Once again apply 08.2016 chart style to solve system
Convert to equivalent
(x1=>x2) v (x3=>x4)=1
(x3=>x4) ^ (x5=>x6)=0
(x5=>x6) v (x7=>x8)=1
(x7=>x8) ^ (x9=>x10)=0
Original idea was proposed in http://kpolyakov.spb.ru/download/mea-2016-8.pdf
Now consider system which seems hard to to solve at first sight
Convert to equivalent
(x1=>x2) v (x3=>x4)=1
(x3=>x4) ^ (x5=>x6)=0
(x5=>x6) v (x7=>x8)=1
(x7=>x8) ^ (x9=>x10)=0
Original idea was proposed in http://kpolyakov.spb.ru/download/mea-2016-8.pdf
Now consider system which seems hard to to solve at first sight
((x1=>x2)=>x3)v((x4=>x5)=>x6)=1
((x4=>x5)=>x6)^((x7=>x8)=>x9)=0
((x7=>x8)=>x9)v((x10=>x11)=>x12)=1
((x10=>x11)=>x12)^((x13=>x14)=>x15)=0
However, forking 08.2016 chart solves problem in a couple of minutes
Classic pairs solution for original system. Needless to comment. Yes 08.2016 has obvious limitations However in USE's 23-s they appear to be gone pretty often by some mysterious reasons. More over the particular task even doesn't require cross-reference tables generation. See post
https://vk.com/bderzhavets?w=wall209645472_496%2Fall
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