Mapping Method ( 10.2019 ) VERSUS Stream 35 provided by Informatik BU System 190 (Noskin ege23.doc)

Original system

    Solution provided by BU may be seen via Video
    https://www.youtube.com/watch?v=9TV1GVFakEc
    ( 2 hr 2 min)

Original source

which is equivalent to system below
    (x5^y5)=>(x4^y4)=1
    (x4^y4)=>(x3^y3)=1
    (x3^y3)=>(x2^y2)=1
    (x2^y2)=>(x1^y1)=1
    x2^y4 = 0
Now build 10.2019 chart to solve the system above
 

Solution task 22 Variant 1 "Krylov && Churkina 20 sample exam options for USE Informatics 2020"

   Building graphs to solve the task

 

         Answer is 12*30 =360

Variant 24112019 from Eugene Dzhobs task 27b

Original solution provided

 Now fixing errors and trace the module

var i, N, x, max: integer;
// Error fix 'a'
 a : array[0..9] of integer;       
begin
 readln(N);
 for i:=1 to N do begin
 readln(x);
 while x > 0 do begin
 a[ x mod 10 ] := a[ x mod 10] + 1;
 x := x div 10;
 end;
 end;
for i:=0 to 9 do begin
writeln(i,' has been selected ', a[i] ,' times');
end;
 max := 0;
 for i := 0 to 9 do
 if a[i] > max then max := a[i];
 write('max = '); writeln(max);
 for i := 9 downto 0 do
// Error fix  'i'
 if a[i] = max then writeln( i ,' was submitted ', max ,' times');
end.

[boris@ServerCentOS8 Djobs]$ ./prg27b2411
5
687
374
897
608
711
0 has been selected 1 times
1 has been selected 2 times
2 has been selected 0 times
3 has been selected 1 times
4 has been selected 1 times
5 has been selected 0 times
6 has been selected 2 times
7 has been selected 4 times
8 has been selected 3 times
9 has been selected 1 times
max = 4  7 was submitted 4 times
[boris@ServerCentOS8 Djobs]$ ./prg27b2411
10
5678
4563
789
7654
8576
9073
3201
4563
9876
3454
0 has been selected 2 times
1 has been selected 1 times
2 has been selected 1 times
3 has been selected 5 times
4 has been selected 5 times
5 has been selected 6 times
6 has been selected 6 times
7 has been selected 6 times
8 has been selected 4 times
9 has been selected 3 times
max = 6  7 was submitted 6 times
6 was submitted 6 times
5 was submitted 6 times
[boris@ServerCentOS8 Djobs]$ vi control.txt
[boris@ServerCentOS8 Djobs]$ cat  control.txt|grep "6"|wc -l
6
[boris@ServerCentOS8 Djobs]$ cat  control.txt|grep "5"|wc -l
6
[boris@ServerCentOS8 Djobs]$ cat  control.txt|grep "7"|wc -l
6

Алгебра предикатов и Задание 18 из Варианта Евгения Джобса 24112019

https://www.youtube.com/watch?v=9t2adabEuV4

( 1 час 43 мин )

Ниже мы используем следующее
Утверждение 01
*****************************************************
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)
******************************************************
Допустим  ∃ y : (P(y)=1)^(Q(y) = 0 ) =1. Тогда P(y) => Q(y) = False
Что противоречит условию (*) и $(P) вложено в $(Q)
Отсюда также следует , что максимальная область истинности P ($(P)) есть область истинности Q ($(Q)), поскольку при Q(z)=1, мы можем не теряя общности считать P(z)=1, а минимальная область истинности Q ($(Q)) есть область истинности P ($(P)).

Определим предикаты Р(х) и Q(x) следующим образом
P(x)={1, x ∈ [-8,8]
      0, x!∈ [-8,8]
      }
Q(x)={1, x ∈ [-6,8]
      0, x!∈ [-6,8]
      }
 где -6,8 корни уравнения x^2-2*x-48 =0
Найти наименьшую область истинности предиката А(х)  такого,что   (A=>P)^(Q=>A) ≡1

Решение
Откуда следует система
A=>P ≡1
Q=>А ≡1
Следовательно области истинности предикатов P и Q 
$(P) и $(Q) должны удовлетворять условиям
$(Q) ⊂ $(A) ⊂ $(P)
то есть
[-6,8] ⊂ $(A) ⊂ [-8,8]
Откуда Min $(A) = [-6,8]
Длина которой равна 14

Официальное решение задачи Р-45 из файла ege23.doc VERSUS решение, предложенное Информатиком БУ в Стриме 32 (2020)

Решение БУ && логика БУ изложены https://www.youtube.com/watch?v=QsC5rHfUb44   ( 2 часа 54 сек )

 Одно из официальных решений той же задачи Р-45 ( 14/04/2019 ) в файле ege23.doc сайт  К.Ю.Полякова

Решение

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