(x1 => x2) v (x3 =>x4) = 1 D1
(x3 => x4) ^ (x5 =>x6) =0 D2
(x5 => x6) v (x7 =>x8) = 1 D1
(x7 => x8) ^ (x9 =>x10) = 0
Thursday, April 26, 2018
Switching fork diagrams and Mapping Metod for system per request 26/04/2018 (BU's Queue picked up)
(x1 => x2) v (x3 =>x4) = 1 D1
(x3 => x4) ^ (x5 =>x6) =0 D2
(x5 => x6) v (x7 =>x8) = 1 D1
(x7 => x8) ^ (x9 =>x10) = 0
Friday, April 20, 2018
Complicated system of boolean equations kind of P-40 (ege23.pdf) && Mapping method
Use case bellow has obvious typos as missing right hand side values
of equations. Just assume that right hand side of equations bellow has 0
(zerro) values and attempt to create approach to generate matrix with
given values.
In case of values 1-th it is obvious that both diagrams will be updated corespondently and will generate different matrix. Same schema would work
for values sequence 1,0,1,0,1,0,1,0 ..... with another set of fork diagrams.
((x1^y1 = x3^y3)) => (x2^y2) = 1
((x2^y2 = x4^y4)) => ¬(x3^y3) = 1
((x3^y3 = x5^y5)) => (x4^y4) = 1
((x4^y4 = x6^y6)) => ¬(x5^y5) = 1
((x5^y5 = x7^y7)) => (x6^y6) = 1
((x6^y6 = x8^y8)) => ¬(x7^y7) = 1
zj=zj^yj
Just repeat. If they are 1-th (on right hand side) or changing 1 with 0 on regular basis, same schema would work as well with corespondently updated fork diagrams per right hand side value of first and second equations.
I mean there should be just two diagrams generating matrix.
Each outgoing value of zj having value 0 should mutiply cell's value
content by 3. Here situation looks like P-40 outgoing values 0 due to in P-40
has zj=xj^yj . Setup Helen's Mironchick may be viewed in ege23.pdf close to top of document
Consider "0" right hand size case
((x1^y1 = x3^y3)) => (x2^y2) = 0
((x2^y2 = x4^y4)) => ¬(x3^y3) = 0
((x3^y3 = x5^y5)) => (x4^y4) = 0
((x4^y4 = x6^y6)) => ¬(x5^y5) = 0
((x5^y5 = x7^y7)) => (x6^y6) = 0
((x6^y6 = x8^y8)) => ¬(x7^y7) = 0
zj=xj^yj
((x1vy1 = x3vy3)) => (x2vy2) = 0
((x2vy2 = x4vy4)) => ¬(x3vy3) = 0
((x3vy3 = x5vy5)) => (x4vy4) = 0
((x4vy4 = x6vy6)) => ¬(x5vy5) = 0
((x5vy5 = x7vy7)) => (x6vy6) = 0
((x6vy6 = x8vy8)) => ¬(x7vy7) = 0
((x7vy7 = x9vy9)) => ¬(x8vy8) = 0
zj=xjvyj
In case of values 1-th it is obvious that both diagrams will be updated corespondently and will generate different matrix. Same schema would work
for values sequence 1,0,1,0,1,0,1,0 ..... with another set of fork diagrams.
((x1^y1 = x3^y3)) => (x2^y2) = 1
((x2^y2 = x4^y4)) => ¬(x3^y3) = 1
((x3^y3 = x5^y5)) => (x4^y4) = 1
((x4^y4 = x6^y6)) => ¬(x5^y5) = 1
((x5^y5 = x7^y7)) => (x6^y6) = 1
((x6^y6 = x8^y8)) => ¬(x7^y7) = 1
zj=zj^yj
Just repeat. If they are 1-th (on right hand side) or changing 1 with 0 on regular basis, same schema would work as well with corespondently updated fork diagrams per right hand side value of first and second equations.
I mean there should be just two diagrams generating matrix.
Each outgoing value of zj having value 0 should mutiply cell's value
content by 3. Here situation looks like P-40 outgoing values 0 due to in P-40
has zj=xj^yj . Setup Helen's Mironchick may be viewed in ege23.pdf close to top of document
Consider "0" right hand size case
((x1^y1 = x3^y3)) => (x2^y2) = 0
((x2^y2 = x4^y4)) => ¬(x3^y3) = 0
((x3^y3 = x5^y5)) => (x4^y4) = 0
((x4^y4 = x6^y6)) => ¬(x5^y5) = 0
((x5^y5 = x7^y7)) => (x6^y6) = 0
((x6^y6 = x8^y8)) => ¬(x7^y7) = 0
zj=xj^yj
((x1vy1 = x3vy3)) => (x2vy2) = 0
((x2vy2 = x4vy4)) => ¬(x3vy3) = 0
((x3vy3 = x5vy5)) => (x4vy4) = 0
((x4vy4 = x6vy6)) => ¬(x5vy5) = 0
((x5vy5 = x7vy7)) => (x6vy6) = 0
((x6vy6 = x8vy8)) => ¬(x7vy7) = 0
((x7vy7 = x9vy9)) => ¬(x8vy8) = 0
zj=xjvyj
Tuesday, April 17, 2018
Another (like P-40) Complicated system of boolean equations && Mapping method
Use case bellow has obvious typos as missing right hand side values
of equations. Just assume that right hand side of equations bellow has 0
(zerro) values and attempt to create approach to generate matrix with
given values.
In case of values 1-th it is obvious that both diagrams will be updated corespondently and will generate different matrix. Same schema would work
for values sequence 1,0,1,0,1,0,1,0 ..... with another set of fork diagrams.
Assume right hand values are equal 0 (all), otherwise regenerate fork diagrams correspodently
((x1 ^ y1) ~ (x2 ^ y2)) => (x3 ^ y3) = 1
((x2 ^ y2) v ¬(x3 ^ y3)) => (x4 ^ y4) = 1
((x3 ^ y3) ~ (x4 ^ y4)) => (x5 ^ y5) = 1
((x4 ^ y4) v ¬(x5^ y5)) => (x6 ^ y6) = 1
Just repeat. If they are 1-th (on right hand side) or changing 1 with 0 on regular basis, same schema would work as well with corespondently updated fork diagrams. Per right hand side value of first and second equations.
I mean there should be just two diagrams generating matrix.
Each outgoing value of zj having value 0 should mutiply cell's value
content by 3. Here situation is exactly like in P-40 :- zj=xj^yj . Setup with no
changing forking diagrams belongs to Helen Mironchick and may be viewed in ege23.pdf close to top of document
Solution zj = xj^yj
Case "1"
(z1~z2) =>z3 = 1
(z3=>z2) => z4 = 1
(z3~z4) => z5 = 1
(z5=>z4)=>z6 = 1
Case "0"
((x1 ^ y1) ~ (x2 ^ y2)) => (x3 ^ y3) = 0
((x2 ^ y2) v ¬(x3 ^ y3)) => (x4 ^ y4) = 0
((x3 ^ y3) ~ (x4 ^ y4)) => (x5 ^ y5) = 0
((x4 ^ y4) v ¬(x5^ y5)) => (x6 ^ y6) = 0
zj=xj^yj Solution
(z1~z2) =>z3 = 0
(z3=>z2) => z4 = 0
(z3~z4) => z5 = 0
(z5=>z4)=>z6 = 0
In case of values 1-th it is obvious that both diagrams will be updated corespondently and will generate different matrix. Same schema would work
for values sequence 1,0,1,0,1,0,1,0 ..... with another set of fork diagrams.
((x1 ^ y1) ~ (x2 ^ y2)) => (x3 ^ y3) = 1
((x2 ^ y2) v ¬(x3 ^ y3)) => (x4 ^ y4) = 1
((x3 ^ y3) ~ (x4 ^ y4)) => (x5 ^ y5) = 1
((x4 ^ y4) v ¬(x5^ y5)) => (x6 ^ y6) = 1
Just repeat. If they are 1-th (on right hand side) or changing 1 with 0 on regular basis, same schema would work as well with corespondently updated fork diagrams. Per right hand side value of first and second equations.
I mean there should be just two diagrams generating matrix.
Each outgoing value of zj having value 0 should mutiply cell's value
content by 3. Here situation is exactly like in P-40 :- zj=xj^yj . Setup with no
changing forking diagrams belongs to Helen Mironchick and may be viewed in ege23.pdf close to top of document
Solution zj = xj^yj
Case "1"
(z1~z2) =>z3 = 1
(z3=>z2) => z4 = 1
(z3~z4) => z5 = 1
(z5=>z4)=>z6 = 1
Case "0"
((x1 ^ y1) ~ (x2 ^ y2)) => (x3 ^ y3) = 0
((x2 ^ y2) v ¬(x3 ^ y3)) => (x4 ^ y4) = 0
((x3 ^ y3) ~ (x4 ^ y4)) => (x5 ^ y5) = 0
((x4 ^ y4) v ¬(x5^ y5)) => (x6 ^ y6) = 0
zj=xj^yj Solution
(z1~z2) =>z3 = 0
(z3=>z2) => z4 = 0
(z3~z4) => z5 = 0
(z5=>z4)=>z6 = 0
Wednesday, April 11, 2018
One Complicated system of boolean equations && Mapping method
Use case bellow has obvious typos as missing right hand side values of equations. Just assume that right hand side of equations bellow has 0 (zerro) values and attempt to create approach to generate matrix with given values.
In case of values 1-th it is obvious that both diagrams will be updated corespondently and will generate different matrix. Same schema would work
for values sequence 1,0,1,0,1,0,1,0 ..... with another set of fork diagrams.
((x1 v y1) ~ (x2 v y2)) => (x3 v y3) = 1
((x2 v y2) v ¬(x3 v y3)) => (x4 v y4) = 1
((x3 v y3) ~ (x4 v y4)) => (x5 v y5) = 1
((x4 v y4) v ¬(x5 v y5)) => (x6 v y6) = 1
((x5 v y5) ~ (x6 v y6)) => (x7 v y7) = 1
((x6 v y6) v ¬(x7 v y7)) => (x8 v y8) = 1
((x7 v y7) ~ (x8 v y8)) => (x9 v y9) = 1
zj = xj v yj (j=1,2,3,.....,9)
Just repeat. If they are 1-th (on right hand side) or changing 1 with 0 on regular basis, same schema would work as well with corespondently updated fork diagrams. Per right hand side value of first and second equations.
I mean there should be just two diagrams generating matrix.
Each outgoing value of zj having value 1 should mutiply cell's value
content by 3. Here situation looks like P-40 outgoing values 0 due to in P-40
has zj=xj^yj . Those setup belongs Helen Mironchick and may be viewed in ege23.pdf close to top of document
(z1 ~ z2) => z3 =1
(z2 v ¬z3) => z4 =1
(z3 ~ z4 ) => z5 =1
(z4 v ¬z5) => z6 =1
(z5 ~ z6) => z7 =1
(z6 v ¬z7) => z8 =1
(z7 ~ z8) => z9 =1
((x1 v y1) ~ (x2 v y2)) => (x3 v y3) = 1 (z1z2 z2z3 -diagram1)
((x2 v y2) v ¬(x3 v y3)) => (x4 v y4) = 1 (z2z3 z3z4 -diagram2)
((x3 v y3) ~ (x4 v y4)) => (x5 v y5) = 1 diagram1
((x4 v y4) v ¬(x5 v y5)) => (x6 v y6) = 1 diagram2
((x5 v y5) ~ (x6 v y6)) => (x7 v y7) = 1 diagram1
((x6 v y6) v ¬(x7 v y7)) => (x8 v y8) = 1 diagram2
((x7 v y7) ~ (x8 v y8)) => (x9 v y9) = 1 diagram1
Loading original system matches Polyakov's calculator result
Another sample
((x1 v y1) ~ (x2 v y2)) => (x3 v y3) = 1
((x2 v y2) ~ (x3 v y3)) => ¬(x4 v y4) = 1
((x3 v y3) ~ (x4 v y4)) => (x5 v y5) = 1
((x4 v y4) ~ (x5 v y5)) => ¬(x6 v y6) = 1
((x5 v y5) ~ (x6 v y6)) => (x7 v y7) = 1
((x6 v y6) ~ (x7 v y7)) => ¬(x8 v y8) = 1
zj=xj v yj
(z1 ~ z2) => z3 = 1
(z2 ~ z3) => ¬z4 = 1
(z3 ~ z4) => z5 = 1
(z4 ~ z5) => ¬z6 = 1
(z5 ~ z6) => z7 = 1
(z6 ~ z7) =>¬z8 = 1
In case of values 1-th it is obvious that both diagrams will be updated corespondently and will generate different matrix. Same schema would work
for values sequence 1,0,1,0,1,0,1,0 ..... with another set of fork diagrams.
((x1 v y1) ~ (x2 v y2)) => (x3 v y3) = 1
((x2 v y2) v ¬(x3 v y3)) => (x4 v y4) = 1
((x3 v y3) ~ (x4 v y4)) => (x5 v y5) = 1
((x4 v y4) v ¬(x5 v y5)) => (x6 v y6) = 1
((x5 v y5) ~ (x6 v y6)) => (x7 v y7) = 1
((x6 v y6) v ¬(x7 v y7)) => (x8 v y8) = 1
((x7 v y7) ~ (x8 v y8)) => (x9 v y9) = 1
zj = xj v yj (j=1,2,3,.....,9)
Just repeat. If they are 1-th (on right hand side) or changing 1 with 0 on regular basis, same schema would work as well with corespondently updated fork diagrams. Per right hand side value of first and second equations.
I mean there should be just two diagrams generating matrix.
Each outgoing value of zj having value 1 should mutiply cell's value
content by 3. Here situation looks like P-40 outgoing values 0 due to in P-40
has zj=xj^yj . Those setup belongs Helen Mironchick and may be viewed in ege23.pdf close to top of document
(z1 ~ z2) => z3 =1
(z2 v ¬z3) => z4 =1
(z3 ~ z4 ) => z5 =1
(z4 v ¬z5) => z6 =1
(z5 ~ z6) => z7 =1
(z6 v ¬z7) => z8 =1
(z7 ~ z8) => z9 =1
((x1 v y1) ~ (x2 v y2)) => (x3 v y3) = 1 (z1z2 z2z3 -diagram1)
((x2 v y2) v ¬(x3 v y3)) => (x4 v y4) = 1 (z2z3 z3z4 -diagram2)
((x3 v y3) ~ (x4 v y4)) => (x5 v y5) = 1 diagram1
((x4 v y4) v ¬(x5 v y5)) => (x6 v y6) = 1 diagram2
((x5 v y5) ~ (x6 v y6)) => (x7 v y7) = 1 diagram1
((x6 v y6) v ¬(x7 v y7)) => (x8 v y8) = 1 diagram2
((x7 v y7) ~ (x8 v y8)) => (x9 v y9) = 1 diagram1
Loading original system matches Polyakov's calculator result
((x1 v y1) ~ (x2 v y2)) => (x3 v y3) = 1
((x2 v y2) ~ (x3 v y3)) => ¬(x4 v y4) = 1
((x3 v y3) ~ (x4 v y4)) => (x5 v y5) = 1
((x4 v y4) ~ (x5 v y5)) => ¬(x6 v y6) = 1
((x5 v y5) ~ (x6 v y6)) => (x7 v y7) = 1
((x6 v y6) ~ (x7 v y7)) => ¬(x8 v y8) = 1
zj=xj v yj
(z1 ~ z2) => z3 = 1
(z2 ~ z3) => ¬z4 = 1
(z3 ~ z4) => z5 = 1
(z4 ~ z5) => ¬z6 = 1
(z5 ~ z6) => z7 = 1
(z6 ~ z7) =>¬z8 = 1
Monday, April 9, 2018
Once again Bitwise2 shooting 18-th task (per bit conjunction) VK's News Wire as of 09/04/2018
I still cannot realize the reason for hiding Bitwise2 technique from public.
Those ones who could understand Bitwise2 idea became automatically armed with suggested approach. The rest would reproduce Informatic's "BU" logic
uncountable number of times ( each time for every case ).
Code bellow follows core ideas of A. V. Zdvizhkova
See http://kpolyakov.spb.ru/download/bitwise2.pdf
(page 6)
(¬Z(28)v¬Z(45)) => (Z(48) => ¬A) = 1
¬(Z(28)^Z(45)) => (¬Z(48)v¬ A) = 1
Z(28)^Z(45) v ¬((Z(48)^A) = 1
Z(48)^A => Z(28)^Z(45)
Z(48)^A => Z(28 OR 45)
*********************************
Per bit disjunction 28 v 45 = 61
********************************
45 = 101101
v
28 = 011100
=================
61 = 111101
48 = 110000
=================
A = 001101 = 13 (decimal)
Solution suggested at VK News Wire for same task
https://vk.com/informatics_100?z=photo-40390768_456255272%2Falbum-40390768_251859617
Sunday, April 8, 2018
Решение задач на стратегию ( 67 ) из ege26-C3.pdf
(ТТ ЕЕЕЕ НН ИИИ КК АА) (ХЧСЯБР)
1 2 3 4 5 6
Скобка (1) нессимметрична позиция из-за вхождения блока (4).
Петя 1-ым ходом убирает И из (4) и передает Ване
симметиричную позицию. Выигрышная стратегия
у Пети - в дальнейшем после каждого хода Вани,Петя
делает позизицию симметиричной
Ссылки
1 https://kpolyakov.spb.ru/download/ege26-C3.doc
Wednesday, April 4, 2018
Насколько эффективным может быть Метод Отображений - Два решения Задачи 14 из документа "Сколько различных решений имеет система уравнений.pdf"
Первое :-
https://www.youtube.com/watch?v=KNpJsN3smQg
Второе:-
Применяем метод отображений для решения первой системы из четырех уравнений и затем генерируем матрицу с стартовым значением Х1Х2 по строке "10" равным 0 определяем число решений с (Х9 => X10) равным 0 ( строка "10"). Таким образом определяем число решений исходной системы из 4 уравнений, таких что
(Х9 => X10) v X1^¬X2 = 0
Третье:-
Четвертое - Метод битовых масок :-
(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
z1 = x1 =>x2
z2 = x3 =>x4
z3 = x5 =>x6
z4 = x7 => x8
z5 = x10 => x9
z2 => z1 =1
z3 => z2 =1
z4 => z3 =1
z5 => z4 =1
z1 => z5 =1
z5 z4 z3 z2 z1
===========
1 1 1 1 1 3^5
0 0 0 0 0 3^0
Результат 243 + 1 = 244
Monday, April 2, 2018
Решение Р-42 из ege23.pdf методом отображений
(x1^y1) => ¬(x2^y2) =1
Базовая диаграмма и генерация матрицы с учетом Х2 =0 и Y4=0
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