Solution of one system of boolean equations via reverse pass per Helen Mironchick (II)

"Ignorance of the mapping method is still half the misfortune ..."
  B.D. 

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  I do emphasize again that core idea of reverse pass to center belongs 
 to Helen Mironchick (unpublished manuscript) .
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Original system

(x1^x2=>x3)^x4=>x5=1
(y1^y2=>y3)^y4=>y5=1
(z1^z2=>z3)^z4=>z5=1
(x1=>y3)=>z4=1

#1 - x1
#2 - x1^x2
#3 - (x1^x2=>x3)
#4 - (x1^x2=>x3)^x4
#5 - (x1^x2=>x3)^x4=>x5)

   Passing Polyakov's Control

Solution of one system of boolean equations via reverse pass per Helen Mironchick

"Ignorance of the mapping method is still half the misfortune ..."
  B.D.

Down here we rely on technique proposed in 
https://mapping-metod.blogspot.com/2019/03/blog-post.html
Original system

((x1^x2=>x3)^x4=>x5)^x6=>x7=1
((y1^y2=>y3)^y4=>y5)^y6=>y7=1
x1=>y7=1

#1 - x1
#2 - x1^x2
#3 - (x1^x2=>x3)
#4 - (x1^x2=>x3)^x4
#5 - (x1^x2=>x3)^x4=>x5)
#6 - (x1^x2=>x3)^x4=>x5)^x6
#7 - ((x1^x2=>x3)^x4=>x5)^x6=>x7=1

   Passing Polyakov's Control

Another system solved via pass to center per Helen Mironchick unpublished manuscript 

((x1^x2=>x3)^x4=>x5)^x6=>x7=1
((y1^y2=>y3)^y4=>y5)^y6=>y7=1
x1=>y5=1

#1 - x1
#2 - x1^x2
#3 - (x1^x2=>x3)
#4 - (x1^x2=>x3)^x4
#5 - (x1^x2=>x3)^x4=>x5)
#6 - (x1^x2=>x3)^x4=>x5)^x6
#7 - ((x1^x2=>x3)^x4=>x5)^x6=>x7=1

    Passing Polyakov's Control

Метод Отображений - тестирование проход к центру и обратный проход

Оригинальна схема прохода к центру предложена 
Еленой А. Мирончик . Смотри 
https://vk.com/doc6125348_505417587?hash=44fb5048df995959ea&dl=210d2b189eb483566a

Одно из основных преимуществ метода - это толерантность к числу уравнений системы

Исходная система

((((x1=>x2)=>x3)=>x4)=>x5)=>x6 =1
((((y1=>y2)=>y3)=>y4)=>y5)=>y6 =1
((((z1=>z2)=>z3)=>z4)=>z5)=>z6 =1
(x3=>y4)=>z5 =1

                                  X3 Y4 Z5

                            0    23  19  27
                            1    20  24  16

 Контроль по Полякову

   Ссылки

  1. https://vk.com/doc6125348_505417587?hash=44fb5048df995959ea&dl=210d2b189eb483566a

Solution of one problem 23 from USE in Informatics 2019

Original task

Solution

System below is equivalent to original one.
The previous system is supposed to make convertion itself
a real headaches generator

(x1=>x2)^(y2=>y1)^(x1 v y1)=1
(x2=>x3)^(y3=>y2)^(x2 v y2)=1
(x3=>x4)^(y4=>y3)^(x3 v y3)=1
(x4=>x5)^(y5=>y4)^(x4 v y4)=1
(x5=>x6)^(y6=>y5)^(x5 v y5)=1
x6 v y6=1

Proceed with building fork diagram with transition pair (x2,y2) &&  forking matrix

   Passing Polyakov's Control