Solution of one task type 23 from VK's News Wire 10.03.18 via mapping method

     Attempt to work via mapping method
 
    ((x1~x3)v(x2~x4))^(¬((x1~x3)^¬(x2~x4))) = 0
    ((x2~x4)v(x5~x7))^(¬((x2~x4)^¬(x5~x7))) = 0
    ((x5~x7)v(x6~x8))^(¬((x5~x7)^¬(x6~x8))) = 0
    ((x6~x8)v(x9~x10))^(¬((x6~x8)^¬(x9~x10))) = 0
    ((x1~x3)=>(x2~x4))=>((x6~x8)v¬(x9~x10)) = 1

   ((x1~x3)v(x2~x4))^((¬(x1~x3)v(x2~x4))) = 0
   ((x2~x4)v(x5~x7))^((¬(x2~x4)v(x5~x7))) = 0
   ((x5~x7)v(x6~x8))^((¬(x5~x7)v(x6~x8))) = 0
   ((x6~x8)v(x9~x10))^((¬(x6~x8)v(x9~x10))) = 0
   ((x1~x3)=>(x2~x4))=>((x6~x8)v¬(x9~x10)) = 1

  ((x1~x3)v(x2~x4))^((x1~x3)=>(x2~x4)) = 0
  ((x2~x4)v(x5~x7))^((x2~x4)=>(x5~x7)) = 0
  ((x5~x7)v(x6~x8))^((x5~x7)=>(x6~x8)) = 0
  ((x6~x8)v(x9~x10))^((x6~x8)=>(x9~x10)) = 0
  ((x1~x3)=>(x2~x4))=>((x9~x10)=>(x6~x8)) = 1


    ((x1 = x3) v (x2 = x4)) ^ ((x1 = x3) => (x2 = x4)) =0


    Build diagrama based on first equation which would work
   for first four equations. Notice that transition pairs are x2x4 , x5x7 , x6x8
   In fact adding 
         ((x1~x3)=>(x2~x4))=>((x9~x10)=>(x6~x8)) = 1 
   would add  ( 0 => 0) => (0 =>0)  =1 on lines "01" and "10"
   e.g. True => True = 1 . So adding  this 5-th equation won't decrease number      of solutions of original sytem 

   

    Passing Polyakov's Control

  

   

  Thus answer is 64