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Geometric representation in .NET Attach QR in .NET Geometric representation VS .NET Code 128B

Geometric representation using barcode printer for none control to generate, create none image in none applications.create code 128 c# Unate functions have none for none several interesting properties, which are best illustrated by a geometrical representation. An n-cube contains 2n vertices, each of which represents an assignment of values to the n variables and thus corresponds to a minterm. A line is drawn between every pair of vertices that differ in just one variable, and no other lines are drawn.

The vertices corresponding to true minterms, that is, for which the function assumes the value 1, are called true vertices while those for which the function assumes the value 0 are called false vertices. The analogy between the n-cube and map methods for representing switching functions is evident. Example The three-cube representation of the function f = x y + xz is shown in Fig.

7.8. The bolder lines connecting the two pairs of true vertices, i.

e., the pair (1, 1, 1) and (1, 0, 1) and the pair (0, 0, 1) and (0, 0, 0), represent the cubes xz and x y , respectively..

iPhone OS (1,1,1). (1,1,0). (0,1,1). (1,0,1). (0,1,0). (1,0,0). (0,0,1). (0,0,0). Fig. 7.8 A three-cube (23 -vertex) representation of f = x y + xz. 7.2 Synthesis of threshold networks It is convenient to de ne a partial-ordering relation between vertices of the n-cube, such that (a1 , a2 , . . .

, an ) (b1 , b2 , . . .

, bn ) if and only if, for all i, ai bi . As shown in 2, this partially ordered set of vertices is a lattice and the vertices (0, 0, . .

. , 0) and (1, 1, . .

. , 1) are, respectively, the least vertex and the greatest vertex of the lattice. As in any partial ordering, some pairs of vertices may be incomparable, for example, (0, 0, .

. . , 0, 1) and (1, 0, .

. . , 0, 0).

Without loss of generality, we shall subsequently restrict our attention to unate functions that are positive in all their variables, that is, functions without any complemented variable. Such a restriction is justi ed because every complemented variable in a unate function may be relabeled, so that xi yi , etc., and obviously, the resulting function is unate if and only if the original one is.

For example, the unate function x1 x2 x3 + x2 x3 x4 may be converted to x1 x2 x3 + x2 x3 x4 , using the relabelings x1 x1 and x3 x3 . By reconverting the latter function it is possible to determine the original one. Theorem 7.

1 A switching function f (x1 , x2 , . . .

, xn ) is unate if and only if it is not a tautology2 and the above partial ordering exists, such that, for every pair of vertices (a1 , a2 , . . .

, an ) and (b1 , b2 , . . .

, bn ), if (a1 , a2 , . . .

, an ) is a true vertex and (b1 , b2 , . . .

, bn ) (a1 , a2 , . . .

, an ) then (b1 , b2 , . . .

, bn ) is also a true vertex of f . Proof Suppose that f is unate. Let us nd an expression that represents f as a positive function in all its variables.

Obviously, is not a tautology. If (a1 , a2 , . .

. , an ) is a true vertex then it represents an assignment of values to input variables that causes some prime implicant of to be true. If (b1 , b2 , .

. . , bn ) (a1 , a2 , .

. . , an ) then, for every ai = 1, the corresponding bi = 1.

Therefore, since is positive, (b1 , b2 , . . .

, bn ) also represents an assignment of values of the input variables, which causes at least the previously mentioned prime implicant to be true. This proves the only if part of the theorem. Now suppose that f is not a tautology and that, for every pair of its vertices (a1 , a2 , .

. . , an ) and (b1 , b2 , .

. . , bn ), if (a1 , a2 , .

. . an ) is true and (b1 , b2 , .

. . , bn ) (a1 , a2 , .

. . , an ) then (b1 , b2 , .

. . , bn ) is also true.

Since f is not a tautology, (0, 0, . . .

, 0) is a false vertex. Consider the k vertices S1 , S2 , . .

. , Sk , which are the minimal3 true vertices of the lattice. To each vertex Si there corresponds a product term that consists of just those uncomplemented literals whose corresponding value in Si is 1; for example, if for a function f (x1 , x2 , x3 , x4 ) we have Si = (0, 1, 0, 1) then the corresponding product term is x2 x4 .

The expression formed by the disjunction of the k product terms, which. A tautology is a fun none none ction which is equal to 1 for all combinations of its variables. A true vertex Si is said to be minimal if no other true vertex Sj < Si . A false vertex Si is said to be maximal if no other false vertex Sj > Si .

(See Section 2.3.).

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