Solving Combinatorial Problems Using Boolean Equations Christian Posthoff The University of The West Indies St. Augustine Campus Trinidad & Tobago email: christian@posthoff.de Bernd Steinbach Freiberg University of Mining and Technology Institute of Computer Science D-09596 Freiberg, Germany email: steinb@informatik.tu-freiberg.de This paper honors the excellent scientist Arkadij D. Zakrevskij who significantly extended the knowledge in the Boolean Domain. Abstract The recent developments indicate that Boolean equations get a very general and powerful instrument for the solution of discrete problems. Several classical areas (very often related to search) are now replaced by solution methods for Boolean equations, and also other areas of Discrete Mathematics (graph problems, coloring, planning etc.) can be solved by using Boolean equations. The problem only needs a correct Boolean model, the rest will be done by very efficient algorithms and programs which will be available. We show several problem classes and the way how they can be solved. 1 Introduction This paper has been planned as a common paper with Arkadij, he suggested the title, and now we try to continue this work and to implement the discussed ideas. In the long period of common work published together the results of many special problems [14 17, 20]. Fully aware of the importance of Boolean equations, Arkadij Zakrevskij commonly published with Dieter Bochmann and Christian Posthoff a book to this topic [2]. It was our pleasure to contribute to this book with two chapters [6, 18]. The wide field of applications of Boolean (logic) equations impress our text-book [7] and the associated book about examples and exercises of logic functions and equations [5]. In this paper, we take ideas and problems from several fields and want to show how far-reaching this methodology can be applied. 2 Ternary Vectors - the Main Data Structure Sometime in the seventies when we started our research in this area we saw the book of Arkadij Zakrevskij [19] where he used ternary vectors as a data structure. We found this structure useful and used it as well in a rather intuitive way. Later on we explored it more in detail, utilized it as data structure for calculations, and found many very useful properties. Definition 1 Let x = (x 1, x 2,..., x n ), x i {0, 1, }, then x is a ternary vector.
Table 1: Coding of ternary values character bit combinations 0 10 1 11 00 At the beginning we processed the three symbols as characters, thereafter we represented the characters by two bits using the coding of Table 1. The combination (01) is not used, the first bit equal to 1 indicates that a value has been set. In this case the second bit indicates the value itself. The empty memory (00) indicates that it is filled with. This coding enabled the parallel processing of 32, 64 or even more components of the vectors. Definition 2 Let be given two ternary vectors x = (x 1, x 2,..., x n ) and y = (y 1, y 2,..., y n ) two ternary vectors of the same length. Then x is orthogonal to y if there is for at least one component i the combination x i = 0, y i = 1 or vice versa. 3 Set-related Problems Many problems in Discrete Mathematics deal with finite sets. Let M = (m 1,..., m n ) a finite set of n elements, P (M) the set of all subsets, the power set of M. Then each subset S M can be defined by a binary vector (x 1, x 2,..., x n 1, x n ), x i {0, 1}. If x i = 0 then m i / S, if x i = 1 then m i S. On one side we have P (M) with the set operations (S,, ), on the other side the set of all binary vectors B n. It is not very difficult to transfer the set operations: The complement of a set S with regard to M is given by transforming single bits using not: 0 1, 1 0. The intersection S 1 S 2 uses the logical and ( ) component by component. The union S 1 S 2 uses the logical or ( ) component by component. So we can state that (P (M), S,, ) and (B n, x,, ) are isomorphic structures. If necessary, the symmetric difference also can be transformed by using the antivalence ( ), the complement of the symmetric difference can be transformed by means of the equivalence ( ). This use of binary vectors is related to the concept of characteristic functions; each binary vector is the characteristic function of the respective subset. Now we extend the possibilities by the following extended interpretation: the third value indicates that the value of this component is not yet known or not yet defined, and can be replaced (possibly later) by the values 0 and 1. One vector with one can be replaced by two vectors with a 0 and a 1 in this position. In this way one ternary vector defines 2, 4, 8,... binary vectors depending on the number of components with a, i.e., a set of binary vectors. We mention the programming system XBOOLE [5, 7, 11] which implements all the set operations for sets given by ternary vectors efficiently and comfortably. Additionally this understanding is already a powerful mechanism for the creation of very complex and correct models. Together with correct algorithms and correct programs extremely complex problems can be solved. Example 1 The first steps to solve a problem is there specification by an adequate model. We take as example the modeling of a well-known combinatorial problem, the placement of 8 queens on a chessboard, which is traditionally solved by searching. We transform it into the area of logic equations in the following way. As shown in Table 2, we use vectors of 64 components corresponding to the 64 fields of the board.
Table 2: Ternary vectors to describe the assignment of a queen to one field and all consequent restrictions for other fields of the board a1 a2 a3... b1 c1... b2 c3... b4 b5... 1 0 0... 0 0... 0 0...... 0 1 0... 0... 0...... Table 3: Intersection of two ternary components first component second component result 0 0 0 0 1 no result 0 0 1 0 no result 1 1 1 1 1 0 0 1 1 The setting of a queen on the field a1 can be seen by the value 1 for this field, but the additional effect of this setting is also included into this vector: it is not allowed to set a queen on other fields of the same column and the first row and the respective diagonal, but by it is also indicated that there is no knowledge with regard to this field. So the eight vectors for the column (a) are considered as an of these possibilities. Now we do the same for the second column (b), and the two columns are combined by, i.e., by. We must use possible settings for the first as well as for the second column. The intersection of the sets of two ternary vectors is implemented using Table 3. The most interesting situations are the combinations (0 1) or (1 0). In this case the combination of these two vectors makes no contribution to the final result. The search procedures are no longer necessary to find the final result, the solution of a Boolean equation delivers all solutions. Figure 1 shows one of the 92 solutions of the board 8 8. So we have a very general procedure for problems such as: sets of fault combinations, representation of state sets and verification of sequential machines, computation of prime implicants, the design of encoders, decoders, multiplexers, adders, comparators etc. 8 0Z0Z0Z0L 7 Z0Z0ZQZ0 6 0Z0L0Z0Z 5 ZQZ0Z0Z0 4 0Z0Z0ZQZ 3 Z0Z0L0Z0 2 0ZQZ0Z0Z 1 L0Z0Z0Z0 a b c d e f g h Figure 1: Eight queens on a chess board which does not threaten each other.
2 2 2 1 3 5 1 3 5 1 3 5 4 4 4 (a) (b) (c) Figure 2: Analysis regarding Hamiltonian Circuits: (a) graph without a Hamiltonian Circuit; (b) graph that contain a Hamiltonian Circuit; (c) Hamiltonian Circuit of the graph given in (b). Example 2 Very often there is a binary description just from the beginning, but sometimes a careful modeling is required. Sudoku [8, 12], for instance, is a purely combinatorial problem and needs the binary modeling. Here we use another interpretation of a ternary vector. If the component x i = 0 then we include the variable x i into a conjunction, if the component x i = 1, then we include x i, and for x i =, the variable x i is omitted. In this way a ternary vector can be transformed into a conjunction, and a set of ternary vectors corresponds to a disjunctive form. { 1 if the value on the field (i, j) = k x ijk = 0 if the value on the field (i, j) k The requirements and constraints can be stated by one single conjunction C ijk for each number on each field: C 111 = x 111 x 112 x 113 x 114 x 115 x 116 x 117 x 118 x 119 x 121 x 131 x 141 x 151 x 161 x 171 x 181 x 191 x 211 x 311 x 411 x 511 x 611 x 711 x 811 x 911 x 221 x 231 x 321 x 331. This conjunction describes completely the setting of 1 on the field (1, 1) and all the consequences. Hence, for each field we need 9 conjunctions for the 9 different possibilities. These conjunctions have the advantage that they are orthogonal to each other, for one field one and only one can be equal to 1. The Boolean equation C 111 C 112 C 113 C 114 C 115 C 116 C 117 C 118 C 119 = 1 describes the 9 possibilities which exist for one field ((111) is only an example) and all the consequences resulting from a given setting. In principle, 729 of such conjunctions are possible, however, each given setting (such as x 1,9 = 5) selects one out of 9 conjunctions. Now it is only necessary to write a tiny program for the input of some given values and the transformation into a conjunctive form of a Boolean equation. 4 The Description of Graphs Many problems can be expressed by graphs - planning problems, accessibility, path problems, coloring of edges and nodes in a graph and many more. We only mention the Birkhoff Diamond and the passing of bridges in Koenigsberg (solved by L. Euler), and explain the basic model for the problem of Hamiltonian Circuits. Figure 2 shows examples of graphs without and with an Hamiltonian Circuit. The edges can be expressed by Boolean variables: { 1 if the edge is used from node i to node j, x ij = 0 otherwise. Using this encoding the rules of the graph problem can be expressed by Boolean equations. Table 4 describes all rules which must be hold by a Hamiltonian Circuit and gives the concrete Boolean equation for a selected edge in the graph of Figure 2 (b). Again without any searching
Table 4: Rules of Hamiltonian Circuits applied to the edge from node 3 to node 5 in the graph of Figure 2 (c) rule example An edge from node i to node j, i.e., x ij = 1, prohibits (x 35 = x 53 ) = 1 that the reverse edge is used, i.e., x ji = 0. An edge from node i to node j, i.e., x ij = 1, prohibits (x 35 = x 32 ) all edges to other destination nodes d l, i.e., x idl = 0. (x 35 = x 34 ) = 1 An edge from node i to node j, i.e., x ij = 1, prohibits (x 35 = x 45 ) = 1 all edges from other source nodes s m, i.e., x smj = 0. procedure all solutions of the problem will be found as solution of created system of Boolean equations. Alternatively the nodes of the graph can be encoded by Boolean variables and additional bits can express the colors of a coloring problem. An edge is expressed in this model by a conjunction of variables that encode both the source node and the destination node. 5 Rule-based Problems Rule-based problems are very important, at a given point of time they were a crucial point of Artificial Intelligence. A lot of efforts have been spent to build efficient SAT-solvers [1, 9]. Again search procedures were a central point, such as breadth-first search, depth-first search, heuristic search etc. Again we transform the rules into a Boolean equation, only a set of correct rules is necessary, afterwards the system takes over. Any rule has the format or in a logical way of speaking if condition then conclusion, x y. Each rule x y can now be transformed into a disjunction: x y = x y. The validity of several rules can be ensured by using the conjunction. By means of ternary vectors as the basic data structure, we are able to solve SAT-problems directly. We will use the following small example: (x 1 x 2 x 3 )(x 2 x 4 x 5 )(x 1 x 4 x 5 )(x 2 x 3 x 5 ) = 1. (1) This equation is equivalent to four single equations: x 1 x 2 x 3 = 1, x 2 x 4 x 5 = 1, x 1 x 4 x 5 = 1, x 2 x 3 x 5 = 1. As suggested by Arkadij Zakrevskij, the first equation now will be transformed into a ternary matrix (a set or list of ternary vectors): 1 0 0 This matrix shows all the vectors that satisfy the first equation. If x 1 = 1, then the values of the other variables are not important. If x 1 = 0, then x 2 must be equal to 1, i.e., x 2 = 0. Finally, if x 1 = 0 and x 2 = 1, then x 3 must be equal to 0. Double solutions cannot exist. It is very characteristic that each vector of the matrix includes more information than the previous
vectors. The number of vectors in the resulting matrix is equal to the number of variables in the disjunction. In the example each disjunction has three variables (an example of a 3-SATproblem). If we repeat this procedure for the other three equations, then we get the following four matrices: M 1 = 1 0 0, M 2 = 1 0 0 0 1 0, M 3 = 0 1 1 1 0 1, M 4 = 1 0 1 0 0 0 In order to get the final solution, these four matrices have to be combined by intersection (see above). Each line of one matrix has to be combined with each line of the next matrix, empty intersections can be omitted. For the first and second matrix, we get M 1 M 2 as follows:. 1 0 0 1 0 0 0 1 0 = 1 1 1 0 0 1 0 1 0 0 0 0 0 0 1 0. Now we calculate (M 1 M 2 ) M 3 : 0 1 1 1 0 1 1 1 1 0 0 1 0 1 0 0 0 0 0 0 1 0 = 0 0 0 0 0 1 0 1 1 1 1 0 1 0 1 1 0 1 1 0 0 1, and (M 1 M 2 ) M 3 ) M 4 will return the final result: 1 0 1 0 0 0 0 0 0 0 0 1 0 1 1 1 1 0 1 0 1 1 0 1 1 0 0 1 = 0 0 0 0 1 0 1 1 1 1 1 1 0. This matrix of ternary vectors represents all solutions of the original SAT-problem (1). Since the value represents 0 as well as 1, the equation has 18 solutions. 6 Combinatorial Design In this area the binary description of problems is well-known. It is necessary to transform the requirements of a special problem into the area of binary equations. We use as tiny example the circuit of Figure 3. x 1, x 2, x 3, x 4 are input variables, g 1, g 2, g 3 are intermediate variables, finally f 1, f 2 are the output variables.
x 1 g 1 f 1 x 2 g 2 f 2 x 3 g 3 x 4 Figure 3: Combinational circuit. The global task to be solved is the analysis the circuit. That requires the calculation of the behavior that may be expressed by a target list of ternary vectors. The local behavior of each gate can be expressed by a simple Boolean equation: g 1 = x 1, g 2 = x 1 x 2, g 3 = x 3 x 4, f 1 = g 1 g 2, f 2 = f 1 g 3. Using the orthogonal solutions of each single equation: M 1 = x 1 g 1 0 1 1 0, M 2 = x 1 x 2 g 2 0 0 1 0 0 1 1 1, M 3 = x 3 x 4 g 3 0 0 0 1 1 0 1 1, M 4 = g 1 g 2 f 1 0 0 0 0 1 1 1 0 1 1 1 0, M 5 = f 1 g 3 f 2 0 0 1 0 0 1 1 1, the needed global behavior results form the intersections of these five ternary vector lists (TVLs): 5 M i = i=1 x 1 x 2 x 3 x 4 g 1 g 2 g 3 f 1 f 2 0 0 0 1 0 0 1 0 0 1 0 1 0 1 1 1 0 1 1 0 1 1 1 1 0 0 0 0 0 0 0 0 1 0 1 0 0 0 1 0 0 1 0 1 0 0 1 0 0 1 1 0 0 0 1 0 1 0 1 1 1 0 0 1 1 1 1 1 1 1 0 1 1 1 1. This approach applies to all digital circuits. Because of the importance of this field there are many design systems and design languages on the market or the property of different companies. It might be the responsibility or the desire of these places to improve their algorithms and their systems. 7 Rectangle-free Four-colorable Grids The solution of this problem was one of the highlights of our research [10, 13]. The definition of the problem is rather short: A two-dimensional grid is a set G m,n = [m] [n]. A grid G m,n is c-colorable if there is a function χ m,n : G m,n [c] such that there are no rectangles with all four corners of the same color. It is known by some theorems [3] that rectangle-free 4-colorable grids exist up to certain sizes of the grid. As an example we can state that the grids G 22,4, G 22,5,..., G 22,10 are rectangle-free 4-colorable, but the grids G 22,11, G 22,12,... does not have this property. It follows from this theory that there must be a rectangle-free 4-colorable grid G 16,20, however, an example for such a grid was not shown before. To our best knowledge only one grid G 17,17 was known that is almost 4-colorable (with only one incorrect position). This grid was constructed by the student Rohan Puttagunta. It can be restricted to a rectangle-free 4-colorable grid G 16,16.
Table 5: Mapping of 4-valued color x to two Boolean variables a and b x a b 1 0 0 2 1 0 3 0 1 4 1 1 It was unknown whether there are rectangle-free 4-colorable grids of the sizes 17 17, 17 18, 18 17, and 18 18. At the first glance this problem seems to be easy. It is only necessary to fix a color for the 18 18 = 324 grid points and to check whether all possible rectangles do not have the same color at their corners. However, when the first check fails, we have to take into account the next color pattern and so on, until the test is successful or all patterns have been tested. The grid G 18,18 has 18 18 = 324 positions, and one of four colors must be selected for each of them. Hence, there are 4 324 = 1.16798 10 195 different patterns altogether in which one of four colors has been assigned to each of the 18 18 positions of the grid. When we assume that we are able to evaluate one pattern in one nano-second (10 9 ) and to spend 100 years (3 10 9 ) then we must repeat the job 3.8 10 176 times in order to know whether there is an valid color pattern and if YES which is the valid color pattern. So we see that we are going to solve an extremely complex problem. The condition that the four corners of the rectangle selected by the rows r i and r j and by the columns c k and c l do not show only one of the colors 1, 2, 3, 4 can be expressed as For the whole grid G m,n, we get: m 1 m f ec (x ri,c k, x ri,c l, x rj,c k, x rj,c l ) = 0. (2) n 1 n i=1 j=i+1 k=1 l=k+1 f ec (x ri,c k, x ri,c l, x rj,c k, x rj,c l ) = 0. (3) This model uses 324 variables, and each variable can take 4 values. Because of the intention to apply our previous research in the area of Boolean equations, we transformed the 4-valued model into a Boolean (binary) model. This can be done by coding the four color values by two Boolean values. Table 5 shows the mapping which has been used. Function (4) depends on eight Boolean variables and has a Boolean result equal to 1 if the colors in the corners of the rectangle selected by the rows r i and r j and by the columns c k and c l are identical: f ecb (a ri,c k, b ri,c k, a ri,c l, b ri,c l, a rj,c k, b rj,c k, a rj,c l, b rj,c l ) = (a ri,c k b ri,c k a ri,c l b ri,c l a rj,c k b rj,c k a rj,c l b rj,c l ) (a ri,c k b ri,c k a ri,c l b ri,c l a rj,c k b rj,c k a rj,c l b rj,c l ) (a ri,c k b ri,c k a ri,c l b ri,c l a rj,c k b rj,c k a rj,c l b rj,c l ) (a ri,c k b ri,c k a ri,c l b ri,c l a rj,c k b rj,c k a rj,c l b rj,c l ) (4) The conditions of the rectangle-free 4-color problem on a grid G m,n are met when the function f ecb (4) is equal to 0 for all rectangles which can be expressed by m 1 m n 1 n i=1 j=i+1 k=1 l=k+1 f ecb (a ri,c k, b ri,c k, a ri,c l, b ri,c l, a rj,c k, b rj,c k, a rj,c l, b rj,c l ) = 0. (5) It is not possible to represent here all the steps we tried and all the difficulties that we met. Finally we restricted our investigations to one color and checked whether this single color uses 25 % of the grid positions in a correct way. We could find one solution for a single color, and this solution could be used three more times after the configuration has been rotated by 90 degrees. Figure 4 shows one of these extremely rare cyclically reusable, rectangle-free coloring of grid G 18,18.
8 Conclusions 1 0 0 1 0 0 0 1 0 1 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1 0 1 0 0 0 1 0 1 1 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 1 1 0 0 0 1 1 0 0 0 1 0 0 0 0 1 0 0 0 1 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0 1 0 0 1 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 1 1 0 0 0 0 0 0 0 0 1 1 0 0 0 1 0 1 0 0 0 0 1 1 1 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 1 0 1 0 0 0 1 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 1 1 0 1 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 1 0 1 0 1 0 0 1 0 0 0 0 0 0 0 0 Figure 4: Cyclically reusable, rectangle-free coloring of grid G 18,18. A matrix containing the ternary elements (0,1,-) was suggested and called by Arkadij Zakrevskij as troiqna matrica. It is a very helpful data structure in the Boolean domain. Boolean equations are a powerful instrument for many problems of Discrete Mathematics, not only for digital systems, but for many other problems as well. It is sufficient to create correct models, the appropriate software is available. Both the domain specific software XBOOLE [5, 7, 11] and the more specialized SAT-solvers [4] can be used. The complexity of the problems to be solved is extremely high. The simple hint that the problem has exponential complexity is no longer sufficient to leave the problem back, without intention to solve it. There is a strong relation between Mathematics and Engineering to solve these problems in a cooperative way. Most likely a course of one semester is not sufficient, two semesters would be better and obviously necessary. Very often it is understandable that traditional research areas or available instruments are no longer useful. Very good programming skills are required in each engineering field. References [1] A. Biere et al., eds. Handbook of Satisfiability. Vol. 185. Frontiers in Artificial Intelligence and Applications. IOS Press, Feb. 2009, p. 980. isbn: 978-1-58603-929-5. [2] D. Bochmann, A. Zakrevskij, and C. Posthoff. Boolesche Gleichungen: Theorie, Anwendungen, Algorithmen (in German - Boolean Equations: Theory, Applications, Algorithms). Wien, New York: Springer, Feb. 1985. isbn: 978-3211958155. [3] S. Fenner et al. Rectangle Free Coloring of Grids. 2009. url: http://www.cs.umd.edu/ ~gasarch/papers/grid.pdf. [4] M. Gebser et al. clasp: A Conflict-Driven Answer Set Solver. In: 9th International Conference on Logic Programming and Nonmonotonic Reasoning. Vol. LNAI 4483. LPNMR. Tempe, AZ, USA: Springer, 2007, pp. 260 265.
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