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Solution of one system in Boolean variables via Classic Mapping Method
dbaxps: Original system ((x1vy1)^z1)⊕((x2=>y2)=>z2)=1 ((x2vy2)^z2)⊕((x3=>y3)=>z3)=1 (x1≡y2)=>(y2≡yz3) =1 Solving system (1) ((x1vy1)^z1)⊕((x2=>y2)=>z2)=1 ((x2vy2)^z2)⊕((x3=>y3)=>z3)=1 Result 140 Solving system (2) ((x1vy1)^z1)⊕((x2=>y2)=>z2)=1 ((x2vy2)^z2)⊕((x3=>y3)=>z3)=1 (x1≡y2)=>(y2≡yz3) = 0 Two additional passes with predefined variables same fork diagram based x1=1 ; y2=1 ; z3=0 gives 28 x1=0 ; y2=0 ; z3=1 gives 16 Total 44 Answer is 140-44 = 96 Verification
Ответов - 4
dbaxps: @Attention of MEA I have no any idea how to make it smart without six passes over to exclude ((x1≡y2)=>(y2≡z3)) =>z4=0 solutions ((x1vy1)^z1)⊕((x2=>y2)=>z2)=1 ((x2vy2)^z2)⊕((x3=>y3)=>z3)=1 ((x3vy3)^z3)⊕((x4=>y4)=>z4)=1 ((x1≡y2)=>(y2≡z3)) =>z4=1
MEA: Эта система шибко "закручена", конечно. dbaxps пишет: six passes Меньше трех проходов придумать не могу. Но подозреваю, что принципиально Ваши шесть не отличаются от моих трех. Подробное решение
dbaxps: Елена Александровна, спасибо . Дело даже не в числе проходов, а в самом мышлении при анализе задачи.
dbaxps: Once again I strongly believe that technique proposed in http://kpolyakov.spb.ru/download/mea-2016-8.pdf seems to be underestimated either not well understood until date. Regardless Unified State Examination Polyakov's forum contains more then enough samples demonstrating advantages of 08.2016 technique. Consider system which is more complicated then P149 :- ((x1 v y1)^z1)=> ((x2=>y2)=>z2)=1 ((x2 v y2)^z2)=> ((x3=>y3)=>z3)=1 ((x3 v y3)^z3)=> ((x4=>y4)=>z4)=1 ((x4 v y4)^z4)=> ((x5=>y5)=>z5)=1 Solve this system via same approach Passing Polyakov's Control Obviously it might be solved via traditional 8x8 truth table per 2013 original MM design, what will cause painful procedure of building truth table in general having 64 rows to analyse.
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