Hard 3-CNF-SAT problems are in P -- A first step in proving NP=P
AA first step in proving ‘NP=P’ :Hard 3-CNF-SAT problems are in P
Prof. Marcel R´emon and Johan Barth´elemy
Abstract
The relationship between the complexity classes P and NP is an unsolved question in thefield of theoretical computer science. In the first part of this paper, a lattice framework isproposed to handle the 3-CNF-SAT problems, known to be in NP . In the second section,we define a multi-linear descriptor function H ϕ for any 3-CNF-SAT problem ϕ of size n , inthe sense that H ϕ : { , } n → { , } n is such that Im H ϕ is the set of all the solutions of ϕ .A new “merge” operation H ϕ (cid:86) H ψ is defined, where ψ is a single 3-CNF clause. Given H ϕ [but this can be of exponential complexity], the complexity needed for the computation of Im H ϕ , the set of all solutions, is shown to be polynomial for “hard” 3-CNF-SAT problems,i.e. the one with few ( ≤ k ) or no solutions. The third part uses the relation between H ϕ andthe indicator function S ϕ for the set of solutions, to develop a greedy polynomial algorithmto solve “hard” 3-CNF-SAT problems. Index Terms
Algorithm Complexity, P − NP problem, 3-CNF-SAT problem IntroductionI. Lattice framework for 3-CNF-SAT problems
A. The 3-CNF-SAT problem, a NP reference problemBoolean formulae are built in the usual way from propositional variables x i and three logicalconnectives ∧ , ∨ and ¬ , which are interpreted as conjunction, disjunction, and negation,respectively. A literal is a propositional variable or the negation of a propositional variable,and a clause is a disjunction of literals. A Boolean formula is in conjunctive normal form ifand only if it is a conjunction of clauses. M.R´emon, Department of Mathematics, Namur University, Belgium; [email protected]´elemy, SMART Infrastructure Facility, University of Wollongong, Australia; [email protected] a r X i v : . [ c s . CC ] J a n ARD 3-CNF-SAT PROBLEMS ARE IN P . 2 A ϕ is a Boolean formula in conjunctive normal form with exactly threeliterals per clause, like ϕ := ( x ∨ x ∨ ¬ x ) ∧ ( ¬ x ∨ x ∨ ¬ x ) := ψ ∧ ψ . A is composed of n propositional variables x i and m clauses ψ j .The is to decide whether there exists or notlogical values for the propositional variables, so that ϕ can be true. Until now, we do notknow whether it is possible or not to check the satisfiability of any given formula ϕ in a polynomial time with respect of n , as the problem is known to belong tothe hardest problems in the class NP . See [2] for details. B. A matrix representation of the set of solutions for a 3-CNF formula
B.1 DefinitionsThe size of a 3-CNF formula ϕ is defined as the size of the corresponding Boolean circuit ,i.e. the number of logical connectives in ϕ . Let us note the following property : size( ϕ ) = O ( m ) = O (∆ × n ) (1)where ∆ = m/n is the ratio of clauses with respect to variables. It seems that ∆ ≈ . ϕ ( x , x , · · · , x n ) be a 3-CNF formula. The set S ϕ of all satisfying solutions is S ϕ = { ( x , · · · , x n ) ∈ { , } n | ϕ ( x , · · · , x n ) = 1 } (2)Let Σ ϕ = S ϕ and ¯ s , · · · , ¯ s Σ ϕ be the sorted [with respect to the binary order] elements of S ϕ . For 1 ≤ j ≤ Σ ϕ : ¯ s j = ( s j , · · · , s ij , · · · , s nj ). We define the S ϕ -matrix representation of S ϕ as [ S ϕ ] : [ S ϕ ] = x x i x n s · · · s n ... s ij ... s ϕ · · · s n Σ ϕ (3) January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 3 B.2 ExamplesThe set of solutions for any single clause ψ i will be represented by a 7 × S ψ ] = [ S x ∨ x ∨¬ x ] = x x x and [ S ψ ] = [ S ¬ x ∨ x ∨¬ x ] = x x x S ψ ∧ ψ will be represented by a 12 × S ψ ∧ ψ ] = [ S ( x ∨ x ∨¬ x ) ∧ ( ¬ x ∨ x ∨¬ x ) ] = x x x x C. First properties for S ϕ -matrices C.1 Extension to new variablesLet A be a S ϕ -matrix, A can be extended to new propositional variables by adding columnsfilled with the neutral sign “.”, meaning that the corresponding variable can be set either to January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 4 A is equivalent to A . A = x x x a a a a j a j a j a ϕ a ϕ a ϕ ≡ x x x x a a . [ ] a a j a j . a j a ϕ a ϕ . a ϕ = A (4)C.2 The join operation of S ϕ -matricesLet A and B be two S ϕ -matrices and { x , · · · , x n } the union of their support variables. Let A and B be their extensions over { x , · · · , x n } . Then we define the join operation of A and B by A ∨ B = x · · · x n AB (5)Of course, this new matrix should be reordered so that the lines are in a ascending binaryorder, which can yield sometimes in replacing a line with a neutral sign by two lines with aone and a zero.C.3 The meet operation of S ϕ -matricesLet A and B be two S ϕ -matrices, A and B their extensions to the joint set of propositionalvariables. Let A k and B l be the one line matrices such that : A = Σ A (cid:95) k =1 A k and B = Σ B (cid:95) l =1 B l (6)We define the meet operation of A and B as A ∧ B ≡ A ∧ B = Σ A (cid:95) k =1 A k ∧ Σ B (cid:95) l =1 B l = Σ A (cid:95) k =1 Σ B (cid:95) l =1 (cid:0) A k ∧ B l (cid:1) = Σ A (cid:95) k =1 Σ B (cid:95) l =1 C k,l (7)where C k,l = (cid:32) x x i x n a k a ik a nk (cid:33) ∧ (cid:32) x x i x n b l b il b nl (cid:33) = ∅ if ∃ c im = “NaN” (cid:32) x x i x n c m c im c nm (cid:33) otherwise (8) January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 5 with c im = a ik if a ik = b il a ik if b il = “ · ” b il if a ik = “ · ” “NaN” otherwise (9)C.4 The empty and full S ϕ -matricesLet us call ∅ , the empty matrix , with no line at all. The empty matrix is neutral for the joinoperator ∨ and absorbing for the meet operator ∧ .Let us define Ω, the full matrix , as a one line matrix with only neutral signs “ · ” in it. Thefull matrix is neutral for ∧ and absorbing for ∨ .C.5 Lattice structure of S ϕ -matricesA semi-lattice ( X, ∨ ) is a pair consisting of a set X and a binary operation ∨ which isassociative, commutative, and idempotent.Let us note A the set of all the S ϕ -matrices. Then ( A , ∨ ) and ( A , ∧ ) are both semi-lattices,respectively called join and meet semi-lattices.Let us define the two absorption laws as x = x ∨ ( x ∧ y ) and its dual x = x ∧ ( x ∨ y ). A lattice is an algebra ( X, ∨ , ∧ ) satisfying equations expressing associativity, commutativity,and idempotence of ∨ and ∧ , and satisfying the two absorption equations.Therefore, ( A , ∨ , ∧ ) is a lattice over the set of S ϕ -matrices with respect to the join andmeet operators. Indeed, S ϕ -matrices satisfy the absorption equations as S ϕ = S ϕ ∨ ( ϕ ∧ ϕ (cid:48) ) = S ϕ ∧ ( ϕ ∨ ϕ (cid:48) ) .Moreover, ( A , ∨ , ∧ ) is a distributive bounded lattice as ∧ is distributive with respect to ∨ and A ∨ Ω = Ω & A ∧ ∅ = ∅ ∀ A ∈ A . See [1] for more details over lattices. January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 6 D. “Hard” 3-CNF-SAT problemsDefinition I.1:
A “hard” 3-CNF-SAT problem ϕ is defined in this paper as a problemwith a small or limited set of solutions, in the sense that the number of solutions is bounded: Σ ϕ = 2 k = 2 O (1) [for some k ] (10)Note : the problem is said to be “hard” in the sense that the probability to get a solutionat random [= Σ ϕ n ] tends to zero as n tends to infinity. The hardiest 3-CNF-SAT problemsare the one without solution. This paper only considers “hard” 3-CNF-SAT problems. II. The multi-linear descriptor function H ϕ A. Characterization theorem of S ϕ via the descriptor function H ϕ Theorem II.1:
Every non empty S ϕ -matrix of n literals can be characterizedby a single n -dimensional descriptor function H ϕ : { , } n → { , } n such that Im H ϕ = S ϕ . ∀ [ S ϕ ] = x x i x n s · · · s n ... s ij ... s ϕ · · · s n Σ ϕ (cid:54) = ∅ , ∃ n functions h i : { , } i → { , } such that[ S ϕ ] = (cid:95) ( α , ··· ,α n ) ∈{ , } n (cid:32) x · · · x i · · · x n h ( α ) · · · h i ( α , · · · , α i ) · · · h n ( α , · · · , α n ) (cid:33) (11) notation ≡ (cid:104) h ( α ) . . . h n ( α , · · · , α n ) (cid:105) notation ≡ (cid:104) H ϕ ( α , · · · , α n ) (cid:105) (12)So, the knowledge of H ϕ ( α , · · · , α n ) characterizes fully [ S ϕ ]. H ϕ ( α , · · · , α n ) is called thedescriptor function of S ϕ . All operations are done in a mod(2) framework.
Beforeproving the existence of such a function, let us consider some examples.
January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 7 Examples of H ϕ : • ϕ = ( ¬ x ∨ ¬ x ∨ ¬ x ) (13)[ S ϕ ] = (cid:95) ( α , ··· ,α ) ∈{ , } (cid:32) x x x α α α α α + α (cid:33) (mod 2) ≡ (cid:104) α α α α α + α (cid:105) ≡ (cid:104) H ϕ (cid:105) [ S ϕ ] = α α α H ϕ ( 0 0 0 ) H ϕ ( 0 0 1 ) H ϕ ( 0 1 0 ) H ϕ ( 0 1 1 ) H ϕ ( 1 0 0 ) H ϕ ( 1 0 1 ) H ϕ ( 1 1 0 ) H ϕ ( 1 1 1 ) = x x x = x x x • ϕ = ( x ∨ x ∨ ¬ x ) ∧ ( ¬ x ∨ x ∨ ¬ x ) (14)[ S ϕ ] = (cid:95) ( α , ··· ,α ) ∈{ , } (cid:32) x x x x α α ( α + 1)( α + 1) α + α α ( α + 1) α + α (cid:33) (mod 2) ≡ (cid:104) α α ( α + 1)( α + 1) α + α α ( α + 1) α + α (cid:105) ≡ (cid:104) H ϕ (cid:105) = α α α α H ϕ ( 0 0 0 0 ) H ϕ ( 0 0 0 1 ) H ϕ ( 0 0 1 0 ) H ϕ ( 0 0 1 1 ) H ϕ ( 0 1 0 0 ) H ϕ ( 0 1 0 1 ) H ϕ ( 0 1 1 0 ) H ϕ ( 0 1 1 1 ) H ϕ ( 1 0 0 0 ) H ϕ ( 1 0 0 1 ) H ϕ ( 1 0 1 0 ) H ϕ ( 1 0 1 1 ) H ϕ ( 1 1 0 0 ) H ϕ ( 1 1 0 1 ) H ϕ ( 1 1 1 0 ) H ϕ ( 1 1 1 1 ) = x x x x = x x x x January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 8 Proof: (Existence of H ϕ ) [ Remember : all operations in mod(2) ] • The theorem is satisfied for n = 1 as (cid:34) x (cid:35) = (cid:32) x h ( α ) ≡ (cid:33) ; (cid:34) x (cid:35) = (cid:32) x h ( α ) ≡ (cid:33) ; x = (cid:95) α ∈{ , } (cid:32) x α (cid:33) • Let the theorem be true for n − S ] be a S ϕ -matrix of dimension n . There exist two S ϕ -matrices [ S ] and [ S ] of size n − S ] = (cid:34) x x · · · x n S ] (cid:35) ∨ (cid:34) x x · · · x n S ] (cid:35) as [ S ] can be divided in two sets of lines, the ones beginning with 0 and the ones with 1.Using the recurrence hypothesis :[ S ] = (cid:95) α i ∈{ , } (cid:32) x x · · · x n f ( α ) · · · f n ( α , · · · , α n ) (cid:33) (cid:95) α i ∈{ , } (cid:32) x x · · · x n g ( α ) · · · g n ( α , · · · , α n ) (cid:33) Thus [ S ] = (cid:95) α i ∈{ , } (cid:32) x · · · x n h ( α ) · · · h n ( α , · · · , α n ) (cid:33) where h ( α ) = α h i ( α , · · · , α i ) = ( α + 1) f i ( α , · · · , α i ) (cid:124) (cid:123)(cid:122) (cid:125) for lines where x = 0 + α g i ( α , · · · , α i ) (cid:124) (cid:123)(cid:122) (cid:125) for lines where x = 1 for i (cid:54) = 1 Definition II.2:
Length of H ϕ and h i ( α , · · · , α i ) . Let len ( h i ) be defined as the number of terms in h i ( α , · · · , α i ) . Let len ( H ϕ ) be defined as the maximum length of h i ( α , · · · , α i ) : len ( H ϕ ) = max i len ( h i ) Corollary II.1:
The descriptor function H ϕ is a n -dimensional modulo-2 multi-linear combination of α i . January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 9 Proof:
This is a mere consequence of the definition of h i ( α , · · · , α i ) in Theorem II.1 . Corollary II.2:
Let A ⊆ { α , · · · , α n } , with α i ∈ { , } and h i ∈ combi ( A ) notation ⇔ h i is amulti-linear combination of α i ∈ A , modulo 2, then h i ( α , · · · , α i ) ∈ combi ( { α , · · · , α i } ) , len ( h i ) ≤ i and len ( H ϕ ) ≤ n Example : h ( α , α ) ∈ combi ( { α , α } ) ⇒ h ( α , α ) = δ (0 , α α + δ (1 , α + δ (0 , α + δ (1 , α α ⇒ len ( h ) ≤ ( { α ,α } ) = 2 B. Computation of H ϕ Theorem II.3:
Simple characterization theorem (one-clause 3-CNF formula)
Consider the 3-CNF formula, consisting of only one clause ψ ≡ [ ¬ ] x r ∨ [ ¬ ] x s ∨ [ ¬ ] x t where 1 ≤ r < s < t ≤ n . [ S ψ ] can be characterized by the following [ H ψ ] ≡ [ h i ( α , · · · , α i )] descriptorfunction where : h i ( α , · · · , α i ) = α i ∀ i (cid:54) = t (1 ≤ i ≤ n ) h t ( α r , α s , α t ) = ( α r + 1)( α s + 1)( α t + 1) + α t if ψ = x r ∨ x s ∨ x t ( α r + 1)( α s + 1) α t + α t if ψ = x r ∨ x s ∨ ¬ x t ( α r + 1) α s ( α t + 1) + α t if ψ = x r ∨ ¬ x s ∨ x t ( α r + 1) α s α t + α t if ψ = x r ∨ ¬ x s ∨ ¬ x t α r ( α s + 1)( α t + 1) + α t if ψ = ¬ x r ∨ x s ∨ x t α r ( α s + 1) α t + α t if ψ = ¬ x r ∨ x s ∨ ¬ x t α r α s ( α t + 1) + α t if ψ = ¬ x r ∨ ¬ x s ∨ x t α r α s α t + α t if ψ = ¬ x r ∨ ¬ x s ∨ ¬ x t (15) Proof:
The proof is straightforward. See (13) for an example.
January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 10 Theorem II.4:
General descriptor function theorem
The descriptor function H ϕ ∧ ψ ( α , · · · , α n ) , for the conjunction of a 3-CNF formulae ϕ with F ϕ as descriptor function and a 3-CNF clause ψ associated to G ψ can be computed via ageneral algorithm.Let Λ = { ( α , · · · , α n ) ∈ { , } n : F ϕ ( α , · · · , α n ) = G ψ ( α , · · · , α n ) } . Then the followingalgorithm will give the exact H ϕ ∧ ψ ( α , · · · , α n ) : ∀ ( α , · · · , α n ) ∈ Λ : H ϕ ∧ ψ ( α , · · · , α n ) := F ψ ( α , · · · , α n ) = G ψ ( α , · · · , α n ) ∀ ( α , · · · , α n ) (cid:54)∈ Λ : H ϕ ∧ ψ ( α , · · · , α n ) := H ϕ ∧ ψ ( α ∗ , · · · , α ∗ n ) for some ( α ∗ , · · · , α ∗ n ) ∈ ΛThe algorithm defines ( α ∗ , · · · , α ∗ n ) as the “nearest” line of ( α , · · · , α n ) in Λ. This dependson the clause ψ . Let ψ = [ ¬ ] x r ∨ [ ¬ ] x s ∨ [ ¬ ] x t (1 ≤ r < s < t ≤ n ), then( α ∗ , · · · , α ∗ n ) := ( α , · · · , α t + 1 , · · · , α n ) if ( α , · · · , α t + 1 , · · · , α n ) ∈ Λelse ( α , · · · , α t − + 1 , α t , · · · , α n ) if ( α , · · · , α t − + 1 , α t , · · · , α n ) ∈ Λelse · · ·
For instance, if[ S ϕ ] = α α α F ϕ ( 0 0 0 ) F ϕ ( 0 0 1 ) F ϕ ( 0 1 0 ) F ϕ ( 0 1 1 ) F ϕ ( 1 0 0 ) F ϕ ( 1 0 1 ) F ϕ ( 1 1 0 ) F ϕ ( 1 1 1 ) = x x x = x x x and G ψ ≡ (cid:104) g ( α ) g ( α , α ) g ( α , α , α ) (cid:105) ≡ (cid:104) α α α α + α α + α α α (cid:105) (16) January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 11 ⇒ [ S ψ ] = α α α G ψ ( 0 0 0 ) G ψ ( 0 0 1 ) G ψ ( 0 1 0 ) G ψ ( 0 1 1 ) G ψ ( 1 0 0 ) G ψ ( 1 0 1 ) G ψ ( 1 1 0 ) G ψ ( 1 1 1 ) = x x x = x x x Remark : The forbidden values ( α ∗ r , α ∗ s , α ∗ t ) for ψ are (0 , 0 , 1) . then [ S ϕ ∧ ψ ] = α α α H ϕ ∧ ψ ( 0 0 0 ) H ϕ ∧ ψ ( 0 0 1 ) H ϕ ∧ ψ ( 0 1 0 ) H ϕ ∧ ψ ( 0 1 1 ) H ϕ ∧ ψ ( 1 0 0 ) H ϕ ∧ ψ ( 1 0 1 ) H ϕ ∧ ψ ( 1 1 0 ) H ϕ ∧ ψ ( 1 1 1 ) = x x x F ϕ (0 , ,
0) = 0 1 1 F ϕ (0 , ,
0) = 0 1 1 F ϕ (0 , ,
0) = 0 1 1 F ϕ (0 , ,
1) = 0 1 1 F ϕ (1 , ,
0) = 1 0 0 F ϕ (1 , ,
1) = 1 0 1 F ϕ (1 , ,
0) = 1 1 1 F ϕ (1 , ,
1) = 1 1 1 (1 st nearest line)(2 nd nearest line) = x x x Proof:
The merging of [ S ϕ ] and [ S ψ ] should correspond to the intersection of the sets of solutions[ S ϕ ] ∩ [ S ψ ]. In terms of S ϕ -matrices, this means that only the lines common to [ S ϕ ] and [ S ψ ]should be retained in [ S ϕ ∧ ψ ]. As these lines are, by definition of the descriptor function, theelements of Im F ϕ and Im G ψ , we have to get Im H ϕ ∧ ψ = Im F ϕ ∩ Im G ψ .Let ψ = [ ¬ ] x r ∨ [ ¬ ] x s ∨ [ ¬ ] x t where 1 ≤ r < s < t ≤ n be a 3-CNF clause and ( α ∗ r , α ∗ s , α ∗ t )be the unique triplet of non satisfying values for x r , x s and x t . The clause ψ puts a soleconstraint over ϕ , in the sense that we have to discard the lines of [ S ϕ ] including the forbiddenvalues ( α ∗ r , α ∗ s , α ∗ t ). Three situations can occur. • Situation A : All lines of [ S ϕ ] can be kept to build [ S ϕ ∧ ψ ], as none of them includes theforbidden values ( α ∗ r , α ∗ s , α ∗ t ). So, ψ does not introduce any new constraint with respect to ϕ and [ S ϕ ∧ ψ ] := [ S ϕ ] or H ϕ ∧ ψ := F ϕ . January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 12 Let us look at situations where some lines of [ S ϕ ] includes the forbidden values ( α ∗ r , α ∗ s , α ∗ t ).We define the “nearest” line in { , } n as the line having all the same α i , except for α t where we have α t + 1. Let us look at a line containing the forbidden values ( α ∗ r , α ∗ s , α ∗ t ). Twosituations can occur : • Situation B : The image by F ϕ of its nearest line does not include the forbidden values( α ∗ r , α ∗ s , α ∗ t ). All the α i ( i (cid:54) = t ) are the same as in the original line, except α t that becomes α t + 1. The algorithm give us the solution : H ϕ ∧ ψ ( α , · · · , α t , · · · , α n ) := F ψ ( α , · · · , α t +1 , · · · , α n ) • Situation C : The image by F ϕ of the nearest line does include the forbidden values( α ∗ r , α ∗ s , α ∗ t ). This corresponds to the above example where the first nearest line of the linein grey [ F ϕ (0 , ,
0) = (0 , , F ϕ (0 , ,
1) = (0 , , second nearest line , defined as the line having the same α i as the originalline, except for α t − where it is the opposite. In our example, this is the third line, where F ϕ (0 , ,
0) = (0 , , But it might be necessary to look atsuccessive nearest lines before finding a line without the forbidden values and thus a solutionfor H ϕ ∧ ψ . Otherwise, the algorithm stops and the 3-CNF formula ϕ ∧ ψ is without solution. The three situations can be summarized. The solution for H ϕ ∧ ψ will correspond to the January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 13 following algorithm : H ϕ ∧ ψ ( · ) := F ϕ ( α , · · · , α t , · · · , α n ) (situation A)when ( f r ( · ) , f s ( · ) , f t ( · )) (cid:54) = ( α ∗ r , α ∗ s , α ∗ t ) see (16) ⇔ g t ( f r ( α , · · · , α r ) , f s ( α , · · · , α s ) , f t ( α , · · · , α t )) = f t ( α , · · · , α t ) F ϕ ( α , · · · , α t + 1 , · · · , α n ) (situation B)when ( f r ( · ) , f s ( · ) , f t ( · · · , α t )) = ( α ∗ r , α ∗ s , α ∗ t )and ( f r ( · ) , f s ( · ) , f t ( · · · , α t + 1)) (cid:54) = ( α ∗ r , α ∗ s , α ∗ t ) see (16) ⇔ g t ( f r ( α , · · · , α r ) , f s ( α , · · · , α s ) , f t ( α , · · · , α t )) (cid:54) = f t ( α , · · · , α t )and g t ( f r ( · ) , f s ( · ) , f t ( α , · · · , α t + 1)) = f t ( α , · · · , α t + 1) F ϕ ( α , · · · , α j + 1 , · · · , α t , · · · , α n ) (situation C)when ( f r ( · ) , f s ( · ) , f t ( · · · , α t )) = ( α ∗ r , α ∗ s , α ∗ t )and ( f r ( · ) , f s ( · ) , f t ( · · · , α t + 1)) = ( α ∗ r , α ∗ s , α ∗ t )and ( f r ( · ) , f s ( · ) , f t ( · · · , α j + 1 , · · · , α t )) (cid:54) = ( α ∗ r , α ∗ s , α ∗ t )for some α j ( j < t ) so that situation A or B arises. We take theunique highest such α j . If not existing, ϕ ∧ ψ has no solution. Theorem II.5:
General computation theorem for 3-CNF formulaThe descriptor function for the conjunction of a 3-CNF formulae ϕ and a 3-CNFclause ψ = [ ¬ ] x r ∨ [ ¬ ] x s ∨ [ ¬ ] x t (1 ≤ r < s < t ≤ n ) can be numerically computed asthe merging of their descriptor functions : [ H ϕ ∧ ψ ] = [ F ϕ ] ∧ [ G ψ ] . More precisely, if [ F ϕ ] ≡ t f ( α )... f n ( α , · · · , α n ) and [ G ψ ] see (15) ≡ t g ( α ) = α ... g t ( α r , α s , α t )... g n ( α , · · · , α n ) = α n January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 14 Then [ H ϕ ∧ ψ ] ≡ t h ( α )... h n ( α , · · · , α n ) notation = t f ( α ) ∧ g ( α )... f n ( α , · · · , α n ) ∧ g n ( α , · · · , α n ) where, in a loop for l going from n to 1 :Let β i notation ≡ f i ( α , · · · , α i ) , [Situations A and B] h l ( α , · · · , α l ) := ( α l + 1) · { [ f l ( α , · · · , α l − ,
0) + g l ( β , · · · , β l − , · [ f l ( α , · · · , α l − , · g l ( β , · · · , β l − , f l ( α , · · · , α l − , · g l ( β , · · · , β l − , } + α l · { [ f l ( α , · · · , α l − ,
1) + g l ( β , · · · , β l − , · [ f l ( α , · · · , α l − ,
0) + g l ( β , · · · , β l − , f l ( α , · · · , α l − ,
1) + g l ( β , · · · , β l − , · [ f l ( α , · · · , α l − , · g l ( β , · · · , β l − , f l ( α , · · · , α l − , · g l ( β , · · · , β l − , } Moreover if there exists a j < l , related to the highest α j (j is thus unique), such that : g ∗ j ( α , · · · , α j ) ≡ [ f l ( α , · · · , α l − ,
0) + g l ( β , · · · , β l − , · [Situation C][ f l ( α , · · · , α l − ,
1) + g l ( β , · · · , β l − , = 1 (18)[(18) corresponds to situations where F ϕ ( α , · · · , α l − + 1 , α l )includes the forbidden values. We look at the nearest l -uplenot including the forbidden values, where we put α j := α j + 1] ⇒ then replace f j ( α , · · · , α j ) in [ F ϕ ]by this new merging f j ( α , · · · , α j ) ∧ { g ∗ j ( α , · · · , α j ) + α j } (19)computed by using a recursive call to definition (17) . January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 15 Recursivity will end as soon as there is no longer such g ∗ j ( α , · · · , α j ) = 1. When g ∗ j ( α , · · · , α j )is no longer a function of α i , that means that the 3-CNF-SAT ϕ ∧ ψ has no solution.At the end of the loop, we replace α i by h i ( α , · · · , α i ) in h j ( · ) where 1 ≤ i < j ≤ n . (20) Proof:
This computational formula gives the same answer for H ϕ ∧ ψ as the algorithm in Theorem II.4 . Remember that ( α ∗ r , α ∗ s , α ∗ t ) are the forbidden values for ψ . More formally, four situations in equation (17) should be considered for the index t : • f t ( α , · · · ,
0) = g t ( β r , β s ,
0) and f t ( α , · · · ,
1) = g t ( β r , β s ,
1) [as in situation A] ⇓ h t ( α , · · · , α t ) := ( α t + 1) · { [ f t ( · ,
0) + g t ( · , · [ f t ( · , · g t ( · , f t ( · , · g t ( · , } + α t · { [ f t ( · ,
1) + g t ( · , · [ f t ( · ,
0) + g t ( · , f t ( · ,
1) + g t ( · , · [ f t ( · , · g t ( · , f t ( · , · g t ( · , } := ( α t + 1) · f t ( · ,
0) + α t · f t ( · ,
1) [as f t () + g t () = 0 ; f t () · g t () = f t () = f t ()]:= f t ( α , · · · , α t )[ h t () is thus the merging of the cells f t () and g t () which are equivalent.] January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 16 • f t ( α , · · · ,
0) = g t ( β r , β s ,
0) but f t ( α , · · · , (cid:54) = g t ( β r , β s ,
1) [as in situation B] ⇓ h t ( α , · · · , α t ) := ( α t + 1) · f t ( · ,
0) + α t · f t ( · ,
0) [as f t ( · ,
1) + g t ( · ,
1) = 1 and f t ( · , · g t ( · ,
1) = 0]:= f t ( α , · · · , h t () sends α t to a value where the cells f t () and g t () are the same in [ F ϕ ] and [ G ψ ]] • f t ( α , · · · ,
1) = g t ( β r , β s ,
1) but f t ( α , · · · , (cid:54) = g t ( β r , β s ,
0) [as in situation B] ⇓ h t ( α , · · · , α t ) := ( α t + 1) · f t ( · ,
1) + α t · f t ( · ,
1) [as f t ( · ,
0) + g t ( · ,
0) = 1 and f t ( · , · g t ( · ,
0) = 0]:= f t ( α , · · · , h t () sends α t to a value where the cells f t () and g t () are the same in [ F ϕ ] and [ G ψ ]] • f t ( α , · · · , (cid:54) = g t ( β r , β s ,
1) and f t ( α , · · · , (cid:54) = g t ( β r , β s ,
0) [as in situation C] ⇓ [Impossibility to find a common cell between f t () and g t () ] ⇓ [No constraint for h t ( · , α t ) but a induced constraintover some predecessor α j ( j < t )] h t ( α , · · · , α t ) := α t [as f t () + g t () = 1 and f t () · g t () = 0] but [ f t ( · ,
0) + g t ( · , · [ f t ( · ,
1) + g t ( · , α , · · · , α j ) = 1 and g j ( α , · · · , α j ) ← [ f t ( · ,
0) + g t ( · , · [ f t ( · ,
1) + g t ( · , g j ( α , · · · , α j )[New additional constraint over g j ( · , α j ) ( j < t ) as g j ( · , α j ) ← g j ( · , α j ) + 1 . A descending order with respect of t for the computations ensures usthat the new additional constraint over g j ( · , α j ) has no repercussion over thealready-computed function h l ( · , α l ) [ l ≥ t ], as α l is not involved in g j ( · , α j ).] January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 17 Note :
The code for this merging operation is available at https://github.com/3cnf/ in the descriptor-solver directory.
C. Examples of computation Example of a simple merging of two clauses
Let ϕ = ( x ∨ x ∨ ¬ x ) and ψ = ( ¬ x ∨ x ∨ ¬ x ) (see 14)[ F ϕ ] = t α α ( α + 1)( α + 1) α + α α = t β β β β [ G ψ ] = t g ( · ) g ( · ) g ( · ) g ( · ) = t α α α α ( α + 1) α + α ⇒ [ H ϕ ∧ ψ ] = t h ( · ) h ( · ) h ( · ) h ( · ) = t α α ( α + 1)( α + 1) α + α α ( α + 1) α + α Computations for h t ( · ) [Descending order for t ] : • t = 4 : g ( β i ) = α · { [( α + 1)( α + 1) α + α ] + 1 } · α + α = α α α + α α + α h ( · ) (17) = ( α + 1) { [0 + 0] · [1 · ( α α + α + 1)] + [0 · } + α { [1 + α α + α + 1] · [0 + 0] + [1 + α α + α + 1] · [0 ·
0] + [1 · ( α α + α + 1)] } = α α α + α α + α = g ( · )Indeed, the merging of f ( · ) and g ( · ) should give g ( · ) as f ( · ) puts no constraint over x . • t = 3 : h ( α , α , α ) = f ( α , α , α ) as g ( β i ) = β = f ( · )Here, g ( · ) puts no constraint over x . Idem for h ( · ) and h ( · ). January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 18 Example of a uniformly distributed 3-CNF-SAT problem :
Let ϕ := (cid:94) i =1 ψ i = (cid:94) x ∨ ¬ x ∨ ¬ x x ∨ x ∨ ¬ x ¬ x ∨ ¬ x ∨ ¬ x ¬ x ∨ x ∨ ¬ x x ∨ ¬ x ∨ x x ∨ x ∨ x ¬ x ∨ ¬ x ∨ x ¬ x ∨ x ∨ x (21)We have here : Step h ( · ) h ( · ) h ( · ) {S ϕ } ϕ = ψ α α ( α + 1) α α + α ϕ = ψ ∧ ψ α α α α ϕ = ∧ i =1 ψ i α α α α α + α α ϕ = ∧ i =1 ψ i α α ϕ = ∧ i =1 ψ i α α α ϕ = ∧ i =1 ψ i α ϕ = ∧ i =1 ψ i ϕ = ∧ i =1 ψ i (cid:64) (cid:64) (cid:64) Property II.1:
Let
A, B ⊆ { α , · · · , α n } . If f t ( · ) ∈ combi ( A ) and g t ( · ) ∈ combi ( B ), then f t ( · ) ∧ g t ( · ) ∈ combi ( A ∪ B ) (22)and g ∗ j ( · ) ∈ combi ([ A ∪ B ] \ { α j +1 , · · · , α n } ) (23) Proof:
This is straightforward from the definition of f t ( · ) ∧ g t ( · ) in (17) and g ∗ j ( · ) in (18). III. Complexity analysis for computing H ϕ , the descriptor function A. Some definitionsDefinition III.1:
Sorted clauses (to ensure the descendant order of t ) January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 19 Let ϕ = (cid:86) mk =1 ψ k be a 3-CNF formula. We suppose, without any loss of generality, thatthese m sorted , in the following way : ϕ = m (cid:94) k =1 ψ k where ψ k = [ ¬ ] x r k ∨ [ ¬ ] x s k ∨ [ ¬ ] x t k (1 ≤ r k < s k < t k ≤ n )and t k < t k (cid:48) or t k = t k (cid:48) where ψ k = [ ¬ ] x r k ∨ [ ¬ ] x s k ∨ ¬ x t k while ψ k (cid:48) = [ ¬ ] x r k (cid:48) ∨ [ ¬ ] x s k (cid:48) ∨ x t k (cid:48) ⇒ k < k (cid:48) (24) Definition III.2:
Set of predecessors P ( t )Let us compute f t ( · ) ∧ g t ( · ). We define the indice j of the highest α j found in the recursivecall (18) as the predecessor of t , t being the successor of j . In the same way, x j and α j arecalled the predecessors of x t and α t . We denote them by : j def = pred ( t ) and t def = succ ( j ) x j def = pred ( x t ) and α j def = pred ( α t ) j (cid:48) = pred ( t ) iff j (cid:48) = pred ( pred ( t )) and so on.Let us define P ( t ) as the set of the predecessors of t : P ( t ) = { j < t | ∃ k : j = pred k ( t ) } (25) Definition III.3:
Set of connected variables V ( x t ) and W ( x t ) January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 20 We define for all t in { , · · · , n } : Cl ( x t ) = { ψ k where the variable x t appears with the highest indice } (26) V ( x t ) = { α i ( i ≤ t ) | x i appears in some ψ k ∈ Cl ( x t ) } (27) V | α i ( x t ) = V ( x t ) \ { α i +1 , · · · } (28) W ∗ ( x t ) = (cid:91) u>t : t ∈ P ( u ) V ( x u ) ∪ V ( x t ) (29) W ( x t ) = W ∗ ( x t ) \ { α t +1 , · · · , α n } [ with W ( x n ) def = V ( x n ) ] (30) V ( x t ) is the set of α i corresponding to the variables connected to x t by at least one clause ψ k where x t is the highest indexed variable. W ( x t ) is the union of α i connected to thesuccessors of x t , excluding α i with indice i higher than t . Definition III.4:
Sub-problem ϕ ( L ) We define the sub-problem ϕ ( L ) associated to the subset L ⊆ { , · · · , m } by : ϕ ( L ) = (cid:94) k ∈ L ψ k (31)and Cl ( L ) ( x t ), V ( L ) ( x t ) and W ( L ) ( x t ) being the respective sets Cl ( x t ) , V ( x t ) and W ( x t ) for ϕ ( L ) . In the same way, we define P ( L ) ( t ) as the set of the predecessors of t for ϕ ( L ) . See(25).Finally, we define : H ϕ ( L ) = h ( L )1 ( · )... h ( L ) n ( · ) B. Complexity theorem for computing H ϕ Theorem III.5: H ϕ The complexity of the descriptor approach for a 3-CNF problem ϕ with m clauses and n January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 21 propositional variables is O ( m n max ≤ t ≤ n max ≤ l ≤ m len ( h ( { l, ··· ,m } ) t ) )See (II.2) for the definition of len ( h t ). Proof:
Let us compute the complexity of f t ( · ) ∧ g t ( · ) in (17) for the most general case. First of all,one has to compute the four functions in square brackets : [ f t ( · , g t ( · , f t ( · , g t ( · , f t ( · , · g t ( · , f t ( · , · g t ( · , len ( f t ( · , ≤ len ( f t ( · , α t )) and len ( f t ( · , ≤ len ( f t ( · , α t )) [ ≡ len ( f t )] len ( g t ( · , ≤ len ( g t ( · , α t )) and len ( g t ( · , ≤ len ( g t ( · , α t )) [ ≡ len ( g t )] len ( f t + g t ) ≤ len ( f t ) + len ( g t ) ≤ len ( f t ) · len ( g t ) when len ( f t ) > len ( g t ) > O ( len ( f t ) · len ( g t )).The complexity for computing h t ( · ) in (17) is : O ( 3 · [( len ( f t ) · len ( g t )) + ( len ( f t ) · len ( g t ))] + 2 · [2( len ( f t ) · len ( g t )) + ( len ( f t ) · len ( g t ))])= O (7 · ( len ( f t ) · len ( g t )) + 5 · ( len ( f t ) · len ( g t )))= O ([ len ( f t ) · len ( g t )] ) for large len ( f t ) · len ( g t ) (32)Note : it needs three runs over the formula in the brackets to do the product with ( α t + 1) :one to compute the formula, one to multiply it by α t and one to add both results. Similarly,it takes two runs to compute the product with α t .Using the same argumentation, we have : len ( h t ) = O ([ len ( f t ) · len ( g t )] ) for large len ( f t ) · len ( g t ) (33)and for the recursive call with j < t [see (18)] len ( g ∗ j ) = O ([ len ( f t ) · len ( g t )] ) for large len ( f t ) · len ( g t ) (34)To solve the 3-CNF problem, one should compute all n functional descriptors h ( { l, ··· ,m } ) t ( · )at each step of integration of the m clauses, i.e. for each ϕ ( { l, ··· ,m } ) with l decreasing from January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 22 m to 1. Each h ( { l, ··· ,m } ) t ( · ) could yield to at most n recursive calls with similar complexity.So, using the equivalence between (33) and (32), the overall complexity of the functionalapproach to 3-CNF problem will be of order O ( m n max ≤ t ≤ n max ≤ l ≤ m len ( h ( { l, ··· ,m } ) t ) ) C. Cluster effect
From numerical tests, we see that a cluster effect appears in the middle of the algorithm.This is understandable as the variables with smaller indices are subject to more and moreconstraints coming from the first treated clauses. When W ( n − t +1) ( x t ) = { α , · · · , α t } , we seea linear decrease of W ( n − t +1) ( x t ) as t decreases from ∼ n to 1. This cluster effect is atthe hart of hard Figure 1 for the 3-CNF-SAT problem uuf50-02.cnf taken from ∼ hoos/SATLIB/benchm.html . Both graphs shows the complexity (inlog scale and in normal scale) as the descriptor algorithm goes from the first clause to thelast one (in the sorted 3-CNF-SAT problem). Exact log [ len (h (n-‐t+1)t ) ] for 50 variables Exact len (h (n-‐t+1)t ) for 50 variables Figure 1 : log len ( h t ( · )) and len ( h t ( · )) for 3-CNF-SAT problem uuf50-02.cnf IV. Complexity theorems for listing the solutions, given H ϕ Theorem IV.1:
The complexity for listing the solutions of a “hard” 3-CNF-SAT ϕ ,given H ϕ , is polynomial .We consider a “hard” 3-CNF-SAT problem ϕ with n propositional variables. We suppose H ϕ is computed and available. Let Σ ϕ = S ϕ = 2 O (1) be the number of solutions for ϕ . January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 23 Then the complexity needed to list all Σ ϕ solutions from H ϕ is O (2 n Σ ϕ ) = n O (1) = O ( n ). Proof:
Let us note that if S ϕ = ∅ (no solution), H ϕ does not exist. If there is only one solution { ¯ s } , H ϕ is a constant function, as Im H ϕ = { ¯ s } .Let S ϕ = { ¯ s , · · · , ¯ s Σ ϕ } be the set of solutions with ¯ s j = ( s j , · · · , s nj ) and s ij ∈ { , } . We candescribe the solutions as leafs of a tree. No solution (1, ··· ) h ( α ) ≡ h (0 , α ) = 1 h (0 , , α ) ≡ s = (0 , , No solution (0,1,0, ··· ) h (0 , α ) = 0 h (0 , , α ) = 1¯ s = (0 , , h (0 , , α ) = 0¯ s = (0 , , ϕ For each node, one needs to find whether
Im h t ( · · · , α t ) = { , } , { } or { } , where “ · · · ”represents the branch to the node. This takes O (2 × n nodes. So, themaximal number of nodes for Σ ϕ solutions is n × Σ ϕ . Therefore, the complexity for listingthe solutions of a “hard” 3-CNF-SAT problem ϕ is O (2 n Σ ϕ ) = O ( n ) as Σ ϕ = 2 O (1) . Theorem IV.2:
The complexity for listing the solutions common to many “solutionstrees” is bound by their minimal complexity.
Proof:
The complexity for listing the solutions of the sub-problem ϕ ( L ) is O (2 n Σ ϕ ( L ) ) . To list the
January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 24 Figure 4 : Merging of two solutions trees Σ ϕ ( L ) and Σ ϕ ( L (cid:48) ) , without solution for the red branches.solutions common to several “solutions trees”, one needs to follow the paths in common (inred on figure 4). The number of paths in common will be less or equal to the minimumnumber of paths among the different sub-problems, as the common paths should belong tothis “solutions tree”. So, the complexity is O (2 n min L Σ ϕ ( L ) ). V. The indicator function S ϕ ( x , · · · , x n ) of the set of solutions S ϕ A. Another description of S ϕ = Im H ϕ . Let consider a solution ( x , · · · , x n ) for the 3-CNF-SAT problem ϕ :( x , · · · , x n ) ∈ S ϕ ⇔ H ϕ ( x , · · · , x n ) = ( x , · · · , x n ) by construction of H ϕ ⇔ t h ,ϕ ( x )... h n,ϕ ( x , · · · , x n ) = t x ... x n (mod 2) ⇔ t h ,ϕ ( x ) + x ... h n,ϕ ( x , · · · , x n ) + x n = t (mod 2) ⇔ n (cid:89) i =1 [ h i,ϕ ( x , · · · , x i ) + x i + 1] = 1 (35) January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 25 Theorem V.1:
Indicator function S ϕ ( x , · · · , x n ) and descriptor function H ϕ Consider the 3-CNF-SAT problem ϕ and its descriptor function H ϕ , if it exists. The indicatorfunction of the set of solutions S ϕ is given by : S ϕ ( x , · · · , x n ) ≡ n (cid:89) i =1 [ h i,ϕ ( x , · · · , x i ) + x i + 1] if H ϕ exists0 otherwise (36) Proof: If H ϕ does not exist, this means that there is no solution for ϕ : S ϕ = ∅ .From (35), it follows that :( x , · · · , x n ) (cid:54)∈ S ϕ (mod 2) ⇔ n (cid:89) i =1 [ h i,ϕ ( x , · · · , x i ) + x i + 1] = 0So, the definition given in (36) corresponds to the indicator function of S ϕ . B. Properties • Let ϕ and ϕ (cid:48) be two 3-CNF-SAT problems, and S ϕ ∩ S ϕ (cid:48) the set of common solutions.Then, S ϕ ∧ ϕ (cid:48) = S ϕ ∩ S ϕ (cid:48) and S ϕ ∧ ϕ (cid:48) ( · ) = S ϕ ( · ) × S ϕ (cid:48) ( · ), following the normal properties ofindicator functions. • Let S ϕ = { ( s , · · · , s n ) } [ ϕ has only one solution] then S ϕ ( x , · · · , x n ) = n (cid:89) i =1 x s i i ( x i + 1) ( s i +1) (mod 2) (37)as S ϕ ( x , · · · , x n ) = 1 ⇐⇒ ∀ i : x s i i ( x i + 1) ( s i +1) = 1 ⇐⇒ ∀ i : x i = s i . • Let S ϕ have many solutions, then S ϕ ( x , · · · , x n ) = (cid:88) ( s , ··· ,s n ) ∈S ϕ n (cid:89) i =1 x s i i ( x i + 1) ( s i +1) (mod 2) (38)The proof is easily done by recurrence. January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 26 • Let ϕ be a 3-CNF formula. ϕ = m (cid:94) k =1 ψ k where ψ k = [ ¬ ] δ rk x r k ∨ [ ¬ ] δ sk x s k ∨ [ ¬ ] δ tk x t k where [ ¬ ] δ i x i = (cid:40) x i if δ i = 0 ¬ x i if δ i = 1Then S ϕ ( · ) = m (cid:89) k =1 { x δ rk r k ( x r k + 1) ( δ rk +1) x δ sk s k ( x s k + 1) ( δ sk +1) x δ tk t k ( x t k + 1) ( δ tk +1) + 1 } (39)The proof follows directly from (15) and (35), considering that : S ϕ ( x , · · · , x n ) = m (cid:89) k =1 S ψk ( x , · · · , x n ) (40)Let us remark that (39) could be of exponential complexity, as it is a multi-linear product of m sums of at least two terms, with m = O ( n ). Numerical results show here again a clustereffect when computing (39). VI. A greedy polynomial algorithm for “hard” 3-CNF-SAT problems
A. The sub-problems ϕ ⊕ ( x t ) and ϕ (cid:9) ( x t ) Definition VI.1:
Let ϕ ⊕ ( x t ) be the sorted sub-problem of ϕ , restricted to the clauses in Cl ( x t ) having x t as the highest indexed positive variable : ϕ ⊕ ( x t ) = m ⊕ t (cid:94) k =1 [ ¬ ] x r k ∨ [ ¬ ] x s k ∨ x t = m ⊕ t (cid:94) k =1 ψ k where r k < s k < t and ϕ (cid:9) x t with the clauses in Cl ( x t ) having x t as the highest indexed negative variable : ϕ (cid:9) ( x t ) = m (cid:9) t (cid:94) k =1 [ ¬ ] x r k ∨ [ ¬ ] x s k ∨ ¬ x t k = m (cid:9) t (cid:94) k =1 ψ k where r k < s k < t We get ϕ = n (cid:94) t =1 [ ϕ ⊕ ( x t ) ∧ ϕ (cid:9) ( x t ) ] and m = n (cid:88) t =1 ( m ⊕ t + m (cid:9) t ) January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 27 Considering together the definition (40) of S ϕ ( x , · · · , x n ) and (36), we get : S ϕ ( x , · · · , x n ) = n (cid:89) t =1 [ S ϕ ⊕ ( xt ) ( x , ··· ,x n ) · S ϕ (cid:9) ( xt ) ( x , ··· ,x n ) ]= n (cid:89) t =1 t (cid:89) i =1 [ h i,ϕ ⊕ ( xt ) ( x , · · · , x i ) + x i + 1] · n (cid:89) t =1 t (cid:89) i =1 [ h i,ϕ (cid:9) ( xt ) ( x , · · · , x i ) + x i + 1] (43) = n (cid:89) t =1 [ h t,ϕ ⊕ ( xt ) ( x , · · · , x t ) + x t + 1] · n (cid:89) t =1 [ h t,ϕ (cid:9) ( xt ) ( x , · · · , x t ) + x t + 1] (41)where len ( h t,ϕ ⊕ ( xt ) ( x , · · · , x t ) + x t + 1) = O (2 ∆ ) and len ( h t,ϕ (cid:9) ( xt ) ( x , · · · , x t ) + x t + 1) = O (2 ∆ ).See (46). Theorem VI.2: The computation of H ϕ ⊕ ( xt ) and H ϕ (cid:9) ( xt ) is O ( n k ) for 3-CNF-SAT prob-lems where ∆ = mn = O (1) .Proof: For each ψ k in ϕ ⊕ ( x t ) , we get h t,ψ k ( α r k , α s k ,
1) = h t,ψ k ( α , · · · , α t − ,
1) = 1 and h i,ψ k ( α , α , · · · , α i ) = α i for i < t . See (15).Remember the computation algorithm (see Theorem II.5 ). By (17), we get : h t,ψ ∧ ψ ( α , · · · , α t ) := ( α t + 1) · { [ h t,ψ ( α , · · · , α t − ,
0) + h t,ψ ( α , · · · , α t − , · [ h t,ψ ( α , · · · , α t − , · h t,ψ ( α , · · · , α t − , h t,ψ ( α , · · · , α t − , · h t,ψ ( α , · · · , α t − , } + α t · { [ h t,ψ ( α , · · · , α t − ,
1) + h t,ψ ( α , · · · , α t − , · [ h t,ψ ( α , · · · , α t − ,
0) + h t,ψ ( α , · · · , α t − , h t,ψ ( α , · · · , α t − ,
1) + h t,ψ ( α , · · · , α t − , · [ h t,ψ ( α , · · · , α t − , · h t,ψ ( α , · · · , α t − , h t,ψ ( α , · · · , α t − , · h t,ψ ( α , · · · , α t − , } := ( α t + 1) · { [ h t,ψ ( α , · · · , α t − ,
0) + h t,ψ ( α , · · · , α t − , h t,ψ ( α , · · · , α t − , · h t,ψ ( α , · · · , α t − , } + α t · h i,ψ ∧ ψ ( α , · · · , α i ) := α i ∀ i < t as h i,ψ ( α , · · · , α i ) = h i,ψ ( α , · · · , α i ) = α i January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 28 (After step one) = ⇒ (cid:40) h t,ψ ∧ ψ ( α , · · · , α t − ,
1) = 1 ∀ ( α , · · · , α t − ) ∈ { , } t − h i,ψ ∧ ψ ( α , · · · , α i ) = α i ∀ i (cid:54) = t ∀ ( α , · · · , α i ) ∈ { , } i (42)For the next step of the algorithm [the computation of h t, ( ψ ∧ ψ ) ∧ ψ ( α , · · · , α t )], we will getthe same property (42) as h t,ψ ∧ ψ ( α , · · · , α t ) will replace h t,ψ ( α , · · · , α t ) in the formula(17) and h t,ψ ( α , · · · , α t ) the term h t,ψ ( α , · · · , α t ).Let us note that there will be no recursive call in the algorithm, as the condition (18) neveroccurs. Indeed :[ h t,ψ ( α r , α s ,
0) + h t,ψ ( α r , α s , · [ h t,ψ ( α r , α s ,
1) + h t,ψ ( α r , α s , (cid:54) = 1as h t,ψ ( α r , α s ,
1) = 1 and h t,ψ ( α r , α s ,
1) = 1.Therefore, at the end of the algorithm, we get H ⊕ ϕ ( xt ) = [ h ,ϕ ⊕ ( xt ) ( · ) · · · h t,ϕ ⊕ ( xt ) ( · )] :with h t,ϕ ⊕ ( xt ) ( α , · · · , α t − ,
1) = 1 ∀ ( α , · · · , α t − ) ∈ { , } t − h i,ϕ ⊕ ( xt ) ( α , · · · , α i ) = α i ∀ i (cid:54) = t ∀ ( α , · · · , α i ) ∈ { , } i (43)Similarly, we get H (cid:9) ϕ ( xt ) = [ h ,ϕ (cid:9) ( xt ) ( · ) · · · h t,ϕ (cid:9) ( xt ) ( · )] :with h t,ϕ (cid:9) ( xt ) ( α , · · · , α t − ,
0) = 0 ∀ ( α , · · · , α t − ) ∈ { , } t − h i,ϕ (cid:9) ( xt ) ( α , · · · , α i ) = α i ∀ i (cid:54) = t ∀ ( α , · · · , α i ) ∈ { , } i (44)Theorem III.5 states that the complexity to compute H ⊕ ϕ ( xt ) is : O ( m ⊕ t n max ≤ i ≤ t len ( h i,ϕ ⊕ ( xt ) ( · )) ) = O ( m ⊕ t n len ( h t,ϕ ⊕ ( xt ) ( α , · · · , α t − , ∀ i < t : len ( h i,ϕ ⊕ ( xt ) ( α , · · · , α i )) = len ( α i ) = len ( h t,ϕ ⊕ ( xt ) ( α , · · · , α t − , len (1) = 1.All together, the m ⊕ t clauses of ϕ ⊕ ( x t ) concern at most (2 · m ⊕ t ) + 1 variables. Therefore, len ( h t,ϕ ⊕ ( xt ) ( α , · · · , α t − , ≤ (2 · m ⊕ t )+1 = O (2 ∆+1 ) (46)where ∆ = mn is the ratio of the 3-CNF-SAT problem, see (1). Without loss of generality,one can relabel the variables so that the minimal index 1 is attributed to the most frequentvariable and so on for the remaining variables. An exact uniform distribution of the variablesyields to m ⊕ t + m (cid:9) t ≤
3∆ for all t . An extreme non uniform distribution, i.e. when each January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 29 variable occurs only once except for one variable, relabeled x , that occurs 3 m − ( n − m ⊕ t + m (cid:9) t = 1 for all t .If the variables are at random in ϕ , ¬ x t and x t should occur approximately mn times,and m ⊕ t ≈ m (cid:9) t ≈ ∆2 for large n . This ratio is a good indicator of the hardness of the 3-CNF-SAT problem (see [3]).∆ being a constant with respect to n , the complexity for H ⊕ ϕ ( xt ) given in (45) is thus : O ( ∆2 · n · ∆ ) = O ( n ) (47)The same arguments are used to prove that the complexity to get H (cid:9) ϕ ( xt ) is O ( n ). B. Complexity theorem for 3-CNF-SAT problem ϕ with one or zero solutionTheorem VI.3: Necessary and sufficient condition for satisfiabilitywhen S ϕ ≤ S ϕ ≤
1, then ϕ is satisfiable [ S ϕ = 1] if and only if n (cid:89) t =1 [ h t,ϕ ⊕ ( xt ) ( x , · · · , x t ) + x t + 1] · n (cid:89) t =1 [ h t,ϕ (cid:9) ( xt ) ( x , · · · , x t ) + x t + 1]= (cid:32) n (cid:89) t =1 x t (cid:33) + E ( x , · · · , x n ) (48)where E ( x , · · · , x n ) is a linear combination of rank strictly less than n Proof:
From the hypothesis, ϕ is satisfiable if and only if S ϕ = 1. Let ( s , · · · , s n ) be thisonly solution of ϕ . Thus, S ϕ ( x , · · · , x n ) (41) = n (cid:89) t =1 [ h t,ϕ ⊕ ( xt ) ( x , · · · , x t ) + x t + 1] · n (cid:89) t =1 [ h t,ϕ (cid:9) ( xt ) ( x , · · · , x t ) + x t + 1] (37) = n (cid:89) t =1 x s t t ( x t + 1) ( s t +1) = (cid:89) t † s t =1 ( x t ) · (cid:89) t † s t =0 ( x t + 1)= (cid:89) t † s t =1 x t · (cid:89) t † s t =0 x t + (cid:88) j † s j =0 (cid:89) t (cid:54) = j † s t =0 x t + · · · = (cid:32) n (cid:89) t =1 x t (cid:33) + E ( x , · · · , x n ) January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 30 E ( x , · · · , x n ) depends on the value of ( s , · · · , s n ) : E ( x , · · · , x n ) = 0 when ( s , · · · , s n ) =(1 , · · · ,
1) and E ( x , · · · , x n ) = (cid:81) nt =1 ( x t + 1) − (cid:81) nt =1 ( x t ) when ( s , · · · , s n ) = (0 , · · · , (cid:81) nt =1 x t ) appears in (41).As[ S ϕ = ∅ ] ⇔ n (cid:89) t =1 [ h t,ϕ ⊕ ( xt ) ( x , · · · , x t ) + x t + 1] · n (cid:89) t =1 [ h t,ϕ (cid:9) ( xt ) ( x , · · · , x t ) + x t + 1] ≡ n (cid:89) t =1 x t ) in (41) is equivalent to prove that S ϕ = 1. Example :
Let us consider the following 3-CNF-SAT problem with 6 variables and nosolution : ϕ = ( x ∨ x ∨ x ) ∧ ( x ∨ x ∨ ¬ x ) ∧ ( ¬ x ∨ ¬ x ∨ x ) ∧ ( ¬ x ∨ ¬ x ∨ x ) ∧ ( ¬ x ∨ ¬ x ∨ ¬ x ) ∧ ( ¬ x ∨ x ∨ ¬ x ) ∧ ( x ∨ x ∨ x ) ∧ ( x ∨ ¬ x ∨ ¬ x ) ∧ ( x ∨ ¬ x ∨ ¬ x ) ∧ ( ¬ x ∨ x ∨ ¬ x ) ∧ ( ¬ x ∨ x ∨ ¬ x ) ∧ ( ¬ x ∨ x ∨ ¬ x ) ∧ ( x ∨ ¬ x ∨ x ) ∧ ( ¬ x ∨ ¬ x ∨ x ) ∧ ( ¬ x ∨ ¬ x ∨ ¬ x ) ∧ ( x ∨ x ∨ ¬ x ) ∧ ( ¬ x ∨ x ∨ ¬ x ) January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 31 This gives us : h ⊕ ,ϕ ( x ( · ) = x h ⊕ ,ϕ ( x ( · ) = x h ⊕ ,ϕ ( x ( · ) = x + x + x x + x x + x x x h ⊕ ,ϕ ( x ( · ) = x + x + x x + x x + x x x h ⊕ ,ϕ ( x ( · ) = 1 + x + x + x x + x x + x x + x x x + x x x + x x x x h ⊕ ,ϕ ( x ( · ) = 1 + x + x x + x x + x x + x x x + x x x and h (cid:9) ,ϕ ( x ( · ) = x h (cid:9) ,ϕ ( x ( · ) = x h (cid:9) ,ϕ ( x ( · ) = x + x x + x x x h (cid:9) ,ϕ ( x ( · ) = x + x x + x x x + x x x x h (cid:9) ,ϕ ( x ( · ) = x + x x + x x x x x h (cid:9) ,ϕ ( x ( · ) = x + x x + x x + x x x + x x + x x x + x x x + x x x x + x x x + x x x x + x x x x + x x x x x + x x x + x x x + x x x x + x x x + x x x x + x x x x + x x x x x + x x x x x + x x x x x + x x x x x x S ϕ ( x , · · · , x n ) = n (cid:89) t =1 [ h t,ϕ ⊕ ( xt ) ( x , · · · , x t ) + x t + 1] · n (cid:89) t =1 [ h t,ϕ (cid:9) ( xt ) ( x , · · · , x t ) + x t + 1]= 2 x + 16 x x + 10 x x + 2 x x + 10 x x + 4 x x + 2 x x + 2 x x + 242 x x x + 160 x x x + 80 x x x + 68 x x x + 10 x x x + 38 x x x + 56 x x x + 6 x x x + 26 x x x + 90 x x x + 6 x x x + 2 x x x + 24 x x x + 6 x x x + 8 x x x + 3 152 x x x x + 950 x x x x + 2 122 x x x x + 624 x x x x + 2 396 x x x x + 1 352 x x x x + 30 x x x x + 176 x x x x + 508 x x x x + 268 x x x x + 10 x x x x + 76 x x x x + 26 x x x x + 102 x x x x + 18 950 x x x x x + 75 450 x x x x x + 25 916 x x x x x + 26 252 x x x x x + 1 344 x x x x x + 384 x x x x x + 1 086 442 x x x x x x mod = 0 January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 32 Let us consider the similar 3-CNF-SAT problem ϕ (cid:48) without the 13th clause : ( x ∨¬ x ∨ x ). ϕ (cid:48) has a unique solution : ( s , · · · , s ) = (0 , , , , , S ϕ ( x , · · · , x n ) = n (cid:89) t =1 [ h t,ϕ ⊕ ( xt ) ( x , · · · , x t ) + x t + 1] · n (cid:89) t =1 [ h t,ϕ (cid:9) ( xt ) ( x , · · · , x t ) + x t + 1] mod = 1 x x x + 291 x x x x + 3 x x x x + 13 x x x x + 3 633 x x x x x +8 293 x x x x x + 125 x x x x x + 257 641 x x x x x x mod = (1 + x ) x x (1 + x ) x (1 + x ) Theorem VI.4: The 3-CNF-SAT problems, with
S ≤ and ∆ = mn = O (1) , are in P Proof : (cid:13) h t,ϕ ⊕ ( xt ) ( x , · · · , x t ) + x t + 1 is a linear combination of the variables x i present in ϕ ⊕ ( x t ) , and len ( h t,ϕ ⊕ ( xt ) ( x , · · · , x t ) + x t + 1) = O (2 ∆ ) . See (46). The same is true for h t,ϕ (cid:9) ( xt ) ( x , · · · , x t ) + x t + 1. Therefore, the computation of g t ( x , · · · , x t ) def = [ h t,ϕ ⊕ ( xt ) ( x , · · · , x t ) + x t + 1] · [ h t,ϕ (cid:9) ( xt ) ( x , · · · , x t ) + x t + 1]is O (2 ) for all t and g t ( x , · · · , x t ) is a linear combination of the variables x i present in ϕ ⊕ ( x t ) ( x , · · · , x t ) or ϕ (cid:9) ( x t ) ( x , · · · , x t ), that is V ( x t ) as defined in (27).2 (cid:13) Let us consider g t ( x , · · · , x t ) : g t ( x , · · · , x t ) = (cid:88) ( δ , ··· ,δ t ) ∈{ , } t c t ( δ , ··· ,δ t ) x δ · · · x δ t t We have just seen that the computation of all the coefficients c t ( δ , ··· ,δ t ) is O (2 ). Let usnote that g t ( s , · · · , s t ) corresponds to the total of the terms in g t ( x , · · · , x t ) not includ-ing x j such that s j = 0. For example, if g ( x , x , x ) = 1 + x + 3 x x + 7 x x x , we get g (1 , ,
0) = 5 mod = 1. It takes O (2 ∆ ) to get this result, as g t ( x , · · · , x t ) is a linear combina-tion of O (∆) variables, as explained in (47). Therefore, we have access, in polynomial time,to any of the coefficients c t ( δ , ··· ,δ t ) and to any value for g t ( s , · · · , s t ).3 (cid:13) We have now to compute the coefficient of ( n (cid:89) t =1 x t ) in n (cid:89) t =1 g t ( x , · · · , x t ). But the entire January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 33 computation of n (cid:89) t =1 g t ( x , · · · , x t ) is O (2 n ∆ ) as each factor has O (2 ∆ ) terms. Definition VI.5: • Let us define the sub-product : G [ i : i ] ( x , · · · , x i ) ≡ i (cid:89) j = i g j ( x , · · · , x j ) • And C [ i : i ]( δ , ··· ,δ i ) as (cid:88) ( δ , ··· ,δ i ) ∈{ , } i C [ i : i ]( δ , ··· ,δ i ) x δ · · · x δ i i = G [ i : i ] ( x , · · · , x i )What we are looking for is C [1: n ](1 , ··· , . Lemma VI.1: C [1: n ]( δ , ··· ,δ n ) = δ (cid:88) ξ =0 δ (cid:88) ζ = δ − ξ · · · δ n − (cid:88) ξ n − =0 δ n − (cid:88) ζ n − = δ n − − ξ n − c n ( ξ , ··· ,ξ n − ,δ n ) C [1: n − ζ , ··· ,ζ n − ) C [1: n ](1 , ··· , = (cid:88) ( ξ , ··· ,ξ n − ) ∈{ , } n − c n ( ξ , ··· ,ξ n − ,
1) 1 (cid:88) ζ =1 − ξ · · · (cid:88) ζ n − =1 − ξ n − C [1: n − ζ , ··· ,ζ n − ) Proof of the lemma : G [1: n ] ( x , · · · , x n ) = g n ( x , · · · , x n ) · G [1: n − ( x , · · · , x n − )= (cid:88) ( ξ , ··· ,ξ n ) ∈{ , } n c n ( ξ , ··· ,ξ n ) x ξ · · · x ξ n n × (cid:88) ( ζ , ··· ,ζ n − ) ∈{ , } n − C [1: n − ζ , ··· ,ζ n − ) x ζ · · · x ζ n − n − = (cid:88) ( δ , ··· ,δ n ) ∈{ , } n δ (cid:88) ξ =0 δ (cid:88) ζ = δ − ξ · · · δ n − (cid:88) ξ n − =0 δ n − (cid:88) ζ n − = δ n − − ξ n − c n ( ξ , ··· ,ξ n − ,δ n ) C [1: n − ζ , ··· ,ζ n − ) x δ · · · x δ n n The coefficient C [1: n ](1 , ··· , is given by the sum of the terms where x ξ j j x ζ j j = x j , i.e. ( ζ j , ξ j ) ∈{ (0 , , (1 , , } as x mod = x , for j ≤ i . (cid:4) January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 34 (cid:13) Therefore, C [1:2](1 , = c , c + c , ( c + c ) C [1:2](0 , = c , c C [1:2](1 , = c , c + c , ( c + c ) C [1:2](0 , = c , c C [1:3](1 , , = c , , C [1:2](1 , + c , , (cid:16) C [1:2](1 , + C [1:2](1 , (cid:17) + c , , (cid:16) C [1:2](0 , + C [1:2](1 , (cid:17) + c , , (cid:16) C [1:2](0 , + C [1:2](1 , + C [1:2](0 , + C [1:2](1 , (cid:17) C [1:3](0 , , = c , , C [1:2](0 , + c , , (cid:16) C [1:2](0 , + C [1:2](0 , (cid:17) C [1:3](0 , , = c , , C [1:2](0 , and so on · · · Each c i ( δ , ··· ,δ i − ,δ i ) is multiplied by 2 ( δ + ··· + δ i − ) terms C [1: i − · ) , as every “1” in c i ( δ , ··· ,δ i − ,δ i ) , except for δ i , is replaced by a “0” and a “1” in the factors C [1: i − δ , ··· ,δ i − ) . So, toget C [1: i ]( δ , ··· ,δ i ) , one should compute :3 ( δ + ··· + δ i − ) = δ + ··· + δ i − (cid:88) j =0 (cid:18) δ + · · · + δ i − i (cid:19) j terms, (49)and to get all the C [1: i ]( · ) , this needs :2 · ( i − = 2 · i − (cid:88) j =0 (cid:18) i − j (cid:19) j terms, considering δ i ∈ { , } . (50)5 (cid:13) But:
1. A very limited number of c i ( · ) are non zero in a hard 3-CNF-SAT problem.2. g i ( x , · · · , x i ) is a linear combination of V ( x i ) variables, with V ( x i ) = O (∆).3. So, the maximum number of non-zero c i ( · ) is O (2 ∆ ) in g i ( x , · · · , x i ).4. And the maximum total number of non-zero c i ( · ) for all 1 ≤ i ≤ n is O ( n · ∆ ). January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 35 Lemma VI.2:
The complexity to compute C [1: i ]( δ , ··· ,δ i ) from C [1: i − δ , ··· ,δ i − ) is O (3 ∆ ) Proof of the lemma :
Without loss of generality, one can suppose that V ( x i ) = { x i − k ∆ , · · · , x i } , as the number ofvariables in g i ( x , · · · , x i ) is O (∆). Thus, c i ( δ , ··· ,δ i ) = 0 if δ j = 1 for some j < i − k ∆The formula (49) to get C [1: i ]( δ , ··· ,δ i ) should be modified :3 ( δ i − k ∆ + ··· + δ i − ) = δ i − k ∆ + ··· + δ i − (cid:88) j =0 (cid:18) δ i − k ∆ + · · · + δ i − i (cid:19) j (51)= O (3 ∆ ) as ( δ i − k ∆ + · · · + δ i − ) ≤ k ∆ (cid:4) Unfortunately, the formula (50) to get all C [1: i ]( δ , ··· ,δ i ) is still O (2 i ).2 ( i − k ∆) · ( k ∆) = 2 i − k ∆ · k ∆ (cid:88) j =0 (cid:18) k ∆ j (cid:19) j (52)as ( δ i − k ∆ + · · · + δ i − ) ≤ k ∆and ( δ , · · · , δ i − k ∆ − ) ∈ { , } ( i − k ∆ − and ( δ i ) ∈ { , } = O (2 i )So, the complexity to get all C [1: n ]( δ , ··· ,δ n ) from C [1: n − δ , ··· ,δ n − ) is O (2 n ).6 (cid:13) BUT (this is the turning point of the proof ) if we only need to know C [1: n ](1 , ··· , :
1. The complexity to compute C [1: n ](1 , ··· , from C [1: n − δ , ··· ,δ n − ) is O (3 ∆ ), by Lemma VI.2
2. To compute C [1: n ](1 , ··· , , one need to know O (2 ∆ ) terms C [1: n − δ , ··· ,δ n − ) . Indeed, if January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 36 V ( x n ) = { x n − k ∆ , · · · , x n } :[ δ i = 1 for i < n − k ∆] ⇒ c n ( δ , ··· ,δ n − k ∆ , ··· ,δ n ) = 0 ⇓ Only C [1: n − , ··· , ,δ n − k ∆ , ··· ,δ n − ) with ( δ n − k ∆ , · · · , δ n ) ∈ { , } k ∆ can occur in C [1: n ](1 , ··· , , depending of the values for c n (0 , ··· , ,δ n − k ∆ , ··· ,δ n ) . AND to compute these O (2 ∆ ) terms C [1: n − , ··· , ,δ n − k ∆ , ··· ,δ n − ) , the complexity givenby (52) is not exponential but O (4 ∆ ) :2 · ( k ∆ − = 2 · k ∆ − (cid:88) j =0 (cid:18) k ∆ − j (cid:19) j (53)as ( δ n − k ∆ + · · · + δ n − ) ≤ k ∆ − δ , · · · , δ i − k ∆ − ) = (0 , · · · ,
0) and ( δ n − ) ∈ { , } = O (4 ∆ )(instead of O (3 ∆ ) for the previous step for only one calculation)4. And we still need to know only O (2 ∆ ) terms C [1: n − δ , ··· ,δ n − ) to compute the O (2 ∆ )terms C [1: n − , ··· , ,δ n − k ∆ , ··· ,δ n − ) . Indeed, g n − ( x , · · · , x n − ) is a linear combination of O (2 ∆ ) variables, the ones in V ( x n − ) . So, δ i = 1 for i : x i (cid:54)∈ V ( x n − ) ⇒ c n − δ , ··· ,δ n − ) = 0 ⇓ Only C [1: n − δ , ··· ,δ n − ) with δ i = 0 for i : x i (cid:54)∈ V ( x n − )can occur in C [1: n − , ··· , ,δ n − k ∆ , ··· ,δ n − ) , depending of the values for c n − δ , ··· ,δ n − ) . Thus, at most O (2 ∆ ) terms C [1: n − δ , ··· ,δ n − ) will be needed.5. This step is similar to step (3) : the complexity to compute these O (2 ∆ ) terms January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 37 C [1: n − δ , ··· ,δ n − ) with δ i = 0 for i : x i (cid:54)∈ V ( x n − ), is once again O (4 ∆ ) :2 · ( k ∆ − = 2 · k ∆ − (cid:88) j =0 (cid:18) k ∆ − j (cid:19) j as ( δ + · · · + δ n − ) ≤ [ V ( x n − ) −
1] = k ∆ − δ i = 0 for i : x i (cid:54)∈ V ( x n − ) and ( δ n − ) ∈ { , } = O (4 ∆ )6. This step is similar to step (4) : the same arguments with V ( x n − ) yields tothe same conclusion, that one need to know at most O (2 ∆ ) terms C [1: n − δ , ··· ,δ n − ) tocompute the previous step.7. And so on, till g ( x ). In conclusion, the complexity to compute C [1: n ](1 , ··· , is the sum of the computingcomplexity of each even step : O (3 ∆ ) + ( n − O (4 ∆ ) = O ( n ) (54) (cid:4) Example :
Let us consider this situation : • V ( x i ) = { x i − , x i − , x i } for 3 ≤ i ≤ n • g i ( x , · · · , x i ) = 1 + x i − + x i − + x i + x i − x i − + x i − x i + x i − x i + x i − x i − x i ⇒ c i ( δ , ··· ,δ i ) = 0 if δ j (cid:54) = 0 for j < i − c i (0 , ··· , ,δ i − ,δ i − ,δ i ) = 1 otherwise January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 38 Then, C [3: n ](1 , ··· , , , = c n (0 , ··· , , δn − , δn − , δn (cid:0) C [3: n − , ··· , , δn − , δn − (cid:1) + c n (0 , ··· , , , , (cid:0) C [3: n − , ··· , , , + C [3: n − , ··· , , , (cid:1) + c n (0 , ··· , , , , (cid:0) C [3: n − , ··· , , , + C [3: n − , ··· , , , (cid:1) + c n (0 , ··· , , , , (cid:0) C [3: n − , ··· , , , + C [3: n − , ··· , , , + C [3: n − , ··· , , , + C [3: n − , ··· , , , (cid:1) = 4 C [3: n − , ··· , , , + 2 C [3: n − , ··· , , , + 2 C [3: n − , ··· , , , + C [3: n − , ··· , , , mod = C [3: n − , ··· , , , C [3: n − , ··· , δn − , δn − , δn − = c n − , ··· , , δn − , δn − , δn − (cid:0) C [3: n − , ··· , , δn − , δn − (cid:1) + c n − , ··· , , δn − , δn − , δn − (cid:0) C [3: n − , ··· , , δn − , δn − + C [3: n − , ··· , , δn − , δn − (cid:1) = 2 C [3: n − , ··· , , , + C [3: n − , ··· , , , mod = C [3: n − , ··· , , , C [3: n − , ··· , δn − , δn − , δn − = c n − , ··· , , δn − , δn − , δn − (cid:0) C [3: n − , ··· , , δn − , δn − (cid:1) + c n − , ··· , , δn − , δn − , δn − (cid:0) C [3: n − , ··· , , δn − , δn − + C [3: n − , ··· , , δn − , δn − (cid:1) = 2 C [3: n − , ··· , , , + C [3: n − , ··· , , , mod = C [3: n − , ··· , , , ⇒ C [3: n ](1 , ··· , , , = C [3:3](1 , , = c , , = 1 Corollary VI.1:
A 3-CNF-SAT problem ϕ with ∆ = O (1) such that [ C [1: n ](1 , ··· , = 1 ⇔ S ϕ (cid:54) = ∅ ] is P . Proof:
As the computation of C [1: n ](1 , ··· , is O ( n ) when ∆ = O (1), the satisfiability problem January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 39 of ϕ is O ( n ). Corollary VI.2:
A 3-CNF-SAT problem ϕ with ∆ = O (1) and S ϕ = 2 k + 1 ( k ∈ N ) is P . Proof:
From (38), we get : S ϕ ( x , · · · , x n ) = (cid:88) ( s , ··· ,s n ) ∈ S ϕ n (cid:89) i =1 x s i i ( x i + 1) ( s i +1) (mod 2) (48) = k +1 (cid:88) j =1 (cid:32) n (cid:89) t =1 x t + E j ( x , · · · , x n ) (cid:33) = (2 k + 1) (cid:32) n (cid:89) t =1 x t (cid:33) + k +1 (cid:88) j =1 E j ( x , · · · , x n ) mod = (cid:32) n (cid:89) t =1 x t (cid:33) + E ( x , · · · , x n )Therefore, the hypotheses of Corollary VI.1 are satisfied, and the 3-CNF-SAT problem is P C. Complexity theorems for 3-CNF-SAT problems with S ϕ ≤ k We have found a polynomial algorithm to solve the satisfiability of any 3-CNF-SAT problem,assuming only zero or an odd number of solutions can occur. What about 3-CNF-SATproblems with a possible even number of solutions ?1 (cid:13)
Let us begin with ϕ such that S ϕ is zero or 2 and S ϕ = { ( s , · · · , s n ) , ( s (cid:48) , · · · , s (cid:48) n ) } or ∅ . January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 40 • Suppose ∃ ! j ∈ { , · · · , n } such that s j (cid:54) = s (cid:48) j . S ϕ ( x , · · · , x n ) = n (cid:89) i =1 x s i i (1 + x i ) s i + n (cid:89) i =1 x s (cid:48) i i (1 + x i ) s (cid:48) i = (cid:89) i (cid:54) = j x s i i (1 + x i ) s i (cid:0) x s j j (1 + x j ) s j + x s (cid:48) j j (1 + x j ) s (cid:48) j (cid:1) = (cid:89) i (cid:54) = j x s i i (1 + x i ) s i (cid:0) x j + (1 + x j ) (cid:1) = (cid:89) i (cid:54) = j x s i i (1 + x i ) s i = (cid:0) n (cid:89) i =1 i (cid:54) = j x i (cid:1) + E ( x , · · · , x j − , x j +1 , · · · , x n ) ⇒ ϕ is satisfiable ⇔ ∃ C [1: n ]( δ , ··· ,δ n ) (cid:54) = 0 with (cid:80) δ i = n − δ i equals 0, i.e. δ j = 0] . • Let us consider the general situation : I (1) = { i : s i = 0 } and I (2) = { i : s (cid:48) i = 0 } . Let I (1) = { i (1)1 , · · · , i (1) m (1) } , I (2) = { i (2)1 , · · · , i (2) m (2) } and µ ( j ) def = { l : j ∈ I ( l ) } . S ϕ ( x , · · · , x n ) = (cid:89) i (cid:54)∈ I (1) x i (cid:89) i ∈ I (1) (1 + x i ) + (cid:89) i (cid:54)∈ I (2) x i (cid:89) i ∈ I (2) (1 + x i ) (cid:89) i (cid:54)∈ I (1) x i (cid:89) i ∈ I (1) (1 + x i ) = ( (cid:89) i (cid:54)∈ I (1) x i ) ( (cid:88) ( δ , ··· ,δ m (1) ) ∈{ , } m (1) m (1) (cid:89) j =1 x δ j i j )= n (cid:89) i =1 x i + n (cid:89) i =1 , i (cid:54) = i (1)1 x i + (cid:89) i (cid:54) = i (1)2 x i + · · · (cid:89) i (cid:54)∈ I (2) x i (cid:89) i ∈ I (2) (1 + x i ) = n (cid:89) i =1 x i + (cid:89) i (cid:54) = i (2)1 x i + (cid:89) i (cid:54) = i (2)2 x i + · · · mod = (cid:88) j : µ ( j )=1 (cid:0) n (cid:89) i =1 i (cid:54) = j x i (cid:1) + E ( x , · · · , x n ) (cid:2) ( S ϕ = 2) ⇒ { j : µ ( j ) = 1 } (cid:54) = ∅ (cid:3) ϕ is satisfiable ⇔ ∃ C [1: n ]( δ , ··· ,δ n ) (cid:54) = 0 with (cid:80) δ j = n − δ j = 0 with µ ( j ) = 1] . So, the unsatisfiability of any 3-CNF-SAT ϕ with maximum 2 solutions will be proved onlyif one gets : C [1: n ](1 , ··· , = 0 and C [1: n ]( δ , ··· ,δ n ) = 0 for the (cid:0) n (cid:1) situations where n (cid:88) j =1 δ j = n − January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 41 Using the same arguments as to get (54), the complexity to compute any C [1: n ]( δ , ··· ,δ n ) with (cid:80) δ i = n − O ( n ). So the general complexity for any 3-CNF-SAT problem with S ϕ ≤ O ( n ) + (cid:0) n (cid:1) O ( n ) = O ( n ).2 (cid:13) Let us consider ϕ such that S ϕ = K = 2 k and S ϕ = { ( s (1)1 , · · · , s (1) n ) , · · · , ( s ( K )1 , · · · , s ( K ) n ) } .Let I ( l ) = { i : s ( l ) i = 0 } for 1 ≤ l ≤ KI ( l ) = { i ( l )1 , · · · , i ( l ) m ( l ) } and µ ( { j , · · · , j p } ) def = { l : { j , · · · , j p } ⊆ I ( l ) } . S ϕ ( x , · · · , x n ) = K (cid:88) l =1 (cid:0) (cid:89) i (cid:54)∈ I ( l ) x i (cid:89) i ∈ I ( l ) (1 + x i ) (cid:1) (cid:89) i (cid:54)∈ I ( l ) x i (cid:89) i ∈ I ( l ) (1 + x i ) = ( (cid:89) i (cid:54)∈ I ( l ) x i ) ( (cid:88) ( δ , ··· ,δ m ( l ) ) ∈{ , } m ( l ) m ( l ) (cid:89) j =1 x δ j i j )= n (cid:89) i =1 x i + (cid:88) j ∈ I ( l ) (cid:89) i (cid:54) = j x i + (cid:88) { j ,j } ⊆ I ( l ) (cid:89) i (cid:54)∈{ j ,j } x i + · · · (cid:88) { j , ··· ,j m ( l ) } ⊆ I ( l ) (cid:89) i (cid:54)∈{ j , ··· ,j m ( l ) } x i mod = (cid:88) j : µ ( { j } )=1 (cid:89) i (cid:54) = j x i + (cid:88) { j ,j } : µ ( { j ,j } )=1 (cid:89) i (cid:54)∈{ j ,j } x i + · · · + (cid:88) { j , ··· ,j p } : µ ( { j , ··· ,j p } )=1 (cid:89) i (cid:54)∈{ j , ··· ,j p } x i + E ( x , · · · , x n ) S ϕ ≤ k ⇒ (cid:8) { j , · · · , j k } : µ ( { j , · · · , j k } ) = 1 (cid:9) (cid:54) = ∅ It is possible that : (cid:8) { j , · · · , j p } : µ ( { j , · · · , j p } ) = 1 (cid:9) = ∅ for p < k For example, S ϕ = { (1 , , , , , (1 , , , , , (1 , , , , , (1 , , , , , (1 , , , , , (1 , , , , , (1 , , , , , (1 , , , , }⇒ (cid:8) { j } : µ ( { j } ) = 1 (cid:9) = (cid:8) { j , j } : µ ( { j , j } ) = 1 (cid:9) = ∅⇒ (cid:8) { j , j , j } : µ ( { j , j , j } ) = 1 (cid:9) (cid:54) = ∅ Assuming S ϕ ≤ k : ϕ is satisfiable ⇔ ∃ C [1: n ]( δ , ··· ,δ n ) (cid:54) = 0 with (cid:80) δ j ≤ n − k. January 6, 2020—1 : 28 am DRAFTARD 3-CNF-SAT PROBLEMS ARE IN P . 42 So, assuming that our 3-CNF-SAT problems have at most k solutions, the unsatisfiabilityof any such 3-CNF-SAT ϕ will be proved only if one gets : C [1: n ]( δ , ··· ,δ n ) = 0 for the k (cid:88) i =0 (cid:18) ni (cid:19) situations where n (cid:88) j =1 δ j ≤ n − k Using the same arguments as to get (54), the complexity to compute any C [1: n ]( δ , ··· ,δ n ) with (cid:80) δ i ≤ n is shown to be O ( n ). In conclusion, on the assumption of a k limit for the number of solutions, thegeneral complexity for any 3-CNF-SAT problem with ∆ = mn = O (1) is k (cid:88) i =0 (cid:18) ni (cid:19) O ( n ) ≤ ( n + 1) k O ( n ) = O ( n k ) for large n wrt k . Therefore, the “hard” 3-CNF-SAT problems are in P , as their number of so-lutions is limited by k . VII. Remarks and further researches
It is important to stress that this is not a heuristic proof. The fact that our polynomialalgorithm does not deliver any solution , but only states whether they exist or not, isa key issue for downgrading the complexity level in our paper. This is a very high price topay, perhaps further researches could lighten this price.It is also essential to underline that this is not a proof that NP = P . It is a firstinsight in the complex question of the boundary between P and NP . The search of apolynomial algorithm for easy S ϕ (cid:54) = O (2 k )] is under way, but themain issue seems to be how to distinguish between easy and hard 3-CNF-SAT problems and if it is possible to do that in a polynomial complexity. An algorithm, freely available, exists and was heavily used to check the theoretical resultsof this paper.
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