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