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wi xi using barcode generator for .net framework control to generate, create qr code jis x 0510 image in .net framework applications. SQL Server 2000/2005/2008/2012 T <T then f = 1, then f = 0. (7.5). When V2 replaces V1 and xj replaces xj , we nd that if wj xj + wi xi T wj < T wj then g = 1, then g = 0,. (7.6). where g is the function real VS .NET qrcode ized by element V2 . To prove that g and f are identical functions, let xj = 0 so that xj = 1.

Then Eqs. (7.5) and (7.

6) become identical. Next, let xj = 1 so that xj = 0. Again, Eqs.

(7.5) and (7.6) become identical.

Consequently, both f and g assume identical values for each input combination and are thus identical functions. The above property leads to several important conclusions. If a function is realizable by a single threshold element then, by an appropriate selection of complemented and uncomplemented input variables, it is possible to obtain a realization by an element whose weights have any desired sign distribution.

Therefore, if a function is realizable by a single threshold element then it is realizable by an element with only positive weights. Clearly, this assertion is valid only if the input variables are available in both complemented and uncomplemented forms. We shall next show that if a function f (x1 , x2 , .

. . , xn ) is realizable by a single threshold element whose weight threshold vector is V1 = {w1 , w2 , .

. . , wn ; T } then its complement f (x1 , x2 , .

. . , xn ) is realizable by a single threshold element whose weight threshold vector is V2 = { w1 , w2 , .

. . , wn ; T }, under a given condition.

1. The condition requires us to restrict the values of the weights and threshold such that for no input combination will the weighted sum be exactly equal to T .. 7.2 Synthesis of threshold networks From the inequalities in Eq. (7.1) and from V1 we obtain wi xi > T when f = 1,. i=1 n wi xi < T when f = 0. (7.7). Multiplying both sides of Eq. (7.7) by 1 yields wi xi < T when f = 1 or f = 0,. i=1 n wi xi > T when f = 0 or f = 1. (7.8). Clearly, the inequalities in Eq. (7.8) demonstrate that f is realizable by the threshold element whose weight threshold vector is V2 .

. 7.2 Synthesis of threshold networks Our principal goal in this s ection is the development of methods for the identi cation and realization of threshold functions as well as for the synthesis of networks of threshold elements, called threshold networks. Before proceeding with this general study, we shall present a number of properties of threshold functions that provide the theoretical background necessary for the development of simpler and more effective synthesis methods. We shall be concerned with the synthesis of threshold functions as well as the realization of nonthreshold functions with a network of threshold elements.

. Unate functions A function f (x1 , x2 , . . .

, xn ) is said to be positive in a variable xi if there exists a disjunctive or conjunctive expression for the function in which xi appears only in uncomplemented form. Analogously, f (x1 , x2 , . .

. , xn ) is said to be negative in xi if there exists a disjunctive or conjunctive expression for f in which xi appears only in complemented form. If f is either positive or negative in xi then it is said to be unate in xi .

Example The function f = x1 x2 + x2 x3 is positive in x1 and negative in x3 but is not unate in x2 . If a function f (x1 , x2 , . .

. , xn ) is unate in each of its variables then it is called unate. Thus, a function is unate if it can be represented by a disjunctive or conjunctive expression in which no variable appears in both its complemented and uncomplemented forms.

. Threshold logic for nanotechnologies Example The function f = x1 x2 + x1 x2 x3 is unate because a disjunctive expression for f exists that satis es the above de nition, namely, f = x1 x2 + x2 x3 . However, the function f = x1 x2 + x1 x2 is clearly not unate in either of its variables. If f (x1 , x2 , .

. . , xn ) is positive in xi then it can be expressed as f (x1 , x2 , .

. . , xn ) = xi g1 (x1 , .

. . , xi 1 , xi+1 , .

. . , xn ) + h1 (x1 , .

. . , xi 1 , xi+1 , .

. . , xn ).

f (x1 , x2 , . . .

, xn ) = xi g2 (x1 , . . .

, xi 1 , xi+1 , . . .

, xn ) + h2 (x1 , . . .

, xi 1 , xi+1 , . . .

, xn ). (7.9).

Similarly, if f (x1 , x2 , . qr barcode for .NET .

. , xn ) is negative in xi then it can be expressed as (7.10).

By de nition, if a function f can be expressed by Eq. (7.9) (Eq.

(7.10)) then it is positive (negative) in xi . Hence, the existence of two such functions, g1 and h1 (g2 and h2 ), is a necessary and suf cient condition for f to be positive (negative) in xi .

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