Linear Programming Formulation of Boolean Satisfiability Problem

Linear Programming Formulation of Boolean Satisfiability Problem Maknickas Algirdas Antano Abstract With using of multi-nary logic 2 SAT boolean satisfiability was formulated as linear programming optimization problem. The same linear programming formulation was extended to finding of 3 SAT and k SAT boolean satisfiability for k greater than 3. Using new analytic formulas and linear programming formulation of boolean satisfiability proposition that k SAT is in P and could be solved in linear time was proved. Key words: Boolean satisfiability, conjunctive normal form, multivalued logic, linear programming, time complexity, NP problem 1 Introduction The Boolean satisfiability (SAT) problem [1] is defined as follows: given a Boolean formula, check whether an assignment of Boolean values to the propositional variables in the formula exists, such that the formula evaluates to true. If such an assignment exists, the formula is said to be satisfiable; otherwise, it is unsatisfiable. For a formula with n variables and m clauses the conjunctive normal form (CNF) (X 1 X 2 ) (X 3 X 4 ) (X n 1 X n ) (1) is most the frequently used for representing Boolean formulas, where X i are independent and in most cases m n. In CNF, the variables of the formula appear in literals (e.g., x) or their negation (e.g., x (logical NOT )). Literals are grouped into clauses, which represent a disjunction (logical OR ) of the literals they con- Maknickas Algirdas Antano Vilnius Gediminas Technical University, Sauletekio al. 11, Vilnius, Lithuania, e-mail: algirdas.maknickas@vgtu.lt 1

2 Maknickas Algirdas Antano tain. A single literal can appear in any number of clauses. The conjunction (logical AND ) of all clauses represents a formula. Several algorithms are known for solving the 2 - satisfiability problem; the most efficient of them take linear time [2, 3, 4]. Instances of the 2-satisfiability or 2SAT problem are typically expressed as 2-CNF or Krom formulas [2] SAT was the first known NP-complete problem, as proved by Cook and Levin in 1971 [1] [5]. Until that time, the concept of an NP-complete problem did not even exist. The problem remains NP-complete even if all expressions are written in conjunctive normal form with 3 variables per clause (3-CNF), yielding the 3SAT problem. This means the expression has the form: (X 1 X 2 X 3 ) (X 4 X 5 X 6 ) (X n 2 X n 1 X n ) (2) NP-complete and it is used as a starting point for proving that other problems are also NP-hard. This is done by polynomial-time reduction from 3-SAT to the other problem. In 2010, Moustapha Diaby provided two further proofs for P=NP. His papers Linear programming formulation of the vertex colouring problem and Linear programming formulation of the set partitioning problem give linear programming formulations for two well-known NP-hard problems [6, 7]. 2 Expressions of multi-valued logic Let describe integer discrete logic units as i, where i Z +. Let describe discrete function g n k (a), a R, k {0,1,2,...,n 1} as g n k (a)= a +k (mod n) (3) As mentioned in [8, 9] function g n k (a) is one variable multi-nary logic generation function for multi-nary set {0,1,2,..,n 1} The are n n one variable logic functions: i 0 i 1 i 2 ρ (a)=... i n 1 a 0 g n i 0 (a) 1 g n i 1 (a) 2 g n i 2 (a)...... n 1 g n i n 1 (a) The are n n2 two variables logic functions:, i 0 i 1 i 2... i n 1 {0,1,2,...,n 1} (4)

Linear Programming Formulation of Boolean Satisfiability Problem 3 i 0,0 i 0,1 i 0,2... i 0,n 1 i 1,0 i 1,1 i 1,2... i 1,n 1 µ i 2,0 i 2,1 i 2,2... i 2,n 1 =............... i n 1,0 i n 1,1 i n 1,2... i n 1,n 1 a \ b 0 1 2... n 1 0 g n i 0,0 (a b) g n i 0,1 (a b) g n i 0,2 (a b)... g n i 0,n 1 (a b) 1 2 g n i 1,0 (a b) g n i 1,1 (a b) g n i 1,2 (a b)... g n i 1,n 1 (a b) g n i 2,0 (a b) g n i 2,1 (a b) g n i 2,2 (a b)... g n, i 2,n 1 (a b).................. n 1 g n i n 1,0 (a b) g n i n 1,1 (a b) g n i n 1,2 (a b)... g n i n 1,n 1 (a b) i 0,0 i 0,1 i 0,2... i 0,n 1 i 1,0 i 1,1 i 1,2... i 1,n 1 i 2,0 i 2,1 i 2,2... i 2,n 1 {0,1,2,...,n 1} (5)............... i n 1,0 i n 1,1 i n 1,2... i n 1,n 1 Using of this expressions for logic of higher dimensions with non deterministic Turing machine enables to make counter of linear time complexity [9] The goal of this paper is the proof of proposition that ksat is in P and could be solved in linear time. For this purpose will be used one formula of multi-valued second order logic for function of two variables described above. 3 2SAT is in P Theorem 1. If all variables are unique, equation where maxβ 2 (X 1,X 2,...,X n 1,X n ) (6) β 2 (X 1,X 2,...,X n 1,X n )= µ(x 1 + X 2, µ(x 3 + X 4,..., µ(x n 3 + X n 2,X n 1 + X n ))) µ(a,b)= a \ b 0 1 2 0 0 0 0 1 0 1 1 2 0 1 1 and+ is algebraic summation could be solved for X i {0,1} in O(m). Proof. Let start from investigation of (7)

4 Maknickas Algirdas Antano f(x 1,x 2,...,x m )= in hyper-cube of sides[0,1]. This function is convex, because m i=1 m i=1 x i, x i R (8) x i m i=1 x iα i (9) 1 = m i=1 αi (10) It is obvious, on right side of (9) we have hyper-plane which goes through vertexes (0,0,0,..,0) and (1,1,1,..,1) of investigating hyper-cube. This hyper-plane has global maximum like a function f(x 1,x 2,...,x m ) and it equals 1. So we could resume that finding of global maximum of f(x 1,x 2,...,x m ) could be replaced by finding of global maximum of this hyper plane or m max i = i=1x 1 n n max x i (11) i=1 Let start to investigate (6) X i R,i {1,2,...,n}. Function β 2 could be calculated within O(m) [10]. According to (7) equation (6) could be rewritten as follow max µ(x 1 + X 2, µ(x 3 + X 4,..., µ(x n 3 + X n 2,X n 1 + X n )))= max µ(x 1 + X 2,max µ(x 3 + X 4,...,max µ(x n 3 + X n 2,X n 1 + X n ))) (12) So, m local partial maximums must satisfy equalities max µ(x k 1 + X k )=1 to avoid 0 result of global maximum. This leads to inequalities 1 (X k 1 + X k ) 2 but if we want to use (11) we must leave (X k 1 + X k ) = 1 Now we could start to solve satisfiability problem as global maximum of (6). If we have all variables unique, (12) could be solved repeating solving of system of LP equations max k i=k 1 X i X k + X k 1 = 1 (13) Each of them has not zero max and could be solved using best known algorithm of linear programming [11] in O ( n 3.5) or in O(2 3.5 ). So all clauses should be optimized in O ( 2 3.5 m ) 3.1 Special cases Theorem 2. If some of unique variables are negations X i, equation maxβ 2 (X 1,X 2,...,X n 1,X n ) (14)

Linear Programming Formulation of Boolean Satisfiability Problem 5 could be solved for X i {0,1} in O(m). Proof. If we have all variables unique, let replace all negations X k with Xḱ and all others just renamed by X i. Now (14) could be solved repeating solving of system of LP equations max k i=k 1 X i Xḱ+ Xḱ 1 = 1 0 Xḱ 1 1 (15) 0 Xḱ 1 Each of them has not zero max and could be solved in O(2 3.5 ). Going back to old variables do not change complexity of each solution. So all clauses should be optimized in O ( 2 3.5 m ) Theorem 3. If some of variables in different clauses are not unique, equation could be solved for X i {0,1} in O(m). maxβ 2 (X 1,X 2,...,X n 1,X n ) (16) Proof.. If X n,x n 1 are unique, let start to solve (16) starting from a system of LP equations max n i=n 1 X i X n + X n 1 = 1 (17) 0 X n 1 1 0 X n 1 Now we know two unique variables. So maximum is reached and system of equation is solved in O(2 3.5 ). If X n = X n 1, X n = 1 X n 1 = 1. All other sub-equation could be solved as follow: if solving sub-equation has one variable with earlier found value, LP equations max k i=k 1 X i 1 X k + X k 1 2 (18) could be solved by reducing of (18) within inserting of earlier found variable value resigned to 1 (lower and upper bound could be increased in case the broken plane result to 1 earlier and than is flat in 2D space); if all variables of solving sub-equations (18) aren t unique, these variables could be reassigned to 1 and solving repeated within next clause; if solving sub-equation has two unique variables, LP equations max k i=k 1 X i X k + X k 1 = 1 (19) could be solved. So all clauses should be optimized in O ( 2 3.5 m )

6 Maknickas Algirdas Antano Theorem 4. If some of variables in different clauses are not unique and are negation each other, equation maxβ 2 (X 1,X 2,...,X n 1,X n ) (20) could be solved for X i {0,1} in O(v+e). Proof.. Let mark all variables they are unique or not starting from the end of CNF. Now cycle through n variables must be repeated to rename them to Xḱ so that new replaced variables marked as not unique will be negations which must be replaced with 1 Xḱ, if they found second time. In kind first time found not unique variable could be assigned to 0. This lead to value of 1 for negation. It could be done in O(2n). Now we start solving process for last clause. If Xń = Xń 1, Xń = 1. If Xń,Xń 1 are unique, let start to solve (20) from a system of LP equations max n i=n 1 X i Xń+ Xń 1=1 0 Xń 1 1 (21) 0 Xń 1 Now we know two or one unique variables. Aspvall, Plass and Tarjan [12] found the linear time procedure for solving 2- satisfiability instances, based on the notion of strongly connected components from graph theory. There are several efficient linear time algorithms for finding the strongly connected components of a graph, based on depth first search: Tarjan s strongly connected components algorithm [13] and the path-based strong component algorithm [14] each perform a single depth first search. Let choose the clause walking path as depth first search algorithm of the strongly connected components of a graph, where graph is formed as interconnected clauses with the not unique variables. So, other sub-equation could be solved as follow: if solving sub-equation has one variable with earlier found value, LP equations max k i=k 1 X i 1 X k + X k 1 2 (22) could be solved by reducing of (22) within inserting of earlier found variable value reassigned to 1 (lower and upper bound could be increased in case the broken plane result to 1 earlier and than is flat in 2D space); if all variables of solving sub-equations aren t unique and are negations of earlier found values in a different clauses, they values could be assigned to 0 and values of residual variables leads to 1; if all variables of solving sub-equations aren t unique and are negations of earlier found values in the same clause, one of variables must be assigned to 1 and other to 0; if at least two clauses, where X i X i, CNF is not satisfiable ;if solving sub-equation has two unique variables, LP equations

Linear Programming Formulation of Boolean Satisfiability Problem 7 max k i=k 1 X i X k + X k 1 = 1 (23) could be solved. Each sub-system of equation is solved in O(2 3.5 ) to reach global maximum 1. Finally, all clauses should be optimized at most v+e (v - amount of vertexes, e - amount of edges) times or in O ( 2 3.5 (v+e) ). 4 3SAT is in P Theorem 5. If all variables are unique, equation where β 3 (X 1,X 2,...,X n 1,X n )= maxβ 3 (X 1,X 2,...,X n 1,X n ) (24) µ(x 1 + X 2 + X 3, µ(x 4 + X 5 + X 6,..., µ(x n 5 + X n 4 + X n 3,X n 2 + X n 1 + X n ))) (25) µ(a,b)= a\b 0 1 2 3 0 0 0 0 0 1 0 1 1 1 2 0 1 1 1 3 0 1 1 1 (26) and+ is algebraic summation could be solved for X i {0,1} in O(m). Proof. Let start to investigate (24) when X i R,i {1,2,...,n}. Function β 3 could be calculated within O(m) [10]. According to (25) equation (24) could be rewritten as follow max µ(x 1 + X 2 + X 3, µ(x 4 + X 5 + X 6,..., µ(x n 5 + X n 4 + X n 3,X n 2 + X n 1 + X n ))) (27) So, m local partial maximums must satisfy equalities max µ(x k 2 +(X k 1 + X k )=1 to avoid 0 result of global maximum. This leads to inequalities 1 (X k 2 + X k 1 + X k ) 3 but if we want to use (11) we must leave (X k 2 + X k 1 + X k )=1 If we have all variables unique, (24) could be solved repeating solving of system of LP equations

8 Maknickas Algirdas Antano max k i=k 2 X i X k + X k 1 + X k 2 = 1 0 X k 2 1 (28) Each of them has not zero max and could be solved in O(3 3.5 ). So all clauses should be optimized in O ( 3 3.5 m ) 4.1 Special cases Theorem 6. If some of unique variables are negations X i, equation could be solved for X i {0,1} in O(m). maxβ 3 (X 1,X 2,...,X n 1,X n ) (29) Proof. If we have all variables unique, let replace all negations X k with Xḱ and all others just renamed by X i. Now (29) could be solved repeating solving of system of LP equations max k i=k 2 X i Xḱ+ Xḱ 1+ Xḱ 1 = 1 0 Xḱ 2 1 (30) 0 Xḱ 1 1 0 Xḱ 1 Each of them has not zero max and could be solved in O(3 3.5 ). Going back to old variables do not change complexity of each solution. So all clauses should be optimized in O ( 3 3.5 m ) Theorem 7. If some of variables in different clauses are not unique, equation could be solved for X i {0,1} in O(m). maxβ 3 (X 1,X 2,...,X n 1,X n ) (31) Proof. If X n,x n 1,X n 2 are unique, let start to solve (31) starting from a system of LP equations max n i=n 2 X i X n + X n 1 + X n 2 = 1 0 X n 2 1 (32) 0 X n 1 1 0 X n 1 Now we know three unique variables. Maximum is reached and system of equation is solved in O(3 3.5 ). If X n = X n 1, X n = 1 X n 1 = 1 X n 2 = 1. All other subequation could be solved as follow: if solving sub-equation has one or two variables

Linear Programming Formulation of Boolean Satisfiability Problem 9 with earlier found value, LP equations max k i=k 2 X i 1 X k + X k 1 + X k 2 3 0 X k 2 1 (33) could be solved by reducing of (33) with inserting of earlier found variables value reassigned to 1; if solving sub-equation three unique variables, LP equations max k i=k 2 X i X k + X k 1 + X k 2 = 1 0 X k 2 1 (34) could be found. So all clauses should be optimized in O ( 3 3.5 m ) Theorem 8. If some of variables in different clauses are not unique and are negation each other, equation maxβ 3 (X 1,X 2,...,X n 1,X n ) (35) could be solved for X i {0,1} in O(v+e). Proof. Let mark all variables they are unique or not starting from the end of CNF. Now cycle through n variables must be repeated to rename them to Xḱ so that new replaced variables marked as not unique will be negations which must be replaced with 1 Xḱ, if they found second time. In kind first time found not unique variable could be assigned to 0. This lead to value of 1 for negation. It could be done in O(2n). Now we start solving process for last clause. If not all variables are unique in the first clause, after sorting of variables so that third one will be negation and second one will be unique, LP equations maxxń+ Xń 1+1 Xń Xń+ Xń 1+1 Xń 2=1 0 Xń 2 1 (36) 0 Xń 1 1 0 Xń 1 could be solved. If Xń,Xń 1,Xń 2 are unique, let start to solve (25) from a system of LP equations max n i=n 2 X i Xń+ Xń 1+ Xń 2=1 0 Xń 2 1 (37) 0 Xń 1 1 0 Xń 1

10 Maknickas Algirdas Antano Now we know three or two unique variables. Let choose the clause walking path as depth first search algorithm of the strongly connected components of a graph, where graph is formed as interconnected clauses with the not unique variables. So, other sub-equation could be solved as follow: if solving sub-equation has one or two variable with earlier found value, LP equations max k i=k 2 X i ` 1 X + X `k 1 + X `k 2 3 0 X `k 2 1 (38) 0 X `k 1 1 0 X `k 1 where one or two of three variables X `k 2,X`k 1,X`k are equal 1 X q could be solved by reducing of (38) within inserting of earlier found variable value reassigned to 1; if all variables of solving sub-equations aren t unique and are negations of earlier found values in a different clauses, they values could be assigned to 0 and values of residual variables leads to 1; if all variables of solving sub-equations aren t unique and are negations of earlier found values in the same clause, one of variables must be assigned to 1 and other to 0; if solving sub-equation has three unique variables, LP equations max k i=k 2 X i X k + X k 1 + X k 2 = 1 0 X k 2 1 (39) could be solved. Each sub-system of equation is solved in O(3 3.5 ) to reach global maximum 1. Finally, all clauses should be optimized at most v+e (v - amount of vertexes, e - amount of edges) times or in O ( 3 3.5 (v+e) ). 5 ksat is in P Theorem 9. If all variables are unique, equation maxβ k (X 1,X 2,...,X n 1,X n ) (40) where β k (X 1,X 2,...,X n 1,X n )= k µ( X i, µ i=1 ( 2k X i,..., µ i=k+1 ( n k i=n 2k+1 X i, n i=n k+1 X i ))) (41)

Linear Programming Formulation of Boolean Satisfiability Problem 11 µ(a,b)= a\b 0 1 2... n 1 0 0 0 0... 0 1 0 1 1... 1 2 0 1 1... 1... n 1 0 1 1... 1 (42) and is algebraic summation could be solved for X i {0,1} in O ( k 3.5 m ). Proof.. Let start to investigate 40 when X i R,i {1,2,...,n}. Function β k could be calculated within O(m) [10]. According to 7 equation 40 could be rewritten as follow ( ( k 2k n k max µ( X i, µ X i,..., µ i=1 i=k+1 i=n 2k+1 X i, n i=n k+1 X i ))) If we have all variables unique, (41) could be solved repeating solving of system of LP equations max 2k i=k+1 X i 2k i=k+1 X i = 1 (44) 0 X i 1 i {1,2,...,n} Each of them has not zero max and could be solved in O(k 3.5 ). So all clauses should be optimized in O ( k 3.5 m ) (43) 5.1 Special cases Theorem 10. If some of unique variables are negations X i, equation could be solved for X i {0,1} in O ( k 3.5 m ). maxβ k (X 1,X 2,...,X n 1,X n ) (45) Proof.. If we have all variables unique, let replace all negations X k with Y k and all others just renamed by Y i. Now (45) could be solved repeating solving of system of LP equations max 2k i=k+1 Y i i=k+1 2k Y i = 1 (46) 0 Y i 1 i {1,2,...,n} Each of them has not zero max and could be solved in O(k 3.5 ). Going back to old variables do not change complexity of each solution. So all clauses should be optimized in O ( k 3.5 m ) Theorem 11. If some of variables in different clauses are not unique, equation

12 Maknickas Algirdas Antano could be solved for X i {0,1} in O ( k 3.5 m ). maxβ k (X 1,X 2,...,X n 1,X n ) (47) Proof.. If X n,x n 1,...,X n k+1 are unique, let start to solve (47) starting from a system of LP equations max n i=n k+1 X i n i=n k+1 X i = 1 (48) 0 X i 1 i {1,2,...,n} Now we know k unique variables. Maximum is reached and system of equation is solved in O(k 3.5 ). If X n = X n 1, X n = 1 X n 2 = 1,... All other sub-equation could be solved as follow: if solving sub-equation has at least one variable with earlier found value, LP equations max n i=n k+1 X i 1 k i=n k+1 X i k (49) 0 X i 1 i {1,2,...,n} could be solved by reducing of (49) with inserting of earlier found variables value reassigned to 1; if solving sub-equation k unique variables, LP equations max k i=n k+1 X i k i=n k+1 X i = 1 (50) 0 X i 1 i {1,2,...,n} could be solved. So all clauses should be optimized in O ( k 3.5 m ) Theorem 12. If some of variables in different clauses are not unique and are negation each other, equation could be solved for X i {0,1} in O ( k 3.5 (v+e) ). maxβ k (X 1,X 2,...,X n 1,X n ) (51) Proof.. Let mark all variables they are unique or not starting from the end of CNF. Now cycle through n variables must be repeated to rename them to Xḱ so that new replaced variables marked as not unique will be negations which must be replaced with 1 Xḱ, if they found second time. In kind first time found not unique variable could be assigned to 0. This lead to value of 1 for negation. It could be done in O(2n). Now we start solving process for last clause. If not all variables are unique in the first clause, after sorting of variables so that negations occurs at the end of the list, LP equations max l i=n k+1 X i + ( ) n i=l+1 1 X i l i=n k+1 X i + ( n i=l+1 1 X i) = 1 (52) 0 X i 1 i {1,2,...,n}

Linear Programming Formulation of Boolean Satisfiability Problem 13 could be solved. If Xń,Xń 1,...,Xń k+1 are unique, let start to solve (51) from a system of LP equations max n i=n k+1 X i n i=n k+1 X i = 1 (53) 0 X i 1 i {1,2,...,n} Now we know at most k unique variables. Let choose the clause walking path as depth first search algorithm of the strongly connected components of a graph, where graph is formed as interconnected clauses with the not unique variables. So, other sub-equation could be solved as follow: if solving sub-equation has at least one variable with earlier found value, LP equations max l i=k+1 X i + ( ) 2k i=l+1 1 X i 1 l i=k+1 X i + ( 2k i=l+1 1 X i) k (54) 0 X i 1 i {1,2,...,n} could be solved by reducing of (54) within inserting of earlier found variables values reassigned to 1; if all variables of solving sub-equations aren t unique and are negations of earlier found values in a different clauses, they values could be assigned to 0 and values of residual variables leads to 1; if all variables of solving sub-equations aren t unique and are negations of earlier found values in the same clause, one of variables must be assigned to 1 and other to 0; if solving sub-equation has k unique variables, LP equations max 2k i=k+1 X i 2k i=k+1 X i = 1 (55) 0 X i 1 i {1,2,...,n} could be solved. Each sub-system of equation is solved in O(k 3.5 ) to reach global maximum 1. Finally, all clauses should be optimized at most v+e (v - amount of vertexes, e - amount of edges) times or in O ( k 3.5 (v+e) ). 6 Conclusions Every 2 SAT problem is solvable using linear programming formulation in linear time. Every 3 SAT problem is solvable using linear programming formulation in linear time. Every k SAT problem is solvable using linear programming formulation in linear time for k>3. Every NP mathematical problem is reducible in polynomial time to k SAT problem and solvable in linear time if exist full, appropriate and correct knowledge

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