aa r X i v : . [ c s . L O ] A ug CONNECTION AND DISPERSION OF COMPUTATION
KOBAYASHI, KOJI
Abstract.
In this paper, we describe the impact on the computational com-plexity of Connection and Dispersion of CNF. In previous paper [1], we toldabout structural differences in the P-complete problems and NP-completeproblems. In this paper, we clarify the CNF’s dispersion and HornCNF’sconnection, and shows the difference between CNFSAT HornSAT. First we fo-cus on the MUC decision problem. We clarify the relationship between MUCand the classifying of the truth value assignment. Next, we clarify the clausescorrelation and orthogonal by using the two inner product of clauses. BecauseHornMUC has higher orthogonal, its orthogonal MUC is polynomial size. Be-cause MUC has higher correlation, its orthogonal MUC is not polynomial size.And, HornMUC whereas only be a large polynomial is at most its size evenif orthogonal than orthogonal high, MUC will be fit to size polynomial in thesize and orthogonalized using HornCNF more highly correlated shown. So DP = P , and NP = P . CNF’s classification and CNFSAT
We show the relationship between CNFSAT and CNF’s classification. We showthe relationship between MUC decision problem and CNFSAT. And We show therelationship that determined by the CNF. And We show the relationship betweenCNF’s classification and MUC dicition problem.1.1.
MUC decision problem.
Describes the MUC decision problem. MUC de-cision problem is the problem to decide the CNF is MUC (Minimum UnsatisfiableCore) or not. MUC is the unsatisfiable CNF. And it changes MUC to satisfiableCNF that remove one of the MUC’s clause. MUC decision problem is combina-tion problem of coNP-complete and P-complete problem. And HornCNF’s MUCdicision problem is P-complete because of P = coP .The relationship between the DP-complete and P-complete is; Theorem 1. If P = DP then P = N P . So if we can not reduce MUC dicisionproblem to HornMUC dicision problem in polynomial time, then P = N P .Proof. If P = N P then
N P = coN P and P = DP . So MUC dicision problem canreduce HornMUC dicisition problem in polynomial time. So take the contraposi-tive, if we can not reduce MUC dicision problem to HornMUC dicision problem inpolynomial time, then P = N P . (cid:3) CNF classification.
Describe the relationship that define the CNF. CNFclauses value corresponds to either true or false. Clauses are the rules that mapseach truth value assignment to truth value. This is equivalence relation that classifyeach truth values to equivalent class.
Definition 2.
Clauses equivalence relation is the truth value assignments relationthat equal the clauses value. Similarly, CNF equivalence relation is the truth valueassignments relation that equal the all clauses value set.I will use the term “CNF classification” to the truth value assignments equiva-lence class of CNF, and “CNF equivalence class” to the equivalence class of CNFclassification. And “Logical value assignment” to the set of the clauses values. And“Cyclic value assignment” to the logical value assignment that have only one falsevalue clause, and “All true assignment” to the logical value assignment that have nofalse value. Number of cyclic value assignment matches the number of the clauses.All true assignment is only one. The combined truth value assignment and logicalvalue assignment is the truth value table of the clauses. Especially, I will use theterm “Logical value table” to the truth value table.1.3.
CNF classification and MUC dicision problem.
MUC decision problemis a matter to determine the truth as an input CNF, can be thought as the problemsdealing with CNF classification. The problem is the decision problem that thelogical value assignment includes all the cyclic value assignment and excludes alltru assignment. So MUC decision problem can be divided into two calculations,CNF classification and decision of the logical value assignments.CNF classification includes the difference of MUC decision problem and Horn-MUC decision problem. In decision of logical value assignment, There is no dif-ference between MUC decision problem and HornMUC decision problem. Andthese can be determined in polynomial time either. Especially, all logical valueassignments is cyclic value assignment, the decision of logical value assignment candetermined in polynomial time of the number of the clauses.
Theorem 3.
Decision of logical value assignment can be done in polynomial timeeither MUC decision problem or HornMUC decision problem. Thus, the differencein computational complexity of the MUC decision problem and HornMUC decisionproblem will appear in size of the logic value assignments of CNF classification.Proof.
The decision of logical value assignment of MUC decision problem and Horn-MUC decision problem is the computation that the logical value assignment includesall cyclic value assignment and excludes all truth assignment. We can determinethe decision by determine the all logical value assignment. Thus, We can determinethe decision only the polynomial time of the logical value assignment.And we can handle in polynomial time of the logical value assignment that reducefrom MUC decision problem and HornMUC decision problem. So, if they have thecoputation complexity difference between MUC decision problem and HornMUCdecision problem, the difference included in CNF classification. (cid:3)
Therefore, What has to be noticed is the CNF classification.2.
MUC as Periodic function
In view of periodic function, let us then consider MUC. CNF clauses classificationhave the periodicity in truth value table. Thus, we can deal with CNF as periodicfunction.The truth table grows larger the type of variables included in the clauses, andtruth value assignment is changed and the clause is false by variable’s positiveand negative. Thus, in periodic function of clauses, variable is cycle and variable’s
ONNECTION AND DISPERSION OF COMPUTATION 3 positive/negative is phase. And CNF is the periodic function that put many clausesperiodic function. Clauses is the notation in the frequency domain, logical valuetable is the notation in the time domain.
Definition 4.
I will use the term “Clause cycle” to the number of the variables inclauses. The number of Clause cycle is equal to the number of all cases of changethe clauses variable’s positive/negative.I will use the term “Clause phase” to the positive/negative configuration of thevariables in clauses. Clause phase is the position of the truth value assignmentthat the clause make false in truth value table. The clause phase difference of twoclauses is equal the minimum Hamming distance of these truth value assignmentthat make these clauses false.In MUC, there is a equivalence that makes only each clause false, and the truthvalue assignment does not belong to the false equivalence of other clauses. In otherwords, the equivalence class is not a combination of other clauses of MUC. Thus,between the clause and the other clauses, there is the orthogonal that could not savethe logical value in transposition. To put it the other way round, any false clausein the same equivalence can transpose another clause each other. Thus, betweenthe clause and the another, there is the correlation that can save the logical valuein transposition.
Theorem 5.
Each clauses of the MUC have the orthogonality that can not replacewith a combination of other clauses in the MUC. If MUC have the equivalence classthat have some false clauses, the clauses have the correlation that can replace eachother.Proof.
We prove the clauses orthogonality of MUC using proof by contradiction.We assume Clauses C without orthogonality is included in the MUC. Because C does not have the orthogonality, the truth value assignments that C is false willbe false in another clauses. But this is inconsistent with the terms of the MUC(there are truth value assignment clause that clause is false). Thus From the proofby contradiction, clauses of MUC have orthogonality.It is clear that the clauses have the correlation if the clauses have the sameequivalence class. (cid:3) Logic value assignment of MUC and orthogonality/correlation has a deep rela-tionship. Logical value Assignment represents the relationship between the someclauses which is false in truth value assignment. In other words, the same valueclauses in same logical value assignment have correlation. To put it the other wayround, the different value clauses in same logical value assignment have orthogo-nality. In addition, there are the cycric value assignment to each clauses in MUC,there must also be orthogonal of the each clauses. Therefore, to increase the or-thogonality between MUC clauses, it is necessary that all logical value assignmentis cyclic value assignment in MUC.
Theorem 6.
All clauses in MUC that all logical value assignment is cyclic valueassignmet is orthogonal. And any truth value assignment is false at only one clause.Proof.
It is clear from the definition of the cyclic value assignment. (cid:3)
In addition, MUC is the CNF that is false of all truth value assignment, and thereis always a clause corresponding to the equivalence classes. In other words, MUCis complete system based on the equivalence classes of the truth value assignment.
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Theorem 7.
MUC is complete system based on the equivalence classes of the truthvalue assignment.Proof.
It is clear from the definition of the equivalence classes of the truth valueassignment. (cid:3)
Thus, MUC thats all logical value assignments are cyclic value assignments iscomplete orthogonal function.
Theorem 8.
MUC thats all logical value assignments are cyclic value assignmentsis complete orthogonal function. And every clauses are orthogonal basis of thecomplete orthogonal function. I will use the term “Orthogonal MUC” to the MUC.Proof.
Judging from the above67, this is clear. (cid:3)
Thus, by reducing logical value assignments to all cyclic value assignments withkeeping the orthogonality, we can express the MUC size in the number of clauses.3.
Orthogonalization of MUC
We reduce MUC to orthogonal MUC under the constraints of the HornMUC.First, we define the two inner product of clauses, and define the orthogonality ofclauses. Secondly, we clarify the limitations HornMUC. And we show how to reducethe MUC to the orthogonal MUC.And we show that we can reduce HornMUC to orthogonal MUC in polynomialtime from its connectivity, and can not reduce MUC.I will use the term “Fact clause” to the HornCNF’s clauses that include onlypositive variable. “Goal clause” to the HornCNF’s clauses that include only negativevariable. “Rule clause” to the HornCNF’s clauses that is not fact clause and goalclause. “Case clause” to the HorncNF’s clauses that is fact clauses and rule clauses.3.1.
Clauses inner product and inner harmony.
Let us start with definingthe inner product and inner harmony, and considering the the these orthogonality.However, considering the duality of the conjunction and disjunction, and considerthe dual inner product.
Definition 9.
I will use the term “Inner product” as follow; h C , C i = h C ⊥ C i = W ( C ( x i ) ∧ C ( x i )) = ∃ x i ( C ( x i ) ∧ C ( x i )) I will use the term “Inner harmony” as the duality of the inner product. h C , C i d = h C ⊤ C i = W (cid:16) C ( x i ) ∧ C ( x i ) (cid:17) = V ( C ( x i ) ∨ C ( x i ))= ¬∃ x i (cid:16) C ( x i ) ∧ C ( x i ) (cid:17) = ∀ x i ( C ( x i ) ∨ C ( x i )) W () , V () is the disjunction and conjunction of all truth value assignment value.It also defines the inner product and inner harmony of more than two. And italso defines the inner product and inner harmony of CNF.Also, we can define the orthogonality of inner product and inner harmony.depending on whether that is false in the inner section and can be determinedfor each of the orthogonality. Defined as follows: Note that the duality of innerproduct replace true and false. Definition 10.
When the inner product is false, the clauses are orthogonal at innerproduct. I will use “ C ⊥ C ” to orthogonal at inner product. Orthogonal CNF atinner product is unsatisfiable CNF. ONNECTION AND DISPERSION OF COMPUTATION 5
When the inner harmony is true, the clauses are orthogonal at inner harmony.I will use “ C ⊤ C ” to orthogonal at inner harmony. Orthogonal CNF at innerharmony is the CNF that all logical value assignment are cyclic value assignment.I will use term “Clauses orthogonalization” to reducing two clauses to orthogonalclauses at inner harmony.When the inner harmony is orthognal, the CNF that include these clauses areorthogonal at inner harmony. I will use “ C ⊤ C ” to orthogonal at inner harmony.3.2. HornMUC constraints.
Describes the HornMUC’s constraints. HornMUCis the CNF that each clauses have at most one positive variables. This restrict thephase difference of clauses. Therefore, it is also restrict the inner harmony.First, we show the restriction of the phase difference in the HornMUC clauses.
Theorem 11.
Phase difference between the two clauses of HornMUC would be atmost one. In other words, HornMUC clauses are connected together.Proof.
We assume that HornMUC have two clauses C = ( x ∨ x · · · ) , C =( x ∨ x · · · ) . These clauses are true when T = ( x , x · · · ) = ( ⊥ , ⊥ · · · ) . Thus,HornMUC must include the clause that is false at truth value assignment T . Butin order to satisfy this condition, HornCNF include ( x ∨ x ) or be false regardless ( x , x ) . This is contradicts the assumption that HornMUC include C , C . ThusFrom the proof by contradiction, HornMUC do not include the clauses these phasediffernet more than two. (cid:3) We can see from the HornMUC restrict what the structure of HornCNF is re-stricted. HornMUC’s phase difference is at most one, HornMUC structure has beenconnected not only at whole but also each part. Thus, HornMUC can not constructthe structure with the non-connected part. HornMUC constitutes a partial orderof phases. And clause cycle do not affect to the HornMUC’s partial order.
Theorem 12.
HornMUC can not construct the structure with the non-connectedpart. And HornMUC constitutes a partial order of phases.Proof.
Shows the HornMUC’s connection. From mentioned above 11, all clausesare connecting or crossing. We assume that there are HornMUC that can be dividedinto two subsets of non-connected to each other. We assume that we can split theHornMUC into two subsets that is not connected. The subsets have no commonvariables or have only common negative variables. But if the subsets have nocommon variables, HornMUC’s unsatisfiability do not change to delete one of thesesubset. So, This is contradicts the assumption of HornMUC. And if the subsetshave only common negative variables, HornMUC’s unsatisfiability do not changeto delete the clauses that include that negative variables. So, from the proof bycontradiction, we can not divide HornMUC into two subsets of non-connected toeach other.Shows the HornMUC’s structure of partially ordered. About HornCNF’s upperclause C U and lower clause C L ; · · · ≥ C U ≥ C k ≥ C k − ≥ · · · ≥ C ≥ C L ≥ · · · ⇄ · · · ≥ ( x U ∨ x k · · · ) ≥ ( x k ∨ x k − ∨ · · · ) ≥ · · · ≥ ( x ∨ x L ∨ · · · ) ≥ ( x L ∨ · · · ) ≥ · · · It is clear that reflexive is satisfied. It is clear that transitivity is satisfied fromHornCNF’s constraints (clause include at most one positive variable.)
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We can see the antisymmetric from the following; ∀ C , C (( C ≤ C ) ∧ ( C ≤ C ))= ∀ F i =1 , , , ( ∀ x , x (( x ∨ F ) ∧ ( x ∨ x ∨ F ) ∧ ( x ∨ F ) ∧ ( x ∨ x ∨ F )))= ∀ x , x (( x = x ) ∧ ( x ∨ x ) ∧ ( x ∨ x )) = ∀ x , x ( x = x ) F i =1 , , , : CN F
So HornMUC clauses constitutes a partial order. (cid:3)
This constrain HornMUC’s inner harmony. HornMUC constitutes a partial orderfrom empty clause leading to fact clause, rule clause, goal clause, and finally emptyclause. Thus, we want to split the HornMUC, we can only cut into upper and lowerpart of this partial order.
Theorem 13.
When we split the HornMUC, we can only cut into upper and lowerpart of this partial order. Each part is a partial order. Cutting clause exist onlyone part.Proof.
It is clear that HornMUC is a partial order. (cid:3)
For example, if you divide the ( x ) ∧ ( x ) by MUC; ( x ) ∧ ( x ∨ x ) ∧ ( x ∨ x ∨ x ) ∧ ( x ∨ x ∨ x ) ∧ ( x ∨ x ) ∧ ( x ∨ x )( x ∨ x ) is divide to ( x ∨ x ∨ x ∨ x ) ∧ ( x ∨ x ∨ x ∨ x ) .But because HornMUC can not be greater than the distance of two clauses, wecan not divide any clauses.3.3. Clause Orthogonalization by using HornMUC.
Describes the way toOrthogonalize clause by using HornMUC. For mentioned above 1213, we are con-strained when split MUC by using HornMUC because of HornMUC’s partial order.And we must cut the clauses when we orthogonalize clauses to orthogonal part andcorrelation part.For example, if you orthogonalize ( x ∨ x ∨ x ) and ( x ∨ x ∨ x ) , we must cutfollows; ( 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 ) repeat that cutting target clause to orthogonal part and correlation part. Andfinally correlation clause is absorbed to another clauses which match base clause.In addition, this reduction is added only HornMUC, so we can keep CNF’s satisfiespossibility. Definition 14.
I will use term “Clause cut” to divide clause by using HornMUC.
Theorem 15.
By cutting correlation clauses to orthogonal part and correlationpart, we can orthogonalize the clauses.Proof.
It is clear that we can achieved by generalizing the above procedure. So Iomit. (cid:3)
HornMUC Orthogonalization.
Describes that we can reduce HornMUCto orthogonalize MUC in polynomial time. For mentioned above 15, we can reduceHornMUC to orthogonalize MUC by cutting all correlation parts. And we can re-duce more easier HornMUC to orthogonalize MUC by using HornCNF’s constraint.
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Specifically, we can reduce that we cut the clause at the negative variable corre-spond to the positive variable of lower clause that is nearer from fact clause. Thus,we can absorb every cutting.For example to Orthogonalize ( x ) ∧ ( x ∨ x ) ∧ ( x ∨ x ) ∧ ( x ) , cutting ( x ∨ x ) at x , cutting ( x ) at x , and cutting ( x ∨ x ) at 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 ) Because the number of clauses do not increase and the number of variables ineach clauses increase at most polynomial size, so this reduction increase at mostpolynomial size. And HornMUC clauses constitutes a partial order and it is enoughto orthogonalize each clause at lower clauses. So the reduction HornMUC to or-thogonal MUC takes at most polynomial time.
Theorem 16.
We can reduce HornMUC to orthogonal MUC at most polynomialsize and time.Proof.
It is clear that we can achieved by generalizing the above procedure. So Iomit. (cid:3)
Theorem 17.
The clauses and variables of the orthogonal MUC that reduced fromHorn MUC constitute total order.Proof.
From the above procedure, the clause of orthogonal MUC include the nega-tive variable that’s positive variable is included in the lower clauses. So, all clauseshave order, and orthogonal MUC constitute total order. (cid:3)
Theorem 18.
If we can not reduce MUC to orthogonal MUC in polynomial timeby using clause cutting, then P = N P .Proof.
From the above 17, if MUC is HornMUC, then we can reduce the MUCto orthogonal MUC by using clause cutting in polynomial time and size. And if P = N P , then we can reduce MUC to HornMUC in polynomial time and size. So,if P = N P and we can reduce MUC to HornMUC in polynomial time, then we canreduce the MUC to orthogonal MUC in polynomial time and size.Taking the contrapositive, if we can not reduce MUC to orthogonal MUC inpolynomial time and size, then P = N P or we can not reduce MUC to HornMUCin polynomial time and size. So, from the above 1, if we can not reduce MUCto orthogonal MUC in polynomial time and size by using clause cutting, then P = N P . (cid:3) MUC Orthogonalization.
Describes that we can not reduce MUC to or-thogonalize MUC in polynomial time. Unlike HornMUC, MUC can be set to anyphase difference between the clauses. So MUC clauses have high dispartion andcorrelation. But for mentioned above 12, HornMUC connected each clauses. So wemust use many HornMUC each other to cut every dispart areas to orthogonal partand correlation part. And because of HornMUC’s connection, we can not use sameHornMUC to cut different areas.
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For example to this, we think MUC include follow clauses; O ( x , x , x ) ∧ E ( x , x , x ) O ( x , x , x ) = ( x ∨ x ∨ x ) ∧ ( x ∨ x ∨ x ) ∧ ( x ∨ x ∨ x ) ∧ ( x ∨ x ∨ x ) E ( x , x , x ) = ( x ∨ x ∨ x ) ∧ ( x ∨ x ∨ x ) ∧ ( x ∨ x ∨ x ) ∧ ( x ∨ x ∨ x ) In other words, O is true when it contains an odd trues in truth value assign-ment, E is true when it contains even trues in truth value assignment. O and E divide the truth value assignment. So if we want to orthogonalize the MUC thatinclude O and E , we must cut every area by using every other HornCNF. Thisorthogonalize MUC is the MUC that expanded to CNF; O ( x , x , x ) ∧ ( E ( x , x , x ) ∨ (( x ∨ x ∨ x ) ∧ ( x ∨ x ∨ x ) ∧ ( x ∨ x ∨ x ) ∧ ( x ∨ x ∨ x )) Theorem 19.
When we orthogonalize CNF that disparted some area, We must cuteach area by using every another HornMUC.Proof.
To assume that we can orthogonalize two area by using one HornMUC. Thistime, there is other closure area between the divided area. From assuming, we canorthogonalize these area by using one HornMUC. But we must cross the HornMUCat the closure that divided area. This is contradicts the assumption that we canorthogonalize by using one HornMUC. Thus, From the proof by contradiction, wecan not orthogonalize two area by using one HornMUC. (cid:3)
As a example, let us construct the MUC that have many divided area amountof non-polynomial size of MUC. We should notice that the truth value assignmentshaving same even-odd of true value do not connect each other.By combining proper O and E above, it is possible to configure the CNF that istrue when the truth value assignment having same even-odd of true. For example; O ( x , x , x , x ) = O ( x, x , x ) ∧ E ( x, x , x ) This CNF is true only if the truth value having odd of true value.This can be extended easily for any number of variables. O n +1 ( x , · · · , x n ) = O n ( x, x , · · · , x n − ) ∧ E ( x, x n − , x n ) Nunber of O n clause is polynomial size of variables. And truth value assignmenthaving odd of true value is amount of half of all truth value assignment, and everytruth value assignment having same even-odd of true do not connect each other.So O n have many divided area amount of non-polynomial size. And there is MUCthat we can not reduce to orthogonal MUC in polynomial size. Theorem 20. there is MUC that we can not reduce to orthogonal MUC in poly-nomial size by using HornMUC.Proof.
For mentioned above examples, there is CNF that divide the truth valueassignment into the same even-odd of true value. And these CNF can be a part ofthe MUC. So there is the MUC that divided area amount of non-polynomial size.For mentioned above 19, we must cut each area to orthogonalize MUC. So there isMUC that we can not reduce to orthogonal MUC in polynomial size. (cid:3) DP is not P and NP is not P
Describes the difference between MUC decision problem and HornMUC decisionproblem. For mentioned above 3, the difference of the MUC decision problem andHornMUC decision problem will appear in CNF classification. And orthogonal basisof inner harmony is different between MUC and HornMUC. For mentiond above
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