Solution of System in Boolean variables kind of P233 with {x,y,z} variables in 08/2016 format

Исходная система

((((x1 =>y1) =>z1) =>x2 )=>y2) =>z2 =1
((((x2 =>y2) =>z2) =>x3) =>y3) =>z3 =1
((((x3 =>y3) =>z3) =>x4) =>y4) =>z4 =1
((((x4 =>y4) =>z4) =>x5) =>y5) =>z5 =1
((((x5 =>y5) =>z5) =>x6) =>y6) =>z6 =1
((((x6 =>y6) =>z6) =>x7) =>y7) =>z7 =1

Линии "0" и "1" определены по (x1=>y1)=>z1=0  и (x1=>y1)=>z1=1
Тройки бит {000,010,110} - линия "0" ; {001,011,100,101,111} - линия "1".  
Закономерность переходов в диаграмме, по факту определяется на битовых тройках. 
Но как только  правильная закономерность получена на битовых тройках можно      комфортно реплицировать одну и ту же процедуру в графике.

     Контроль по Полякову

  

Solution of System in Boolean variables cloning P233 to {x,y,z} variables in 08/2016 format

Исходная система

(((x1 =>y1 =>z1) =>x2 )=>y2) =>z2 =1
(((x2 =>y2 =>z2) =>x3) =>y3) =>z3 =1
(((x3 =>y3 =>z3) =>x4) =>y4) =>z4 =1
(((x4 =>y4 =>z4) =>x5) =>y5) =>z5 =1
(((x5 =>y5 =>z5) =>x6) =>y6) =>z6 =1
(((x6 =>y6 =>z6) =>x7) =>y7) =>z7 =1



Забываем то что сделано раньше
Эксперимент  Линии "0" и "1" определены по (x1=>y1)=>z1=0  и (x1=>y1)=>z1=1
Тройки бит {000,010,110} - линия "0" ; {001,011,100,101,111} - линия "1".  
Закономерность переходов в диаграмме, по факту определяется на битовых тройках. 
Но как только  правильная закономерность получена на битовых тройках можно        комфортно реплицировать одну и ту же процедуру в графике.

Solution of onе System in Boolean Variables in 08/2016 format from Informatics_100 VK's News Wire as of 27/08/2019

Once again a bunch of 23rd USE Informatics problems might be solved via 08/2016 technique rather then bits chains way 

 Original Source
 (x1 v x2)^(¬x1 v ¬x2)^(x1 v y1)=1
 (x2 v x3)^(¬x2 v ¬x3)^(x2 v y2)=1
 (x3 v x4)^(¬x3 v ¬x4)^(x3 v y3)=1
 (x4 v x5)^(¬x4 v ¬x5)^(x4 v y4)=1
 (x5 v x6)^(¬x5 v ¬x6)^(x5 v y5)=1
 (x6 v x7)^(¬x6 v ¬x7)^(x6 v y6)=1
 (x7 v y7)=1

Good knowledge of Algebra of Logic is required pretty often to simplify original system to the one which is obviously good for 08/2016 treatment.

Convert to equivalent due to  (x1 v x2)^(¬x1 v ¬x2) =  (x1^¬x2)v(x2^¬x1)=x1⊕x2
 (x1⊕x2)^(x1 v y1)=1
 (x2⊕x3)^(x2 v y2)=1
 (x3⊕x4)^(x3 v y3)=1
 (x4⊕x5)^(x4 v y4)=1
 (x5⊕x6)^(x5 v y5)=1
 (x6⊕x7)^(x6 v y6)=1
 (x7 v y7)=1

Fork  08/2016 chart

   

Predicates for segments on real axis

Теорема 1
Пусть P и Q два одноместных предиката, определенных
На множестеве Х любой природы.
Если ∀ x ∈ Х : P(x) => Q(x) = True (*),то область истинности 
предиката $(P) вложена в область истинности предиката $(Q)

Решение

   Определим предикаты
   Р(x) ={ 1: x ∈ [8,50] ;
                0: x !∈ [8,50] ;
       }
   Q(x) ={ 1: x ∈ [27;76];
                0: x !∈ [27;76];
       }
Найти наименьшую область истинности предиката А(x), 
такого что
∀ x∈ R : ¬A(x) => ¬(¬P(x)=>Q(x)) =1
∀ x∈ R : (¬P(x)=>Q(x)) => A(x)  =1
∀ x∈ R : (P(x) v Q(x)) => A(x)   = 1
По Теореме 1 импликация тождественна тогда и только
тогда, когда множество истинности $(PvQ) ⊂  $(A).
То есть минимальное $(A) = $(PvQ) 
 $(A)=$(PvQ) =$(P)∪$(Q) = [8,50]∪[27;76]= [8,76]

Solution of one new system of Boolean Equations with no transition triples in 08/2016 style as of 26/08/2019

Original system

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

Fork 08/2016 chart
  

Решение одной известной системы в Булевских переменных в формате 08.2106

Рассмотри следующую систему

(x1+x2)*(x1*x2->y1)=1
(x2+x3)*(x2*x3->y2)=1
(x3+x4)*(x3*x4->y3)=1
(x4+x5)*(x4*x5->y4)=1
(x5+x6)*(x5*x6->y5)=1
(x6+x7)*(x6*x7->y6)=1
(x7+x8)*(x7+y7) =1

(x8+y8)=1

Строим диаграммы в стиле 08.2016

  Сравните трудо-затрыты с классически подходом 
 https://mapping-metod.blogspot.com/2019/05/8.html
 

Solution of one system of Boolean equations in 08.2016 format

Original system
((x1=>y1) =>z1) ^ ((x2 v y2 v z2) =>((x1≡y1)≡z1)) =1
((x2=>y2) =>z2) ^ ((x3 v y3 v z3) =>((x2≡y2)≡z2)) =1
((x3=>y3 )=>z3) ^ ((x4 v y4 v z4) =>((x3≡y3)≡z3)) =1
((x4=>y4) =>z4) ^ ((x5 v y5 v z5) =>((x4≡y4)≡z4)) =1
((x5=>y5) =>z5) ^ ((x6 v y6 v z6) =>((x5≡y5)≡z5)) =1
((x6=>y6) =>z6) ^ ((x7 v y7 v z7) =>((x6≡y6)≡z6)) =1
(x7≡y7)≡z7=1
****************************************
Consider the transmission on the line "0"
****************************************
((xj=yj)=zj) ======> !xj^!y^!z
True triples
((xj=yj)=zj) ===*3==> !xj*!yj*!zj
001                                 ==>110 
010                                 ==>101 
100                                 ==>011
111                                 ==>000
We obtain exactly ((xj = yj) = zj) four false triples
110 
101 
100 
111 
Thus, on the line "0" we transmit false triples
**************************************
Consider the transmission on the line "1"
**************************************
In the lower right node has total 24 triples. Notice that the left upper node gives diagonally 12 true bit triples, and the lower left node gives 12 false ones. The lower left node generates 12 triples that are not supposed to be transmitted (they are false). 12 bit triples from the upper left node give true triples and they are transmitted, 12 from the lower left node give 0 (stop on line 1)

Three "3" is the power of intersection of the truth regions (xj => yj) => zj and (xj≡yj) ≡zj
this is {001,100,111} = {001,011,100,101,111} ∩ {010,001,100,111}
Convert  implication     (x2 v y2 v z2) =>((x1≡y1)≡z1)  into  ((x1≡y1)≡z1) v  !x2^!y2^!z2
and so on

    All underlined passed Polyakov's control

   

Predicates for discrete sets and segments current problems on Informatics_100 News Wire as of 27/08/19

******************************************************************
Теорема 1
Пусть P и Q два одноместных предиката, определенных
На множестеве Х любой природы.
Если ∀ x ∈ Х : P(x) => Q(x) = True (*),то область истинности 
предиката $(P) вложена в область истинности предиката $(Q)

******************************************************************

 Решение
   Определим предикаты
   Р(x) ={ 1: x ∈ {2,4,6,8,10,12.14,16,18,20 } ;
        0: x !∈ {2,4,6,8,10,12.14,16,18,20 }
      }
   Q(x) ={ 1: x ∈ {3,6,9,12.15,18,21,24,27,30} ;
              0: x !∈ {3,6,9,12.15,18,21,24,27,30}
      }
Известно,что  ∀ x∈ N : (P(x) =>A(x)) v (!A(x) =>!Q(x))=1
Предполагая,что А предикат, имеющий дискретное множество истинности
найти наименьшую возможную сумму его элементов
Имеем 
      ∀ x∈ N : (P(x) =>A(x)) v (Q(x) =>A(x))=1
      ∀ x∈ N : (!P(x) v A(x)) v (!Q(x) v A(x))=1
      ∀ x∈ N : !P(x) v !Q(x) v A(x)=1
      ∀ x∈ N : !(P(x) ^ Q(x)) v A(x)=1
      ∀ x∈ N : (P(x) ^ Q(x))=>A(x)=1
Импликация тожественна тогда и только
тогда, когда множество истинности $(P^Q) ⊂  $(A).
То есть минимальное $(A) = $(P^Q) 
Пересечем множества 
{2,4,6,8,10,12,14,16,18,20 } ∩ {3,6,9,12.15,18,21,24,27,30} =
= {6,12,18}

Ответ 36

Сравни URL первого рабора https://vk.com/doc342403709_514914240?hash=2093efa522f66c8650&dl=8c479daad6e0542733
и решение полученное в Алгебре Предикатов , где Р(х) и Q(x) имеют , соответсвующие области истинности. Минимальная область истинности А(х) определяется решением.

Решение

   Определим предикаты
   Р(x) ={ 1: x ∈ [8,50] ;
                0: x !∈ [8,50] ;
       }
   Q(x) ={ 1: x ∈ [27;76];
                0: x !∈ [27;76];
       }
Найти наименьшую область истинности предиката А(x), 
такого что
∀ x∈ R : ¬A(x) => ¬(¬P(x)=>Q(x)) =1
∀ x∈ R : (¬P(x)=>Q(x)) => A(x)  =1
∀ x∈ R : (P(x) v Q(x)) => A(x)   = 1
По Теореме 1 импликация тождественна тогда и только
тогда, когда множество истинности $(PvQ) ⊂  $(A).
То есть минимальное $(A) = $(PvQ) 
 $(A)=$(PvQ) =$(P)∪$(Q) = [8,50]∪[27;76]= [8,76]

Answer is 68

Solution of systems of Boolean equations with three variables via format 08/2016

Система

(((x1≡y1)≡z1) v ((x2≡y2)≡z2)) ^ (x1 v y1 v z1)=1
(((x2≡y2)≡z2) v ((x3≡y3)≡z3)) ^ (x2 v y2 v z2)=1
(((x3≡y3)≡z3) v ((x4≡y4)≡z4)) ^ (x3 v y3 v z3)=1
(((x4≡y4)≡z4) v ((x5≡y5)≡z5)) ^ (x4 v y4 v z4)=1
(((x5≡y5)≡z5) v ((x6≡y6)≡z6)) ^ (x5 v y5 v z5)=1
(((x6≡y6)≡z6) v ((x7≡y7)≡z7)) ^ (x6 v y6 v z6)=1
(x7 v y7 v z7)=1

Решение в формате 08.2016

    Кортроль по Полякову

Решение в формате 08.2016 Задания #Т4888 ( тип 23 ЕГЭ Информатика ) из очереди для стрима Информатика БУ

Задание

Решение системы формате 08.2016

((x1 v y1)=>(x2 v y2)) ^ (x1=>y1)=1
((x2 v y2)=>(x3 v y3)) ^ (x2=>y2)=1
((x3 v y3)=>(x4 v y4)) ^ (x3=>y3)=1
((x4 v y4)=>(x5 v y5)) ^ (x4=>y4)=1
((x5 v y5)=>(x6 v y6)) ^ (x5=>y5)=1
((x6 v y6)=>(x7 v y7)) ^ (x6=>y6)=1
((x7 v y7)=>(x8 v y8)) ^ (x7=>y7)=1
(x8=>y8)=1

  

Solution of systems of Boolean equations kind of P204 with three variables via format 08/2016

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. A poor understanding of the ideology 08.2016 in essence for today means a lack of understanding what  kind of flexibility and power Mapping Method provides for USE in  Informatics developers in meantime.   See for instance (having active connection to VK)

https://vk.com/bderzhavets?w=wall209645472_433/all

Original system 

(((x1≡y1)≡z1) =>((x2≡y2)≡z2)) ^ (x1 v y1 v z1)=1
(((x2≡y2)≡z2) =>((x3≡y3)≡z3)) ^ (x2 v y2 v z2)=1
(((x3≡y3)≡z3) =>((x4≡y4)≡z4)) ^ (x3 v y3 v z3)=1
(((x4≡y4)≡z4) =>((x5≡y5)≡z5)) ^ (x4 v y4 v z4)=1
(((x5≡y5)≡z5) =>((x6≡y6)≡z6)) ^ (x5 v y5 v z5)=1
(((x6≡y6)≡z6) =>((x7≡y7)≡z7)) ^ (x6 v y6 v z6)=1
(x7 v y7 v z7)=1

As soon as we have a conditions that allow us to determine the bits of x(j),y(j),z(j)  for the arrows outgoing from the node of the 08/2016 chart, we can start. In particular case, we are ready to go ahead with  08/2016 chart style.  In other words, we have to know bits combination {x,y,z} associated with arrows connecting  neighboring nodes in chart . 

Cloning to {x,y,z} systems like P175 or P116 is quite straight forward
Consider second use case

((x1 v y1 v z1) =>(x2 v y2 v z2)) ^ ((x1⊕y1) =>z1) =1
((x2 v y2 v z2) =>(x3 v y3 v z3)) ^ ((x2⊕y2) =>z2) =1
((x3 v y3 v z3) =>(x4 v y4 v z4)) ^ ((x3⊕y3) =>z3) =1
((x4 v y4 v z4) =>(x5 v y5 v z5)) ^ ((x4⊕y4) =>z4) =1
((x5 v y5 v z5) =>(x6 v y6 v z6)) ^ ((x5⊕y5) =>z5) =1
((x6 v y6 v z6) =>(x7 v y7 v z7)) ^ ((x6⊕y6) =>z6) =1
((x7⊕y7)=>z7)=1

Solution via 08/2016 format

Consider third use case

(((x1≡y1)≡z1) v ((x2≡y2)≡z2)) ^ (x1 v y1 v z1)=1
(((x2≡y2)≡z2) v ((x3≡y3)≡z3)) ^ (x2 v y2 v z2)=1
(((x3≡y3)≡z3) v ((x4≡y4)≡z4)) ^ (x3 v y3 v z3)=1
(((x4≡y4)≡z4) v ((x5≡y5)≡z5)) ^ (x4 v y4 v z4)=1
(((x5≡y5)≡z5) v ((x6≡y6)≡z6)) ^ (x5 v y5 v z5)=1
(((x6≡y6)≡z6) v ((x7≡y7)≡z7)) ^ (x6 v y6 v z6)=1
(x7 v y7 v z7)=1

Solution via 08/2016 format

    Passing Polyakov's control



Consider fourth use case

(((x1≡y1)≡z1) =>((x2≡y2)≡z2)) ^ ((x1 =>y1) =>z1)=1
(((x2≡y2)≡z2) =>((x3≡y3)≡z3)) ^ ((x2 =>y2) =>z2)=1
(((x3≡y3)≡z3) =>((x4≡y4)≡z4)) ^ ((x3 =>y3) =>z3)=1
(((x4≡y4)≡z4) =>((x5≡y5)≡z5)) ^ ((x4 =>y4) =>z4)≡1
(((x5≡y5)≡z5) =>((x6≡y6)≡z6)) ^ ((x5 =>y5) =>z5)=1
(((x6≡y6)≡z6) =>((x7≡y7)≡z7)) ^ ((x6 =>y6) =>z6)=1
(((x7≡y7)≡z7) =>((x8≡y8)≡z8)) ^ ((x7 =>y7) =>z7)=1
((x8 =>y8) =>z8)=1

Solution via 08/2016 format

Solution of System P204,P116,P175 from ege23.pdf in 08/2016 format

Original

((x1≡y1) =>(x2≡y2)) ^ (x1 v y1)=1
((x2≡y2) =>(x3≡y3)) ^ (x2 v y2)=1
((x3≡y3) =>(x4≡y4)) ^ (x3 v y3)=1
((x4≡y4) =>(x5≡y5)) ^ (x4 v y4)=1
((x5≡y5) =>(x6≡y6)) ^ (x5 v y5)=1
((x6≡y6) =>(x7≡y7)) ^ (x6 v y6)=1
((x7≡y7) =>(x8≡y8)) ^ (x7 v y7)=1
x8 v y8=1

As soon as we have a conditions that allow us to determine the bits of x(j),y(j) for the arrows outgoing from the node of the 08/2016 chart, we can start. In particular case, we are ready to go ahead with  08/2016 diagram style.  In other words, we have to know bits combination associated with arrows sequence connecting  neighboring nodes in chart . 

Here you might want to compare solution for P175 proposed by Informatic BU . See

https://www.youtube.com/watch?v=R7UqPN3-l4k

solution of the P175 provided via  08/2016 chart 

http://kpolyakov.spb.ru/download/mea-2016-8.pdf

    Control P116  via Polyakov's calculator
    

Solution of one 23rd problems pending stream from Informatic BU in 08/2016 format

Original source

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

As soon as we have a conditions that allow us to determine the bits of x(j) for the arrows outgoing from the node of the 08/2016 chart, we can start. In particular case, we are ready to go ahead with  08/2016  diagram style.

Fork two 08/2016 diagram. The first one in details , the second one in brief.

Решение одной задачи 23 , ожидающей стрима БУ в формате 08/2016

Условие

Решение

 
    Проверка по Полякову
   

   

Solution of one system in Boolean Variables from VK's Informatics_100 23rd database via 08/2016 technique

Original System



Convert system as follows

(x1=>y1)=>(¬x2^y2)=1

(x2=>y2)=>(¬x3^y3)=1

(x3=>y3)=>(¬x4^y4)=1

(x4=>y4)=>(¬x5^y5)=1

(x5=>y5)=>(¬x6^y6)=1

(x6=>y6)=>(¬x7^y7)=1

Fork 08/2016 diagram

The recent sample from Informatics_100 News Wire solved via 08/2016 technique

Original system
https://vk.com/informatics_100?z=photo-40390768_457273821%2Fwall-40390768_193394

Convert system to equivalent
 (x2=>x1)^(x2=>x3)^(x4=>x3)^(x4=>x5)=1
 (y2=>y1)^(y2=>y3)^(y4=>y3)^(y4=>y5)=1
 x1^y1= 1

Just fork 08/2016 diagrams right away  && Neither one word is supposed to be told 
presuming you understand "GRAPHS AND SYSTEMS OF LOGIC EQUATIONS"