Solution task №140 from ege18.pdf via Algebra of predicates {D(k)} VS Stream 32 issued by Informatik BU

Original task

BU's logic is explained here https://www.youtube.com/watch?v=QsC5rHfUb44
(48 min 15 sec )

   Now solve same task following "ALGEBRA OF PREDICATES AND RELATED GEOMETRIC MODELS CREATION IN REGARDS OF UNIFIED STATE EXAM IN INFORMATICS (RUSSIAN EGE)" Informatics at school #3 2019

 https://vk.com/club180658320?w=wall-180658320_65%2Fall

(D(A)^¬D(36)) => ¬D(12)≡1
¬D(A) V D(36) V ¬D(12) ≡1
Due to :-
  D(12) = D(2^2)^D(3)
¬D(12) = ¬D(2^2) V ¬D(3) 
So we get
 ¬D(A) V D(2^2)^D(3^2) V ¬D(2^2) V ¬D(3) ≡1
Suppress D(2^2) in conjunction
¬D(A) v D(2^2)^D(3^2) V ¬D(2^2) V ¬D(3) ≡1
¬D(A) V D(3^2) V ¬D(12) ≡1
Thus A(min) = 9

Solution three tasks from Eugene Dzhobs recent Video (06/01/20) per bit conjunction related via Algebra of predicates {E(k)}

https://vk.com/inform_web?z=video-184870282_456239124%2F85fc69eef8cd9338ce%2Fpl_wall_-184870282


Below we follow technique proposed in 
http://kpolyakov.spb.ru/download/mea18bit.pdf

 (1)
¬E(21) V (¬E(11)=>E(A)) =1
¬E(16)^¬E(4)^¬E(1) V E(8) V E(2) V E(1) + E(A) =1
Suppress ¬E(1) in conjunction
¬E(16)^¬E(4) V E(11) V E(A) ≡1
¬E(20) V E(11) V E(A) ≡1
Thus A(min) = 20
(2)
E(58)=E(32) V E(16) V E(8) V E(2)
E(22)=E(16) V E(4) V E(2)
¬E58)=¬E(32)^¬E(16)^¬E(8)^¬E(2)
¬E(58) V E(22) V ¬E(A) =1
Suppress ¬E(16) and ¬E(2) in conjunction
¬E(32)^¬E(16)^¬E(8)^¬E(2) V E(16) V E(4) V E(2) V E(A) ≡1
¬E(40) V E(22) V E(A) ≡1
Thus A(min)=40
(3)
(x&A=0) V ((x&69 = 4) => (x&118 = 6)) ≡1
(x&69 = 4)=¬E(64)^E(4)^¬E(1)
(x&69 ! = 4)=E(64) V ¬E(4) V E(1)
(x&118 = 6)=¬E(64)^¬E(32)^¬E(16)^E(4)^E(2)
E(64)  V ¬E(4) V E(1) V ¬E(32)^¬E(16)^E(2) V ¬E(A) ≡1
Suppress ¬E(64) and E(4) in conjunction
E(65) V ¬E(4) V ¬E(32)^¬E(16)^E(2) V ¬E(A) ≡1
Thus A(max)=65

Original idea proposed by Helen Mironchick

Solution equation kind of
(x&67 != 3) v ((x&55 = 7) => (x&A =0))
via Algebra of predicates {E(k)}
(x&67 = 3) = ¬E(64)^E(2)^E(1)
(x&67 != 3) = E(64) V ¬E(2) V ¬E(1)
(x&55 = 7) = ¬E(32)^¬E(16)^E(4)^E(2)^E(1)
(x&55 != 7) = E(32) V E(16) V ¬E(4) V ¬E(2) V ¬E(1)
E(64) V ¬E(2) V ¬E(1) V E(32) V E(16) V
V ¬E(4) V ¬E(2) V ¬E(1) V ¬E(A) ≡1
E(112) V ¬(E(4)^E(2)^E(1)) V ¬E(A) ≡1
Thus A(max) = 112
 

Solution of Assignment # T2951 (oncoming and reverse pass) Yandex Tutor of the Unified State Examination Informatics 2020

Original source

(x2=>x1)^(x3=>x2)^(x4=>x3)^(x5=>x4)=1
(x8=>(x6≡x7))^(x11=>(x9≡x10))^(x14=>(x12≡x13))=1
x2=>x8=1

Demonstration the technology of the oncoming and reverse passage 10.2019 on a simple example of Yandex # T2951 USE Informatics 2020

Решение Задания #T2951 (встречным и обратным проходом) Yandex Репетитор ЕГЭ Информатика 2020

Исходная система

https://yandex.ru/tutor/subject/problem/?problem_id=T2951

(x2=>x1)^(x3=>x2)^(x4=>x3)^(x5=>x4)=1
(x8=>(x6≡x7))^(x11=>(x9≡x10))^(x14=>(x12≡x13))=1
x2=>x8=1

Демонстрация техники встречного и обратного прохода 10.2019 на простом примере Yandex #T2951 ЕГЭ Информатика

Решение Задания #T2948,#T2949 (встречным проходом) Yandex Репетитор ЕГЭ Информатика 2020

Исходная система

 https://yandex.ru/tutor/subject/problem/?problem_id=T2948

 Демонстрация техники встречного прохода 10.2019 на простом примере Yandex #T2948 ЕГЭ Информатика

Исходная система

 https://yandex.ru/tutor/subject/problem/?problem_id=T2949

  Демонстрация техники встречного прохода 10.2019 на простом примере Yandex #T2949 ЕГЭ Информатика

Cross Reference Table Setup ( double predicates based ) and Solution of one standard assignment 23 of USE Informatics 2020

Original system per VK's News Wire Informatics_100

https://vk.com/doc201442577_528394217?hash=800896655a26dfa264&dl=b3b920ffa6ded30d68

First notice

The key place is a detailed description of building a cross-reference table 
when moving to a new line of the system. The chart generation is just a consequence.
First consider system
   (1) F1(x1,y1)=>F2(x2,y2) =1
   (2) F1(x2,y2)=>F2(x3,y3) =1
   (3) F1(x3,y3)=>F2(x4,y4) =1
   (4) F1(x4,y4)=>F2(x5,y5) =1
   (6) F1(x5,y5)=>F2(x6,y6) =1
   (7) F1(x6,y6)=>F2(x7,y7) =1

where F1 and F2 are double predicates
Denote card(N) the power of set N
Denote n1,n2,m1,m2,s1,s2
    n1=card (falseSet_F2 ∩ falseSet_F1)
    n2=card (falseSet_F2 ∩ truthSet_F1)
    m1=card (truthSet_F2 ∩ falseSet_F1)
    m2=card (truthSet_F2 ∩ truthSet_F1)
    s1=card (falseSet_F1)
    s2=card (truthSet_F1)
See Setting up a cross-reference table in 08/2016 approach of Helen Mironchick when moving to a new line of the system of Boolean equations 
for more details  Then following 10.2019 diagram would show

   Now fork 10.2019 chart
  

Решение Задания #T2947 Yandex Репетитор ЕГЭ Информатика 2020

Исходная система

https://yandex.ru/tutor/subject/problem/?problem_id=T2947

Демонстрация техники Метода Отображения версия 10.2019 на простом примере Yandex #T2947 ЕГЭ Информатика

 Собственно, все #T2949,#T2950,#T2951 решаются корректировкой узлов 2-го графа и удалением ложных решений. В терминологии Бориса Трушина - это задачи ни о чем.