x1x2 00 1 1 1 0 01 0 0 0 0 11 1 1 0 0 10 1 1 1 0 17.3.14 T. Evartson 45 x1 1 1 x3 1 y x2 1 & 1 x4 17.3.14 T. Evartson 46 Lihtsustada ja koostada loogikaskeem X1 ( X2 + X3X4 + X1 X2) + X1X3X4 +X2X4 = X1X2 + X2X4 + X1X3 X1 ( X2 + X3X4 + X1 X2) + X1X3X4 +X2X4 = X1X2 + X2X4 X1 ( X2 + X3X4 + X1 X2) + X1X2X4 +X2X4 = X1 + X2X4 X1 ( X3 + X3X4 + X1 X2) + X1X3X4 +X2X4 = X1 + X2X4 X3 ( X2 + X3X4 + X1 X2) + X1X3X4 +X2X4 = X1X3 + X2X3 + X2X4 17.3.14 T. Evartson 47
(xx 1 v x3 v xx 4) (xx 3 v x4)) (xx 1x3 v x1xx 4 v x1xx 3x4 v x3xx 4) v (xx 1xx 4 v xx 1x3 v x1xx 3x4 v x3xx 4) ((x1 v xx 3) (xx 1 v x4) (xx 1 v x3 v xx 4) (xx 3 v x4)) = ((x1 v xx 3x4)(xx 1xx 3 v xx 1x4 v x3x4 v xx 3xx 4))(xx 1x3 v x1xx 4 v x1xx 3x4 v x3xx 4) v (xx 1xx 4 v xx 1x3 v x1xx 3x4 v x3xx 4) ((x1x4 v xx 1xx 3 v xx 3x4)(xx 1xx 3 v xx 1x4 v x3x4 v xx 3xx 4)) = (x1x3x4 v x1xx 3xx 4 v xx 1xx 3x4)(xx 1x3 v x1xx 4 v x1xx 3x4 v x3xx 4) v (xx 1xx 4 v xx 1x3 v x1xx 3x4 v x3xx 4)(x1x3x4 v xx 1xx 3 v xx 1xx 3x4 v xx 1xx 3xx 4) = x1xx 3xx 4 v xx 1xx 3xx 4 10.3.x3 järgi: f = f(x1,x2,0,x4) f(x1,x2,1,x4) = (xx 1xx 2xx 4 v xx 10 v x1x2xx 4 v x11x4 v 0xx 4) (xx 1xx 2xx 4 v xx 11 v x1x2xx 4 v x10x4 v 1xx 4) = (xx 1xx 2xx 4 v x1x2xx 4 v x1x4) (xx 1xx 2xx 4 v xx 1 v x1x2xx 4 v xx 4) = (xx 1xx 2xx 4 v x1x2xx 4 v x1x4) (xx 1 v xx 4) =
x3 = x2 x3 (x1 x2 )(x1 x3 x2 )= (x1 x2 x1 x2 )( x1 x3 x2 )= (x1 x2 x1 x2 )( x1 x3 x2 )= x1 x2 x1 x1 x2 x3 x1 x2 x2 x1 x2 x1 x1 x2 x3 x1 x2 x2 = x1 x2 x1 x2 x3 0 0 x1 x2 x3 x1 x2 = x1 x2 x1 x2 3.Lihtsustada etteantud loogikaavaldis DNK-ks põhiseoste ja tehete asendusseoaste abil: (x1x2x3x4 x3x4 x1x3 )(x1 x4 )(x1 x4 )= (x1x2x3x4 x3x4 x1x3 )(x1 x4 )(x1 x4 )= =(x1x2x3x4 x3x4 x1x3 )(0 x4x1 x1x4 0)= x3x4x1 x1x3x4 = x1x3x4 x1 x4 x3 (x1 x2)(x1x2 x4 )= x1 x4 x3 (x2 x1 )(x1 x2 x4 )= x1 x4 x3 x2x1 (x1 x2 x4 )= = x1 x4 x1 x2 x3 x1 x2 x2 x1 x2 x1 x1 x2 x1 x4 x4 x3 x4 x2 x4 x1 x4 =x1 x2 x4 x1 x2 x3 x3 x4 x2 x4 x1 x4 x1 x1 x2 ( x1 x3 ) x1 x1 x2 x3 x4 x1 = x1 ( x1 x3 ) x1 x1 x2 x3 x4 x1 = (x1 x1x1 x1 x3 x1 x2 x3 x4 ) x1 = =(x1 x1 x1 x3 x1 x2 x3 x4 )x1 =(x1 x1 x2 x3 x4 )x1 = x1 x1 x1 x1 x2 x3 x4 =00=0 (x1x3 )x1 (x1 x2 )(x1 x3 )(x3 x2 )=( x1 x3 x1 x3)x1 (x1 x2 )(x1 x3 )(x3 x2)=
) (3/7) 0-11 (13 1101 ) (10/11) 101- 3 (7) 0111 (13/15) 11-1 (11) 1011 3 (7/15) -111 4 (15 1111 ) (11/15)Tallinn 1-11University of Technology 2012 7 - II . 1000 1100 1101 1111 0111 0010 --11 1-00 -01- 11-1 110- -01- 0 0 0 0 0 1 X3X4 X1X3X4 X2X3 X1X2X4 --11 0 0 0 1 1 0 X1X2X3 1-00 1 1 0 0 0 0 10-0 1 0 0 0 0 0 00-1 0 0 0 0 0 0 X3X4vX1X3X4vX2X3vX1X2X4 110- 0 1 1 0 0 0 11-1 0 0 1 1 0 0 Tallinn University of Technology 2012 8 - I 000 0 0 (0) 0000 010
(8/10) 1 0 -- 0 (10/11) 1 0 1 -- (4/5/6/7/4/6/5/7) 0 1 -- -- (3/11) -- 0 1 1 (3/11/7/15/3/7/11/15) -- -- 1 1 (4/5/6/7/4/6/5/7) 0 1 -- -- . . (3/11/7/15/3/7/11/15) -- -- 1 1 . -- X1X2 v X3X4 v X1X2X3 v X1X2X4 v X2X3X4 v X1X3X4 · II 0100 0101 0011 0110 1011 0111 1. 0--00 1 0 0 0 0 0 --000 0 0 0 0 0 0 10--0 0 0 0 0 0 0 01---- 1 1 0 1 0 0 101-- 0 0 0 0 1 0
´x 3 v ´x 2x4 v ´x 1 ´x 3 v x2 ´x 3 v ´x v ´x 3x4 v ´x 1x4 v x2x4 v ´x 3x4 v x4)(x3 v ´x 4) = x1x2 x3 v x1x3x4 v ´x ´x 3 1 x v ´x 2x3x4 v ´x 1x3x4 v x2x3x4 v x3x4 v v x1x2 ´x
= (x2xx4 V xx2xx3 V xx3xx4) (xx2 xx3 V x2x4) V (xx2 x3 V x2x4)(x2xx4 V xx2x3 V x3xx4) = = xx2xx3 V xx2xx3xx4 V xx2x3 V xx2x3xx4 = xx2xx3 V xx2x3 = xx2 6 Muutuja x2 järgi: δ f ( x1x 2x 3 x4) δx2 = (x1x3 V x1x3) ⊕ (x4) = (xx4) (xx1x3 V x1xx3) V (x4)= = (xx4) (xx1x3 V x1xx3) V (x4)[(x1 V xx3)(xx1 V x3)] = xx1x3xx4 V x1xx3xx4V x1x3x4 V xx1xx3x4 Muutuja x3 järgi: δ f ( x1x 2x 3 x4) δx3 = (xx1xx2 V x2x4) ⊕ (x1xx2 V x2x4) = = (x1xx2 V x2x4) V (xx1xx2 V x2x4) = = [(x1 V x2)(xx2 V xx4)]( x1xx2 V x2x4) V (xx1xx2 V x2x4) [(xx1 V x2)(xx2 V xx4)] = = x1xx2 V x1xx2xx4 Vxx1xx2 V xx1xx2xx4 = x1xx2 V xx1xx2 = xx2 Muutuja x4 järgi: δ f ( x1x 2x 3 x4) δx4 = (xx1xx2x3 V x1xx2xx3 V x2) ⊕ (xx1xx2x3 V x1xx2xx3) = = (xx1 xx2x3 V x1xx2 xx3) V (xx1 xx2x3 V x1xx2xx3 V x2)=
Tallinna Tehnikaülikool Diskreetne matemaatika KODUTÖÖ Kristjan Keskküla 093540 IASB Tallinn 2009 ÜLESANNE 1 Leida oma martiklinumbrile vastav 4-muutuja loogikafunktsioon f(x1, x2, x3, x4) = (2,4,8,9,14,15) (6,11,13) _ (järgnevalt kui funktsioon) 1 ÜLESANNE 2 Leida MDNK ja MKNK, mis sobiksid martiklinumbrist leitud osaliselt määratud 4-muutuja funktsiooni esitamiseks Kuna minu martiklinumber on paarisarvuline leian: MKNK Karnaugh' kaardiga ja MDNK McCluskey' meetodiga. 1) Leian MKNK Karnaugh' kaardiga MKNK leidmiseks joonestan Karnaugh' kaardi, kuhu kannan peale funktsiooni 1d, 0d ja määramatused. x3x400 01 11 10 x1x2 00 0 0 0 1 01 1 0 0 - 11 0 - 1 1 10 1 1 - 0 Tegu ...
X1X4 2,1796326299 X2X3 6,2780697533 X2X4 2,9855420624 X3X4 2,5142605106 X1X2X3 7,3475809864 X1X3X4 3,4729028416 X1X2X4 4,1457010814 X2X3X4 4,2014352048 X1X2X3X4 4,9987527278 SUMMARY OUTPUT Regression Statistics Multiple R 0,5624358723
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