/j)^j ( j n ) ≤ ( e n / j ) j ( 2 e n / j ) j (2en/j)^j ( 2 e n / j ) j 1 ≤ j ≤ n 1 \le j \le n 1 ≤ j ≤ n
∑ j = 0 q 2 j ( n j ) ≤ ( q + 1 ) ( 2 e n q ) q . \sum_{j=0}^{q} 2^j \binom{n}{j} \le(q+1)\left(\frac{2en}{q}\right)^q. j = 0 ∑ q 2 j ( j n ) ≤ ( q + 1 ) ( q 2 e n ) q . Since q = Θ ( n R / Q ) q = \Theta(nR/Q) q = Θ ( n R / Q ) O ( n ( R / Q ) log ( 2 Q / R ) ) O(n(R/Q)\log(2Q/R)) O ( n ( R / Q ) log ( 2 Q / R )) (13) therefore gives
O ( n [ R Q log 2 Q R + τ log ( 2 D + 1 ) + 1 r ] ) . O\left(n\left[\frac{R}{Q}\log\frac{2Q}{R}+\tau\log(2D+1)+\frac{1}{r}\right]\right). O ( n [ Q R log R 2 Q + τ log ( 2 D + 1 ) + r 1 ] ) . Each term is small under (12) :
R Q log 2 Q R = O ( log log n log n ⋅ log log n ) , \frac{R}{Q}\log\frac{2Q}{R}
=O\left(\frac{\log\log n}{\sqrt{\log n}}\cdot\log\log n\right), Q R log R 2 Q = O ( log n log log n ⋅ log log n ) , τ log ( 2 D + 1 ) = O ( 1 r log n ⋅ log n ) = O ( 1 / r ) , \tau\log(2D+1)=O\left(\frac{1}{r\log n}\cdot\log n\right)=O(1/r), τ log ( 2 D + 1 ) = O ( r log n 1 ⋅ log n ) = O ( 1/ r ) , 1 r = O ( ( log log n ) 2 log n ) . \frac{1}{r}=O\left(\frac{(\log\log n)^2}{\sqrt{\log n}}\right). r 1 = O ( log n ( log log n ) 2 ) . Thus the bounded-occurrence instance can be counted in
n O ( 1 ) 2 O ( n ( log log n ) 2 / log n ) (14) n^{O(1)}2^{O\left(n(\log\log n)^2/\sqrt{\log n}\right)} \tag*{(14)} n O ( 1 ) 2 O ( n ( l o g l o g n ) 2 / l o g n ) (14) bit operations and polynomial space. The finitely many smaller inputs are handled by exhaustive search.
The number of points Q Q Q r r r J = O ( 1 + n R / Q ) J=O(1+nR/Q) J = O ( 1 + n R / Q ) n n n J J J
For the bit-complexity and space bounds, see Appendix § 8 .§ 4 .
Constructing the low-degree replacement linear forms
Consider one occurrence of X v X_v X v T = 0 T=0 T = 0 X v X_v X v t = 1 , … , Q t=1,\ldots,Q t = 1 , … , Q Q Q Q R R R Q Q Q
Suppose a term in the replacement uses a variable outside the clause’s assigned group only at t = 2 t=2 t = 2 1 − T / 2 1-T/2 1 − T /2 t = 2 t=2 t = 2
Step 1: Set up the auxiliary coordinates.
At each evaluation point, an occurrence ( e , v ) (e,v) ( e , v ) v v v e e e Q Q Q Q Q Q
B = ( [ r ] × [ r ] ) Q , D = ∣ B ∣ = r 2 Q . \mathcal{B} = ([r] \times[r])^{Q}, \qquad D = |\mathcal{B}| = r^{2Q}. B = ([ r ] × [ r ] ) Q , D = ∣ B ∣ = r 2 Q . Write an index as
u = ( ( a u ( 1 ) , d u ( 1 ) ) , … , ( a u ( Q ) , d u ( Q ) ) ) . u = \bigl((a_u(1),d_u(1)),\ldots,(a_u(Q),d_u(Q))\bigr). u = ( ( a u ( 1 ) , d u ( 1 )) , … , ( a u ( Q ) , d u ( Q )) ) . We will construct one replacement row for every such index, so it will be available whatever group choices we later make for the formula.
Let ζ ⃗ = ( ζ w ) w ∈ B \vec{\zeta} = (\zeta_w)_{w\in\mathcal{B}} ζ = ( ζ w ) w ∈ B t t t ζ w \zeta_w ζ w a w ( t ) a_w(t) a w ( t ) ζ u \zeta_u ζ u d u ( t ) d_u(t) d u ( t )
S t , d u ( t ) , S t , g : = { w ∈ B : a w ( t ) = g } . S_{t,d_u(t)}, \qquad S_{t,g} := \{w \in\mathcal{B} : a_w(t) = g\}. S t , d u ( t ) , S t , g := { w ∈ B : a w ( t ) = g } . In particular, ζ w \zeta_w ζ w a w ( t ) = d u ( t ) a_w(t)=d_u(t) a w ( t ) = d u ( t ) S t , g S_{t,g} S t , g B \mathcal{B} B D / r D/r D / r
Our auxiliary goal is to construct coefficient rows P ⃗ u ( T ) ∈ Q [ T ] D \vec{P}_u(T) \in\mathbb{Q}[T]^D P u ( T ) ∈ Q [ T ] D
C ∈ { 0 , 1 , − 1 } c × D , c ≤ D τ 3 r , C \in\{0,1,-1\}^{c\times D}, \qquad c \le\frac{D\tau}{3r}, C ∈ { 0 , 1 , − 1 } c × D , c ≤ 3 r D τ , such that
⟨ P ⃗ u ( 0 ) , ζ ⃗ ⟩ = ζ u , deg P ⃗ u ≤ R − 1 , \langle\vec{P}_u(0),\vec{\zeta}\rangle= \zeta_u, \qquad\deg\vec{P}_u \le R-1, ⟨ P u ( 0 ) , ζ ⟩ = ζ u , deg P u ≤ R − 1 , and, at every t ∈ [ Q ] t \in[Q] t ∈ [ Q ] ⟨ P ⃗ u ( t ) , ζ ⃗ ⟩ \langle\vec{P}_u(t),\vec{\zeta}\rangle ⟨ P u ( t ) , ζ ⟩
ζ ⃗ ∣ S t , d u ( t ) and C ζ ⃗ . (15) \vec{\zeta}\big|_{S_{t,d_u(t)}} \quad\text{and}\quad C\vec{\zeta}. \tag*{(15)} ζ S t , d u ( t ) and C ζ . (15) The same measurements C ζ ⃗ C\vec{\zeta} C ζ § 2.4 .
Step 2: Construct the constant matrix.
We start from the identity
ζ u = ( ( I D − M 0 ) ζ ⃗ ) u + ∑ w ∈ B M 0 [ u , w ] ζ w . \zeta_u = \bigl((I_D-M_0)\vec{\zeta}\bigr)_u + \sum_{w\in\mathcal{B}} M_0[u,w]\zeta_w. ζ u = ( ( I D − M 0 ) ζ ) u + w ∈ B ∑ M 0 [ u , w ] ζ w . We want the first term to be determined by C ζ ⃗ C\vec{\zeta} C ζ
E u , w = { t ∈ [ Q ] : a w ( t ) ≠ d u ( t ) } . E_{u,w} = \{t \in[Q] : a_w(t) \ne d_u(t)\}. E u , w = { t ∈ [ Q ] : a w ( t ) = d u ( t )} . The two requirements on M 0 ∈ Q D × D M_0 \in\mathbb{Q}^{D\times D} M 0 ∈ Q D × D
M 0 ξ ⃗ = ξ ⃗ for every ξ ⃗ ∈ ker C , M_0\vec{\xi} = \vec{\xi}\quad\text{for every }\vec{\xi}\in\ker C, M 0 ξ = ξ for every ξ ∈ ker C , M 0 [ u , w ] ≠ 0 ⟹ ∣ E u , w ∣ ≤ R − 1. (16) M_0[u,w] \ne0 \quad\Longrightarrow\quad|E_{u,w}| \le R-1. \tag*{(16)} M 0 [ u , w ] = 0 ⟹ ∣ E u , w ∣ ≤ R − 1. (16) The first says that I D − M 0 I_D-M_0 I D − M 0 ker C \ker C ker C C C C ( I D − M 0 ) ζ ⃗ (I_D-M_0)\vec{\zeta} ( I D − M 0 ) ζ C ζ ⃗ C\vec{\zeta} C ζ
Restricting the support by keeping f f f Let Π s \Pi_s Π s s s s Π s 2 = I D \Pi_s^2=I_D Π s 2 = I D
V 0 = ∏ s = 1 Q Π s , W 0 = 1 2 ∑ s = 1 Q ( I D − Π s ) , M 0 = V 0 f ( W 0 ) , V_0=\prod_{s=1}^{Q}\Pi_s,\qquad W_0=\frac{1}{2}\sum_{s=1}^{Q}(I_D-\Pi_s),\qquad M_0=V_0f(W_0), V 0 = s = 1 ∏ Q Π s , W 0 = 2 1 s = 1 ∑ Q ( I D − Π s ) , M 0 = V 0 f ( W 0 ) , where we will choose a polynomial f f f R − 1 R-1 R − 1 V 0 V_0 V 0 u u u w w w
( a w ( s ) , d w ( s ) ) = ( d u ( s ) , a u ( s ) ) for every s . (a_w(s),d_w(s))=(d_u(s),a_u(s))\quad\text{for every }s. ( a w ( s ) , d w ( s )) = ( d u ( s ) , a u ( s )) for every s . In particular, E u , w E_{u,w} E u , w
Multiplying by f ( W 0 ) f(W_0) f ( W 0 ) f f f f ( W 0 ) f(W_0) f ( W 0 ) R − 1 R-1 R − 1 Π s 2 = I D \Pi_s^2=I_D Π s 2 = I D V 0 V_0 V 0
M 0 = ∑ S ⊆ [ Q ] ∣ S ∣ ≤ R − 1 γ S ∏ s ∉ S Π s for some γ S ∈ Q . (17) M_0=\sum_{\substack{S\subseteq[Q]\\ |S|\leq R-1}}\gamma_S\prod_{s\notin S}\Pi_s\quad\text{for some }\gamma_S\in\mathbb{Q}. \tag*{(17)} M 0 = S ⊆ [ Q ] ∣ S ∣ ≤ R − 1 ∑ γ S s ∈ / S ∏ Π s for some γ S ∈ Q . (17) The permutation in the summand indexed by S S S S S S a w ( s ) = d u ( s ) a_w(s)=d_u(s) a w ( s ) = d u ( s ) s ∉ S s\notin S s ∈ / S E u , w ⊆ S E_{u,w}\subseteq S E u , w ⊆ S ∣ E u , w ∣ > R − 1 |E_{u,w}|>R-1 ∣ E u , w ∣ > R − 1 (16) for any polynomial f f f
Choosing f f f We now choose f f f M 0 M_0 M 0 [ r ] × [ r ] [r]\times[r] [ r ] × [ r ] e ⃗ a b \vec e_{ab} e ab ( a , b ) (a,b) ( a , b )
e ⃗ a a ( a ∈ [ r ] ) , e ⃗ a b + e ⃗ b a ( a < b ) , eigenvalue + 1 , \vec e_{aa}\quad(a\in[r]),\qquad\vec e_{ab}+\vec e_{ba}\quad(a<b),\qquad\text{eigenvalue }+1, e aa ( a ∈ [ r ]) , e ab + e ba ( a < b ) , eigenvalue + 1 , e ⃗ a b − e ⃗ b a ( a < b ) , eigenvalue − 1. \vec e_{ab}-\vec e_{ba}\quad(a<b),\qquad\text{eigenvalue }-1. e ab − e ba ( a < b ) , eigenvalue − 1. There are r ( r + 1 ) / 2 r(r+1)/2 r ( r + 1 ) /2 + 1 +1 + 1 r ( r − 1 ) / 2 r(r-1)/2 r ( r − 1 ) /2 − 1 -1 − 1 { 0 , 1 , − 1 } \{0,1,-1\} { 0 , 1 , − 1 }
For each position s ∈ [ Q ] s\in[Q] s ∈ [ Q ] ψ ⃗ s \vec\psi_s ψ s ψ ⃗ ∈ Q B \vec\psi\in\mathbb{Q}^{B} ψ ∈ Q B
ψ u = ∏ s = 1 Q ψ s ( a u ( s ) , d u ( s ) ) . \psi_u=\prod_{s=1}^{Q}\psi_s(a_u(s),d_u(s)). ψ u = s = 1 ∏ Q ψ s ( a u ( s ) , d u ( s )) . These D = ( r 2 ) Q D=(r^2)^Q D = ( r 2 ) Q Q Q Q Π s \Pi_s Π s ψ ⃗ \vec\psi ψ ψ ⃗ s \vec\psi_s ψ s j j j − 1 -1 − 1
W 0 ψ ⃗ = j ψ ⃗ , V 0 ψ ⃗ = ( − 1 ) j ψ ⃗ , M 0 ψ ⃗ = ( − 1 ) j f ( j ) ψ ⃗ . W_0\vec\psi=j\vec\psi,\qquad V_0\vec\psi=(-1)^j\vec\psi,\qquad M_0\vec\psi=(-1)^jf(j)\vec\psi. W 0 ψ = j ψ , V 0 ψ = ( − 1 ) j ψ , M 0 ψ = ( − 1 ) j f ( j ) ψ . Thus we make M 0 M_0 M 0 f ( j ) = ( − 1 ) j f(j)=(-1)^j f ( j ) = ( − 1 ) j
For a uniformly chosen basis vector, the number j j j
W ∼ Bin ( Q , p ) , p = r − 1 2 r , W\sim\operatorname{Bin}(Q,p),\qquad p=\frac{r-1}{2r}, W ∼ Bin ( Q , p ) , p = 2 r r − 1 , because at each position a fraction p p p r 2 r^2 r 2 − 1 -1 − 1 j j j Q p Qp Qp I \mathcal I I R R R Q p Qp Qp f ( j ) = ( − 1 ) j f(j)=(-1)^j f ( j ) = ( − 1 ) j j ∈ I j\in\mathcal I j ∈ I deg f ≤ R − 1 \deg f\leq R-1 deg f ≤ R − 1 M 0 M_0 M 0 j ∈ I j\in\mathcal I j ∈ I { 0 , … , Q } \{0,\ldots,Q\} { 0 , … , Q }
Measuring the exceptional directions. Use the transposes of the remaining basis vectors as the rows of C C C { 0 , 1 , − 1 } \{0,1,-1\} { 0 , 1 , − 1 } ker C \ker C ker C (16) .
We can now see how to choose R R R W W W Q p ( 1 − p ) = Θ ( Q ) \sqrt{Qp(1-p)}=\Theta(\sqrt{Q}) Qp ( 1 − p ) = Θ ( Q ) R R R Q \sqrt{Q} Q R R R Q Q Q 1 + ⌊ 3 n R / Q ⌋ = o ( n ) 1+\lfloor3nR/Q\rfloor=o(n) 1 + ⌊ 3 n R / Q ⌋ = o ( n )
c = D Pr [ W ∉ I ] ≤ 2 D exp ( − R 2 8 Q ) ≤ D τ 3 r , c=D\Pr[W\notin I]\le2D\exp\left(-\frac{R^2}{8Q}\right)\le\frac{D\tau}{3r}, c = D Pr [ W ∈ / I ] ≤ 2 D exp ( − 8 Q R 2 ) ≤ 3 r D τ , as required. Thus low degree gives restricted support, while interpolation makes the correction I D − M 0 I_D-M_0 I D − M 0
Step 3: Define the auxiliary replacement rows.
We now turn the decomposition of ζ u \zeta_u ζ u M 0 [ u , w ] ≠ 0 M_0[u,w]\ne0 M 0 [ u , w ] = 0
φ u , w ( T ) = ∏ s ∈ E u , w ( 1 − T / s ) . \varphi_{u,w}(T)=\prod_{s\in E_{u,w}}(1-T/s). φ u , w ( T ) = s ∈ E u , w ∏ ( 1 − T / s ) . This polynomial equals one at zero and vanishes at every point where ζ w \zeta_w ζ w P ⃗ u ( T ) \vec{P}_u(T) P u ( T )
⟨ P ⃗ u ( T ) , ζ ⃗ ⟩ = ( ( I D − M 0 ) ζ ⃗ ) u + ∑ w ∈ B M 0 [ u , w ] ≠ 0 M 0 [ u , w ] φ u , w ( T ) ζ w . (18) \langle\vec{P}_u(T),\vec{\zeta}\rangle=((I_D-M_0)\vec{\zeta})_u+\sum_{\substack{w\in B\\M_0[u,w]\ne0}}M_0[u,w]\varphi_{u,w}(T)\zeta_w. \tag*{(18)} ⟨ P u ( T ) , ζ ⟩ = (( I D − M 0 ) ζ ) u + w ∈ B M 0 [ u , w ] = 0 ∑ M 0 [ u , w ] φ u , w ( T ) ζ w . (18) This directly gives all three required properties:
At T = 0 T=0 T = 0 ζ u \zeta_u ζ u
Each multiplier has degree ∣ E u , w ∣ ≤ R − 1 |E_{u,w}|\le R-1 ∣ E u , w ∣ ≤ R − 1 deg P ⃗ u ≤ R − 1 \deg\vec{P}_u\le R-1 deg P u ≤ R − 1
At T = t T=t T = t C ζ ⃗ C\vec{\zeta} C ζ a w ( t ) = d u ( t ) a_w(t)=d_u(t) a w ( t ) = d u ( t ) S t , d u ( t ) S_{t,d_u(t)} S t , d u ( t )
Thus the replacement uses exactly the kinds of data allowed in (15) .
Step 4: Substitute the original variables.
The auxiliary rows are now constructed. To use them for the formula, choose schedules
a v ∈ [ r ] Q ( v ∈ [ n ] ) , d e ∈ [ r ] Q ( e ∈ F ) . a_v\in[r]^Q\quad(v\in[n]),\qquad d_e\in[r]^Q\quad(e\in\mathcal{F}). a v ∈ [ r ] Q ( v ∈ [ n ]) , d e ∈ [ r ] Q ( e ∈ F ) . At point t t t v v v a v ( t ) a_v(t) a v ( t ) e e e d e ( t ) d_e(t) d e ( t )
S t , g = { v : a v ( t ) = g } , σ e ( t ) = d e ( t ) . S_{t,g}=\{v:a_v(t)=g\},\qquad\sigma_e(t)=d_e(t). S t , g = { v : a v ( t ) = g } , σ e ( t ) = d e ( t ) . The occurrence ( e , v ) (e,v) ( e , v )
u ( e , v ) = ( ( a v ( s ) , d e ( s ) ) ) s ∈ [ Q ] ∈ B . u(e,v)=\bigl((a_v(s),d_e(s))\bigr)_{s\in[Q]}\in\mathcal{B}. u ( e , v ) = ( ( a v ( s ) , d e ( s )) ) s ∈ [ Q ] ∈ B . Let h h h h h h Lemma 4.5 chooses the schedules deterministically in polynomial time so that
∣ S t , g ∣ ≤ 2 n / r , h ≤ 2 M / D , |S_{t,g}|\le2n/r,\qquad h\le2M/D, ∣ S t , g ∣ ≤ 2 n / r , h ≤ 2 M / D , provided
12 D 2 ≤ M , 2 Q r 2 ≤ n . (19) 12D^2\le M,\qquad2Qr^2\le n. \tag*{(19)} 12 D 2 ≤ M , 2 Q r 2 ≤ n . (19) The idea is to derandomize a random choice of schedules. If all a v a_v a v d e d_e d e [ r ] Q [r]^Q [ r ] Q M / D M/D M / D S t , g S_{t,g} S t , g n / r n/r n / r (19) , the standard method of conditional expectations derandomizes this choice in polynomial time, ensuring that every label has multiplicity at most 2 M / D 2M/D 2 M / D 2 n / r 2n/r 2 n / r
Different occurrences can receive the same label while requiring different original variables at that coordinate. Use h h h B B B j ( e , v ) ∈ [ h ] j(e,v) \in[h] j ( e , v ) ∈ [ h ] ( e , v ) (e,v) ( e , v ) X v X_v X v u ( e , v ) u(e,v) u ( e , v ) ζ ⃗ ( j ) ( X ⃗ ) \vec{\zeta}^{(j)}(\vec{X}) ζ ( j ) ( X ) j j j X v X_v X v
Apply the auxiliary row to the copy containing the occurrence:
⟨ P ⃗ e , v ( T ) , X ⃗ ⟩ = ⟨ P ⃗ u ( e , v ) ( T ) , ζ ⃗ ( j ( e , v ) ) ( X ⃗ ) ⟩ . (20) \langle\vec{P}_{e,v}(T),\vec{X}\rangle=\langle\vec{P}_{u(e,v)}(T),\vec{\zeta}^{(j(e,v))}(\vec{X})\rangle. \tag*{(20)} ⟨ P e , v ( T ) , X ⟩ = ⟨ P u ( e , v ) ( T ) , ζ ( j ( e , v )) ( X )⟩ . (20) At zero the right side reads the coordinate containing X v X_v X v R − 1 R-1 R − 1 t t t v ′ v' v ′ a v ′ ( t ) = d e ( t ) a_{v'}(t)=d_e(t) a v ′ ( t ) = d e ( t ) S t , d e ( t ) S_{t,d_e(t)} S t , d e ( t ) e e e
The remaining dependence is through the h c hc h c C ζ ⃗ ( j ) ( X ⃗ ) C\vec{\zeta}^{(j)}(\vec{X}) C ζ ( j ) ( X ) h h h x 1 + x 2 x_1+x_2 x 1 + x 2 x 1 x_1 x 1 x 2 x_2 x 2 ℓ ( X ⃗ ) \ell(\vec{X}) ℓ ( X )
ℓ ( X ⃗ ) = ∑ g = 1 r ℓ ∣ S t , g ( X ⃗ ∣ S t , g ) . \ell(\vec{X})=\sum_{g=1}^{r}\ell|_{S_{t,g}}(\vec{X}|_{S_{t,g}}). ℓ ( X ) = g = 1 ∑ r ℓ ∣ S t , g ( X ∣ S t , g ) . and collect the coefficient rows of these restrictions into C t , g C_{t,g} C t , g g g g S t , g S_{t,g} S t , g (7) .
There are r h c rhc r h c h D ≤ 2 M ≤ 6 n hD\le2M\le6n h D ≤ 2 M ≤ 6 n
∑ g = 1 r c t , g = r h c ≤ r h D τ 3 r ≤ 2 τ n . \sum_{g=1}^{r}c_{t,g}=rhc\le rh\frac{D\tau}{3r}\le2\tau n. g = 1 ∑ r c t , g = r h c ≤ r h 3 r D τ ≤ 2 τ n . Each row of C C C D D D { 0 , 1 , − 1 } \{0,1,-1\} { 0 , 1 , − 1 } C t , g C_{t,g} C t , g D D D ( r , τ , D ) (r,\tau,D) ( r , τ , D )
Step 5: Verify the promised parameter bounds.
The construction needs the two balancing inequalities (19) and the tail bound 2 e − R 2 / ( 8 Q ) ≤ τ / ( 3 r ) 2e^{-R^2/(8Q)}\le\tau/(3r) 2 e − R 2 / ( 8 Q ) ≤ τ / ( 3 r ) (12) and verify these requirements.
Put λ = ⌈ log 2 n ⌉ \lambda=\lceil\log_2 n\rceil λ = ⌈ log 2 n ⌉ ν = ⌈ log 2 λ ⌉ \nu=\lceil\log_2\lambda\rceil ν = ⌈ log 2 λ ⌉
r = ⌈ λ ν 2 ⌉ , Q = ⌈ λ 128 ν ⌉ , R = 4 ⌈ λ ⌉ , τ = 1 r λ , D = r 2 Q . (21) \begin{aligned}
r&=\left\lceil\frac{\sqrt{\lambda}}{\nu^2}\right\rceil,\qquad Q=\left\lceil\frac{\lambda}{128\nu}\right\rceil,\qquad R=4\lceil\sqrt{\lambda}\rceil,\\
\tau&=\frac{1}{r\lambda},\qquad D=r^2Q. \tag*{(21)}
\end{aligned} r τ = ⌈ ν 2 λ ⌉ , Q = ⌈ 128 ν λ ⌉ , R = 4 ⌈ λ ⌉ , = r λ 1 , D = r 2 Q . (21) These have the orders stated in (12) . For sufficiently large n n n r ≥ 2 r\ge2 r ≥ 2 R ≥ 32 R\ge32 R ≥ 32 6 R ≤ Q 6R\le Q 6 R ≤ Q R / Q = O ( ν / λ ) → 0 R/Q=O(\nu/\sqrt{\lambda})\to0 R / Q = O ( ν / λ ) → 0
A 0 = ⌈ Q p − R − 1 2 ⌉ , I = { A 0 , … , A 0 + R − 1 } . A_0=\left\lceil Qp-\frac{R-1}{2}\right\rceil,\qquad\mathcal{I}=\{A_0,\ldots,A_0+R-1\}. A 0 = ⌈ Qp − 2 R − 1 ⌉ , I = { A 0 , … , A 0 + R − 1 } . Since Q / 4 ≤ Q p ≤ Q / 2 Q/4 \le Qp \le Q/2 Q /4 ≤ Qp ≤ Q /2 6 R ≤ Q 6R \le Q 6 R ≤ Q { 0 , … , Q } \{0,\ldots,Q\} { 0 , … , Q } R ≥ 32 R \ge32 R ≥ 32 R / 4 R/4 R /4 Q p Qp Qp Pr [ W ∉ I ] ≤ 2 e − R 2 / ( 8 Q ) \Pr[W \notin I] \le2e^{-R^2/(8Q)} Pr [ W ∈ / I ] ≤ 2 e − R 2 / ( 8 Q )
log 2 D = 2 Q log 2 r ≤ λ / 128 , D ≤ ( 2 n ) 1 / 128 . \log_2 D = 2Q\log_2 r \le\lambda/128,\qquad D \le(2n)^{1/128}. log 2 D = 2 Q log 2 r ≤ λ /128 , D ≤ ( 2 n ) 1/128 . For sufficiently large n n n
12 D 2 ≤ 12 ( 2 n ) 1 / 64 ≤ n ≤ M , 2 Q r 2 = O ( λ 2 ) ≤ n , 12D^2 \le12(2n)^{1/64} \le n \le M,\qquad2Qr^2 = O(\lambda^2) \le n, 12 D 2 ≤ 12 ( 2 n ) 1/64 ≤ n ≤ M , 2 Q r 2 = O ( λ 2 ) ≤ n , as required for Step 4.
2 e − R 2 / ( 8 Q ) ≤ 2 e − 256 ν ≤ 1 3 λ 2 ≤ 1 3 r 2 λ = τ 3 r . 2e^{-R^2/(8Q)} \le2e^{-256\nu} \le\frac{1}{3\lambda^2} \le\frac{1}{3r^2\lambda} = \frac{\tau}{3r}. 2 e − R 2 / ( 8 Q ) ≤ 2 e − 256 ν ≤ 3 λ 2 1 ≤ 3 r 2 λ 1 = 3 r τ . Thus Step 2 supplies the auxiliary measurement bound used in Step 4.
Finally, all construction dimensions are polynomially bounded in n n n h D ≤ 6 n hD \le6n h D ≤ 6 n R − 1 ≤ R R-1 \le R R − 1 ≤ R ( r , τ , D ) (r,\tau,D) ( r , τ , D ) t ∈ [ Q ] t \in[Q] t ∈ [ Q ] § 2.6 ; Lemma 4.8 records the parameter check for the more general setting of the full proof.
A weaker algorithm for general #k-SAT
The algorithm above already gives a subexponential algorithm for general #k k k
Fix k ≥ 2 k \ge2 k ≥ 2 [28 Appendix A] of sparsification [39 ] , with quantitative bounds from [19 Section 6] and [18 Lemma 5.14 and equation (5.4)] . As recorded in Lemma 6.3, this expresses the input count as the sum of the counts of at most poly k ( n , σ − 1 ) 2 σ n \operatorname{poly}_k(n,\sigma^{-1})2^{\sigma n} poly k ( n , σ − 1 ) 2 σ n G G G G G G n n n
B = O k ( ( σ − 1 log ( 2 / σ ) ) k − 1 ) B = O_k\left((\sigma^{-1}\log(2/\sigma))^{k-1}\right) B = O k ( ( σ − 1 log ( 2/ σ ) ) k − 1 ) occurrences per variable, and hence at most B n Bn B n
First turn G G G z = u ∨ v z=u\lor v z = u ∨ v
( ¬ u ∨ z ) ∧ ( ¬ v ∨ z ) ∧ ( u ∨ v ∨ ¬ z ) . (\neg u \lor z) \land(\neg v \lor z) \land(u \lor v \lor\neg z). ( ¬ u ∨ z ) ∧ ( ¬ v ∨ z ) ∧ ( u ∨ v ∨ ¬ z ) . Here u , v u,v u , v z z z G G G G G G O ( k ) O(k) O ( k )
Next reduce the occurrences. If a variable x x x s ≥ 2 s \ge2 s ≥ 2 x 1 , … , x s x_1,\ldots,x_s x 1 , … , x s [64 p. 86] :
( ¬ x 1 ∨ x 2 ) ∧ ⋯ ∧ ( ¬ x s − 1 ∨ x s ) ∧ ( ¬ x s ∨ x 1 ) . (\neg x_1 \lor x_2) \land\cdots\land(\neg x_{s-1} \lor x_s) \land(\neg x_s \lor x_1). ( ¬ x 1 ∨ x 2 ) ∧ ⋯ ∧ ( ¬ x s − 1 ∨ x s ) ∧ ( ¬ x s ∨ x 1 ) . The cycle forces all copies to agree. Each copy appears in one clause inherited from the 3-CNF and two cycle clauses, hence in exactly three clauses. Variables with at most one occurrence are kept unchanged, including unused declared variables. This standard variable-copy reduction [Gol08, Proposition 4.7] therefore also gives a bijection between satisfying assignments. The final formula G ^ \widehat{G} G # G ^ = # G \#\widehat{G}=\#G # G = # G
n ≤ N : = # { variables of G ^ } = O k ( B n ) . n \le N := \#\{\text{variables of }\widehat{G}\}=O_k(Bn). n ≤ N := # { variables of G } = O k ( B n ) . All its clauses have size at most three, including the two-literal cycle clauses, and every variable occurs at most three times. Thus it is exactly an instance of the setting of this overview.
Apply (14) to each G ^ \widehat{G} G σ − 1 \sigma^{-1} σ − 1 log n \log n log n B B B
( n + m ) O k ( 1 ) 2 O k ( σ n + B n ( log log n ) 2 / log n ) . (n+m)^{O_k(1)}2^{O_k(\sigma n+Bn(\log\log n)^2/\sqrt{\log n})}. ( n + m ) O k ( 1 ) 2 O k ( σ n + B n ( l o g l o g n ) 2 / l o g n ) . For example, choosing σ \sigma σ ( log n ) − 1 / ( 2 k ) (\log n)^{-1/(2k)} ( log n ) − 1/ ( 2 k )
( n + m ) O k ( 1 ) 2 O k ( n ( log log n ) k + 1 / ( log n ) 1 / ( 2 k ) ) = ( n + m ) O k ( 1 ) 2 o ( n ) (n+m)^{O_k(1)}2^{O_k(n(\log\log n)^{k+1}/(\log n)^{1/(2k)})}=(n+m)^{O_k(1)}2^{o(n)} ( n + m ) O k ( 1 ) 2 O k ( n ( l o g l o g n ) k + 1 / ( l o g n ) 1/ ( 2 k ) ) = ( n + m ) O k ( 1 ) 2 o ( n ) with polynomial space, processing the sparsification tree and the reduced instances one at a time. The parameter σ \sigma σ
A faster algorithm for # k \#k # k F 2 \mathbb{F}_2 F 2
The running-time exponent in § 2.8 is O k ( σ n + B ρ ( n ) ) O_k(\sigma n+B\rho(n)) O k ( σ n + B ρ ( n ))
ρ ( n ) = n ( log log n ) 2 log n . \rho(n)=\frac{n(\log\log n)^2}{\sqrt{\log n}}. ρ ( n ) = log n n ( log log n ) 2 . If sparsification gave B = O k ( 1 ) B=O_k(1) B = O k ( 1 ) σ = Θ ( ρ ( n ) / n ) \sigma=\Theta(\rho(n)/n) σ = Θ ( ρ ( n ) / n ) 2 O k ( ρ ( n ) ) 2^{O_k(\rho(n))} 2 O k ( ρ ( n )) # k \#k # k O k ( n ) O_k(n) O k ( n ) σ \sigma σ B = ( log n ) O k ( 1 ) B=(\log n)^{O_k(1)} B = ( log n ) O k ( 1 )
We obtain the 2 O k ( ρ ( n ) ) 2^{O_k(\rho(n))} 2 O k ( ρ ( n )) O ( n ) O(n) O ( n ) F 2 \mathbb{F}_2 F 2 k k k Theorem 2.2 ). Second, even if a leaf G G G B n Bn B n n n n G G G
A randomized reduction to sparse polynomial equations. Write v e ( x ) = 1 v_e(x)=1 v e ( x ) = 1 x x x e e e 0 0 0 parity test on a set S S S ∑ e ∈ S v e ( x ) = 0 \sum_{e\in S}v_e(x)=0 ∑ e ∈ S v e ( x ) = 0 F 2 \mathbb{F}_2 F 2 x x x S S S G G G m G ≤ n m_G\le n m G ≤ n m G > n m_G>n m G > n
A short test can easily miss a single violated clause, but all assignments that violate only that clause give identical answers. Thus there are only m G m_G m G 8 n 8n 8 n L = 2 B ⌈ log 2 ( 2 m G ) ⌉ L=2B\lceil\log_2(2m_G)\rceil L = 2 B ⌈ log 2 ( 2 m G )⌉ 1 / 2 1/2 1/2 m G ≤ B n m_G\le Bn m G ≤ B n w ≥ 1 w\ge1 w ≥ 1 u = L w / m G u=Lw/m_G u = L w / m G 1 / 2 1/2 1/2
p w = 1 2 ( 1 − ( 1 − w m G ) L ) ≥ 1 2 ( 1 − e − u ) . p_w=\frac{1}{2}\left(1-\left(1-\frac{w}{m_G}\right)^L\right)\ge\frac{1}{2}(1-e^{-u}). p w = 2 1 ( 1 − ( 1 − m G w ) L ) ≥ 2 1 ( 1 − e − u ) . We count sets for few violations and assignments for many:
If u ≤ 1 u\le1 u ≤ 1 p w ≥ u / 4 p_w\ge u/4 p w ≥ u /4 m G w m_G^w m G w w w w m G w e − 8 n p w ≤ m G w e − 2 n L w / m G ≤ 16 − w m_G^w e^{-8np_w}\le m_G^w e^{-2nLw/m_G}\le16^{-w} m G w e − 8 n p w ≤ m G w e − 2 n L w / m G ≤ 1 6 − w w ≥ 1 w\ge1 w ≥ 1 1 / 15 1/15 1/15
If u > 1 u>1 u > 1 p w > 1 / 4 p_w>1/4 p w > 1/4 2 n 2^n 2 n 2 n ( 3 / 4 ) 8 n < 1 / 4 2^n(3/4)^{8n}<1/4 2 n ( 3/4 ) 8 n < 1/4
Thus with probability greater than 2 / 3 2/3 2/3 G G G v e v_e v e k k k x j x_j x j 1 + x j 1+x_j 1 + x j k k k 2 k L 2^kL 2 k L n n n § 6 constructs short tests with the same solution set deterministically. We first use parity checks from an expander code to ensure that any nonsatisfying assignment still passing must violate many clauses. A random short test then catches each such assignment with probability at least 1 / 4 1/4 1/4
Counting the common zeros. Recall from § 2.2 that Φ e ( x ) \Phi_e(x) Φ e ( x ) x x x e e e S i S_i S i this test are violated, using ordinary integer addition:
L i ( x ) = ∑ e ∈ S i v e ( x ) = ∑ e ∈ S i ( 1 − Φ e ( x ) ) . L_i(x)=\sum_{e\in S_i}v_e(x)=\sum_{e\in S_i}(1-\Phi_e(x)). L i ( x ) = e ∈ S i ∑ v e ( x ) = e ∈ S i ∑ ( 1 − Φ e ( x )) . The test passes exactly when L i ( x ) L_i(x) L i ( x ) ℓ = n + 1 \ell=n+1 ℓ = n + 1 2 n 2^n 2 n 2 ℓ 2^\ell 2 ℓ p ℓ p_\ell p ℓ 2 ℓ 2^\ell 2 ℓ S i S_i S i p ℓ ( 2 ) ≡ 1 ( m o d 2 ℓ ) p_\ell(2)\equiv1\pmod{2^\ell} p ℓ ( 2 ) ≡ 1 ( mod 2 ℓ ) m t e s t m_{\mathrm{test}} m test
∑ x ∏ e ∈ G Φ e ( x ) ⟶ ∑ x ∏ i = 1 m t e s t p ℓ ( L i ( x ) ) . \sum_{x}\prod_{e\in G}\Phi_e(x)\longrightarrow\sum_x\prod_{i=1}^{m_{\mathrm{test}}}p_\ell(L_i(x)). x ∑ e ∈ G ∏ Φ e ( x ) ⟶ x ∑ i = 1 ∏ m test p ℓ ( L i ( x )) . The left side is # G \#G # G 2 ℓ 2^\ell 2 ℓ # G \#G # G P i ( x ) = 0 P_i(x)=0 P i ( x ) = 0 F 2 \mathbb{F}_2 F 2 L i ( x ) L_i(x) L i ( x )
To reuse the Taylor step, we need a product of input degree O ( n ) O(n) O ( n ) m t e s t m_{\mathrm{test}} m test O ( k m t e s t ℓ ) O(km_{\mathrm{test}}\ell) O ( k m test ℓ ) n n n across all tests together . § 5.1 writes p ℓ ( y ) p_\ell(y) p ℓ ( y ) e j ( y ) h ( y ) j e_j(y)h(y)^j e j ( y ) h ( y ) j e j e_j e j h ( y ) = y ( y − 1 ) h(y)=y(y-1) h ( y ) = y ( y − 1 ) ℓ \ell ℓ h h h 2 ℓ 2^\ell 2 ℓ ℓ \ell ℓ m t e s t m_{\mathrm{test}} m test ℓ − 1 \ell-1 ℓ − 1 L i L_i L i deg L i ≤ k \deg L_i\leq k deg L i ≤ k k ( m t e s t + 2 ( ℓ − 1 ) ) = O ( n ) k(m_{\mathrm{test}}+2(\ell-1))=O(n) k ( m test + 2 ( ℓ − 1 )) = O ( n )
For the replacement forms, what matters is which variables each factor reads. Where § 2.3 used one form for each variable in a clause, we now use one for each variable in a test, reusing it throughout that test. Each sparse equation reads only polylogarithmically many variables. Branching on the values of O ( n / log n ) O(n/\sqrt{\log n}) O ( n / log n ) § 4 . On each branch, fixing the shared measurements again lets us sum one group at a time. Because the truncated product has input degree O ( n ) O(n) O ( n ) R R R deg T H = O ( n R ) \deg_T H=O(nR) deg T H = O ( n R ) O ( 1 + n R / Q ) O(1+nR/Q) O ( 1 + n R / Q ) T = 0 T=0 T = 0 0 , 1 , … , ℓ − 1 0,1,\ldots,\ell-1 0 , 1 , … , ℓ − 1 h h h 2 ℓ 2^\ell 2 ℓ
Theorem 2.2 (Counting common zeros of sparse polynomials). Fix integers k ≥ 2 k\geq2 k ≥ 2 c ≥ 1 c\geq1 c ≥ 1 P 1 , … , P m t e s t P_1,\ldots,P_{m_{\mathrm{test}}} P 1 , … , P m test F 2 \mathbb{F}_2 F 2 n ≥ 20 n\geq20 n ≥ 20 m t e s t ≤ c n m_{\mathrm{test}}\leq cn m test ≤ c n P i P_i P i
as a list of at most ( log n ) c (\log n)^c ( log n ) c k k k { 0 , 1 } n \{0,1\}^n { 0 , 1 } n 2 O k , c ( n ( log log n ) 2 / log n ) 2^{O_{k,c}(n(\log\log n)^2/\sqrt{\log n})} 2 O k , c ( n ( l o g l o g n ) 2 / l o g n ) n O k , c ( 1 ) n^{O_{k,c}(1)} n O k , c ( 1 )
Sections 3 and 4 supply the counting routine and the replacement forms; Section 5 uses them to prove the theorem directly for polynomial equations.
For the final system constructed for G G G # G \#G # G 2 ( log n ) O k ( 1 ) 2^{(\log n)^{O_k(1)}} 2 ( l o g n ) O k ( 1 ) σ \sigma σ ( log log n ) 2 / log n (\log\log n)^2/\sqrt{\log n} ( log log n ) 2 / log n B B B 2 O k ( n ( log log n ) 2 / log n ) 2^{O_k(n(\log\log n)^2/\sqrt{\log n})} 2 O k ( n ( l o g l o g n ) 2 / l o g n )
A fast counting algorithm via shared linear summaries
Sections 3–5 prove Theorem 2.2. We begin with the sum from Section 2.2: for each Boolean assignment, multiply the factors, then add the products over all assignments. Here the factors can be any rational polynomials. We show how to compute the sum when the replacement forms and measurements are given. Section 4 constructs them using only the variables read by each factor. Section 5 supplies one factor per sparse polynomial equation.
Polynomial factors and the value to recover
Let F \mathcal{F} F I f ⊆ [ n ] I_f \subseteq[n] I f ⊆ [ n ] Φ f \Phi_f Φ f
Z = ∑ x ∈ { 0 , 1 } n ∏ f ∈ F Φ f ( ( x v ) v ∈ I f ) . (22) Z = \sum_{x\in\{0,1\}^n}\prod_{f\in\mathcal{F}}\Phi_f((x_v)_{v\in I_f}). \tag*{(22)} Z = x ∈ { 0 , 1 } n ∑ f ∈ F ∏ Φ f (( x v ) v ∈ I f ) . (22) For clause satisfaction indicators, Z Z Z Q \mathbb{Q} Q A \mathcal{A} A A \mathcal{A} A A = Q \mathcal{A}=\mathbb{Q} A = Q
Different factors may use different replacements for the same variable. Give each incidence ( f , v ) (f,v) ( f , v ) Y f , v Y_{f,v} Y f , v
deg { Y f , v } ( ∏ f ∈ F Φ f ( ( Y f , v ) v ∈ I f ) ) ≤ d . (23) \deg_{\{Y_{f,v}\}}\left(\prod_{f\in\mathcal{F}}\Phi_f((Y_{f,v})_{v\in I_f})\right)\le d. \tag*{(23)} deg { Y f , v } f ∈ F ∏ Φ f (( Y f , v ) v ∈ I f ) ≤ d . (23) This is the degree of the whole product computed over A \mathcal{A} A P ⃗ f , v ( T ) \vec{P}_{f,v}(T) P f , v ( T ) R R R P ⃗ f , v ( 0 ) = δ v \vec{P}_{f,v}(0)=\delta_v P f , v ( 0 ) = δ v
H ( T ) = ∑ x ∈ { 0 , 1 } n ∏ f ∈ F Φ f ( ( ⟨ P ⃗ f , v ( T ) , x ⟩ ) v ∈ I f ) ∈ A [ T ] . (24) H(T)=\sum_{x\in\{0,1\}^n}\prod_{f\in\mathcal{F}}\Phi_f((\langle\vec{P}_{f,v}(T),x\rangle)_{v\in I_f})\in\mathcal{A}[T]. \tag*{(24)} H ( T ) = x ∈ { 0 , 1 } n ∑ f ∈ F ∏ Φ f ((⟨ P f , v ( T ) , x ⟩ ) v ∈ I f ) ∈ A [ T ] . (24) The replacements restore the inputs at zero and multiply the degree by at most R R R H ( 0 ) = Z H(0)=Z H ( 0 ) = Z deg T H ≤ d H : = d R \deg_T H\le d_H:=dR deg T H ≤ d H := d R d = O ( n ) d=O(n) d = O ( n )
The required rows, partitions, and measurements
We use the same design as in Section 2.4, with a group assignment for each factor in place of each clause. The final condition below records the coefficient bound needed for exact arithmetic. Here ξ ~ \widetilde{\xi} ξ n n n
Definition 3.1. For the factor scopes ( I f ) f ∈ F (I_f)_{f\in\mathcal{F}} ( I f ) f ∈ F r ≥ 2 r\ge2 r ≥ 2 D ≥ 1 D\ge1 D ≥ 1 1 ≤ R < Q 1\le R<Q 1 ≤ R < Q 0 < τ ≤ 1 / 2 0<\tau\le1/2 0 < τ ≤ 1/2
For each t ∈ [ Q ] t\in[Q] t ∈ [ Q ] [ n ] = S t , 1 ∪ ⋯ ∪ S t , r [n]=S_{t,1}\cup\cdots\cup S_{t,r} [ n ] = S t , 1 ∪ ⋯ ∪ S t , r ∣ S t , g ∣ ≤ 2 n / r |S_{t,g}|\le2n/r ∣ S t , g ∣ ≤ 2 n / r
For every factor f ∈ F f\in\mathcal{F} f ∈ F σ f ∈ [ r ] Q \sigma_f\in[r]^Q σ f ∈ [ r ] Q σ f ( t ) \sigma_f(t) σ f ( t ) f f f t t t
(3) For every t ∈ [ Q ] t \in[Q] t ∈ [ Q ] g ∈ [ r ] g \in[r] g ∈ [ r ] C t , g ∈ Z [ c t , g ] × S t , g C_{t,g} \in\mathbb{Z}^{[c_{t,g}] \times S_{t,g}} C t , g ∈ Z [ c t , g ] × S t , g c t , g c_{t,g} c t , g D D D
∑ g = 1 r c t , g ≤ 2 τ n . \sum_{g=1}^{r} c_{t,g} \le2\tau n. g = 1 ∑ r c t , g ≤ 2 τ n . For an assignment x x x C t , g ( x ∣ S t , g ) C_{t,g}(x|_{S_{t,g}}) C t , g ( x ∣ S t , g ) g g g
(4) For every f ∈ F f \in\mathcal{F} f ∈ F v ∈ I f v \in I_f v ∈ I f P ⃗ f , v ( T ) ∈ Q [ T ] 1 × n \vec{P}_{f,v}(T) \in\mathbb{Q}[T]^{1 \times n} P f , v ( T ) ∈ Q [ T ] 1 × n P ⃗ f , v ( 0 ) = δ v \vec{P}_{f,v}(0) = \delta_v P f , v ( 0 ) = δ v deg P ⃗ f , v ≤ R \deg\vec{P}_{f,v} \le R deg P f , v ≤ R t ∈ [ Q ] t \in[Q] t ∈ [ Q ] g ∈ [ r ] ∖ { σ f ( t ) } g \in[r] \setminus\{\sigma_f(t)\} g ∈ [ r ] ∖ { σ f ( t )} ξ ~ ∈ Q S t , g \tilde{\xi} \in\mathbb{Q}^{S_{t,g}} ξ ~ ∈ Q S t , g
C t , g ξ ~ = 0 ⃗ ⟹ ⟨ P ⃗ f , v ( t ) , ξ ~ ⟩ = 0. C_{t,g}\tilde{\xi} = \vec{0} \quad\Longrightarrow\quad\langle\vec{P}_{f,v}(t),\tilde{\xi}\rangle= 0. C t , g ξ ~ = 0 ⟹ ⟨ P f , v ( t ) , ξ ~ ⟩ = 0. where ξ ~ ∈ Q n \tilde{\xi} \in\mathbb{Q}^n ξ ~ ∈ Q n ξ ~ \tilde{\xi} ξ ~
(5) With d d e n = ( R − 1 ) ! Q ! d_{\mathrm{den}} = (R-1)!Q! d den = ( R − 1 )! Q ! d d e n P ⃗ f , v d_{\mathrm{den}}\vec{P}_{f,v} d den P f , v O ( Q log Q + log D ) O(Q\log Q+\log D) O ( Q log Q + log D )
To interpret the kernel condition, fix t t t f f f v ∈ I f v \in I_f v ∈ I f g = σ f ( t ) g=\sigma_f(t) g = σ f ( t ) x h = x ∣ S t , h x_h=x|_{S_{t,h}} x h = x ∣ S t , h x , x ′ ∈ Q n x,x' \in\mathbb{Q}^n x , x ′ ∈ Q n S t , h S_{t,h} S t , h h ≠ g h\ne g h = g C t , h x h = C t , h x h ′ C_{t,h}x_h=C_{t,h}x'_h C t , h x h = C t , h x h ′ x h ′ − x h ∈ ker C t , h x'_h-x_h\in\ker C_{t,h} x h ′ − x h ∈ ker C t , h
⟨ P ⃗ f , v ( t ) , x ′ ⟩ = ⟨ P ⃗ f , v ( t ) , x ⟩ . \langle\vec{P}_{f,v}(t),x'\rangle=\langle\vec{P}_{f,v}(t),x\rangle. ⟨ P f , v ( t ) , x ′ ⟩ = ⟨ P f , v ( t ) , x ⟩ . Applying this one group at a time shows that the tuple ( ⟨ P ⃗ f , v ( t ) , x ⟩ ) v ∈ I f (\langle\vec{P}_{f,v}(t),x\rangle)_{v\in I_f} (⟨ P f , v ( t ) , x ⟩ ) v ∈ I f x g x_g x g ( C t , h x h ) h ≠ g (C_{t,h}x_h)_{h\ne g} ( C t , h x h ) h = g x g x_g x g Φ f \Phi_f Φ f A A A
Each measurement of a Boolean vector lies in [ − D , D ] ∩ Z [-D,D]\cap\mathbb{Z} [ − D , D ] ∩ Z C t , g C_{t,g} C t , g { 0 , 1 , − 1 } \{0,1,-1\} { 0 , 1 , − 1 } ( 1 , 2 , 4 , … , 2 n − 1 ) (1,2,4,\ldots,2^{n-1}) ( 1 , 2 , 4 , … , 2 n − 1 ) 2 n 2^n 2 n
Separating the sum by fixing measurements
Fixing the measurements makes every factor depend on only its assigned group. The following lemma states the resulting calculation over a commutative ring, so it applies equally to rational-valued factors and to factors in A A A x i = 1 x_i=1 x i = 1
Lemma 3.2. Let [ n ] = S 1 ⊔ ⋯ ⊔ S r [n]=S_1\sqcup\cdots\sqcup S_r [ n ] = S 1 ⊔ ⋯ ⊔ S r be a partition, allowing empty parts. For each g ∈ [ r ] g\in[r] g ∈ [ r ] let C g C_g C g be an integer matrix with c g c_g c g rows and columns indexed by S g S_g S g Suppose each row has sum of absolute values at most an integer D ≥ 0 D\ge0 D ≥ 0 and write c = ∑ g c g c=\sum_g c_g c = ∑ g c g Let A A A be a commutative ring with identity, and let A g : Q n → A A_g:\mathbb{Q}^n\to A A g : Q n → A satisfy
A g ( x + ξ ) = A g ( x ) ( x ∈ Q n , ξ ∈ ker C h , h ≠ g ) , A_g(x+\xi)=A_g(x)\qquad(x\in\mathbb{Q}^n,\ \xi\in\ker C_h,\ h\ne g), A g ( x + ξ ) = A g ( x ) ( x ∈ Q n , ξ ∈ ker C h , h = g ) , where ξ \xi ξ is extended by zero outside S h S_h S h For any E ⊆ [ n ] E\subseteq[n] E ⊆ [ n ] the sum
Z E = ∑ x ∈ { 0 , 1 } n x i = 1 ( i ∈ E ) ∏ g = 1 r A g ( x ) Z_E=\sum_{\substack{x\in\{0,1\}^n\\x_i=1\ (i\in E)}}\prod_{g=1}^{r}A_g(x) Z E = x ∈ { 0 , 1 } n x i = 1 ( i ∈ E ) ∑ g = 1 ∏ r A g ( x ) can be evaluated by examining at most
2 ( 2 D + 1 ) c ∑ g = 1 r 2 ∣ S g ∣ 2(2D+1)^c\sum_{g=1}^{r}2^{|S_g|} 2 ( 2 D + 1 ) c g = 1 ∑ r 2 ∣ S g ∣ Boolean vectors, with at most one A g A_g A g evaluation, poly ( n , r , c , log ( D + 2 ) ) \operatorname{poly}(n,r,c,\log(D+2)) poly ( n , r , c , log ( D + 2 )) additional bit operations, and O ( 1 ) O(1) O ( 1 ) ring operations per vector. Apart from the evaluators’ own storage and O ( 1 ) O(1) O ( 1 ) ring elements, the procedure uses poly ( n , r , c , log ( D + 2 ) ) \operatorname{poly}(n,r,c,\log(D+2)) poly ( n , r , c , log ( D + 2 )) bits of space.
Proof. For y g ∈ Z c g y_g \in\mathbb{Z}^{c_g} y g ∈ Z c g
X g ( y g ) = { u ∈ { 0 , 1 } S g : C g u = y g } , Y = ∏ g = 1 r { − D , … , D } c g . \mathcal{X}_g(y_g) = \{u \in\{0,1\}^{S_g}: C_g u = y_g\}, \qquad\mathcal{Y} = \prod_{g=1}^{r} \{-D,\ldots,D\}^{c_g}. X g ( y g ) = { u ∈ { 0 , 1 } S g : C g u = y g } , Y = g = 1 ∏ r { − D , … , D } c g . Every tuple of measurements of a Boolean assignment belongs to Y \mathcal{Y} Y ∣ Y ∣ = ( 2 D + 1 ) c |\mathcal{Y}| = (2D+1)^c ∣ Y ∣ = ( 2 D + 1 ) c y ⃗ = ( y 1 , … , y r ) \vec{y} = (y_1,\ldots,y_r) y = ( y 1 , … , y r ) X g ( y g ) \mathcal{X}_g(y_g) X g ( y g ) x ˉ g ∈ X g ( y g ) \bar{x}_g \in\mathcal{X}_g(y_g) x ˉ g ∈ X g ( y g ) g g g
X g , E ( y g ) = { u ∈ X g ( y g ) : u i = 1 for all i ∈ E ∩ S g } . \mathcal{X}_{g,E}(y_g) = \{u \in\mathcal{X}_g(y_g): u_i = 1 \text{ for all } i \in E \cap S_g\}. X g , E ( y g ) = { u ∈ X g ( y g ) : u i = 1 for all i ∈ E ∩ S g } . If x h ∈ X h ( y h ) x_h \in\mathcal{X}_h(y_h) x h ∈ X h ( y h ) h h h x h − x ˉ h ∈ ker C h x_h-\bar{x}_h \in\ker C_h x h − x ˉ h ∈ ker C h h ≠ g h \ne g h = g
A g ( ∑ h = 1 r x h ) = A g ( x g + ∑ h ≠ g x ˉ h ) . A_g\left(\sum_{h=1}^{r}x_h\right) = A_g\left(x_g+\sum_{h\ne g}\bar{x}_h\right). A g ( h = 1 ∑ r x h ) = A g x g + h = g ∑ x ˉ h . The assignments with measurement tuple y ⃗ \vec{y} y E E E ∏ g X g , E ( y g ) \prod_g \mathcal{X}_{g,E}(y_g) ∏ g X g , E ( y g )
Σ g ( y ⃗ ) = ∑ x g ∈ X g , E ( y g ) A g ( x g + ∑ h ≠ g x ˉ h ) . (25) \Sigma_g(\vec{y}) = \sum_{x_g \in\mathcal{X}_{g,E}(y_g)} A_g\left(x_g+\sum_{h\ne g}\bar{x}_h\right). \tag*{(25)} Σ g ( y ) = x g ∈ X g , E ( y g ) ∑ A g x g + h = g ∑ x ˉ h . (25) Distributivity shows that these assignments contribute ∏ g Σ g ( y ⃗ ) \prod_g \Sigma_g(\vec{y}) ∏ g Σ g ( y ) Z E Z_E Z E X g , E ( y g ) \mathcal{X}_{g,E}(y_g) X g , E ( y g ) Σ g ( y ⃗ ) = 0 \Sigma_g(\vec{y})=0 Σ g ( y ) = 0 x ˉ g \bar{x}_g x ˉ g E E E
Z E = ∑ y ⃗ ∈ Y ∏ g = 1 r Σ g ( y ⃗ ) , Z_E = \sum_{\vec{y}\in\mathcal{Y}}\prod_{g=1}^{r}\Sigma_g(\vec{y}), Z E = y ∈ Y ∑ g = 1 ∏ r Σ g ( y ) , where the product is taken to be zero if any X g ( y g ) \mathcal{X}_g(y_g) X g ( y g )
The algorithm evaluates this identity one measurement tuple at a time, starting with Z = 0 A Z=0_A Z = 0 A y ⃗ ∈ Y \vec{y}\in\mathcal{Y} y ∈ Y { 0 , 1 } S g \{0,1\}^{S_g} { 0 , 1 } S g x ˉ g ∈ X g ( y g ) \bar{x}_g\in\mathcal{X}_g(y_g) x ˉ g ∈ X g ( y g ) p = 1 A p=1_A p = 1 A Σ g ( y ⃗ ) \Sigma_g(\vec{y}) Σ g ( y ) s = 0 A s=0_A s = 0 A u ∈ { 0 , 1 } S g u\in\{0,1\}^{S_g} u ∈ { 0 , 1 } S g
C g u = y g , u i = 1 ( i ∈ E ∩ S g ) , C_g u = y_g,\qquad u_i=1\quad(i\in E\cap S_g), C g u = y g , u i = 1 ( i ∈ E ∩ S g ) , the algorithm performs the update
s ← s + A g ( u + ∑ h ≠ g x ˉ h ) . s \leftarrow s+A_g\left(u+\sum_{h\ne g}\bar{x}_h\right). s ← s + A g u + h = g ∑ x ˉ h . At the end of this scan, s = Σ g ( y ⃗ ) s=\Sigma_g(\vec{y}) s = Σ g ( y ) p ← p s p\leftarrow ps p ← p s Z ← Z + p Z\leftarrow Z+p Z ← Z + p y ⃗ \vec{y} y Z = Z E Z=Z_E Z = Z E
For each y ⃗ \vec{y} y Σ g ( y ⃗ ) \Sigma_g(\vec{y}) Σ g ( y ) ∑ g 2 ∣ S g ∣ \sum_g 2^{|S_g|} ∑ g 2 ∣ S g ∣ poly ( n , r , c , log ( D + 2 ) ) \operatorname{poly}(n,r,c,\log(D+2)) poly ( n , r , c , log ( D + 2 )) D D D O ( log ( D + 2 ) ) O(\log(D+2)) O ( log ( D + 2 )) y ⃗ \vec{y} y ( x ˉ 1 , … , x ˉ r ) (\bar{x}_1,\ldots,\bar{x}_r) ( x ˉ 1 , … , x ˉ r ) g g g u u u s s s p p p Z Z Z y ⃗ \vec{y} y c c c [ − D , D ] [-D,D] [ − D , D ] n n n s , p , Z ∈ A s,p,Z\in A s , p , Z ∈ A
To evaluate (24) at a fixed t t t σ f ( t ) \sigma_f(t) σ f ( t ) A A A E = ∅ E=\varnothing E = ∅ Definition 3.1 gives its invariance hypothesis, so the sum is H ( t ) H(t) H ( t )
Taylor coefficients from fixed Boolean substitutions
The Q Q Q H ( 1 ) , … , H ( Q ) H(1),\ldots,H(Q) H ( 1 ) , … , H ( Q ) H H H
q = ⌊ d H Q ⌋ , J = q + 1. (26) q=\left\lfloor\frac{d_H}{Q}\right\rfloor,\qquad J=q+1. \tag*{(26)} q = ⌊ Q d H ⌋ , J = q + 1. (26) Since Q J > d H QJ>d_H Q J > d H
H t ( z ) = ∑ j = 0 J − 1 ( [ z j ] H ( t + z ) ) z j ( t ∈ [ Q ] ) . (27) \mathcal{H}_t(z)=\sum_{j=0}^{J-1}([z^j]H(t+z))z^j\qquad(t\in[Q]). \tag*{(27)} H t ( z ) = j = 0 ∑ J − 1 ([ z j ] H ( t + z )) z j ( t ∈ [ Q ]) . (27) Hermite interpolation recovers H H H Z = H ( 0 ) Z=H(0) Z = H ( 0 ) Q J QJ Q J A \mathcal{A} A Q J QJ Q J Q \mathbb{Q} Q A \mathcal{A} A
Fix t t t t + z t+z t + z t t t
Δ P ⃗ f , v ( z ) = P ⃗ f , v ( t + z ) − P ⃗ f , v ( t ) . \Delta\vec{P}_{f,v}(z)=\vec{P}_{f,v}(t+z)-\vec{P}_{f,v}(t). Δ P f , v ( z ) = P f , v ( t + z ) − P f , v ( t ) . Every entry is divisible by z z z x ∈ Q n x\in\mathbb{Q}^n x ∈ Q n X ⃗ \vec{X} X
⟨ P ⃗ f , v ( t ) , x ⟩ + ⟨ Δ P ⃗ f , v ( z ) , X ⃗ ⟩ . (28) \langle\vec{P}_{f,v}(t),x\rangle+\langle\Delta\vec{P}_{f,v}(z),\vec{X}\rangle. \tag*{(28)} ⟨ P f , v ( t ) , x ⟩ + ⟨ Δ P f , v ( z ) , X ⟩ . (28) Let
G x ( X ⃗ , z ) = ∏ f ∈ F Φ f ( ( ⟨ P ⃗ f , v ( t ) , x ⟩ + ⟨ Δ P ⃗ f , v ( z ) , X ⃗ ⟩ ) v ∈ I f ) . (29) G_x(\vec{X},z)=\prod_{f\in\mathcal{F}}\Phi_f\left(\left(\langle\vec{P}_{f,v}(t),x\rangle+\langle\Delta\vec{P}_{f,v}(z),\vec{X}\rangle\right)_{v\in I_f}\right). \tag*{(29)} G x ( X , z ) = f ∈ F ∏ Φ f ( ( ⟨ P f , v ( t ) , x ⟩ + ⟨ Δ P f , v ( z ) , X ⟩ ) v ∈ I f ) . (29) Setting X ⃗ = x \vec{X}=x X = x ∑ x G x ( x , z ) = H ( t + z ) \sum_xG_x(x,z)=H(t+z) ∑ x G x ( x , z ) = H ( t + z ) X ⃗ \vec{X} X x x x t t t X ⃗ \vec{X} X z z z
deg X ⃗ ( [ z j ] G x ( X ⃗ , z ) ) ≤ j . (30) \deg_{\vec{X}}([z^j]G_x(\vec{X},z))\le j. \tag*{(30)} deg X ([ z j ] G x ( X , z )) ≤ j . (30) The next identity recovers the substitution X ⃗ = x \vec{X}=x X = x
Lemma 3.3. Let q ≥ 0 q\ge0 q ≥ 0 A \mathcal{A} A x ∈ { 0 , 1 } n x\in\{0,1\}^n x ∈ { 0 , 1 } n p x ( X 1 , … , X n ) p_x(X_1,\ldots,X_n) p x ( X 1 , … , X n ) A \mathcal{A} A q q q
∑ x ∈ { 0 , 1 } n p x ( x ) = ∑ E ⊆ [ n ] ∣ E ∣ ≤ q ∑ U ⊆ E ( − 1 ) ∣ E ∣ − ∣ U ∣ ∑ x ∈ { 0 , 1 } n x i = 1 ( i ∈ E ) p x ( 1 U ) . (31) \sum_{x\in\{0,1\}^n}p_x(x)=\sum_{\substack{E\subseteq[n]\\|E|\le q}}\sum_{U\subseteq E}(-1)^{|E|-|U|}\sum_{\substack{x\in\{0,1\}^n\\x_i=1\ (i\in E)}}p_x(\mathbf{1}_U). \tag*{(31)} x ∈ { 0 , 1 } n ∑ p x ( x ) = E ⊆ [ n ] ∣ E ∣ ≤ q ∑ U ⊆ E ∑ ( − 1 ) ∣ E ∣ − ∣ U ∣ x ∈ { 0 , 1 } n x i = 1 ( i ∈ E ) ∑ p x ( 1 U ) . (31) Here 1 U \mathbf{1}_U 1 U U U U
Proof. For fixed x x x X i X_i X i X i X_i X i
p ~ x ( X ⃗ ) = ∑ E ⊆ [ n ] c x , E ∏ i ∈ E X i , c x , E = ∑ U ⊆ E ( − 1 ) ∣ E ∣ − ∣ U ∣ p x ( 1 U ) . \widetilde{p}_x(\vec{X})=\sum_{E\subseteq[n]}c_{x,E}\prod_{i\in E}X_i,\qquad c_{x,E}=\sum_{U\subseteq E}(-1)^{|E|-|U|}p_x(\mathbf{1}_U). p x ( X ) = E ⊆ [ n ] ∑ c x , E i ∈ E ∏ X i , c x , E = U ⊆ E ∑ ( − 1 ) ∣ E ∣ − ∣ U ∣ p x ( 1 U ) . The degree bound gives c x , E = 0 c_{x,E}=0 c x , E = 0 ∣ E ∣ > q |E|>q ∣ E ∣ > q X ⃗ = x \vec{X}=x X = x x x x ∏ i ∈ E x i \prod_{i\in E}x_i ∏ i ∈ E x i x i = 1 x_i=1 x i = 1 i ∈ E i\in E i ∈ E
Apply Lemma 3.3 to [ z j ] G x [z^j]G_x [ z j ] G x A \mathcal{A} A j < J j<J j < J U ⊆ E U\subseteq E U ⊆ E ∣ E ∣ ≤ q |E|\le q ∣ E ∣ ≤ q
R = A [ z ] / ( z J ) . \mathcal{R}=\mathcal{A}[z]/(z^J). R = A [ z ] / ( z J ) . For a fixed pair ( E , U ) (E,U) ( E , U ) Φ f , U ( x ; z ) \Phi_{f,U}(x;z) Φ f , U ( x ; z ) f f f (29) after setting X ⃗ = 1 U \vec{X}=\mathbf{1}_U X = 1 U
A g , U ( x ; z ) = ∏ f : σ f ( t ) = g Φ f , U ( x ; z ) . (32) A_{g,U}(x;z)=\prod_{f:\sigma_f(t)=g}\Phi_{f,U}(x;z). \tag*{(32)} A g , U ( x ; z ) = f : σ f ( t ) = g ∏ Φ f , U ( x ; z ) . (32) Their product is G x ( 1 U , z ) \mathcal{G}_x(\mathbf{1}_U,z) G x ( 1 U , z ) h ≠ g h\ne g h = g x x x ker C t , h \ker C_{t,h} ker C t , h x i = 1 x_i=1 x i = 1 i ∈ E i\in E i ∈ E
2 r ( 2 D + 1 ) 2 τ n 2 2 n / r (33) 2r(2D+1)^{2\tau n}2^{2n/r} \tag*{(33)} 2 r ( 2 D + 1 ) 2 τ n 2 2 n / r (33) Boolean vectors. Combining all the pairs gives
H ( t + z ) ≡ ∑ E ⊆ [ n ] ∣ E ∣ ≤ q ∑ U ⊆ E ( − 1 ) ∣ E ∣ − ∣ U ∣ ∑ x ∈ { 0 , 1 } n x i = 1 ( i ∈ E ) G x ( 1 U , z ) ( m o d z J ) . (34) H(t+z)\equiv\sum_{\substack{E\subseteq[n]\\|E|\le q}}\sum_{U\subseteq E}(-1)^{|E|-|U|}\sum_{\substack{x\in\{0,1\}^n\\x_i=1\ (i\in E)}}\mathcal{G}_x(\mathbf{1}_U,z)\pmod{z^J}. \tag*{(34)} H ( t + z ) ≡ E ⊆ [ n ] ∣ E ∣ ≤ q ∑ U ⊆ E ∑ ( − 1 ) ∣ E ∣ − ∣ U ∣ x ∈ { 0 , 1 } n x i = 1 ( i ∈ E ) ∑ G x ( 1 U , z ) ( mod z J ) . (34) The group products and sums are formed in R \mathcal{R} R H t ( z ) H_t(z) H t ( z ) t t t n n n X ⃗ = 1 U \vec{X}=\mathbf{1}_U X = 1 U J J J A \mathcal{A} A
The three exponential costs
For each t ∈ [ Q ] t\in[Q] t ∈ [ Q ] U ⊆ E ⊆ [ n ] U\subseteq E\subseteq[n] U ⊆ E ⊆ [ n ] ∣ E ∣ ≤ q |E|\le q ∣ E ∣ ≤ q
C q = ∑ s = 0 min ( q , n ) 2 s ( n s ) . (35) C_q=\sum_{s=0}^{\min(q,n)}2^s\binom{n}{s}. \tag*{(35)} C q = s = 0 ∑ m i n ( q , n ) 2 s ( s n ) . (35) In particular, C 0 = 1 C_0=1 C 0 = 1 C q = 3 n C_q=3^n C q = 3 n q ≥ n q\ge n q ≥ n 0 < η ≤ 1 0<\eta\le1 0 < η ≤ 1
η q C q ≤ ∑ s = 0 n ( n s ) ( 2 η ) s = ( 1 + 2 η ) n , (36) \eta^qC_q\le\sum_{s=0}^{n}\binom{n}{s}(2\eta)^s=(1+2\eta)^n, \tag*{(36)} η q C q ≤ s = 0 ∑ n ( s n ) ( 2 η ) s = ( 1 + 2 η ) n , (36) because η q ≤ η s \eta^q\le\eta^s η q ≤ η s s ≤ q s\le q s ≤ q
Proposition 3.4 (Common counting routine). Let A \mathcal{A} A Q \mathbb{Q} Q (22) , computed over A \mathcal{A} A (23) , and that the replacement rows, partitions, and measurements satisfy Definition 3.1. With q , J q,J q , J (26) , the routine computes Z Z Z
2 Q r C q ( 2 D + 1 ) 2 τ n 2 2 n / r 2QrC_q(2D+1)^{2\tau n}2^{2n/r} 2 Q r C q ( 2 D + 1 ) 2 τ n 2 2 n / r Boolean vectors. For each vector it evaluates at most ∣ F ∣ |\mathcal{F}| ∣ F ∣ A [ z ] / ( z J ) \mathcal{A}[z]/(z^J) A [ z ] / ( z J ) A \mathcal{A} A Q \mathbb{Q} Q n n n ∣ F ∣ |\mathcal{F}| ∣ F ∣ D D D Q Q Q R R R J J J r r r
Proof. There are Q Q Q C q C_q C q (33) Boolean vectors per pair. Forming each group product uses at most ∣ F ∣ |\mathcal{F}| ∣ F ∣ z J z^J z J O ( J 2 ) O(J^2) O ( J 2 ) A \mathcal{A} A Q J QJ Q J § 5.4 .
When d = O ( n ) d=O(n) d = O ( n ) d H = O ( n R ) d_H=O(nR) d H = O ( n R ) q = O ( n R / Q ) q=O(nR/Q) q = O ( n R / Q ) η = R / Q \eta=R/Q η = R / Q (36) gives
log C q ≤ q log ( Q / R ) + n log ( 1 + 2 R / Q ) = O ( n R Q log 2 Q R ) . \log C_q \leq q\log(Q/R)+n\log(1+2R/Q)=O\left(n\frac{R}{Q}\log\frac{2Q}{R}\right). log C q ≤ q log ( Q / R ) + n log ( 1 + 2 R / Q ) = O ( n Q R log R 2 Q ) . Consequently, if factor evaluation and coefficient arithmetic have polynomial bit cost, the running time is a polynomial factor times
2 O ( n [ R Q log 2 Q R + log ( 2 D + 1 ) + 1 r ] ) (37) 2^{O\left(n\left[\frac{R}{Q}\log\frac{2Q}{R}+\log(2D+1)+\frac{1}{r}\right]\right)} \tag*{(37)} 2 O ( n [ Q R l o g R 2 Q + l o g ( 2 D + 1 ) + r 1 ] ) (37) These are exactly the three costs from the overview. The parameter choice in Lemma 4.8 makes their sum O ( ( log log n ) 2 / log n ) O((\log\log n)^2/\sqrt{\log n}) O (( log log n ) 2 / log n ) log n \log n log n d d d q = ⌊ d R / Q ⌋ q=\lfloor dR/Q\rfloor q = ⌊ d R / Q ⌋
We have so far assumed that the design is supplied. The next section constructs it from the factor scopes. The remaining two sections use the resulting routine first to count the zeros of a supplied polynomial system and then to choose such a system for a formula.
Constructing low-degree replacement forms
The counting routine needs one replacement form for each pair ( f , v ) (f,v) ( f , v ) f f f v v v § 5 , f f f P ⃗ f , v ( T ) \vec{P}_{f,v}(T) P f , v ( T ) C t , g C_{t,g} C t , g Definition 3.1 , together with the partitions S t , g S_{t,g} S t , g σ f \sigma_f σ f deg P ⃗ f , v ≤ R \deg\vec{P}_{f,v}\leq R deg P f , v ≤ R R R R Q Q Q H H H R R R R / Q R/Q R / Q T = 0 , 1 , … , Q T=0,1,\ldots,Q T = 0 , 1 , … , Q Q Q Q
The construction first gives polynomial rows on a complete array of auxiliary coordinates. We then place each factor–variable pair in a copy of that array. Several positions can contain the same Boolean variable x v x_v x v
The auxiliary coordinates
We first specify the larger coordinate set B \mathcal{B} B P ⃗ u ( T ) \vec{P}_u(T) P u ( T ) ( f , v ) (f,v) ( f , v ) ( I f ) f ∈ F (I_f)_{f\in\mathcal{F}} ( I f ) f ∈ F I f ⊆ [ n ] I_f\subseteq[n] I f ⊆ [ n ]
A = { ( f , v ) : f ∈ F , v ∈ I f } , M = ∣ A ∣ , \mathcal{A}=\{(f,v):f\in\mathcal{F},\ v\in I_f\},\qquad M=|\mathcal{A}|, A = {( f , v ) : f ∈ F , v ∈ I f } , M = ∣ A ∣ , ∣ I f ∣ ≤ w ( f ∈ F ) , # { f : v ∈ I f } ≤ b ( v ∈ [ n ] ) . (38) |I_f|\leq w\quad(f\in\mathcal{F}),\qquad\#\{f:v\in I_f\}\leq b\quad(v\in[n]). \tag*{(38)} ∣ I f ∣ ≤ w ( f ∈ F ) , # { f : v ∈ I f } ≤ b ( v ∈ [ n ]) . (38) Thus w w w b b b w w w { v } \{v\} { v } b b b
n ≥ 1 , b , w ≥ 1 , n ≤ M ≤ b n . n\geq1,\qquad b,w\geq1,\qquad n\leq M\leq bn. n ≥ 1 , b , w ≥ 1 , n ≤ M ≤ bn . Throughout this section, r ≥ 2 r\geq2 r ≥ 2 1 ≤ R < Q 1\leq R<Q 1 ≤ R < Q 0 < τ ≤ 1 / 2 0<\tau\leq1/2 0 < τ ≤ 1/2
At each point t ∈ [ Q ] t\in[Q] t ∈ [ Q ] x v x_v x v v v v
a v ∈ [ r ] Q ( v ∈ [ n ] ) , d f ∈ [ r ] Q ( f ∈ F ) , a_v\in[r]^Q\quad(v\in[n]),\qquad d_f\in[r]^Q\quad(f\in\mathcal{F}), a v ∈ [ r ] Q ( v ∈ [ n ]) , d f ∈ [ r ] Q ( f ∈ F ) , where a v ( t ) a_v(t) a v ( t ) x v x_v x v d f ( t ) d_f(t) d f ( t ) f f f ( f , v ) (f,v) ( f , v ) ( a v , d f ) (a_v,d_f) ( a v , d f ) x v x_v x v a v a_v a v
Before choosing a v a_v a v d f d_f d f
B = ( [ r ] Q ) 2 , u = ( a u , d u ) , D = ∣ B ∣ = r 2 Q . (39) \mathcal{B} = ([r]^{Q})^2,\qquad u=(a_u,d_u),\qquad D=|\mathcal{B}|=r^{2Q}. \tag*{(39)} B = ([ r ] Q ) 2 , u = ( a u , d u ) , D = ∣ B ∣ = r 2 Q . (39) An auxiliary vector ζ ∈ Q B \zeta\in\mathbb{Q}^{\mathcal{B}} ζ ∈ Q B ζ u \zeta_u ζ u u u u u u u ζ \zeta ζ t ∈ [ Q ] t\in[Q] t ∈ [ Q ] g ∈ [ r ] g\in[r] g ∈ [ r ]
S t , g B = { u ∈ B : a u ( t ) = g } . (40) S^{\mathcal{B}}_{t,g}=\{u\in\mathcal{B}:a_u(t)=g\}. \tag*{(40)} S t , g B = { u ∈ B : a u ( t ) = g } . (40) For fixed t t t B \mathcal{B} B r r r D / r D/r D / r
For each u ∈ B u\in\mathcal{B} u ∈ B P ⃗ u ( T ) ∈ Q [ T ] 1 × B \vec{P}_u(T)\in\mathbb{Q}[T]^{1\times\mathcal{B}} P u ( T ) ∈ Q [ T ] 1 × B R − 1 R-1 R − 1 P ⃗ u ( 0 ) = δ u \vec{P}_u(0)=\delta_u P u ( 0 ) = δ u T = t T=t T = t ⟨ P ⃗ u ( t ) , ζ ⟩ \langle\vec{P}_u(t),\zeta\rangle ⟨ P u ( t ) , ζ ⟩
ζ ∣ S t , d u ( t ) B and C B ζ , \zeta|_{S^{\mathcal{B}}_{t,d_u(t)}}\qquad\text{and}\qquad C^{\mathcal{B}}\zeta, ζ ∣ S t , d u ( t ) B and C B ζ , where C B C^{\mathcal{B}} C B u , t u,t u , t { 0 , 1 , − 1 } \{0,1,-1\} { 0 , 1 , − 1 } u f , v = ( a v , d f ) u_{f,v}=(a_v,d_f) u f , v = ( a v , d f ) f f f d f ( t ) d_f(t) d f ( t ) a v , d f a_v,d_f a v , d f § 4.4 and § 4.5 , respectively.
The matrices M 0 M_0 M 0 C B C^{\mathcal{B}} C B
To construct the rows P ⃗ u ( T ) \vec{P}_u(T) P u ( T ) M 0 ∈ Q B × B M_0\in\mathbb{Q}^{\mathcal{B}\times\mathcal{B}} M 0 ∈ Q B × B ker C B \ker C^{\mathcal{B}} ker C B
M 0 [ u , y ] ≠ 0 ⟹ ∣ { s ∈ [ Q ] : a y ( s ) ≠ d u ( s ) } ∣ ≤ R − 1. (41) M_0[u,y]\ne0\quad\Longrightarrow\quad\left|\{s\in[Q]:a_y(s)\ne d_u(s)\}\right|\le R-1. \tag*{(41)} M 0 [ u , y ] = 0 ⟹ ∣ { s ∈ [ Q ] : a y ( s ) = d u ( s )} ∣ ≤ R − 1. (41) The identity
ζ u = ( ( I − M 0 ) ζ ) u + ∑ y ∈ B M 0 [ u , y ] ζ y (42) \zeta_u=((I-M_0)\zeta)_u+\sum_{y\in\mathcal{B}}M_0[u,y]\zeta_y \tag*{(42)} ζ u = (( I − M 0 ) ζ ) u + y ∈ B ∑ M 0 [ u , y ] ζ y (42) explains both requirements. If M 0 M_0 M 0 ker C B \ker C^{\mathcal{B}} ker C B C B ζ C^{\mathcal{B}}\zeta C B ζ T T T M 0 [ u , y ] M_0[u,y] M 0 [ u , y ] t t t a y ( t ) ≠ d u ( t ) a_y(t)\ne d_u(t) a y ( t ) = d u ( t ) (41) permits degree at most R − 1 R-1 R − 1 T = t T=t T = t y ∈ S t , d u ( t ) B y\in S^{\mathcal{B}}_{t,d_u(t)} y ∈ S t , d u ( t ) B
For s ∈ [ Q ] s\in[Q] s ∈ [ Q ] ϑ s : B → B \vartheta_s:\mathcal{B}\to\mathcal{B} ϑ s : B → B a u ( s ) a_u(s) a u ( s ) d u ( s ) d_u(s) d u ( s ) Π s \Pi_s Π s
( Π s ζ ) u = ζ ϑ s ( u ) . (\Pi_s\zeta)_u=\zeta_{\vartheta_s(u)}. ( Π s ζ ) u = ζ ϑ s ( u ) . The exchanged pair ϑ s ( u ) \vartheta_s(u) ϑ s ( u ) B \mathcal{B} B [ r ] Q [r]^Q [ r ] Q Π s \Pi_s Π s Π s 2 = I \Pi_s^2=I Π s 2 = I
V 0 = ∏ s = 1 Q Π s V_0=\prod_{s=1}^{Q}\Pi_s V 0 = s = 1 ∏ Q Π s exchanges the complete sequences a u a_u a u d u d_u d u V 0 [ u , y ] ≠ 0 V_0[u,y]\ne0 V 0 [ u , y ] = 0 a y ( s ) = d u ( s ) a_y(s)=d_u(s) a y ( s ) = d u ( s ) s s s (41) with no mismatches. We will modify V 0 V_0 V 0
Define
W 0 = 1 2 ∑ s = 1 Q ( I − Π s ) . (43) W_0=\frac{1}{2}\sum_{s=1}^{Q}(I-\Pi_s). \tag*{(43)} W 0 = 2 1 s = 1 ∑ Q ( I − Π s ) . (43) On a common eigenvector ψ \psi ψ Π s \Pi_s Π s j j j s s s Π s ψ = − ψ \Pi_s\psi=-\psi Π s ψ = − ψ
W 0 ψ = j ψ , V 0 ψ = ( − 1 ) j ψ , V 0 p ( W 0 ) ψ = ( − 1 ) j p ( j ) ψ W_0\psi=j\psi,\qquad V_0\psi=(-1)^j\psi,\qquad V_0p(W_0)\psi=(-1)^jp(j)\psi W 0 ψ = j ψ , V 0 ψ = ( − 1 ) j ψ , V 0 p ( W 0 ) ψ = ( − 1 ) j p ( j ) ψ for every polynomial p p p V 0 p ( W 0 ) V_0p(W_0) V 0 p ( W 0 ) ψ \psi ψ p ( j ) = ( − 1 ) j p(j)=(-1)^j p ( j ) = ( − 1 ) j j j j Pr [ W = j ] \Pr[W=j] Pr [ W = j ] W ∼ Bin ( Q , ( r − 1 ) / ( 2 r ) ) W\sim\operatorname{Bin}(Q,(r-1)/(2r)) W ∼ Bin ( Q , ( r − 1 ) / ( 2 r ))
For an interval I = { A 0 , … , A 0 + R − 1 } ⊆ { 0 , … , Q } \mathcal{I}=\{A_0,\ldots,A_0+R-1\}\subseteq\{0,\ldots,Q\} I = { A 0 , … , A 0 + R − 1 } ⊆ { 0 , … , Q } f I ( j ) = ( − 1 ) j f_{\mathcal{I}}(j)=(-1)^j f I ( j ) = ( − 1 ) j j ∈ I j\in\mathcal{I} j ∈ I R − 1 R-1 R − 1
f I ( z ) = ( − 1 ) A 0 ∑ j = 0 R − 1 ( − 2 ) j ( z − A 0 j ) , M 0 = V 0 f I ( W 0 ) , (44) f_{\mathcal{I}}(z)=(-1)^{A_0}\sum_{j=0}^{R-1}(-2)^j\binom{z-A_0}{j},\qquad M_0=V_0f_{\mathcal{I}}(W_0), \tag*{(44)} f I ( z ) = ( − 1 ) A 0 j = 0 ∑ R − 1 ( − 2 ) j ( j z − A 0 ) , M 0 = V 0 f I ( W 0 ) , (44) where
( z − A 0 j ) = 1 j ! ∏ a = 0 j − 1 ( z − A 0 − a ) . \binom{z-A_0}{j}=\frac{1}{j!}\prod_{a=0}^{j-1}(z-A_0-a). ( j z − A 0 ) = j ! 1 a = 0 ∏ j − 1 ( z − A 0 − a ) . and the empty product is one. For 0 ≤ q < R 0\leq q<R 0 ≤ q < R f I ( A 0 + q ) = ( − 1 ) A 0 ( 1 − 2 ) q = ( − 1 ) A 0 + q f_{\mathcal{I}}(A_0+q)=(-1)^{A_0}(1-2)^q=(-1)^{A_0+q} f I ( A 0 + q ) = ( − 1 ) A 0 ( 1 − 2 ) q = ( − 1 ) A 0 + q
Lemma 4.1. For B \mathcal{B} B I \mathcal{I} I M 0 M_0 M 0 C B ∈ { 0 , 1 , − 1 } [ c B ] × B C^{\mathcal{B}}\in\{0,1,-1\}^{[c_{\mathcal{B}}]\times\mathcal{B}} C B ∈ { 0 , 1 , − 1 } [ c B ] × B
c B ≤ D Pr [ W ∉ I ] , W ∼ Bin ( Q , r − 1 2 r ) , c_{\mathcal{B}}\leq D\Pr[W\notin\mathcal{I}],\qquad W\sim\operatorname{Bin}\left(Q,\frac{r-1}{2r}\right), c B ≤ D Pr [ W ∈ / I ] , W ∼ Bin ( Q , 2 r r − 1 ) , and M 0 ξ = ξ M_0\xi=\xi M 0 ξ = ξ ξ ∈ ker C B \xi\in\ker C^{\mathcal{B}} ξ ∈ ker C B
Proof. For ( α , δ ) ∈ [ r ] 2 (\alpha,\delta)\in[r]^2 ( α , δ ) ∈ [ r ] 2 e α δ e_{\alpha\delta} e α δ
e α α ( α ∈ [ r ] ) , e α δ + e δ α , e α δ − e δ α ( 1 ≤ α < δ ≤ r ) . e_{\alpha\alpha}\quad(\alpha\in[r]),\qquad e_{\alpha\delta}+e_{\delta\alpha},\qquad e_{\alpha\delta}-e_{\delta\alpha}\quad(1\leq\alpha<\delta\leq r). e α α ( α ∈ [ r ]) , e α δ + e δ α , e α δ − e δ α ( 1 ≤ α < δ ≤ r ) . The first two types have eigenvalue + 1 +1 + 1 − 1 -1 − 1 r ( r + 1 ) / 2 r(r+1)/2 r ( r + 1 ) /2 + 1 +1 + 1 r ( r − 1 ) / 2 r(r-1)/2 r ( r − 1 ) /2 − 1 -1 − 1
For each s ∈ [ Q ] s\in[Q] s ∈ [ Q ] ψ s : [ r ] 2 → { 0 , 1 , − 1 } \psi_s:[r]^2\to\{0,1,-1\} ψ s : [ r ] 2 → { 0 , 1 , − 1 }
ψ ( u ) = ∏ s = 1 Q ψ s ( a u ( s ) , d u ( s ) ) ( u ∈ B ) . \psi(u)=\prod_{s=1}^{Q}\psi_s(a_u(s),d_u(s))\qquad(u\in\mathcal{B}). ψ ( u ) = s = 1 ∏ Q ψ s ( a u ( s ) , d u ( s )) ( u ∈ B ) . The D = r 2 Q D=r^{2Q} D = r 2 Q ψ s \psi_s ψ s Q B \mathbb{Q}^{\mathcal{B}} Q B { 0 , 1 , − 1 } \{0,1,-1\} { 0 , 1 , − 1 } j j j ψ s \psi_s ψ s − 1 -1 − 1 W 0 ψ = j ψ W_0\psi=j\psi W 0 ψ = j ψ V 0 ψ = ( − 1 ) j ψ V_0\psi=(-1)^j\psi V 0 ψ = ( − 1 ) j ψ M 0 M_0 M 0 ψ \psi ψ j ∈ I j\in\mathcal{I} j ∈ I
To make ker C B \ker C^{\mathcal{B}} ker C B M 0 M_0 M 0 C B C^{\mathcal{B}} C B j ∉ I j\notin\mathcal{I} j ∈ / I ( r − 1 ) / ( 2 r ) (r-1)/(2r) ( r − 1 ) / ( 2 r ) − 1 -1 − 1 D D D Pr [ W ∉ I ] \Pr[W\notin\mathcal{I}] Pr [ W ∈ / I ] C B ξ = 0 ⃗ C^{\mathcal{B}}\xi=\vec{0} C B ξ = 0 ξ \xi ξ j ∈ I j\in\mathcal{I} j ∈ I M 0 M_0 M 0 □ \square □
For a bound ϵ > 0 \epsilon>0 ϵ > 0
Pr [ Bin ( Q , r − 1 2 r ) ∉ I ] ≤ ϵ (45) \Pr\left[\operatorname{Bin}\left(Q,\frac{r-1}{2r}\right)\notin\mathcal{I}\right]\leq\epsilon \tag*{(45)} Pr [ Bin ( Q , 2 r r − 1 ) ∈ / I ] ≤ ϵ (45) thus gives c B ≤ ϵ D c_{\mathcal{B}}\leq\epsilon D c B ≤ ϵD ϵ = τ / ( r b ) \epsilon=\tau/(rb) ϵ = τ / ( r b ) M 0 M_0 M 0
Lemma 4.2. For all u , y ∈ B u,y\in\mathcal{B} u , y ∈ B M 0 [ u , y ] ≠ 0 M_0[u,y]\neq0 M 0 [ u , y ] = 0
∣ { s ∈ [ Q ] : a y ( s ) ≠ d u ( s ) } ∣ ≤ R − 1. \left|\{s\in[Q]:a_y(s)\neq d_u(s)\}\right|\leq R-1. ∣ { s ∈ [ Q ] : a y ( s ) = d u ( s )} ∣ ≤ R − 1. Proof. Expand f I ( W 0 ) f_I(W_0) f I ( W 0 ) Π s \Pi_s Π s deg f I ≤ R − 1 \deg f_I \le R-1 deg f I ≤ R − 1 Π s 2 = I \Pi_s^2=I Π s 2 = I ∏ s ∈ J Π s \prod_{s\in J}\Pi_s ∏ s ∈ J Π s J ⊆ [ Q ] J\subseteq[Q] J ⊆ [ Q ] ∣ J ∣ ≤ R − 1 |J|\le R-1 ∣ J ∣ ≤ R − 1 V 0 V_0 V 0
V 0 ∏ s ∈ J Π s = ∏ s ∉ J Π s . V_0\prod_{s\in J}\Pi_s=\prod_{s\notin J}\Pi_s. V 0 s ∈ J ∏ Π s = s ∈ / J ∏ Π s . A nonzero ( u , y ) (u,y) ( u , y ) a y ( s ) = d u ( s ) a_y(s)=d_u(s) a y ( s ) = d u ( s ) s ∉ J s\notin J s ∈ / J R − 1 R-1 R − 1 R − 1 R-1 R − 1 ( u , y ) (u,y) ( u , y ) M 0 [ u , y ] = 0 M_0[u,y]=0 M 0 [ u , y ] = 0
The polynomial rows
We now use M 0 M_0 M 0 P ⃗ u ( T ) \vec{P}_u(T) P u ( T ) § 4.1 . Lemma 4.1 makes ( I − M 0 ) ζ (I-M_0)\zeta ( I − M 0 ) ζ C B ζ C^{\mathcal B}\zeta C B ζ Lemma 4.2 bounds the number of points at which a nonzero term M 0 [ u , y ] ζ y M_0[u,y]\zeta_y M 0 [ u , y ] ζ y S t , d u ( t ) B S^{\mathcal B}_{t,d_u(t)} S t , d u ( t ) B
For M 0 [ u , y ] ≠ 0 M_0[u,y]\ne0 M 0 [ u , y ] = 0
E u , y = { s ∈ [ Q ] : a y ( s ) ≠ d u ( s ) } , φ u , y ( T ) = ∏ s ∈ E u , y ( 1 − T s ) . (46) E_{u,y}=\{s\in[Q]:a_y(s)\ne d_u(s)\},\qquad
\varphi_{u,y}(T)=\prod_{s\in E_{u,y}}\left(1-\frac{T}{s}\right). \tag*{(46)} E u , y = { s ∈ [ Q ] : a y ( s ) = d u ( s )} , φ u , y ( T ) = s ∈ E u , y ∏ ( 1 − s T ) . (46) The empty product is one. Then deg φ u , y ≤ R − 1 \deg\varphi_{u,y}\le R-1 deg φ u , y ≤ R − 1 φ u , y ( 0 ) = 1 \varphi_{u,y}(0)=1 φ u , y ( 0 ) = 1 φ u , y ( t ) = 0 \varphi_{u,y}(t)=0 φ u , y ( t ) = 0 a y ( t ) ≠ d u ( t ) a_y(t)\ne d_u(t) a y ( t ) = d u ( t )
P ⃗ u ( T ) = δ u ( I − M 0 ) + ∑ y ∈ B M 0 [ u , y ] ≠ 0 M 0 [ u , y ] φ u , y ( T ) δ y . (47) \vec{P}_u(T)=\delta_u(I-M_0)+\sum_{\substack{y\in\mathcal B\\M_0[u,y]\ne0}}M_0[u,y]\varphi_{u,y}(T)\delta_y. \tag*{(47)} P u ( T ) = δ u ( I − M 0 ) + y ∈ B M 0 [ u , y ] = 0 ∑ M 0 [ u , y ] φ u , y ( T ) δ y . (47) This row acts as a linear form in ζ \zeta ζ T T T
Lemma 4.3. The rows in (47) have degree at most R − 1 R-1 R − 1 P ⃗ u ( 0 ) = δ u \vec{P}_u(0)=\delta_u P u ( 0 ) = δ u u ∈ B u\in\mathcal B u ∈ B
For t ∈ [ Q ] t\in[Q] t ∈ [ Q ] g ∈ [ r ] g\in[r] g ∈ [ r ] C t , g B C^{\mathcal B}_{t,g} C t , g B C B C^{\mathcal B} C B S t , g B S^{\mathcal B}_{t,g} S t , g B u , t , g u,t,g u , t , g g ≠ d u ( t ) g\ne d_u(t) g = d u ( t ) ξ ∈ Q S t , g B \xi\in\mathbb Q^{S^{\mathcal B}_{t,g}} ξ ∈ Q S t , g B ξ ~ \widetilde{\xi} ξ B \mathcal B B
C t , g B ξ = 0 ⃗ ⟹ ⟨ P ⃗ u ( t ) , ξ ~ ⟩ = 0. C^{\mathcal B}_{t,g}\xi=\vec{0}\quad\Longrightarrow\quad
\langle\vec{P}_u(t),\widetilde{\xi}\rangle=0. C t , g B ξ = 0 ⟹ ⟨ P u ( t ) , ξ ⟩ = 0. More generally, ⟨ P ⃗ u ( t ) , ζ ⟩ \langle\vec{P}_u(t),\zeta\rangle ⟨ P u ( t ) , ζ ⟩ ζ ∣ S t , d u ( t ) B \zeta|_{S^{\mathcal B}_{t,d_u(t)}} ζ ∣ S t , d u ( t ) B C B ζ C^{\mathcal B}\zeta C B ζ
Proof. The degree bound follows from that of φ u , y \varphi_{u,y} φ u , y
P ⃗ u ( 0 ) = δ u ( I − M 0 ) + δ u M 0 = δ u . \vec{P}_u(0)=\delta_u(I-M_0)+\delta_uM_0=\delta_u. P u ( 0 ) = δ u ( I − M 0 ) + δ u M 0 = δ u . Applying (47) to ζ \zeta ζ
⟨ P ⃗ u ( T ) , ζ ⟩ = ( ( I − M 0 ) ζ ) u + ∑ y ∈ B M 0 [ u , y ] ≠ 0 M 0 [ u , y ] φ u , y ( T ) ζ y . (48) \langle\vec{P}_u(T),\zeta\rangle=((I-M_0)\zeta)_u+
\sum_{\substack{y\in\mathcal B\\M_0[u,y]\ne0}}M_0[u,y]\varphi_{u,y}(T)\zeta_y. \tag*{(48)} ⟨ P u ( T ) , ζ ⟩ = (( I − M 0 ) ζ ) u + y ∈ B M 0 [ u , y ] = 0 ∑ M 0 [ u , y ] φ u , y ( T ) ζ y . (48) At T = t T=t T = t a y ( t ) ≠ d u ( t ) a_y(t)\ne d_u(t) a y ( t ) = d u ( t ) ξ ~ \widetilde{\xi} ξ S t , g B S^{\mathcal B}_{t,g} S t , g B g ≠ d u ( t ) g\ne d_u(t) g = d u ( t ) ξ ~ \widetilde{\xi} ξ C t , g B ξ = 0 ⃗ C^{\mathcal B}_{t,g}\xi=\vec{0} C t , g B ξ = 0 C B ξ ~ = 0 ⃗ C^{\mathcal B}\widetilde{\xi}=\vec{0} C B ξ = 0 ( I − M 0 ) ξ ~ = 0 ⃗ (I-M_0)\widetilde{\xi}=\vec{0} ( I − M 0 ) ξ = 0 Lemma 4.1 . This proves the implication.
For the final assertion, two vectors with the same measurements under C B C^{\mathcal B} C B ( I − M 0 ) ζ (I-M_0)\zeta ( I − M 0 ) ζ C B ζ C^{\mathcal B}\zeta C B ζ (48) at T = t T=t T = t ζ ∣ S t , d u ( t ) B \zeta|_{S^{\mathcal B}_{t,d_u(t)}} ζ ∣ S t , d u ( t ) B
Fix t t t ζ g = ζ ∣ S t , g B \zeta_g=\zeta|_{S^{\mathcal B}_{t,g}} ζ g = ζ ∣ S t , g B C B C^{\mathcal B} C B
C B ζ = ∑ g = 1 r C t , g B ζ g . (49) C^{\mathcal B}\zeta=\sum_{g=1}^{r}C^{\mathcal B}_{t,g}\zeta_g. \tag*{(49)} C B ζ = g = 1 ∑ r C t , g B ζ g . (49) Fixing each vector C t , g B ζ C^{B}_{t,g}\zeta C t , g B ζ C B B ζ C^{B}_{B}\zeta C B B ζ r r r C t , g B ζ C^{B}_{t,g}\zeta C t , g B ζ ⟨ P ⃗ u ( t ) , ζ ⟩ \langle\vec{P}_{u}(t),\zeta\rangle ⟨ P u ( t ) , ζ ⟩ ζ ∣ S t , d u ( t ) B \left.\zeta\right|_{S^{B}_{t,d_{u}(t)}} ζ ∣ S t , d u ( t ) B ( f , v ) (f,v) ( f , v ) ( a v , d f ) (a_v,d_f) ( a v , d f )
Remark 4.4. We illustrate how (47) confines dependence outside the assigned group to one linear measurement. Take Q = R = 2 Q=R=2 Q = R = 2 R < Q R<Q R < Q r = 2 r=2 r = 2 { 0 , 1 } \{0,1\} { 0 , 1 } I = { 0 , 1 } I=\{0,1\} I = { 0 , 1 } f I ( z ) = 1 − 2 z f_I(z)=1-2z f I ( z ) = 1 − 2 z ( 00 , 11 ) (00,11) ( 00 , 11 )
ζ 0 = ζ ( 00 , 11 ) , ζ 1 = ζ ( 10 , 01 ) , ζ 2 = ζ ( 01 , 10 ) , ζ 12 = ζ ( 11 , 00 ) . \zeta_0=\zeta_{(00,11)},\qquad\zeta_1=\zeta_{(10,01)},\qquad\zeta_2=\zeta_{(01,10)},\qquad\zeta_{12}=\zeta_{(11,00)}. ζ 0 = ζ ( 00 , 11 ) , ζ 1 = ζ ( 10 , 01 ) , ζ 2 = ζ ( 01 , 10 ) , ζ 12 = ζ ( 11 , 00 ) . Both exchanges preserve this set of four coordinates, so we can restrict the calculation to it. On this restriction, M 0 = Π 1 + Π 2 − Π 1 Π 2 M_0=\Pi_1+\Pi_2-\Pi_1\Pi_2 M 0 = Π 1 + Π 2 − Π 1 Π 2
c ( ζ ) = ζ 0 − ζ 1 − ζ 2 + ζ 12 . c(\zeta)=\zeta_0-\zeta_1-\zeta_2+\zeta_{12}. c ( ζ ) = ζ 0 − ζ 1 − ζ 2 + ζ 12 . The row indexed by u = ( 00 , 11 ) u=(00,11) u = ( 00 , 11 )
⟨ P ⃗ u ( T ) , ζ ⟩ = ζ 0 − T 2 ζ 1 − T ζ 2 . \langle\vec{P}_{u}(T),\zeta\rangle=\zeta_0-\frac{T}{2}\zeta_1-T\zeta_2. ⟨ P u ( T ) , ζ ⟩ = ζ 0 − 2 T ζ 1 − T ζ 2 . It equals ζ 0 \zeta_0 ζ 0
⟨ P ⃗ u ( 1 ) , ζ ⟩ = c ( ζ ) + 1 2 ζ 1 − ζ 12 , ⟨ P ⃗ u ( 2 ) , ζ ⟩ = c ( ζ ) − ζ 2 − ζ 12 . \langle\vec{P}_{u}(1),\zeta\rangle=c(\zeta)+\frac{1}{2}\zeta_1-\zeta_{12},\qquad\langle\vec{P}_{u}(2),\zeta\rangle=c(\zeta)-\zeta_2-\zeta_{12}. ⟨ P u ( 1 ) , ζ ⟩ = c ( ζ ) + 2 1 ζ 1 − ζ 12 , ⟨ P u ( 2 ) , ζ ⟩ = c ( ζ ) − ζ 2 − ζ 12 . At t = 1 t=1 t = 1 ζ 1 , ζ 12 \zeta_1,\zeta_{12} ζ 1 , ζ 12 a y ( 1 ) = 1 = d u ( 1 ) a_y(1)=1=d_u(1) a y ( 1 ) = 1 = d u ( 1 ) t = 2 t=2 t = 2 ζ 2 , ζ 12 \zeta_2,\zeta_{12} ζ 2 , ζ 12 a y ( 2 ) = 1 = d u ( 2 ) a_y(2)=1=d_u(2) a y ( 2 ) = 1 = d u ( 2 ) c ( ζ ) c(\zeta) c ( ζ ) ⟨ P ⃗ u ( t ) , ζ ⟩ \langle\vec{P}_{u}(t),\zeta\rangle ⟨ P u ( t ) , ζ ⟩ c ( ζ ) c(\zeta) c ( ζ ) C B B ζ C^{B}_{B}\zeta C B B ζ u ∈ B u\in B u ∈ B ϵ D \epsilon D ϵD
Choosing the variable and factor groups
We now choose the families ( a v ) v ∈ [ n ] (a_v)_{v\in[n]} ( a v ) v ∈ [ n ] ( d f ) f ∈ F (d_f)_{f\in\mathcal{F}} ( d f ) f ∈ F § 4.1 , and set
u f , v = ( a v , d f ) ( ( f , v ) ∈ A ) . u_{f,v}=(a_v,d_f)\qquad((f,v)\in\mathcal{A}). u f , v = ( a v , d f ) (( f , v ) ∈ A ) . These choices have two requirements. First, if a label occurs c u c_u c u c u c_u c u
c u = # { ( f , v ) ∈ A : u f , v = u } , h = max u ∈ B c u . c_u=\#\{(f,v)\in\mathcal{A}:u_{f,v}=u\},\qquad h=\max_{u\in\mathcal{B}}c_u. c u = # {( f , v ) ∈ A : u f , v = u } , h = u ∈ B max c u . Second, the actual variable group { v : a v ( t ) = g } \{v:a_v(t)=g\} { v : a v ( t ) = g } h h h
Lemma 4.5. Let ( I f ) f ∈ F (I_f)_{f\in\mathcal{F}} ( I f ) f ∈ F (38) , with n ≤ M ≤ b n n\le M\le bn n ≤ M ≤ bn D = r 2 Q D=r^{2Q} D = r 2 Q r ≥ 2 r\ge2 r ≥ 2 Q ≥ 1 Q\ge1 Q ≥ 1
2 ( b + w ) D 2 ≤ M , 2 Q r 2 ≤ n . (50) 2(b+w)D^2\le M,\qquad2Qr^2\le n. \tag*{(50)} 2 ( b + w ) D 2 ≤ M , 2 Q r 2 ≤ n . (50) One can deterministically choose a v , d f ∈ [ r ] Q a_v,d_f\in[r]^Q a v , d f ∈ [ r ] Q d f ( t ) = 1 d_f(t)=1 d f ( t ) = 1
max u ∈ B # { ( f , v ) ∈ A : ( a v , d f ) = u } ≤ 2 M D , \max_{u\in\mathcal{B}}\#\{(f,v)\in\mathcal{A}:(a_v,d_f)=u\}\le\frac{2M}{D}, u ∈ B max # {( f , v ) ∈ A : ( a v , d f ) = u } ≤ D 2 M , # { v ∈ [ n ] : a v ( t ) = g } ≤ 2 n r ( t ∈ [ Q ] , g ∈ [ r ] ) . \#\{v\in[n]:a_v(t)=g\}\le\frac{2n}{r}\qquad(t\in[Q],\ g\in[r]). # { v ∈ [ n ] : a v ( t ) = g } ≤ r 2 n ( t ∈ [ Q ] , g ∈ [ r ]) . The construction takes poly ( n , M , ∣ F ∣ , D , Q , r , b , w ) \operatorname{poly}(n,M,|\mathcal{F}|,D,Q,r,b,w) poly ( n , M , ∣ F ∣ , D , Q , r , b , w )
Proof. We first estimate the effect of random labels; we will then fix their values deterministically. For this estimate, all a v a_v a v d f d_f d f I f I_f I f [ r ] Q [r]^Q [ r ] Q I f I_f I f d f d_f d f u f , v = ( a v , d f ) u_{f,v}=(a_v,d_f) u f , v = ( a v , d f ) B \mathcal{B} B n t , g = # { v : a v ( t ) = g } n_{t,g}=\#\{v:a_v(t)=g\} n t , g = # { v : a v ( t ) = g }
Φ = ∑ u ∈ B ( c u − M / D ) 2 , Ξ = ∑ t = 1 Q ∑ g = 1 r ( n t , g − n / r ) 2 . \Phi=\sum_{u\in\mathcal{B}}(c_u-M/D)^2,\qquad\Xi=\sum_{t=1}^{Q}\sum_{g=1}^{r}(n_{t,g}-n/r)^2. Φ = u ∈ B ∑ ( c u − M / D ) 2 , Ξ = t = 1 ∑ Q g = 1 ∑ r ( n t , g − n / r ) 2 . Their expansions as sums of indicators are
Φ = ∑ ( f , v ) , ( f ′ , v ′ ) ∈ A 1 { ( a v , d f ) = ( a v ′ , d f ′ ) } − M 2 D , \Phi=\sum_{(f,v),(f',v')\in\mathcal{A}}\mathbf{1}_{\{(a_v,d_f)=(a_{v'},d_{f'})\}}-\frac{M^2}{D}, Φ = ( f , v ) , ( f ′ , v ′ ) ∈ A ∑ 1 {( a v , d f ) = ( a v ′ , d f ′ )} − D M 2 , Ξ = ∑ t = 1 Q ∑ v , v ′ ∈ [ n ] 1 { a v ( t ) = a v ′ ( t ) } − Q n 2 r . \Xi=\sum_{t=1}^{Q}\sum_{v,v'\in[n]}\mathbf{1}_{\{a_v(t)=a_{v'}(t)\}}-\frac{Qn^2}{r}. Ξ = t = 1 ∑ Q v , v ′ ∈ [ n ] ∑ 1 { a v ( t ) = a v ′ ( t )} − r Q n 2 . Two incidence labels are independent unless they share a factor or a variable. For each ( f , v ) (f,v) ( f , v ) b + w b+w b + w f f f v v v ( f , v ) (f,v) ( f , v ) 1 / D 1/D 1/ D
E Φ ≤ M ( b + w ) . \mathbb{E}\Phi\leq M(b+w). E Φ ≤ M ( b + w ) . For each t t t v ≠ v ′ v\ne v' v = v ′ a v ( t ) , a v ′ ( t ) a_v(t),a_{v'}(t) a v ( t ) , a v ′ ( t ) [ r ] [r] [ r ]
E Ξ = Q n ( 1 − 1 / r ) ≤ Q n . \mathbb{E}\Xi=Qn(1-1/r)\leq Qn. E Ξ = Q n ( 1 − 1/ r ) ≤ Q n . To obtain the labels deterministically, we fix the a v a_v a v d f d_f d f I f I_f I f
Φ M ( b + w ) + Ξ Q n . \frac{\Phi}{M(b+w)}+\frac{\Xi}{Qn}. M ( b + w ) Φ + Q n Ξ . Such a choice exists because the current expectation is the average over the possible values of the next a v a_v a v d f d_f d f
c u ≤ M D + 2 M ( b + w ) , n t , g ≤ n r + 2 Q n . c_u\leq\frac{M}{D}+\sqrt{2M(b+w)},\qquad n_{t,g}\leq\frac{n}{r}+\sqrt{2Qn}. c u ≤ D M + 2 M ( b + w ) , n t , g ≤ r n + 2 Q n . The two inequalities in (50) make the square-root terms at most M / D M/D M / D n / r n/r n / r poly ( n , M , ∣ F ∣ , D , Q , r , b , w ) \operatorname{poly}(n,M,|\mathcal{F}|,D,Q,r,b,w) poly ( n , M , ∣ F ∣ , D , Q , r , b , w ) K = r Q K=r^Q K = r Q D = K 2 D=K^2 D = K 2 Φ \Phi Φ a v , d f , a v ′ , d f ′ a_v,d_f,a_{v'},d_{f'} a v , d f , a v ′ , d f ′ Ξ \Xi Ξ a v , a v ′ a_v,a_{v'} a v , a v ′ K 4 = D 2 K^4=D^2 K 4 = D 2 K 4 K^4 K 4 M 2 + Q n 2 M^2+Qn^2 M 2 + Q n 2 M ( b + w ) M(b+w) M ( b + w ) Q n Qn Q n
O ( log ( 2 D ) + log ( 2 M ) + log ( 2 n ) + log ( 2 Q ) + log ( 2 + b + w ) ) O(\log(2D)+\log(2M)+\log(2n)+\log(2Q)+\log(2+b+w)) O ( log ( 2 D ) + log ( 2 M ) + log ( 2 n ) + log ( 2 Q ) + log ( 2 + b + w )) bits. There are at most n + ∣ F ∣ n+|\mathcal{F}| n + ∣ F ∣ K K K K 4 = D 2 K^4=D^2 K 4 = D 2 poly ( n , M , ∣ F ∣ , D , Q , r , b , w ) \operatorname{poly}(n,M,|\mathcal{F}|,D,Q,r,b,w) poly ( n , M , ∣ F ∣ , D , Q , r , b , w ) □ \square □
Placing copies of the variables in the auxiliary arrays
We now transfer the auxiliary rows P ⃗ u ( T ) \vec{P}_u(T) P u ( T ) C B \mathcal{C}^{B} C B P ⃗ f , v ( T ) \vec{P}_{f,v}(T) P f , v ( T ) C t , g C_{t,g} C t , g Definition 3.1 . Assume (50) and apply Lemma 4.5 . With h = max u c u h=\max_u c_u h = max u c u h h h [ h ] × B [h]\times\mathcal{B} [ h ] × B ( f , v ) ∈ A (f,v)\in\mathcal{A} ( f , v ) ∈ A c ( f , v ) ∈ [ h ] c(f,v)\in[h] c ( f , v ) ∈ [ h ] ( c ( f , v ) , u f , v ) (c(f,v),u_{f,v}) ( c ( f , v ) , u f , v ) h h h
h D ≤ 2 M ≤ 2 b n . hD\le2M\le2bn. h D ≤ 2 M ≤ 2 bn . Define the linear map ι : Q n ⟶ Q [ h ] × B \iota:\mathbb{Q}^{n}\longrightarrow\mathbb{Q}^{[h]\times\mathcal{B}} ι : Q n ⟶ Q [ h ] × B
( ι x ) c , u = { x v , ( c , u ) = ( c ( f , v ) , u f , v ) for some ( f , v ) ∈ A , 0 , otherwise . (\iota x)_{c,u}=
\begin{cases}
x_v, & (c,u)=(c(f,v),u_{f,v})\text{ for some }(f,v)\in\mathcal{A},\\
0, & \text{otherwise}.
\end{cases} ( ι x ) c , u = { x v , 0 , ( c , u ) = ( c ( f , v ) , u f , v ) for some ( f , v ) ∈ A , otherwise . The occupied positions are distinct, so this definition is unambiguous. Different positions can contain the same x v x_v x v c ∈ [ h ] c\in[h] c ∈ [ h ]
π c : Q [ h ] × B ⟶ Q B , ( π c η ) u = η c , u . \pi_c:\mathbb{Q}^{[h]\times\mathcal{B}}\longrightarrow\mathbb{Q}^{\mathcal{B}},\qquad
(\pi_c\eta)_u=\eta_{c,u}. π c : Q [ h ] × B ⟶ Q B , ( π c η ) u = η c , u . Now put
P ⃗ f , v ( T ) = P ⃗ u f , v ( T ) π c ( f , v ) , v , \vec{P}_{f,v}(T)=\vec{P}_{u_{f,v}}(T)\pi_{c(f,v),v}, P f , v ( T ) = P u f , v ( T ) π c ( f , v ) , v , S t , g = { v ∈ [ n ] : a v ( t ) = g } , σ f ( t ) = d f ( t ) . (51) S_{t,g}=\{v\in[n]:a_v(t)=g\},\qquad\sigma_f(t)=d_f(t). \tag*{(51)} S t , g = { v ∈ [ n ] : a v ( t ) = g } , σ f ( t ) = d f ( t ) . (51) The row P ⃗ f , v ( T ) \vec{P}_{f,v}(T) P f , v ( T ) x ∈ Q n x\in\mathbb{Q}^{n} x ∈ Q n T = 0 T=0 T = 0 ( c ( f , v ) , u f , v ) (c(f,v),u_{f,v}) ( c ( f , v ) , u f , v ) x v x_v x v P ⃗ f , v ( 0 ) = δ v \vec{P}_{f,v}(0)=\delta_v P f , v ( 0 ) = δ v R − 1 R-1 R − 1 Lemma 4.5 gives ∣ S t , g ∣ ≤ 2 n / r |S_{t,g}|\le2n/r ∣ S t , g ∣ ≤ 2 n / r
Define the measurement matrices by
C ~ = diag ( C B , … , C B ⏟ h copies ) , C ^ = C ~ ι , C t , g = C ^ ∣ S t , g . (52) \widetilde{C}=\operatorname{diag}(\underbrace{\mathcal{C}^{B},\ldots,\mathcal{C}^{B}}_{h\ \text{copies}}),\qquad
\widehat{C}=\widetilde{C}_{\iota},\qquad
C_{t,g}=\widehat{C}\big|_{S_{t,g}}. \tag*{(52)} C = diag ( h copies C B , … , C B ) , C = C ι , C t , g = C S t , g . (52) The restriction keeps the columns in S t , g S_{t,g} S t , g C ^ x \widehat{C}x C x x x x C t , g C_{t,g} C t , g { 0 , 1 , − 1 } \{0,1,-1\} { 0 , 1 , − 1 } x v x_v x v C ~ \widetilde{C} C D D D 1 1 1 − 1 -1 − 1 ι \iota ι C t , g C_{t,g} C t , g D D D [ − D , D ] ∩ Z [-D,D]\cap\mathbb{Z} [ − D , D ] ∩ Z
If (45) holds, each C t , g C_{t,g} C t , g c t , g = h c B c_{t,g}=hc_B c t , g = h c B
∑ g = 1 r c t , g = r h c B ≤ r h ϵ D ≤ 2 r ϵ M ≤ 2 r ϵ b n . (53) \sum_{g=1}^{r}c_{t,g}=rhc_B\le rh\epsilon D\le2r\epsilon M\le2r\epsilon bn. \tag*{(53)} g = 1 ∑ r c t , g = r h c B ≤ r h ϵD ≤ 2 r ϵ M ≤ 2 r ϵ bn . (53) Setting ϵ = τ / ( r b ) \epsilon=\tau/(rb) ϵ = τ / ( r b ) 2 τ n 2\tau n 2 τ n
To verify separation, fix t ∈ [ Q ] t\in[Q] t ∈ [ Q ] g ∈ [ r ] g\in[r] g ∈ [ r ] ( f , v ) ∈ A (f,v)\in\mathcal{A} ( f , v ) ∈ A g ≠ σ f ( t ) g\ne\sigma_f(t) g = σ f ( t ) ξ ∈ Q S t , g \xi\in\mathbb{Q}^{S_{t,g}} ξ ∈ Q S t , g C t , g ξ = 0 ⃗ C_{t,g}\xi=\vec{0} C t , g ξ = 0 ξ ~ ∈ Q n \widetilde{\xi}\in\mathbb{Q}^{n} ξ ∈ Q n (52) gives C ^ ι ξ ~ = 0 ⃗ \widehat{C}\iota\widetilde{\xi}=\vec{0} C ι ξ = 0 c ∈ [ h ] c\in[h] c ∈ [ h ]
C B π c ι ξ ~ = 0 ⃗ . \mathcal{C}^{B}\pi_c\iota\widetilde{\xi}=\vec{0}. C B π c ι ξ = 0 . Every nonzero coordinate of π c ι ξ ~ \pi_c\iota\widetilde{\xi} π c ι ξ v ′ v' v ′ a v ′ ( t ) = g a_{v'}(t)=g a v ′ ( t ) = g a v ′ a_{v'} a v ′ S t , g B S^B_{t,g} S t , g B Lemma 4.3 , applied in copy c ( f , v ) c(f,v) c ( f , v ) d u f , v ( t ) = d f ( t ) = σ f ( t ) d_{u_{f,v}}(t)=d_f(t)=\sigma_f(t) d u f , v ( t ) = d f ( t ) = σ f ( t )
⟨ P ⃗ f , v ( t ) , ξ ~ ⟩ = ⟨ P ⃗ u f , v ( t ) , π c ( f , v ) ι ξ ~ ⟩ = 0. \langle\vec{P}_{f,v}(t),\widetilde{\xi}\rangle
=
\langle\vec{P}_{u_{f,v}}(t),\pi_{c(f,v)}\iota\widetilde{\xi}\rangle=0. ⟨ P f , v ( t ) , ξ ⟩ = ⟨ P u f , v ( t ) , π c ( f , v ) ι ξ ⟩ = 0. This is the kernel condition in Definition 3.1 . All inputs of factor f f f d f ( t ) d_f(t) d f ( t ) n n n
Coefficient bounds and the completed construction
The substitution rows P ⃗ f , v ( T ) ∈ Q [ T ] 1 × n \vec{P}_{f,v}(T) \in\mathbb{Q}[T]^{1\times n} P f , v ( T ) ∈ Q [ T ] 1 × n (51) have polynomial entries with rational coefficients. We now bound the bits needed to store those coefficients; these bounds will control exact arithmetic in § 5.4 . We first give a common denominator and coefficient bounds for the auxiliary rows P ⃗ u ( T ) \vec{P}_u(T) P u ( T ) (47) . Passing from P ⃗ u ( T ) \vec{P}_u(T) P u ( T ) P ⃗ f , v ( T ) \vec{P}_{f,v}(T) P f , v ( T ) (51) only adds coefficients in common columns or discards unused columns, so the same denominator and bound on the sum of absolute values apply.
Lemma 4.6. Let d d e n = ( R − 1 ) ! Q ! d_{\mathrm{den}}=(R-1)!Q! d den = ( R − 1 )! Q ! d d e n P ⃗ u ( T ) ∈ Z [ T ] 1 × D d_{\mathrm{den}}\vec{P}_u(T)\in\mathbb{Z}[T]^{1\times D} d den P u ( T ) ∈ Z [ T ] 1 × D u ∈ B u\in B u ∈ B d d e n P ⃗ u ( T ) d_{\mathrm{den}}\vec{P}_u(T) d den P u ( T ) 2 O ( Q log Q ) 2^{O(Q\log Q)} 2 O ( Q l o g Q ) d d e n P ⃗ f , v ( T ) d_{\mathrm{den}}\vec{P}_{f,v}(T) d den P f , v ( T ) M 0 M_0 M 0 C B C^B C B P ⃗ u ( T ) \vec{P}_u(T) P u ( T ) poly ( D , Q , r ) \operatorname{poly}(D,Q,r) poly ( D , Q , r )
Proof. For 0 ≤ j < R 0\le j<R 0 ≤ j < R f I ( W 0 ) f_I(W_0) f I ( W 0 )
( − 1 ) A 0 ( − 2 ) j ( W 0 − A 0 I j ) = ( − 1 ) A 0 + j j ! ∏ a = 0 j − 1 ( 2 W 0 − 2 ( A 0 + a ) I ) . (-1)^{A_0}(-2)^j\binom{W_0-A_0I}{j}
=
\frac{(-1)^{A_0+j}}{j!}\prod_{a=0}^{j-1}\bigl(2W_0-2(A_0+a)I\bigr). ( − 1 ) A 0 ( − 2 ) j ( j W 0 − A 0 I ) = j ! ( − 1 ) A 0 + j a = 0 ∏ j − 1 ( 2 W 0 − 2 ( A 0 + a ) I ) . Each matrix factor on the right is integral because 2 W 0 = Q I − ∑ s Π s 2W_0=QI-\sum_s\Pi_s 2 W 0 = Q I − ∑ s Π s j ! j! j ! ( R − 1 ) ! (R-1)! ( R − 1 )! V 0 V_0 V 0 ( R − 1 ) ! M 0 (R-1)!M_0 ( R − 1 )! M 0 φ u , y \varphi_{u,y} φ u , y ∏ s ∈ E u , y s \prod_{s\in E_{u,y}}s ∏ s ∈ E u , y s Q ! Q! Q ! (47) proves the asserted common denominator.
For the coefficient estimates, we use the matrix norm ∥ A ∥ ∞ = max i ∑ j ∣ A [ i , j ] ∣ \lVert A\rVert_\infty=\max_i\sum_j|A[i,j]| ∥ A ∥ ∞ = max i ∑ j ∣ A [ i , j ] ∣ ∥ ∑ j a j T j ∥ 1 = ∑ j ∣ a j ∣ \lVert\sum_j a_jT^j\rVert_1=\sum_j|a_j| ∥ ∑ j a j T j ∥ 1 = ∑ j ∣ a j ∣ ∥ W 0 ∥ ∞ ≤ Q \lVert W_0\rVert_\infty\le Q ∥ W 0 ∥ ∞ ≤ Q 0 ≤ A 0 + a ≤ Q 0\le A_0+a\le Q 0 ≤ A 0 + a ≤ Q 2 W 0 − 2 ( A 0 + a ) I 2W_0-2(A_0+a)I 2 W 0 − 2 ( A 0 + a ) I 4 Q 4Q 4 Q
∥ M 0 ∥ ∞ ≤ R ( 4 Q ) R − 1 , ∥ φ u , y ∥ 1 ≤ ∏ s ∈ E u , y ( 1 + 1 / s ) ≤ 2 Q . \lVert M_0\rVert_\infty\le R(4Q)^{R-1},\qquad
\lVert\varphi_{u,y}\rVert_1\le\prod_{s\in E_{u,y}}(1+1/s)\le2^Q. ∥ M 0 ∥ ∞ ≤ R ( 4 Q ) R − 1 , ∥ φ u , y ∥ 1 ≤ s ∈ E u , y ∏ ( 1 + 1/ s ) ≤ 2 Q . The sum of the absolute values of all coefficients in the row d d e n P ⃗ u d_{\mathrm{den}}\vec{P}_u d den P u
d d e n ( 1 + ( 1 + 2 Q ) R ( 4 Q ) R − 1 ) . d_{\mathrm{den}}\bigl(1+(1+2^Q)R(4Q)^{R-1}\bigr). d den ( 1 + ( 1 + 2 Q ) R ( 4 Q ) R − 1 ) . Its logarithm is O ( Q log Q ) O(Q\log Q) O ( Q log Q ) R < Q R<Q R < Q u = u f , v u=u_{f,v} u = u f , v d d e n P ⃗ u ( T ) d_{\mathrm{den}}\vec{P}_u(T) d den P u ( T ) π c ( f , v ) \pi_{c(f,v)} π c ( f , v ) (51) , the resulting row is d d e n P ⃗ f , v ( T ) d_{\mathrm{den}}\vec{P}_{f,v}(T) d den P f , v ( T )
Enumerating the eigenbasis in Lemma 4.1 , evaluating f I ( W 0 ) f_I(W_0) f I ( W 0 ) φ u , y \varphi_{u,y} φ u , y poly ( D , Q , r ) \operatorname{poly}(D,Q,r) poly ( D , Q , r ) O ( Q log Q ) O(Q\log Q) O ( Q log Q ) poly ( D , Q , r ) \operatorname{poly}(D,Q,r) poly ( D , Q , r )
The construction has three quantitative requirements. The two inequalities in (50) bound the number of auxiliary copies and the number of actual variables in each group. The tail bound with ϵ = τ / ( r b ) \epsilon=\tau/(rb) ϵ = τ / ( r b ) (53) . We collect the result for use in the counting algorithm.
Lemma 4.7. Let n , b , w ≥ 1 n,b,w\ge1 n , b , w ≥ 1 ( I f ) f ∈ F (I_f)_{f\in\mathcal{F}} ( I f ) f ∈ F [ n ] [n] [ n ] w w w v ∈ [ n ] v\in[n] v ∈ [ n ] b b b M = ∑ f ∣ I f ∣ M=\sum_f|I_f| M = ∑ f ∣ I f ∣ n ≤ M ≤ b n n\le M\le bn n ≤ M ≤ bn r ≥ 2 r\ge2 r ≥ 2 1 ≤ R < Q 1\le R<Q 1 ≤ R < Q 0 < τ ≤ 1 / 2 0<\tau\le1/2 0 < τ ≤ 1/2 D = r 2 Q D=r^{2Q} D = r 2 Q
2 ( b + w ) D 2 ≤ M , 2 Q r 2 ≤ n , 2(b+w)D^2\le M,\qquad2Qr^2\le n, 2 ( b + w ) D 2 ≤ M , 2 Q r 2 ≤ n , and an interval I = { A 0 , … , A 0 + R − 1 } ⊆ { 0 , … , Q } \mathcal{I}=\{A_0,\ldots,A_0+R-1\}\subseteq\{0,\ldots,Q\} I = { A 0 , … , A 0 + R − 1 } ⊆ { 0 , … , Q }
Pr [ W ∉ I ] ≤ τ r b , W ∼ Bin ( Q , r − 1 2 r ) . \Pr[W\notin\mathcal{I}]\le\frac{\tau}{rb},\qquad W\sim\operatorname{Bin}\left(Q,\frac{r-1}{2r}\right). Pr [ W ∈ / I ] ≤ r b τ , W ∼ Bin ( Q , 2 r r − 1 ) . One can deterministically construct all the data in Definition 3.1 in poly ( n , M , ∣ F ∣ , D , Q , r , b , w ) \operatorname{poly}(n, M, |\mathcal{F}|, D, Q, r, b, w) poly ( n , M , ∣ F ∣ , D , Q , r , b , w )
Proof. Lemma 4.1 constructs C B C^{B} C B c B ≤ τ D / ( r b ) c_{B} \le\tau D/(rb) c B ≤ τ D / ( r b ) h D ≤ 2 M hD \le2M h D ≤ 2 M ∣ S t , g ∣ ≤ 2 n / r |S_{t,g}| \le2n/r ∣ S t , g ∣ ≤ 2 n / r (51) and (52) then give rows with P ⃗ f , v ( 0 ) = δ v \vec{P}_{f,v}(0)=\delta_{v} P f , v ( 0 ) = δ v R − 1 ≤ R R-1\le R R − 1 ≤ R C t , g C_{t,g} C t , g D D D (53) gives ∑ g = 1 r c t , g ≤ 2 τ n \sum_{g=1}^{r}c_{t,g}\le2\tau n ∑ g = 1 r c t , g ≤ 2 τ n t ∈ [ Q ] t\in[Q] t ∈ [ Q ] poly ( n , M , ∣ F ∣ , D , Q , r , b , w ) \operatorname{poly}(n, M, |\mathcal{F}|, D, Q, r, b, w) poly ( n , M , ∣ F ∣ , D , Q , r , b , w ) poly ( n , M , ∣ F ∣ , D , Q , r , b , w ) \operatorname{poly}(n, M, |\mathcal{F}|, D, Q, r, b, w) poly ( n , M , ∣ F ∣ , D , Q , r , b , w )
Choosing the parameters
In the application, both the largest scope w w w b b b log n \log n log n
R Q log 2 Q R , τ log ( 2 D + 1 ) , 1 r . \frac{R}{Q}\log\frac{2Q}{R},\qquad\tau\log(2D+1),\qquad\frac{1}{r}. Q R log R 2 Q , τ log ( 2 D + 1 ) , r 1 . They account respectively for recovering Taylor coefficients, enumerating measurement values, and enumerating assignments in one group. The following choices bound all three at once.
Lemma 4.8. Fix an integer β ≥ 1 \beta\ge1 β ≥ 1 n ≥ 20 n\ge20 n ≥ 20 λ = ⌈ log 2 n ⌉ \lambda=\lceil\log_{2}n\rceil λ = ⌈ log 2 n ⌉ ν = ⌈ log 2 λ ⌉ \nu=\lceil\log_{2}\lambda\rceil ν = ⌈ log 2 λ ⌉ b , w ≤ λ β b,w\le\lambda^{\beta} b , w ≤ λ β n ≤ M ≤ b n n\le M\le bn n ≤ M ≤ bn n n n
r = ⌈ λ ν 2 ⌉ , Q = ⌈ λ 128 ν ⌉ , R = ( β + 3 ) ⌈ λ ⌉ , (54) r=\left\lceil\frac{\sqrt{\lambda}}{\nu^{2}}\right\rceil,\qquad
Q=\left\lceil\frac{\lambda}{128\nu}\right\rceil,\qquad
R=(\beta+3)\lceil\sqrt{\lambda}\rceil, \tag*{(54)} r = ⌈ ν 2 λ ⌉ , Q = ⌈ 128 ν λ ⌉ , R = ( β + 3 ) ⌈ λ ⌉ , (54) τ = 1 r λ , ϵ = 1 b r 2 λ , D = r 2 Q , \tau=\frac{1}{r\lambda},\qquad
\epsilon=\frac{1}{br^{2}\lambda},\qquad
D=r^{2Q}, τ = r λ 1 , ϵ = b r 2 λ 1 , D = r 2 Q , together with the interval specified in the proof, satisfy all numerical hypotheses of Lemma 4.7. They also satisfy
R ≥ 32 , 6 R ≤ Q , D ≤ 2 λ / 128 , R\ge32,\qquad6R\le Q,\qquad D\le2^{\lambda/128}, R ≥ 32 , 6 R ≤ Q , D ≤ 2 λ /128 , and
R Q log 2 Q R + τ log ( 2 D + 1 ) + 1 r = O β ( ν 2 λ ) . (55) \frac{R}{Q}\log\frac{2Q}{R}+\tau\log(2D+1)+\frac{1}{r}
=O_{\beta}\left(\frac{\nu^{2}}{\sqrt{\lambda}}\right). \tag*{(55)} Q R log R 2 Q + τ log ( 2 D + 1 ) + r 1 = O β ( λ ν 2 ) . (55) Given the numerical values n , M , b , w n,M,b,w n , M , b , w poly β ( log n ) \operatorname{poly}_{\beta}(\log n) poly β ( log n ) β \beta β n n n
Proof. For sufficiently large n n n
r = Θ ( λ ν 2 ) , Q = Θ ( λ ν ) , R = Θ β ( λ ) . r=\Theta\left(\frac{\sqrt{\lambda}}{\nu^{2}}\right),\qquad
Q=\Theta\left(\frac{\lambda}{\nu}\right),\qquad
R=\Theta_{\beta}(\sqrt{\lambda}). r = Θ ( ν 2 λ ) , Q = Θ ( ν λ ) , R = Θ β ( λ ) . In particular, eventually r ≥ 2 r\ge2 r ≥ 2 R ≥ 32 R\ge32 R ≥ 32 6 R ≤ Q 6R\le Q 6 R ≤ Q log 2 r ≤ ν / 2 \log_{2}r\le\nu/2 log 2 r ≤ ν /2
log 2 D = 2 Q log 2 r ≤ 2 λ 128 ν ν 2 = λ 128 . \log_{2}D=2Q\log_{2}r\le2\frac{\lambda}{128\nu}\frac{\nu}{2}=\frac{\lambda}{128}. log 2 D = 2 Q log 2 r ≤ 2 128 ν λ 2 ν = 128 λ . The two balance inequalities follow for sufficiently large n n n
2 ( b + w ) D 2 ≤ 4 λ β 2 λ / 64 ≤ n ≤ M , 2 Q r 2 ≤ n . 2(b+w)D^{2}\le4\lambda^{\beta}2^{\lambda/64}\le n\le M,\qquad2Qr^{2}\le n. 2 ( b + w ) D 2 ≤ 4 λ β 2 λ /64 ≤ n ≤ M , 2 Q r 2 ≤ n . For the second, Q , r Q,r Q , r λ \lambda λ n > 2 λ − 1 n>2^{\lambda-1} n > 2 λ − 1
Put
μ = Q ( r − 1 ) 2 r , A 0 = ⌈ μ − R − 1 2 ⌉ , I = { A 0 , … , A 0 + R − 1 } . \mu=\frac{Q(r-1)}{2r},\qquad
A_{0}=\left\lceil\mu-\frac{R-1}{2}\right\rceil,\qquad
\mathcal{I}=\{A_{0},\ldots,A_{0}+R-1\}. μ = 2 r Q ( r − 1 ) , A 0 = ⌈ μ − 2 R − 1 ⌉ , I = { A 0 , … , A 0 + R − 1 } . Here Q / 4 ≤ μ ≤ Q / 2 Q/4 \le\mu\le Q/2 Q /4 ≤ μ ≤ Q /2 R ≤ Q / 6 R \le Q/6 R ≤ Q /6 { 0 , … , Q } \{0,\ldots,Q\} { 0 , … , Q } μ \mu μ R ≥ 32 R \ge32 R ≥ 32 R / 4 R/4 R /4 μ \mu μ W ∼ Bin ( Q , ( r − 1 ) / ( 2 r ) ) W \sim\operatorname{Bin}(Q,(r-1)/(2r)) W ∼ Bin ( Q , ( r − 1 ) / ( 2 r ))
Pr [ W ∉ I ] ≤ 2 e − R 2 / ( 8 Q ) ≤ 2 e − 16 ( β + 3 ) 2 ν ≤ 2 − ( β + 2 ) ν ≤ λ − ( β + 2 ) ≤ 1 b r 2 λ . \Pr[W \notin\mathcal{I}] \le2e^{-R^2/(8Q)} \le2e^{-16(\beta+3)^2\nu} \le2^{-(\beta+2)\nu} \le\lambda^{-(\beta+2)} \le\frac{1}{br^2\lambda}. Pr [ W ∈ / I ] ≤ 2 e − R 2 / ( 8 Q ) ≤ 2 e − 16 ( β + 3 ) 2 ν ≤ 2 − ( β + 2 ) ν ≤ λ − ( β + 2 ) ≤ b r 2 λ 1 . The last inequality uses b ≤ λ β b \le\lambda^\beta b ≤ λ β r 2 ≤ λ r^2 \le\lambda r 2 ≤ λ (45) with ϵ = τ / ( r b ) \epsilon=\tau/(rb) ϵ = τ / ( r b ) 0 < τ ≤ 1 / 2 0<\tau\le1/2 0 < τ ≤ 1/2 r ≥ 2 r\ge2 r ≥ 2
Finally,
R Q = O β ( ν / λ ) , log ( 2 Q / R ) = O ( ν ) , τ log ( 2 D + 1 ) = O ( 1 / r ) . \frac{R}{Q}=O_\beta(\nu/\sqrt{\lambda}),\qquad\log(2Q/R)=O(\nu),\qquad\tau\log(2D+1)=O(1/r). Q R = O β ( ν / λ ) , log ( 2 Q / R ) = O ( ν ) , τ log ( 2 D + 1 ) = O ( 1/ r ) . Together with 1 / r = O ( ν 2 / λ ) 1/r=O(\nu^2/\sqrt{\lambda}) 1/ r = O ( ν 2 / λ ) (55) .
We finish by showing that the parameters in (54) can be constructed exactly within the stated bit bound. The algorithm finds λ = ⌈ log 2 n ⌉ \lambda=\lceil\log_2 n\rceil λ = ⌈ log 2 n ⌉ n ≤ 2 λ n\le2^\lambda n ≤ 2 λ ν = ⌈ log 2 λ ⌉ \nu=\lceil\log_2\lambda\rceil ν = ⌈ log 2 λ ⌉ r r r r 2 ν 4 ≤ λ r^2\nu^4\le\lambda r 2 ν 4 ≤ λ Q Q Q ⌈ λ ⌉ \lceil\sqrt{\lambda}\rceil ⌈ λ ⌉ R R R r < 2 r<2 r < 2 Q ≤ R Q\le R Q ≤ R τ , ϵ , D \tau,\epsilon,D τ , ϵ , D I \mathcal{I} I I ⊆ { 0 , … , Q } \mathcal{I}\subseteq\{0,\ldots,Q\} I ⊆ { 0 , … , Q } R ≥ 32 R\ge32 R ≥ 32 6 R ≤ Q 6R\le Q 6 R ≤ Q (45) has denominator ( 2 r ) Q (2r)^Q ( 2 r ) Q poly β ( log n ) \operatorname{poly}_\beta(\log n) poly β ( log n ) n n n
Counting common zeros of sparse polynomial equations
We now prove Theorem 2.2. The input is a system P \mathcal{P} P m t e s t ≤ c n m_{\mathrm{test}}\le cn m test ≤ c n P i ( x ) = 0 P_i(x)=0 P i ( x ) = 0 F 2 \mathbb{F}_2 F 2 n n n P i P_i P i ( log n ) c (\log n)^c ( log n ) c k k k k , c k,c k , c
N ( P ) = # { x ∈ { 0 , 1 } n : P i ( x ) = 0 for every i } . (56) N(\mathcal{P})=\#\{x\in\{0,1\}^n:P_i(x)=0\text{ for every }i\}. \tag*{(56)} N ( P ) = # { x ∈ { 0 , 1 } n : P i ( x ) = 0 for every i } . (56) First we build a product with one factor per equation whose value on the Boolean cube records the common-zero indicator. The difficulty is keeping its total input degree linear in n n n § 3 to sum the product. § 6 uses the resulting algorithm for arbitrary supplied systems, including systems that still accept nonsatisfying assignments of a formula.
An integer polynomial for one equation
On Boolean inputs we may replace each positive power x v j x_v^j x v j x v x_v x v M i \mathcal{M}_i M i P i P_i P i
L i ( X ⃗ ) = ∑ M ∈ M i ∏ v ∈ M X v . L_i(\vec{X})=\sum_{M\in\mathcal{M}_i}\prod_{v\in M}X_v. L i ( X ) = M ∈ M i ∑ v ∈ M ∏ X v . The empty monomial is 1 1 1 0 0 0 L i ( x ) L_i(x) L i ( x ) P i ( x ) = 0 P_i(x)=0 P i ( x ) = 0 L i ( x ) L_i(x) L i ( x ) I i = ⋃ M ∈ M i M I_i=\bigcup_{M\in\mathcal{M}_i}M I i = ⋃ M ∈ M i M L i L_i L i k ( log n ) c k(\log n)^c k ( log n ) c k k k
Set ℓ = n + 1 \ell=n+1 ℓ = n + 1 0 ≤ N ( P ) ≤ 2 n < 2 ℓ 0\le N(\mathcal{P})\le2^n<2^\ell 0 ≤ N ( P ) ≤ 2 n < 2 ℓ 2 ℓ 2^\ell 2 ℓ m t e s t m_{\mathrm{test}} m test
Put h ( Y ) = Y ( Y − 1 ) h(Y)=Y(Y-1) h ( Y ) = Y ( Y − 1 ) Y Y Y h h h h h h
Lemma 5.1. For every integer ℓ ≥ 1 \ell\ge1 ℓ ≥ 1 e 0 , … , e ℓ − 1 ∈ Z [ Y ] e_0,\ldots,e_{\ell-1} \in\mathbb{Z}[Y] e 0 , … , e ℓ − 1 ∈ Z [ Y ]
p ℓ ( Y ) = ∑ j = 0 ℓ − 1 e j ( Y ) h ( Y ) j , p ℓ ( y ) ≡ 1 { 2 ∣ y } ( m o d 2 ℓ ) ( y ∈ Z ) . (57) p_{\ell}(Y)=\sum_{j=0}^{\ell-1}e_j(Y)h(Y)^j,\qquad p_{\ell}(y)\equiv1_{\{2\mid y\}}\pmod{2^\ell}\quad(y\in\mathbb{Z}). \tag*{(57)} p ℓ ( Y ) = j = 0 ∑ ℓ − 1 e j ( Y ) h ( Y ) j , p ℓ ( y ) ≡ 1 { 2 ∣ y } ( mod 2 ℓ ) ( y ∈ Z ) . (57) The coefficients of the e j e_j e j poly ( ℓ ) \operatorname{poly}(\ell) poly ( ℓ ) poly ( ℓ ) \operatorname{poly}(\ell) poly ( ℓ ) poly ( ℓ ) \operatorname{poly}(\ell) poly ( ℓ )
Proof. Let F ℓ F_\ell F ℓ [12 Section 2.3] . Chan and Williams apply F ℓ F_\ell F ℓ [22 Theorem 2.5 and proof of Theorem 1.6] . We need the indicator of even parity, so we take its complement:
p ℓ ( Y ) = 1 − F ℓ ( Y ) = ( 1 − Y ) ℓ ∑ j = 0 ℓ − 1 ( ℓ + j − 1 j ) Y j . (58) p_{\ell}(Y)=1-F_{\ell}(Y)=(1-Y)^\ell\sum_{j=0}^{\ell-1}\binom{\ell+j-1}{j}Y^j. \tag*{(58)} p ℓ ( Y ) = 1 − F ℓ ( Y ) = ( 1 − Y ) ℓ j = 0 ∑ ℓ − 1 ( j ℓ + j − 1 ) Y j . (58) The finite sum is the inverse of ( 1 − Y ) ℓ (1-Y)^\ell ( 1 − Y ) ℓ Y ℓ Y^\ell Y ℓ Y j Y^j Y j ( 1 + Y + Y 2 + ⋯ ) ℓ (1+Y+Y^2+\cdots)^\ell ( 1 + Y + Y 2 + ⋯ ) ℓ j j j ℓ \ell ℓ ( ℓ + j − 1 j ) \binom{\ell+j-1}{j} ( j ℓ + j − 1 )
p ℓ ( Y ) ≡ 1 ( m o d Y ℓ ) , p ℓ ( Y ) ≡ 0 ( m o d ( 1 − Y ) ℓ ) . p_{\ell}(Y)\equiv1\pmod{Y^\ell},\qquad p_{\ell}(Y)\equiv0\pmod{(1-Y)^\ell}. p ℓ ( Y ) ≡ 1 ( mod Y ℓ ) , p ℓ ( Y ) ≡ 0 ( mod ( 1 − Y ) ℓ ) . For even y y y p ℓ ( y ) ≡ 1 ( m o d 2 ℓ ) p_{\ell}(y)\equiv1\pmod{2^\ell} p ℓ ( y ) ≡ 1 ( mod 2 ℓ ) y y y p ℓ ( y ) ≡ 0 ( m o d 2 ℓ ) p_{\ell}(y)\equiv0\pmod{2^\ell} p ℓ ( y ) ≡ 0 ( mod 2 ℓ )
The polynomial p ℓ p_{\ell} p ℓ 2 ℓ 2\ell 2 ℓ h = Y 2 − Y h=Y^2-Y h = Y 2 − Y ∑ j < ℓ e j h j \sum_{j<\ell}e_jh^j ∑ j < ℓ e j h j e j e_j e j poly ( ℓ ) \operatorname{poly}(\ell) poly ( ℓ )
Keeping the product degree linear
The full product ∏ i = 1 m t e s t p ℓ ( L i ( X ⃗ ) ) \prod_{i=1}^{m_{\mathrm{test}}}p_{\ell}(L_i(\vec{X})) ∏ i = 1 m test p ℓ ( L i ( X )) k m t e s t ( 2 ℓ − 1 ) km_{\mathrm{test}}(2\ell-1) k m test ( 2 ℓ − 1 ) h ( y ) h(y) h ( y ) ℓ \ell ℓ 2 ℓ 2^\ell 2 ℓ h h h ℓ \ell ℓ V V V V V V § 3.1 :
A ℓ = Q [ V ] / ( V ℓ ) . \mathcal{A}_{\ell}=\mathbb{Q}[V]/(V^\ell). A ℓ = Q [ V ] / ( V ℓ ) . In this algebra every multiplication discards powers V s V^s V s s ≥ ℓ s\ge\ell s ≥ ℓ
S ℓ ( ∑ s = 0 ℓ − 1 b s V s ) = ∑ s = 0 ℓ − 1 b s . S_{\ell}\left(\sum_{s=0}^{\ell-1}b_sV^s\right)=\sum_{s=0}^{\ell-1}b_s. S ℓ ( s = 0 ∑ ℓ − 1 b s V s ) = s = 0 ∑ ℓ − 1 b s . This map need not preserve products, so it is applied only after the factors have been multiplied. We use the same notation for polynomial coefficients b s b_s b s
Lemma 5.2. Let m t e s t ≥ 0 m_{\mathrm{test}}\ge0 m test ≥ 0 ℓ ≥ 1 \ell\ge1 ℓ ≥ 1 y 1 , … , y m t e s t y_1,\ldots,y_{m_{\mathrm{test}}} y 1 , … , y m test
B i ( V ) = ∑ j = 0 ℓ − 1 e j ( y i ) h ( y i ) j V j . B_i(V)=\sum_{j=0}^{\ell-1}e_j(y_i)h(y_i)^jV^j. B i ( V ) = j = 0 ∑ ℓ − 1 e j ( y i ) h ( y i ) j V j . Then
S ℓ ( ∏ i = 1 m t e s t B i ( V ) m o d V ℓ ) ≡ ∏ i = 1 m t e s t 1 { 2 ∣ y i } ( m o d 2 ℓ ) . S_{\ell}\left(\prod_{i=1}^{m_{\mathrm{test}}}B_i(V)\bmod V^\ell\right)\equiv\prod_{i=1}^{m_{\mathrm{test}}}1_{\{2\mid y_i\}}\pmod{2^\ell}. S ℓ ( i = 1 ∏ m test B i ( V ) mod V ℓ ) ≡ i = 1 ∏ m test 1 { 2 ∣ y i } ( mod 2 ℓ ) . For any nonnegative integer d ∗ d_* d ∗ y i y_i y i Q \mathbb{Q} Q d ∗ d_* d ∗ V s V^s V s d ∗ ( m t e s t + 2 ( ℓ − 1 ) ) d_*(m_{\mathrm{test}}+2(\ell-1)) d ∗ ( m test + 2 ( ℓ − 1 ))
Proof. Before truncation, a term contributing to the coefficient of V s V^s V s
∏ i = 1 m t e s t e j i ( y i ) h ( y i ) j i , ∑ i = 1 m t e s t j i = s . \prod_{i=1}^{m_{\mathrm{test}}}e_{j_i}(y_i)h(y_i)^{j_i},\qquad\sum_{i=1}^{m_{\mathrm{test}}}j_i=s. i = 1 ∏ m test e j i ( y i ) h ( y i ) j i , i = 1 ∑ m test j i = s . At integer arguments it is divisible by 2 s 2^s 2 s s ≥ ℓ s\geq\ell s ≥ ℓ 2 ℓ 2^\ell 2 ℓ ∏ i p ℓ ( y i ) \prod_i p_\ell(y_i) ∏ i p ℓ ( y i ) ∑ i j i ≤ ℓ − 1 \sum_i j_i\leq\ell-1 ∑ i j i ≤ ℓ − 1 deg e j i ≤ 1 \deg e_{j_i}\leq1 deg e j i ≤ 1 deg h = 2 \deg h=2 deg h = 2
∑ i = 1 m t e s t d ∗ ( 1 + 2 j i ) ≤ d ∗ ( m t e s t + 2 ( ℓ − 1 ) ) . \sum_{i=1}^{m_{\mathrm{test}}}d_*(1+2j_i)\leq d_*(m_{\mathrm{test}}+2(\ell-1)). i = 1 ∑ m test d ∗ ( 1 + 2 j i ) ≤ d ∗ ( m test + 2 ( ℓ − 1 )) . The bound holds coefficient by coefficient, as well as for their sum. □
To see why the truncation uses the total order across all equations, take ℓ = 2 \ell=2 ℓ = 2
p 2 ( Y ) = 1 − 3 Y 2 + 2 Y 3 = ( 1 − Y ) + ( 2 Y − 1 ) h ( Y ) . p_2(Y)=1-3Y^2+2Y^3=(1-Y)+(2Y-1)h(Y). p 2 ( Y ) = 1 − 3 Y 2 + 2 Y 3 = ( 1 − Y ) + ( 2 Y − 1 ) h ( Y ) . For two equations, multiplication modulo V 2 V^2 V 2 S 2 S_2 S 2
( 1 − y 1 ) ( 1 − y 2 ) + ( 2 y 1 − 1 ) h ( y 1 ) ( 1 − y 2 ) + ( 1 − y 1 ) ( 2 y 2 − 1 ) h ( y 2 ) . (1-y_1)(1-y_2)+(2y_1-1)h(y_1)(1-y_2)+(1-y_1)(2y_2-1)h(y_2). ( 1 − y 1 ) ( 1 − y 2 ) + ( 2 y 1 − 1 ) h ( y 1 ) ( 1 − y 2 ) + ( 1 − y 1 ) ( 2 y 2 − 1 ) h ( y 2 ) . Its degree is at most four. The omitted term ( 2 y 1 − 1 ) ( 2 y 2 − 1 ) h ( y 1 ) h ( y 2 ) (2y_1-1)(2y_2-1)h(y_1)h(y_2) ( 2 y 1 − 1 ) ( 2 y 2 − 1 ) h ( y 1 ) h ( y 2 )
Counting with supplied replacement forms
For equation i i i § 3 with the factor
B i ( ( X v ) v ∈ I i , V ) = ∑ j = 0 ℓ − 1 e j ( L i ( X ⃗ ) ) h ( L i ( X ⃗ ) ) j V j in A ℓ [ ( X v ) v ∈ I i ] . (59) B_i((X_v)_{v\in I_i},V)=\sum_{j=0}^{\ell-1}e_j(L_i(\vec{X}))h(L_i(\vec{X}))^jV^j\quad\text{in }A_\ell[(X_v)_{v\in I_i}]. \tag*{(59)} B i (( X v ) v ∈ I i , V ) = j = 0 ∑ ℓ − 1 e j ( L i ( X )) h ( L i ( X ) ) j V j in A ℓ [( X v ) v ∈ I i ] . (59) Within a factor, every occurrence of X v X_v X v X v X_v X v L i L_i L i k k k A ℓ A_\ell A ℓ d = k ( m t e s t + 2 ( ℓ − 1 ) ) d=k(m_{\mathrm{test}}+2(\ell-1)) d = k ( m test + 2 ( ℓ − 1 )) (23) of the common routine. Adding factors equal to one with scope { v } \{v\} { v }
Let H ( T ) ∈ A ℓ [ T ] H(T)\in A_\ell[T] H ( T ) ∈ A ℓ [ T ] (24) for these factors. Its V V V T = 0 T=0 T = 0
deg T H ≤ d H : = k R ( m t e s t + 2 ( ℓ − 1 ) ) , S ℓ ( H ( 0 ) ) ≡ N ( P ) ( m o d 2 ℓ ) . (60) \deg_T H\leq d_H:=kR(m_{\mathrm{test}}+2(\ell-1)),\qquad S_\ell(H(0))\equiv N(\mathcal{P})\pmod{2^\ell}. \tag*{(60)} deg T H ≤ d H := k R ( m test + 2 ( ℓ − 1 )) , S ℓ ( H ( 0 )) ≡ N ( P ) ( mod 2 ℓ ) . (60) Consequently the least nonnegative residue of S ℓ ( H ( 0 ) ) S_\ell(H(0)) S ℓ ( H ( 0 )) T = 0 T=0 T = 0
Cost of evaluating the factors
Proposition 5.3. Let a list of degree-at-most-k k k n n n Definition 3.1 for their factors be given. Let S ≥ 1 S\geq1 S ≥ 1 ℓ \ell ℓ 2 ℓ > 2 n 2^\ell>2^n 2 ℓ > 2 n d H d_H d H (60) and q , J q,J q , J (26) , N ( P ) N(\mathcal{P}) N ( P )
poly ( n , m t e s t , ℓ , D , Q , R , J , r , k , S ) C q ( 2 D + 1 ) 2 r n 2 n / r \operatorname{poly}(n,m_{\mathrm{test}},\ell,D,Q,R,J,r,k,S)\,C_q(2D+1)^{2^r n}2^{n/r} poly ( n , m test , ℓ , D , Q , R , J , r , k , S ) C q ( 2 D + 1 ) 2 r n 2 n / r bit operations and poly ( n , m t e s t , ℓ , D , Q , R , J , r , k , S ) \operatorname{poly}(n,m_{\mathrm{test}},\ell,D,Q,R,J,r,k,S) poly ( n , m test , ℓ , D , Q , R , J , r , k , S )
Proof. Apply Proposition 3.4 with coefficients in A ℓ \mathcal{A}_{\ell} A ℓ t t t A ℓ [ z ] / ( z ℓ ) \mathcal{A}_{\ell}[z]/(z^{\ell}) A ℓ [ z ] / ( z ℓ ) J ℓ J\ell J ℓ i i i X ⃗ = 1 U \vec{X} = \mathbf{1}_{U} X = 1 U (28) , first evaluate each listed monomial by multiplying its at most k k k L i L_i L i (59) , successively multiplying by h ( L i ) h(L_i) h ( L i ) I i I_i I i e j e_j e j
For bit complexity, the rows have common denominator d d e n = ( R − 1 ) ! Q ! d_{\mathrm{den}}=(R-1)!Q! d den = ( R − 1 )! Q ! Δ = d d e n k \Delta=d_{\mathrm{den}}^k Δ = d den k Γ = Δ 2 ℓ − 1 \Gamma=\Delta^{2\ell-1} Γ = Δ 2 ℓ − 1 Ω = Γ m t e s t \Omega=\Gamma^{m_{\mathrm{test}}} Ω = Γ m test § 8 bounds the bit lengths in these sums by
B s u m = poly ( n , m t e s t , ℓ , k , Q , log ( D + 2 ) , log ( S + 2 ) ) . (61) B_{\mathrm{sum}}=\operatorname{poly}(n,m_{\mathrm{test}},\ell,k,Q,\log(D+2),\log(S+2)). \tag*{(61)} B sum = poly ( n , m test , ℓ , k , Q , log ( D + 2 ) , log ( S + 2 )) . (61) For each 0 ≤ s < ℓ 0\le s<\ell 0 ≤ s < ℓ [ V s ] H ( T ) [V^s]H(T) [ V s ] H ( T ) Q J QJ Q J (60) . The O ( Q J ℓ ) O(QJ\ell) O ( Q J ℓ )
Constructing the forms and completing the proof
Each sparse polynomial reads few variables, but one variable might occur in many equations. We first fix the values of the few variables with the largest numbers of occurrences. On each resulting branch we can construct the design required by Proposition 5.3.
Proof of Theorem 2.2 . Let P P P 2 n 2^n 2 n 1 = 0 1=0 1 = 0 0 = 0 0=0 0 = 0 { v } \{v\} { v } v v v
M = ∑ f ∈ F ∣ I f ∣ M=\sum_{f\in\mathcal{F}} |I_f| M = f ∈ F ∑ ∣ I f ∣ for the total number of factor–variable incidences. The input bounds imply M = n ( log n ) O k , c ( 1 ) M=n(\log n)^{O_{k,c}(1)} M = n ( log n ) O k , c ( 1 ) ( log n ) O k , c ( 1 ) (\log n)^{O_{k,c}(1)} ( log n ) O k , c ( 1 )
Put λ = ⌈ log 2 n ⌉ \lambda=\lceil\log_2 n\rceil λ = ⌈ log 2 n ⌉
b 0 = ⌈ M n ⌉ ⌈ λ ⌉ . b_0=\left\lceil\frac{M}{n}\right\rceil\left\lceil\sqrt{\lambda}\right\rceil. b 0 = ⌈ n M ⌉ ⌈ λ ⌉ . Let S ⊆ [ n ] S\subseteq[n] S ⊆ [ n ] b 0 b_0 b 0 b ∈ { 0 , 1 } S b\in\{0,1\}^{S} b ∈ { 0 , 1 } S x ∣ S = b x|_{S}=b x ∣ S = b [ n ] ∖ S [n]\setminus S [ n ] ∖ S ∣ S ∣ ≤ M / b 0 ≤ n / λ |S|\le M/b_0\le n/\sqrt{\lambda} ∣ S ∣ ≤ M / b 0 ≤ n / λ 2 n / λ 2^{n/\sqrt{\lambda}} 2 n / λ n ′ n' n ′
n ′ ≥ n − n λ ≥ n 2 ( n ≥ 20 ) . n'\ge n-\frac{n}{\sqrt{\lambda}}\ge\frac{n}{2}\qquad(n\ge20). n ′ ≥ n − λ n ≥ 2 n ( n ≥ 20 ) . Substitute the fixed bits in every P i P_i P i F 2 \mathbb{F}_2 F 2 N ( P ) \mathcal{N}(P) N ( P )
To apply § 4 , remove the fixed variables from the scopes and relabel the free variables by [ n ′ ] [n'] [ n ′ ] I f ′ ⊆ [ n ′ ] I'_f\subseteq[n'] I f ′ ⊆ [ n ′ ] M ′ = ∑ f ∣ I f ′ ∣ M'=\sum_f|I'_f| M ′ = ∑ f ∣ I f ′ ∣ b b b w = max f ∣ I f ′ ∣ w=\max_f|I'_f| w = max f ∣ I f ′ ∣ b ≤ b 0 b\le b_0 b ≤ b 0 b b b w w w log n ′ \log n' log n ′ M ′ ≥ n ′ M'\ge n' M ′ ≥ n ′
λ ′ = ⌈ log 2 n ′ ⌉ , ν ′ = ⌈ log 2 λ ′ ⌉ , \lambda'=\lceil\log_2 n'\rceil,\qquad\nu'=\lceil\log_2\lambda'\rceil, λ ′ = ⌈ log 2 n ′ ⌉ , ν ′ = ⌈ log 2 λ ′ ⌉ , and choose the smallest integer β ≥ 1 \beta\ge1 β ≥ 1 b , w ≤ ( λ ′ ) β b,w \le(\lambda')^\beta b , w ≤ ( λ ′ ) β k , c k,c k , c β \beta β n n n Lemma 4.8 to the restricted scopes ( I f ′ ) (I'_f) ( I f ′ ) [ n ′ ] [n'] [ n ′ ] M ′ M' M ′
r = ⌈ λ ′ ( ν ′ ) 2 ⌉ , Q = ⌈ λ ′ 128 ν ′ ⌉ , R = ( β + 3 ) ⌈ λ ′ ⌉ . r=\left\lceil\frac{\sqrt{\lambda'}}{(\nu')^2}\right\rceil,\qquad Q=\left\lceil\frac{\lambda'}{128\nu'}\right\rceil,\qquad R=(\beta+3)\lceil\sqrt{\lambda'}\rceil. r = ⌈ ( ν ′ ) 2 λ ′ ⌉ , Q = ⌈ 128 ν ′ λ ′ ⌉ , R = ( β + 3 ) ⌈ λ ′ ⌉ . If r < 2 r<2 r < 2 Q ≤ R Q\le R Q ≤ R τ \tau τ ϵ \epsilon ϵ I \mathcal{I} I
τ = 1 r λ ′ , ϵ = 1 b r 2 λ ′ , D = r 2 Q , \tau=\frac{1}{r\lambda'},\qquad\epsilon=\frac{1}{br^2\lambda'},\qquad D=r^{2Q}, τ = r λ ′ 1 , ϵ = b r 2 λ ′ 1 , D = r 2 Q , and carry out the remaining feasibility checks from that lemma. For all sufficiently large n n n Lemma 4.7 supplies the required rows and measurement matrices, with
r = Θ ( log n ′ ( log log n ′ ) 2 ) , Q = Θ ( log n ′ log log n ′ ) , R = O k , c ( log n ′ ) , D ≤ ( 2 n ′ ) 1 / 128 . r=\Theta\left(\frac{\sqrt{\log n'}}{(\log\log n')^2}\right),\qquad Q=\Theta\left(\frac{\log n'}{\log\log n'}\right),\qquad R=O_{k,c}(\sqrt{\log n'}),\qquad D\le(2n')^{1/128}. r = Θ ( ( log log n ′ ) 2 log n ′ ) , Q = Θ ( log log n ′ log n ′ ) , R = O k , c ( log n ′ ) , D ≤ ( 2 n ′ ) 1/128 . On every branch we may use the original precision ℓ = n + 1 \ell=n+1 ℓ = n + 1 2 ℓ 2^\ell 2 ℓ n ′ ≥ n / 2 n'\ge n/2 n ′ ≥ n /2 ℓ \ell ℓ m t e s t m_{\mathrm{test}} m test O c ( n ′ ) O_c(n') O c ( n ′ ) d H = O k , c ( n ′ R ) d_H=O_{k,c}(n'R) d H = O k , c ( n ′ R ) J = O k , c ( n ′ R / Q ) + 1 J=O_{k,c}(n'R/Q)+1 J = O k , c ( n ′ R / Q ) + 1 (37) , in dimension n ′ n' n ′
O k , c ( n ′ ( log log n ′ ) 2 log n ′ ) = O k , c ( n ( log log n ) 2 log n ) . O_{k,c}\left(\frac{n'(\log\log n')^2}{\sqrt{\log n'}}\right)=O_{k,c}\left(\frac{n(\log\log n)^2}{\sqrt{\log n}}\right). O k , c ( log n ′ n ′ ( log log n ′ ) 2 ) = O k , c ( log n n ( log log n ) 2 ) . All substitution data, coefficient arrays, and interpolation systems have total representation length n O k , c ( 1 ) n^{O_{k,c}(1)} n O k , c ( 1 ) 2 n / λ 2^{n/\sqrt{\lambda}} 2 n / λ
To implement the construction for every input size, the algorithm checks its numerical hypotheses using integer arithmetic. In addition to the preliminary guards above, it checks 2 ( b + w ) D 2 ≤ M ′ 2(b+w)D^2\le M' 2 ( b + w ) D 2 ≤ M ′ 2 Q r 2 ≤ n ′ 2Qr^2\le n' 2 Q r 2 ≤ n ′ Lemma 4.5 , the endpoints of I \mathcal{I} I R ≥ 32 R\ge32 R ≥ 32 6 R ≤ Q 6R\le Q 6 R ≤ Q (45) , which can be tested exactly. If any check fails, the algorithm abandons the branched computation and checks the original equations directly on all 2 n 2^n 2 n k , c k,c k , c Lemma 4.8 show that a failure can occur only for finitely many n n n
To combine the branch counts within the stated space bound, we process the branches one at a time. We initialize an integer accumulator Z = 0 Z=0 Z = 0 n O k , c ( 1 ) n^{O_{k,c}(1)} n O k , c ( 1 ) 0 ≤ Z ≤ 2 n 0\le Z\le2^n 0 ≤ Z ≤ 2 n N ( P ) N(P) N ( P )
Reducing #k-SAT to counting common zeros
Given a sparse k k k F F F B n Bn B n B B B n n n O ( n ) O(n) O ( n ) n n n B n Bn B n Theorem 2.2 both to choose these equations and to obtain the final count.
Let m m m v e ( x ) v_e(x) v e ( x ) e e e F 2 \mathbb{F}_2 F 2 v e v_e v e 1 + x v 1+x_v 1 + x v x v x_v x v v e = 0 v_e=0 v e = 0 v e = 1 v_e=1 v e = 1 deg v e ≤ k \deg v_e\le k deg v e ≤ k v e v_e v e 2 k 2^k 2 k v ( x ) = ( v 1 ( x ) , … , v m ( x ) ) v(x)=(v_1(x),\ldots,v_m(x)) v ( x ) = ( v 1 ( x ) , … , v m ( x )) A A A m m m i i i
P i ( x ) : = ∑ e : A i e = 1 v e ( x ) = 0 over F 2 . (62) P_i(x):=\sum_{e:A_{ie}=1}v_e(x)=0\quad\text{over }\mathbb{F}_2. \tag*{(62)} P i ( x ) := e : A i e = 1 ∑ v e ( x ) = 0 over F 2 . (62) Write P A \mathcal{P}_A P A N ( A ) = N ( P A ) N(A)=N(\mathcal{P}_A) N ( A ) = N ( P A ) N ( A ) = # F + D ( A ) N(A)=\#F+D(A) N ( A ) = # F + D ( A ) D ( A ) D(A) D ( A ) # F \#F # F N N N D D D
We represent a test row by a list of pairs ( e , c ) (e,c) ( e , c ) e ∈ [ m ] e\in[m] e ∈ [ m ] c ∈ { 0 , 1 } c\in\{0,1\} c ∈ { 0 , 1 } s s s k s ks k s
Initial tests that exclude small violation sets
If an assignment violates only one of many clauses, a test sampling a short list of clauses is unlikely to notice it. We therefore first construct a matrix A ( 0 ) A^{(0)} A ( 0 )
We use the standard edge-based Tanner code: bits are assigned to the edges of an expander, and the bits incident to each vertex must form a word of a prescribed local linear code (see Tanner [63 Section II] for the local constraints and Sipser and Spielman [62 Sections IV and VI] for the expander construction). The usual local-distance and expansion argument gives the needed lower bound on the weight of a nonzero codeword. The proof below constructs the graph and local code deterministically, with parameters that meet our bounds on the number and length of the tests.
Lemma 6.1. Let n n n m m m B B B m ≤ B n m\le Bn m ≤ B n
μ = ⌊ log 2 m ⌋ + 1 , P = 20000 B 2 μ 2 , D 0 = 320000 B 2 μ 3 . \mu=\lfloor\log_{2}m\rfloor+1,\qquad P=20000B^{2}\mu^{2},\qquad D_{0}=320000B^{2}\mu^{3}. μ = ⌊ log 2 m ⌋ + 1 , P = 20000 B 2 μ 2 , D 0 = 320000 B 2 μ 3 . There is a deterministic construction of a binary matrix A ( 0 ) A^{(0)} A ( 0 ) m m m 3 n 3n 3 n D 0 D_{0} D 0
z ∈ { 0 , 1 } m ∖ { 0 ⃗ } , A ( 0 ) z = 0 ⃗ ( m o d 2 ) ⟹ ∣ supp z ∣ ≥ m / P . z\in\{0,1\}^{m}\setminus\{\vec{0}\},\qquad A^{(0)}z=\vec{0}\pmod2\quad\Longrightarrow\quad|\operatorname{supp}z|\ge m/P. z ∈ { 0 , 1 } m ∖ { 0 } , A ( 0 ) z = 0 ( mod 2 ) ⟹ ∣ supp z ∣ ≥ m / P . The construction uses exp ( poly ( B , log ( m + 2 ) ) ) \exp(\operatorname{poly}(B,\log(m+2))) exp ( poly ( B , log ( m + 2 ))) poly ( n , m , B ) \operatorname{poly}(n,m,B) poly ( n , m , B )
Proof. We construct a bipartite graph with between m m m 2 m 2m 2 m m m m
d G = min ( m , D 0 ) , q G = min { 2 a : a ∈ Z ≥ 0 , 2 a d G ≥ m } . d_G=\min(m,D_0),\qquad q_G=\min\{2^a:a\in\mathbb{Z}_{\ge0},\ 2^a d_G\ge m\}. d G = min ( m , D 0 ) , q G = min { 2 a : a ∈ Z ≥ 0 , 2 a d G ≥ m } . Then m ≤ q G d G < 2 m m\le q_Gd_G<2m m ≤ q G d G < 2 m q G = 2 ℓ G q_G=2^{\ell_G} q G = 2 ℓ G { 0 , 1 } ℓ G \{0,1\}^{\ell_G} { 0 , 1 } ℓ G u u u j ∈ [ d G ] j\in[d_G] j ∈ [ d G ] u u u u ⊕ s j u\oplus s_j u ⊕ s j ⊕ \oplus ⊕ s 1 , … , s d G ∈ { 0 , 1 } ℓ G s_1,\ldots,s_{d_G}\in\{0,1\}^{\ell_G} s 1 , … , s d G ∈ { 0 , 1 } ℓ G ( u , j ) (u,j) ( u , j ) j j j d G d_G d G
To obtain the mixing estimate (64) , we choose the shifts so that every parity of a nonempty set of coordinates is nearly balanced. For a nonzero vector b ∈ { 0 , 1 } ℓ G b\in\{0,1\}^{\ell_G} b ∈ { 0 , 1 } ℓ G b ⋅ s j b\cdot s_j b ⋅ s j s j s_j s j b b b ε = 1 / ( 200 B μ ) \varepsilon=1/(200B\mu) ε = 1/ ( 200 B μ )
∣ ∑ j = 1 d G ( − 1 ) b ⋅ s j ∣ ≤ ε d G ( b ∈ { 0 , 1 } ℓ G ∖ { 0 ⃗ } ) . (63) \left|\sum_{j=1}^{d_G}(-1)^{b\cdot s_j}\right|\le\varepsilon d_G\qquad\left(b\in\{0,1\}^{\ell_G}\setminus\{\vec{0}\}\right). \tag*{(63)} j = 1 ∑ d G ( − 1 ) b ⋅ s j ≤ ε d G ( b ∈ { 0 , 1 } ℓ G ∖ { 0 } ) . (63) The sum in (63) is the number of indices j j j b b b ( 1 − ε ) / 2 (1-\varepsilon)/2 ( 1 − ε ) /2 ε \varepsilon ε [52 Definitions 2.1–2.2] .
If q G = 1 q_G=1 q G = 1 d G = D 0 d_G=D_0 d G = D 0 b b b ( − 1 ) b ⋅ s j (-1)^{b\cdot s_j} ( − 1 ) b ⋅ s j { − 1 , 1 } \{-1,1\} { − 1 , 1 } (63) for this b b b 2 exp ( − ε 2 d G / 2 ) 2\exp(-\varepsilon^2d_G/2) 2 exp ( − ε 2 d G /2 ) b b b
2 q G exp ( − ε 2 d G / 2 ) = 2 q G exp ( − 4 μ ) < 1 , 2q_G\exp(-\varepsilon^2d_G/2)=2q_G\exp(-4\mu)<1, 2 q G exp ( − ε 2 d G /2 ) = 2 q G exp ( − 4 μ ) < 1 , where q G < 2 m < 2 μ + 1 q_G<2m<2^{\mu+1} q G < 2 m < 2 μ + 1 (63) therefore finds a valid tuple in exp ( O ( d G log ( m + 2 ) ) ) \exp(O(d_G\log(m+2))) exp ( O ( d G log ( m + 2 )))
We next derive the mixing estimate from (63) . Let A G A_G A G q G × q G q_G\times q_G q G × q G A G [ u , v ] = ∣ { j ∈ [ d G ] : u ⊕ s j = v } ∣ A_G[u,v]=|\{j\in[d_G]:u\oplus s_j=v\}| A G [ u , v ] = ∣ { j ∈ [ d G ] : u ⊕ s j = v } ∣ u u u v v v { 0 , 1 } ℓ G \{0,1\}^{\ell_G} { 0 , 1 } ℓ G χ b ( u ) = ( − 1 ) b ⋅ u \chi_b(u)=(-1)^{b\cdot u} χ b ( u ) = ( − 1 ) b ⋅ u b , u ∈ { 0 , 1 } ℓ G b,u\in\{0,1\}^{\ell_G} b , u ∈ { 0 , 1 } ℓ G
( A G χ b ) ( u ) = ∑ j = 1 d G χ b ( u ⊕ s j ) = χ b ( u ) ∑ j = 1 d G ( − 1 ) b ⋅ s j . (A_G\chi_b)(u)=\sum_{j=1}^{d_G}\chi_b(u\oplus s_j)=\chi_b(u)\sum_{j=1}^{d_G}(-1)^{b\cdot s_j}. ( A G χ b ) ( u ) = j = 1 ∑ d G χ b ( u ⊕ s j ) = χ b ( u ) j = 1 ∑ d G ( − 1 ) b ⋅ s j . Thus each χ b \chi_b χ b χ 0 ⃗ \chi_{\vec0} χ 0 d G d_G d G (63) bounds all other eigenvalues in absolute value by ε d G \varepsilon d_G ε d G U U U V V V e G ( U , V ) e_G(U,V) e G ( U , V ) U , V U,V U , V
e G ( U , V ) ≤ d G ∣ U ∣ ∣ V ∣ q G + ε d G ∣ U ∣ ∣ V ∣ . (64) e_G(U,V)\le\frac{d_G|U||V|}{q_G}+\varepsilon d_G\sqrt{|U||V|}. \tag*{(64)} e G ( U , V ) ≤ q G d G ∣ U ∣∣ V ∣ + ε d G ∣ U ∣∣ V ∣ . (64) To specify the local code, we put
c G = ⌈ d G 8 B ⌉ , δ = 1 100 B μ . c_G=\left\lceil\frac{d_G}{8B}\right\rceil,\qquad\delta=\frac{1}{100B\mu}. c G = ⌈ 8 B d G ⌉ , δ = 100 B μ 1 . We need a binary matrix A l o c A_{\mathrm{loc}} A loc c G c_G c G d G d_G d G 2 2 2 δ d G \delta d_G δ d G 2 − c G 2^{-c_G} 2 − c G h = ⌈ δ d G ⌉ − 1 h=\lceil\delta d_G\rceil-1 h = ⌈ δ d G ⌉ − 1 h = 0 h=0 h = 0 δ d G \delta d_G δ d G
∑ j = 1 h ( d G j ) ≤ ( h + 1 ) ( 2 d G ) h ≤ ( 2 d G ) 2 h < 2 d G / ( 25 B ) . \sum_{j=1}^{h}\binom{d_G}{j}\le(h+1)(2d_G)^h\le(2d_G)^{2h}<2^{d_G/(25B)}. j = 1 ∑ h ( j d G ) ≤ ( h + 1 ) ( 2 d G ) h ≤ ( 2 d G ) 2 h < 2 d G / ( 25 B ) . For the last inequality, use h < δ d G h<\delta d_G h < δ d G d G ≤ m < 2 μ d_G\le m<2^\mu d G ≤ m < 2 μ μ + 1 ≤ 2 μ \mu+1\le2\mu μ + 1 ≤ 2 μ c G ≥ d G / ( 8 B ) c_G\ge d_G/(8B) c G ≥ d G / ( 8 B ) A l o c A_{\mathrm{loc}} A loc 2 c G d G 2^{c_Gd_G} 2 c G d G δ d G \delta d_G δ d G exp ( poly ( B , log ( m + 2 ) ) ) \exp(\operatorname{poly}(B,\log(m+2))) exp ( poly ( B , log ( m + 2 )))
At every vertex on both sides, impose the c G c_G c G A l o c A_{\mathrm{loc}} A loc C l o c = ker F 2 A l o c C_{\mathrm{loc}}=\ker_{\mathbb{F}_2}A_{\mathrm{loc}} C loc = ker F 2 A loc U , V U,V U , V h ′ h' h ′ U ∪ V U\cup V U ∪ V δ d G \delta d_G δ d G h ′ ≥ δ d G max ( ∣ U ∣ , ∣ V ∣ ) h'\ge\delta d_G\max(|U|,|V|) h ′ ≥ δ d G max ( ∣ U ∣ , ∣ V ∣ )
δ d G ∣ U ∣ ∣ V ∣ ≤ h ′ ≤ e G ( U , V ) ≤ d G ∣ U ∣ ∣ V ∣ q G + ε d G ∣ U ∣ ∣ V ∣ . \delta d_G\sqrt{|U||V|}\le h'\le e_G(U,V)\le\frac{d_G|U||V|}{q_G}+\varepsilon d_G\sqrt{|U||V|}. δ d G ∣ U ∣∣ V ∣ ≤ h ′ ≤ e G ( U , V ) ≤ q G d G ∣ U ∣∣ V ∣ + ε d G ∣ U ∣∣ V ∣ . Both U U U V V V d G ∣ U ∣ ∣ V ∣ d_G\sqrt{|U||V|} d G ∣ U ∣∣ V ∣ ∣ U ∣ ∣ V ∣ ≥ ( δ − ε ) q G \sqrt{|U||V|}\ge(\delta-\varepsilon)q_G ∣ U ∣∣ V ∣ ≥ ( δ − ε ) q G
h ′ ≥ δ ( δ − ε ) q G d G = q G d G 20000 B 2 μ 2 ≥ m P . h'\ge\delta(\delta-\varepsilon)q_Gd_G=\frac{q_Gd_G}{20000B^2\mu^2}\ge\frac{m}{P}. h ′ ≥ δ ( δ − ε ) q G d G = 20000 B 2 μ 2 q G d G ≥ P m . Shorten this Tanner code by keeping any m m m q G d G q_Gd_G q G d G [ m ] [m] [ m ] A ( 0 ) A^{(0)} A ( 0 ) A ( 0 ) A^{(0)} A ( 0 ) d G ≤ D 0 d_G \le D_0 d G ≤ D 0 2 q G c G 2q_Gc_G 2 q G c G q G = 1 q_G=1 q G = 1 d G = m d_G=m d G = m
2 c G ≤ m 4 B + 2 ≤ n 4 + 2 ≤ 3 n . 2c_G \le\frac{m}{4B}+2 \le\frac{n}{4}+2 \le3n. 2 c G ≤ 4 B m + 2 ≤ 4 n + 2 ≤ 3 n . If q G > 1 q_G>1 q G > 1 d G = D 0 ≥ 8 B d_G=D_0\ge8B d G = D 0 ≥ 8 B
2 q G c G ≤ q G d G 4 B + 2 q G ≤ q G d G 2 B < m B ≤ n . 2q_Gc_G \le\frac{q_Gd_G}{4B}+2q_G \le\frac{q_Gd_G}{2B}<\frac{m}{B}\le n. 2 q G c G ≤ 4 B q G d G + 2 q G ≤ 2 B q G d G < B m ≤ n . The searches over shifts, local matrices, and test words are sequential. Storing the current choices and the final rows uses poly ( n , m , B ) \operatorname{poly}(n,m,B) poly ( n , m , B )
Choosing additional equations by counting common zeros
After the initial checks, every nonsatisfying assignment still passing violates at least m / P m/P m / P [55 ] .
Lemma 6.2. With P P P D 0 D_0 D 0 A ( 0 ) A^{(0)} A ( 0 ) Lemma 6.1 , a deterministic procedure appends 3 n 3n 3 n P P P A A A N ( A ) = # F N(A)=\#F N ( A ) = # F 3 n ( 2 m ) P 3n(2m)^P 3 n ( 2 m ) P 6 n 6n 6 n 2 k max ( D 0 , P ) 2^k\max(D_0,P) 2 k max ( D 0 , P ) k k k
Proof. At a stage with current matrix A A A
γ = ( ( e 1 , c 1 ) , … , ( e P , c P ) ) ∈ C : = ( [ m ] × { 0 , 1 } ) P . \gamma=((e_1,c_1),\ldots,(e_P,c_P))\in C:=([m]\times\{0,1\})^P. γ = (( e 1 , c 1 ) , … , ( e P , c P )) ∈ C := ([ m ] × { 0 , 1 } ) P . The list γ \gamma γ h γ h_\gamma h γ
( h γ ) e = ( ∑ j : e j = e c j ) m o d 2 ( e ∈ [ m ] ) . (h_\gamma)_e=\left(\sum_{j:e_j=e}c_j\right)\bmod2 \qquad(e\in[m]). ( h γ ) e = j : e j = e ∑ c j mod 2 ( e ∈ [ m ]) . There are ( 2 m ) P (2m)^P ( 2 m ) P A ; h γ A;h_\gamma A ; h γ h γ h_\gamma h γ A A A
Fix a nonsatisfying assignment x x x A A A A A A ∣ supp v ( x ) ∣ ≥ m / P |\operatorname{supp}v(x)|\ge m/P ∣ supp v ( x ) ∣ ≥ m / P γ \gamma γ e 1 , … , e P e_1,\ldots,e_P e 1 , … , e P
( 1 − 1 / P ) P ≤ e − 1 < 1 / 2. (1-1/P)^P\le e^{-1}<1/2. ( 1 − 1/ P ) P ≤ e − 1 < 1/2. Conditional on the indices hitting the support, choose one hit position j j j c j c_j c j c j c_j c j h γ ⋅ v ( x ) h_\gamma\cdot v(x) h γ ⋅ v ( x ) 2 2 2 x x x x x x
Summing this bound over the nonsatisfying assignments still passing, we find
E γ D ( A ; h γ ) ≤ 3 4 D ( A ) . \mathbb{E}_{\gamma}D(A;h_\gamma)\le\frac{3}{4}D(A). E γ D ( A ; h γ ) ≤ 4 3 D ( A ) . For every candidate, compute N ( A ; h γ ) N(A;h_\gamma) N ( A ; h γ ) N ( A ; h γ ) = # F + D ( A ; h γ ) N(A;h_\gamma)=\#F+D(A;h_\gamma) N ( A ; h γ ) = # F + D ( A ; h γ ) γ \gamma γ 3 4 D ( A ) \frac{3}{4}D(A) 4 3 D ( A ) 3 n 3n 3 n ( 3 / 4 ) 3 n 2 n = ( 27 / 32 ) n < 1 (3/4)^{3n}2^n=(27/32)^n<1 ( 3/4 ) 3 n 2 n = ( 27/32 ) n < 1 D ( A ) D(A) D ( A ) # F \#F # F
The initial matrix has at most 3 n 3n 3 n 3 n 3n 3 n s s s (62) , to at most 2 k s 2^ks 2 k s F 2 \mathbb{F}_2 F 2
Counting one sparse formula
Suppose n ≥ 20 n \ge20 n ≥ 20 1 ≤ m ≤ B n 1 \le m \le Bn 1 ≤ m ≤ B n B B B log n \log n log n k k k Lemma 6.1 . This uses exp ( ( log n ) O k ( 1 ) ) \exp((\log n)^{O_k(1)}) exp (( log n ) O k ( 1 ) ) P P P D 0 D_0 D 0 ( log n ) O k ( 1 ) (\log n)^{O_k(1)} ( log n ) O k ( 1 ) Lemma 6.2 is an instance of Theorem 2.2 with a single fixed c = c ( k ) c=c(k) c = c ( k ) k k k
At most 3 n ( 2 m ) P + 1 = exp ( ( log n ) O k ( 1 ) ) 3n(2m)^P+1=\exp((\log n)^{O_k(1)}) 3 n ( 2 m ) P + 1 = exp (( log n ) O k ( 1 ) ) 2 O k ( n ( log log n ) 2 / log n ) 2^{O_k(n(\log\log n)^2/\sqrt{\log n})} 2 O k ( n ( l o g l o g n ) 2 / l o g n ) log n \log n log n o ( n ( log log n ) 2 / log n ) o(n(\log\log n)^2/\sqrt{\log n}) o ( n ( log log n ) 2 / log n )
( n + m ) O k ( 1 ) 2 O k ( n ( log log n ) 2 / log n ) . (65) (n+m)^{O_k(1)}2^{O_k(n(\log\log n)^2/\sqrt{\log n})}. \tag*{(65)} ( n + m ) O k ( 1 ) 2 O k ( n ( l o g l o g n ) 2 / l o g n ) . (65) The final count is # F \#F # F
Returning to a general formula
We use disjoint sparsification to partition the satisfying assignments among sparse formulas on the same n n n σ \sigma σ B B B B B B log n \log n log n σ \sigma σ
Lemma 6.3. Fix k ≥ 2 k \ge2 k ≥ 2 k k k n n n m m m 0 < σ ≤ 1 0<\sigma\le1 0 < σ ≤ 1 k k k L σ L_\sigma L σ σ \sigma σ
poly k ( n , σ − 1 ) 2 σ n \operatorname{poly}_k(n,\sigma^{-1})\,2^{\sigma n} poly k ( n , σ − 1 ) 2 σ n nodes and depth poly k ( n , σ − 1 ) \operatorname{poly}_k(n,\sigma^{-1}) poly k ( n , σ − 1 ) B B B B B B
B = O k ( [ σ − 1 log ( 2 / σ ) ] k − 1 ) . B=O_k\left(\left[\sigma^{-1}\log(2/\sigma)\right]^{k-1}\right). B = O k ( [ σ − 1 log ( 2/ σ ) ] k − 1 ) . Input normalization uses poly k ( n , m ) \operatorname{poly}_k(n,m) poly k ( n , m ) poly k ( L σ ) \operatorname{poly}_k(L_\sigma) poly k ( L σ ) poly k ( n , log ( 2 / σ ) ) \operatorname{poly}_k(n,\log(2/\sigma)) poly k ( n , log ( 2/ σ ))
Appendix § 7 proves the lemma, including the disjointness that justifies adding the leaf counts.
Proposition 6.4. For each fixed k ≥ 2 k\ge2 k ≥ 2 k k k n ≥ 20 n\ge20 n ≥ 20 m m m ( n + m ) O k ( 1 ) 2 O k ( ρ ( n ) ) (n+m)^{O_k(1)}2^{O_k(\rho(n))} ( n + m ) O k ( 1 ) 2 O k ( ρ ( n )) ( n + m ) O k ( 1 ) (n+m)^{O_k(1)} ( n + m ) O k ( 1 )
Proof. We first normalize the input in ( n + m ) O k ( 1 ) (n+m)^{O_k(1)} ( n + m ) O k ( 1 ) n n n 2 n 2^n 2 n n < 20 n<20 n < 20 n ≥ 20 n\ge20 n ≥ 20
λ n = ⌈ log 2 n ⌉ , ν n = ⌈ log 2 λ n ⌉ , s = ⌊ λ n ν n 2 ⌋ . \lambda_n=\lceil\log_2 n\rceil,\qquad\nu_n=\lceil\log_2\lambda_n\rceil,\qquad s=\left\lfloor\frac{\sqrt{\lambda_n}}{\nu_n^2}\right\rfloor. λ n = ⌈ log 2 n ⌉ , ν n = ⌈ log 2 λ n ⌉ , s = ⌊ ν n 2 λ n ⌋ . We compute s s s s 2 ν n 4 ≤ λ n s^2\nu_n^4\le\lambda_n s 2 ν n 4 ≤ λ n s < 2 s<2 s < 2 λ n / ν n 2 \sqrt{\lambda_n}/\nu_n^2 λ n / ν n 2 σ = 1 / s \sigma= 1/s σ = 1/ s 2 n / s poly k ( n , s ) 2^{n/s}\operatorname{poly}_k(n,s) 2 n / s poly k ( n , s )
B = O k ( ( s log s ) k − 1 ) = ( log n ) O k ( 1 ) . B = O_k\left((s\log s)^{k-1}\right) = (\log n)^{O_k(1)}. B = O k ( ( s log s ) k − 1 ) = ( log n ) O k ( 1 ) . Each leaf uses the original n n n B n Bn B n § 6.3 , whose running time is given by (65) . Leaves with no clauses or containing an empty clause are handled directly. The clause count in that application is the number of clauses in the leaf; m m m
For s ≥ 2 s \ge2 s ≥ 2
n s = O ( n n λ 2 λ n ) = O ( ρ ( n ) ) . \frac{n}{s} = O\left(\frac{n\sqrt{n_\lambda^2}}{\sqrt{\lambda_n}}\right) = O(\rho(n)). s n = O ( λ n n n λ 2 ) = O ( ρ ( n )) . Thus both the number of leaves and the time to count each leaf have exponent O k ( ρ ( n ) ) O_k(\rho(n)) O k ( ρ ( n )) ( n + m ) O k ( 1 ) (n+m)^{O_k(1)} ( n + m ) O k ( 1 )
( n + m ) O k ( 1 ) 2 O k ( ρ ( n ) ) . (n+m)^{O_k(1)}2^{O_k(\rho(n))}. ( n + m ) O k ( 1 ) 2 O k ( ρ ( n )) . The finite exhaustive-search cases are absorbed in the constants. All parameters, initial checks, and additional rows are computed by the deterministic procedures above; every counting query uses the common-zero algorithm of Theorem 2.2.
To bound the space used by the full algorithm, we traverse the sparsification tree depth first and count one leaf at a time. The current path has poly k ( n , s ) \operatorname{poly}_k(n,s) poly k ( n , s ) poly k ( n ) \operatorname{poly}_k(n) poly k ( n ) s ≤ ⌈ log 2 n ⌉ s \le\lceil\sqrt{\log_2 n}\rceil s ≤ ⌈ log 2 n ⌉ poly k ( n , m ) \operatorname{poly}_k(n,m) poly k ( n , m ) 2 n 2^n 2 n n + 1 n+1 n + 1 poly k ( n , m ) \operatorname{poly}_k(n,m) poly k ( n , m )
Appendix A. Partitioning the satisfying assignments
Fix k ≥ 2 k \ge2 k ≥ 2 0 < σ ≤ 1 0 < \sigma\le1 0 < σ ≤ 1 n n n [19 Section 6] . The refinement that makes the two cases disjoint is standard in exact counting [28 Appendix A] .
Throughout this appendix, a clause is a set of literals and denotes their disjunction. Its width is the number of literals it contains. We delete tautological clauses, repeated clauses, and clauses containing another clause. Thus the remaining clauses form an antichain under inclusion. We keep all n n n 2 n 2^n 2 n
A.1. The branching rule. Fix an integer M ≥ 2 M \ge2 M ≥ 2 B = O k ( M k − 1 ) B = O_k(M^{k-1}) B = O k ( M k − 1 ) M M M poly k ( n , σ − 1 ) 2 σ n \operatorname{poly}_k(n,\sigma^{-1})2^{\sigma n} poly k ( n , σ − 1 ) 2 σ n
θ j = M j ( 0 ≤ j < k ) . \theta_j = M^j \qquad(0 \le j < k). θ j = M j ( 0 ≤ j < k ) . Writing F F F
deg c ( H ) = # { C ∈ F : ∣ C ∣ = c , H ⊆ C } \operatorname{deg}_c(H) = \#\{C \in F : |C| = c,\ H \subseteq C\} deg c ( H ) = # { C ∈ F : ∣ C ∣ = c , H ⊆ C } for each literal set H H H H H H c c c
deg c ( H ) ≥ θ c − ∣ H ∣ . (66) \operatorname{deg}_c(H) \ge\theta_{c-|H|}. \tag*{(66)} deg c ( H ) ≥ θ c − ∣ H ∣ . (66) The algorithm chooses the smallest possible c c c ∣ H ∣ |H| ∣ H ∣ H H H j = c − ∣ H ∣ j = c - |H| j = c − ∣ H ∣ M j M^j M j H ∪ P 1 , … , H ∪ P z H \cup P_1,\ldots,H \cup P_z H ∪ P 1 , … , H ∪ P z c c c H H H G G G j = c − ∣ H ∣ j=c-|H| j = c − ∣ H ∣ P a P_a P a H H H a a a P a P_a P a j j j z ≥ θ j z \ge\theta_j z ≥ θ j
F A = G ∧ H , F B = G ∧ ⋀ a = 1 z P a ∧ ⋀ ℓ ∈ H { ¬ ℓ } . (67) F_A=G\wedge H,\qquad F_B=G\wedge\bigwedge_{a=1}^{z}P_a\wedge\bigwedge_{\ell\in H}\{\neg\ell\}. \tag*{(67)} F A = G ∧ H , F B = G ∧ a = 1 ⋀ z P a ∧ ℓ ∈ H ⋀ { ¬ ℓ } . (67) We restore the antichain condition in each child by removing clauses made redundant by its new clauses. The sets P a P_a P a F A F_A F A H H H F B F_B F B H H H
Lemma A.1. The children in (A.2) partition the satisfying assignments of their parent. Every new clause has width less than c c c A A A B B B θ j \theta_j θ j j j j
Proof. Every assignment satisfies exactly one of the conditions some literal of H H H H H H G ∧ H G\wedge H G ∧ H P a P_a P a { ¬ ℓ } \{\neg\ell\} { ¬ ℓ }
The old antichain contains no clause included in H H H P a P_a P a H ∪ P a H\cup P_a H ∪ P a H H H P a P_a P a P a P_a P a H H H { ¬ ℓ } \{\neg\ell\} { ¬ ℓ } ℓ ∈ H \ell\in H ℓ ∈ H P a P_a P a { ¬ ℓ } \{\neg\ell\} { ¬ ℓ } P a P_a P a H H H A A A P a P_a P a B B B
A.2. Counting insertions. To bound the tree, we count every clause introduced along one path, including clauses later deleted. Clauses present in the input are excluded from this count. By Lemma A.1 , each step inserts a clause, so bounding the total number of insertions also bounds the path length.
For this accounting, a clause is marked when a branch introduces it. First we bound how many marked clauses can still be present and contain a given literal set. The smallest-width priority controls those present before a branch; the largest-∣ H ∣ |H| ∣ H ∣
Lemma A.2. At every node, for 1 ≤ i < j < k 1\le i<j<k 1 ≤ i < j < k S S S i i i 2 θ j − i 2\theta_{j-i} 2 θ j − i j j j S S S
Proof. Initially there are no marked clauses. It suffices to consider a branch that inserts clauses of width j j j c c c j j j θ j − i \theta_{j-i} θ j − i j j j S S S j j j
Choosing child A A A B B B P a P_a P a j j j P a P_a P a S S S j j j S S S S S S H H H P a P_a P a S S S H ∪ P a H\cup P_a H ∪ P a H ∪ S H\cup S H ∪ S 0 < i < j 0<i<j 0 < i < j H ∪ S H\cup S H ∪ S H H H c c c
# { a : S ⊆ P a } = deg c ( H ∪ S ) < θ c − ∣ H ∪ S ∣ = θ j − i . \#\{a:S\subseteq P_a\}=\deg_c(H\cup S)<\theta_{c-|H\cup S|}=\theta_{j-i}. # { a : S ⊆ P a } = deg c ( H ∪ S ) < θ c − ∣ H ∪ S ∣ = θ j − i . The other new clauses { ¬ ℓ } \{\neg\ell\} { ¬ ℓ } j j j
Lemma A.3. Along any path or path prefix, the total number T j T_j T j j j j T j ≤ 4 ( 3 M ) j − 1 n T_j \le4(3M)^{j-1}n T j ≤ 4 ( 3 M ) j − 1 n 1 ≤ j < k 1 \le j < k 1 ≤ j < k
Proof. The case n = 0 n=0 n = 0 n ≥ 1 n \ge1 n ≥ 1 L j L_j L j j j j D j D_j D j j j j T j = L j + D j T_j=L_j+D_j T j = L j + D j T 1 ≤ 2 n ≤ 4 n T_1 \le2n \le4n T 1 ≤ 2 n ≤ 4 n j ≥ 2 j \ge2 j ≥ 2 Lemma A.2 over the 2 n 2n 2 n L j ≤ 4 n θ j − 1 L_j \le4n\theta_{j-1} L j ≤ 4 n θ j − 1
Whenever a marked clause of width j j j Lemma A.1 guarantees such a subset. If the new clause has width i < j i<j i < j Lemma A.2 shows that it was contained in fewer than 2 θ j − i 2\theta_{j-i} 2 θ j − i j j j i i i i i i D j ≤ 2 ∑ i = 1 j − 1 θ j − i T i D_j \le2\sum_{i=1}^{j-1}\theta_{j-i}T_i D j ≤ 2 ∑ i = 1 j − 1 θ j − i T i
T j = L j + D j ≤ 4 n θ j − 1 + 2 ∑ i = 1 j − 1 θ j − i T i . (68) T_j=L_j+D_j \le4n\theta_{j-1}+2\sum_{i=1}^{j-1}\theta_{j-i}T_i. \tag*{(68)} T j = L j + D j ≤ 4 n θ j − 1 + 2 i = 1 ∑ j − 1 θ j − i T i . (68) The first term counts clauses still present; the sum counts those that were deleted. Dividing by n M j − 1 nM^{j-1} n M j − 1
T j n M j − 1 ≤ 4 + 2 ∑ i = 1 j − 1 4 ⋅ 3 i − 1 = 4 ⋅ 3 j − 1 . \frac{T_j}{nM^{j-1}} \le4+2\sum_{i=1}^{j-1}4\cdot3^{i-1}=4\cdot3^{j-1}. n M j − 1 T j ≤ 4 + 2 i = 1 ∑ j − 1 4 ⋅ 3 i − 1 = 4 ⋅ 3 j − 1 . as required.
The size of the tree
We now convert Lemma A.3 into a bound on the number of tree nodes. It controls both the length of a path and the number of times that path chooses the B B B C n Cn C n Γ n / M \Gamma n/M Γ n / M
C = 4 ∑ j = 0 k − 2 ( 3 M ) j , Γ = 4 ∑ j = 0 k − 2 3 j = 2 ( 3 k − 1 − 1 ) . C=4\sum_{j=0}^{k-2}(3M)^j,\qquad\Gamma=4\sum_{j=0}^{k-2}3^j=2(3^{k-1}-1). C = 4 j = 0 ∑ k − 2 ( 3 M ) j , Γ = 4 j = 0 ∑ k − 2 3 j = 2 ( 3 k − 1 − 1 ) . Lemma A.3 gives depth at most C n Cn C n B B B ∣ P a ∣ = j |P_a|=j ∣ P a ∣ = j M j M^j M j j j j 4 ⋅ 3 j − 1 n / M 4\cdot3^{j-1}n/M 4 ⋅ 3 j − 1 n / M B B B Γ n / M \Gamma n/M Γ n / M
A path with few choices of B B B B B B A A A M ≥ Γ M\ge\Gamma M ≥ Γ
( C n + 1 ) ∑ a ≤ ⌊ Γ n / M ⌋ ( C n a ) ≤ ( C n + 1 ) 2 ( Γ n / M ) log 2 ( 4 C M ) . (Cn+1)\sum_{a\le\lfloor\Gamma n/M\rfloor}\binom{Cn}{a}\le(Cn+1)2^{(\Gamma n/M)\log_2(4CM)}. ( C n + 1 ) a ≤ ⌊ Γ n / M ⌋ ∑ ( a C n ) ≤ ( C n + 1 ) 2 ( Γ n / M ) l o g 2 ( 4 C M ) . Here Γ n / M ≤ n ≤ C n / 4 \Gamma n/M\le n\le Cn/4 Γ n / M ≤ n ≤ C n /4 a a a ( C n a ) ≤ ( 4 C n / a ) a \binom{Cn}{a}\le(4Cn/a)^a ( a C n ) ≤ ( 4 C n / a ) a a ≤ C n / 4 a\le Cn/4 a ≤ C n /4 C n + 1 Cn+1 C n + 1 Γ n / M < 1 \Gamma n/M<1 Γ n / M < 1
We choose M M M σ n \sigma n σ n M ≥ Γ M\ge\Gamma M ≥ Γ C C C h = ⌈ log 2 ( 4 C M ) ⌉ h=\lceil\log_2(4CM)\rceil h = ⌈ log 2 ( 4 C M )⌉
Γ h ≤ σ M . \Gamma h\le\sigma M. Γ h ≤ σ M . All comparisons are exact. Since C = O k ( M k − 2 ) C=O_k(M^{k-2}) C = O k ( M k − 2 ) Γ \Gamma Γ k k k
M = O k ( σ − 1 log ( 2 / σ ) ) . M=O_k\left(\sigma^{-1}\log(2/\sigma)\right). M = O k ( σ − 1 log ( 2/ σ ) ) . The tree therefore has at most ( C n + 1 ) 2 2 σ n (Cn+1)2^{2\sigma n} ( C n + 1 ) 2 2 σ n
At a leaf, (A.1) fails for every nonempty proper subset of a clause. A fixed literal belongs to fewer than θ c − 1 \theta_{c-1} θ c − 1 c c c c ≥ 2 c \ge2 c ≥ 2
B = 2 ∑ j = 0 k − 1 M j ≤ 4 M k − 1 = O k ( [ σ − 1 log ( 2 / σ ) ] k − 1 ) . B = 2\sum_{j=0}^{k-1} M^j \le4M^{k-1} = O_k\left(\left[\sigma^{-1}\log(2/\sigma)\right]^{k-1}\right). B = 2 j = 0 ∑ k − 1 M j ≤ 4 M k − 1 = O k ( [ σ − 1 log ( 2/ σ ) ] k − 1 ) . Thus every leaf has the required bound on occurrences per variable. The repeated two-case partition also shows that each original satisfying assignment occurs at exactly one leaf, so summing the leaf counts gives the original count.
To implement the branching rule at a node, we compute deg c ( H ) \deg_c(H) deg c ( H ) c c c H H H ( c , H ) (c,H) ( c , H ) O k ( n k ) O_k(n^k) O k ( n k ) 2 k 2^k 2 k O k ( log ( n + 2 ) ) O_k(\log(n+2)) O k ( log ( n + 2 )) O k ( log M ) O_k(\log M) O k ( log M ) poly k ( n , log M ) \operatorname{poly}_k(n,\log M) poly k ( n , log M ) M M M log M = O k ( log ( 2 / σ ) ) \log M=O_k(\log(2/\sigma)) log M = O k ( log ( 2/ σ ))
With L σ L_\sigma L σ O k ( L σ ) O_k(L_\sigma) O k ( L σ ) O k ( L σ ) O_k(L_\sigma) O k ( L σ ) poly k ( L σ ) \operatorname{poly}_k(L_\sigma) poly k ( L σ )
Exact arithmetic and space
Section 4 bounds the rational coefficients of the substitution rows P ⃗ f , v ( T ) \vec{P}_{f,v}(T) P f , v ( T ) H H H G = ∑ α c α X ⃗ α G=\sum_\alpha c_\alpha\vec{X}^{\alpha} G = ∑ α c α X α ∥ G ∥ 1 = ∑ α ∣ c α ∣ \|G\|_1=\sum_\alpha|c_\alpha| ∥ G ∥ 1 = ∑ α ∣ c α ∣
Integer coefficients of the lifted indicator
We first bound the coefficients of the parity polynomials supplied by Lemma 5.1. They occur in every equation factor B i B_i B i (59) , so their sizes enter the bit-cost bounds for evaluating those factors. For an integer ℓ ≥ 1 \ell\ge1 ℓ ≥ 1 h ( Y ) = Y ( Y − 1 ) h(Y)=Y(Y-1) h ( Y ) = Y ( Y − 1 )
p ℓ ( Y ) = ( 1 − Y ) ℓ ∑ j = 0 ℓ − 1 ( ℓ + j − 1 j ) Y j = ∑ j = 0 ℓ − 1 e j ( Y ) h ( Y ) j , deg e j ≤ 1. p_\ell(Y)=(1-Y)^\ell\sum_{j=0}^{\ell-1}\binom{\ell+j-1}{j}Y^j=\sum_{j=0}^{\ell-1}e_j(Y)h(Y)^j,\qquad\deg e_j\le1. p ℓ ( Y ) = ( 1 − Y ) ℓ j = 0 ∑ ℓ − 1 ( j ℓ + j − 1 ) Y j = j = 0 ∑ ℓ − 1 e j ( Y ) h ( Y ) j , deg e j ≤ 1. That lemma proves the parity congruence and constructs the e j e_j e j h h h
Lemma B.1. Write e j ( Y ) = a j + b j Y e_j(Y)=a_j+b_jY e j ( Y ) = a j + b j Y for 0 ≤ j < ℓ 0\le j<\ell 0 ≤ j < ℓ Then a j , b j ∈ Z a_j,b_j\in\mathbb{Z} a j , b j ∈ Z and
∣ a j ∣ , ∣ b j ∣ ≤ 2 C ℓ 2 ( 0 ≤ j < ℓ ) |a_j|,|b_j|\le2^{C\ell^2}\qquad(0\le j<\ell) ∣ a j ∣ , ∣ b j ∣ ≤ 2 C ℓ 2 ( 0 ≤ j < ℓ ) for an absolute constant C C C p ℓ p_\ell p ℓ h h h poly ( ℓ ) \operatorname{poly}(\ell) poly ( ℓ ) bit operations and bits of working space.
Proof. The coefficient-norm inequality ∥ f g ∥ 1 ≤ ∥ f ∥ 1 ∥ g ∥ 1 \|fg\|_1\le\|f\|_1\|g\|_1 ∥ f g ∥ 1 ≤ ∥ f ∥ 1 ∥ g ∥ 1
∑ j = 0 ℓ − 1 ( ℓ + j − 1 j ) = ( 2 ℓ − 1 ℓ − 1 ) \sum_{j=0}^{\ell-1}\binom{\ell+j-1}{j}=\binom{2\ell-1}{\ell-1} j = 0 ∑ ℓ − 1 ( j ℓ + j − 1 ) = ( ℓ − 1 2 ℓ − 1 ) gives
∥ p ℓ ∥ 1 ≤ 2 ℓ ( 2 ℓ − 1 ℓ − 1 ) ≤ 2 3 ℓ . \|p_\ell\|_1\le2^\ell\binom{2\ell-1}{\ell-1}\le2^{3\ell}. ∥ p ℓ ∥ 1 ≤ 2 ℓ ( ℓ − 1 2 ℓ − 1 ) ≤ 2 3 ℓ . Monic division by h = Y 2 − Y h=Y^{2}-Y h = Y 2 − Y O ( ℓ 2 ) O(\ell^{2}) O ( ℓ 2 ) 2 O ( ℓ 2 ) 2^{O(\ell^{2})} 2 O ( ℓ 2 ) a j , b j a_j,b_j a j , b j O ( ℓ 2 ) O(\ell^{2}) O ( ℓ 2 ) O ( ℓ ) O(\ell) O ( ℓ ) O ( ℓ 2 ) O(\ell^{2}) O ( ℓ 2 ) O ( ℓ 2 ) O(\ell^{2}) O ( ℓ 2 ) poly ( ℓ ) \operatorname{poly}(\ell) poly ( ℓ )
Bit complexity of the sums over Boolean assignments
We now bound every intermediate value in the computation of H t ( z ) = H ( t + z ) m o d z J \mathcal{H}_t(z)=H(t+z)\bmod z^J H t ( z ) = H ( t + z ) mod z J (34) . Starting with the substitution rows P ⃗ f , v ( T ) \vec{P}_{f,v}(T) P f , v ( T ) T t T_t T t P \mathcal{P} P n n n m t e s t m_{\mathrm{test}} m test k k k ℓ \ell ℓ 2 ℓ > 2 n 2^\ell>2^n 2 ℓ > 2 n P ⃗ f , v \vec{P}_{f,v} P f , v S t , g S_{t,g} S t , g C t , g C_{t,g} C t , g Definition 3.1 , with parameters Q Q Q R R R r r r D D D τ \tau τ H H H
d H = k R ( m t e s t + 2 ( ℓ − 1 ) ) , J = ⌊ d H Q ⌋ + 1. d_H=kR(m_{\mathrm{test}}+2(\ell-1)),\qquad J=\left\lfloor\frac{d_H}{Q}\right\rfloor+1. d H = k R ( m test + 2 ( ℓ − 1 )) , J = ⌊ Q d H ⌋ + 1. All polynomials in the Taylor variable z z z V V V ( z J , V ℓ ) (z^J,V^\ell) ( z J , V ℓ )
The substitution rows have common denominator
d d e n = ( R − 1 ) ! Q ! . d_{\mathrm{den}}=(R-1)!Q!. d den = ( R − 1 )! Q ! . We use an integer B P ≥ 1 B_P\geq1 B P ≥ 1 d d e n P ⃗ f , v ( T ) d_{\mathrm{den}}\vec{P}_{f,v}(T) d den P f , v ( T ) 2 B P 2^{B_P} 2 B P Definition 3.1 gives B P = O ( Q log Q + log D ) B_P=O(Q\log Q+\log D) B P = O ( Q log Q + log D ) t ∈ [ Q ] t\in[Q] t ∈ [ Q ] U ⊆ [ n ] U\subseteq[n] U ⊆ [ n ] f f f v v v
θ f , v ( z ; x ) = ⟨ P ⃗ f , v ( t ) , x ⟩ + ⟨ P ⃗ f , v ( t + z ) − P ⃗ f , v ( t ) , 1 U ⟩ , x ∈ { 0 , 1 } n . \theta_{f,v}(z;x)=\langle\vec{P}_{f,v}(t),x\rangle+\left\langle\vec{P}_{f,v}(t+z)-\vec{P}_{f,v}(t),\mathbf{1}_U\right\rangle,\qquad x\in\{0,1\}^{n}. θ f , v ( z ; x ) = ⟨ P f , v ( t ) , x ⟩ + ⟨ P f , v ( t + z ) − P f , v ( t ) , 1 U ⟩ , x ∈ { 0 , 1 } n . Let L i ( z ; x ) L_i(z;x) L i ( z ; x ) i i i d d e n θ i , v ( z ; x ) d_{\mathrm{den}}\theta_{i,v}(z;x) d den θ i , v ( z ; x )
B 0 = d d e n + 3 n ( R + 1 ) 2 B P ( Q + 1 ) R . B_0=d_{\mathrm{den}}+3n(R+1)2^{B_P}(Q+1)^R. B 0 = d den + 3 n ( R + 1 ) 2 B P ( Q + 1 ) R . Indeed, each of the three row evaluations in θ i , v \theta_{i,v} θ i , v n ( R + 1 ) 2 B P ( Q + 1 ) R n(R+1)2^{B_P}(Q+1)^R n ( R + 1 ) 2 B P ( Q + 1 ) R
Put Δ = d d e n k \Delta=d_{\mathrm{den}}^k Δ = d den k u ≤ k u\leq k u ≤ k Δ \Delta Δ d d e n k − u B 0 u ≤ B 0 k d_{\mathrm{den}}^{k-u}B_0^u\leq B_0^k d den k − u B 0 u ≤ B 0 k S S S
B = max { Δ , S B 0 k } satisfies ∥ Δ L i ( z ; x ) ∥ 1 ≤ B . B=\max\{\Delta,SB_0^k\}\quad\text{satisfies}\quad\|\Delta L_i(z;x)\|_1\leq B. B = max { Δ , S B 0 k } satisfies ∥Δ L i ( z ; x ) ∥ 1 ≤ B . Consequently log d d e n = O ( Q log ( Q + 2 ) ) \log d_{\mathrm{den}}=O(Q\log(Q+2)) log d den = O ( Q log ( Q + 2 )) log B 0 = O ( Q log ( Q + 2 ) + log ( n D + 2 ) ) \log B_0=O(Q\log(Q+2)+\log(nD+2)) log B 0 = O ( Q log ( Q + 2 ) + log ( n D + 2 )) log B = O ( k log B 0 + log ( S + 2 ) ) \log B=O(k\log B_0+\log(S+2)) log B = O ( k log B 0 + log ( S + 2 ))
The substituted factor of equation i i i
B i ( z , V ; x ) = ∑ j = 0 ℓ − 1 e j ( L i ( z ; x ) ) h ( L i ( z ; x ) ) j V j ( m o d ( z J , V ℓ ) ) . \mathcal{B}_i(z,V;x)=\sum_{j=0}^{\ell-1}e_j(L_i(z;x))h(L_i(z;x))^jV^j\pmod{(z^J,V^\ell)}. B i ( z , V ; x ) = j = 0 ∑ ℓ − 1 e j ( L i ( z ; x )) h ( L i ( z ; x ) ) j V j ( mod ( z J , V ℓ )) . Since e j e_j e j Δ \Delta Δ h ( L i ) j h(L_i)^j h ( L i ) j Δ j \Delta^j Δ j
Γ = Δ 2 ℓ − 1 , \Gamma=\Delta^{2\ell-1}, Γ = Δ 2 ℓ − 1 , every coefficient of Γ B i ( z , V ; x ) \Gamma B_i(z,V;x) Γ B i ( z , V ; x )
E ℓ = max { 1 , ∥ e 0 ∥ 1 , … , ∥ e ℓ − 1 ∥ 1 } , B 1 = ℓ E ℓ ( B + Δ + 2 ) 2 ℓ − 1 . E_{\ell}=\max\{1,\lVert e_0\rVert_1,\ldots,\lVert e_{\ell-1}\rVert_1\},\qquad B_1=\ell E_{\ell}(B+\Delta+2)^{2\ell-1}. E ℓ = max { 1 , ∥ e 0 ∥ 1 , … , ∥ e ℓ − 1 ∥ 1 } , B 1 = ℓ E ℓ ( B + Δ + 2 ) 2 ℓ − 1 . For A i = Δ L i A_i=\Delta L_i A i = Δ L i Δ e j ( L i ) \Delta e_j(L_i) Δ e j ( L i ) A i ( A i − Δ ) A_i(A_i-\Delta) A i ( A i − Δ ) E ℓ ( B + Δ ) E_{\ell}(B+\Delta) E ℓ ( B + Δ ) B ( B + Δ ) B(B+\Delta) B ( B + Δ ) Γ \Gamma Γ
∥ Γ B i ∥ 1 ≤ B 1 , B 1 ≥ Γ , log B 1 = O ( ℓ 2 + ℓ log ( B + Δ + 2 ) + log ( ℓ + 1 ) ) . \lVert\Gamma B_i\rVert_1\le B_1,\qquad B_1\ge\Gamma,\qquad\log B_1=O\left(\ell^2+\ell\log(B+\Delta+2)+\log(\ell+1)\right). ∥ Γ B i ∥ 1 ≤ B 1 , B 1 ≥ Γ , log B 1 = O ( ℓ 2 + ℓ log ( B + Δ + 2 ) + log ( ℓ + 1 ) ) . We next apply these estimates to the actual group sums. Write S g = S t , g S_g=S_{t,g} S g = S t , g C g = C t , g C_g=C_{t,g} C g = C t , g c g c_g c g C g C_g C g c = ∑ g c g ≤ 2 τ n c=\sum_g c_g\le2\tau n c = ∑ g c g ≤ 2 τ n C g x g C_gx_g C g x g [ − D , D ] ∩ Z [-D,D]\cap\mathbb{Z} [ − D , D ] ∩ Z m t e s t , g m_{\mathrm{test},g} m test , g g g g t t t ∑ g m t e s t , g = m t e s t \sum_gm_{\mathrm{test},g}=m_{\mathrm{test}} ∑ g m test , g = m test A g , U ( x ; z ) A_{g,U}(x;z) A g , U ( x ; z ) (32) is represented in Q [ z , V ] / ( z J , V ℓ ) \mathbb{Q}[z,V]/(z^J,V^\ell) Q [ z , V ] / ( z J , V ℓ ) Γ m t e s t , g \Gamma^{m_{\mathrm{test},g}} Γ m test , g B 1 m t e s t , g B_1^{m_{\mathrm{test},g}} B 1 m test , g
For a pair U ⊆ E ⊆ [ n ] U\subseteq E\subseteq[n] U ⊆ E ⊆ [ n ] ∣ E ∣ < J |E|<J ∣ E ∣ < J y ⃗ = ( y 1 , … , y r ) \vec{y}=(y_1,\ldots,y_r) y = ( y 1 , … , y r ) x ˉ h \bar{x}_h x ˉ h C h x ˉ h = y h C_h\bar{x}_h=y_h C h x ˉ h = y h (25) is
Σ g ( y ⃗ ) = ∑ x g ∈ { 0 , 1 } S g C g x g = y g ( x g ) v = 1 ( v ∈ E ∩ S g ) A g , U ( x g + ∑ h ≠ g x ˉ h ; z ) . \Sigma_g(\vec{y})=
\sum_{\substack{x_g\in\{0,1\}^{S_g}\\ C_gx_g=y_g\\
(x_g)_v=1\ (v\in E\cap S_g)}}
A_{g,U}\left(x_g+\sum_{h\ne g}\bar{x}_h;z\right). Σ g ( y ) = x g ∈ { 0 , 1 } S g C g x g = y g ( x g ) v = 1 ( v ∈ E ∩ S g ) ∑ A g , U x g + h = g ∑ x ˉ h ; z . Here vectors on individual groups are extended by zero. The sum has at most 2 ∣ S g ∣ 2^{|S_g|} 2 ∣ S g ∣
∥ Γ m t e s t , g Σ g ( y ⃗ ) ∥ 1 ≤ 2 ∣ S g ∣ B 1 m t e s t , g . \left\lVert\Gamma^{m_{\mathrm{test},g}}\Sigma_g(\vec{y})\right\rVert_1\le2^{|S_g|}B_1^{m_{\mathrm{test},g}}. ∥ Γ m test , g Σ g ( y ) ∥ 1 ≤ 2 ∣ S g ∣ B 1 m test , g . For 1 ≤ g ≤ r 1\le g\le r 1 ≤ g ≤ r Π g ( y ⃗ ) = ∏ h = 1 g Σ h ( y ⃗ ) \Pi_g(\vec{y})=\prod_{h=1}^{g}\Sigma_h(\vec{y}) Π g ( y ) = ∏ h = 1 g Σ h ( y )
Γ ∑ h ≤ g m t e s t , h . \Gamma^{\sum_{h\le g}m_{\mathrm{test},h}}. Γ ∑ h ≤ g m test , h . Consequently every full contribution Π r ( y ⃗ ) \Pi_r(\vec{y}) Π r ( y )
Ω = Γ m t e s t , \Omega=\Gamma^{m_{\mathrm{test}}}, Ω = Γ m test , and it satisfies
∥ Ω Π r ( y ⃗ ) ∥ 1 ≤ 2 n B 1 m t e s t . \left\lVert\Omega\Pi_r(\vec{y})\right\rVert_1\le2^nB_1^{m_{\mathrm{test}}}. ∥ Ω Π r ( y ) ∥ 1 ≤ 2 n B 1 m test . Let T t ( z , V ) T_t(z,V) T t ( z , V ) ( − 1 ) ∣ E ∣ − ∣ U ∣ (-1)^{|E|-|U|} ( − 1 ) ∣ E ∣ − ∣ U ∣ (34) , its final value is the A ℓ A_{\ell} A ℓ H t ( z ) H_t(z) H t ( z ) M ∗ M_* M ∗
M ∗ ≤ ( ∑ s = 0 min ( J − 1 , n ) 2 s ( n s ) ) ( 2 D + 1 ) c ≤ 3 n ( 2 D + 1 ) c . M_*\le\left(\sum_{s=0}^{\min(J-1,n)}2^s\binom{n}{s}\right)(2D+1)^c\le3^n(2D+1)^c. M ∗ ≤ s = 0 ∑ m i n ( J − 1 , n ) 2 s ( s n ) ( 2 D + 1 ) c ≤ 3 n ( 2 D + 1 ) c . Every intermediate value of T t T_t T t
∥ Ω T t ∥ 1 ≤ M ∗ 2 n B 1 m t e s t . \left\lVert\Omega T_t\right\rVert_1\le M_*2^nB_1^{m_{\mathrm{test}}}. ∥ Ω T t ∥ 1 ≤ M ∗ 2 n B 1 m test . Each coefficient of the integer polynomial Ω T t \Omega T_t Ω T t
O ( log ( M ∗ + 1 ) + n + m t e s t log B 1 ) O\left(\log(M_*+1)+n+m_{\mathrm{test}}\log B_1\right) O ( log ( M ∗ + 1 ) + n + m test log B 1 ) bits. This estimate also bounds the integer numerators of all group sums and partial products with the fixed denominators given above. The denominator satisfies
log Ω = k m t e s t ( 2 ℓ − 1 ) log d d e n = O ( k ( m t e s t ℓ + 1 ) Q log ( Q + 2 ) ) . \log\Omega=km_{\mathrm{test}}(2\ell-1)\log d_{\mathrm{den}}=O\left(k(m_{\mathrm{test}}\ell+1)Q\log(Q+2)\right). log Ω = k m test ( 2 ℓ − 1 ) log d den = O ( k ( m test ℓ + 1 ) Q log ( Q + 2 ) ) . Each contribution multiplies at most m t e s t m_{\mathrm{test}} m test
This is why exponentially many iterations contribute only log ( M ∗ + 1 ) \log(M_*+1) log ( M ∗ + 1 ) J ℓ ≤ ( d H + 1 ) ℓ J\ell\le(d_H+1)\ell J ℓ ≤ ( d H + 1 ) ℓ T t T_t T t (61) and polynomial bit cost per rational operation. The Q J ℓ QJ\ell Q J ℓ poly ( n , m t e s t , ℓ , D , Q , R , J , r , k , S ) \operatorname{poly}(n,m_{\mathrm{test}},\ell,D,Q,R,J,r,k,S) poly ( n , m test , ℓ , D , Q , R , J , r , k , S )
Interpolation and the integer answer
For 0 ≤ s < ℓ 0 \le s < \ell 0 ≤ s < ℓ H s ( T ) = [ V s ] H ( T ) ∈ Q [ T ] H_s(T)=[V^s]H(T)\in\mathbb{Q}[T] H s ( T ) = [ V s ] H ( T ) ∈ Q [ T ] [ z j V s ] H t ( z ) = [ z j ] H s ( t + z ) [z^jV^s]H_t(z)=[z^j]H_s(t+z) [ z j V s ] H t ( z ) = [ z j ] H s ( t + z ) t ∈ [ Q ] t\in[Q] t ∈ [ Q ] 0 ≤ j < J 0\le j<J 0 ≤ j < J s s s ℓ \ell ℓ M H = Q J > d H M_H=QJ>d_H M H = Q J > d H 1 , T , … , T M H − 1 1,T,\ldots,T^{M_H-1} 1 , T , … , T M H − 1
V ( t , j ) , u = { ( u j ) t u − j , u ≥ j , 0 , u < j , t ∈ [ Q ] , 0 ≤ j < J , 0 ≤ u < M H . V_{(t,j),u}=
\begin{cases}
\binom{u}{j}t^{u-j}, & u\ge j,\\
0, & u<j,
\end{cases}
\qquad t\in[Q],\quad0\le j<J,\quad0\le u<M_H. V ( t , j ) , u = { ( j u ) t u − j , 0 , u ≥ j , u < j , t ∈ [ Q ] , 0 ≤ j < J , 0 ≤ u < M H . It is nonsingular by the Hermite interpolation statement in Section 3. Its entries have O ( M H log ( Q + 2 ) ) O(M_H\log(Q+2)) O ( M H log ( Q + 2 )) y \mathbf{y} y [ z j V s ] H t ( z ) [z^jV^s]H_t(z) [ z j V s ] H t ( z ) ( t , j ) (t,j) ( t , j ) h \mathbf{h} h H s H_s H s M H M_H M H V h = y V\mathbf{h}=\mathbf{y} V h = y b = Ω y \mathbf{b}=\Omega\mathbf{y} b = Ω y B d a t a B_{\mathrm{data}} B data
B d a t a ≤ poly ( n , m t e s t , ℓ , k , Q , log ( D + 2 ) , log ( S + 2 ) ) . B_{\mathrm{data}}\le\operatorname{poly}(n,m_{\mathrm{test}},\ell,k,Q,\log(D+2),\log(S+2)). B data ≤ poly ( n , m test , ℓ , k , Q , log ( D + 2 ) , log ( S + 2 )) . Only the constant coefficient h 0 = H s ( 0 ) h_0=H_s(0) h 0 = H s ( 0 ) V ( 0 ) V^{(0)} V ( 0 ) V V V u = 0 u=0 u = 0 b \mathbf{b} b
N H = det V ( 0 ) , D H = det V ≠ 0. N_H=\det V^{(0)},\qquad D_H=\det V\ne0. N H = det V ( 0 ) , D H = det V = 0. Cramer’s rule for V ( Ω h ) = b V(\Omega\mathbf{h})=\mathbf{b} V ( Ω h ) = b
H s ( 0 ) = h 0 = N H Ω D H . (69) H_s(0)=h_0=\frac{N_H}{\Omega D_H}. \tag*{(69)} H s ( 0 ) = h 0 = Ω D H N H . (69) We compute these two determinants by fraction-free Gaussian elimination [10 Sections I–II and IV–V] . For an M H × M H M_H\times M_H M H × M H O ( M H 3 ) O(M_H^3) O ( M H 3 ) V V V V ( 0 ) V^{(0)} V ( 0 ) D H D_H D H N H N_H N H ( V ∣ b ) (V\mid\mathbf{b}) ( V ∣ b )
2 O ( M H ( B d a t a + M H log ( Q + 2 ) + log ( M H + 1 ) ) ) . 2^{O\left(M_H\left(B_{\mathrm{data}}+M_H\log(Q+2)+\log(M_H+1)\right)\right)}. 2 O ( M H ( B data + M H l o g ( Q + 2 ) + l o g ( M H + 1 ) ) ) . Indeed, its determinant is a sum of at most M H ! M_H! M H ! M H M_H M H
O ( M H ( B d a t a + M H log ( Q + 2 ) + log ( M H + 1 ) ) ) O\left(M_H\left(B_{\mathrm{data}}+M_H\log(Q+2)+\log(M_H+1)\right)\right) O ( M H ( B data + M H log ( Q + 2 ) + log ( M H + 1 ) ) ) bits in signed binary notation. Computing N H , D H N_H,D_H N H , D H poly ( M H , B d a t a ) \operatorname{poly}(M_H,B_{\mathrm{data}}) poly ( M H , B data )
At T = 0 T=0 T = 0 H s ( 0 ) H_s(0) H s ( 0 ) s s s ℓ \ell ℓ u = ∑ s = 0 ℓ − 1 H s ( 0 ) u=\sum_{s=0}^{\ell-1}H_s(0) u = ∑ s = 0 ℓ − 1 H s ( 0 ) (60) , the least nonnegative residue
u − 2 ℓ ⌊ u 2 ℓ ⌋ = N ( P ) u-2^\ell\left\lfloor\frac{u}{2^\ell}\right\rfloor=N(\mathcal{P}) u − 2 ℓ ⌊ 2 ℓ u ⌋ = N ( P ) is the exact count. Reusing the same workspace for successive coordinates, all ℓ \ell ℓ poly ( M H , B d a t a , log ( Ω + 1 ) , ℓ ) \operatorname{poly}(M_H,B_{\mathrm{data}},\log(\Omega+1),\ell) poly ( M H , B data , log ( Ω + 1 ) , ℓ )
B.4. Arithmetic for the technique overview. We also justify the bit-complexity and space bounds for the bounded-occurrence algorithm in § 2 . Lemma 4.6 bounds the bit lengths of the replacement coefficients by a polynomial in n n n § 2.4 gives measurement coefficients λ ⃗ e , v , h \vec{\lambda}_{e,v,h} λ e , v , h δ \delta δ δ \delta δ M ≤ 3 n M \le3n M ≤ 3 n δ 3 n \delta^{3n} δ 3 n
Appendix C. Additional Consequences of Refuting ETH
Theorem 1.1 gives a circuit lower bound and subexponential-time algorithms for inverting Boolean maps with constant output locality. The fixed-density case gives the quantitative circuit lower bound in Corollary 1.4 . Both circuit arguments use only the decision version of our 3-SAT algorithm. Exact preimage counting uses its full counting guarantee.
Circuit lower bounds for EN P ^{\mathrm{NP}} NP
For the qualitative bound below, write SIZE ( O ( n ) ) \operatorname{SIZE}(O(n)) SIZE ( O ( n )) O ( n ) O(n) O ( n )
Corollary C.1. We have
E N P ⊈ SIZE ( O ( n ) ) . \mathrm{E}^{\mathrm{NP}} \not\subseteq\operatorname{SIZE}(O(n)). E NP ⊆ SIZE ( O ( n )) . Proof. Williams proved that if ETH is false, then EN P ^{\mathrm{NP}} NP [71 Theorem 6.1] . Apply this implication to the deterministic subexponential-time 3-SAT algorithm supplied by Theorem 1.1 . □
The connection is an instance of deriving circuit lower bounds from faster satisfiability algorithms. To encode a Boolean circuit with s s s n n n O ( n + s ) O(n+s) O ( n + s ) 2 o ( n ) 2^{o(n)} 2 o ( n ) N P ^{\mathrm{NP}} NP
Proof of Corollary 1.4. We use Williams’s witness-circuit argument [71 Theorem 6.1] with the local reduction of Jahanjou, Miles, and Viola [40 Theorem 1 and Section 6] .
By Žák’s nondeterministic time hierarchy [75 ] , in the form stated in [71 Theorem 2.1 in the full version] , fix a unary language H ∈ NTIME ( 2 n ) ∖ NTIME ( 2 n / 2 ) H \in\operatorname{NTIME}(2^n) \setminus\operatorname{NTIME}(2^{n/2}) H ∈ NTIME ( 2 n ) ∖ NTIME ( 2 n /2 ) n n n 1 n 1^n 1 n r = 2 r=2 r = 2 T = 2 n n O ( 1 ) T=2^n n^{O(1)} T = 2 n n O ( 1 ) q = n + O ( log n ) q=n+O(\log n) q = n + O ( log n ) φ n \varphi_n φ n 2 q 2^q 2 q 1 n ∈ H 1^n \in H 1 n ∈ H
Each of the decoder’s O ( n ) O(n) O ( n ) w w w w w w O ( n ) O(n) O ( n )
Pad each variable label to d n dn d n d d d 2 d n 2^{dn} 2 d n ∣ i ∣ = d n |i|=dn ∣ i ∣ = d n n n n φ n \varphi_n φ n W ( i ) W(i) W ( i ) i i i W ( i ) = 0 W(i)=0 W ( i ) = 0 i i i n n n W W W 2 q 2^q 2 q 2 d n 2^{dn} 2 d n 2 O ( n ) 2^{O(n)} 2 O ( n ) W ∈ E N P W\in\mathrm{E}^{\mathrm{NP}} W ∈ E NP
Put a ( t ) = log 2 t a(t)=\sqrt{\log_2 t} a ( t ) = log 2 t Theorem 1.1 is positive decides 3-SAT with V V V 5 V 5V 5 V V b 2 A V / a ( V ) V^{b2^{AV/a(V)}} V b 2 A V / a ( V ) A , b > 0 A,b>0 A , b > 0 A A A V O ( 1 ) V^{O(1)} V O ( 1 ) b b b
Fix a positive rational η < 1 / ( 12 A ) \eta<1/(12A) η < 1/ ( 12 A ) s n = ⌊ η n ⌊ a ( n ) ⌋ ⌋ s_n=\lfloor\eta n\lfloor a(n)\rfloor\rfloor s n = ⌊ η n ⌊ a ( n )⌋⌋ n ≥ 2 n\ge2 n ≥ 2 n n n ⌊ log 2 n ⌋ \lfloor\log_2 n\rfloor ⌊ log 2 n ⌋ C W ( d n ) ≤ s n C_W(dn)\le s_n C W ( d n ) ≤ s n n n n H H H 1 n 1^n 1 n d n dn d n s n s_n s n 1 n ∈ H 1^n\in H 1 n ∈ H W W W d n dn d n 1 n ∉ H 1^n\notin H 1 n ∈ / H
The test circuit has q q q 3 s n + O ( n ) 3s_n+O(n) 3 s n + O ( n ) U n = 3 s n + B n U_n=3s_n+Bn U n = 3 s n + B n 5 U n 5U_n 5 U n B B B U n = ( 3 η + o ( 1 ) ) n a ( n ) U_n=(3\eta+o(1))n a(n) U n = ( 3 η + o ( 1 )) na ( n ) a ( U n ) ∼ a ( n ) a(U_n)\sim a(n) a ( U n ) ∼ a ( n ) 2 ( 3 A η + o ( 1 ) ) n 2^{(3A\eta+o(1))n} 2 ( 3 A η + o ( 1 )) n H H H 2 n / 2 2^{n/2} 2 n /2 n n n H H H C W ( d n ) > s n C_W(dn)>s_n C W ( d n ) > s n n n n s n ∼ η n log 2 n s_n\sim\eta n\sqrt{\log_2 n} s n ∼ η n log 2 n d d d L = W L=W L = W 0 < c < η / ( d log 2 ) 0<c<\eta/(d\sqrt{\log2}) 0 < c < η / ( d log 2 ) (2) . □
Inverting Boolean maps with constant output locality
Fix a constant d d d G : { 0 , 1 } n → { 0 , 1 } m G:\{0,1\}^n\to\{0,1\}^m G : { 0 , 1 } n → { 0 , 1 } m i i i
G ( x ⃗ ) i = g i ( x ⃗ ∣ S i ) , S i ⊆ [ n ] , ∣ S i ∣ ≤ d . G(\vec{x})_i=g_i(\vec{x}|_{S_i}),\qquad S_i\subseteq[n],\quad|S_i|\le d. G ( x ) i = g i ( x ∣ S i ) , S i ⊆ [ n ] , ∣ S i ∣ ≤ d . The input description of G G G S i S_i S i g i g_i g i S i S_i S i b ⃗ ∈ { 0 , 1 } m \vec{b}\in\{0,1\}^m b ∈ { 0 , 1 } m x ⃗ \vec{x} x G ( x ⃗ ) = b ⃗ G(\vec{x})=\vec{b} G ( x ) = b
Corollary C.2. For every fixed d d d n ≥ 20 n\ge20 n ≥ 20 ∣ G − 1 ( b ⃗ ) ∣ |G^{-1}(\vec{b})| ∣ G − 1 ( b ) ∣
( n + m ) O d ( 1 ) 2 d O d ( ρ ( n ) ) (n+m)^{O_d(1)}2^{d\,O_d(\rho(n))} ( n + m ) O d ( 1 ) 2 d O d ( ρ ( n )) and space ( n + m ) O d ( 1 ) (n+m)^{O_d(1)} ( n + m ) O d ( 1 )
Proof. For each i i i α ⃗ ∈ { 0 , 1 } S i \vec{\alpha}\in\{0,1\}^{S_i} α ∈ { 0 , 1 } S i g i ( α ⃗ ) ≠ b i g_i(\vec{\alpha})\ne b_i g i ( α ) = b i
⋁ j ∈ S i ℓ j , α j , ℓ j , 0 = X j , ℓ j , 1 = ¬ X j . \bigvee_{j\in S_i}\ell_{j,\alpha_j},\qquad\ell_{j,0}=X_j,\quad\ell_{j,1}=\neg X_j. j ∈ S i ⋁ ℓ j , α j , ℓ j , 0 = X j , ℓ j , 1 = ¬ X j . This clause forbids exactly the local assignment α ⃗ \vec{\alpha} α S i S_i S i n n n d d d 2 d m 2^d m 2 d m G − 1 ( b ⃗ ) G^{-1}(\vec{b}) G − 1 ( b ) Theorem 1.1 with k = max { 2 , d } k=\max\{2,d\} k = max { 2 , d } n n n
Conditional counts also allow an exactly uniform preimage to be sampled, when the preimage set is nonempty, with the same bound in expected time up to polynomial factors [41 ] . In particular, the corollary applies to constant-locality versions of Goldreich’s candidate one-way functions, which evaluate local predicates on overlapping subsets of input bits [32 ] . For polynomially many output bits, it gives a 2 o ( n ) 2^{o(n)} 2 o ( n ) 2 Ω ( n ) 2^{\Omega(n)} 2 Ω ( n )
A sharper bound for exact counting
We prove Theorem 1.1 by contracting Boolean tensor networks. The two running times reflect the number of wires in the networks. A formula with m ≤ D n m \le Dn m ≤ D n O k , D ( n ) O_{k,D}(n) O k , D ( n ) § 6 . The degree bound of Lemma 5.2 lets us index a monomial retained in the product of their polynomial lifts with O k ( n log log n ) O_k(n \log\log n) O k ( n log log n )
Boolean tensor contraction
A Boolean tensor network is a bipartite multigraph G G G v v v δ ( v ) \delta(v) δ ( v ) T v T_v T v
Z = ∑ x ∈ { 0 , 1 } E ( G ) ∏ v ∈ V ( G ) T v ( x ∣ δ ( v ) ) . Z = \sum_{x \in\{0,1\}^{E(G)}} \prod_{v \in V(G)} T_v(x|_{\delta(v)}). Z = x ∈ { 0 , 1 } E ( G ) ∑ v ∈ V ( G ) ∏ T v ( x ∣ δ ( v ) ) . An isolated vertex contributes its scalar entry. We measure the size of the network by its number of edges.
Theorem D.1 (Boolean tensor contraction). Fix positive constants C C C A A A b b b s ≥ 2 s \ge2 s ≥ 2 L \mathcal{L} L C s Cs C s ⌈ s 9 / 10 ⌉ \lceil s^{9/10} \rceil ⌈ s 9/10 ⌉ s s s
B P 0 , ℓ = ( Z / 2 P 0 Z ) [ V ] / ( V ℓ ) , 1 ≤ P 0 , ℓ ≤ C s . \mathcal{B}_{P_0,\ell} = (\mathbb{Z}/2^{P_0}\mathbb{Z})[V]/(V^\ell), \qquad1 \le P_0,\ell\le Cs. B P 0 , ℓ = ( Z / 2 P 0 Z ) [ V ] / ( V ℓ ) , 1 ≤ P 0 , ℓ ≤ C s . Suppose, uniformly for every vertex and Boolean choice of its ports, that its entry can be computed from the description in time ( s + L ) b 2 A s 9 / 10 (s+\mathcal{L})^{b}2^{As^{9/10}} ( s + L ) b 2 A s 9/10 ( s + L ) b (s+\mathcal{L})^b ( s + L ) b
( s + L ) O ( 1 ) 2 O ( s / log s ) (s+\mathcal{L})^{O(1)}2^{O(s/\sqrt{\log s})} ( s + L ) O ( 1 ) 2 O ( s / l o g s ) bit operations and ( s + L ) O ( 1 ) (s+\mathcal{L})^{O(1)} ( s + L ) O ( 1 ) ℓ = 1 \ell=1 ℓ = 1
The algorithm in Section D.4 replaces a single global Taylor expansion by recursion on the wires. At each call it deforms the wire identifications so that, at a chosen interpolation node, only ports in the same group interact. Near that node, interactions between groups expand in variables called modes. A mode of weight j j j 2 j 2^j 2 j M M M u u u 2 O ( M / u ) 2^{O(M/\sqrt{u})} 2 O ( M / u ) M / 8 M/8 M /8 d d d M = O ( s 8 − d ) M = O(s8^{-d}) M = O ( s 8 − d ) u = Θ ( ( log s ) / 2 d ) u = \Theta((\log s)/2^d) u = Θ (( log s ) / 2 d ) O ( s / log s ) O(s/\sqrt{\log s}) O ( s / log s )
Fixed clause density
Proof. of the fixed-density bound in Theorem 1.1 . Give each clause a vertex carrying its Boolean satisfaction table, with one port for each distinct variable in the clause. A tautology may be dropped, and an empty clause gives count zero. For a variable with d ≥ 1 d \ge1 d ≥ 1 d = 1 d = 1 d = 1 ( 1 , 1 ) (1,1) ( 1 , 1 ) d = 2 d = 2 d = 2 d ≥ 3 d \ge3 d ≥ 3 d d d d − 2 d-2 d − 2
There are at most k m km k m O k ( n + m ) O_k(n+m) O k ( n + m ) max { k , 3 } \max\{k,3\} max { k , 3 }
With s = n + m s=n+m s = n + m Theorem D.1 over Z / 2 n + 1 Z \mathbb{Z}/2^{n+1}\mathbb{Z} Z / 2 n + 1 Z P 0 = n + 1 P_0=n+1 P 0 = n + 1 ℓ = 1 \ell=1 ℓ = 1 k k k s s s u u u 2 u 2^u 2 u 2 n + 1 2^{n+1} 2 n + 1 # F \#F # F 0 ≤ # F ≤ 2 n 0 \le\#F \le2^n 0 ≤ # F ≤ 2 n s O k ( 1 ) 2 O k ( s / log s ) s^{O_k(1)}2^{O_k(s/\sqrt{\log s})} s O k ( 1 ) 2 O k ( s / l o g s ) m ≤ D n m \le Dn m ≤ D n n ≤ s ≤ ( 1 + D ) n n \le s \le(1+D)n n ≤ s ≤ ( 1 + D ) n O k , D ( n / log n ) O_{k,D}(n/\sqrt{\log n}) O k , D ( n / log n )
Arbitrary clause density
The sparsification and greedy row selection in § 6 reduce this case to counting common zeros of sparse polynomial equations. We retain the parity lift of § 5 . Its degree bound lets us encode a monomial by a subset of test–variable incidences using O ( n log log n ) O(n\log\log n) O ( n log log n )
Lemma D.2 (Counting sparse polynomial systems by contraction). Fix integers k , c ≥ 1 k,c \ge1 k , c ≥ 1 P 1 , … , P a P_1,\ldots,P_a P 1 , … , P a F 2 \mathbb{F}_2 F 2 n ≥ 20 n \ge20 n ≥ 20 a ≤ c n a \le cn a ≤ c n P i P_i P i ⌈ log 2 n ⌉ c \lceil\log_2 n\rceil^c ⌈ log 2 n ⌉ c k k k n O k , c ( 1 ) 2 O k , c ( n log log n / log n ) n^{O_{k,c}(1)}2^{O_{k,c}(n\log\log n/\sqrt{\log n})} n O k , c ( 1 ) 2 O k , c ( n l o g l o g n / l o g n ) n O k , c ( 1 ) n^{O_{k,c}(1)} n O k , c ( 1 )
Proof. Put λ = ⌈ log 2 n ⌉ \lambda=\lceil\log_2 n\rceil λ = ⌈ log 2 n ⌉ ν = ⌈ log 2 λ ⌉ \nu=\lceil\log_2\lambda\rceil ν = ⌈ log 2 λ ⌉ i i i k λ c k\lambda^c k λ c N 0 N_0 N 0 ( i , v ) (i,v) ( i , v ) v v v i i i N 0 ≤ a k λ c N_0 \le ak\lambda^c N 0 ≤ ak λ c
D ∗ = max { 1 , ⌊ N 0 / n ⌋ } ⌈ λ ⌉ D_*=\max\{1,\lfloor N_0/n\rfloor\}\lceil\sqrt{\lambda}\rceil D ∗ = max { 1 , ⌊ N 0 / n ⌋} ⌈ λ ⌉ scopes. There are at most n / λ n/\sqrt{\lambda} n / λ 2 n / λ 2^{n/\sqrt{\lambda}} 2 n / λ I i I_i I i i i i k λ c k\lambda^c k λ c D ∗ = λ O k , c ( 1 ) D_*=\lambda^{O_{k,c}(1)} D ∗ = λ O k , c ( 1 ) a a a n n n ( i , v ) (i,v) ( i , v ) v ∈ I i v\in I_i v ∈ I i z i v z_{iv} z i v z i = ( z i v ) v ∈ I i z_i=(z_{iv})_{v\in I_i} z i = ( z i v ) v ∈ I i L ~ i ( z i ) \widetilde L_i(z_i) L i ( z i ) z i v z_{iv} z i v x v x_v x v
B = ( Z / 2 p Z ) [ V ] / ( V ℓ ) , p = ℓ = n + 1. \mathcal{B}=(\mathbb{Z}/2^p\mathbb{Z})[V]/(V^\ell), \qquad p=\ell=n+1. B = ( Z / 2 p Z ) [ V ] / ( V ℓ ) , p = ℓ = n + 1. Take h ( Y ) = Y ( Y − 1 ) h(Y)=Y(Y-1) h ( Y ) = Y ( Y − 1 ) e j e_j e j 2 p 2^p 2 p
B i ( z i , V ) = ∑ j = 0 ℓ − 1 e j ( L ~ i ( z i ) ) h ( L ~ i ( z i ) ) j V j . B_i(z_i,V)=\sum_{j=0}^{\ell-1}e_j(\widetilde{L}_i(z_i))h(\widetilde{L}_i(z_i))^jV^j. B i ( z i , V ) = j = 0 ∑ ℓ − 1 e j ( L i ( z i )) h ( L i ( z i ) ) j V j . Let B ‾ i \overline{B}_i B i z i z_i z i V V V B i B_i B i z i z_i z i F ( z , V ) = ∏ i B ‾ i ( z i , V ) F(z,V)=\prod_i\overline{B}_i(z_i,V) F ( z , V ) = ∏ i B i ( z i , V ) B [ z ] \mathcal{B}[z] B [ z ] F F F x v x_v x v v v v
The coefficient of V j V^j V j B ‾ i \overline{B}_i B i k ( 1 + 2 j ) k(1+2j) k ( 1 + 2 j ) deg e j ≤ 1 \deg e_j\leq1 deg e j ≤ 1 deg h = 2 \deg h=2 deg h = 2 B \mathcal{B} B ∑ i j i ≤ ℓ − 1 \sum_i j_i\leq\ell-1 ∑ i j i ≤ ℓ − 1 F F F
K = k ( a + 2 ( ℓ − 1 ) ) = O k , c ( n ) . (70) K=k\bigl(a+2(\ell-1)\bigr)=O_{k,c}(n). \tag*{(70)} K = k ( a + 2 ( ℓ − 1 ) ) = O k , c ( n ) . (70) Let S ℓ S_\ell S ℓ 1 , V , … , V ℓ − 1 1,V,\ldots,V^{\ell-1} 1 , V , … , V ℓ − 1 B \mathcal{B} B z i v = x v z_{iv}=x_v z i v = x v B ‾ i \overline{B}_i B i B i B_i B i L ~ i ( ( x v ) v ∈ I i ) \widetilde{L}_i((x_v)_{v\in I_i}) L i (( x v ) v ∈ I i ) P i P_i P i x x x 2 p 2^p 2 p
S ℓ ( ∑ x F ( ( z i v = x v ) i v , V ) ) , (71) S_\ell\left(\sum_x F\bigl((z_{iv}=x_v)_{iv},V\bigr)\right), \tag*{(71)} S ℓ ( x ∑ F ( ( z i v = x v ) i v , V ) ) , (71) where x x x
Indexing the monomials. Partition the tests and the free variables separately into J = ⌈ n 1 / 4 ⌉ J=\lceil n^{1/4}\rceil J = ⌈ n 1/4 ⌉ T g T_g T g X h X_h X h O ( n 3 / 4 ) O(n^{3/4}) O ( n 3/4 ) n 3 / 4 λ O k , c ( 1 ) n^{3/4\lambda}O_{k,c}(1) n 3/4 λ O k , c ( 1 ) N g h N_{gh} N g h T g T_g T g X h X_h X h N = ∑ g , h N g h ≤ N 0 N=\sum_{g,h}N_{gh}\leq N_0 N = ∑ g , h N g h ≤ N 0
For an incidence monomial let d g h d_{gh} d g h g , h g,h g , h ( d g h ) (d_{gh}) ( d g h ) 0 ≤ d g h ≤ N g h 0\leq d_{gh}\leq N_{gh} 0 ≤ d g h ≤ N g h ∑ g , h d g h ≤ K \sum_{g,h}d_{gh}\leq K ∑ g , h d g h ≤ K
( K + 1 ) J 2 = 2 O k , c ( n log n ) (K+1)^{J^2}=2^{O_{k,c}(\sqrt{n}\log n)} ( K + 1 ) J 2 = 2 O k , c ( n l o g n ) arrays, which can be enumerated using J 2 J^2 J 2
C g h = ( N g h d g h ) C_{gh}=\binom{N_{gh}}{d_{gh}} C g h = ( d g h N g h ) subsets of the specified size for this pair. Draw ⌈ log 2 C g h ⌉ \lceil\log_2 C_{gh}\rceil ⌈ log 2 C g h ⌉ C g h = 1 C_{gh}=1 C g h = 1
The union of the subsets over all pairs specifies a squarefree monomial of degree at most K K K N N N ( N + K K ) \binom{N+K}{K} ( K N + K ) M M M
M ≤ J 2 + log 2 ∏ g , h C g h ≤ J 2 + log 2 ( N + K K ) ≤ J 2 + K log 2 ( e ( N + K ) K ) = O k , c ( n ν ) . (72) \begin{aligned}
M &\le J^{2}+\log_{2}\prod_{g,h} C_{gh}\le J^{2}+\log_{2}\binom{N+K}{K}\\
&\le J^{2}+K\log_{2}\left(\frac{e(N+K)}{K}\right)=O_{k,c}(n\nu). \tag*{(72)}
\end{aligned} M ≤ J 2 + log 2 g , h ∏ C g h ≤ J 2 + log 2 ( K N + K ) ≤ J 2 + K log 2 ( K e ( N + K ) ) = O k , c ( n ν ) . (72) Here K = Θ k , c ( n ) K=\Theta_{k,c}(n) K = Θ k , c ( n ) N ≤ n λ O k , c ( 1 ) N\le n^{\lambda}O_{k,c}(1) N ≤ n λ O k , c ( 1 ) ⌈ log 2 C g h ⌉ ≤ N g h \lceil\log_{2}C_{gh}\rceil\le N_{gh} ⌈ log 2 C g h ⌉ ≤ N g h n 3 / 4 λ O k , c ( 1 ) n^{3/4\lambda}O_{k,c}(1) n 3/4 λ O k , c ( 1 )
Tensor entries and contraction. A valid assignment of wire bits determines a subset S i ⊆ I i S_i\subseteq I_i S i ⊆ I i z i , S i = ∏ v ∈ S i z i v z_{i,S_i}=\prod_{v\in S_i}z_{iv} z i , S i = ∏ v ∈ S i z i v g g g ∏ i ∈ T g B i ‾ ( z i , V ) \prod_{i\in T_g}\overline{B_i}(z_i,V) ∏ i ∈ T g B i ( z i , V ) z i z_i z i
∏ i ∈ T g [ z i , S i ] B i ‾ ( z i , V ) in B . \prod_{i\in T_g}[z_i,S_i]\overline{B_i}(z_i,V)\qquad\text{in }\mathcal{B}. i ∈ T g ∏ [ z i , S i ] B i ( z i , V ) in B . For a variable group h h h b v = ∣ { i : v ∈ S i } ∣ b_v=\lvert\{i:v\in S_i\}\rvert b v = ∣{ i : v ∈ S i }∣ v ∈ X h v\in\mathcal{X}_h v ∈ X h
∑ x ∈ { 0 , 1 } X h ∏ v ∈ X h x v b v = 2 ∣ { v ∈ X h : b v = 0 } ∣ ( m o d 2 p ) . \sum_{x\in\{0,1\}^{\mathcal{X}_h}}\prod_{v\in\mathcal{X}_h}x_v^{b_v}=2^{\lvert\{v\in\mathcal{X}_h:b_v=0\}\rvert}\pmod{2^p}. x ∈ { 0 , 1 } X h ∑ v ∈ X h ∏ x v b v = 2 ∣{ v ∈ X h : b v = 0 }∣ ( mod 2 p ) . Indeed a positive exponent requires x v = 1 x_v=1 x v = 1 x v x_v x v
Each monomial of F F F F F F v v v B \mathcal{B} B S ℓ S_{\ell} S ℓ
To evaluate one test-side coefficient, let 1 T \mathbf{1}_T 1 T I i I_i I i T ⊆ S i T\subseteq S_i T ⊆ S i 1 T \mathbf{1}_T 1 T T T T B i B_i B i
[ z i , S i ] B i ‾ ( z i , V ) = ∑ T ⊆ S i ( − 1 ) ∣ S i ∣ − ∣ T ∣ B i ( 1 T , V ) in B . [z_i,S_i]\overline{B_i}(z_i,V)=\sum_{T\subseteq S_i}(-1)^{\lvert S_i\rvert-\lvert T\rvert}B_i(\mathbf{1}_T,V)\qquad\text{in }\mathcal{B}. [ z i , S i ] B i ( z i , V ) = T ⊆ S i ∑ ( − 1 ) ∣ S i ∣ − ∣ T ∣ B i ( 1 T , V ) in B . Each term costs polynomially many bit operations. Evaluate L ~ i ( 1 T ) \widetilde{L}_i(\mathbf{1}_T) L i ( 1 T ) e j e_j e j ℓ \ell ℓ B i ( 1 T , V ) B_i(\mathbf{1}_T,V) B i ( 1 T , V ) 2 p 2^p 2 p 2 ∣ S i ∣ ≤ 2 k λ c 2^{\lvert S_i\rvert}\le2^{k\lambda c} 2 ∣ S i ∣ ≤ 2 k λ c B \mathcal{B} B 2 λ O k , c ( 1 ) 2^{\lambda O_{k,c}(1)} 2 λ O k , c ( 1 )
Apply Theorem D.1 with size parameter s = n ν s=n^\nu s = n ν p = n + 1 p=n+1 p = n + 1 ℓ = n + 1 \ell=n+1 ℓ = n + 1 2 J 2J 2 J O k , c ( s ) O_{k,c}(s) O k , c ( s ) n n n ⌈ s 9 / 10 ⌉ \lceil s^{9/10}\rceil ⌈ s 9/10 ⌉ O ( s ) O(s) O ( s ) 2 O k , c ( s 9 / 10 ) 2^{O_{k,c}(s^{9/10})} 2 O k , c ( s 9/10 ) n O k , c ( 1 ) 2 O k , c ( s / log s ) n^{O_{k,c}(1)}2^{O_{k,c}(s/\sqrt{\log s})} n O k , c ( 1 ) 2 O k , c ( s / l o g s ) s / log s = O ( n ν / λ ) s/\sqrt{\log s}=O(n^\nu/\sqrt{\lambda}) s / log s = O ( n ν / λ )
2 n ν / λ + O k , c ( n log n ) . 2^{n^\nu/\sqrt{\lambda}+O_{k,c}(\sqrt{n}\log n)}. 2 n ν / λ + O k , c ( n l o g n ) . Both terms in this exponent are O k , c ( n ν / λ ) O_{k,c}(n^\nu/\sqrt{\lambda}) O k , c ( n ν / λ ) B \mathcal{B} B S ℓ S_{\ell} S ℓ 2 n + 1 2^{n+1} 2 n + 1 2 n 2^n 2 n
Proof of the arbitrary-density bound in Theorem 1.1 . Normalize the formula on its original n n n § 6.4 . Empty clauses and formulas with no clauses have immediate answers. For n ≥ 20 n \ge20 n ≥ 20 Lemma 6.3 with σ = 1 / λ \sigma= 1/\sqrt{\lambda} σ = 1/ λ λ = ⌈ log 2 n ⌉ \lambda= \lceil\log_2 n \rceil λ = ⌈ log 2 n ⌉ n O k ( 1 ) 2 O k ( n / λ ) n^{O_k(1)}2^{O_k(n/\sqrt{\lambda})} n O k ( 1 ) 2 O k ( n / λ ) B = O k ( ( σ − 1 log ( 2 / σ ) ) k − 1 ) = λ O k ( 1 ) B = O_k((\sigma^{-1}\log(2/\sigma))^{k-1}) = \lambda^{O_k(1)} B = O k (( σ − 1 log ( 2/ σ ) ) k − 1 ) = λ O k ( 1 ) B n Bn B n
For a nontrivial leaf with m G m_G m G Lemma 6.1 and the greedy procedure of Lemma 6.2 . Each initial row has support size at most D 0 D_0 D 0 P P P λ O k ( 1 ) \lambda^{O_k(1)} λ O k ( 1 ) (62) , every row gives a degree-at-most-k k k 2 k max { P , D 0 } 2^k\max\{P,D_0\} 2 k max { P , D 0 } 6 n 6n 6 n c = c ( k ) c=c(k) c = c ( k ) λ \lambda λ n n n Lemma D.2 to compute N ( A ; h γ ) N(A;h_\gamma) N ( A ; h γ ) h γ h_\gamma h γ N ( A ) N(A) N ( A ) Lemma 6.2 .
Constructing the initial checks and enumerating candidates costs 2 λ O k ( 1 ) 2^{\lambda^{O_k(1)}} 2 λ O k ( 1 ) 3 n ( 2 m G ) P + 1 = 2 λ O k ( 1 ) 3n(2m_G)^P+1=2^{\lambda^{O_k(1)}} 3 n ( 2 m G ) P + 1 = 2 λ O k ( 1 ) λ \lambda λ o ( n log log n / log n ) o(n\log\log n/\sqrt{\log n}) o ( n log log n / log n )
( n + m ) O k ( 1 ) 2 O k ( n log log n / log n ) . (n+m)^{O_k(1)}2^{O_k(n\log\log n/\sqrt{\log n})}. ( n + m ) O k ( 1 ) 2 O k ( n l o g l o g n / l o g n ) . Here m m m ( n + 1 ) (n+1) ( n + 1 )
A single algorithm attains both bounds. With λ = ⌈ log 2 n ⌉ \lambda=\lceil\log_2 n\rceil λ = ⌈ log 2 n ⌉ ν = ⌈ log 2 λ ⌉ \nu=\lceil\log_2\lambda\rceil ν = ⌈ log 2 λ ⌉ m ≤ n ν m\le n\nu m ≤ n ν n ≤ s = n + m = O ( n log log n ) n\le s=n+m=O(n\log\log n) n ≤ s = n + m = O ( n log log n ) O k ( n log log n / log n ) O_k(n\log\log n/\sqrt{\log n}) O k ( n log log n / log n ) D D D n n n m ≤ D n m\le Dn m ≤ D n s = O D ( n ) s=O_D(n) s = O D ( n )
Proof of the contraction theorem
Throughout the proof of Theorem D.1 , log \log log ln \ln ln
The coefficient ring and exact arithmetic
Write λ = ⌈ log 2 s ⌉ \lambda=\lceil\log_2 s\rceil λ = ⌈ log 2 s ⌉ e = 4 λ e=4\lambda e = 4 λ f ∈ Z [ U ] f\in\mathbb{Z}[U] f ∈ Z [ U ] e e e { 0 , 1 } \{0,1\} { 0 , 1 } s s s 2 e 2^e 2 e F 2 \mathbb{F}_2 F 2 e / 2 e/2 e /2
O = Z ( 2 ) [ U ] / ( f ) , K = Q [ U ] / ( f ) , A = O [ V ] / ( V ℓ ) , A p = A / ( 2 p ) . \mathcal{O}=\mathbb{Z}_{(2)}[U]/(f),\qquad
\mathcal{K}=\mathbb{Q}[U]/(f),\qquad
\mathcal{A}=\mathcal{O}[V]/(V^\ell),\qquad
\mathcal{A}_p=\mathcal{A}/(2^p). O = Z ( 2 ) [ U ] / ( f ) , K = Q [ U ] / ( f ) , A = O [ V ] / ( V ℓ ) , A p = A / ( 2 p ) . Here Z ( 2 ) \mathbb{Z}_{(2)} Z ( 2 ) Q \mathbb{Q} Q K \mathcal{K} K K \mathcal{K} K v 2 v_2 v 2 1 , U , … , U e − 1 1,U,\ldots,U^{e-1} 1 , U , … , U e − 1 O \mathcal{O} O F 2 e \mathbb{F}_{2^e} F 2 e
v 2 ( x y ) = v 2 ( x ) + v 2 ( y ) ( x , y ≠ 0 ) . v_2(xy)=v_2(x)+v_2(y)\qquad(x,y\ne0). v 2 ( x y ) = v 2 ( x ) + v 2 ( y ) ( x , y = 0 ) . Every element of O \mathcal{O} O Z ( 2 ) \mathbb{Z}_{(2)} Z ( 2 )
Arithmetic in A p \mathcal{A}_p A p s , p , ℓ s,p,\ell s , p , ℓ 1 , U , … , U e − 1 1,U,\ldots,U^{e-1} 1 , U , … , U e − 1 1 , V , … , V ℓ − 1 1,V,\ldots,V^{\ell-1} 1 , V , … , V ℓ − 1 2 p − 1 2^p-1 2 p − 1 [31 ] . Divisibility and precision for elements of A \mathcal{A} A
The binary-coordinate lifts of the 2 e 2^e 2 e A p \mathcal{A}_p A p s 4 s^4 s 4 B P 0 , ℓ \mathcal{B}_{P_0,\ell} B P 0 , ℓ A P 0 \mathcal{A}_{P_0} A P 0
When the precision increases, we lift each tensor entry separately. For an entry originally specified modulo 2 p 2^p 2 p canonical lift is the element of A \mathcal{A} A p c p_c p c P 0 P_0 P 0 2 p c 2^{p_c} 2 p c
Parameters and collision packing
All estimates below hold for sufficiently large s s s
H = ⌈ s ∣ λ ∣ ⌉ , r = 64 , ε = 1 128 ⋅ 1024 , Δ = ⌈ s 9 / 10 ⌉ . H = \left\lceil\frac{s}{\lvert\sqrt{\lambda}\rvert} \right\rceil,\qquad r=64,\qquad\varepsilon=\frac{1}{128\cdot1024},\qquad\Delta=\left\lceil s^{9/10}\right\rceil. H = ⌈ ∣ λ ∣ s ⌉ , r = 64 , ε = 128 ⋅ 1024 1 , Δ = ⌈ s 9/10 ⌉ . A call with M ≤ H M \le H M ≤ H d d d
u = u d = ⌈ ε λ 2 d ⌉ , a = a d = 2 ∑ j = 0 d u j , a − 1 = 0 , L = 2 a , D a r r = r 2 a . (73) u=u_d=\left\lceil\frac{\varepsilon\lambda}{2^d}\right\rceil,\qquad a=a_d=2\sum_{j=0}^{d}u_j,\qquad a_{-1}=0,\qquad L=2^a,\qquad D_{\mathrm{arr}}=r^{2a}. \tag*{(73)} u = u d = ⌈ 2 d ε λ ⌉ , a = a d = 2 j = 0 ∑ d u j , a − 1 = 0 , L = 2 a , D arr = r 2 a . (73) We will arrange that every child has at most M / 8 M/8 M /8 Δ \Delta Δ
H < M ≤ C s 8 2 − d , d ≤ log 2 λ 6 + O C ( 1 ) . H<M\le Cs^8 2^{-d},\qquad d\le\frac{\log_2\lambda}{6}+O_C(1). H < M ≤ C s 8 2 − d , d ≤ 6 log 2 λ + O C ( 1 ) . Consequently
u = Θ C ( λ / 2 d ) , λ 4 / 5 ≤ u ≤ M , a ≤ 4 ε λ , L , D a r r ≤ s 1 / 1000 (74) u=\Theta_C(\lambda/2^d),\qquad\lambda^{4/5}\le u\le M,\qquad a\le4\varepsilon\lambda,\qquad L,D_{\mathrm{arr}}\le s^{1/1000} \tag*{(74)} u = Θ C ( λ / 2 d ) , λ 4/5 ≤ u ≤ M , a ≤ 4 ε λ , L , D arr ≤ s 1/1000 (74) for all sufficiently large s s s
M ≥ 2 D a r r 2 Δ , M ≥ L D a r r , 2 L ≤ Δ . (75) M\ge2D_{\mathrm{arr}}^2\Delta,\qquad M\ge LD_{\mathrm{arr}},\qquad2L\le\Delta. \tag*{(75)} M ≥ 2 D arr 2 Δ , M ≥ L D arr , 2 L ≤ Δ. (75) For example, the right side of the first inequality is O ( s 9 / 10 + 2 / 1000 ) = o ( H ) O(s^{9/10+2/1000})=o(H) O ( s 9/10 + 2/1000 ) = o ( H )
Lemma D.3 (Packing into packets). Under (D.6), label the vertices on each side of the network by elements of [ r ] a [r]^a [ r ] a L L L
(1) There are at most 3 M / L 3M/L 3 M / L
(2) In every packet and every coordinate, the unordered pair of row and column symbols is fixed. When these symbols differ, bit i i i i i i
(3) For every coordinate and every symbol, at most 3 M / r 3M/r 3 M / r
The construction uses polynomial time and space in s + L s+L s + L
Proof. For each pair of labels let X γ X_\gamma X γ
E ∑ γ ( X γ − M D a r r ) 2 ≤ 2 M Δ . \mathbb{E}\sum_{\gamma}\left(X_\gamma-\frac{M}{D_{\mathrm{arr}}}\right)^2\leq2M\Delta. E γ ∑ ( X γ − D arr M ) 2 ≤ 2 M Δ. Indeed, expansion expresses this expectation as a sum over ordered pairs of edges of their label-agreement probability minus 1 / D a r r 1/D_{\mathrm{arr}} 1/ D arr 2 M Δ 2M\Delta 2 M Δ r a r^a r a [55 ] .
Every load is at most M / D a r r + 2 M Δ M/D_{\mathrm{arr}}+\sqrt{2M\Delta} M / D arr + 2 M Δ H p a d H_{\mathrm{pad}} H pad L L L
H p a d ≤ M / D a r r + 2 M Δ + L ≤ 3 M / D a r r . H_{\mathrm{pad}}\leq M/D_{\mathrm{arr}}+\sqrt{2M\Delta}+L\leq3M/D_{\mathrm{arr}}. H pad ≤ M / D arr + 2 M Δ + L ≤ 3 M / D arr . Fix the unordered pair of symbols in each coordinate. If a ′ a' a ′ 2 a ′ 2^{a'} 2 a ′ 2 a ′ − a 2^{a'-a} 2 a ′ − a H p a d H_{\mathrm{pad}} H pad L L L 0 0 0 L − 1 L-1 L − 1 i i i i i i
There are H p a d D a r r / L ≤ 3 M / L H_{\mathrm{pad}}D_{\mathrm{arr}}/L\leq3M/L H pad D arr / L ≤ 3 M / L 1 / r 1/r 1/ r H p a d D a r r / r ≤ 3 M / r H_{\mathrm{pad}}D_{\mathrm{arr}}/r\leq3M/r H pad D arr / r ≤ 3 M / r
A Cauchy deformation and integral local coordinates
D.4.3. A Cauchy deformation and integral local coordinates . Fix an active call, its packets, and its entrywise lifted tensors. Use
S = { − L , − L + 1 , … , L − 1 } , C ( T ) = ∏ t ∈ S ( T − t ) , α = 2 a + 1 . S=\{-L,-L+1,\ldots,L-1\},\qquad C(T)=\prod_{t\in S}(T-t),\qquad\alpha=2^{a+1}. S = { − L , − L + 1 , … , L − 1 } , C ( T ) = t ∈ S ∏ ( T − t ) , α = 2 a + 1 . Within a packet, for its row slot x x x y y y
W x y ( T ) = C ( T ) C ′ ( 0 ) ( T − ( y − x ) ) . (76) W_{xy}(T)=\frac{C(T)}{C'(0)(T-(y-x))}. \tag*{(76)} W x y ( T ) = C ′ ( 0 ) ( T − ( y − x )) C ( T ) . (76) The factor in the denominator cancels a numerator factor, since y − x ∈ S y-x\in S y − x ∈ S Q \mathbb{Q} Q 2 L − 1 2L-1 2 L − 1 W ( 0 ) = I W(0)=I W ( 0 ) = I
Let A ( Y ) A(Y) A ( Y ) B ( X ) B(X) B ( X ) Y , X ∈ { 0 , 1 } M Y,X\in\{0,1\}^M Y , X ∈ { 0 , 1 } M l 0 ( z ) = 1 − z l_0(z)=1-z l 0 ( z ) = 1 − z l 1 ( z ) = z l_1(z)=z l 1 ( z ) = z
F ( T ) = ∑ X , Y ∈ { 0 , 1 } M A ( Y ) B ( X ) ∏ row ports x l x ( ( W ( T ) X ) x ) . (77) F(T)=\sum_{X,Y\in\{0,1\}^M}A(Y)B(X)\prod_{\text{row ports }x}l_{x}\bigl((W(T)X)_x\bigr). \tag*{(77)} F ( T ) = X , Y ∈ { 0 , 1 } M ∑ A ( Y ) B ( X ) row ports x ∏ l x ( ( W ( T ) X ) x ) . (77) Its coefficients lie in K [ V ] / ( V ℓ ) \mathbb{K}[V]/(V^\ell) K [ V ] / ( V ℓ ) M ( 2 L − 1 ) M(2L-1) M ( 2 L − 1 ) F ( 0 ) F(0) F ( 0 ) F ( 0 ) F(0) F ( 0 )
Lemma D.4. For every t ∈ S t\in S t ∈ S W ( t + α z ) W(t+\alpha z) W ( t + α z ) O [ z ] \mathcal{O}[z] O [ z ] F ( t + α z ) F(t+\alpha z) F ( t + α z ) A [ z ] \mathcal{A}[z] A [ z ] e e e ζ \zeta ζ 2 p 2^p 2 p W x y ( t + α ζ ) W_{xy}(t+\alpha\zeta) W x y ( t + α ζ ) 2 p 2^p 2 p s , p s,p s , p
Proof. The grid S S S 2 a + 1 2^{a+1} 2 a + 1 t ≠ q t \ne q t = q S S S v 2 ( t − q ) ≤ a v_2(t-q) \le a v 2 ( t − q ) ≤ a t t t v 2 ( C ′ ( t ) ) = v 2 ( C ′ ( 0 ) ) v_2(C'(t)) = v_2(C'(0)) v 2 ( C ′ ( t )) = v 2 ( C ′ ( 0 ))
C ( t + α z ) C ′ ( 0 ) = α z ∏ q ∈ S ∖ { t } ( t − q + α z ) C ′ ( 0 ) , \frac{C(t+\alpha z)}{C'(0)}=\frac{\alpha z\prod_{q\in S\setminus\{t\}}(t-q+\alpha z)}{C'(0)}, C ′ ( 0 ) C ( t + α z ) = C ′ ( 0 ) α z ∏ q ∈ S ∖ { t } ( t − q + α z ) , each factor indexed by q ≠ t q \ne t q = t 2 v 2 ( t − q ) 2^{v_2(t-q)} 2 v 2 ( t − q ) 2 a + 1 2^{a+1} 2 a + 1 y − x y-x y − x y − x = t y-x=t y − x = t α z \alpha z α z α z \alpha z α z 2 v 2 ( t − y + x ) 2^{v_2(t-y+x)} 2 v 2 ( t − y + x ) v 2 ( t − y + x ) ≤ a v_2(t-y+x) \le a v 2 ( t − y + x ) ≤ a F F F
For the evaluation, first divide each factor t − q + α z t-q+\alpha z t − q + α z q ≠ t q\ne t q = t 2 v 2 ( t − q ) 2^{v_2(t-q)} 2 v 2 ( t − q ) 2 p 2^p 2 p ζ \zeta ζ ζ = 0 \zeta=0 ζ = 0 e p ep e p O ( L ) O(L) O ( L ) s , p s,p s , p
Tapered Hermite recovery
The next lemma isolates the precision estimate. All power-series truncations in its statement are ordinary truncations in z z z
Lemma D.5 (Tapered Hermite recovery). Let integers u , M u,M u , M satisfy 1 ≤ u ≤ a 1\le u\le a 1 ≤ u ≤ a and M ≥ u M\ge u M ≥ u and put h = ⌈ 3 M / u ⌉ h=\lceil3M/u\rceil h = ⌈ 3 M / u ⌉ For t ∈ S ∖ { 0 } t\in S\setminus\{0\} t ∈ S ∖ { 0 } set
i = v 2 ( t ) , q t = q ( i ) = h max ( 0 , u − i ) , D t = L / 2 i . i=v_2(t),\qquad q_t=q(i)=h\max(0,u-i),\qquad D_t=L/2^i. i = v 2 ( t ) , q t = q ( i ) = h max ( 0 , u − i ) , D t = L / 2 i . Suppose F ∈ ( K [ V ] / ( V ℓ ) ) [ T ] F\in(K[V]/(V^\ell))[T] F ∈ ( K [ V ] / ( V ℓ )) [ T ] has degree at most M ( 2 L − 1 ) M(2L-1) M ( 2 L − 1 ) F ( 0 ) ∈ A F(0)\in A F ( 0 ) ∈ A and F ( t + α z ) ∈ A [ z ] F(t+\alpha z)\in A[z] F ( t + α z ) ∈ A [ z ] for every t ∈ S t\in S t ∈ S For p ≥ 1 p\ge1 p ≥ 1 its value F ( 0 ) F(0) F ( 0 ) modulo 2 p 2^p 2 p can be recovered from
F ( t + α z ) ( m o d 2 p t ) , p t = p + 8 h D t + 8 M ( a + 1 ) , q t > 0. F(t+\alpha z)\pmod{2^{p_t}},\qquad p_t=p+8hD_t+8M(a+1),\qquad q_t>0. F ( t + α z ) ( mod 2 p t ) , p t = p + 8 h D t + 8 M ( a + 1 ) , q t > 0. Only the first q t q_t q t coefficients of each polynomial are needed. Recovery takes polynomial time and space in M , L , ℓ , e M,L,\ell,e M , L , ℓ , e and the precisions.
Proof. There are L L L S S S ∑ t ≠ 0 q t ≥ L h u ≥ 3 M L > M ( 2 L − 1 ) \sum_{t\ne0}q_t\ge Lhu\ge3ML>M(2L-1) ∑ t = 0 q t ≥ L h u ≥ 3 M L > M ( 2 L − 1 ) h ≤ 4 M / u h\le4M/u h ≤ 4 M / u q t ≤ 4 M q_t\le4M q t ≤ 4 M t t t
E t ( T ) = ∏ q ∈ S ∖ { 0 , t } ( T − q t − q ) q q . E_t(T)=\prod_{q\in S\setminus\{0,t\}}\left(\frac{T-q}{t-q}\right)^{q_q}. E t ( T ) = q ∈ S ∖ { 0 , t } ∏ ( t − q T − q ) q q . Hermite interpolation gives
F ( 0 ) = ∑ t : q t > 0 E t ( 0 ) ( [ F ( t + α z ) E t ( t + α z ) ] < q t ) ( − t / α ) . (78) F(0)=\sum_{t:q_t>0}E_t(0)\left(\left[\frac{F(t+\alpha z)}{E_t(t+\alpha z)}\right]_{<q_t}\right)(-t/\alpha). \tag*{(78)} F ( 0 ) = t : q t > 0 ∑ E t ( 0 ) ( [ E t ( t + α z ) F ( t + α z ) ] < q t ) ( − t / α ) . (78) For completeness, replace zero in the right side by an indeterminate T T T − t / α -t/\alpha − t / α ( T − t ) / α (T-t)/\alpha ( T − t ) / α ∑ q t \sum q_t ∑ q t F F F ∏ t ≠ 0 ( T − t ) q t \prod_{t\ne0}(T-t)^{q_t} ∏ t = 0 ( T − t ) q t [31 Section 5.6] .
Each factor of E t ( t + α z ) E_t(t+\alpha z) E t ( t + α z ) 1 + α z / ( t − q ) 1+\alpha z/(t-q) 1 + α z / ( t − q ) q t q_t q t
Fix i = v 2 ( t ) i=v_{2}(t) i = v 2 ( t ) D = D t D=D_{t} D = D t
v 2 ( E t ( 0 ) ) = ∑ q ≠ 0 , t q q ( v 2 ( q ) − v 2 ( t − q ) ) = D ∑ j = 1 a − i j 2 j ( q ( i + j ) − q ( i ) ) ≥ − 6 D h . (79) v_{2}(E_{t}(0))=\sum_{q\ne0,t}q_{q}\left(v_{2}(q)-v_{2}(t-q)\right)
= D\sum_{j=1}^{a-i}\frac{j}{2^{j}}\left(q(i+j)-q(i)\right)\ge-6Dh. \tag*{(79)} v 2 ( E t ( 0 )) = q = 0 , t ∑ q q ( v 2 ( q ) − v 2 ( t − q ) ) = D j = 1 ∑ a − i 2 j j ( q ( i + j ) − q ( i ) ) ≥ − 6 D h . (79) To justify the second equality, points of valuation less than i i i 1 ≤ j ≤ a − i 1\le j\le a-i 1 ≤ j ≤ a − i D / 2 j D/2^{j} D / 2 j v 2 ( q ) = i + j v_{2}(q)=i+j v 2 ( q ) = i + j v 2 ( t − q ) = i + j v_{2}(t-q)=i+j v 2 ( t − q ) = i + j v 2 ( q ) = i v_{2}(q)=i v 2 ( q ) = i t t t a a a q ( i + j ) − q ( i ) ≥ − h j q(i+j)-q(i)\ge-hj q ( i + j ) − q ( i ) ≥ − hj ∑ j ≥ 1 j 2 / 2 j = 6 \sum_{j\ge1}j^{2}/2^{j}=6 ∑ j ≥ 1 j 2 / 2 j = 6
Evaluation of a polynomial of degree below q t q_{t} q t − t / α -t/\alpha − t / α 4 M ( a + 1 ) 4M(a+1) 4 M ( a + 1 ) 2 p t 2^{p_{t}} 2 p t 2 p A 2^{p}A 2 p A F ( 0 ) F(0) F ( 0 ) 2 p A 2^{p}A 2 p A
One can perform this last combination by rational arithmetic in the specified bases. The multipliers E t ( 0 ) ( − t / α ) j E_{t}(0)(-t/\alpha)^{j} E t ( 0 ) ( − t / α ) j j < q t j<q_{t} j < q t O ( a ) O(a) O ( a ) M , L , a M,L,a M , L , a 2 p 2^{p} 2 p
At a call with input precision p p p t t t F ( t + α z ) F(t+\alpha z) F ( t + α z ) 2 p t 2^{p_{t}} 2 p t M ( 2 L − 1 ) + 1 M(2L-1)+1 M ( 2 L − 1 ) + 1 A p t A_{p_{t}} A p t s 4 s^{4} s 4 p t p_{t} p t
The input and child precisions satisfy
p ≤ s λ O ( 1 ) 2 a d − 1 , p t = O C ( M D t / u ) . (80) p\le s^{\lambda^{O(1)}2^{a_{d-1}}},\qquad p_{t}=O_{C}(MD_{t}/u). \tag*{(80)} p ≤ s λ O ( 1 ) 2 a d − 1 , p t = O C ( M D t / u ) . (80) The first assertion holds at the root since P 0 ≤ C s P_{0}\le Cs P 0 ≤ C s 8 h D t + 8 M ( a + 1 ) 8hD_{t}+8M(a+1) 8 h D t + 8 M ( a + 1 ) h ≤ 4 M / u h\le4M/u h ≤ 4 M / u M ≤ C s M\le Cs M ≤ C s D t ≤ 2 a d D_{t}\le2^{a_{d}} D t ≤ 2 a d O ( log λ ) O(\log\lambda) O ( log λ ) q t > 0 q_{t}>0 q t > 0 i < u i<u i < u
D t ≥ 2 a d − u = 2 a d − 1 + u . D_{t}\ge2^{a_{d}-u}=2^{a_{d-1}+u}. D t ≥ 2 a d − u = 2 a d − 1 + u . Since M > H M>H M > H u ≥ λ 4 / 5 u\ge\lambda^{4/5} u ≥ λ 4/5 2 u 2^{u} 2 u λ \lambda λ p p p 8 M ( a + 1 ) 8M(a+1) 8 M ( a + 1 ) O C ( M D t / u ) O_{C}(MD_{t}/u) O C ( M D t / u ) 8 h D t 8hD_{t} 8 h D t s s s a d ≤ 4 e λ a_{d}\le4e\lambda a d ≤ 4 e λ
The paper accounts for the factor 1 / u 1/u 1/ u q ∗ = Θ ( M ) q_{*}=\Theta(M) q ∗ = Θ ( M ) u u u i = u − 1 i=u-1 i = u − 1
− q ∗ D t ∑ j = 1 a − i j 2 j = − Θ ( M D t ) . -q_{*}D_{t}\sum_{j=1}^{a-i}\frac{j}{2^{j}}=-\Theta(MD_{t}). − q ∗ D t j = 1 ∑ a − i 2 j j = − Θ ( M D t ) . Our multiplicities instead change by only h = Θ ( M / u ) h=\Theta(M/u) h = Θ ( M / u ) O ( M D t / u ) O(MD_{t}/u) O ( M D t / u ) a d a_{d} a d 2 u 2^{u} 2 u
Modes for interactions between groups
Fix an active t t t ζ ∈ O \zeta\in\mathcal{O} ζ ∈ O z z z T = t + α ζ T=t+\alpha\zeta T = t + α ζ 2 p t 2^{p_t} 2 p t i = v 2 ( t ) < u i=v_2(t)<u i = v 2 ( t ) < u D = L / 2 i D=L/2^i D = L / 2 i i i i W ( T ) W(T) W ( T )
This choice of coordinate makes W ( t ) W(t) W ( t ) W x y ( t ) = 0 W_{xy}(t)=0 W x y ( t ) = 0 y − x = t y-x=t y − x = t i i i i i i i i i t t t t + α ζ t+\alpha\zeta t + α ζ
In a packet whose two symbols in coordinate i i i x x x y y y i i i
y = y 0 + 2 i + 1 j , 0 ≤ y 0 < 2 i + 1 . y=y_0+2^{i+1}j,\qquad0\le y_0<2^{i+1}. y = y 0 + 2 i + 1 j , 0 ≤ y 0 < 2 i + 1 . The column group is constant on a tile. For a row crossing to that tile,
v 2 ( t − y 0 + x ) ≤ i . v_2(t-y_0+x)\le i. v 2 ( t − y 0 + x ) ≤ i . Otherwise y 0 y_0 y 0 x + 2 i x+2^i x + 2 i 2 i + 1 2^{i+1} 2 i + 1 i i i α ζ \alpha\zeta α ζ
W x y ( T ) = ∑ b ≥ 0 b + 1 < p t C ( T ) C ′ ( 0 ) ( T − y 0 + x ) ( 2 i + 1 T − y 0 + x ) b j b ( m o d 2 p t ) . (81) W_{xy}(T)=\sum_{\substack{b\ge0\\b+1<p_t}}\frac{C(T)}{C'(0)(T-y_0+x)}
\left(\frac{2^{i+1}}{T-y_0+x}\right)^b j^b
\pmod{2^{p_t}}. \tag*{(81)} W x y ( T ) = b ≥ 0 b + 1 < p t ∑ C ′ ( 0 ) ( T − y 0 + x ) C ( T ) ( T − y 0 + x 2 i + 1 ) b j b ( mod 2 p t ) . (81) Indeed the first fraction has valuation at least a + 1 − i ≥ 1 a+1-i\ge1 a + 1 − i ≥ 1 b b b 2 p t 2^{p_t} 2 p t
For each such packet, tile, and b b b Z Z Z w Z = b + 1 w_Z=b+1 w Z = b + 1
L Z ( X ) = ∑ real columns y in the tile j b X y . L_Z(X)=\sum_{\text{real columns }y\text{ in the tile}}j^bX_y. L Z ( X ) = real columns y in the tile ∑ j b X y . As usual j 0 = 1 j^0=1 j 0 = 1 j = 0 j=0 j = 0 Z Z Z 2 w Z 2^{w_Z} 2 w Z
( 3 M / L ) 2 i + 1 = 6 M / D (82) (3M/L)2^{i+1}=6M/D \tag*{(82)} ( 3 M / L ) 2 i + 1 = 6 M / D (82) variables of any one weight.
In the product of the affine factors l Y x ( ( W ( T ) X ) x ) l_{Y_x}((W(T)X)_x) l Y x (( W ( T ) X ) x ) Z β Z^\beta Z β 2 ∑ Z w Z β Z 2^{\sum_Z w_Z\beta_Z} 2 ∑ Z w Z β Z
∑ Z w Z β Z < p t (83) \sum_Z w_Z\beta_Z<p_t \tag*{(83)} Z ∑ w Z β Z < p t (83) need be considered. The number of these tuples is
exp ( O ( M p t / D ) ) = exp ( O C ( M / u ) ) . (84) \exp\!\left(O\!\left(\sqrt{Mp_t/D}\right)\right)
=\exp\!\left(O_{\mathcal{C}}(M/\sqrt{u})\right). \tag*{(84)} exp ( O ( M p t / D ) ) = exp ( O C ( M / u ) ) . (84) Here is the complete counting estimate. For any θ > 0 \theta>0 θ > 0
exp ( θ p t + 6 M D ∑ j ≥ 1 − ln ( 1 − e − θ j ) ) . \exp\!\left(\theta p_t+\frac{6M}{D}\sum_{j\ge1}-\ln(1-e^{-\theta j})\right). exp ( θ p t + D 6 M j ≥ 1 ∑ − ln ( 1 − e − θ j ) ) . Expanding the logarithm and summing the geometric series gives
∑ j ≥ 1 − ln ( 1 − e − θ j ) = ∑ h ≥ 1 1 h ( e θ h − 1 ) ≤ 1 θ ∑ h ≥ 1 1 h 2 . \sum_{j\ge1} -\ln(1-e^{-\theta j})=\sum_{h\ge1}\frac{1}{h(e^{\theta h}-1)}\le\frac{1}{\theta}\sum_{h\ge1}\frac{1}{h^2}. j ≥ 1 ∑ − ln ( 1 − e − θ j ) = h ≥ 1 ∑ h ( e θ h − 1 ) 1 ≤ θ 1 h ≥ 1 ∑ h 2 1 . Take θ = M D p t \theta=\sqrt{\frac{M}{Dpt}} θ = D pt M
The child networks and their descriptions
For one tuple β \beta β g ∈ [ r ] g\in[r] g ∈ [ r ] g g g 2 p t 2^{p_t} 2 p t Y Y Y X X X g g g
To define this new packet tensor, form the product of the row factors l Y X l_{Y X} l Y X g g g β Z \beta_Z β Z g g g
∏ Z in this packet source ( Z ) = g L Z ( X ) β Z . \prod_{\substack{Z\text{ in this packet}\\ \operatorname{source}(Z)=g}}L_Z(X)^{\beta_Z}. Z in this packet source ( Z ) = g ∏ L Z ( X ) β Z . Empty products equal one. If there are no target rows, the extracted coefficient is zero whenever a target exponent is positive. If the source tile of Z Z Z L Z ( X ) = 0 L_Z(X)=0 L Z ( X ) = 0 β Z = 0 \beta_Z=0 β Z = 0
Every mode has one target and one source. Coefficient extraction therefore factors over the packet-group row products, and the substitution for each mode depends only on bits in its source group. Distributing the product in (D.8), and then summing the independent bits within groups, proves the exact identity
F ( t + α ζ ) = ∑ β : ∑ Z β Z ≤ p t ∏ g = 1 r Z g , β ( m o d 2 p t ) , (85) F(t+\alpha\zeta)=\sum_{\beta:\,\sum_Z\beta_Z\le pt}\prod_{g=1}^{r}Z_{g,\beta}\pmod{2^{pt}}, \tag*{(85)} F ( t + α ζ ) = β : ∑ Z β Z ≤ pt ∑ g = 1 ∏ r Z g , β ( mod 2 pt ) , (85) where Z g , β Z_{g,\beta} Z g , β
By Lemma D.3 , a child has at most 6 M / r = 3 M / 32 < M / 8 6M/r=3M/32<M/8 6 M / r = 3 M /32 < M /8 2 L ≤ Δ 2L\le\Delta 2 L ≤ Δ
All new tensors are given implicitly. To evaluate a packet tensor entry, expand its product of at most L L L s s s
exp ( O ( L log s ) ) \exp(O(L\log s)) exp ( O ( L log s )) time and polynomial space. The factors L Z ( X ) β Z L_Z(X)^{\beta_Z} L Z ( X ) β Z A p t \mathbb{A}_{p_t} A p t L ≤ s 1 / 1000 L\le s^{1/1000} L ≤ s 1/1000 exp ( O ( s 9 / 10 ) ) \exp(O(s^{9/10})) exp ( O ( s 9/10 ))
A description of a new tensor consists of its packet data, the tuple β \beta β
Time and space bounds.
Proof of Theorem D.1 . Embed the original entries into A P 0 A_{P_0} A P 0 F ( t + α z ) F(t+\alpha z) F ( t + α z ) Lemma D.5 . At a leaf, sum directly over all Boolean edge assignments. Lemma D.4 , the recursion identity, and the recovery lemma prove correctness by induction from the leaves.
The number of interpolation nodes, values, and groups at a call is polynomial in s s s L \mathcal{L} L d d d
s O ( 1 ) exp ( O C ( s λ ( 2 8 ) d ) ) . s^{O(1)}\exp\left(O_C\left(\frac{s}{\sqrt{\lambda}}\left(\frac{\sqrt{2}}{8}\right)^d\right)\right). s O ( 1 ) exp O C λ s ( 8 2 ) d . Indeed M ≤ C S s 8 − d M\leq C_Ss8^{-d} M ≤ C S s 8 − d u d = Θ C ( λ / 2 d ) u_d=\Theta_C(\lambda/2^d) u d = Θ C ( λ / 2 d ) O C ( log λ ) O_C(\log\lambda) O C ( log λ )
exp ( O C ( s / λ ) + O C ( λ log λ ) ) = exp ( O C ( s / λ ) ) \exp\left(O_C(s/\sqrt{\lambda})+O_C(\lambda\log\lambda)\right)=\exp(O_C(s/\sqrt{\lambda})) exp ( O C ( s / λ ) + O C ( λ log λ ) ) = exp ( O C ( s / λ )) calls in total. At a leaf, direct summation uses at most 2 H 2^H 2 H s + L s+\mathcal{L} s + L
exp ( O C ( H ) + O C , A ( s 9 / 10 ) ) = exp ( O C , A ( s / λ ) ) . \exp\left(O_C(H)+O_{C,A}(s^{9/10})\right)=\exp(O_{C,A}(s/\sqrt{\lambda})). exp ( O C ( H ) + O C , A ( s 9/10 ) ) = exp ( O C , A ( s / λ )) . Local arithmetic, packing, descriptions, and recovery use polynomial time per generated term, apart from the already bounded entry computations. The description-dependent factor bounds the work per call. Since the branching bound is independent of L \mathcal{L} L s + L s+\mathcal{L} s + L
For space, traverse the recursion depth first, accumulating one sum over β \beta β s + L s+\mathcal{L} s + L O C ( log λ ) O_C(\log\lambda) O C ( log λ ) Lemma D.5 also uses only polynomial-bit integers. Each call accumulates its contraction terms modulo its own fixed precision. Original entry routines use polynomial space by hypothesis, and the new routines have the same property. This proves the space bound. Finally extract the base coordinates in A P 0 A_{P_0} A P 0 B P 0 , ℓ B_{P_0,\ell} B P 0 , ℓ
