A Collection of Problems in Propositional Logic

Transcription

1 A Collection of Problems in Propositional Logic Hans Kleine Büning SS 2016

2 Problem 1: SAT (respectively SAT) Instance: A propositional formula α (for SAT in CNF). Question: Is α satisfiable? The problems are NP-complete. The proof for SAT is based on a representation of calculations of non-deterministic, polynomial time bounded Turing machines by propositional formulas. A polynomial reduction from SAT to SAT is given, for example, by the algorithm TREE-CNF.

3 Problem 2: 3-SAT Instance: α 3-CNF. Question: Is α satisfiable? The problem is NP-complete. The satisfiability problem for the class 2-CNF is solvable in linear time. The algorithm SPLIT-CLAUSES transforms, for k = 3, a formula in CNF in polynomial time into a satisfiability equivalent formula in 3-CNF. If two initial formulas are satisfiability equivalent, then the formulas obtained using SPLIT-CLAUSES are still satisfiability equivalent. A transformation which preserves logical equivalence therefore requires so much information about the initial formula that (under the assumption P conp) it can only be computed with superpolynomial effort.

4 Problem 3: READ-3 3-CNF SAT Instance: α 3-CNF, in which every atom occurs at most three times. Question: Is α satisfiable? The problem is NP-complete. If every atom occurs at most twice and if the formula is in CNF, then the satisfiability problem is solvable in linear time. Dropping the restriction to formulas in CNF in this case, the problem again becomes NP-complete.

5 Problem 4: Monotone 3-SAT Instance: α β 3-CNF, where α only has positive literals and β only has negative literals. Question: Is α β satisfiable? The problem Monotone 3-SAT is NP-complete. A reduction of 3-SAT is possible using the following mapping: Replace (A B C) by (A Y ) ( Y B C) and replace (A B C) by (A B Y ) ( Y C), where for each clause Y is a new atom. The other clauses remain unchanged.

6 Problem 5: 4-CSAT Instance: α 4-CNF such that for every clause (L 1 L 2 L 3 L 4 ) in α the clause ( L 1 L 2 L 3 L 4 ) also occurs in α. Question: Is α satisfiable? The problem is NP-complete. A reduction of 3-SAT is as follows: For α we choose a new atom X and replace every clause (L 1 L 2 L 3 ) in α by (X L 1 L 2 L 3 ) and ( X L 1 L 2 L 3 ). Then α is satisfiable if and only if the resulting formula is satisfiable.

7 Problem 6: 3-CSAT Instance: α 3-CNF such that for each clause (L 1 L 2 L 3 ) α the clause ( L 1 L 2 L 3 ) also occurs in α. Question: Is α satisfiable? The problem is NP-complete. A reduction of 4-CSAT is as follows: For every clause pair we choose a new atom X and replace the clauses (L 1 L 2 L 3 L 4 ) and ( L 1 L 2 L 3 L 4 ) by (X L 1 L 2 ), ( X L 1 L 2 ), ( X L 3 L 4 ) and (X L 3 L 4 ). Then the initial formula is in 4-CSAT if and only if the resulting formula is in 3-CSAT.

8 Problem 7: Monotone 3-CSAT Instance: α β 3-CNF such that every clause in α has only positive literals and every clause in β only has negative literals. Furthermore, (A 1 A 2 A 3 ) α if and only if ( A 1 A 2 A 3 ) β. Question: Is α β satisfiable? The problem is NP-complete by virtue of the following reduction of 3-CSAT: We replace every clause pair (A B C), ( A B C) by (A B X ), ( X C), ( A B X ), (X C) with a corresponding new atom X. The positive and negative clauses remain unchanged. Then the initial formula is in 3-CSAT if and only if the resulting formula is in Monotone 3-CSAT.

9 Problem 8: Monotone (3,2)-SAT Instance: α β CNF such that α 3-CNF has only positive literals and β 2-CNF has only negative literals. Question: Is α β satisfiable? The problem is NP-complete. A reduction of Monotone 3-CSAT can be given using the following mapping: Replace each clause ( B 1 B 2 B 3 ) by (X 1 X 2 X 3 ), ( X 1 B 1 ), ( X 2 B 2 ), ( X 3 B 3 ), where X 1, X 2, X 3 are corresponding new atoms. The other clauses remain unchanged.

10 Problem 9: 1-3-SAT Instance: α 3-CNF with only positive literals. Question: Does there exist a truth assignment such that in each clause exactly one literal is true? The problem 1-3-SAT is NP-complete by. A reduction of 3-SAT can be given as follows: For α 3-CNF with atoms X 1,..., X n we choose for each clause (L 1 L 2 L 3 ) in α new atoms A 1,..., A 6 and replace the clause by (L 1 A 1 A 4 ), (L 2 A 2 A 4 ), (A 1 A 2 A 5 ), (A 3 A 4 A 6 ) and (L 3 A 3 ). In the next step we choose for each X i new atoms Bi 1, B2 i, B3 i, X + i and X i and replace every occurrence of X i by X + i and every occurrence of X i by X i. Finally we add the clauses (X + i X i Bi 1 + ), (X i X i Bi 2) and (B1 i Bi 2 Bi 3 ). The resulting formula is in 1-3-SAT if and only if α is satisfiable. The formula has only positive literals and is in 3-CNF.

11 Problem 10: NOT-ALL-EQUAL-3-SAT Instance: α 3-CNF. Question: Is there a truth assignment of α such that every clause contains at least one true and one false literal? The NP-completeness of this problem follows immediately from a reduction of 3-CSAT.

12 Problem 11: 1-Satisfiability Instance: α CNF. Question: Does the formula α have at most one satisfying truth assignment? With a reduction the conp-completeness of this problem can be shown. The complement of the problem, namely 1-Satisfiability, the class of formulas in CNF with at least two satisfying truth assignments, is obviously in NP. It remains to be shown that 1-Satisfiability is conp-hard. For this we use the following reduction of SAT: To every formula β = β 1... β k CNF with atoms X 1,..., X n is assigned a formula β 1-Sat = (X 0 X 1... X n ) ( X 0 β) where X 0 is a new atom. Then we have: β SAT β 1-Sat has exactly one satisfying truth assignment.

13 Problem 12: 3,4-SAT Instance: α CNF with exactly three literals per clause and every atom occurs at most four times. Question: Is α satisfiable? The problem is NP-complete. If a formula in CNF consists of clauses each with exactly r literals and every atom occurs at most r times in α, then α is satisfiable. (Multiple occurrences of literals in one clause are in this case not allowed.)

14 Problem 13: 2/2/4-SAT Instance: α CNF over m atoms and with m clauses each with exactly four literals per clause and at most four occurrences per atom in α, two negated and two unnegated. Question: Does there exist a satisfying truth assignment for α such that in each clause exactly two literals have the value true? The problem is NP-complete. The proof is established by reducing the problem 1-3-SAT to this problem.

15 Problem 14: MAX-SAT Instance: α = {α 1,..., α n } CNF and k IN. Question: Are there at least k clauses α i1,..., α ik that {α i1,..., α ik } is satisfiable? in α such The problem is NP-complete. For the class of all 2-HORN formulas, i.e. Horn formulas with at most two literals per clause, and for the class of formulas which, except for the unit clauses, contain a positive literal in each clause, the problem MAX-SAT remains NP-complete. The reduction is carried out using the Vertex Cover Problem. The problem is polynomial time decidable for the class of 2-HORN formulas which contain in each clause two literals, one positive and one negative.

16 Problem 15: #SAT Instance: α CNF. Question: How many satisfying truth assignments does α have? This problem is complete in the class #P, the class of problems for which there exists a non-deterministic polynomial Turing machine that yields the number of accepting computations for a given input.

17 Problem 16: SAT (Parity-SAT) Instance: α CNF. Question: Does α have an odd number (respectively even number) of satisfying truth assignments? This problem is complete in the class P, the class of problems for which there exists a non-deterministic polynomial Turing machine that accepts an input precisely when the number of accepting computations for that input is odd.

18 Problem 17: Consequence Instance: α CNF and L a literal. Question: Does α = L hold? The problem is conp-complete by virtue of the following observations: For L literals (α) we have α = L if and only if α L is inconsistent. If L literals (α), then α = L if and only if α is inconsistent.

19 Problem 18: Equivalence Instance: α, β CNF. Question: Does α β hold? The problem is conp-complete. Since for β = A A the formula α is equivalent to β if and only if α is inconsistent, we have a reduction of the problem SAT. With the help of a non-deterministic polynomial procedure we can test for consequence every clause from β with respect to α and every clause from α with respect to β. For formulas in 3-CNF we can test equivalence to a formula consisting of just one unit clause in linear time, as well as equivalence to a fixed formula from CNF.

20 Problem 19: Monotone 3-CNF EQUIV 3-DNF Instance: (α, β) such that α 3-CNF, β 3-DNF and only positive literals occur in α and β. Question: Does α β hold? The problem is conp-complete. Membership of conp is trivial. Completeness can be established by a reduction of the problem Monotone 3-CSAT to the complement. For suppose α β is an input for the problem Monotone 3-CSAT, then α β is satisfiable if and only if α β holds.

21 Problem 20: k-closure (for fixed k 3) Instance: π is a (k + 1)-clause and α k-cnf such that α = {δ δ is a k-clause, atoms (δ) atoms (α), and α = δ}. Question: Does α = π hold? The problem k-closure conp-complete.

22 Problem 21: Minimum Length Instance: Φ CNF and k IN. Question: Does there exist a formula Ψ CNF equivalent to Φ with Ψ k ( number of literals)? If Φ and Ψ are Horn formulas, then the problem is NP-complete. The equivalence of Horn formulas can be decided in quadratic time. For every formula Φ of length k which is equivalent to a Horn formula there exists an equivalent Horn formula of length less than or equal to k, yet consisting of Horn subclauses of the clauses of Φ. Choosing and testing such Horn subformulas of the initial formula leads to an NP-algorithm. Completeness can be established by a reduction of the Set Covering Problem. From this it follows that the minimum length problem for CNF is NP-hard. From the following reduction of the problem SAT it follows that the problem is also conp-hard. For this we choose for a formula α CNF two new atoms A and B. Then α is inconsistent if and only if for the formula α A B an equivalent formula of length less than or equal to 2 exists.

23 Problem 22: Number of Atoms Instance: Φ CNF Question: Do any/all formulas equivalent to Φ in CNF satisfy some conditions on upper and/or lower bounds of the number of atoms? For polynomially computable functions f : IN IN with 1 f (m) m we define V min f := {Φ CNF ϕ CNF with ϕ Φ : atoms (ϕ) f ( atoms (Φ) )}. For every equivalent representation of a formula in Vf min atoms at least f (m) atoms are required. over m V max f := {Φ CNF ϕ CNF with ϕ Φ : atoms (ϕ) f ( atoms (Φ) )}. For at least one representation of a formula in Vf max at most f (m) atoms are necessary. over m atoms

24 The class V f := Vf min Vf max contains all formulas Φ for which each equivalent representation requires at least f ( atoms (Φ) ) atoms and at least one representation requires exactly this number of atoms. The classes V f1,f 2 := Vf min 1 Vf max 2 are defined analogously, but with two different bounds. For a given constant k > 1 we use k to denote the constant function, i.e. k(m) := k for all m IN. Then we have (a) Vf min NP and Vk min is NP-complete for k > 1, (b) Vf max conp and Vk max is conp-complete, (c) V k is conp-complete, (d) V f D P, (e) V f4,f 2 is D P -complete for f 4 (m) = m 4 and f 2(m) = m 2.

25 Problem 23: All Atoms (Exact Atom Number) Instance: Φ CNF. Question: Does every formula in CNF which is equivalent to Φ contain all atoms of Φ? The problem can be described using the terminology of problem 22 as V id such that id(m) = m for all m. The problem is in NP, since V id = Vid min. Completeness is a consequence of a reduction of the problem SAT. Let Φ CNF with atoms A 1,..., A n and let A n+1 be a new atom for Φ. Furthermore, let I 1 be the truth assignment which gives all atoms the value 1.

26 Problem 24: Minimum Disjunctive Normal Form Instance: Set of atoms U = {A 1,..., A n }, set of truth assignments I = {I 1,..., I r } over U and k 1. Question: Does there exist a formula Φ DNF with atoms from U and at most k clauses such that for every truth assignment I over U we have (I(Φ) = 1 I I )? In [?] by using a reduction of the Minimum Cover Problem the NP-completeness of this problem is established. It remains NP-complete if the input also contains the complement of I with respect to U, i.e. the input is U, I, {I I truth assignment over U} \ I and k 1.

27 Problem 25: Not All Redundant Instance: Φ = {ϕ 1,..., ϕ n } CNF. Question: Does there exist i {1,..., n} such that Φ \ {ϕ i } Φ? The problem is NP-complete. A reduction of SAT is possible with the mapping which assigns to each formula Φ the formula Φ A A, where A is a new atom.

28 Problem 26: Maximum Clauses Instance: Φ CNF and k IN. Question: Does there exist a formula equivalent to Φ containing at most k clauses? The problem is conp-hard. This follows immediately from the following reduction of SAT. For a formula Φ with atoms X 1,..., X m we define Φ i = Φ[X 1 /X1 i,..., X n/xn] i for i = 1, 2, 3, where the Xj i are new atoms. For k = 2 we obviously have Φ SAT if and only if for Φ 1 Φ 2 Φ 3 there is no equivalent formula with at most two clauses.

29 Problem 27: Minimal CNF Instance: Φ = {ϕ 1,..., ϕ n } CNF. Question: Is Φ minimal, i.e. does the statement Φ \ {ϕ i } Φ hold for all 1 i n? The problem is NP-complete. Membership of NP is obvious. Completeness can be established by a reduction of 3-SAT. Let Φ = {ϕ 1,..., ϕ n } be a formula in 3-CNF with clauses ϕ i = (L i,1 L i,2 L i,3 ) and let X 1,..., X n be new atoms; furthermore, let ψ i = (X 1... X i 1 X i+1... X n ). We now construct a corresponding formula Ψ = (ϕ i ψ i ) ( L i,j ψ i X i ) 1 i n 1 i<j n 1 i n,1 j 3 ( X i X j ) (X 1... X n ). Then Ψ is inconsistent and furthermore Φ is in 3-SAT if and only if Ψ \ {σ} Ψ for all clauses σ in Ψ.

30 Problem 28: SAT-UNSAT Instance: (α, β) CNF CNF. Question: Does α SAT and β SAT hold? The problem is D P -complete. Membership of D P can be seen by the construction X = {(α, β) CNF CNF α SAT} and Y = {(α, β) CNF CNF β SAT}. For every problem M = X Y in D P with X NP and Y conp there are polynomial time computable functions f and g such that for each input x for M we have x M (f (x), g(x)) SAT-UNSAT.

31 Problem 29: Minimal Unsatisfiability (Critical Satisfiability) Instance: α = {α 1,..., α n } 3-CNF. Question: Is α inconsistent and is α \ {α i } satisfiable for all 1 i n? The Minimal Unsatisfiability Problem is D P -complete. Let Y = SAT and X = {α = {α 1,..., α n } i : α \ {α i } satisfiable}, then Y conp and X NP. So the problem is in D P. Completeness can be shown by a reduction of the D P -complete problem SAT-UNSAT.

32 Problem 30: Not All Redundant UNSAT Instance: Φ = {ϕ 1,..., ϕ m } CNF. Question: Does Φ SAT and 1 i m : (Φ \ {φ i } SAT) hold? The problem is D P -complete. A proof can be given by reducing the D P -complete problem Minimal Unsatisfiability. For every formula Φ = {ϕ 1,..., ϕ m } with atoms X 1,..., X n we construct the formula f (Φ) as follows: For 1 i, j m and 1 k n let X i,k and Y i,j be new atoms; furthermore, let ϕ i j be a copy of ϕ j with X 1,..., X n replaced by X i,1,..., X i,n. We define Φ i = (ϕ i j Y i,j ) (ϕ i j X i,j ) ( Y i,j X i,j ) 1 j m 1 j m 1 j i m for 1 i m. From the m individual formulas we construct f (Φ): f (Φ) = Φ i ( Y 1,1... Y m,m ). 1 i m Then Φ is in Minimal Unsatisfiability if and only if f (Φ) is in Not All Redundant UNSAT.

33 Problem 31: Unique-SAT Instance: α CNF. Question: Does the formulas α have exactly one satisfying truth assignment? Unique-SAT is conp-hard. To establish this we give a reduction of SAT. To every formula β = β 1... β k CNF with atoms X 1,..., X n is assigned the formula U(β) = (X 0 X 1... X n ) ( X 0 β), where X 0 is a new atom. Then we have β SAT if and only if U(β) has exactly one satisfying truth assignment. Unique-SAT is D P -complete if and only if SAT is polynomially reducible to Unique-SAT. This equivalence follows from the D P -completeness of SAT-UNSAT.

34 Problem 32: Satisfiability Equivalence Instance: α, β CNF. Question: Are α and β not satisfiability equivalent? The problem is in D P. Let X = {(π, τ) π SAT or τ SAT} and Y = {(π, τ) π SAT or τ SAT}, then X is in NP and Y in conp. Furthermore, we have (α, β) X Y if and only if α and β are not satisfiability equivalent.

35 Problem 33: Renaming DHORN Instance: α HORN. Question: Does there exist a renaming f such that f (α) is a definite Horn formula? By a renaming f we understand a mapping which to each atom A assigns either A or A. If we apply this renaming to a formula Φ, then every atom A is replaced by f (A). (Double negations are omitted.) In f (Φ) for some atoms the property negated/unnegated is swapped in all occurrences. The problem can be shown to be NP-complete by a reduction of the problem 1-3-SAT.

36 Problem 34: Partial Horn Renaming Instance: α = {α 1,..., α n } CNF, m n. Question: Does there exist a renaming f of the atoms in α so that at least m clauses α i1,..., α im in α are Horn clauses after renaming, i.e. f ({α i1,..., α im }) is a Horn formula? The NP-completeness of this problem can be established by a reduction of the problem 1-3-SAT.

37 Problem 35: Renaming MAX-2-POS Instance: Φ CNF. Question: Does there exist a renaming f of the atoms in Φ, such that for every clause in f (Φ) at most two positive literals occur (i.e. f (Φ) MAX-2-POS)? The problem is NP-complete. The reduction of the problem 1-3-SAT can be carried out as follows: Let Φ = {ϕ 1,..., ϕ n } 3-CNF with every clause containing only positive literals, and let ϕ i = (A i,1 A i,2 A i,3 ). We add to the formula Φ the new atoms A, B and P to give the following formula: Φ = (A B P) (A B P) ( A B P) ( A B P) ((A i,1 A i,2 A i,3 P) ( A i,1 A i,2 A i,3 )). 1 i n By contrast, if we ask whether there exists a renaming into a Horn formula, i.e. a formula with at most one positive literal in each clause, then the problem is solvable in linear time.

38 Problem 36: HORN-SAT with additional units Instance: Sets of atoms {A 1, B 1 },..., {A n, B n } and a Horn formula α. Question: Do there exist X 1 {A 1, B 1 },..., X n {A n, B n } such that α {X 1,..., X n } is inconsistent? By means of a reduction of 3-SAT, NP-completeness is established.

39 Problem 37: Minimal Input Set Instance: An acyclic definite Horn formula α, atom A and k IN. Question: Does there exist a minimal input set S for A with S k? An input set for A is a subset S of the atoms which only occur in α as negative literals such that α S = A holds. The problem is NP-complete. This is established by means of a reduction of the Vertex Cover Problem.

40 Problem 38: Simple Formula Instance: α CNF. Question: Does there exists a formula β equivalent to α with a simple syntactic structure? In detail the problem can be subdivided into the following questions: Given α CNF, does there exist a formula β with α β and (a) β HORN? (c) β k-cnf (k IN, k 1 fixed)? (b) β DHORN? (d) β READ-1 3-CNF? Each of the problems (a) (d) is conp-complete.

41 Problem 39: Restricted Equivalence Horn Instance: Formulas α, β HORN, a set of atoms R. Question: Does α R β hold? The problem is conp-complete. This also holds for the question α = R β.

42 Problem 40: Loop-Test Prolog Instance: Propositional Prolog program π. Question: Does there exist a query clause γ such that the program does not terminate with this query clause? The problem is NP-complete. The equivalence problem for propositional Prolog programs is conp-complete.

43 Problem 41: HORN Partition Instance: Φ CNF and k IN, k 1. Question: Do there exist atoms A 1,..., A k in Φ, such that for all ɛ 1,..., ɛ k {0, 1} we have Φ[A 1 /ɛ 1,..., A k /ɛ k ] HORN? The problem can be shown to be NP-complete by a reduction of the Vertex Cover Problem.

44 Problem 42: Satisfiability of Boolean Expressions Instance: A set M of atoms, a subset J of the 16 possible binary boolean operators and a statement E built up using atoms from M and the connectives in J. Question: Is E satisfiable? The problem is NP-complete. The problem remains NP-complete for every complete set of boolean operators and every incomplete set of operations containing one of the sets { }, { }, {, } or {, }. Otherwise, the problem is solvable in polynomial time.