Tuesday, June 18, 2019

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

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
  



Monday, June 17, 2019

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

Saturday, June 15, 2019

Solution of one problem 23 from USE in Informatics 2019 or addressing one ugly 23-rd problem

Original task




Down here we intend to demonstrate one more time well known axiom of well known USE's in Informatics developer ( sure not myself ), which states any system of boolean equations could be solved via Mapping Method. Would you expirience trouble in doing so that appears to be the limitation of yours skilset rather then the limitation of the method.

You might want to see the most impressive and up to date Mapping
Method solution of more or less complicated  system at
Helen Mironchick's Solution of one sysem of boolean equations from USE in Informatics forum (16.05.2019)

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
  


Wednesday, June 12, 2019

Mapping Method and solution of system of boolean equations as of 12/06/2019

Original system

(x1⊕y1) => (x2⊕y2) =1
(x1⊕y1) => (x2⊕y2) =1
(x2⊕y2) => (x3⊕y3) =1
(x3⊕y3) => (x4⊕y4) =1
(x1⊕x4) => (y1⊕y4) =1

Solve first system 1

(x1⊕y1) => (x2⊕y2) =1
(x1⊕y1) => (x2⊕y2) =1
(x2⊕y2) => (x3⊕y3) =1
(x3⊕y3) => (x4⊕y4) =1

Solve second system 2

(x1⊕y1) => (x2⊕y2) =1
(x1⊕y1) => (x2⊕y2) =1
(x2⊕y2) => (x3⊕y3) =1
(x3⊕y3) => (x4⊕y4) =1
(x1⊕x4) => (y1⊕y4) =0

and deduct from number of solutions
system 1 number of solutions system 2



    Passing Polyakov's control