The 23 -rd USE's Informatics task casual support on VK Informatics_100 on the go

Task1

https://vk.com/informatics_100?z=photo-40390768_457277148%2Fwall-40390768_199869

  Fork two 08.2016 charts

Task 2

https://vk.com/informatics_100?z=photo-40390768_457277144%2Fwall-40390768_199867 

Fork 08.2016 charts

  

   Task 3

   https://vk.com/informatics_100?z=photo-40390768_457277134%2Fwall-40390768_199890

Fork 08.2016 charts

   

The 23 -rd USE's Informatics task casual support on VK Informatics_100

Original source

Convert  to equivalent

(x1⊕x2) v (y1⊕y2)=1
(x2⊕x3) v (y2⊕y3)=1
(x3⊕x4) v (y3⊕y4)=1
(x4⊕x5) v (y4⊕y5)=1
x5≡y5=1

Just exercises for kids 08.2016 style to solve USE Informatics systems of Boolean equations

Original system

¬(x1≡x2) v ¬(x1≡x3)^(¬x2 v x3) =1
¬(x3≡x4) v ¬(x3≡x5)^(¬x4 v x5) =1
¬(x5≡x6) v ¬(x5≡x7)^(¬x6 v x7) =1
¬(x7≡x8) v ¬(x7≡x9)^(¬x8 v x9) =1

Convert to equivalent (verification basic knowledge of Boolean Algebra ) 

(x1≡x2) => (x1⊕x3)^(x2=>x3) =1
(x3≡x4) => (x3⊕x5)^(x4=>x5) =1
(x5≡x6) => (x5⊕x7)^(x6=>x7) =1
(x7≡x8) => (x7⊕x9)^(x8=>x9) =1

 Here we have transition variables x3,x5,x7 rather then transition pairs
 between equations. Thus we would manage via 08.2016 charts.

Solution of one USE Informatics system of Boolean equations in 08.2016 style

Original system

Orinal system
¬(x1≡x2)v¬(x1≡x3)^(x2≡x3)=1
¬(x3≡x4)v¬(x3≡x5)^(x4≡x5)=1
¬(x5≡x6)v¬(x5≡x7)^(x6≡x7)=1
¬(x7≡x8)v¬(x7≡x9)^(x8≡x9)=1

Convert to equivalent
  (x1≡x2) => (x1⊕x3)^(x2≡x3) =1
  (x3≡x4) => (x3⊕x5)^(x4≡x5) =1
  (x5≡x6) => (x5⊕x7)^(x6≡x7) =1
  (x7≡x8) => (x7⊕x9)^(x8≡x9) =1

   Here we have transition variables x3,x5,x7 rather then transition pairs
   between equations. Thus we would manage via 08.2016 charts.

Consider a bit more complex sample of similar system

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

Transition variables are x4 and x7. Fork 08.2016 chart for this system

   

Solution of System in Boolean variables cloning P233 to x,y,z variables (just for fun)

Original system

(((x1 =>y1 =>z1) =>x2 )=>y2) =>z2=1
(((x2 =>y2 =>z2) =>x3) =>y3) =>z3=1
(((x3 =>y3 =>z3) =>x4) =>y4) =>z4=1
(((x4 =>y4 =>z4) =>x5) =>y5) =>z5=1
(((x5 =>y5 =>z5) =>x6) =>y6) =>z6=1
(((x6 =>y6 =>z6) =>x7) =>y7) =>z7=1

Solution of one system in Boolean Variables from Informatics_100 23rd database via 08/2016 technique

Original System

Convert system as follows

(x1=>y1)=>(¬x2^y2)=1

(x2=>y2)=>(¬x3^y3)=1

(x3=>y3)=>(¬x4^y4)=1

(x4=>y4)=>(¬x5^y5)=1

(x5=>y5)=>(¬x6^y6)=1

(x6=>y6)=>(¬x7^y7)=1

Fork 08/2016 diagram

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