ÜLESANNE 5 Leida vabaltvalitul viisil punktis 2 saadud MKNK-ga loogiliselt võrdne Täielik KNK (x1Vx2Vx3)&( x1V x 4 )&( x 1 V x 2 Vx3)&( x 1 V x2V x 3 ) = (x1Vx2Vx3Vx4)& &(x1Vx2Vx3V x 4 )&( x 1 V x 2 Vx3Vx4)&( x 1 V x 2 Vx3V x 4 )&( x 1 V x2V x 3 Vx4) &( x 1 V x2V x 3 V x 4 )&(x1V x 4 Vx3)&(x1V x 4 V x3 ) = (x1Vx2Vx3Vx4)& &(x1Vx2Vx3V x 4 )&( x 1 V x 2 Vx3Vx4)&( x 1 V x 2 Vx3V x 4 )&( x 1 V x2V x 3 Vx4) & &( x 1 V x2V x 3 V x 4 )&(x1Vx2V x3V x 4 )& (x1V x 2 Vx3V x 4 )&(x1Vx2V x3 V x 4 ) &(x1V x 2 V x3 V x 4 ) TKNK f(x1,x2,x3,x4) = (x1Vx2Vx3Vx4)&(x1Vx2Vx3V x 4 )&( x 1 V x 2 Vx3Vx4)&( x 1 V x 2 Vx3V x 4 ) &( x 1 V x2V x 3 Vx4)&( x 1 V x2V x 3 V x 4 )&(x1Vx2V x3V x 4 )& (x1V x 2 Vx3V x 4 ) &(x1Vx2V x3 V x 4 )&(x1V x 2 V x3 V x 4 ) ÜLESANNE 6 Teha punktis 2 saadud MDNK-le Shannoni disjunktiivne arendus muutuja xi järgi, mida esineb MDNK-s kõige rohkem MDNK f = x1x2x3Vx1 x 2 x3 V x1 x2 x 4 V x1 x3 x 4
0-0- 1 1 0 1 0 0 0--0 10 1 0 0 0 1 -0-1 0 0 1 0 0 0 -0-1 0 0 1 0 0 1 -0-1 0 0 1 0 0 1 -11- 0 0 0 0 1 1 1-10 0 0 0 0 1 0 101- 0 0 0 0 0 0 Tallinn University of Technology 2012 11 {&, ¬} X3X4vX1X3X4vX2X3vX1X2X4 = (X4vX2)X3v(X3X4vX2X4)X1 = (X4vX2)X3(X3=X4XvX 4X22X X43)X X31X4X2X4X1 (X3vX1)(X2vX4)(X1vX3vX4) = (X3vX1)(X2vX4)(X1vX3vX4) = (X3X1)(X2X4)(X1X3X4) Tallinn University of Technology 2012 12 {} X4X2X3X3X4X2X4X=1 (X4X2X3) )=
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