Chains of tie nodes of a multi-part graph && reverse pass to center per Helen Mironchick's "The Mapping method with a reverse calculation" - Computer science at school №10 2019 (awaiting publication)

Original systems

System 1

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


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 direct XOR
3) Getting y4  via reverse pass to center
    Here we calculate 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 direct XOR
5) Completing Z-LINE having z4  (on line-0 112 , on line-1-200)

 
System 2

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

*******************
Chaining order
*******************
1) We get x3 via reverse pass to center
2) Getting y3 via  x3 v y3 =1 basic chart
3) Getting y4  via reverse pass to center
    Here we calculate 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 x4 v y4 =1 basic chart
5) Completing Z-LINE having z4  (on line-0 160 , on line-1-456)

    System 3

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

 
    See also
    1.  https://informatics-ege.blogspot.com/2019/11/chaining-order-reverse-pass-to-center.html
    2.  https://informatics-ege.blogspot.com/2019/11/unleash-power-of-reverse-pass-to-center.html

Chains of tie nodes of a multi-part graph && reverse pass to center per Helen Mironchick's "The Mapping method with a reverse calculation" - Computer science at school #10 2019

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 into 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)

Another sample

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

*******************
Chaining order
*******************
1) We get x3 via reverse pass to center
2) Getting y3 via direct XOR
3) Getting y4  via reverse pass to center
    Here we calculate 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 direct XOR
5) Completing Z-LINE having z4  (on line-0 112 , on line-1-200)

Unleash the power of reverse pass to center per Helen Mironchick's "The Mapping method with a reverse calculation" - Computer Science at School №10 ( supposedly ) 2019

This post has been done in relation with previous one. It is kind of vs "Algebra of Logic from Informatik BU" https://www.youtube.com/watch?v=zeWrJbWEfmg (2hr. 17 min ) and demonstrates that bit-mask's BU approach is pretty much limited

See also 
"Chaining order && reverse pass to center per Helen Mironchick's "The Mapping method with a reverse calculation"
https://informatics-ege.blogspot.com/2019/11/chaining-order-reverse-pass-to-center.html

Original system

((((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

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

    Passing Polyakov's Control
 

 

Метод обратного прогона VS Стрим #20. ЕГЭ по информатике 2019 Информатика БУ

Исходная система

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

С техникой Информатика БУ можно ознакомиться здесь
https://www.youtube.com/watch?v=zeWrJbWEfmg
(2 часа 17 мин. )

Решение той же задачи техникой , предложенной Мирончик Еленой Александровной в работе "Метод отображения с обратным просчетом" - Информатика в школе №10 , 2019 .
Препринт
https://vk.com/doc6125348_500219375?hash=98bd7718a258b37a51&dl=7cbbc2f9e20b7f5178

Очевидное - невероятное. Алгебра Предикатов {E(K)} VS Стрим 22 2020 Иформатика БУ

Детально с логикой БУ можно ознакомиться https://www.youtube.com/watch?v=VwGfuTnZnbM (2ч. 05 мин.)
С Алгеброй Предикатов {E(K)} по http://kpolyakov.spb.ru/download/mea18bit.pdf .              Отмечу , что второй подход универсален и решает любое уравнение с побитной конъюнкцией уровня ЕГЭ Информатика либо более сложной. Это не имеет значения в силу универсальности концепсии разложения по базисным предикатам

Решение с применением Алгебры {E(K)}
Конвертируем выражение в термины алгебры {Е(К)}
(E(28) v E(45))=>(¬E(17)=>E(A)) ≡1

По Де Моргану

¬E(28)^¬E(45) v E(17) v E(A) ≡1

Разложим по базисным предикатам

E(28)=E(16) v E(8) v E(4)

E(45)=E(32) v E(8) v E(4) v Е(1)

По Де Моргану

¬Е(28)=¬Е(16)^¬Е(8)^¬E(4)

¬E(45)=¬E(32)^¬E(8)^¬E(4)^¬E(1)

Откуда

¬Е(28)^¬Е(45)=¬Е(32)^¬Е(16)^¬Е(8)^¬Е(4)^¬Е(1)

Таким образом

¬Е(32)^¬Е(16)^¬Е(8)^¬Е(4)^¬Е(1) v Е(16) v Е(1) v 

v Е(А) ≡1

Очевидно,что ¬Е(16) и ¬Е(1) будут подавлены в 

конъюнкции и мы получим

¬Е(32)^¬Е(8)^¬Е(4) v Е(16) v Е(1) v Е(А) ≡1

¬(Е(32) v Е(8) v Е(4)) v Е(17) v Е(А) ≡1

¬Е(44) v Е(17) v Е(А)≡1

Откуда А(min)=44

   

08.2016 Mapping Method Style VS Stream #17 provided by Informatick BU

First see https://www.youtube.com/watch?v=hh3ALpjxqfA ( 3 hr 00 min )

Original system

Generation of 08.2016 chart

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

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

   

Running cross-reference tables with systems of Boolean Equations VS 18-th Stream provided by Informatick BU

Once again a poor understanding of 08.2016 technique proposed by Helen Mironchick is just a half of disaster. As far as to my knowledge advanced concepts of significantly pushed forward ideas based on mentioned technique are currently pending publication in "Informatics at school" magazine.   
 Problem  #149 from ege23.pdf might be solved pretty shortly vs classic solution via triples connectivity.  Just compare suggested solution based on building cross-reference table and Demo provided by Informatik BU   for #149  https://www.youtube.com/watch?v=W1oGIwgpw8A  (1 hr 33 minutes )


Now build cross-reference table and fork 08.2016 chart

Running cross-reference tables with systems of Boolean Equations of triple predicates

Original system

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

    In particular,  #149 from ege23.pdf might be solved same way pretty shortly vs classical solution via triples connectivity.
  You might also want to compare suggested approach with Demo provided by Informatik BU   for #149  https://www.youtube.com/watch?v=W1oGIwgpw8A  (1 hr  33 minutes )

     Running cross-reference tables based on  08.2016 Charts style VERSUS classic BU's calculation
   

      

   Polyakov's Control for system (1)

  

Another 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

    Polyakov's Control

   

The 23 -rd USE's Informatics task casual support on VK Informatics_100 as of 14/11/2019

Original task

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

Here we build cross-reference table at cartridge return and fork 08.2016 chart

Solution task #131 from ege23.doc via Mapping Method ( Classic 10/2013)

Original system

(x1^x2) v (x1 v x3)^(x1 v y1)=1
(x2^x3) v (x2 v x4)^(x2 v y2)=0
(x3^x4) v (x3 v x5)^(x3 v y3)=1
(x4^x5) v (x4 v x6)^(x4 v y4)=0
(x5^x6) v (x5 v x7)^(x5 v y5)=1
(x6^x7) v (x6 v x8)^(x6 v y6)=0

Notice that we would have two arrows charts . The first on for equations with odd numbers and the second one for equations with even numbers

Solution problem #113 from ege23.doc in 08/2016 style vs Video for 23-rd by Informatik BU

Original source for #113

Building meta chart and fork generation 08.2016 chart providing an answer

    You might want to compare 08/2016 chart generation with well known video been recorded by Informatik BU   https://vk.com/inf_bu?z=video-89501371_456239128%2Fvideos-89501371%2Fpl_-89501371_-2

The solution to problems № 14 of "How many different solutions does the system of equations.pdf have" in the style of 08.2016

Original system ( compare with  https://informatics-ege.blogspot.com/2018/01/13-14-pdf.html  )
Source  1Сколько различных решений имеет система уравнений.pdf

Equivalent system

          (x3 => x4) => (x1=>x2) =1

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

I do remember that about two years ago Helen Mironchick warned that classical pairs originally used by myself was not the best way to solve task 14 and even sent to me in private a couple of charts wich looked to me at those time as a kind of "Chinese manuscript". One more time I have to state that poor understanding on 08.2016 is just a half of disaster.

Now fork a couple of 08.2016 charts required to solve the task

Let's stop being afraid of predicates and pretend that everything is fine even without them in relation to the classical clones of problem 18 USE in Informatics

Tеорема 01

*******************************************************************************
Пусть W и V два одноместных предиката, определенных
На множестеве Х любой природы.
Если ∀ x ∈ Х : W(x) => V(x) = True (*),то область истинности
предиката $(W) вложена в область истинности предиката $(V)
*******************************************************************************
Допустим  ∃ y : (W(y)=1)^(V(y) = 0 ) =1. Тогда W(y) => V(y) = False
Что противоречит условию (*) и $(W) вложено в $(V)
Отсюда также следует , что максимальная область истинности W ($(W)) есть область истинности V ($(V)), поскольку при V(z)=1, мы можем не теряя общности считать W(z)=1, а минимальная область истинности V ($(V)) есть область истинности  $(W).

Исходная задача

 Теперь сформулируем задачу несколько иначе -
 Определим одноместные предикаты

    P(x) = { 1  if  (x ∈ P) ;
                0  if  (x! ∈ P)
               }
    Q(x)={ 1  if  (x ∈ Q);
               0   if  (x! ∈ Q)
              }
    Найдите минимальную область истинности  A(x) такую,что

    ∀ x∈R : P(x)=>(Q(x)^¬A(x)=> ¬P(x))=1

  Решение 

  ∀ x∈R :  ¬P(x) v ¬(Q(x)^¬A(x)) v ¬P(x) =1

  ∀ x∈R : ¬P(x) v ¬Q(x) v A(x) =1

  ∀ x∈R : ¬(P(x)^Q(x)) v A(x) =1

  ∀ x∈R : (P(x)^Q(x)) => A(x) =1

Откуда следует (Теорема 1)  Min($(A))= $(P^Q)=[150;171]

Once again 08.2016 technique and solving simple 23-rd task for USE in Informatics as of 02/11/2019

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

((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