{"paper_meta":{"paper_id":"arxiv:0704.0108","title":"0704.0108","authors":[],"year":null,"published":null,"updated":null,"venue_or_source":"arxiv-papers-shard","primary_category":"cs","secondary_categories":[],"doi":null,"license":"unknown","source_type":"pdf+shard","source_url":null,"pdf_url":null},"abstract":{"raw":"","cleaned":"","offsets":[]},"full_paper_text":{"raw_ordered_text":"arXiv:0704.0108v1 [cs.CC] 1 Apr 2007\n1\nAbstract\nDescription of a polynomial time reduction of SAT to 2-SAT of\npolynomial size.\n1\n\nReducing SAT to 2-SAT\nSergey Gubin\nNovember 4, 2018\n1\nIntroduction\nAmong all dimensions, 2-SAT possesses many special properties unique in\nthe sense of computational complexity [1, 2, 3, 4, 5]. But in light of works\n[6, 8, 7, 9] a problem arose: either those properties are accidental or there\nare polynomial time reductions of SAT to 2-SAT of polynomial size. This\narticle describes one such reduction.\n2\nPresenting SAT with XOR\nIn [6] was described one of the ways to present SAT with a conjunction of\nXOR. Let us summarize it.\nLet Boolean formula f define a given SAT instance:\nf = c1 ∧c2 ∧. . . ∧cm.\n(1)\nClauses ci are disjunctions of literals:\nci = Li1 ∨Li2 ∨. . . ∨Lini, i = 1, 2, . . . , m\n- where ni is the number of literals in clause ci; and Lij are the literals. Using\ndistributive laws, formula (1) can be rewritten in disjunctive form:\nf = d1 ∨d2 ∨. . . dp, p = n1n2 . . . nm.\nClauses dk in this presentation are conjunctions of m literals - one literal\nfrom each clause ci, i = 1, 2, . . . , m:\ndk = L1k1 ∧L2k2 ∧. . . ∧Lmkm, k = 1, 2, . . . , p.\n(2)\n∗Author’s email: sgubin@genesyslab.com\n2\n\nIt is obvious that formula (1) is satisfiable iffthere are clauses without com-\nplimentary literals amongst conjunctive clauses (2). Disjunction of all those\nclauses is the disjunctive normal form of formula (1). Thus, formula (1) is\nsatisfiable iffthere are members in its disjunctive normal form.\nThere is a generator for conjunctive clauses (2):\ng =\nm\n^\ni=1\n(ξi1 ⊕ξi2 ⊕. . . ⊕ξini) = true,\n(3)\n- where Boolean variable ξμν indicates whether literal Lμν participates in con-\njunction (2). Solutions of equation (3) generate conjunctive clauses (2). Let’s\ncall the variables ξ the indicators. To select from all solutions of equation (3)\nthose without complimentary clauses, let’s use another Boolean equation.\nFor each of the combination of clauses (ci, cj), 1 ≤i < j ≤m, let’s build\na set of all couples of literals participating in the clauses:\nAij = { (Liμ, Ljν) | ci = Liμ ∨. . . ; cj = Ljν ∨. . . }.\nLet Bij be a set of such couples of indicators (ξiμ, ξjν), that the literals they\npresent are complimentary:\nBij = { (ξiμ, ξjν) | (Liμ, Ljν) ∈Aij, Liμ = ̄Ljν }.\nThere are C2\nm sets Bij, 1 ≤i < j ≤m, and\n|Bij| ≤min{ni, nj}.\nLet’s mention that some of the sets can be empty. Then, the following equa-\ntion will select from all solutions of equation (3) those without complimentary\nclauses:\nh =\n^\n1≤i<j≤m\n^\n(ξ,ζ)∈Bij\n( ̄ξ ∨ ̄ζ) = true.\n(4)\nDue to the above estimations of the number of sets Bij and of their sizes, the\nnumber of clauses in formula (4) is\nn = O(t2m2),\n- where t2 is the second number in the row of clauses’ sizes sorted by value:\nt1 = max{n1, n2, . . . , nm}, t2 = max\ni<j min{ni, nj}, . . .\nBecause satisfiability of formula (1) means that the disjunctive normal\nform of formula (1) has conjunctive clauses, formula (1) is satisfiable iffthe\nfollowing formula/equation is satisfiable:\ng ∧h = true.\n(5)\n3\n\nThe reasons for replacing formula (1) with formula (5) are explained in\n[6]. The number of true-strings in truth-tables of XOR clauses of formula\n(3) is linear over initial input. The number of true-strings in truth-tables of\ndisjunctive clauses of formula (4) is just 3. The number of all clauses in (5)\nis cubic over initial input. It can be estimated as\nm + n = O(t2m2).\nThus, application of the simplified compatibility matrices method [6] to equa-\ntion (5) will produce a polynomial time algorithm for SAT. But let’s return\nto the reduction.\n3\nSAT vs. 2-SAT\nLet’s apply the simplified method of compatibility matrices [6] to equation\n(5). The method consists of sequential Boolean transformations of compat-\nibility matrices of equation (5). Let’s mention that after m iterations, due\nto the allocation of formula (4) at the end of formula (5), there will only be\ncompatibility matrices of equation (4) left in play. They will be grouped in\nan upper triangular box matrix\nS = (Fm+μ,m+ν)1≤μ<ν≤n.\n(6)\nThe matrix is displayed below:\nFm+1,m+2\nFm+1,m+3\n. . .\nFm+1,m+n\nFm+2,m+3\n. . .\nFm+2,m+n\n...\n...\nFm+n−1,m+n\nIf there are no complimentary literals in different clauses of formula (1),\nthen formula (4) is just missing. The size of matrix (6) is 0 × 0. In this case,\nformula (1) is reducible to 1-SAT instance\nω1 ∧ω2 ∧. . . ∧ωm,\n- where\nωi = ξi1 ⊕ξi2 ⊕. . . ⊕ξini, i = 1, 2, . . . , m.\nThis singularity belongs to the set of all 2-SAT instances.\nIf, during the first m iterations, a pattern of unsatisfiability arises (one of\nthe compatibility matrices becomes filled with false entirely), then formulas\n4\n\n(5) and (1) are both unsatisfiable [6]. This case may be thought of as a case\nof formula (1) being reduced to an unsatisfiable formula\nfalse.\nLet’s include this singularity in the set of all 2-SAT instances.\nOtherwise, boxes Fm+μ,m+ν in matrix (6) are what is left of the compati-\nbility matrices of equation (4) after the first m iterations of the method.\nDue to their construction [6], the boxes are 3 × 3 matrices:\nFm+μ,m+ν = (xij)3×3, 1 ≤μ < ν ≤n\n(7)\n- where xij ∈{false, true}. The number of boxes is C2\nn. Thus, the number\nof all elements in matrix (6) is\ne = 9C2\nn = O(t2\n2m4).\nLet’s enumerate the elements arbitrarily:\ny1, y2, . . . , ye.\nThen, distribution of true/false in matrix (6) can be described with a 1-SAT\nformula/equation\nw = η1 ∧η2 . . . ∧ηe = true,\n(8)\n- where ηi are literals over a set of Boolean variables\n{ b1, b2, . . . , be }.\nThe literals are\nηi =\n \nbi,\nyi = true\n ̄bi,\nyi = false , i = 1, 2, . . . , e.\nLet’s take the following 2-SAT instance:\nh ∧w.\n(9)\nBox matrix (6) is an initialization of the modified method of compatibility\nmatrices [6] for formula (9): compatibility matrices of formula (4) are de-\npleted to satisfy equation (8). Thus, continuation of the simplified method\nof compatibility matrices for equation (5) from its Step m + 1 to its finish is\nan application of the modified method of compatibility matrices to system\n(9) from its Step 1 to its finish [6]. After n−2 iterations, both methods must\nresult with the same version of satisfiability of formula (1). Thus, formulas\n(5) and (1) are satisfiable iff2-SAT formula (9) is satisfiable. The number of\nclauses in formula (9) is\ne + n = O(t2\n2m4).\nAccording to [6], the time to deduce formula (9) can be safely estimated as\nO(t4\n1t4\n2m6)\n5\n\n4\nSAT vs. 1-SAT\nLet’s take one step further. Applying to formula (1)/(5) either of the varia-\ntions of the compatibility matrices method [6] will produce a Boolean matrix.\nLet it be a matrix R:\nR = (rij)a×b.\nSize of the matrix depends on the method’s variation and the order of clauses\nin formula (1). The size can be changed if permute the clauses and repeat the\nmethod [6]. The formula (1) is satisfiable iffmatrix R contains true-elements\n[6] (elements which are true). The existence/absence of the true-elements is\nthe only invariant.\nIf formula (1) is unsatisfiable, then that formula is reducible to formula\n“false”. Otherwise, formula (1) is reducible to a 1-SAT instance.\nProof. Let’s enumerate elements of matrix R in arbitrarily order:\nz1, z2, . . . , zab.\nLet B be a set of t = ab Boolean variables:\nB = { bi ∈{false, true} | i = 1, 2, . . . , t }.\nThen the following 1-SAT formula describes distribution of true/false in\nmatrix R:\nθ1 ∧θ2 ∧. . . ∧θt,\n(10)\n- where literals θi are\nθi =\n \nbi,\nzi = true\n ̄bi,\nzi = false , i = 1, 2, . . . , t.\nThus, the compatibility matrices method reduces satisfiable formula (1) to\n1-SAT formula (10).\nIn its turn, formula (10) can be rewritten as SAT of any dimension by\nappropriate substitution of variables.\nIf use the simplified method of compatibility matrices, then matrix R is a\n3 × 3 Boolean matrix [6]. Let there be two clauses shorter than 3 in formula\n(1). Let’s permute all clauses and make those shortest clauses to be the last\nones in formula (1). Then, result of the modified method [6] will be a matrix\nR of size less than 3 × 3. That proves the following theorem.\nTheorem 1. Any SAT instance is reducible to a 1-SAT instance with 9\nvariables or less. A SAT instance is unsatisfiable iffits 1-SAT presentation\nis “false” - there is not any variables in its 1-SAT presentation.\n6\n\n5\nConclusions\nFormula (1) may be thought of as a “Business Requirements”.\nAnd any\nappropriate computer program may be thought of as a solution of the SAT\ninstance. Then, theorem 1 can be an explanation of the remarkable efficiency\nof the “natural programs”. From this point of view, the iterations of the\nmethod of compatibility matrices may be thought of as a learning/modeling\nof the business domain. In the artificial programming, the calculation of the\ncompatibility matrices - a virtual business domain - could be a conclusion of\nthe stage “Business Requirements Analysis/Mathematical Modeling”. That\nwould improve the programs’ performance. The resulting compatibility ma-\ntrices may be thought of as a fussy logic’s tables of rules for the domain.\nThe whole solution of formula (1) can be achieved, with one of the fol-\nlowing approaches, for example. ANN approach is the applying of the com-\npatibility matrices method backward, starting from matrix R. An example\nof that can be found in [7]. DTM approach is the looping trough of the\nfollowing three steps: selection of any true-element from matrix R; substi-\ntution of the appropriate true-assignments in formula (1); and repeating of\nthe compatibility matrices method. The last method is an implication of the\nself-reducibility property of SAT [5].\nIn certain sense, theorem 1 may be seen as an answer to the Feasibility\nThesis [2].\nReferences\n[1] Stephen Cook. The complexity of theorem-proving procedures. In Con-\nference Record of Third Annual ACM Symposium on Theory of Com-\nputing. p.151-158, 1971\n[2] Stephen\nCook.\nThe\nP\nversus\nNP\nproblem.\nhttp://www.claymath.org/millennium/P_vs_NP/pvsnp.pdf\n[3] Richard M. Karp.\nReducibility Among Combinatorial Problems.\nIn\nComplexity of Computer Computations, Proc. Sympos. IBM Thomas\nJ. Watson Res. Center, Yorktown Heights, N.Y. New York: Plenum,\np.85-103, 1972.\n[4] M.R. Garey and D.S. Johnson. Computers and Intractability, a Guide to\nthe Theory of NP-Completeness. W.H. Freeman and Co. San Francisco,\n1979.\n7\n\n[5] Lane A. Hemaspaandra, Mitsunori Ogihara. The Complexity Theory\nCompanion. Springer-Verlag Berlin Heidelberg, 2002.\n[6] Sergey\nGubin.\nA\nPolynomial\nTime\nAlgorithm\nfor\nSAT.\nhttp://www.arxiv.org/pdf/cs/0703146\n[7] Sergey Gubin. A Polynomial Time Algorithm for 3-SAT. Examples of\nuse. http://www.arxiv.org/pdf/cs/0703098\n[8] Sergey\nGubin.\nA\nPolynomial\nTime\nAlgorithm\nfor\n3-SAT.\nhttp://www.arxiv.org/pdf/cs/0701023\n[9] Sergey Gubin. A Polynomial Time Algorithm for The Traveling Sales-\nman Problem. http://www.arxiv.org/pdf/cs/0610042\n8","paragraphs":[{"paragraph_id":"p1","order":1,"text":"arXiv:0704.0108v1 [cs.CC] 1 Apr 2007\n1\nAbstract\nDescription of a polynomial time reduction of SAT to 2-SAT of\npolynomial size.\n1"},{"paragraph_id":"p2","order":2,"text":"Reducing SAT to 2-SAT\nSergey Gubin\nNovember 4, 2018\n1\nIntroduction\nAmong all dimensions, 2-SAT possesses many special properties unique in\nthe sense of computational complexity [1, 2, 3, 4, 5]. But in light of works\n[6, 8, 7, 9] a problem arose: either those properties are accidental or there\nare polynomial time reductions of SAT to 2-SAT of polynomial size. This\narticle describes one such reduction.\n2\nPresenting SAT with XOR\nIn [6] was described one of the ways to present SAT with a conjunction of\nXOR. Let us summarize it.\nLet Boolean formula f define a given SAT instance:\nf = c1 ∧c2 ∧. . . ∧cm.\n(1)\nClauses ci are disjunctions of literals:\nci = Li1 ∨Li2 ∨. . . ∨Lini, i = 1, 2, . . . , m\n- where ni is the number of literals in clause ci; and Lij are the literals. Using\ndistributive laws, formula (1) can be rewritten in disjunctive form:\nf = d1 ∨d2 ∨. . . dp, p = n1n2 . . . nm.\nClauses dk in this presentation are conjunctions of m literals - one literal\nfrom each clause ci, i = 1, 2, . . . , m:\ndk = L1k1 ∧L2k2 ∧. . . ∧Lmkm, k = 1, 2, . . . , p.\n(2)\n∗Author’s email: sgubin@genesyslab.com\n2"},{"paragraph_id":"p3","order":3,"text":"It is obvious that formula (1) is satisfiable iffthere are clauses without com-\nplimentary literals amongst conjunctive clauses (2). Disjunction of all those\nclauses is the disjunctive normal form of formula (1). Thus, formula (1) is\nsatisfiable iffthere are members in its disjunctive normal form.\nThere is a generator for conjunctive clauses (2):\ng =\nm\n^\ni=1\n(ξi1 ⊕ξi2 ⊕. . . ⊕ξini) = true,\n(3)\n- where Boolean variable ξμν indicates whether literal Lμν participates in con-\njunction (2). Solutions of equation (3) generate conjunctive clauses (2). Let’s\ncall the variables ξ the indicators. To select from all solutions of equation (3)\nthose without complimentary clauses, let’s use another Boolean equation.\nFor each of the combination of clauses (ci, cj), 1 ≤i < j ≤m, let’s build\na set of all couples of literals participating in the clauses:\nAij = { (Liμ, Ljν) | ci = Liμ ∨. . . ; cj = Ljν ∨. . . }.\nLet Bij be a set of such couples of indicators (ξiμ, ξjν), that the literals they\npresent are complimentary:\nBij = { (ξiμ, ξjν) | (Liμ, Ljν) ∈Aij, Liμ = ̄Ljν }.\nThere are C2\nm sets Bij, 1 ≤i < j ≤m, and\n|Bij| ≤min{ni, nj}.\nLet’s mention that some of the sets can be empty. Then, the following equa-\ntion will select from all solutions of equation (3) those without complimentary\nclauses:\nh =\n^\n1≤i<j≤m\n^\n(ξ,ζ)∈Bij\n( ̄ξ ∨ ̄ζ) = true.\n(4)\nDue to the above estimations of the number of sets Bij and of their sizes, the\nnumber of clauses in formula (4) is\nn = O(t2m2),\n- where t2 is the second number in the row of clauses’ sizes sorted by value:\nt1 = max{n1, n2, . . . , nm}, t2 = max\ni<j min{ni, nj}, . . .\nBecause satisfiability of formula (1) means that the disjunctive normal\nform of formula (1) has conjunctive clauses, formula (1) is satisfiable iffthe\nfollowing formula/equation is satisfiable:\ng ∧h = true.\n(5)\n3"},{"paragraph_id":"p4","order":4,"text":"The reasons for replacing formula (1) with formula (5) are explained in\n[6]. The number of true-strings in truth-tables of XOR clauses of formula\n(3) is linear over initial input. The number of true-strings in truth-tables of\ndisjunctive clauses of formula (4) is just 3. The number of all clauses in (5)\nis cubic over initial input. It can be estimated as\nm + n = O(t2m2).\nThus, application of the simplified compatibility matrices method [6] to equa-\ntion (5) will produce a polynomial time algorithm for SAT. But let’s return\nto the reduction.\n3\nSAT vs. 2-SAT\nLet’s apply the simplified method of compatibility matrices [6] to equation\n(5). The method consists of sequential Boolean transformations of compat-\nibility matrices of equation (5). Let’s mention that after m iterations, due\nto the allocation of formula (4) at the end of formula (5), there will only be\ncompatibility matrices of equation (4) left in play. They will be grouped in\nan upper triangular box matrix\nS = (Fm+μ,m+ν)1≤μ<ν≤n.\n(6)\nThe matrix is displayed below:\nFm+1,m+2\nFm+1,m+3\n. . .\nFm+1,m+n\nFm+2,m+3\n. . .\nFm+2,m+n\n...\n...\nFm+n−1,m+n\nIf there are no complimentary literals in different clauses of formula (1),\nthen formula (4) is just missing. The size of matrix (6) is 0 × 0. In this case,\nformula (1) is reducible to 1-SAT instance\nω1 ∧ω2 ∧. . . ∧ωm,\n- where\nωi = ξi1 ⊕ξi2 ⊕. . . ⊕ξini, i = 1, 2, . . . , m.\nThis singularity belongs to the set of all 2-SAT instances.\nIf, during the first m iterations, a pattern of unsatisfiability arises (one of\nthe compatibility matrices becomes filled with false entirely), then formulas\n4"},{"paragraph_id":"p5","order":5,"text":"(5) and (1) are both unsatisfiable [6]. This case may be thought of as a case\nof formula (1) being reduced to an unsatisfiable formula\nfalse.\nLet’s include this singularity in the set of all 2-SAT instances.\nOtherwise, boxes Fm+μ,m+ν in matrix (6) are what is left of the compati-\nbility matrices of equation (4) after the first m iterations of the method.\nDue to their construction [6], the boxes are 3 × 3 matrices:\nFm+μ,m+ν = (xij)3×3, 1 ≤μ < ν ≤n\n(7)\n- where xij ∈{false, true}. The number of boxes is C2\nn. Thus, the number\nof all elements in matrix (6) is\ne = 9C2\nn = O(t2\n2m4).\nLet’s enumerate the elements arbitrarily:\ny1, y2, . . . , ye.\nThen, distribution of true/false in matrix (6) can be described with a 1-SAT\nformula/equation\nw = η1 ∧η2 . . . ∧ηe = true,\n(8)\n- where ηi are literals over a set of Boolean variables\n{ b1, b2, . . . , be }.\nThe literals are\nηi ="},{"paragraph_id":"p6","order":6,"text":"bi,\nyi = true\n ̄bi,\nyi = false , i = 1, 2, . . . , e.\nLet’s take the following 2-SAT instance:\nh ∧w.\n(9)\nBox matrix (6) is an initialization of the modified method of compatibility\nmatrices [6] for formula (9): compatibility matrices of formula (4) are de-\npleted to satisfy equation (8). Thus, continuation of the simplified method\nof compatibility matrices for equation (5) from its Step m + 1 to its finish is\nan application of the modified method of compatibility matrices to system\n(9) from its Step 1 to its finish [6]. After n−2 iterations, both methods must\nresult with the same version of satisfiability of formula (1). Thus, formulas\n(5) and (1) are satisfiable iff2-SAT formula (9) is satisfiable. The number of\nclauses in formula (9) is\ne + n = O(t2\n2m4).\nAccording to [6], the time to deduce formula (9) can be safely estimated as\nO(t4\n1t4\n2m6)\n5"},{"paragraph_id":"p7","order":7,"text":"4\nSAT vs. 1-SAT\nLet’s take one step further. Applying to formula (1)/(5) either of the varia-\ntions of the compatibility matrices method [6] will produce a Boolean matrix.\nLet it be a matrix R:\nR = (rij)a×b.\nSize of the matrix depends on the method’s variation and the order of clauses\nin formula (1). The size can be changed if permute the clauses and repeat the\nmethod [6]. The formula (1) is satisfiable iffmatrix R contains true-elements\n[6] (elements which are true). The existence/absence of the true-elements is\nthe only invariant.\nIf formula (1) is unsatisfiable, then that formula is reducible to formula\n“false”. Otherwise, formula (1) is reducible to a 1-SAT instance.\nProof. Let’s enumerate elements of matrix R in arbitrarily order:\nz1, z2, . . . , zab.\nLet B be a set of t = ab Boolean variables:\nB = { bi ∈{false, true} | i = 1, 2, . . . , t }.\nThen the following 1-SAT formula describes distribution of true/false in\nmatrix R:\nθ1 ∧θ2 ∧. . . ∧θt,\n(10)\n- where literals θi are\nθi ="},{"paragraph_id":"p8","order":8,"text":"bi,\nzi = true\n ̄bi,\nzi = false , i = 1, 2, . . . , t.\nThus, the compatibility matrices method reduces satisfiable formula (1) to\n1-SAT formula (10).\nIn its turn, formula (10) can be rewritten as SAT of any dimension by\nappropriate substitution of variables.\nIf use the simplified method of compatibility matrices, then matrix R is a\n3 × 3 Boolean matrix [6]. Let there be two clauses shorter than 3 in formula\n(1). Let’s permute all clauses and make those shortest clauses to be the last\nones in formula (1). Then, result of the modified method [6] will be a matrix\nR of size less than 3 × 3. That proves the following theorem.\nTheorem 1. Any SAT instance is reducible to a 1-SAT instance with 9\nvariables or less. A SAT instance is unsatisfiable iffits 1-SAT presentation\nis “false” - there is not any variables in its 1-SAT presentation.\n6"},{"paragraph_id":"p9","order":9,"text":"5\nConclusions\nFormula (1) may be thought of as a “Business Requirements”.\nAnd any\nappropriate computer program may be thought of as a solution of the SAT\ninstance. Then, theorem 1 can be an explanation of the remarkable efficiency\nof the “natural programs”. From this point of view, the iterations of the\nmethod of compatibility matrices may be thought of as a learning/modeling\nof the business domain. In the artificial programming, the calculation of the\ncompatibility matrices - a virtual business domain - could be a conclusion of\nthe stage “Business Requirements Analysis/Mathematical Modeling”. That\nwould improve the programs’ performance. The resulting compatibility ma-\ntrices may be thought of as a fussy logic’s tables of rules for the domain.\nThe whole solution of formula (1) can be achieved, with one of the fol-\nlowing approaches, for example. ANN approach is the applying of the com-\npatibility matrices method backward, starting from matrix R. An example\nof that can be found in [7]. DTM approach is the looping trough of the\nfollowing three steps: selection of any true-element from matrix R; substi-\ntution of the appropriate true-assignments in formula (1); and repeating of\nthe compatibility matrices method. The last method is an implication of the\nself-reducibility property of SAT [5].\nIn certain sense, theorem 1 may be seen as an answer to the Feasibility\nThesis [2].\nReferences\n[1] Stephen Cook. The complexity of theorem-proving procedures. In Con-\nference Record of Third Annual ACM Symposium on Theory of Com-\nputing. p.151-158, 1971\n[2] Stephen\nCook.\nThe\nP\nversus\nNP\nproblem.\nhttp://www.claymath.org/millennium/P_vs_NP/pvsnp.pdf\n[3] Richard M. Karp.\nReducibility Among Combinatorial Problems.\nIn\nComplexity of Computer Computations, Proc. Sympos. IBM Thomas\nJ. Watson Res. Center, Yorktown Heights, N.Y. New York: Plenum,\np.85-103, 1972.\n[4] M.R. Garey and D.S. Johnson. Computers and Intractability, a Guide to\nthe Theory of NP-Completeness. W.H. Freeman and Co. San Francisco,\n1979.\n7"},{"paragraph_id":"p10","order":10,"text":"[5] Lane A. Hemaspaandra, Mitsunori Ogihara. The Complexity Theory\nCompanion. Springer-Verlag Berlin Heidelberg, 2002.\n[6] Sergey\nGubin.\nA\nPolynomial\nTime\nAlgorithm\nfor\nSAT.\nhttp://www.arxiv.org/pdf/cs/0703146\n[7] Sergey Gubin. A Polynomial Time Algorithm for 3-SAT. Examples of\nuse. http://www.arxiv.org/pdf/cs/0703098\n[8] Sergey\nGubin.\nA\nPolynomial\nTime\nAlgorithm\nfor\n3-SAT.\nhttp://www.arxiv.org/pdf/cs/0701023\n[9] Sergey Gubin. A Polynomial Time Algorithm for The Traveling Sales-\nman Problem. http://www.arxiv.org/pdf/cs/0610042\n8"}],"pages":[{"page":1,"text":"arXiv:0704.0108v1 [cs.CC] 1 Apr 2007\n1\nAbstract\nDescription of a polynomial time reduction of SAT to 2-SAT of\npolynomial size.\n1"},{"page":2,"text":"Reducing SAT to 2-SAT\nSergey Gubin\nNovember 4, 2018\n1\nIntroduction\nAmong all dimensions, 2-SAT possesses many special properties unique in\nthe sense of computational complexity [1, 2, 3, 4, 5]. But in light of works\n[6, 8, 7, 9] a problem arose: either those properties are accidental or there\nare polynomial time reductions of SAT to 2-SAT of polynomial size. This\narticle describes one such reduction.\n2\nPresenting SAT with XOR\nIn [6] was described one of the ways to present SAT with a conjunction of\nXOR. Let us summarize it.\nLet Boolean formula f define a given SAT instance:\nf = c1 ∧c2 ∧. . . ∧cm.\n(1)\nClauses ci are disjunctions of literals:\nci = Li1 ∨Li2 ∨. . . ∨Lini, i = 1, 2, . . . , m\n- where ni is the number of literals in clause ci; and Lij are the literals. Using\ndistributive laws, formula (1) can be rewritten in disjunctive form:\nf = d1 ∨d2 ∨. . . dp, p = n1n2 . . . nm.\nClauses dk in this presentation are conjunctions of m literals - one literal\nfrom each clause ci, i = 1, 2, . . . , m:\ndk = L1k1 ∧L2k2 ∧. . . ∧Lmkm, k = 1, 2, . . . , p.\n(2)\n∗Author’s email: sgubin@genesyslab.com\n2"},{"page":3,"text":"It is obvious that formula (1) is satisfiable iffthere are clauses without com-\nplimentary literals amongst conjunctive clauses (2). Disjunction of all those\nclauses is the disjunctive normal form of formula (1). Thus, formula (1) is\nsatisfiable iffthere are members in its disjunctive normal form.\nThere is a generator for conjunctive clauses (2):\ng =\nm\n^\ni=1\n(ξi1 ⊕ξi2 ⊕. . . ⊕ξini) = true,\n(3)\n- where Boolean variable ξμν indicates whether literal Lμν participates in con-\njunction (2). Solutions of equation (3) generate conjunctive clauses (2). Let’s\ncall the variables ξ the indicators. To select from all solutions of equation (3)\nthose without complimentary clauses, let’s use another Boolean equation.\nFor each of the combination of clauses (ci, cj), 1 ≤i < j ≤m, let’s build\na set of all couples of literals participating in the clauses:\nAij = { (Liμ, Ljν) | ci = Liμ ∨. . . ; cj = Ljν ∨. . . }.\nLet Bij be a set of such couples of indicators (ξiμ, ξjν), that the literals they\npresent are complimentary:\nBij = { (ξiμ, ξjν) | (Liμ, Ljν) ∈Aij, Liμ = ̄Ljν }.\nThere are C2\nm sets Bij, 1 ≤i < j ≤m, and\n|Bij| ≤min{ni, nj}.\nLet’s mention that some of the sets can be empty. Then, the following equa-\ntion will select from all solutions of equation (3) those without complimentary\nclauses:\nh =\n^\n1≤i<j≤m\n^\n(ξ,ζ)∈Bij\n( ̄ξ ∨ ̄ζ) = true.\n(4)\nDue to the above estimations of the number of sets Bij and of their sizes, the\nnumber of clauses in formula (4) is\nn = O(t2m2),\n- where t2 is the second number in the row of clauses’ sizes sorted by value:\nt1 = max{n1, n2, . . . , nm}, t2 = max\ni<j min{ni, nj}, . . .\nBecause satisfiability of formula (1) means that the disjunctive normal\nform of formula (1) has conjunctive clauses, formula (1) is satisfiable iffthe\nfollowing formula/equation is satisfiable:\ng ∧h = true.\n(5)\n3"},{"page":4,"text":"The reasons for replacing formula (1) with formula (5) are explained in\n[6]. The number of true-strings in truth-tables of XOR clauses of formula\n(3) is linear over initial input. The number of true-strings in truth-tables of\ndisjunctive clauses of formula (4) is just 3. The number of all clauses in (5)\nis cubic over initial input. It can be estimated as\nm + n = O(t2m2).\nThus, application of the simplified compatibility matrices method [6] to equa-\ntion (5) will produce a polynomial time algorithm for SAT. But let’s return\nto the reduction.\n3\nSAT vs. 2-SAT\nLet’s apply the simplified method of compatibility matrices [6] to equation\n(5). The method consists of sequential Boolean transformations of compat-\nibility matrices of equation (5). Let’s mention that after m iterations, due\nto the allocation of formula (4) at the end of formula (5), there will only be\ncompatibility matrices of equation (4) left in play. They will be grouped in\nan upper triangular box matrix\nS = (Fm+μ,m+ν)1≤μ<ν≤n.\n(6)\nThe matrix is displayed below:\nFm+1,m+2\nFm+1,m+3\n. . .\nFm+1,m+n\nFm+2,m+3\n. . .\nFm+2,m+n\n...\n...\nFm+n−1,m+n\nIf there are no complimentary literals in different clauses of formula (1),\nthen formula (4) is just missing. The size of matrix (6) is 0 × 0. In this case,\nformula (1) is reducible to 1-SAT instance\nω1 ∧ω2 ∧. . . ∧ωm,\n- where\nωi = ξi1 ⊕ξi2 ⊕. . . ⊕ξini, i = 1, 2, . . . , m.\nThis singularity belongs to the set of all 2-SAT instances.\nIf, during the first m iterations, a pattern of unsatisfiability arises (one of\nthe compatibility matrices becomes filled with false entirely), then formulas\n4"},{"page":5,"text":"(5) and (1) are both unsatisfiable [6]. This case may be thought of as a case\nof formula (1) being reduced to an unsatisfiable formula\nfalse.\nLet’s include this singularity in the set of all 2-SAT instances.\nOtherwise, boxes Fm+μ,m+ν in matrix (6) are what is left of the compati-\nbility matrices of equation (4) after the first m iterations of the method.\nDue to their construction [6], the boxes are 3 × 3 matrices:\nFm+μ,m+ν = (xij)3×3, 1 ≤μ < ν ≤n\n(7)\n- where xij ∈{false, true}. The number of boxes is C2\nn. Thus, the number\nof all elements in matrix (6) is\ne = 9C2\nn = O(t2\n2m4).\nLet’s enumerate the elements arbitrarily:\ny1, y2, . . . , ye.\nThen, distribution of true/false in matrix (6) can be described with a 1-SAT\nformula/equation\nw = η1 ∧η2 . . . ∧ηe = true,\n(8)\n- where ηi are literals over a set of Boolean variables\n{ b1, b2, . . . , be }.\nThe literals are\nηi =\n \nbi,\nyi = true\n ̄bi,\nyi = false , i = 1, 2, . . . , e.\nLet’s take the following 2-SAT instance:\nh ∧w.\n(9)\nBox matrix (6) is an initialization of the modified method of compatibility\nmatrices [6] for formula (9): compatibility matrices of formula (4) are de-\npleted to satisfy equation (8). Thus, continuation of the simplified method\nof compatibility matrices for equation (5) from its Step m + 1 to its finish is\nan application of the modified method of compatibility matrices to system\n(9) from its Step 1 to its finish [6]. After n−2 iterations, both methods must\nresult with the same version of satisfiability of formula (1). Thus, formulas\n(5) and (1) are satisfiable iff2-SAT formula (9) is satisfiable. The number of\nclauses in formula (9) is\ne + n = O(t2\n2m4).\nAccording to [6], the time to deduce formula (9) can be safely estimated as\nO(t4\n1t4\n2m6)\n5"},{"page":6,"text":"4\nSAT vs. 1-SAT\nLet’s take one step further. Applying to formula (1)/(5) either of the varia-\ntions of the compatibility matrices method [6] will produce a Boolean matrix.\nLet it be a matrix R:\nR = (rij)a×b.\nSize of the matrix depends on the method’s variation and the order of clauses\nin formula (1). The size can be changed if permute the clauses and repeat the\nmethod [6]. The formula (1) is satisfiable iffmatrix R contains true-elements\n[6] (elements which are true). The existence/absence of the true-elements is\nthe only invariant.\nIf formula (1) is unsatisfiable, then that formula is reducible to formula\n“false”. Otherwise, formula (1) is reducible to a 1-SAT instance.\nProof. Let’s enumerate elements of matrix R in arbitrarily order:\nz1, z2, . . . , zab.\nLet B be a set of t = ab Boolean variables:\nB = { bi ∈{false, true} | i = 1, 2, . . . , t }.\nThen the following 1-SAT formula describes distribution of true/false in\nmatrix R:\nθ1 ∧θ2 ∧. . . ∧θt,\n(10)\n- where literals θi are\nθi =\n \nbi,\nzi = true\n ̄bi,\nzi = false , i = 1, 2, . . . , t.\nThus, the compatibility matrices method reduces satisfiable formula (1) to\n1-SAT formula (10).\nIn its turn, formula (10) can be rewritten as SAT of any dimension by\nappropriate substitution of variables.\nIf use the simplified method of compatibility matrices, then matrix R is a\n3 × 3 Boolean matrix [6]. Let there be two clauses shorter than 3 in formula\n(1). Let’s permute all clauses and make those shortest clauses to be the last\nones in formula (1). Then, result of the modified method [6] will be a matrix\nR of size less than 3 × 3. That proves the following theorem.\nTheorem 1. Any SAT instance is reducible to a 1-SAT instance with 9\nvariables or less. A SAT instance is unsatisfiable iffits 1-SAT presentation\nis “false” - there is not any variables in its 1-SAT presentation.\n6"},{"page":7,"text":"5\nConclusions\nFormula (1) may be thought of as a “Business Requirements”.\nAnd any\nappropriate computer program may be thought of as a solution of the SAT\ninstance. Then, theorem 1 can be an explanation of the remarkable efficiency\nof the “natural programs”. From this point of view, the iterations of the\nmethod of compatibility matrices may be thought of as a learning/modeling\nof the business domain. In the artificial programming, the calculation of the\ncompatibility matrices - a virtual business domain - could be a conclusion of\nthe stage “Business Requirements Analysis/Mathematical Modeling”. That\nwould improve the programs’ performance. The resulting compatibility ma-\ntrices may be thought of as a fussy logic’s tables of rules for the domain.\nThe whole solution of formula (1) can be achieved, with one of the fol-\nlowing approaches, for example. ANN approach is the applying of the com-\npatibility matrices method backward, starting from matrix R. An example\nof that can be found in [7]. DTM approach is the looping trough of the\nfollowing three steps: selection of any true-element from matrix R; substi-\ntution of the appropriate true-assignments in formula (1); and repeating of\nthe compatibility matrices method. The last method is an implication of the\nself-reducibility property of SAT [5].\nIn certain sense, theorem 1 may be seen as an answer to the Feasibility\nThesis [2].\nReferences\n[1] Stephen Cook. The complexity of theorem-proving procedures. In Con-\nference Record of Third Annual ACM Symposium on Theory of Com-\nputing. p.151-158, 1971\n[2] Stephen\nCook.\nThe\nP\nversus\nNP\nproblem.\nhttp://www.claymath.org/millennium/P_vs_NP/pvsnp.pdf\n[3] Richard M. Karp.\nReducibility Among Combinatorial Problems.\nIn\nComplexity of Computer Computations, Proc. Sympos. IBM Thomas\nJ. Watson Res. Center, Yorktown Heights, N.Y. New York: Plenum,\np.85-103, 1972.\n[4] M.R. Garey and D.S. Johnson. Computers and Intractability, a Guide to\nthe Theory of NP-Completeness. W.H. Freeman and Co. San Francisco,\n1979.\n7"},{"page":8,"text":"[5] Lane A. Hemaspaandra, Mitsunori Ogihara. The Complexity Theory\nCompanion. Springer-Verlag Berlin Heidelberg, 2002.\n[6] Sergey\nGubin.\nA\nPolynomial\nTime\nAlgorithm\nfor\nSAT.\nhttp://www.arxiv.org/pdf/cs/0703146\n[7] Sergey Gubin. A Polynomial Time Algorithm for 3-SAT. Examples of\nuse. http://www.arxiv.org/pdf/cs/0703098\n[8] Sergey\nGubin.\nA\nPolynomial\nTime\nAlgorithm\nfor\n3-SAT.\nhttp://www.arxiv.org/pdf/cs/0701023\n[9] Sergey Gubin. A Polynomial Time Algorithm for The Traveling Sales-\nman Problem. http://www.arxiv.org/pdf/cs/0610042\n8"}]},"section_tree":[],"assets":{"figures":[],"tables":[],"images":[]},"math_expressions":{"inline":[],"block":[{"equation_id":"eq1","equation_number":null,"raw_text":"f = c1 ∧c2 ∧. . . ∧cm.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq2","equation_number":null,"raw_text":"ci = Li1 ∨Li2 ∨. . . ∨Lini, i = 1, 2, . . . , m","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq3","equation_number":null,"raw_text":"f = d1 ∨d2 ∨. . . dp, p = n1n2 . . . nm.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq4","equation_number":null,"raw_text":"from each clause ci, i = 1, 2, . . . , m:","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq5","equation_number":null,"raw_text":"dk = L1k1 ∧L2k2 ∧. . . ∧Lmkm, k = 1, 2, . . . , p.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq6","equation_number":null,"raw_text":"g =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq7","equation_number":null,"raw_text":"i=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq8","equation_number":null,"raw_text":"(ξi1 ⊕ξi2 ⊕. . . ⊕ξini) = true,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq9","equation_number":null,"raw_text":"Aij = { (Liμ, Ljν) | ci = Liμ ∨. . . ; cj = Ljν ∨. . . }.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq10","equation_number":null,"raw_text":"Bij = { (ξiμ, ξjν) | (Liμ, Ljν) ∈Aij, Liμ = ̄Ljν }.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq11","equation_number":null,"raw_text":"h =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq12","equation_number":null,"raw_text":"( ̄ξ ∨ ̄ζ) = true.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq13","equation_number":null,"raw_text":"n = O(t2m2),","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq14","equation_number":null,"raw_text":"t1 = max{n1, n2, . . . , nm}, t2 = max","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq15","equation_number":null,"raw_text":"g ∧h = true.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq16","equation_number":null,"raw_text":"m + n = O(t2m2).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq17","equation_number":null,"raw_text":"S = (Fm+μ,m+ν)1≤μ<ν≤n.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq18","equation_number":null,"raw_text":"ωi = ξi1 ⊕ξi2 ⊕. . . ⊕ξini, i = 1, 2, . . . , m.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq19","equation_number":null,"raw_text":"Fm+μ,m+ν = (xij)3×3, 1 ≤μ < ν ≤n","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq20","equation_number":null,"raw_text":"e = 9C2","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq21","equation_number":null,"raw_text":"n = O(t2","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq22","equation_number":null,"raw_text":"w = η1 ∧η2 . . . ∧ηe = true,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq23","equation_number":null,"raw_text":"ηi =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq24","equation_number":null,"raw_text":"yi = true","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq25","equation_number":null,"raw_text":"yi = false , i = 1, 2, . . . , e.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq26","equation_number":null,"raw_text":"e + n = O(t2","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq27","equation_number":null,"raw_text":"R = (rij)a×b.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq28","equation_number":null,"raw_text":"Let B be a set of t = ab Boolean variables:","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq29","equation_number":null,"raw_text":"B = { bi ∈{false, true} | i = 1, 2, . . . , t }.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq30","equation_number":null,"raw_text":"θi =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq31","equation_number":null,"raw_text":"zi = true","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq32","equation_number":null,"raw_text":"zi = false , i = 1, 2, . . . , t.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"}],"equation_refs":[]},"symbol_table":[],"references":{"in_text_markers":[],"bibliography":[]},"claims":[],"flow_graph":{"nodes":[],"edges":[]},"quality":{"text_char_count":10803,"parse_confidence":0.5,"equation_parse_rate_proxy":1.0,"citation_resolution_rate_proxy":0.0,"structure_coverage_proxy":0.0,"asset_coverage_proxy":0.0,"accepted_for_training":true}}