Unleash the power of reverse pass to center per Helen Mironchick and solution of systems kind of P127 (Mufazzalov D.F.) from ege23.doc

Original system

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

Solution below is based on Helen Mironchick article in meantime pending publication. Intellectual property for idea of "Reverse pass to center" completely belongs Helen Mironchick. We just place in play her technique awaiting publication.

For X-LINE

#2 is (x1=>x2)
#3 is (x1=>x2)=>x3
#4 is ((x1=>x2)=>x3)=x4
#5 is (((x1=>x2)=>x3)=x4)=>x5

For Y-LINE and Z-LINE replace x by y and x by z correspondingly

*******************
Chaining order
*******************
1) We get x3 via reverse pass to center
2) Getting y3 via simple implication
3) Getting y4  via reverse pass to center
    Here we have y1=>y2=>y3 (#3) Y-LINE.
    Perform move from #5 to #4 in the opposite direction.
    Now calculate y4 as usual.
4) Getting z4 via simple implication
5) Completing Z-LINE having z4  (on line-0 213,on line-1-333)

Second system:-

Each one of system's equations are built upon sample 9 from 
http://kpolyakov.spb.ru/download/mea-2013-10.pdf  for x,y and z.
Variables x4,y5,z6 specifically selected to revert different diagrams from mentioned article.
See page 23, which contains original straight forward diagrams.

((x1^x2=>x3)^x4=>x5)^x6=>x7 = 1

((y1^y2=>y3)^y4=>y5)^y6=>y7 = 1

((z1^z2=>z3)^z4=>z5)^z6 =>z7  = 1

(x4=>y5)=>z6=1

Control performed via Polyakov's Calculator

Testing for z6 numbers {64,39}



Third  system :-

 (x1=>x2)^(x2=>x3)^(x3=>x4)^(x4=>x5)^(x5=>x6)=1
 (y1=>y2)^(y2=>y3)^(y3=>y4)^(y4=>y5)^(y5=>y6)=1
 (z1=>z2)^(z2=>z3)^(z3=>z4)^(z4=> z5)^(z5=> z6)=1
 x3^y4^z5=0

Вариации на тему форума ЕГЭ Информатика "Очередное «доказательство» парадигмы Елены Mирончик" - Решение одной системы уравнений в булевских переменных в стиле 08.2016 "ГРАФЫ И СИСТЕМЫ ЛОГИЧЕСКИХ УРАВНЕНИЙ"

Исходная система описана диграммами в стиле 08.2016 и сразу же просчитана

  ((x1≡x2)≡x3)=>((x4=>x5)=>x6)=1
  ((x4≡x5)≡x6)=>((x7=>x8)=>x9)=1
  ((x7≡x8)≡x9)=>((x1=>x2)=>x3)=1

Расчет системы
  ((x1≡x2)≡x3)=>((x4=>x5)=>x6)=1
  ((x4≡x5)≡x6)=>((x7=>x8)=>x9)=1
в верхней половине таблицы

Расчет системы 
  ((x1≡x2)≡x3)=>((x4=>x5)=>x6)=1
  ((x4≡x5)≡x6)=>((x7=>x8)=>x9)=1
  ((x7≡x8)≡x9)=>((x1=>x2)=>x3)=0

в нижней половине таблицы

   В нижней части таблицы мы стартуем прогон выставив единичные значения у 3-ех битовых групп , где (х1=>x2)=>x3=0.  По завершение считаем количество (x7≡x8)≡x9 равных 1 в последней колонке , где вообще говоря, находятся битовые тройки ассоциированные с  (x7=>x8)=>x9 . 

В зеленой группе (3 тройки 001,100,111) и (x7=>x8)=>x9=(x7≡x8)≡x9 
В серой группе (1 тройка 010) и (x7=>x8)=>x9=0, но (x7≡x8)≡x9=1

   Контроль по Полякову

  

    

Контроль системы
   ((x1≡x2)≡x3)=>((x4=>x5)=>x6)=1
   ((x4≡x5)≡x6)=>((x7=>x8)=>x9)=1

   Контроль системы
     ((x1≡x2)≡x3)=>((x4=>x5)=>x6)=1
     ((x4≡x5)≡x6)=>((x7=>x8)=>x9)=1
     ((x7≡x8)≡x9)=>((x1=>x2)=>x3)=0

Ссылки
1.  http://egekp.unoforum.pro/?1-6-0-00000115-000-0-0-1558858221
2.  https://vk.com/club180658320?w=wall-180658320_64%2Fall

Решение одной системы уравнений в булевских переменных в стиле 08.2016 "ГРАФЫ И СИСТЕМЫ ЛОГИЧЕСКИХ УРАВНЕНИЙ"

Исходная система описана диграммами в стиле 08.2016 и сразу же просчитана

  ((x1≡x2)≡x3)=>((x4=>x5)=>x6)=1
  ((x4≡x5)≡x6)=>((x7=>x8)=>x9)=1
  ((x7≡x8)≡x9)=>((x10=>x11)=>x12)=1
  ((x10=>x11)=>x12)=>((x1≡x2)≡x3)=1

Решение (диаграммы и просчет (1) - (3) ) :-

     Прогон системы и получение ответа для исходной

      ((x1≡x2)≡x3)=>((x4=>x5)=>x6)=1

      ((x4≡x5)≡x6)=>((x7=>x8)=>x9)=1
      ((x7≡x8)≡x9)=>((x10=>x11)=>x12)=1
      ((x10=>x11)=>x12)=>((x1≡x2)≡x3)=0

Контроль по Полякову

  

Solution of problems kind of P149 (ege23.doc) via 08.2016 technique "GRAPHS AND SYSTEMS OF LOGICAL EQUATIONS"

"Poor understanding of Mapping Method (08.2016) is still a half of distress" B.D.

First of all be aware that intellectual property of approach been applied completely belongs to Helen Mironchick (unpublished manuscript ). See also PDF attachment to thread on USE in Informatics Polyakov's  forum  http://egekp.unoforum.pro/?1-6-0-00000120-000-0-0-1561979337   MEA's post 229. Posting below is nothing different from exercising technique proposed by MEA in mentioned PDF attachment targeting better understanding of solution been suggested in those  thread.
We would consider a sample P149 and one more system which doesn't look as standard as P149 and requires a bit more experience in building 08.2016 diagrams.



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

  Convert system to equivalent

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

      Traditional solution ( classic version of Mapping Method as of 2013 ) might be found in my old  blog entry  http://mapping-metod.blogspot.com/2017/12/149-ege23pdf.html
  Passing Polyakov's control



Consider a bit more complicated system

  ((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

Now build diagrams required to solve updated problem with implications

     Passing Polyakov's Control

Использование предикатов при решении систем уравнений в булевых переменных ( Версия Метода Отображений 08.2016 )

 Пусть F(x,y,z) трехместный предикат определенный на множестве всех трех битовых цепочек.

 Мощность множества истинности этого предиката равна 5 , мощность области определения 

предиката  очевидно равна 8.  Рассмотрим систему уравнений.

F(x2,y2,z2) => F(x1,y1,z1) =1

F(x3,y3,z3) => F(x2,y2,z2) =1 
F(x4,y4,z4) => F(x3,y3,z3) =1
F(x1,y1,z1) => F(x4,y4,z4) =1

Положим
w1=F(x1,y1,z1)
w2=F(x2,y2,z2)
w3=F(x3,y3,z3)
w4=F(x4,y4,z4)

    Рассмотрим три различных предиката, удовлетворяющих исходному условию

      F(x,y,z)= ((x=>y)=>z) ;

      F(x,y,z)= (x=>y^z);

      F (x,y,z) = ((x v y) => z) ;

    Просчитаем по Полякову каждый случай  ( POC - Proof of concept )

Таким образом,если система может быть сведена к диаграмме типа 08.2016 

для w(j) переменных, результат определяется только мощностью множества 

истинности предиката, а не формулой определяющей F(x,y,z)

Bring into play predicates when solving systems of equations in Boolean variables

This post was initiated by the analysis of the approach of Helen A. Mironchick to the task P-40 from ege23.doc.

Let F (x, y, z) be a triple predicate defined on the set of all three bit chains. The truth set of this predicate is exactly 5 different chains, specific bits included in the chains of truth do not matter.Consider the system of equations.

F (x1, y1, z1) ≡ F (x3, y3, z3) => F (x2, y2, z2) = 1

F (x2, y2, z2) ⊕ F (x4, y4, z4) => F (x3, y3, z3) = 1

F (x3, y3, z3) ≡ F (x5, y5, z5) => F (x4, y4, z4) = 1

F (x4, y4, z4) ⊕ F (x6, y6, z6) => F (x5, y5, z5) = 1

Setw1 = F (x1, y1, z1)

w2 = F (x2, y2, z2)

w3 = F (x3, y3, z3)

w4 = F (x4, y4, z4)

w5 = F (x5, y5, z5)

w6 = F (x6, y6, z6)

Then we plot the diagrams and generateMapping Method table

    Consider three cases. Truth sets of each
    of the predicates considered below consists of 5 three bit
    chains, its own for each predicate constructed

     F (x, y, z) = ((x => y) => z);
     F (x, y, z) = (x => y ^ z);
     F (x, y, z) = ((x v y) => z);

     Calculate according to Polyakov each case (POC - Proof of concept)

 

Thus, for a system that can be reduced to the  traditional matrix of  the  Mapping Method  for w (j) variables, the result is determined only by the power of the truth set of the predicate, and not by the formula defining F (x, y, z)

Solution one system (another draft ) of Boolean equations kind of P40 with different outgoing multipliers for different Mapping Method table rows as of 18/07/19

Original system

  ((x1vy1)^z1 ≡ (x3vy3)^z3)=>((x2vy2)^z2) =1

  ((x2vy2)^z2)⊕ ((x4vy4)^z4))=>((x3vy3)^z3) =1

  ((x3vy3)^z3 ≡ (x5vy5)^z5)=>((x4vy4)^z4) =1

  ((x4vy4)^z4)⊕ ((x6vy6)^z6))=>((x5vy5)^z5) =1

Now make a substitution ( kind of design had been suggested in P-40 from ege23.pdf )

   w1=(x1 v y1)^z1

   w2=(x2 v y2)^z2
   w3=(x3 v y3)^z3
   w4=(x4 v y4)^z4
   w5=(x5 v y5)^z5

   Outgoing multipliers are located  at right hand side of diagrams

   Passing Polyakov's Control