"Ignorance of the mapping method is still half the misfortune ..." B.D.
Original system
((x1vy1)^z1)⊕((x2=>y2)=>z2)=1
((x2vy2)^z2)⊕((x3=>y3)=>z3)=1
(x1≡y2)=>(y2≡z3) =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≡z3) = 0
Two additional passes with predefined variables which are based on the same fork diagram as for system (1). I am targeting to focus yours attention on details of technique had been published in 2013
in http://kpolyakov.spb.ru/download/mea-2013-10.pdf
x1=1 ; y2=1 ; z3=0 gives 28
x1=0 ; y2=0 ; z3=1 gives 16
Total 44
Answer is 140-44 = 96
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Passing Polyakov's Control
A bit more complicated sample
((x1vy1)^z1)⊕((x2=>y2)=>z2)=1
((x2vy2)^z2)⊕((x3=>y3)=>z3)=1
((x3vy3)^z3)⊕((x4=>y4)=>z4)=1
((x1≡y2)=>(y2≡z3)) =>z4=1
Truth Table obviously stays the same
Fork Mapping Method tables (total 6) for the following system
((x1vy1)^z1)⊕((x2=>y2)=>z2)=1
((x2vy2)^z2)⊕((x3=>y3)=>z3)=1
((x3vy3)^z3)⊕((x4=>y4)=>z4)=1
((x1≡y2)=>(y2≡z3)) =>z4=0
Total six passes have been performed due to this and
only this bit's chains matches equation
((x1≡y2)=>(y2≡z3)) =>z4=0
1) x1=1; y2=1; z3=1; z4=0 (1=>1) =1
2) x1=0; y2=0; z3=0; z4=0 (1=>1) =1
------------------------------------------
3) x1=0; y2=1; z3=0; z4=0 (0=>0) =1
4) x1=1; y2=0; z3=1; z4=0 (0=>0) =1
------------------------------------------
5) x1=1; y2=0; z3=0; z4=0 (0=>1) =1
6) x1=0; y2=1; z3=1; z4=0 (0=>1) =1
Predefined zeros have been typed bold in
excel spreadsheets
Total for false of last implication is 60+40+20+16+20+32=168
Passing Polyakov's Control
Original system
((x1vy1)^z1)⊕((x2=>y2)=>z2)=1
((x2vy2)^z2)⊕((x3=>y3)=>z3)=1
(x1≡y2)=>(y2≡z3) =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≡z3) = 0
Two additional passes with predefined variables which are based on the same fork diagram as for system (1). I am targeting to focus yours attention on details of technique had been published in 2013
in http://kpolyakov.spb.ru/download/mea-2013-10.pdf
x1=1 ; y2=1 ; z3=0 gives 28
x1=0 ; y2=0 ; z3=1 gives 16
Total 44
Answer is 140-44 = 96
Click on snapshot to switch to perfect slides show, either collapse
menu of current month blog archive
Passing Polyakov's Control
A bit more complicated sample
((x1vy1)^z1)⊕((x2=>y2)=>z2)=1
((x2vy2)^z2)⊕((x3=>y3)=>z3)=1
((x3vy3)^z3)⊕((x4=>y4)=>z4)=1
((x1≡y2)=>(y2≡z3)) =>z4=1
Truth Table obviously stays the same
((x1vy1)^z1)⊕((x2=>y2)=>z2)=1
((x2vy2)^z2)⊕((x3=>y3)=>z3)=1
((x3vy3)^z3)⊕((x4=>y4)=>z4)=1
((x1≡y2)=>(y2≡z3)) =>z4=0
Total six passes have been performed due to this and
only this bit's chains matches equation
((x1≡y2)=>(y2≡z3)) =>z4=0
1) x1=1; y2=1; z3=1; z4=0 (1=>1) =1
2) x1=0; y2=0; z3=0; z4=0 (1=>1) =1
------------------------------------------
3) x1=0; y2=1; z3=0; z4=0 (0=>0) =1
4) x1=1; y2=0; z3=1; z4=0 (0=>0) =1
------------------------------------------
5) x1=1; y2=0; z3=0; z4=0 (0=>1) =1
6) x1=0; y2=1; z3=1; z4=0 (0=>1) =1
Predefined zeros have been typed bold in
excel spreadsheets
Total for false of last implication is 60+40+20+16+20+32=168
Passing Polyakov's Control
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