1 0 0 1 1 1 Teisendus DNK kujule: 1 0 1 0 1 1 f = (x1 v x3)(x2 v x3 v x4)(xx 1 v xx 3 v xx 4) = (x1x2 v x1x3 1 0 1 1 0 0 v x1x4 v x2x3 v x3x3 v x3x4)(xx 1 v xx 3 v xx 4) = x1x2xx 1 v 1 1 0 0 1 1 x1x2xx 3 v x1x2xx 4 v x1x3xx 1 v x1x3xx 3 v x1x3xx 4 v x1x4xx 1 1 1 0 1 1 1 v x1x4xx 3 v x1x4xx 4 v x2x3xx 1 v x2x3xx 3 v x2x3xx 4 v x3xx 1 v x3xx 3 v x3xx 4 v x3x4xx 1 v x3x4xx 3 v x3x4xx 4 = x1x2xx 3 v 1 1 1 0 1 1 x1x2xx 4 v x1x4xx 3 v xx 1x3 v x3xx 4 1 1 1 1 0 0
=(xx 1 ∨ xx 1 xx 4) ( xx 3 xx 4 ∨ xx 1 xx 4) ∨ (xx 1 ∨ xx 1 xx 4) ( xx 3 xx 4 ∨ xx 1 xx 4) = = xx 1 (xx 1 xx 4 ) ( xx 3 xx 4 ∨ xx 1 xx 4) ∨(xx 1 ∨ xx 1 xx 4) ( xx 3 xx 4) (xx 1 xx 4) = = x1(x1 ∨ x4) ( xx 3 xx 4 ∨ xx 1 xx 4) ∨(xx 1 ∨ xx 1 xx 4)(x3 ∨ x4)(x1 ∨ x4)= = (x1 ∨ x1x4) ( xx 3 xx 4 ∨ xx 1 xx 4) ∨(xx 1 ∨ xx 1 xx 4)( x1x3 ∨ x1x4∨ x4x3 ∨ x4)= =x1 xx 3 xx 4 ∨ x1xx 1 xx 4∨ x1x4xx 3 xx 4 ∨ x1x4 xx 1 xx 4 ∨ xx 1x1x3 ∨ xx 1x1x4∨ xx 1x4x3 ∨ xx 1x4∨ xx 1 xx 4x1x3 ∨ xx 1 xx 4x1x4∨ xx 1 xx 4x4x3 ∨ xx 1 xx 4x4= = x1 xx 3 xx 4 ∨ xx 1x4x3 ∨ xx 1x4 = x1 xx 3 xx 4 ∨ xx 1x4 Leida ja esitada punktis 3 saadud MDNK jaoks tema tuletis muutuja x3 järgi. f(x1,x2,x3,x4) = x2 xx 3 xx 4 ∨ xx 1 xx 2 ∨ xx 1 xx 4 δf (x 1 x 2 x3 x 4 ) δ x3 =f(x1x2*0*x4)f(x1x2*1*x4) = (x2 xx 4 ∨ xx 1 xx 2 ∨ xx 1 xx 4)(xx 1 xx 2 ∨ xx 1 xx 4) =
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