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TIG* 
 
 }/ U'20O Report No. 266 
 
 *>?<-*C-£2f 
 
 CCO- 1469- 0077 
 
 THE SOLUTION OF SIMULTANEOUS BOOLEAN EQUATIONS 
 
 by 
 
 HOWARD STEPHEN SCHWEITZER 
 
 May 28, 1968 
 
 DEPARTMENT OF COMPUTER SCIENCE • UNIVERSITY OF ILLINOIS • URBANA, ILLINOIS 
 
Report No. 266 
 
 THE SOLUTION OF SIMULTANEOUS BOOLEAN EQUATIONS 
 
 by 
 
 HOWARD STEPHEN SCHWEITZER 
 
 May 28, 1968 
 
 Department of Computer Science 
 University of Illinois 
 Urbana, Illinois 618OI 
 
m 
 
 ACKNOWLEDGEMENT 
 
 The author is deeply indebted to his advisor, Professor 
 W. J. Poppelbaum, whose encouragement, understanding, and support 
 have been invaluable in the completion of this project. The author 
 is also extremely grateful to Professor F. E. Hohn, whose cooperation 
 and guidance were essential in the preparation of this paper. 
 
 Thanks are also extended to Mr. M. Payne for many helpful 
 discussions, and to Mrs. Gayle Osterberg for typing the final 
 manuscript. 
 
IV 
 
 TABLE OF CONTENTS 
 
 1. INTRODUCTION 1 
 
 2 . NOTATION , 3 
 
 3. GENERAL SOLUTIONS OF BOOLEAN FUNCTIONAL EQUATIONS 5 
 
 3.1 Basic Theorems 5 
 
 3.2 Logical Algebraic Algorithm 8 
 
 3.3 "Map" Methods Ik 
 
 3.3.1 Svoboda ' s Algorithm .15 
 
 3.3-2 Maitra's Methods 25 
 
 3.3.3 Ledley's Methods 36 
 
 3 . 3 . h Carvallo ' s Method 67 
 
 3.U "Truth-Table-Logic" Method 71 
 
 3 . 5 Parametric Method 75 
 
 3.6 Summary and Applications 8l 
 
 U . PARTI CULAR SOLUTIONS OF GENERAL BOOLEAN EQUATIONS 88 
 
 k.l Logical Algebraic Methods 88 
 
 U.l.l Ashenhurst's Algorithm 88 
 
 k . 1 . 2 Grigor ' yan ' s Algorithm 90 
 
 k,2 "Map" Method 96 
 
 U,3 Pseudo-Boolean Equations ..98 
 
 h.k Summary and Applications 107 
 
 5. RELATED TOPICS 110 
 
 5 . 1 Programmed Methods 110 
 
 5 . 2 Simultaneous Implications Ill 
 
 5.3 Systems Other Than Binary. Ill 
 
 5.U "Physical Mapping" 115 
 
 5.5 Boolean Ring Equations 116 
 
 5 . 6 Sequential Equations 120 
 
 LIST OF REFERENCES .123 
 
1 
 
 1. INTRODUCTION 
 
 In many problems, particularly those involving the logical 
 design of switching circuits, the need arises for the solution of 
 simultaneous Boolean equations. Since the application of Boolean 
 algebra to switching theory is a rather recent development, it is 
 not very surprising to find that the amount of work which has been 
 done in this field is relatively sparse. 
 
 This paper is expository in nature, and is intended as a 
 survey of the pertinent literature, to date, dealing with this 
 subject. Numerous examples, different from those in the actual 
 papers cited, are given, and in many cases are solved by various 
 methods in order to compare the efficiency of the techniques, 
 (For further comparison, see references [1], [3], [9l> [19] 3 and 
 [22] for solutions of the "star-delta- transformations" problem.) 
 
 Inasmuch as obtaining the conditions for the existence of 
 solutions verges upon solving the equations, these conditions will 
 not be investigated in great detail. For all problems, the given 
 system of equations will be linear in bivalent variables. Section 
 three presents seven methods for the solution of simultaneous 
 Boolean functional equations, i.e., the determination of dependent 
 variables as functions of independent variables. Particular 
 solutions (i.e., the assignment of a definite value to a bivalent 
 
variable may always be found by simply exhausting all possible 
 combinations of variables; however, for n variables, there are 2 
 possibilities, and so methods must be devised to minimize the labor 
 by reducing the number of potential solutions. Section four consists 
 of three such techniques for Boolean equations, and one for pseudo- 
 Boolean equations. Throughout these two sections, all analogous 
 procedures are noted, and the merits of each approach are 
 summarized in the final division of each section, along with 
 possible applications of the methods. 
 
 Section five contains topics which are either extensions 
 of the methods of the previous sections, or are related in such a 
 manner as to present an area of future exploration. Finally, the 
 list of references provides an extensive bibliography of a great 
 many papers which would be difficult to locate, due to the diverse 
 and unrelated sources. 
 
2. NOTATION 
 
 The notation of the authors of the many works comprising 
 the "basis of this paper varies a great deal, as might be suspected. 
 Thus, it is necessary to adopt a uniform notation to be used 
 throughout this paper, in order to avoid confusion. Table 2.1 
 below contains the notation to be used henceforth, as well as the 
 other various names and representations which may be found elsewhere, 
 
 Table 2.1 
 
 
 
 
 Other 
 
 Symbol 
 
 Type of Operation 
 
 Name 
 
 Symbols 
 
 n 
 
 Set-theoretical 
 
 Intersection (meet) 
 
 
 u 
 
 Set-theoretical 
 
 Union (join) 
 
 + 
 
 
 Set-theoretical; logical 
 
 Complementation; NOT 
 (negation) 
 
 t 
 
 + 
 
 Logical 
 
 OR (inclusive OR; log. 
 sum; disjunction) 
 
 V, © 
 
 • 
 
 Logical 
 
 AND (logical product; 
 conjunction) 
 
 /\ 
 
 © 
 
 Logical 
 
 Exclusive OR (ring; 
 ring sum) 
 
 o, + 
 
The absence of a symbol between elements will be impli- 
 citly understood as logical AND. 
 
 (Note: Throughout this paper, the symbol (J) represents some solution 
 function, as introduced in Sectiop. 3»1; it ia not to be confused with 
 0, the empty set, which was not used in this paper.) 
 
3. GENERAL SOLUTIONS OF BOOLEAN FUNCTIONAL EQUATIONS 
 
 3.1. Basic Theorems 
 
 Almost a ll methods of finding solutions to simultaneous 
 Boolean equations are based, either explicitly or implicitly, on 
 the following well-known theorem, quoted here from a paper by R. L. 
 Ashenhurst : 
 
 Theorem 3«1 » Given any set of relations 
 
 f, = 
 
 1 °1 
 
 f 2 = S 2 , (3.1) 
 
 f = g 
 
 n n 
 
 where the f and g are switching functions of p variables, they 
 
 K. K. 
 
 can be combined into the single equivalent relation 
 
 n 
 
 * ■ J 1 , < f A + W ■ x (3 - 2) 
 
 k=l 
 
 Proof - It is immediately obvious that (j) = 1 if and only if each 
 of its factors is equal to one; moreover, this is true only when 
 f^ = S v (which is exactly the same as the equivalence statement 
 f g + f g = 1). Thus, (3.2) is true only when (3.1) is true, 
 
and vice versa. Therefore, it follows that (}) is a switching funct. 
 of p variables which has the value one for just those sets of values 
 of the variables which represent solutions' of (3.1). 
 
 It will be shown in Section U#l.l that Theorem. 3. 1 leads 
 directly to a simple method for determing a particular solution by 
 means of logical algebra. 
 
 In many problems, particularly those of circuit synthesis, 
 it is desirable to solve for some of the variables as functions of 
 the remaining variables. The conditions for the existence of 
 solutions to such Boolean functional equations, as well as an 
 indication of the method of solution, are given in another theorem 
 by Ashenhurst .. 
 
 Theorem 3*2 . Given the set of relations (3.1), let the p variables 
 consist of r independent variables x ,x , ...,x , and s dependent 
 variables y-,jy g j«.»,y • Then at least one solution of the form 
 
 y-|_ = .^1 ^ x i ? x p' * * " ,X r ' 
 
 y 2 = y 2 vX l' X 2'' , *' X r' ) ' ^ 3 ° 3 ' 
 
 y s =y s (*v x 2 ,..., x r >, 
 
 r 
 exists if and only if all of the p = 2 complete products of the 
 
 variables x ,x , ...,x appear in the canonical form of the function 
 
(|) given "by (3.2). If t. is the number of terms of (j) containing 
 
 J 
 
 the product of x ,x , ...,x corresponding to the integer j , then 
 the number of distinct solutions of the form (3.3) "will he 
 
 P-1 
 
 TT 
 
 ^ * a 
 
 (Author's note: I have substituted the term "complete product", 
 which is synonymous with "minterm", for "fundamental product", 
 the phrase actually used by Ashenhurst, since the definition of 
 the latter varies among recent authors.) 
 
 Proof: If any minterm does not appear in some term of the canonical 
 form of ({), assigning values to the independent variables x ,x , ...,x 
 in such a way that this missing complete product has the value unity 
 will cause (3-1) to have no solution of the form (3.3) > regardless of 
 the values of the dependent variables. Since all other minterms are 
 different combinations of the binary variables, they would thus equal 
 
 zero. Hence the values of (f> for the combination could not be 1. 
 
 r 
 Therefore, only if all p = 2 minterms appear in the canonical 
 
 form can the system of equations have a solution. Furthermore, if 
 
 P-1 
 all products do appear, the TT t. solutions can be constructed 
 
 3=0 
 
by selecting p terms of (j), one containing each product of x ,x , . ..,x , 
 
 and assigning the coefficients in the disjunctive normal form of the 
 
 functions y (x ,x ,.,.,x ) as or 1, depending on whether the 
 K 1 c. r 
 
 variable y, appears primed or unprimed in the corresponding 
 selection term of (|). 
 
 3.2. Logical Algebraic Algorithm 
 
 This algorithm is the direct consequence of theorems 
 3.1 and 3»2, and the steps described here utilize the simple logical 
 algebraic method of R. L. Ashenhurst. 
 
 Problem : Given a system of n simultaneous Boolean equations in 
 the general form 
 
 f.(x ,y ) = g (x ,y ) i =l,2,...,n (3.4) 
 
 X J K 1 J I 
 
 where x. (j =l,2,...,r) are independent variables 
 J 
 
 y (k =l,2,...,s) are dependent variables 
 and r + s = p = total number of variables 
 
 Solve for y (i.e., Equation- 3.3) • 
 
 This is the basic problem for all methods of Section 3# 
 
Algorithm : 
 
 Step I : Find (J), according to (3*2). 
 
 Step 2 : Form a logical table from (j) by writing each term as a 
 binary number, where primed variables are written as 0, unprimed 
 variables as 1. 
 
 Step 3^ Choose p = 2 terms of (t>, one containing each complete 
 
 p-1 
 
 ;here will be 
 
 solutions, according to Theorem 3*2.) 
 
 product of x ,x , ...,x . (Remember, there will be J]" t. 
 
 r j=0 J 
 
 Step k : Treat this table of p terms as a truth table, and form the 
 Boolean function(s) for all y . 
 
 K. 
 
 Example : Solve the system of Boolean equations 
 
 f 1 (x 1 ,x 2 ,y 1 ,y 2 ) = (x ± + y 1 )x 2 y 2 + x^y.^ + l-^ 2 ) 
 
 = ^! x 2 y 2 + ^1 + x 2 ^ y l^2 + ^ X 2 +y 2' )X i y i 
 ^g 1 (x 1 ,x 2 ,y 1 ,y 2 ) (3>5>1) 
 
10 
 
 f 2 (X !' X 2' y l' y 2 ) S (x i X 2 +X l X 2 +y 2 )y l +(x i y l +y 2 )x 2 +X l X 2 y i y 2 
 
 = ( 5 2 + y 1 )y 2 + ( x 2 yi^2 y i )x i 4X L x 2 y i y 2 
 
 - g 2 (x 1 ,x 2 ,y 1 ,y 2 ) (3.5.2) 
 
 Step 1 : Find (J) from (3.2). 
 
 t = ([(x 1 +y 1 )x 2 y 2 +x 2 (y 1 7 2 +y 1 y 2 )] [x^x^+fx^+xpy^+fx^+y^x^] 
 
 + [<x 1 +y 1 )x 2 y 2 +x 2 (y 1 y 2 +y 1 y 2 )][xij:x 2 y 2 +(x 1 +x 2 )y 1 y 2 +(x 2 +y 2 )x y ]} . . 
 
 ( [ (x 1 x 2 +x 1 x 2 +y 2 ) y 1 +(x 1 y 1 +y 2 ) Xg+x^y^] [ (Xg+y^ y^x^+x^) x ± 
 
 + x i x 2 y i y 2^ + ^ X l X 2 +X l X 2 +y 2 )y l +(x i y i +y 2 )X 2 +X l X 2 y i y 2 ] 
 
 ■ tx 2 + y 1 )y 2 + (x 2 y 1 +x 2 y 1 )x 1 +x 1 x 2 y 1 y 2 ] } (3.6) 
 
 * = ^ X l X 2 y i y 2 +X l X 2 y i y 2 +X 2 y i y 2 +X l X 2 y i y 2 +X l X 2 y i y 2 + f X l y i +X 2 +y 2 3 
 
 [ x" 2 +y 1 y 2 +y 1 y 2 ] [x.j+Xg+y^ [ x^-^+y^ [ x^+x^yj } 
 
11 
 
 {x i X 2 y i y 2 +X l X 2 y i +X l X 2 y i y 2 +X 2 y i y 2 +X l X 2 y i y 2 +X 2 y 2 
 
 +X 2 y i y 2 +X l X 2 y lV [(x i X 2 +X l X 2 )y 2 +y l ] f x 2 +y 2 (x i +y i )] 
 
 [ x ] _+x 2 +y 1 +y 2 ] [y 2 +x 2 y 1 ] [ f^ y^ yj [ x^x^y^y^ 
 
 ♦ = ( X l X 2 y i y 2 +X l X 2 y i y 2 +X 2 y i y 2 +X l X 2 y i y 2 +X l X 2 y i y 2 +X l X 2 y i +X l X 2 y i y 2 
 
 +X i y i y 2 +X l X 2 y i y 2 +X l X 2 y i y 2 +X l X 2 y i y 2 +X 2 y i y 2 +X l X 2 y i y 2 
 
 ^ X l X 2 y i y 2 +X l X 2 y i +X l X 2 y i y 2 +X 2 y l y 2 +X l X 2 y i y 2 +X 2 y 2 +X 2 y i y 2 
 
 +X l X 2 y i y 2 +X l X 2 y i y 2 +X l X 2 y i y 2 +X l X 2 y i y 2 +X l X 2 y i y 2^ 
 
 (j) = x 1 x 2 y 1 y 2 +x 1 x 2 y 1 y 2 +x 1 x 2 y 1 y 2 +x 1 x 2 y 1 y 2 +x 1 x 2 y 1 y 2 +x 1 x 2 y 1 y 2 (3.7) 
 
 Step 2 : Logical table for (}) 
 
 X l X 2 y i y 2 
 
 
 
 
 
 1 
 
 
 
 1 
 
 1 
 
 
 
 
 
 1 
 
 1 
 
 
 
 1 
 
 
 
 1 
 
 1 
 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
12 
 
 Step 3 » Choose p = 2 terms. In this case, r = 2, so only U terms 
 
 are required for a solution. Furthermore, there will be 
 
 3 
 
 \~[ t, = 1#1 # 2"2 = k solutions. Thus, we choose 
 3=0 
 
 (a) 
 
 x l 
 
 X 2 
 
 y l 
 
 y 2 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 1 
 
 1 
 
 
 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 Cb) 
 
 x l 
 
 X 2 
 
 y l 
 
 y 2 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 1 
 
 1 
 
 
 
 
 
 1 
 
 1 
 
 1 
 
 
 
 1 
 
 (c) 
 
 x l 
 
 X 2 
 
 y l 
 
 y 2 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 l 
 
 1 
 
 
 
 1 
 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 (d) 
 
 x l 
 
 X 2 
 
 y l 
 
 y 2 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 1 
 
 1 
 
 
 
 1 
 
 
 
 1 
 
 1 
 
 
 
 1 
 
 Step k : Treating these as truth tables yields the solutions 
 
 (a) y 1 = x x x 2 y 2 = X l + X 2 
 
 (b) 7 X 
 
 y 2 = X l + X 2 
 
13 
 
 (c) y = x x y 2 = x 
 
 (d) y x = x-^ y 2 = x 
 
 2 J 2 2 
 
 The solutions may "be checked by simple substitution into 
 the given equations. For example, checking solution (c) yields; 
 
 f 1 (x 1 ,x 2 ) = (x 1 +x 1 )x 2 x 2 +x 2 (x 1 x 2 +x ;L x 2 ) = x^ 
 
 g 1 (x 1 ,x 2 ) = x 1 x 2 -x 2 +(x 1 +x 2 )x 1 x 2 +(x 2 +x 2 )x ]L x 1 = x x x 2 = f 1 (x 1 ,x 2 ) v/ 
 
 f 2 (x 1? x 2 ) = (x 1 x 2 +x 1 x 2 +x 2 )x 1 +(x 1 x 1 +x 2 )x 2 +x 1 x 2 x 1 x 2 = x 1 x 2 +x 1 x 2 =x. 
 
 g 2 (x 1 ,x 2 ) = (x 2 +x 1 )x 2 +(x 2 x 1 +x 2 x 1 )x 1 +x 1 x 2 x ]L x 2 = x 2 +x 1 x 2 +x 1 x 2 = X, 
 
 = f 2 
 
 (x 1? x 2 ) s/ 
 
 This method is very laborious if the number of terms in 
 each function is greater than 2, and/or the number of equations 
 is greater than 2„ 
 
Ik 
 
 3.3 "Map" Methods 
 
 In order to overcome the tiresome task of logical expansion 
 
 [121 T221 
 in the above algorithm, many authors, including Maitra , Svoboda J 
 
 Ledley 1 - , and Nadler , employ a Veitch L chart, (also called 
 
 a logical matrix, binary map, and Boolean matrix) . 
 
 The binary map commonly known as the Veitch chart was 
 first proposed by Marquand in l88l. It employs straight binary 
 distribution, (as opposed to the Gray code in a Karnough map), thus 
 locating the states consecutively, from left to right, and top to 
 bottom. The bars alongside the left and top of the map indicate the 
 rows and columns, respectively, where the variable has the value 
 unity; those without bars have value zero (i.e., the complement of 
 variable has the value unity) . 
 
 Figure 1 is a four variable map, where the assignment of the 
 variables to the rows and columns is indicated to the right and 
 bottom of the map, respectively. Each cell of the map contains 
 the state index, given as the binary number XiX x x.. , and the 
 corresponding decimal notation is given in the second map. 
 
 All of the following methods are based on Theorem 3«1 
 (and in some cases, Theorem 3=2), and since may be easily 
 formulated with the aid of a Veitch chart, these methods possess a 
 high degree of similarity. 
 
15 
 
 x h x 3 
 
 0000 
 
 0001 
 
 0010 
 
 0011 
 
 0100 
 
 0101 
 
 0110 
 
 0111 
 
 1000 
 
 1001 
 
 1010 
 
 1011 
 
 1100 
 
 1101 
 
 1110 
 
 1111 
 
 IK* 
 
 OJ 
 
 CO 
 
 < 
 CM 
 
 CM 
 
 X 
 
 \ x 3 
 V 3 
 
 4»* 3 
 
 
 
 1 
 
 2 
 
 3 
 
 1+ 
 
 5 
 
 6 
 
 7 
 11 
 
 8 
 
 9 
 
 10 
 
 12 
 
 13 
 
 lU 
 
 15 
 
 Figure 1 , 
 
 3.3.1 Svoboda's Algorithm 
 
 Although basically identical to Ashenhurst's method, 
 Svoboda's algorithm can solve a large number of equations containing 
 functions of many variables without great difficulty. Svoboda's use 
 of the Veitch chart is basic, and a Karnaugh map may be substituted 
 if desired, with no loss in any way. 
 
 The problem is the same as that of Section 3.2: Given the 
 system of Equations (3.^+) ? solve for all solutions of y . 
 
 K 
 
 Algorithm : 
 
 Step 1 : Form n maps of Boolean functions 
 
 = f g + f g 
 k k B k k & k 
 
 (k=l,2, . „ . ,n) 
 
 (3.8) 
 
16 
 
 Y S 
 
 Each map will have 2 columns, and 2 rows, where r and s are the 
 numbers of independent and dependent variables, respectively. 
 
 Step 2 : Form the "discriminant" (J), of the system as the Boolean 
 product of ([) (by finding the intersection of the maps - i.e., 
 logical multiplication of corresponding cells in all maps). 
 
 4 = Vyiv^ 
 
 The "discriminant" (j) is identical to (J) in (3.2), and is 
 more simply found graphically than by logical operations. In the 
 map of (j), cells containing ones correspond to the terms found by 
 expanding (j) logically. 
 
 Step 3 : Calculate the "value" [(j)] of the discriminant as 
 
 P-1 
 [d>] = ] ' t. = T (t . = number of l'e per column) 
 
 as in Theorem 3.2. The value T is the product of the numbers of 
 l's in the columns of the maps of (j), and gives the number of 
 distinct solutions. 
 
IT 
 
 Step k : Decompose the map of (j) into all T possible different Boolean 
 
 components (j) j t = 1,2,...,T, where each (j) has only a single 1 per 
 
 12 T 
 column. <|) = (p + (j) +....+ ()) (logical addition for each cell). 
 
 T 
 
 This is equivalent to choosing all sets of 2 complete 
 product terms of (j), as in Step 3 of Ashenhurst's method. 
 
 Step 5 : From each ()) , develop s functions 
 
 Y^ = $\ (k=l,2,...,s) 
 
 by logical multiplication of the graphical representation of y and 
 (j) . This selects those l's in the (j) -map that are also in the 
 1-region of the y -map. 
 
 Step 6 : Transform each Y into y by filling with l's every column 
 
 of Y. containing a 1. 
 k 
 
 Step 7 ; Express each y as a Boolean function by means of the y 
 
 maps, 
 
 These last three steps are equivalent to the usual method 
 
 of forming the Boolean functions y from the truth tables, as in 
 
 K. 
 
 Ashenhurst's last step. 
 
18 
 
 Remarks on the Algorithm 
 
 n 
 
 1. (J) may also be found as ( 2 9 k )> where the union of all <j) is 
 
 K. = X 
 
 plotted instead of the intersection of all (j) ; thus, only a single 
 map is required, and will result in (f). This procedure shortens 
 Step 1 and eliminates Step 2. 
 
 2. If the positions of the dependent and independent variables 
 were reversed (i.e., rotate the map clockwise 90 ), the (f) maps 
 
 could be treated as truth tables, y may be found by obtaining the 
 
 k 
 
 sum of minterms of x's whose columns contain l's (in Step k) which 
 appear in corresponding rows of y . This simple mental procedure 
 eliminates Steps 5 and 6. 
 
 Equations (3*5) will be solved now by Svoboda's algorithm, 
 following his steps exactly. The reader will see how this process 
 may be shortened, in the light of Remarks 1 and 2. 
 
 Step 1 : Form the maps, of f^g.^, ^,g 2 ,<t> 2 (fcigare S) . 
 Step 2 : Form ()) (Figure^. 
 
 Step 3 : [<|>]= TT t. = 1-2-1-2 = k soluti 
 
 ons 
 
 j=0 J 
 

 
 
 
 
 
 1 
 
 1 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 y 2 y l 
 
 
 1 
 
 
 
 1 
 
 
 1 
 
 1 
 
 
 I 
 
 1 
 
 1 
 
 
 
 1 
 
 
 Y 2 y i 
 
 1 
 
 
 1 
 
 1 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 
 1 
 
 y 2 y l 
 
 1 
 
 1 
 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 1 
 
 
 1 
 
 1 
 
 
 y y 
 
 2*1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 
 
 
 1 
 
 
 1 
 
 1 
 
 
 y y 
 
 2"1 
 
 1 
 
 1 
 
 
 
 1 
 
 1 
 
 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 y 2 Y l 
 
 Figure --2, 
 
 
 
 1 
 
 
 
 
 1 
 
 
 1 
 
 
 
 1 
 1 
 
 
 1 
 
 1 
 
 1 
 
 
 
 
 1 
 
 y 2 y i 
 
 19 
 
 e .3 
 
Step k : Decompose (j) into <|> ', t = 1,2,3,U (Flgus^' I) . 
 
 20 
 
 1 
 
 
 
 
 
 1 
 
 
 
 
 
 _1_ 
 
 1 
 
 ~ 1 
 
 1 
 
 
 1 
 
 
 
 
 
 1 
 
 
 ! 
 
 ... 
 
 1 
 I" 
 
 1 
 
 1 
 
 
 1 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 1 
 
 1 
 
 
 
 1 
 
 
 
 
 1 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 1 
 
 Figure J+. 
 
 Step 5 : Find Y^ = <j> y fe k = 1,2,..,, s (Figure 5). 
 
 '1 
 
 k 0. 
 
 Ill 
 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 — 
 
 A 
 
 
 
 
 
 
 l 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 i 
 
 i 
 
 y, 
 
 y; 
 
 
 
 
 
 
 — 
 
 — 
 
 T 
 
 i 
 
 
 — 
 
 i 
 
 i 
 
 T 
 
 1 
 
 
 
 
 r T~ 
 
 — 
 
 
 
 
 
 
 I 
 i 
 
 
 
 
 ! 
 i~ 
 
 "i 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 i 
 
 
 1 
 
 1" 
 
 1 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 i 
 
 . I j 
 
 i 
 
 Figure* 5. 
 
Step 6 : Y into y (Figure* 6) . 
 
 21 
 
 y. 
 
 y-, 
 
 3 
 
 
 I 1 
 
 t 
 
 1 
 
 1 
 
 ! 
 
 
 1 
 
 J 
 
 1 
 
 I 1 ! 
 
 \ 
 
 n 
 
 [jr 
 
 ~P 
 
 1 
 
 1 
 
 I 1 
 
 1 
 
 y i = X 1 X 2 
 
 7 ± - 
 
 y l = X 1 
 
 y-L - Xl x 2 
 
 y. 
 
 y. 
 
 y; 
 
 y. 
 
 
 jl 
 
 1 
 
 
 1 
 
 1 
 
 
 
 1 
 
 1 
 
 
 1 
 
 1 
 
 
 
 TIT 
 
 Tl 
 
 : 
 
 
 i i 
 
 i 
 
 
 
 rrrr 
 
 i 
 
 ! 
 
 
 i| i 
 
 i| 
 
 I 1 
 
 l 
 
 \T 
 
 1 
 
 1 1 
 
 1 
 
 11 H 
 
 J 2 2 
 
 y 2 - x 1+ x 2 
 
 y 2 = x 2 
 
 J 2 12 
 
 Fi£ure_6, 
 
 Step 7 * Find y as a Boolean function. 
 
22 
 
 Singular Cases 
 
 Svoboda does add an important case not treated by 
 Ashenhurst. As noted earlier, if [<pj = 0, i.e., a column of 9 
 has no units, no solution of the form (3.3) exists. However, there 
 does exist a solution of this form which is combined with a constraint 
 of the form c(x.) = 1, which restricts the domain of possible 
 
 J 
 
 solutions. 
 
 If a column of zeros is present, the combination of 
 binary variables that makes the corresponding minterm of x's equal 
 to 1 must be excluded. I.e., the corresponding minterm must always 
 be zero. The complement of the equation is the constraint mentioned 
 above. 
 
 If more than one column of zeros are present, the sub- 
 sequent constraint equations are simply multiplied to obtain the 
 total constraint. Furthermore, the entries of the columns may be 
 treated as don't-cares for simplification of the solution(s) . 
 
 Example 2 : Given the system of Boolean equations, 
 
 f l ( X l'V y i' y 2 } S (x i +y 2 )? 2 +5r i y i y 2 +X i y i y 2 +5r i X 2 y i y 2 
 
 = y i +X 2 y 2 = 6 1 (x 1 ,x 2 ,y 1 ,y 2 ) 
 
23 
 
 f 2 (x i' X 2' y i' y 2 ) = X l X 2 +y i y 2 +X i y i y 2 +X 2 y i^ 
 
 x l X 2 +x l^ X 2 +y i^ ~ g 2 ^ X l' X 2 jy l' y 2' ) 
 
 The reader may verify that the discriminant is Figure J, 
 
 
 1 
 
 
 
 1 
 
 
 
 
 1 
 
 
 
 
 
 
 
 1 
 
 W= o 
 
 Figure 7. 
 
 Steps J+,5 5 and 6 lead to the y maps (Figure 8) 
 
yn 
 
 
 1 
 
 
 d 
 
 l 
 
 1 
 
 JL 
 
 1 
 
 
 d 
 
 l 
 
 1 
 
 
 d 
 
 l 
 
 1 
 
 
 d 
 
 1 
 
 y-, = 
 
 x i + X 2 
 
 :l- 
 
 
 
 
 — 
 
 
 
 
 d 
 
 1 
 
 1 
 
 
 
 d 
 
 1 
 
 
 
 
 d 
 
 1 
 
 1 
 
 
 
 d 
 
 1 
 
 y 2 =x 2 
 
 Figure 8, 
 
 
 
 d 
 
 1 
 
 
 
 d 
 
 1 
 
 
 
 d 
 
 1 
 
 
 
 d 
 
 1 
 
 y l =X 2 
 
 :ir 
 
 1 
 
 
 d 
 
 1 
 
 1 
 
 
 d 
 
 1 
 
 1 
 
 
 d 
 
 i 
 
 1 
 
 
 d 
 
 1 
 
 y 2 = x 1+ x 2 
 
 Cons 
 
 traint: x_x,_ = or x_ +x^ = 1 (for both solutions) 
 12 12 
 
 2k 
 
 Check: (Solution 1 in Equation l) y- = x , +x y = x x +x = 1 
 
 f l = x i X 2 +X 2 X 2 +X l^ X l +X 2^ X 2 +X l^ X l +X 2^ X 2 +X l X 2^ X l X 2' )X 2 
 
 (x^+x^) (1) = (x 1 x 2 +x 1 x 2 ) (x 1 +x 2 ) = x x x 2 
 
 g l = X lV X 2 X 2 = X 1 X 2 
 
25 
 
 3.3-2 Maitra's Methods 
 
 In many problems (particularly those of circuit design) , 
 it may he necessary to solve n equations of the form 
 
 g k (x 1 ,x 2 ,...,x rnJ x m+1 ,.,x r ) = f k ( yi ,y 2 ,...,y i3 ,x m+1 ,...,x r ) 
 
 (k = 1,2,..., n) (3.9) 
 where the solutions are to have the form 
 
 y i = y 1 ^ v ^""\ ) 
 
 y 2 = y 2 ^ X l' X 2''°' ,X m' ) (3.10) 
 
 • • 
 
 ,; - y (J (x 1 ,x 2 ,...,x m ) 
 
 y 
 
 This is just a special case of (3.^0 • The situation under 
 consideration is shown in Figure 9« 
 
 The general approach to a system of equations like (3-9) 
 is to solve each equation independently, and then find the common 
 solution for all the equations. 
 
 If one allows the y's to be given explicitly, considering 
 k = 1 for (3.9)5 there are, altogether, three basic problem types 
 which may arise: 
 
26 
 
 1. Given the functions f and g explicitly, determine the y.'s. 
 
 J 
 
 2. Given the functions f and y.'s explicitly, determine g. 
 
 J 
 
 3. Given the functions g and y.'s explicitly, determine f. 
 
 J 
 
 X n X...X X , . , X X-, X . o .X 
 
 12m m+1 r 12m 
 
 y l (x i' X 2'-' X J 
 
 I 
 
 y, 
 
 y 2 (x 1 ,x 2 , c ..,xj 
 
 I 
 
 f k ( yi ,...,y ,x m+1 ,...,x r ) 
 
 g k (x 1 ,x 2 ,„ o0 ,x r ) 
 
 Figure £. 
 
27 
 
 K. K. Maitra's approach is also based on Theorem 3.1, and 
 is identical in principle to that of Svoboda. However, since this 
 is a special case of the problems previously considered, it has 
 fewer degrees of freedom. Consequently, a modified Veitch chart 
 in tabular form is used to calculate (j) , and to solve the three 
 problems listed above. 
 
 Solution for problem of type 1 
 
 If k = 1 in (3«9)j since the y's are to be expressed in 
 
 terms of x ,x , ...,x , the single equation f = g must be decomposed 
 
 into a system of equations so that (3.9) holds for all combinations 
 
 of the r - m independent variables x ,,,..., x . Hence, there will 
 
 m+1 r 
 
 r-m 
 result a system of 2 equations f = g ,f, = f n ,...,f = g 
 
 J * o o' 1 1 ' p r-m p r-m 
 
 to be solved simultaneously for y 1? y , ...y . Let the index of the 
 
 resulting functions f, and g, be the decimal equivalent of the 
 
 k k 
 
 binary number corresponding to the complete product of x , ...,x 
 which determines the equation f = g . 
 
 Algorithm : 
 
 r ~m 
 Step 1 : Set up a modified Veitch chart for each of the 2 
 
 equations as follows : 
 
 (a) Designate the columns and rows by the binary numbers corres- 
 
 H m 
 
 ponding to the 2 and 2 combinations of the dependent variables 
 
28 
 
 y ,y , ..o,y/ and independent variables x ,x , ..,x , respectively, 
 or their decimal equivalents. (See Remark 2, Svoboda's algorithm, 
 Section 3.3.1.) 
 
 (b) List the values of the functions f and g , corresponding to the 
 minterms of y ,y , * . . ,y,? and x >x . . jX , along the bottom and right 
 side of the chart, respectively (see Figure 10). 
 
 Step 2 : Form the (b chart by "multiplying" the values of f and 
 g according to the following rules: 
 
 1.1 = 1, 0-0 = 1, 1.0 = 0, 0.1 = (3.11) 
 
 This is precisely due to the equivalence statement of (3.2) , and 
 results in the same maps as those in Step 1 of Svoboda's algorithm. 
 
 Step 3 - Form the (J) chart as the Boolean product of the © (identical 
 
 K. 
 
 to Svoboda ' s Step 2) » 
 
 Step k : Calculate the number of solutions, as [(f) 1 = T (identical 
 to Svoboda's Step 3, but now t. = number of l's per row). 
 
 J 
 
 Maitra does not detail how to obtain the [(j) ] solutions as 
 Boolean functions, but obviously the method parallels that of 
 
29 
 
 Svoboda and Ashenhurst. Thus, "by decomposing <J) into <j) charts, and 
 treating each as a truth table, the functions y ,y , ...,y ) may be 
 found (see Remark 2 in Svoboda' s algorithm). Also, if any row has 
 no units, no solution of the form (3.10) exists. However, this may 
 be used to form a solution with a constraint, as in Svoboda' s 
 treatment of singular cases. 
 
 Example : Given the system of equations 
 
 f 1 (y lS y 2 ,x 1+ ,x 5 ) = y 1 y 2 V5 +y i y 2 ( V5 + V5 )+x i + y 2 (x 5 y l +x 5 y l ) 
 
 +x^ yi (x 5 y 2 +x 5 y 2 ) 
 
 = x 3 (x 1+ x 5 +x 1+ x 5 ) +x 2 x 3 x [ ^(x 1 x 5 +x 1 x 5 ) +x 2 x 3 x ]+ (x 1 x 5 +x 1 x 5 ) +x 3 * u x 5 • 
 
 (xjX^+x^) +x 3 x i ^x" 5 (x 1 x 2 +x 1 x 2 ) = g 1 (x 1 ,x 2 , . . . ,x ) (3.12.1) 
 
 f 2 ( yi ,y 2 ,x u ,x 5 ) h y i y 2 (x 1+ +x 5 )+y 1 y 2 (x^x 5 +x 1+ x 5 )+x 5 y 1 (x 1+ y 2 +x u y 2 ) 
 
 + X 5 y 2^ x ^ y i +X u y i ) ^i^Hv^V 
 

 d)° 
 
 i yi y 2 
 
 (o) (D (2) (3) 
 
 *i 
 
 
 X 1 X 2 X 3 
 
 
 
 00 
 
 1 
 
 01 
 
 10 
 
 11 
 
 
 
 
 
 00 
 
 01 
 
 10 
 
 11 
 
 g 
 
 (o) 
 
 
 
 1 
 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 (1) 
 
 1 
 
 
 
 1 
 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 (2) 
 
 10 
 
 1 
 
 
 
 1 
 
 
 
 
 
 2 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 (3) 
 
 Oil 
 
 
 
 1 
 
 
 
 1 
 
 ]_ 
 
 3 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 M 
 
 10 
 
 1 
 
 
 
 1 
 
 
 
 
 
 k 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 (5) 
 
 10 1 
 
 
 
 1 
 
 
 
 1 
 
 1 
 
 5 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 (6) 
 
 110 
 
 1 
 
 
 
 1 
 
 
 
 
 
 6 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 (7) 
 
 111 
 
 
 
 1 
 
 
 
 1 
 
 1 
 
 7 
 
 1 
 
 T_ 
 
 1 
 
 1 
 
 
 
 10 1 
 
 
 
 Figure 10, 
 
 % 
 
 
 1 
 2 
 3 
 1+ 
 5 
 6 
 7 
 
 1 
 
 
 
 1 
 
 
 
 
 
 1 
 
 
 
 1 
 
 1 
 
 
 
 1 
 
 
 
 
 
 1 
 
 
 
 1 
 
 1 
 
 
 
 1 
 
 
 
 
 
 1 
 
 
 
 1 
 
 1 
 
 
 
 1 
 
 
 
 
 
 1 
 
 
 
 1 
 
 1 
 
 
 
 1 
 
 
 
 1 
 
 
 
 1 
 
 
 
 
 
 1 
 
 2 
 
 3 
 1+ 
 5 
 6 
 
 7 
 
 
 
 1 
 
 1 
 
 1 
 
 
 
 1 
 
 1 
 
 1 
 
 
 
 1 
 
 1 
 
 1 
 
 
 
 1 
 
 1 
 
 1 
 
 
 
 1 
 
 1 
 
 1 
 
 
 
 1 
 
 1 
 
 1 
 
 
 
 1 
 
 1 
 
 1 
 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 1 
 1 
 1 
 1 
 1 
 1 
 
 1 1 
 
*L 
 
 31 
 
 
 
 1 
 2 
 
 3 
 
 k 
 
 5 
 6 
 
 7 
 
 
 
 
 
 1 
 
 
 
 
 
 1 
 
 
 
 1 
 
 
 
 
 
 1 
 
 
 
 
 
 1 
 
 
 
 1 
 
 
 
 
 
 1 
 
 
 
 
 
 1 
 
 
 
 1 
 
 
 
 
 
 1 
 
 
 
 
 
 1 
 
 
 
 1 
 
 Solution matrix 
 for (J).. (Note: 
 
 if only 1 equation 
 
 had been given 
 
 originally, (j>i "would 
 
 represent the 
 
 l*2*l-2*l'2'l-2 = 8 
 
 different solutions). 
 
 1 
 
 1 
 
 
 1 
 2 
 
 3 
 k 
 
 5 
 6 
 
 1 
 
 f-. 
 
 o 
 
 1 
 
 
 
 1 
 
 
 
 
 
 1 
 
 
 
 1 
 
 1 
 
 
 
 1 
 
 
 
 
 
 1 
 
 
 
 1 
 
 1 
 
 
 
 1 
 
 
 
 
 
 1 
 
 
 
 1 
 
 1 
 
 
 
 1 
 
 
 
 
 
 1 
 
 
 
 1 
 
 10 1 
 
 
 
 1 
 
 
 
 1 
 
 
 
 1 
 
 
 
 1 
 
 
 
 1 
 
 1 
 
 1 
 
 \ 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 2 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 3 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 k 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 5 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 6 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 7 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 f l 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 Figure 10. 
 
 
 
 
 
 
• 
 
 
 
 
 r 2 
 
 
 
 
 
 
 1 
 
 2 
 
 3 
 
 g 
 
 
 
 1 
 
 1 
 
 1 
 
 
 
 1 
 
 1 
 
 
 
 
 
 
 
 1 
 
 
 
 2 
 
 1 
 
 1 
 
 1 
 
 
 
 1 
 
 3 
 
 1 
 
 1 
 
 1 
 
 
 
 1 
 
 k 
 
 1 
 
 1 
 
 1 
 
 
 
 1 
 
 5 
 
 
 
 
 
 
 
 1 
 
 
 
 6 
 
 1 
 
 1 
 
 1 
 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 1 
 
 f« 
 
 1 
 
 1 
 
 1 
 
 
 
 
 
 
 1 
 
 2 
 
 3 
 
 • 
 
 
 
 
 
 1 
 
 1 
 
 
 
 
 
 1 
 
 1 
 
 
 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 
 
 
 
 
 
 1 
 
 1 
 
 
 
 
 
 1 
 
 1 
 
 
 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 
 
 fa 
 
 1 
 1 
 1 
 
 1 
 1 
 1 
 
 
 11 
 
 
 
 1 
 
 2 
 
 3 
 
 k 
 
 5 
 6 
 
 7 
 
 
 
 1 
 
 2 
 
 3 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 1 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 1 
 
 
 
 
 
 1 
 
 
 
 
 
 Figure 10, 
 
33 
 
 V VV X 3 V + V5 ( V x 3 } + V X 1 V x 2 V x 3 V 
 
 5 2 (x 1 ,x 2 ,.,.,x 5 ) (3.12.2) 
 
 Steps 1 and 2 : The maps of <j) are formed in Figure 10. Note that 
 this is a given system of two simultaneous equations, but since 
 r-m - 5-3 = 2, there will also be 2 = k simultaneous equations per 
 given equation. (Actually, Maitra treats only this latter set as 
 his simultaneous system, and works with only one given equation,,) 
 
 Step 3: Form (t> (Figure 11). 
 
 y l y 2 
 
 xxx 00 01 10 11 
 
 
 1 
 10 
 Oil 
 10 
 10 1 
 110 
 111 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 1 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 1 
 
 
 
 1 
 
 
 
 
 
 
 Figure 
 
 11 
 
 • 
 
 
3^ 
 
 Step k: [(()] = 3i , l , l'l , l'l*l , l = 1 = T; therefore a unique solution 
 exists. 
 
 Treating Figure 11 as a truth table, the unique answer is 
 immediately seen to be 
 
 y l = V x 3 
 
 V *= X 
 
 y 2 3 
 
 Check: A sample check will be shown for x, = x = 1: 
 
 f l = ( X 2 +X 3^3Y5 + ^2 X 3^3^¥5 + Y5^¥3^5 X 2 +X 5 X 3 +X 5 X 2 X 3^ 
 
 + x^(x 2 +x 3 )(x 5 x 3 +x 5 x 3 ) 
 
35 
 
 = x 2 x 3 +x 2 x 3 +x 3+ x 2 x 3 = 1 
 
 >1 = X 3 +X 1 X 2 X 3 +X 1 X 2 X 3 +X 1 X 2 X 3 +X 1 X 2S = l ^ 
 
 f 2 = ( X 2 +X 3 ) X 3 +X 2 X 3( X 3^ ^ +X 3^ X 2 +X 3) = X 2 X 3 +X 3 = x 2^ 
 
 g 2 = x 2 + x 3 v/ 
 
 Similar checks for x. = 0, x = 1, x, = 1, x = 0, x, = x = 
 show that the solution checks identically. 
 
 Solution for problems of type 2 
 
 Given the functions f and the y.'s, it is desired to 
 
 J 
 
 determine the function g explicitly. The given conditions imply 
 
 that the values of f, and all the elements of the matrices are known, 
 
 k 
 
 (For further insight, see Ledley's solution of this problem.) 
 Furthermore, since the y.'s are independent of x . ,...,x , the 
 
 matrices corresponding to different pairs of f's and g *s 
 
 1 2 9 h 
 (i.e., (J) ,<J) ,<j) ,(|) in Example ) will be identical (for given y.'s). 
 
 J 
 
 Hence, to find the values for g , it is only necessary to multiply 
 the rows of matrix elements by the corresponding elements of f 
 
36 
 
 according to (3. 11). Note that if the value of g obtained for 
 
 each row is not unique, no solution of the original equation 
 
 This problem may also be solved very easily by substitution of the 
 
 y.'s into f, and expanding by the usual logical operations. In this 
 
 way, it may be seen that a solution always exists for this type 
 
 problem. 
 
 Solution for problems of type 3 
 
 Given the functions g and y.'s, it is desired to determine 
 
 J 
 
 the function f explicitly. Once again, the given conditions imply 
 that the values of g and the elements of the single (j) matrix are 
 known. Hence, f is found by the multiplications of columns of 
 matrix elements by the corresponding elements of g , Also, if the 
 value of f obtained for each column is not unique, no solution 
 exists for the original equation* However, a solution with a 
 constraint may be found, but Maitra makes no mention of this. 
 
 3.3.3 Ledley's Methods 
 
 Robert Ledley's methods incorporate many "new" techniques, 
 such as the use of "designation numbers" and solution by matrix 
 computation. However, when closely examined, it may be seen that 
 he is also using the same ideas as those in the previous sections. 
 Also, the three problem types mentioned in Section 3.3.2 will be 
 
37 
 
 solved, and finally non-matrix solutions for general equations will 
 be shown. Before proceeding, we must examine Ledley's notation. 
 
 Terminology : All of the methods are based upon Ledley's digitali- 
 zation of Boolean algebra. Every possible Boolean function (of the 
 given r variables) is assigned a unique binary number of 2 bits; 
 this number is called the "designation number" and is written 
 "#". The designation numbers are first assigned to the r variables, 
 or elements. Such an assignment is called a "basis", and is 
 written "b[ ]". A basis for a 3-element system is 
 
 0123 ^567 
 
 #x = 0101 0101' 
 #x 2 = 0011 0011 
 #x = 0000 1111 
 
 - b[x 1 ,x 2 ,x ] 
 
 (3.13) 
 
 where the decimal numbers designate the column position. Thus, the 
 basis is just a list of all possible binary combinations of the 
 elements. The basis of (3°13) is written b[x-^ ,x ,x ] and is called 
 a "standard" basis, and it will be assumed that all designation 
 numbers refer to standard bases, unless otherwise indicated. Note 
 that the elements are in "natural" order. All standard bases may 
 be easily formed by alternating 0's and l's in the first row, pairs 
 

 of O's and l's in the second, four O's and four l's in the third, 
 etc Observe that a standard basis is nothing more than the BCD 
 code rotated 90 counterclockwise. The significance of the bits 
 is more obscure in a designation number than in a Veitch chart, but 
 the former permits greater ease in many computations. Thus, in a 
 standard basis, the column number in (3. 13) corresponds to the 
 binary representation of the column, when read from bottom to top. 
 Finally, subscripts distinguish between the elements (e.g., x n ,x ), 
 and superscripts denote the column of the bit in a designation number 
 
 (e.g., x 3 = 0, x x = 1) 
 
 Computations with designation numbers 
 
 In order to find the designation number of a given function, 
 it is only necessary to carry out the logical operation on the 
 elements, as indicated by the function. For example, suppose it is 
 desired to find x., + x and x x„ (where complementation is simply 
 the replacing of l's by O's, and vice versa). From (3«13)> 
 
 hi = 0101 
 1 
 
 0101 
 
 #ic = 1010 
 
 1010 
 
 £ g = 1100 
 
 1100 
 
 #x = 0000 
 
 mi 
 
 #(x 1 +x 2 ) = 1101 1101 #(x x ) = 0000 1010 
 
39 
 
 It is most important to note that each designation number corresponds 
 to a unique function (simplification not withstanding). Expressed 
 in another way, this says 
 
 #x = #y iff x=y (3.14) 
 
 Another important rule is #1 = 1111.... 1 (number of bits 
 depends on the basis), where I is the function whose value is 
 always 1. 
 
 Boolean matrix multiplication is denoted by '®", and is 
 equivalent to regular matrix multiplication except logical addition 
 and multiplication are used. 
 
 The last step in all the methods to be developed is 
 changing back from the solution designation number to the Boolean 
 function of the solution. This may be done in any of the following 
 ways: 
 
 1. Disjunctive normal form (first canonical form) - This is simply 
 
 a sum of complete products. For a basis of r elements, there exist 
 
 r 
 2 different complete products. Furthermore, the designation 
 
 number of a complete product always has a single 1. For the basis 
 
 (3.13), we have 
 
1+0 
 
 #(x 1 x 2 x 3 ) 
 
 = 
 
 0000 
 
 0001 
 
 #(V2 X 3 } 
 
 = 
 
 0000 
 
 0010 
 
 #(x 1 x 2 x 3 ) 
 
 = 
 
 0000 
 
 0100 
 
 #(x ] _x 2 x 3 ) 
 
 = 
 
 0000 
 
 1000 
 
 #(x x x 2 x 3 ) 
 
 = 
 
 0001 
 
 0000 
 
 #(x 1 x g x 3 ) 
 
 = 
 
 0010 
 
 0000 
 
 #(x 1 x 2 x ) 
 
 = 
 
 0100 
 
 0000 
 
 #(x x x 2 x 3 ) 
 
 = 
 
 1000 
 
 0000 
 
 Thus, this form is the sum of the complete products which 
 contribute l's to the designation number. 
 
 2. Conjunctive normal form (second cononical form) - This is a 
 product of complete sums. There are 2 different complete sums, 
 each one containing only a single zero (the reader should verify 
 this). The Boolean function is thus formed by taking the product 
 of those complete sums -which contribute a zero to the designation 
 number. 
 
 3. Veitch chart - A Veitch chart may be used quite successfully 
 for the simplification of a Boolean function from its designation 
 number. For example, a l6-bit designation number and its 
 corresponding k- element standard basis are: 
 
hi 
 
 x l 
 
 = 0101 
 
 0101 
 
 0101 
 
 0101 
 
 X 2 
 
 = 0011 
 
 0011 
 
 0011 
 
 0011 
 
 X 3 
 
 = 0000 
 
 1111 
 
 0000 
 
 1111 
 
 x h 
 
 = 0000 
 
 0000 
 
 1111 
 
 1111 
 
 #y = 0111 0111 
 
 1100 
 
 1100 
 
 The corresponding Veitch chart (see explanation below) is Figure 12 
 
 10 
 01 
 11 
 
 00 10 01 1]. 
 
 
 
 
 1 
 
 
 1 
 
 1 
 
 
 
 
 1 
 
 
 1 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 1 
 
 1 
 
 
 1 
 
 
 
 
 
 y = x 2 x l| +x 2 x 1+ +x 1 x 2 
 
 Figure 12. 
 
 By comparing b [x ,x ,x ,x, ] and #y with Figure 1 and Figure 10, it 
 may be seen that the cells of the chart correspond to the successive 
 groups of four binary combinations of the basis. Furthermore, the 
 cells of the chart may be filled by "folding" the designation number, 
 whence each group of four bits becomes a row. The superiority of 
 a Karnaugh map to a Veitch chart for ordinary simplification is 
 lost here, since the former cannot be formed directly from the 
 
1,2 
 
 designation number, as can the latter. Thus, the use of a Veitch 
 chart will occur frequently here. 
 
 ko Other methods - Ledley also demonstrates various methods which 
 are more artificial and complex than those above. Among these are 
 the generation of "mongrel forms", the use of prime implicants, 
 and "elimination pairs" to determine "included" and "non- included" 
 elements • 
 
 Formation of matrices 
 
 Returning to the three problem types of Section 3»3.2 we see 
 that it will be necessary to form matrices corresponding to the 
 given functions, and then calculate the solution. The method of 
 solution is summarized below, 
 lo List the given Boolean functions. 
 2. Find their designation numbers. 
 3o Form corresponding Boolean matrices <, 
 
 k. Compute the Boolean solution matrix with aid of the fundamental 
 formulas » 
 
 5. Find solution designation number. 
 
 6. Derive explicit Boolean function of solution. 
 
 The three matrices of interest are [g ], [f J, and [(K ], 
 which are defined as follows: 
 
h3 
 
 [6^3: (a) Form #g from b[x 1 ,x 2 ,...,x m ,x m+1 ,...,x r J . (This applies 
 to problem types 1 and 3- ) 
 
 (b) Separate the positions of #g into 2 groups of 2 
 bits per group. 
 
 (c) Index the groups by k (k = 0,1,..., 2 -l) . These 
 
 form the rows of [g n ] . 
 
 D km 
 
 (d) Index the bits or positions by m (m = 0,1,.,., 2 -l) . 
 
 These form the columns of [g, ] . 
 
 km 
 
 Note that these steps simply convert the designation number to a 
 Veitch chart, as was shown above. For that example, the indexing 
 would be 
 
 m 0123 0123 0123 0123 
 #g = 0111 0111 1100 1100 
 
 k 1 2 3 
 
 [f /,]: The method is the same as for [g, .', except b [y^,y , ..., 
 yp 9 x ...... ,x 3 is used, (This applies to problem types 
 
 1 and 2.) 
 
 [Of ]: This is for use in problem types 2 and 3. 
 
 (a) Form the #y's from.b [x.. ,x , . .,,x ] ., , and place. In 
 
 successive rows, in numerical order (i.e., #y-,,#ypj...) 
 
Uk 
 
 (b) Index the columns of b [x-,x , ...,x .1 by m(m=0,l, o .., 
 2 m -l) . 
 
 (c) The ^ index is written below each column of the 
 array of #y's, where % is the decimal number equivalent 
 to the binary number formed by the column, when read 
 from bottom to top. 
 
 (d) An element (J),, of [ (JK 1 is a one for all corresponding 
 pairs of the 'I and m indeces found above. So, if Column 3 
 of the basis corresponds to I =1.1 in the #y's array, 
 
 the element d)_ will be a unit, etc. Thus, there will 
 be the 2 "units in [ (L .! , one in each column, all 
 ■'V '• •••'-. other elements being zero. 
 
 Derivation of Boolean function from matrices 
 
 #g: This is found by reversing the procedure for finding 
 
 the [ g, ] matrix. Simply unfold the matrix to obtain 
 km 
 
 the designation number, and then use one of the 
 
 aforementioned procedures. 
 
 #f: Reverse the procedure for finding the [ f » ] matrix. 
 
 -//y's: r ' In this case, multiple sets of solutions are possible. 
 
 A result array consisting of 2 'columns indexed by 
 
 m(m = 0,1, «.„<,2 -l) and i rows indexed by y„,y , ,.., 
 
 y, is formed, (The y.'s may be found immediately by 
 Hi j 
 
 treating [ <b . .' as a Veitch chart,) For the pairs of 
 
^5 
 
 indices $> 9 m for -which typ = 1, place the .'Jth 
 column of b [ y ,y , . . . ,y, J in the mth column of the 
 result array. The rows of the result array will 
 consist of the #y,> ' s, with respect to b [x ,x , ..., 
 x ] , and the explicit Boolean functions may be 
 obtained. 
 The methods of this and the previous section will be 
 demonstrated in subsequent examples. 
 
 Antecedence and Consequence Solutions 
 
 In logical problems , there are two types of solutions to 
 a given equation; these are antecedence and consequence solutions. 
 An antecedence equation implies the given equation; in other words, 
 the truth of an antecedence solution is sufficient for the truth of 
 the given equation. Moreover, an antecedence solution will be false 
 for at least all input conditions for which the given equation will 
 be false, but it is not necessarily equivalent to the given equation,, 
 
 Consequence solutions can be deduced from (i.e., are implied 
 by) the given equation; hence, the truth of a consequence solution 
 is necessary for the truth of the given equation. Also, a 
 consequence solution will be true for at least all input conditions 
 for which the given equation is true, but it is not necessarily an 
 equivalent equation, 
 
U6 
 
 If f is an antecedence solution, f a consequence solution, 
 
 and g the given function (f ,f ,g are functions, as before), the above 
 
 a c 
 
 conditions are expressed as 
 
 antecedence: f — ^ g 
 a 
 
 consequence: f — ^ f 
 If f — y g — ^f , and f = f = f, then f = g. Thus, any solution 
 which is both an antecedence and consequence solution -will be the 
 solution of the given equation. 
 
 The Fundamental Formulas 
 
 Ledley presents three pairs of Boolean matrix equations, an 
 antecedence and consequence formula for each of the three problem types. 
 (For detailed derivation of these formulas, see [10] , pp. U28-M+3.) 
 In the formulas below, the subscripts a and c denote antecedence or 
 consequence solutions, respectively. 
 
 Table 3.1 
 
 iProblem Type Antecedence Solution 
 1 
 
 [f . ]® [g, J = [<k J a 
 !6\l km T ^m 
 
 [t u ] ®K ] ■ ^ 
 
 [ *£» ] 0^k J = [f ik ] a 
 
 Consequence Solution 
 
 [TV. J®Cg. ] =L£, lei 
 #k km r ^m 
 
 [1 u ] ® [ hJ ^km ] ; < 3 - 15) 
 
 Lh -JOCeL, J J*tJ 
 
 an 
 
 ^k 
 
 Jlk 
 
^7 
 
 Properties of the Solutions 
 
 Type 1 problems : The solution will be a set of functions „ It is 
 possible that many sets of functions will exist; however, it is 
 also possible that none may exist. In the latter case, a 
 constraint may be found among the independent variables so that 
 a solution does exist (as in Svoboda's algorithm). The constraint 
 may be found as a Boolean row vector from 
 
 where L is a row vector of all units. Thus, [c 1 = #c. referred 
 
 to b tx ,x , . „.,x 1 . New bases constrained by c (where the 
 
 columns having zeros in #c are disallowed) must now replace 
 
 b [x n ,x . . o . ,x J and b [x n ,x_,....x ] , and the solution is 
 12 r 1 2 m ' 
 
 found as previously described, except m = 0,1,... u-1, where u 
 
 is the number of units in Lc J . 
 
 m 
 
 Type 2 problems : The antecedence and consequence solutions for 
 
 g are the same, so that either equation may be used. For this type 
 
 only, we always have f = g. 
 
 Type 3 problems : There is always at least one antecedence or 
 consequence solution, and it is in the form of a single function. 
 
If there are N solutionis of either type> there will be one which is 
 
 termed "closest" to g. If such a solution is represented by f f; then 
 
 f — ♦ f — !► g (antecedence solutions. n=l,2,... ,Nj 
 n m ' 
 
 g — > f — ^f (consequence solutions, n=l,2,...,N; 
 
 Constraints may also be found between the given functions. The 
 constraint will be found as both an antecedence and consequence 
 solution; if none exists^ the functions are independent. It is 
 found by substituting the single column vector I for [g 1 in 
 the Type 2 fundamental formula a 
 
 Example of T ype 1. Problem 
 
 Equations (3.12) will be solved using Ledley's method, 
 
 Given: Equations 3-12- 
 
 Find: y^y^ 
 
 Step 1 : Find designation numbers of f . ^f^g.. ,g , 
 
 f 
 
 1 
 
 = W^VM^ + V/ : Vl+ X 5 y l } +X 4 y l (x 5 y 2 +X 5 y 2 } 
 
 f 2 = y l y 2 (x i4 +X 5 ) +y l y 2 ( VVV9 +X 5 y l (x uW2 } +X 5 y 2 (x ^ y i + Vl ) 
 
 +y 1 y 2 'x i+ x r; -»-x i4 x 5 ) 
 
^9 
 
 b[ yi ,y 2 , V x 5 ] 
 
 #y = 0101 0101 0101 0101 
 
 #y = 0011 0011 0011 0011 
 
 #x^ = 0000 1111 0000 1111 
 
 #x = 0000 0000 1111 1111 
 
 #f = 0011 1100 0000 0111 
 
 #f = 0011 1110 1111 0101 
 
 §! = x 3 ( x u x 5 + V5 )+x 2 X 3 X U (x l x 5 +x i X 5 )+X 2 X 3 X U (x l X 5 +x l X 5 ) 
 +x x^x (x^+x^+x^x (x^+x^) 
 
 g 2 = x U (x i x 5 +x 3 x 5 )+x U x 5 (x 2 +x 3 )+x 5 (x l x U +X 2 X U +x 3 x )+ ) 
 
50 
 
 Td [x 1 ,x 2 ,x 3 ,x 1| ,x 5 J = 
 
 #x = 0101 0101 0101 0101 0101 0101 0101 0101 
 
 #x = 0011 0011 0011 0011 0011 0011 0011 0011 
 
 #x = 0000 1111 0000 1111 0000 1111 0000 1111 
 
 #x. = 0000 0000 1111 1111 0000 0000 1111 1111 
 
 #x, r = 0000 0000 0000 0000 1111 1111 1111 1111 
 5 
 
 #g = 0000 1111 1111 0000 0000 0000 1111 1111 
 
 #g = 0000 1111 1111 0011 1111 1111 1111 1100 
 
 Step_2: Find [f^] , [f^] , [g^ ] , [ g^ 
 In matrix notation, 
 
 [A..] = f A. .] T 
 
 jk L kj J 
 
 where the T designates the transpose of the matrix, and is formed 
 by interchanging rows and columns. Also, complementation 
 
51 
 
 (denoted by [A ] ) in a Boolean matrix means replacing O's with l's 
 and vice versa. So, we find 
 
 [f 1 ] = 
 
 0011 
 1100 
 0000 
 .0111 
 
 [f 1 ] = 
 
 0100 
 0101 
 1001 
 1001 
 
 
 10lf| 
 1010 I 
 0110' 
 0110 
 
 [f 2 1 
 
 0011 
 1110 
 
 1111 
 
 0101 
 
 r f 2 i 
 ,2k J 
 
 0110 
 0111 
 1110 
 1011 
 
 [f 2 
 
 L1 ^k 
 
 1001 
 1000 
 0001 
 0100 
 
 (The superscripts denote the corresponding equation number.) 
 
 [ g L ] 
 
 0000 
 
 1111 
 
 0000 
 
 1111 
 
 1111 
 
 0000 
 0000 
 
 1111 
 
 [g L ] 
 
 0000 
 
 1111 
 1111 
 1111 
 
 1111 
 
 0011 
 
 I 
 
 1111 
 
 1100 
 
 «L> 
 
 1111 
 
 0000 
 
 1111 
 
 0000 
 
 0000 
 
 1111 
 1111 
 
 0000 
 
 
 1111 
 
 0000 
 
 °km 
 
 0000 
 
 1100 
 
 
 0000 
 
 0000 
 
 
 0000 
 
 0011 
 
52 
 
 Step 3 : Find [ (J) ] , [ (J) ] , by means of the fundamental formulas 
 
 [f £k J ® [ Skm J " [ ^m ] 
 
 [ V®fe*m ] = ^im 1 C 
 
 [f 
 
 £k 
 
 [& 
 
 km 
 
 0100 
 0101 
 1001 
 1001 
 
 4 
 
 ["ion 
 i 
 
 11010 
 0110 
 0110 
 
 O 
 
 ® 
 
 ■where * is logical 
 multiplication of 
 corresponding elements 
 (units appear only 
 where [ $> 1 and [(f)] Save 3l's) # 
 
 1111 
 
 0000 
 
 
 0000 
 
 1111 
 
 
 0000 
 
 1111 
 
 1111 
 1111 
 
 
 0000 
 
 1111 
 
 1111 
 
 0000 
 
 -^l ] a 
 
 0000 
 
 0000 
 
 
 1111 
 
 0000 
 
 
 1 
 
 g km 
 
 
 
 
 0000 
 
 1111 
 
 
 1111 
 
 1111 
 
 
 1111 
 
 0000 
 
 
 0000 
 
 1111 
 
 
 0000 
 
 0000 
 
 = 
 
 1111 
 
 0000 
 
 -if*] c 
 
 1111 
 
 1111 
 
 
 1111 
 
 0000 
 
 
 ] c 
 
 
 ooc 
 
 0000 
 
 
 (Note: If equation 1 
 
 
 
 111 
 
 1 0000 
 
 
 were taken alone, 
 
 ^)j 
 
 
 ooc 
 
 1111 
 
 
 there would be 
 
 its 
 
 
 lOOC 
 
 1111 
 
 
 2.2.2-2 = 8 
 
 
 
 
 
 
 
 so] 
 
 .utions.) 
 
53 
 
 t f L^ 
 
 
 [g 2 ] 
 
 
 
 
 
 "oiio 
 
 
 1111 
 
 0000 
 
 
 0000 
 
 1100 
 
 0111 
 
 1110 
 
 ® 
 
 0000 
 0000 
 
 1100 
 0000 
 
 ■= 
 
 0000 
 
 1111 
 
 1111 
 
 1100 
 
 lOllj 
 
 
 0000 
 
 0011 
 
 
 nn 
 
 0011 
 
 && f 4 
 
 J 
 
 
 1001 
 
 
 0000 
 
 1111 
 
 
 1111 
 
 1111 
 
 1000 
 0001 
 
 <8> 
 
 1111 
 
 1111 
 
 0011 
 
 1111 
 
 - 
 
 0000 
 
 1111 
 
 1111 
 
 1100 
 
 0100 
 
 
 1111 
 
 1100 
 
 
 1111 
 
 0011 
 
 [ *L ] a 
 
 [flj 
 
 em' 
 
 [f. ] = [tf ] . [4,2 ] 
 
 and 
 
 1 J6m a 
 
 % \ 
 
 [ «i ] 
 
 "llll 00 11~ 
 
 joooo 0000 
 1 
 
 [oooo 0000 
 
 1111 0000 
 
 jllll 0000 
 
 • 
 
 1111 0000 
 
 0000 0011 
 
 joooo 0011 
 
 1 
 
 | 0000 0011 
 
 0000 1100 
 
 !oooo iioo 
 
 L . _ 
 
 joooo 1100 
 
51* 
 
 StgeJ*: FlndC*.]- [^.1 -rt 2 . ] 
 
 *«J 
 
 fa, 
 
 0000 
 
 0000 
 
 (Note there will "be 
 
 1111 
 
 0000 
 
 a unique solution 
 
 0000 
 
 0011 
 
 since there is only 
 
 0000 
 
 1100 
 
 1 unit per column.) 
 
 Step 5 : Set up solution array and derive Boolean functions for 
 y 1? y 2 . (Figure* ' 13) . 
 
 b [y-^yg] : 
 
 -8 = 0123 
 #y x = Oioi 
 #y 2 = 11 
 
 (Place the #th column of b [y^ ,y ] in the mth aolvmn for 
 
 indices of units in [fyn ]. ) 
 
 Solution Array 
 
 m 
 
 | 
 
 1 
 
 2 
 
 3 
 
 h 
 
 5 
 
 6 
 
 7 
 
 #y x 
 #y 2 
 
 1 
 
 i 
 
 1 
 
 l 
 
 1 
 
 1 
 
 1 
 
 l 
 
 l 
 
 
 
 
 
 
 
 
 
 
 
 
 
 l 
 
 1 
 
 1 
 
 1 
 
 gj fiyra. 3>3« 
 
55 
 
 Since y*, p = y, ? (x l3 x 3 x ), the explicit Boolean functions for 
 
 y and y are found with a three variable Veitch chart,, (Figure lU) 
 
 x. 
 
 
 00 
 
 X 
 
 10 
 
 1 X 2 
 
 01 
 
 11 
 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 
 
 y l 
 
 = X 
 
 2 + *3 
 
 
 
 1 2 
 
 00 10 01 11 
 
 X, 
 
 o i 
 
 
 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 y^ = x_ 
 y 2 3 
 
 Figure lU„ 
 
 The solutions check with those found by Maitra's methods. 
 
 Remarks on Type 1 Solutions 
 
 It has been shown that 
 
 nJ = thj fl 'ttij 
 
 MR. 
 
 lm J a lm J c 
 
 (3.17) 
 
 where 
 
 [J>, ] = [f, J T ~ r ~ 
 
 #m a 
 
 "kT ** ^knf 
 
 (3*18) 
 
 and 
 
 'J c ■ lt U^ C «Km 3 
 
 (3.19) 
 
56 
 
 Substitution of (3.18) and (3.19) into 3-17 yields 
 
 V ■ J <V «lJ - I ( V Stan' (3 - 20 > 
 
 k K 
 
 By De Morgan's rules, (3.20) is 
 
 k k 
 
 "J ( V + <hJ " (f & + Stan) 
 
 k 
 
 - n (f^Sta + vw> (3 - 22) 
 
 k 
 
 Observe that (3.22) is identical to (3.2). Ledley calls 
 this matrix-multiplication operation the "theta product", and writes 
 it 
 
 tV ■ [f */ e [g kJ (3 - 23) 
 
 Using (3<>22) in the above example would have immediately 
 yielded 
 
57 
 
 Cfl] 
 
 0100 
 0101 
 1001 
 1001 
 
 km 
 
 0000 
 1111 
 0000 
 1111 
 
 1111 
 
 0000 
 0000 
 
 1111 
 
 \1 
 
 r r 
 
 0000 
 
 1111 
 
 0000 
 0000 
 
 0000 
 0000 
 
 1111 
 1111 
 
 tfo 
 
 -1 -1-1 
 
 (Units appear in [<j)y jonly where f = g , i.e., wherever the row 
 of fa is identical to the column of g it is multiplying.) 
 
 Also, by treating [(p? ] as a Veitch chart, the functions 
 may be found immediately. See Figure 15 below. 
 
 -x. 
 
 y 2 y i 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 r 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 1 
 
 
 
 
 
 
 
 
 
 1 
 
 1 
 
 
 
 
 
 Figure 15* 
 
 v = x + x 
 y l 2 3 
 
 y 2 =x 3 
 

 These improvements (i.e., the use of (3.22) and the Veitch 
 chart) result in the methods of Svoboda and Maitra, both of whom 
 reference Ledley. 
 
 Example of Type 2 Problem 
 
 Given: f-^y^y^x^x^ , f^y^y^x^x^ (Equation 3.12), y^g+x ,y g 
 
 Problem: Find g (x ,x ,x, ,x ) (Note: since none of the given 
 
 functions contains x , this variable will not appear in g ,gp.) 
 
 ste^-i: Find [f L J > [ <e ] > [ t ] > rtjJ • 
 
 From the previous example, 
 
 Tf 1 1 
 
 
 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 1 
 
 1 
 
 [f u ] 
 
 
 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 1 
 
 
 
 1 
 
 #x, 
 
 #y, 
 
 12 3 
 10 1 
 11 
 
 1110 
 
 T^s [&J l^J 
 
 Jim 
 
 £m- 
 
 "o 
 
 
 
 
 
 6" 
 
 1 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 1 
 
 
 
 = [4>,J 
 
 nr 
 
 #y 2 = 11 
 113 2 
 
 (Units appear in [^J at (1,0), (1,1), (3,2), (2,3)0 
 
59 
 
 Step 2 : Calculate [g ] and [g ] from the fundamental formula. 
 
 [ 4^w ■ 
 
 11 
 
 
 
 
 
 110 
 
 
 110 
 
 
 
 
 O 
 
 1 
 
 = 
 
 111 
 
 
 10 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 -rO 
 
 [f rf ]0[ W = 
 
 11 
 
 
 1110 
 
 1111 
 
 
 
 10 1 
 
 
 
 
 
 
 
 
 110 
 
 
 1 
 
 
 10 
 
 
 
 
 
 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 tan 
 

 Step 3 : Find g^x^x ,x^,x.) , g 2 (x 2 ,x ,x^,x ) by means of Vettch 
 charts (Figure lU) , 
 
 '5*1* 
 
 
 
 
 
 L 
 
 1 
 
 1 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 X 5 X U 
 
 
 
 
 
 : 
 
 X 
 
 1 
 
 1 
 
 
 
 1 
 
 J. 
 
 i 
 
 i 
 
 1 
 
 1 
 
 1 
 
 
 
 
 Figure 16 
 
 g l = V5 +? 3 X 4 +X 3 x i+ X 5 g 2 = X t + X 5 +X 3 x l + " X 3 X 1| +X 2 x l+ X 5' fx 2 x 5 
 
 (Note: These functions may be obtained by (3.12.1) and (3 . 12. 2), 
 respectively, if minimized by De Morgan's rules.) 
 
 Example of Type 3 Problem 
 
 Given: g. ,g (as in previous problems - Equation 3*12); y, = x +x 
 
 y 2 = x 3 
 
 Problem: Find ^(y^y^^x ) , i 2^ v X ,7 2' X k> X 5^ ' 
 
 Ste^l: Find [g^], [^ [^J, [^] 
 
 From the previous problem, 
 
61 
 
 [g 1 : 
 
 
 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 [g L ] 
 
 n 
 
 11 
 110 1 
 
 11111 
 
 1110 
 
 £ g mk ] 
 
 1 
 
 
 
 1 
 
 —] 
 
 
 1 
 
 
 
 1 
 
 
 
 
 
 1 
 
 1 
 
 
 
 
 
 1 
 
 1 
 
 
 
 fl 
 
 
 
 '- s nk-' 
 
 il 
 
 ! I 
 
 o l o o 
 
 ! ! 
 
 jo o o lj 
 
 tV 
 
 
 
 
 
 
 
 
 
 1 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 p 
 
 
 
 1 
 
 0_ 
 
 — 1 ~2 
 Step 2 : Calculate [£» ], [f, ] from the fundamental formula. 
 
 [*lm ] ® [i mk ] = [? ft ] a- ^erefore, 
 
 [ ^ ] a= 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 1 
 
 
 
 
 
 1 
 
 1 
 
 
 
 
 
 1 
 
 1 
 
 °] 
 
 £**].- 
 
 
 10 
 
 10 1 
 
 1 
 
 r 
 
 [ f U J a ■ 
 
 10 11 
 110 
 
 i 
 
 10 0' 
 
 1111 
 
 Kzh 
 
 1 
 
 
 
 1 
 
 — 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 1 
 
 If f and f were now found with respect to h[y ,y ,x, ,x ], 
 this would imply that no constraint exists between the elements of 
 the basis, i.e., that they are independent. The constraint with 
 respect to the basis is found by 
 

 IhJQK J [c /J 
 
 
 
 
 
 
 
 
 
 
 1 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 1 
 
 0_ 
 
 ® 
 
 ill 
 
 1 
 
 111 
 
 1 
 
 111 
 
 1 
 
 111 
 
 1_ 
 
 b 
 
 
 
 
 
 
 
 i 
 
 i 
 
 1 
 
 1 
 
 i 
 
 i 
 
 1 
 
 .L 
 
 1 
 
 l 
 
 i 
 
 1 
 
 I 
 
 :[c /k ][<? k/ ] 
 
 
 l 
 
 i 
 
 ■ 
 
 
 i 
 
 i 
 
 
 
 i 
 
 i 
 
 
 2 
 
 i 
 
 i 
 
 _ 
 
 If [c A is treated as a Veitch chart, the constraint is 
 found immediately as c = y. +y = 1„ This checks, since 
 x +x +x_ = 1. The f's may now be found from the constrained basis 
 or constrained Veitch chart (binary combinations where #c is zero 
 are disallowed - Figure 17) . 
 
 b c [y l> y 2'V X 5 ] = 
 
 h 1 
 
 = 101 
 
 101 
 
 101 
 
 101 
 
 
 fo 2 
 
 = Oil 
 
 Oil 
 
 Oil 
 
 Oil 
 
 1 
 
 H 
 
 = 000 
 
 111 
 
 000 
 
 111 
 
 5 h 
 
 .2 
 
 #x_ 
 5 
 
 = 000 
 
 000 
 
 111 
 
 > 
 
 111 
 
 
 
 
 
 
 I 
 
 a 
 
 a 
 
 = Oil 
 
 100 
 
 000 
 
 111 
 
 i 
 
 #f 2 
 a 
 
 = on 
 
 110 
 
 111 
 
 101 
 
 > 
 
 1 
 
 i, 
 
 — 1 
 
 y 2 
 
 '4 
 
 
 
 X 
 
 . 
 
 
 1 
 
 -1 
 
 : 
 
 
 
 
 v\ - 
 
 
 
 j 
 
 
 $ 
 
 1 
 
 x r 
 
 
 
 y ? 
 
 — '■' ■ ■ 1 
 
 y i 
 
 t 
 
 
 
 - 
 
 - 
 
 
 I 
 
 
 
 
 /. 
 
 
 
 
 
 
 
 i 
 
 
 
 *£ 
 
 
 
 
 
 2 
 
 1 
 
 :> 
 
 
 
 F 
 
 igure 1 
 
 7. 
 
63 
 
 f a = y 2¥ 5 + ¥A +y 2¥ 5 
 
 f a = y 2 X U +y l y 2 X 5 +y i X 5 +y i y 2 X l| 
 
 At this point, we have f 
 
 1 
 
 5 . It is now necessary 
 
 to use type 2 formulas to see if #f - #g» Thus 
 
 [C,] A [<|>,J = 
 
 lnH, 
 
 £m- 
 
 10 11 
 110 
 
 I 
 
 iooo| 
 
 ® 
 
 '0000 
 
 110 
 
 i 
 
 1 
 
 10 
 
 r° 
 
 
 
 1 
 
 *] 
 
 i 
 
 1 
 
 
 
 
 
 ! ° 
 
 
 
 
 
 
 
 L 1 
 
 1 
 
 1 
 
 li 
 
 = [£j »[i 
 
 'km" 
 
 'km- 
 
 [C.J ® [<|>,J = 
 
 "ki 
 
 Jim- 
 
 10 11 
 
 1110 
 
 1111 
 
 110 1 
 
 © 
 
 
 ! 
 11100 
 
 1 
 
 [0010 
 
 
 
 
 
 1 
 
 1 
 
 jl 
 
 1 
 
 
 
 1 
 
 11 
 
 1 
 
 1 
 
 1 
 
 1 
 
 i 
 
 li 
 
 1 
 
 1 
 
 
 
 [ f kiJ ^ s kiJ 
 
 The above computation may be taken as the substitution of 
 
 y. (x_,x,J and y-fx^.x.) into f and f , or the means of displaying 
 123 223 a a 
 
 that f - g by showing f - 
 s °s a 
 
 Thus , the functions for 1 
 
 JL 
 
 2 
 
 and f are the desired solution (which includes the constraint) 
 
6k 
 
 Ledley's Solution of General Functional Equations 
 
 Ledley's method is the same as Svoboda's, but through the 
 use of charts for the stepwise calculation of the functions f and 
 g , followed by matrix multiplication to find ({), Ledley's method 
 in comparison seems mysterious, and is more laborious. The 
 explanation of the method will be concurrent with the solution of 
 Equations (3»5)* 
 
 Step 1: Set up a chart which lists : 
 
 (a) the designation numbers of the coefficient of each unknown; 
 
 (b) the corresponding unknown; 
 
 (c) the value of the unknown computed for all possible binary 
 combinations 
 
 See Table 3-2 on following page. 
 
 Step 2 : Using Table 3-2, multiply the transpose of each matrix 
 formed in column C by the corresponding matrix formed in column A, 
 This will yield the Boolean matrices [f, ], [g, ], [f p ]> fSp^ wh^ * 1 
 are equivalent to the Veitch charts of these functions found in 
 Svoboda's algorithm (see Figure P.). 
 
65 
 
 Table 3 = 2 
 
 -1< \ 
 
 A. Coefficients and 
 
 B. Corresponding 
 
 C. Binary Combinations 
 
 Designation Numbers 
 
 Unknowns 
 
 of Solution y.Yp 
 
 
 
 00 10 01 11 
 
 V #X3_x 2 = 10 
 
 y 2 
 
 11 
 
 J^ 2 = 110 
 
 y l y 2 
 
 1 
 
 #x 2 = 11 
 
 y i y 2 
 
 10 
 
 #x p = 11 
 
 V 2 
 
 y i y 2 
 
 10 
 
 ' ^Xg = 10 
 
 y 2 
 
 11 
 
 { #(x 1+ x 2 ) = 10 11 
 
 y i y 2 
 
 10 
 
 | #x 1 x 2 = 10 
 
 y l 
 
 1 i 
 
 #x 1 = 10 1 
 
 y l y 2 
 
 10 
 
 f #(x 1 x 2 +x 1 x 2 )= 110 
 
 y i 
 
 10 1 
 
 \#I = 1111 
 
 y i y 2 
 
 10 
 
 itf&^r, = 10 
 
 y l 
 
 10 10 
 
 / #^ 2 = 110 
 
 y 2 
 
 110 
 
 L #x lX2 = 1 
 
 y i y 2 
 
 10 
 
 f U 2 = 110 
 
 y 2 
 
 110 
 
 ] #1 = 1111 
 
 y i y 2 
 
 10 
 
 \#x x x 2 = 1 
 
 y i 
 
 10 1 
 
 #x x x 2 = 1 
 
 y i 
 
 1 1 
 
 Jfx ± x 2 = 10 
 
 y i y 2 
 
 1 
 
66 
 
 
 
 
 
 
 
 ■ 1 
 
 
 
 
 
 
 1 
 
 
 
 1 
 
 
 
 
 
 1 
 
 1 
 
 1 
 
 
 
 q_ 
 
 ® 
 
 
 
 1 
 
 
 
 
 
 1 
 
 1 
 
 
 
 
 
 
 
 
 
 1 
 
 1 
 
 
 
 
 
 1 
 
 L: 
 
 
 
 
 
 
 
 o~~ 
 
 
 
 
 
 1 
 
 i 
 
 
 
 1 
 
 1 
 
 i 
 
 1 
 
 1 
 
 
 
 
 
 -Vjl 
 
 
 
 
 
 1 
 
 
 
 
 
 1 
 
 
 
 
 
 1 
 
 
 
 1 
 
 1 
 
 1 
 
 
 
 
 
 
 
 ® 
 
 
 
 
 
 1 
 
 
 
 1 
 
 
 
 1 
 
 1 
 
 
 
 1 
 
 
 
 
 
 
 
 1 
 
 
 
 1 
 
 
 
 1 
 
 
 
 °1 
 
 1 
 
 
 
 1 
 
 1 
 
 
 
 1 
 
 1 
 
 1 
 
 
 
 
 
 1 
 
 
 
 ■ [8, 
 
 110 
 110 10 
 10 1 
 10 
 
 
 0110^ 
 
 ® 
 
 1111 
 
 
 10 
 
 
 110 
 
 
 1 
 
 1 
 
 1 
 
 
 
 o"" 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 
 
 1 
 
 
 
 1 
 
 1 
 
 
 
 = [f 2 ] 
 
 1110 
 
 
 10 10 
 
 
 
 ® 
 
 10 
 
 
 11 
 
 
 
 
 1 
 
 1 
 
 
 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 
 
 
 
 1 
 
 
 
 1 
 
 
 
 
 
 0_ 
 
 
 
 1 
 
 0_ 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 1 
 
 1 
 
 
 
 = [g 2 ] 
 
67 
 
 In Steps 1 and 2, Ledley has presented a systematic 
 procedure for the mapping of f ,g ,f ,g , which was carried out 
 mentally in Svoboda's algorithm., This procedure is unduly 
 laborious, and seems to present no factor which is superior to 
 Svoboda's use of Veitch charts „ 
 
 The remaining process would consist of finding [(J) ] and 
 
 [<|> ] based upon (3.2), then finding [(()] = [(|) ] ' [(|) J (where • is 
 
 logical multiplication of corresponding elements), and finally 
 
 derivation of the y.' s by the solution array (or a Veitch chart), 
 
 J 
 
 Since this is analogous to Svoboda's solution, completing the 
 solution here would be superfluous „ 
 
 3.3.U Carvallo's Method 
 
 [3] 
 
 Carvallo's method of solution is almost identical to 
 
 Ledley r s alternate method, in the use of a basis and Ledley ' s "9 
 
 matrix multiplication." However, Carvallo's basis is written as a 
 
 r 
 single column matrix of 2 complete products, and is expressed as 
 
 x 
 
 X-, X_« o a oX 
 
 12 r 
 
 X -. X _ • o • X 
 
 1 2 i 
 
 X-, X _ o o o X 
 
 12 r 
 
 'X-, X_o o e oX 
 
 12 r 
 
 1 1....1 
 1 1....0 
 
 • • * 
 
 9 • e 
 
 C.,1 
 
 o.,,,o 
 
 (3.23) 
 
Thus, the 3-element basis equivalent to (3-13) is 
 
 Oi 
 
 (x x ) 
 
 (x 2 ) 
 
 (x 3 ) 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 1 
 
 
 
 1 
 
 1 
 
 
 
 
 
 
 
 1 
 
 1 
 
 
 
 1 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 (3.13') 
 
 The solution is found from (3°2) or (3»22), but since the 
 designation numbers in the f and g matrices are in columns 
 instead of rows, (3*23) must now be written as 
 
 W(.,t) ■ t*J(s,d) 6 [ g] (t,d) 
 
 (3.23') 
 
 a b 
 where s = 2 , t = 2 , (a and b = number of elements in respective bases) 
 
 d = number of #'s in matrix (in Ledley's ease, & -»1) . Also, if the 
 
 number of elements in the bases is small enough (r < 3) so- that the 
 
 matrix multiplication is not unwieldly, the entire system of equations 
 
 may be solved in one step (i.e., it is not necessary to calculate 
 
69 
 
 [(J) ], [$ 2 ],... and finally [(j)] = TT_ [f]). For, if [f] ( d) - 
 [#f 1? #f 2 ,...,#f d ], and [g] (t d) = [#g 1? #g 2 ,...,#g d ], then 
 
 ^(s,d) ^ W(t,d) = [ * 1] (s,t) e [ ^ ] (s,t) - [f]( Sjt) = *( Sjt) 
 
 Carvallo does treat the [§] f ,\ as a veitch chart and thus derives 
 
 Y (s,t) 
 
 his solution function immediately. 
 
 Example : 
 Given : 
 
 f l^ y l' y 2' y 3^ ~ y l y 2 + " y l y 3 = x i X 3 + ^2^3 ~ g 1 ( x l5 x 2 ' x o) (3.24.1) 
 
 f 2 (y 1 ,y 2 ,y 3 ) = y 1 (y 2 +y 3 )+y 1 y 2 y 3 - x^+XgCx-j+x ) = 
 
 g 2 (x l3 x 2 ,x ) (3*24.2) 
 
 f 3 (y 1 ,y 2 ,y 3 ) = y 3 (y 1 +y 2 )+ y x y 2 = x i ( ^ x 2 +x 3' )+x 2 x 3 
 
 ; 3 (x 1? x 2 , x ) (3.2^.3) 
 
 Find: y^x^x^ , y^x^x^ 
 
TO 
 
 Step_l: Find [f] (8 3) , [g] (3>8) 
 
 (s = 2 3 = 8, t = 2 3 = 8, d = 3) 
 
 (#^) (#f 2 ) (#f 3 ) 
 
 [f] = 
 
 r° 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 1 
 
 1 
 
 10110001 
 
 (\) 
 
 1 
 
 1 
 
 
 
 T 
 
 [g] = 
 
 00111101 
 
 (#g 2 ) 
 
 
 
 1 
 
 1 
 
 
 11101000 
 
 (#g 3 ) 
 
 1 
 
 1 
 
 1 
 
 
 
 
 
 
 1 
 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 St 
 
 ep 2 : Find [^(q 8 ) = [f] © [g] (Figure 18) 
 
 y l y 2 y ^ 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 ■ 
 
 
 
 3 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 1 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 iffUrT: 
 
 
 [<)>] 
 
71 
 
 Step 3 : Find y^y , 
 
 By treating [(])] as a Veitch chart, immediately 
 
 y i = X 1 X 2 +X 2 (X 1 +X 3 ) 
 
 y 2 = x 1 x 2X3+ x 3 (x 2+ x 1 ) 
 
 y 3 =X 2 
 
 Note that the variables in the above Veitch chart are the 
 complements of the standard notation used throughout this paper. 
 This was necessitated by the fact that Carvalio's basis is the 
 reverse of the BCD code used by Ledley. 
 
 3.k. "Truth-Table-Logic" Method 
 
 Given the set of relations (3d) , it is also well known 
 that they may be combined into the single equation 
 
 _ n 
 
 <|> = Z (f © g ) = (3o25) 
 
 k=l k k 
 
 which is simply the complement of (3.2), This equation may then be 
 expressed as a sum of minterms of x ,x , „..,x multiplied by 
 coefficients which are functions of y ,y , . . . ,y . This method of 
 
 J_ c. s 
 
72 
 
 attack is the basis of a laborious algebra (whose applications are 
 extended in [h]) which is very similar to the method developed in 
 
 Section 3.5« In order to avoid duplication, only the latter will be 
 
 [151 
 presented, and attention will be given to Phister's more useful 
 
 method o 
 
 Phister's other technique is that of "truth-table-logic", 
 
 It simply defines a systematic procedure for deriving a truth table, 
 
 containing all binary combinations of the independent variables, by 
 
 logical reasoning,, Although this technique is most obvious, it has 
 
 limitations: the equations must be of the form f (y ,y , . ,, ,y , 
 
 J- £- o 
 
 x, ,x , t ..,x ) = g(x ,x , . „ . ,x ); it is only good for simple 
 equations with few unknowns, such as the input equations for 
 memory elements (Phister's examples are only for flip-flops); 
 constraints are necessary for a general solution. The steps are: 
 
 Step 1 : Record all input combinations of the independent variables 
 and the corresponding truth values of g(x ,x , ,.=,x )« 
 
 Step 2 : By logical reasoning, determine truth values (including 
 don't-cares) of the dependent variables. 
 
 Step 3 » Form the general solution of y.'s by regular truth table 
 
 J 
 
 technique* 
 
73 
 
 Step k : By means of a Karnaugh map, find the simplest solution by- 
 assigning specific values to the don't-cares. 
 
 The simple example below was taken from [15], pp. 121-12^-, and the 
 truth table below is Steps 1 and 2 combined* 
 
 Example : Given the equations of an R-S flip-flop 
 
 S + m = g 1 Q + g 2 Q 
 
 RS = 
 
 where g and g represent Boolean functions of whatever variables 
 determine the state of Q; g , g p , and Q, are the independent variables, 
 
 Steps 1 and 2 : The derived truth table is Table 3*3 below. 
 
 Step 3 - Find the general solutions,, 
 From Table 3.3, 
 
 R = ^gjgg^ + g i q + d J+ g l S 2 Q 
 
 S = g 2 Q + d g^Q + d^g^Q 
 
 (These may be checked by substitution into the given equations.) 
 
Table 3*3 
 
 g 1 Q+g 2 Q 
 
 7U 
 
 Q 
 
 = S + RQ 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 1 
 
 
 
 
 
 1 
 
 1 
 
 1 
 
 
 
 
 
 1 
 
 
 
 1 
 
 1 
 
 1 
 
 
 
 1 
 
 1 
 
 1 
 
 
 
 d o 
 
 
 
 
 
 1 
 
 
 
 1 
 
 
 
 1 
 
 
 
 1 
 
 
 
 
 
 \ 
 
 
 
 1 
 
 
 
 d 
 
 1 
 
 
 
 1 
 
 1 
 
 
 
 d 
 
 Step k: Simplify the functions by means of Karnaugh maps (Figure 19) 
 
 lit 
 
 Q 
 
 
 
 
 
 
 
 1 
 
 
 
 d ^ 
 
 d 7 
 
 1 
 
 WW 
 
 Q 
 
 R = S 1 Q 
 
 S = g 2 Q 
 
 Figure 19 
 
75 
 
 3„5 Parametric Method 
 
 So Rude arm 9 obtains the general solution cf 
 functional equations in parametric form, and also expresses the 
 solution as an explicit function of the given independent variables, 
 
 It has been previously shown that a system of equations 
 (3.1) may be expressed as 
 
 f(y 1 jy 29 o..,y s ,x l9 x 2S ...,x r ) =0 (3*26) 
 
 by means of (3°25) ? (the complement cf (3.1)). Furthermore, it 
 is well known (cf„ eg», [3] 9 [15] s [19] ) that a Boolean equation 
 in a single unknown may be written as 
 
 ay + by = (3.27) 
 
 for which 
 
 ab = (3.28) 
 
 is the necessary and sufficient condition for the existence of 
 a solution o The solutions of (3«27) are constrained by 
 
 b < y < a (3°29) 
 
 and all such y's are solutions. 
 
76 
 
 Theorem 3-3' The parametric solution of (3.27) is 
 
 y = b + p (3.30) 
 
 where the parameter p is constrained by 
 
 P<ab (3.31) 
 
 Proof : Every y satisfying (3°30) and (3-31) satisfies (3-29) 
 since b <b + p <t + ab =b + a = a with the aid of the 
 complement of (3.28) . Conversely, every y satisfying (3.29) can 
 be written as (3»30) with a suitably chosen p> for if p = by < a b, 
 then b+p=b+y=y. The algorithm for finding the parametric 
 solution is as follows: 
 
 Step 1 : By means of (3.25), change the system of equations to 
 the form of (3.26) and call this f, (all variables present) , so 
 
 f 1 (y 1 ,y 2 ,.»» ? y s ) = o (3.32) 
 
 where the independent variables are implicitly understood. 
 
77 
 
 Step 2 ; (3-32) is consistent in y i 
 
 ff 
 
 f 2 (y 2 ,...,Y s ) s f 1 (0,y 2 ,...,y s ) f-Jl^, . . .,y g ) =0 (3.33) 
 Thus, in the i-th step, find 
 
 f i+1 (y i+1 ,...,y s ) =f i (o,y i+1 ,...,y s )f i (i,y. +1 ,...,y s ). (3.3*0 
 
 Proceed with the successive elimination of dependent variables 
 until finally, 
 
 f (y ) sf n (0,y )f , (l.y ) - (3-35) 
 
 n V0 V n-l v ,J V n-l v ?J V u J;y 
 
 Step 3 : Solve (3-35) for y by means of (3*30) and (3-31) 
 
 Step h: Substitute the solution for y into f n (y . ,v ) = 
 * — J n n-l u n-l"n' 
 
 to obtain an equation in the single unknown y : 
 
 Wn-l'rt = ° (3 - 36) 
 
 By successive reintroduction of the unknowns (reversal of Step 2) , 
 find y ,y , . . . ,y in parametric form, 
 
 Step 5 ; Find the explicit solutions in the form of (3.3) by 
 stepwise tree-like deductions. 
 
78 
 
 Example : Solve the system of equations 
 
 f a (y i' y 2 } - y l +y 2 X 2 - y l (x l + V X 2 )+y 2 (x l X 2 +y l X l X 2 ) " g a (y l' y 2 } 
 
 f b (y ] _jy 2 ) - y 1 (y 2 + x 2 +x 1 )+x 1 x 2 = y 2 ( yi +x 2 )+y iy2 ( Xl +x 2 ) =g b (y 1 ,y 2 ) 
 
 Step 1 : 
 
 f 1 (y 1 ,y 2 ) 
 
 b 
 
 • Z (f. © g.) 
 
 i=a 1 i 
 
 = [y 1 +y 2 x 2 ] [ (y^^+y^Xg) (y 2 +y 1 x 1 -Hy 1 x 2 +x 1 x 2 +x 1 x 2 ) ]♦ [7 1 (y 2 +x* 2 ) ] 
 
 
 [y-L ( 3C i + y 2 +X 2 ) +y 2 ^ X l X 2 +y l X l X 2^ ) ^ + ^1 ^2 +X l +X 2 ^ +X 1 X 2 ^ 
 
 [(yg+y^g)* x 1 x' 2 (y 1 +y 2 )]+ [(y 1 +y 2 x 1 x 2 )(x 1 +x 2 )] [y^y-^) 
 
 ^ygC^+Xg)] 
 
 = yiL"y 2 ^ x l + V +y 2^ X 1^2 +X l X 2^ +y 1 [y 2 (x 1 x 2 +x 1 x 2 )+y 2 (x 1 +x 2 )=0 
 
79 
 
 Step 2 : 
 
 f 2 (y 2 ) = f^o^f^i,^) 
 
 = [y 2 (x 1 x 2 +x 1 x 2 )+y 2 (x 1 +x 2 ) ]• 
 
 [y 2 (x 1 ^x 2 )+y 2 (x 1 x 2 +x 1 x 2 ) ] 
 
 = x 1 x 2 y 2+ x 1 x 2 y 2 
 
 Step 3 : a = x^ b = x^ p < (x^x^ (x 1 +x 2 ) = x^+x-jX, 
 
 Thus 2 
 
 y 2 = b + P = x x x 2 + p 
 
 p = or x_x_ or x„,x_ or x_x_+x_x_ 
 12 12 1212 
 
 and there are four different solutions for y^. . 
 
 Step k : f 1 (y 19 p) = y 1 [(x^+p) (x^x^ rp^+x^ (x^+x^) ] 
 
 y 1 [(x x g +p) (x 1 x 2 +x 1 x 2 )+p(x 1 +x 2 ) (x 1 +x 2 ) j 
 
BO 
 
 = [(x 1 +T 2 )p+x 1 x 2 p ] y^ [ x 1 x 2 +x 1 x 2 p+p(x 1 x 2 +x 1 x 2 )] y ± 
 
 Thus, y = x 1 x 2 +x 1 x 2 p+p(x 1 x 2 +x 1 x 2 ) + q 
 
 where q ^ a *b = ° 
 
 Step 5 ; Find the explicit functions for y and y , 
 
 (a) For p = 0, y g = x^ 
 
 y l = X l + X 2 
 
 (b) p = x^, y 2 = x 
 
 y l =X 1 
 
 (c) p = x^Xg, y 2 - x 1 
 
 y l ~ X 2 
 
 (d) p = x^g+x^g, y 2 = x 1 + x 2 
 
 y l = X 1 X 2 
 
81 
 
 Remarks: 
 
 lo The parametric form may be avoided by immediately solving for all 
 
 possible values of y and then substituting these into the successive 
 
 equations f . (y . ,y ) , f _(y _,y _ ,y ) , etc This results in 
 ^ n-1 n-1' n ' n-2 n-2 SJ n-l 5 n 3 
 
 a more discernible solution tree G 
 
 2. The labor of Step 1 is equivalent to that of Step 1 in 
 Ashenhurst's algorithm, but the remainder of The solution is easier 
 in the latter methods,, Consequently , the parametric method does 
 not seem to offer any advantages , and its use should be very 
 limited , 
 
 3,6 Summary and Applications 
 
 Of all the methods of Section 3$ Svoboda's algorithm is 
 probably the most efficient , while still maintaining generality,, 
 For the specific type of Equation (3el0), Maitra's methods are 
 simpler than those of Ledley, although Ledley's solutions seem less 
 complex as one practices using them, and " products" ease the 
 number of calculations, as in Carvallo's method „ The techniques of 
 Ashenhurst and Phister, although the most basic, should be 
 employed to solve only the simplest equations with very few 
 unknowns , and the former should be used in preference to Rudeanu's 
 parametric method „ 
 
'& 
 
 The most obvious and widespread use of the solution 
 methods for simultaneous Boolean functional equations is logical 
 circuit design. The three problem types solved by Maitra and 
 Ledley were explicitly shown as circuit design problems, as was 
 the flip-flop problem solved by Raster. Moreover, the general 
 methods of Svoboda and Ashenhurst can be used to solve any type 
 
 of logical circuit design problem* 
 
 [2] 
 George Boole devised Boolean algebra as a symbolic 
 
 interpretation of logical reasoning. There are innumerable everyday 
 
 problems , in such diverse fields as biochemistry, tactical warfare, 
 
 economics, and psychology, to which Boolean equations may be 
 
 applied. These are called word, logic , or sentential problems s and 
 
 are based upon the feasability of expressing a given proposition as 
 
 a Boolean function,, Table 3°*+ below, taken from p 318 of C^]> 
 
 lists the translations of English connectives 9 which are used to 
 
 formulate the problem. 
 
 Table 3 a k 
 
 English connective Logical Translation 
 
 IN O U x\o0ooo09ooeoo006oaooaoooooeoo6oo6oooooooo0<ri 
 
 r\. cLTlGl D o o o o o o o q o o <> o o 4 o o o o 6 o o a c /> o & Q 6 e o a o a o a 6 e o I1. j3> 
 
 A or (inclusive) B; A or B or both. „ „ „ „ ao . „ . „ .A + B 
 
83 
 
 A or (exclusive) B; either A or B. . . . . . . . .A / B, A°B + A°B, A © B 
 
 r\ OU u JJooooo«oooooeooooooooooooooooooeoooo -t\ ]D 
 l\ aJ-IiilOUgn -Doo«ooooooooooooo»oooooooooooo • -H, JD 
 
 A unless B •••.■•.o..B**A, B + A 
 
 A on condition that B» <> <> <> o . . ° « „ » <> <. „ « <> <> <> „ » „B -» A, B + A 
 
 }\ XX .Dooooooeooooooooooooooooooooooooooooo-D ™ xTu g 13 ~>~ r\ 
 
 A implies B; if A, then B.„.,........., t .A-»B, A + B 
 
 A only if B. o o o o o o o o o o o o o o o o o o . . o o o o o o o o o oA -&> B , A + B 
 
 Not unless A then B„ . „ » <. » „ o . » . . . . . . . o . . <> . <>A~-^B, A + B 
 
 A provided that B „ „ „ „ « <, „ <, „ „ « „ » » <> <> <> . » » „ <> . » .B ->■ A , B + A 
 A as well as Bo » . . . .... » . . . a . . ... . « »« . . . . »A*B 
 
 A if and only if B. . . . . . . . . . . . . . . . . „ . . . . . .A = B, A°B + A«B 
 
 Not both A and B . . . . . . . . . . . . . . . . . . . . . „ . . . . aTb , A + B~ 
 
 Neither A nor B. . . . . . . . . . . , . . . . . . . . . . . . . . »A°B 5 A + B 
 
 When A, then B, . . . . . . . . . „ „ „ „ . . . . . . . . . . . . <,„A-»B, A + B 
 
 Problems of this type may be used to investigate 
 consistency, redundancy, or relationships between facts, or to 
 find simultaneous equivalences and implications. (See Section 
 5.2). Ledley's antecedence and consequence solutions are 
 especially useful in problems concerning deduction techniques. 
 
& 
 
 Many fine examples may be found in [10] , pp. l +05-Ul5, U76-U83, and 
 in other sources quoted therein. For a basic discussion, see [20]. 
 The use of simultaneous Boolean equations in the design 
 
 of sequential machines is suggested by the method in a paper by 
 
 [17] 
 N„ Rouche . Briefly, given a system of equations of the form 
 
 (3«3) ? (in previous methods, this was the solution) , Rouche 
 
 transforms 
 
 X l = x i^i^2 soo ' ,V s ) 
 
 X 2 = X 2^ y l' y 2» ,M,0 » y s^ (3.37) 
 
 * c 
 
 a 
 
 
 X r = x r ^ 1 ^2 30O09Y s ) 
 
 into a matrix representation of a sum of minterms. The derivation 
 is based on (3°27) - since cnly the use of the matrix is of 
 importance here, the derivation -will not be discussed, but it 
 should be obvious - and results in a matrix which is actually a 
 Veitch chart mapping x„ = f . (y.. ? y p , . . „ 3 y^) . For example, if 
 r = s = 2 in (3«3)j then the given (mapping) equations are 
 
 y l = b ll x l x 2 + b 12 x l x 2 + b 13 X A + b l4 x l x 2 (3.38.1) 
 
85 
 
 y 2 = b 21 X l X 2 + b 22 X l X 2 + b 23 X l X 2 + b 2^1 X 2 (3 ° 38 ° 2 ) 
 
 and the matrix-Veitch-chart is 
 
 [B] = 
 
 y -i y p 
 
 b ll b 21 
 
 b 12 b 22 
 
 b 13 b 23 
 
 'x 1 
 
 b lU b 2U 
 
 b ll b 2l b 12 b 22 b 13 b 23 b lU b 2^ 
 
 b ll b 21 b 12 b 22 b l 3 b 23 b 1^2l| 
 
 b ll b 21 b 12 b 22 b 13 b 23 b lU b 2U 
 
 (3.39) 
 
 Ledley points out that [B] is only a special case of 
 
 his [R..] matrix (denoted as [{)...] in this paper; see [10] s 
 
 pp. 356-36O5 396-^00) or, in other words, the problem is a, special 
 
 case of the antecedence and consequence problem. It was shown 
 
 that in Type 2 and Type 3 problems (in which the y.'s are given), 
 
 J 
 
 [(p.. J has a single unit per column; this is also true of 
 [b] . Also [bJ is equivalent to the transition matrix in 
 
86 
 
 "The Theory of Nets," by F. Hohn, S. Seshu, and D. Aufenkamp, 
 IR E Transactions on Electronic Computers , VoJLume EC-6, September 
 
 1957, PP. 15^ - 161. 
 
 If the y^'s are interpreted as the next state (i.e„, the 
 state of a machine after an advancing pulse t) and the x. ' s as the 
 present state (before the same pulse t) , then the matrices [bJ and 
 [({>..] represent the "state matrix" of an autonomous sequential 
 machine, and the state diagram for the machine may be drawn. If 
 the machine is in present state n, the next state will correspond 
 to the row subscript of the single unit in column n. 
 
 Example : Find the state diagram for the machine defined by [b] . 
 
 [3] = 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
87 
 
 The equations defining this machine are: 
 
 y n = x n x^ +x.x„ 
 °± 12 13 
 
 y 2 = x 2 x 3 +X;L x 2 x 3 
 
 y 3 = V +X 1 X 3 
 
 Since row 7 has no units, it must be an initial state. 
 The state diagram will he found starting with state 7 (figure 20) 
 
 Figure 20 , 
 

 k. PARTICULAR SOLUTIONS OF GENERAL BOOLEAN EQUATIONS 
 
 The methods of this chapter are concerned with solutions 
 which assign a definite value (0 or l) to each bivalent variable of 
 the simultaneous equations * 
 
 U.l Logical Algebraic Methods 
 
 ll.l.l. Ashenhurst ' s Algorithm 
 
 This method is simply the special case of the algorithm 
 outlined in section 3°2, and is based only on theorem 3«1» It is 
 only necessary to find <J) by (3e2.) and expand it in canonical form. 
 Then, each term of the expansion will yield a solution, -where primed 
 variables are assigned zero values, and unprimed variables have the 
 value unity. 
 
 Since the maximum number of solutions is 2" (for n 
 variables) s this method is most efficient when the number of terms 
 in the canonical expansion is much less than 2" „ For the special 
 cases where g = 1 or 0, the amount of work required to find ty 
 maybe reduced. First note that if g, = 1, then f g, + f g = f . 
 
 j£ K & K K. K 
 
 Secondly, if g fc = 5 then f^ + f^ = I^ 9 and in (3.1), f fc = 0. 
 Thus , any term in the canonical expansion of <j) which contains a 
 product appearing in the expansion of f as an OR-polynomial must 
 
 K. 
 
89 
 
 be zero. Hence, by forming the function <J>*, as in (3° 2) without 
 using any equations of (3d) which are of the form f =0, <b* 
 yields <j) if all the products containing a vanishing term are 
 eliminated from the canonical expansion of $*<. 
 
 Example : Given the equations 
 
 V x 6 = x l +x 3 
 
 x 5 (x 2 +x 3 ) = ° 
 
 X 3 V6 = 1 
 
 Xl +x 3+Xl+ x 6 = x 5+ x 6+ x 2 x 3 
 
 Step 1 : Find <t>* (do not use the second equation) 
 
 t>* = [ (x 2 +x 6 ) (x 1 +x 3 )+x 1 x 2 x 3 x 6 ][ x 3 x u x 6 ][ ( X;L +x 3 +x u x 6 ) (x 5 +x 6 +x 2 x 3 ) 
 
 
'/; 
 
 (x 1 x 2+ x 1 x 6+ x 2 x 3+ x 3 x 6+ x 1 x 2 x 3 x 6 ) (x 3 x u x 6 ) (x^^ 
 
 X 2 X 3 +X 3 X 5 +X 3 X 6 +X U X 5 X 6 +X 1 X 3 X U X 5 X 6 ) 
 
 = x 1 x 2 x 3 x 1+ x 5 x 6+ x 1 x 2 x 3 x u x 6 
 
 ♦* = x l x 2 x 3V 5 x 6 + X 1 X 2 X 3 X U X 5 X 6 + x l 5r 2 X 3 X '4 X 5 X 6 
 
 Step 2 : Eliminate any minterms containing the terms of the second 
 equation. The middle term of §* will vanish, since it contains x x , 
 
 Step 3 ' The solutions are: 
 
 (a) x ± = x h = x 6 = x 2 = x 3 = x 5 = 1 
 
 (b) x 2 = * k = x 6 = Xl = x 3 = x 5 = 1 
 
 ^.1.2. Grigor ' yan ' s Algori thm 
 
 [6] 
 Grigor 'yan has proposed the following algorithm for 
 
 computer solutions of simultaneous Boolean equations. It is based 
 
 on set theory, but it will be shown that the corresponding logic 
 
 theory leads directly to Ashenhurst's method. 
 
91 
 
 Let U. (i=l 3 2, . . . ,k) be any subsets of some set M, which 
 has a system of characteristic functions 
 
 1, if x eU, 
 
 i 
 
 V x) = ) (x€M; i = 1,2,..., k) {k.l) 
 
 0, if x 4\3. 
 
 i 
 
 Suppose we require Y. = ° . , where a p J ?> • a • ' a ±> '"> a y 
 is an arbitrary sequence of 0's and l's; this results in a system 
 of characteristic equations which are satisfied by a given x if and 
 only if 
 
 xrU. whenever c , = 1 
 i i 
 
 and x ^U. whenever a . = 
 
 • i i 
 
 C-+.2) 
 
 Thus ? the given equations Y. = 3. determine a unique subset of M. 
 Indeed, if (*+.2) is true for all xeG, (G c M) , and for no other x., 
 then G will be the solution* A necessary and sufficient condition 
 for the existence of a solution is obviously that G is not the 
 empty set. 
 
 Let q and p be the numbers of Is and 0's, respectively, 
 in the sequence of a 's. U, D, and \ are the set-theoretic 
 
92 
 
 operations of union, intersection, and relative complementation 
 (subtraction), respectively. Then G will be found as 
 
 A \A flB (U.3) 
 
 Q. Q. P 
 
 where 
 
 (o = l): A =Ou. ; (o = 0):B = U U. (4.U) 
 q i r ' P 1 
 
 a=l * p-1 ch p 
 
 Note that depending on the U. sets, G may be empty, finite, or 
 infinite, resulting in no solutions, a finite number of solutions, 
 or on infinitude of solutions, respectively. 
 
 A system of Boolean functions may be uniquely represented 
 in disjunctive normal form, such as 
 
 i 
 
 Y. = 2 F. .N. (i=l,2,...,k; £=2 m -l) (If. 5) 
 
 0=0 
 
 m 
 
 (j= ^ x 2 m - r ) (4.6) 
 
 r=l 
 
 where N. is the minterm whose m binary variables correspond to the 
 J 
 
 decimal number given by 3; each coefficient F. . has a value of or 
 
 1 equal to the value of the function Y. for the value assignment that 
 
 makes the value of N. equal to one. 
 
 J 
 
93 
 
 Every logical function Y. in (4.5) can be put in a one-to- 
 one correspondence with a subset U. cm, where M. = (0,1,2, ... ,2 -l) , 
 and where U. is the set of decimal numbers corresponding to the 
 
 minterms in Y. . Therefore, 
 
 i 
 
 \(j) = 
 
 r i, if 3 eU 
 
 (4.7) 
 
 v o, if j fv ± 
 
 The solution of the set of equations Y. = o. (i=l,2,. . . ,n) may 
 now be found from (4.3) and (4.4). 
 
 Example : Given a system of 6 equations in 100 variables; they will 
 be represented as in (4.5) , but for convenience, only the decimal 
 representation (i.e., j) of the minterm will be used. 
 
 Y 1 = (0, 3, 6, 10, 13, 22, 36, 72, 209, 504) 
 
 Y 2 = (2, 6, 9, 13, 19, 22, 86, 209, 4ll, 1031) 
 
 Y 3 = (1, 3, 6, 7, 13, 22, 100, 158, 209, 1000) 
 
 Y h = (0, 5, 6, 13, 15, 19, 22, 77, 198, 209, 612) 
 
9^ 
 
 Y^ = (5, 6, 13, 28, 91+ , 106, 601, 2 100 -l) 
 
 Y 6 = (6, 25, 63, 111, W+, 980, Uooo) 
 
 CaseJL: Y 1 = Y 2 = Y 3 = Y^ = 1, Y^ = Y^ = 
 
 St ep 1 : Find A and B 
 K — q P 
 
 A^ - (6, 13, 22, 209) 
 
 B = (5, 6, 13, 25, 28, 63, 9*+, 106, 111, Uhk, 601, 980, 4000, 2 100 -l) 
 
 Step 2 : Find G = A \A Hb 
 
 — K — q q P 
 
 G = (6, 13, 22, 209)\(6, 13) = (22, 209) 
 
 These are the two solutions 
 
 Step 3 ' Assign binary values to the variables J 
 
 In "binary, 22 = 0" ••• 0010110 .* x rv: =x r .Q=x or .=l, all other x's = 0, 
 
 9o 9° 99 
 
 209 = •••• 011010001 .% x =x .^x^^Q^l, all other 
 x's = 0, 
 
95 
 
 Cas 
 
 e 2: If Y =Y =Y =Y, =Y =Y,=1, A Hb is the empty set, and 
 123^5 oq,p ' 
 
 G = A . In this case, A =6, which is the unique solution. 
 
 Instead of finding (k.3) by set-theoretic operations, 
 the corresponding logical function may be found as 
 
 g = a q • (a-rTTp) ( i,. 8) 
 
 where 
 
 q p 
 
 a = IT Y b = S Y (U.9) 
 
 q a=l a P [3=1 X q4^ 
 
 If the given equations are in the form of (3.1) , they may 
 be converted to the characteristic functions Y. = 1 by forming 
 
 Y. = f.g. + f.g. = 1 (^.10) 
 
 l ii ii 
 
 and thus b =0. Expanding (^-.8) yields in this case 
 P 
 
 g = a(a +F)=ab = a (U,n) 
 
 q q p q p q v<+.xx/ 
 
 and finally, from (h.9) 9 
 
 g = ~[Y(t & + t ± R ± ) (h.12) 
 
 a=l 
 
96 
 
 which is identical to (3.2). Thus, the method for finding the 
 particular solutions would he that of the preceding section. 
 
 k.2. "Map" Method 
 
 [13] 
 This method by Nadler, is just a simple case of 
 
 Svohoda's method. Nadler assumes equations of the form 
 
 f i (x l3 x 2 ,„..,x r ) = (i=l,2,...,n) (1+.13) 
 
 and maps each function on a Veitch chart (a Karnaugh map may also he 
 used) . The solution is then found as the intersection of the n 
 maps. For the more general equations of the form (3.1), the 
 equivalent method would he to form n maps of the Boolean function 
 (J) , as in (3-8) ; then find the intersection of these maps as (|). 
 These are precisely the first two steps in Svohoda's algorithm. 
 Furthermore, Nadler mentions, as did Svoboda, that <J) may he found 
 on a single chart by mapping the complements of the equations (see 
 remark 1, Svohoda's algorithm). For (3»l) 5 this would be done by 
 mapping f fc © g fe . 
 
97 
 
 Example : Given the equations 
 
 f 1 (x iJ x 2 ,x 3 ,x u )- x 1 x 2 +(x l +x 2 +x 3 )x u +x l x 2 x 3 = x 1 x 2 +(x 2 +x 3 )x u 
 
 g 1 (x 1 ,x 2 ,x 3 ,x J+ ) 
 
 f 2 (x 1 ,x 2 ,x 3? x u )=x 1 x 2 +x i x 2 +x 2 (x 3 x 1+ +x 3 x u ) = x 2 fx- 1 +x 3 +x 1+ ) + 
 
 x l( x 3 V x 2 X 3 x ^ =g 2 (x l ,x 2 ,x 3 5 V 
 
 Step_l: Map f^ g^ f^ §2 ( fig . 2 l) 
 
 x, 
 
 Xn 
 
 
 
 1 
 
 1 
 
 
 
 
 
 1 
 
 
 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 1 
 
 V x 3 
 
 
 
 
 
 
 1 
 
 1 
 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
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 1 
 
 
 
 
 
 
 
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 1 
 
 1 — 1 
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 1 
 
 
 
 
 
 1 
 
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 1 
 
 
 
 1 
 
 1 
 
 1 
 
 
 
 
 
 
 
 1 
 
 1 
 
 
 
 1 
 
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 1 
 
 
 
 1 
 
 1 
 
 
 
 
 
 
 
 1 
 
 1 
 
 g. 
 
 Fig- 21 . 
 
98 
 
 Step 2 : Finding <j) = {) ' <j) ? was shown in Svoboda's example, so this 
 example will demonstrate finding <j) = (J) -kL, by putting a unit in 
 each cell where f / g , f / g (Figure 22). 
 
 0) = 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 1 
 
 1 
 
 _L 
 
 1 
 
 1 
 
 1 
 
 
 
 .1 
 
 1 
 
 
 
 1 
 
 *4 X 3 
 
 Figure 22 . 
 
 St 
 
 ep 3: The solutions are determined from x.-X^x^x, , x_,x x_x. . 
 
 r ' 1 2 3 ^ 1 2 3 r 
 
 X 1 X 2 X 3 X U' 1,e " 
 
 (a) x x = x = 1, x 2 = x^ = 
 
 (b) x 1 = x 2 - x^ = x^ = 1 
 
 (c) x 1 = x 2 = x^ = 1, x = 
 
 U.3. Pseudo-Boolean Equations 
 
 Define B as the two-element Boolean algebra (i.e., the 
 set {0,1] plus the binary operations of logical OR (+) , logical 
 
99 
 
 AND (•), and the logical unary operation of negation or comple- 
 mentation ( ~~" ) .. „ 
 
 A pseudo-Boolean function is one that maps the cartesian 
 product B ( = B X „.. X B n times) into the field of reals R, or 
 
 f : B 2 n -> R (U.13) 
 
 In other words, a function of bivalent variables which assumes some 
 real value (not necessarily or 1 as in a Boolean function) for 
 all possible inputs is a pseudo-Boolean function. Furthermore, if 
 a one-to-one correspondence is established between the real numbers 
 and 1 and the elements of B , then Boolean functions are just a 
 subset of pseudo-Boolean functions, since for every Boolean function 
 
 h : B*—* B 2 (k.lk) 
 
 Pseudo-Boolean functions most frequently have integer values; 
 
 indeed, these functions are called "integer algebraic functions" 
 
 [51 [8 ] 
 
 by R. Fortet J . The following method by P. L. Ivanescu 
 
 ro I 
 and S. Rudeanu for solving simultaneous pseudo-Boolean 
 
 equations might also be used to solve simultaneous Boolean equations, 
 
 but it will become obvious that this would be a trivial special 
 
 case. The method itself is a systematic tree-like construction and 
 
 is derived from Rudeanu' s method of Section :3«^> 
 

 The general form of a linear pseudo -Boolean equation Is 
 
 a l X l +b l X l +a 2 X 2 +b 2 X 2 + **°' +a n X n +b n X n = k ^'^ 
 
 where a., b. (i = l 9 2 so ..,n) and k are given reals, and a.^b. for 
 
 all i. If any a.^ or b."' 0, substitute either 
 l l 
 
 x. - 1 -x„ (for a. <0) (**.l6) 
 
 or 
 
 x = 1 - x (for b. <0) (*+.17) 
 
 into (^-.15) and collect terms,, Tae i-th equation is now written in 
 the "canonical form" 
 
 c^ x. + c^ x. +C...+ c* x. = d x (^18) 
 11 2 2 mm 
 
 where x = x or x ? x„ 9 „ „ . s x„ are the variables of the i-th equation 
 
 1 m 
 
 of the system (of n variables) s and 
 
 cf > c* >...>. c* > (U.19) 
 
 X l X 2~ ^ 
 
101 
 
 
 
 
 
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103 
 
 (Note that the subscript of a coefficient is not necessarily equal 
 to that of its corresponding variable; the former denotes the 
 position of the term in the rearranged equation, while the latter 
 merely differentiates among the variables) . 
 
 Table k.l below (this is table 10 from [8 J p. 39) is a 
 summary of all possible cases of (4.18) and the conclusions which 
 may be drawn in each case. 
 
 Obviously, if one equation has no solutions or two 
 equations result in different values for x. , the system of 
 equations is inconsistent. Depending on the amount of information 
 deduced from the equations, each equation will fall into one of 
 three classes: 
 
 1. Determinate - These are cases 1, 5,2,6, 3 ? 7 in table 4.1. Case 
 1 and 5 are to be considered first, since they conclude that no 
 solutions exist. Cases 2 and 6 are next considered, since in this 
 case all of the appearing variables are fixed. Finally, in cases 
 3 and 7> some of the variables are fixed. 
 
 2. Partially determinate - If there are no determinate equations , 
 those falling into case h would be first treated; in this case, 
 
 the attack is split into p + 1 cases, each with increased information, 
 
 3. Indeterminate - This is case 8, where almost no information is 
 available, and the attack is split into 2 cases (when all of the 
 system equations are indeterminate) . 
 
ioU 
 
 Example : Solve the system of equations 
 
 -x 1 +i+x +2x 2 -2x 1+ +x 1+ +6x -8x 6 = 5 (U.20.1) 
 
 x 1 +x 3 -3x 3 -5x 1+ +x 6 -x 6 +2x 7 = -i (U.20.2) 
 
 -9x' 1 -3x 2 +i+x 2 +i+x 3 +5x 6 -x 7 +x 7 = -k (U.20.3) 
 
 -3x 1 +2x 1 +7x -2x 1+ +2x ]4 -x 5 +2x 6 = k (U.20.U) 
 
 The transformations of (H.l6) and (U.17) change system (U.20) to 
 
 8x.+6x +5x 1 +3x, +2x = 16 (U.21.1) 
 
 5x^+Ur«+2x 6 +2x -hx =8 (U.21.2) 
 
 9oc L +7x 2 +5x 6 +^x 3 +25E 7 = 9 (^-21.3) 
 
 7x +5x +i+x^+2xV-Hx = 10 (lf.21.U) 
 
 Equations (U.20) are the "canonical" form of the system., Note that 
 (U.21.3) is the partially determinate case h 9 while all others 
 
105 
 
 are in the indeterminate case 8. Thus, there will be p + 1 = 2 
 possibilities: ( Of ) v = l, x =x^=x =5L,=0, and ( P )x = 0. 
 
 Case ( a ) : Xj^^x^l, ^ 3 ^=° 
 
 This will eliminate (4.21.3) and change the other equations to 
 
 6x +^ = 6 (4.22.1) 
 
 5*^ = 5 (4.22.2) 
 
 4x, +x = 1 (4.22.4) 
 
 From (4.22.2) , x, = 1, which results in x = 1 from (4.22.1) and 
 (4.22.4). Thus, the total solution is x^=x =x.=x =x=l, x =x,.=0, 
 
 Case ( p ) : x = transforms (4.21) to 
 
 8x" 6 +6x +3x^+2x2 = 11 (4.23.1) 
 
 5x^+4x +2x 6 +2x_ = 8 (4.23.2) 
 
 7i 2 +5x 6 +Ux 3 +25E 7 = 9 (^.23.3) 
 
 7x +4x^+2x 6 +x =5 (4.23.4) 
 
106 
 
 (^.23.^) is the determinate case 3 (all others are indeter- 
 minate) ; consequently, let x =0 (i.e., x =l) . The system becomes 
 
 8x 6 +6x +3x^+2x 2 = 11 (1+.24.1) 
 
 5x^+2x^+2x = 1* (U.2U.2) 
 
 7x 2 +5x 6 +2x 7 =5 (U.2U.3) 
 
 Ux^+2x 6 +x =5 (U.2U.U) 
 
 (U. 2k, 2) and (U.2U.3) are case 3, and (k m 2k.k) is case 7 (note that 
 
 these cases have the same priority) . Since x. = x, in (^.2^.2) and 
 
 X l 
 (k.2k.k) , it will he more convenient to treat one of these first, 
 
 Thus, choosing x, = yields the system 
 
 8x 6 +6x +2x 2 = 8 (U.2541) 
 
 2x 6 +2x = k (^.25.2) 
 
 75^+^+25^ = 5 C^.25.3) 
 
 2x 6 +x = 1 (±J25«X) 
 
io<7 
 
 (4.25.2) is the determinate case 6 which has higher priority than the 
 other equations. Hence, let x^- = x = 1, yielding 
 
 6x +2x 2 =8 (4.26.1) 
 
 7x 2 = (4.26.3) 
 
 x = 1 (4.26.4) 
 
 5 
 
 The values x =1, x_~l. obtained from (4.26.4) and (4.26.3) 
 5 2 
 
 respectively, satisfy (4.26.1) ; therefore, the complete solution 
 is x. = x k =0 > x n =x o =x h~ x f.- x r7-^- - Th e families of solutions found 
 in cases ( QJ ) and ( p ) are summarized in Table 4.2. 
 
 Table 4.2 
 
 X l X 2 X 3 X k X 5 X 6 X 7 
 
 110 110 1 
 
 Note that there are 2 = 128 possibilities for solutions, and the 
 only two solutions were found directly with no wasted effort. 
 
io6 
 
 k.h. Summary and Applications 
 
 As in Section 3, the map method is the most efficient 
 method for normal problems. Ashenhurst's method may be used 
 effectively for very simple systems. Grigor'yan's "set-theory" 
 method is the only one adequate to handle large systems with a 
 great many unknowns, although a computer program may be required. 
 The algorithm for the solution of pseudo -Boolean equations is quite 
 compact and systematic, and is one of the most efficient methods 
 of solution yet devised for problems of this type. 
 
 The methods for particular solutions of Boolean equations 
 are particularly useful in the design of logic circuits, where the 
 low and high potentials are represented by and 1 levels, 
 respectively, 
 
 Pseudo-Boolean equations are especially important in 
 linear programming methods of operations research, used to solve 
 all types of problems, and most frequently, those in economics and 
 business. The solutions of these equations are also beneficial to 
 the theory of graphs and flows in networks, as well as in switching 
 algebra . 
 
 Ivanescu and Rudeanu extend their system to include 
 inequalities. The procedure remains the same: ^ obtain the system 
 in "canonical form," make conclusions based upon determinate, 
 
109 
 
 partially determinate, and indeterminate equations to reduce the system, 
 continue the tree-like construction of the solution until all possible 
 solutions are obtained. A special table, similar to the one for 
 equations, is required, but by including inequalities, the power 
 of the method is greatly increased, especially for use in linear 
 programming. (See [8] , pp. 23-52). 
 
110 
 
 5. RELATED TOPICS 
 
 The folio-wing topics are either extensions of the methods 
 of Sections 3 and k, or are related to the solution of simultaneous 
 Boolean equations in a manner -which engenders them as areas of 
 future study „ 
 
 5.1 Programmed Methods 
 
 In -view of the amount of labor which would be required 
 for equations more complex than those given in the examples, 
 programmed methods should be very valuable, Grigor'yan mentions 
 that a program has already been written for his method; moreover, 
 since it was shown that his algorithm is analogous to Ashenhurst ' s, 
 the latter method is also programmable. 
 
 The map and matrix methods use graphical devices for 
 performing the logical expansion of equations. Since the computer 
 would now do this, these methods would be superfluous; however, it 
 would seem that they too could be easily programmed, and probably 
 with greater ease than the logical algebraic methods. Certainly 
 Ledley's methods would be of great use with a computer manipulating 
 a finite number of designation numbers „ Ivanescu and Rudeanu also 
 mention that their technique is being programmed. Indeed, it would 
 seem that the simultaneous solution of Boolean equations by • 
 computers might become instrumental in the design of new computers. 
 
Ill 
 
 5.2 Simultaneous Implications 
 
 In Section 3*6, it was shown that many word- logic problems 
 consist of propositions in the form of implications, i.e., if 
 A, then B. Furthermore, if A-*"B, Ledley's consequence solution 
 stated that #B must have a unit at least in all positions where 
 #A is a unit. Thus, Ledley's solution of general Boolean 
 functional equations can be modified slightly to solve systems 
 of simultaneous equivalences. 
 
 If f ^ A-^B = g , form the f and g matrices in the 
 
 K K K. K 
 
 usual manner. At this point, it is more convenient to think of 
 the remaining process in terms of Svoboda's algorithm. The map of 
 (j) will be formed by placing a only where a cell in f is 1 
 and the corresponding cell in g is a zero; this is the result 
 
 K 
 
 of the implication. The remainder of the solution is unchanged. 
 
 5.3 Systems Other Than Binary 
 
 Throughout this paper, all methods and problems have 
 considered only bivalent variables, since this is the most common 
 case. However, many of the techniques may be extended to systems 
 of ternary variables, quaternary variables, etc. For the strict 
 map methods, such as those of Svoboda and Nadler, the only 
 requirement is the extension of the Veitch chart to the new bases. 
 
112 
 
 For example, in a ternary system, there will be 3 possible 
 combinations of n variables. A sample map of three ternary 
 variables is shown below in Figure 23. In this chart, x=l, x=2, x=0. 
 
 y «= 1 
 y »= 2 ■= y 
 
 x 2 = 2 * x 2 
 Xg * 1 
 
 x x = 2 = x 1 
 
 X l P 1 
 
 Figure 23 
 
 It is necessary to define the logical operations of all 
 new bases . For radix three, the logical operations of AND, OR, and 
 NOT are defined as follows : 
 
113 
 
 AND 
 
 
 
 1 
 
 2 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 1 
 
 1 
 
 2 
 
 
 
 1 
 
 2 
 
 OR 
 
 
 
 1 
 
 2 
 
 
 
 
 
 1 
 
 2 
 
 1 
 
 1 
 
 1 
 
 2 
 
 2 
 
 2 
 
 2 
 
 2 
 
 NOT 
 
 
 
 
 1 
 
 1 
 
 2 
 
 2 
 
 
 
 Defining operations for logic of any base is easily 
 accomplished by noticing that the value of a OR b is the greater 
 of the values of a and b; also, a AND b takes the smaller value 
 of a and b, while NOT permutes the values cyclically. Hence, 
 for (m+l) -valued logic, the logical operations would be defined as 
 
 AND 
 
 
 
 123 ...ffl 
 
 
 
 
 
 ... 
 
 1 
 
 
 
 1 1 1 ... 1 
 
 2 
 
 
 
 1 2 2 ... 2 
 
 3 
 
 
 
 1 2 3 ... 3 
 
 • 
 
 
 i .'• r o'.'(»y '*"*'■ ■>.. " a 
 
 m 
 
 
 
 1 2 3 ... m 
 
 OR 
 
 
 
 1 
 
 2 
 
 3 
 
 . . . m 
 
 
 
 
 
 1 
 
 2 
 
 3 
 
 . . . m 
 
 1 
 
 1 
 
 1 
 
 2 
 
 3 
 
 . . . m 
 
 2 
 
 2 
 
 2 
 
 2 
 
 3 
 
 . . . m 
 
 3 
 
 3 
 
 3 
 
 3 
 
 3 
 
 . . . m 
 
 • 
 • 
 « 
 
 
 
 
 
 
 m 
 
 m 
 
 rn 
 
 m 
 
 m 
 
 . . . m 
 
 NOT 
 
 
 
 
 1 
 
 1 
 
 2 
 
 2 
 
 3 
 
 3 
 
 k 
 
 * 
 • 
 
 i 
 
 m 
 
 
 
11^ 
 
 Having defined the set of logical operations, Ledley's 
 methods may be extended for any new base. For ternary variables, 
 b [x ,x ] would be 
 
 b [x 1? x 2 J 
 
 #x = 012 012 012 
 
 #x 2 = 000 111 222 
 
 Designation numbers are found as usual, but in this case, 
 the new symbol, x, is possible. For example, 
 
 z = 120 120 120 
 
 I = 201 201 201 
 
 For Type 2 problems, the solution is unchanged, except the 
 new definitions of logical sum and product are used in the matrix 
 multiplication. For Type 3 problems, a new kind of matrix 
 multiplication must first be defined; this results in antecedence, 
 consequence, and "intercedence" problems, whose solutions are of no 
 greater difficulty than those in the binary system. Type 1 
 solutions may be found as before by the matrix product; for (m+l) - 
 
115 
 
 valued logic, an m is recorded in [(J)] , instead of a unit, as in the 
 "binary system. (For further details, see [lOJ, pp. kY9-kSk.) 
 
 Extending the logical algebraic methods would simply 
 entail the use of the new symbol x in the equations, and expansion 
 "by means of the newly defined logical operations. 
 
 5.U "Physical Mapping" 
 
 Instead of using a graph, such as a Veitch chart, to 
 evaluate a function for all possible combinations of variables, 
 Postley describes a method using IBM cards. If the 12 rows 
 and 80 columns of a card are regarded as 2 rows of U80 columns, 
 each card could be punched to represent some combination of n 
 (n < ^80) bivalent variables. 
 
 It has been previously shown that a function may be 
 represented as a sum of minterms. If each "type I" card represents 
 a minterm, then column q is punched in row 1 if x is in the minterm, 
 or in row 2 if x is in the minterm. In this manner, Type I card 
 decks representing each equation may be compiled. 
 
 Type II cards represent the value of each of the n 
 
 variables at time t. If at time t, x, is true, row 1 in column k 
 
 7 k 7 
 
 is punched; if x, is true, row 2 is punched, but in either case, 
 both rows of all other columns are punched. The deck of Type II 
 cards for time t is called the "this time table." 
 
116 
 
 If each Type I card is then sighted against the "this time 
 table," this would be equivalent to evaluating the functions and a 
 "next time table" may be formed. This system was devised for the 
 sequential equations of flip-flops, but could be extended, in a 
 manner similar to Grigor'y an ' s algorithm, for the determination of 
 the particular solution of general equations. 
 
 Represent an equation as the sum of minterms, and let each 
 minterm be represented by the number corresponding to its binary 
 value. The first 256 positions of a card would correspond to all the 
 possible minterms of 8 variables. By punching all positions except 
 those corresponding to the minterms in each equation,' and then sighting 
 the deck of equation cards, the solution is simply the unpunched terms 
 in all equations. This is equivalent to mapping § , and would seem 
 to offer no advantage to a similar mapping on a Veitch chart. 
 
 5.5 Boolean Ring Equations 
 
 It has been shown that a soluable Boolean equation: 
 
 f(x l5 x 2 ,...,x n ) =0 (5.1) 
 
 may be expressed as a sum of minterms and corresponding coefficients, 
 as in (5.2) . 
 
117 
 
 a l X lV ' 'V a 2 X lV • -V- • - +a 2 n X l X 2- " ' X n = ° (5 ' 2) 
 
 This is called the "normal form" of (5.1), and the coefficients are 
 called the "discriminants" of the equation. 
 
 A necessary and sufficient condition for the existence of 
 
 T?S 1 
 a unique solution to (5.2), first obtained by Whitehead , is 
 
 a x a 2 a = (5-3.1) 
 
 and a a. =0 (i / j) (5.3.2) 
 
 The elements of a Boolean algebra form a Boolean ring with 
 respect to the operations e and • . The operator " © " is called 
 "ring sum" in this case, and is defined as usual: 
 
 a © b = ab +ab ( 5 . h) 
 
 A Boolean ring is a ring with unity in which all elements 
 are idempotent. The following relations are true: 
 
 a © a = (5.5) 
 
 a © = a (5.6) 
 
118 
 
 From (5.8) , 
 
 a = 1 © a (5.7) 
 
 a+b = a © b © ab (5.8) 
 
 a © (b © c) = (a © b) © c (5.9) 
 
 a © b = b © a (5-10) 
 
 a(b © c) = ab © ac (5.11) 
 
 a+b = a © b iff.'ab = (5.12) 
 
 Thus, (5.2) may be written as 
 
 a l X l X 2'" X n ® a 2 X l X 2'" X n ® "* ® a n X l X 2*" X n = ° ^.13) 
 
 By means of (5-7) , (5.13) may be written as 
 
 b l X l X 2*" X n ® b 2 X l X 2*" X n-l ® "' ® b n = ° (5.1*0 
 
119 
 
 This equation is called the "O-normal ring form" of (5.1) , and 
 the b's are called the "normal coefficients." It is also easy to 
 sho-w that any Boolean equation may be expressed in the forms 
 
 a lXl x 2 ...x n ea 2 x 1 x 2 ...x n © ... ©a ^^..^ = l (5.15) 
 
 and P l x l X 2" ,X n ® P 2 X lV * ' X n-1 ® ••• ® 6 n = 1 (5.16) 
 
 called the "1-normal logical form" and the 7l -normal ring form," 
 respectively. 
 
 In a lengthy paper , Parker and Bernstein derive 
 numerous conditions for the unique solvability of Boolean ring 
 equations of the forms (5.1*0 and (5.I6) . These criteria give 
 conditions for the normal coefficients and are obtained in several 
 ways: by analogy to the conditions of logical equations, by direct 
 deduction of ring equations, and through the use of determinants 
 or "complexes" (special matrices similar to determinants). No 
 actual methods of solution are given, but the feasibility of solving 
 systems of simultaneous Boolean ring equations is distinctly proven. 
 
 The particular solutions of simultaneous ring equations may 
 be found by the map method of Section k.2. ,• Each equation is mapped on 
 a Veitch chart, and the intersection of the maps is the required 
 solution. The mapping is done by evaluating the equation for the 
 
120 
 
 binary variables of each cell, and placing a unit in each cell which 
 yields an identity (i.e., the equation is true). This is facilitated 
 by considering the ring operator to denote addition modulo 2, and is 
 made still easier if the equations are given in normal logical or 
 normal ring form. 
 
 5.6 Sequential Equations 
 
 Methods for the general and particular solutions of 
 simultaneous Boolean sequential equations have been demonstrated 
 by Phister and Postley, respectively. However, the equations 
 
 involved were of the most elementary type, i.e., the design 
 
 [2k'\ 
 equations of common flip-flops. Wang presents a very thorough 
 
 investigation of means of obtaining a set of explicit functions from 
 
 an implicit Boolean equation. 
 
 Wang adds a sequential or time operator d to the 
 
 ordinary Boolean operators such that x~^x, , dx— ^ x ++i > 
 
 2 
 d x ->• x , , etc. Some properties of this operator are: 
 
 d(A) = (dA), d(AB) = (dA) (dB) , d(A+B) = (dA)+(dB) (5.17) 
 
 d k G(A,„..,B) - G(d k A,...,d k B) . (5-l8) 
 
121 
 
 In the Boolean equation, 
 
 H(i,...,j,x,...,y,d ,...,d ,d , . . . ,d ) =0 (5.19) 
 
 i,....,j are input variables, and x,...,y are output variables. 
 
 The first problem investigated by Wang is to determine if 
 
 dx = f(d i ,...,d , i,...,j, x,...,y) 
 
 dy = g(d , ...,d ,i,...,j,x,...,y) 
 
 (5.20) 
 
 may be found. The functionals (5.20) are called "deterministic 
 
 sequential functionals", and procedures are given for determining 
 
 their existence, as well as obtaining them explicitly. 
 
 "Predictive sequential functionals" are effective solutions 
 
 for d ,...,d which can't be expressed as deterministic functionals. 
 x' y 
 
 In this case, if they exist, 
 
 2 2 Q Q 
 
 dx = f(i,...,j,x,...,y,d ,...,d ,d I,...,d j,...,d i,...,d j) 
 
 ■*- J 
 
 dy = g(i,...,j,x,...,y,d ,...,d ,d i,...,d j,...,d i,o..,d q j) 
 
 J 
 
 (5.21) 
 
122 
 
 and again, procedures are given for the determination of their 
 
 existence and for the explicit functionals. Deterministic and 
 
 predictive sequential functionals both lead to circuit realizations, 
 
 hut not so the final case. 
 
 If a solution for (5.19) exists, but there is no 
 
 effective solution for d ,...,d in terms of predictive functionals, 
 
 x y 
 
 this is termed a "noneffective" solution. A procedure involving a 
 "solution table" which determines the existence of the solution 
 and characterizes it is given. Note that all procedures yield 
 general parametric solutions. 
 
123 
 
 LIST OF REFERENCES 
 
 [1] Ashenhurst, R„ L., "Simultaneous Equations in Switching 
 
 Theory," Report number BL-4, BL-S, Bell Telephone Labs. 
 (Available from Library of Congress) . 
 
 [2] Boole, G.j George Boole's Collected Logical Works (1859) , 
 Volumes 1 and 2, The Open Court Publishing Company, 
 La Salle, Illinois, 1940. 
 
 [3] Carvallo, M., Principles et Applications de 1' Analyse 
 
 Booleene , Gauthier-Vi liars, Paris, 1965, (in French) . 
 
 [4] Follinger, 0., and Weber, W. , "Boolesche Gleichungen-Losung 
 und Anwendungen , " Elektronische Datenverarbeitung, 
 December, 1963, pp. 251-263, February, I96U, pp. 4-13, 
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 [5] Fortet, R., "Application de l'Algebre de Boole en Recherche 
 
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 [6] Grigor'yan, Y., "An Algorithm for the Solution of Systems of 
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 [7] Ivanescu, P. , "Systems of Pseudo-Boolean Equations and 
 
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 [8] Ivanescu, P., and Rudeanu, S., Pseudo-Boolean Methods for 
 
 Bivalent Programming , Springer-Verlag, Berlin-Heidelberg - 
 New York, 1966. 
 
 Klir, J., "Solution of Systems of Boolean Equations," 
 Aplikace Matimatiky , 7, 1962, pp. 265-273j (in Czech), 
 
 Ledley, R., Digital Computer and Control Engineering, 
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 Luce, R., "A Note on Boolean Matrix Theory," Proceedings of 
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 388. 
 
124 
 
 [12] Maitra, K. , "A Map Approach to the Solution of a Class of Boolean 
 Equations," Communications and Electronics, Volume 59> 
 March, 1962, pp. 34-37. 
 
 [13] Nadler, M. , and Elspas, B., "The Solution of Simultaneous Boolean 
 Equations." IRE Transactions on Circuit Theory (Correspondence), 
 Volume CT-7, September i960, pp. 345-346. 
 
 [l4] Parker, W. , and Bernstein, B., On Uniquely Solvable Boolean 
 Equations , University of California Press, Berkeley and 
 Los Angeles, 1955. 
 
 [15] Phister, M. Jr., Logical Design of Digital Computers, Wiley and 
 Sons, New York, 1958. 
 
 [l6] Postley, J., "A Method for the Evaluation of ft System of Boolean 
 Algebraic Equations," Math. Tables and Other Aids for Compu - 
 tation , Volume 9? 1955 j PP. 5-8. 
 
 [17] Rouche, N. , "Some Properties of Boolean Equations," IRE 
 
 Transactions on Electronic Computers , Volume EC-7, Number 4, 
 December 1958, pp. 29I-298. 
 
 [l8] Rudeanu, S., "Boolean Equations and Their Applications to the 
 Study of Bridge Circuits," I. Bull. Math, de la Soc. Sci 
 Math. Phys. de la R. P. Roumaine, 3(51), 1959, PP. 445-473. 
 
 [19] Rudeanu, S., "Irredundant Solutions of Boolean and Pseudo -Boolean 
 Equations, " Rev. Roumaine Math. Pures et Appl ., Volume 11, 
 Number 2, 1966, pp. 183-188. 
 
 [20] Schroeder, E., Algebra der Logik , Volume 1, Teubner Verlags- 
 gesellschaft, Leipzig, 1890. (in German) . 
 
 [21] Schubert, E. , "Simultaneous Logical Equations," Commun i cat i on s 
 and Electronics, Number 46, January i960, pp. IO8O-IO83. 
 
 [22] Svoboda, A., "An Algorithm For Solving Boolean Equations, " IEEE 
 Transactions on Electronic Computers , (Correspondence) , 
 Volume EC-12, October, 1963, pp. 557-559- 
 
 [23] Veitch, E., "A Chart Method for Simplifying Truth Functions," 
 Proc, Conf. Assn. Computing Machinery , May 2-3, 1952, pp. 
 127-133. 
 
125 
 
 [2k] Wang, H., "Circuit Synthesis by Solving Sequential Boolean 
 
 Equations , " Zeitschrift Math. Logik und Grundlagen d. Math . , 
 Volume 5, 1959, PP. 291-322. 
 
 [25] Whitehead, A., "Memoir On the Algebra of Symbolic Logic", 
 
 American Journal of Mathematics, Volume 23, 1901 , pp. 13*+- 
 165, 297-316. 
 
 [26] Zemanek, H„, "Die Losung von Gleichungen in der Schaltalgebra", 
 Arch. Elektr. Ubertragung , Volume 12, Number 1, January 1958, 
 PP. 35-kk. (In German) . 
 
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