Chains of tie nodes of a multi-part graph && reverse pass to center per Helen Mironchick's "The Mapping method with a reverse calculation" - Computer science at school №10 2019 (awaiting publication)

Original systems

System 1

(x1^x2=>x3)^x4=>x5=1
(y1^y2=>y3)^y4=>y5=1
(z1^z2=>z3)^z4=>z5=1
(x3⊕y3)^(y4⊕z4)=1


For X-LINE

#2 is (x1^x2)
#3 is (x1^x2)=>x3
#4 is ((x1^x2)=>x3)^x4
#5 is (((x1^x2)=>x3)^x4)=>x5

For Y-LINE and Z-LINE replace x by y and x by z correspondingly

*******************
Chaining order
*******************
1) We get x3 via reverse pass to center
2) Getting y3 via direct XOR
3) Getting y4  via reverse pass to center
    Here we calculate y1^y2=>y3 (#3) Y-LINE.
    Perform move from #5 to #4 in the opposite direction.
    Now calculate y4 as usual.
4) Getting z4 via direct XOR
5) Completing Z-LINE having z4  (on line-0 112 , on line-1-200)

 
System 2

(x1^x2=>x3)^x4=>x5=1
(y1^y2=>y3)^y4=>y5=1
(z1^z2=>z3)^z4=>z5=1
(x3 v y3)^(y4 v z4) =1

*******************
Chaining order
*******************
1) We get x3 via reverse pass to center
2) Getting y3 via  x3 v y3 =1 basic chart
3) Getting y4  via reverse pass to center
    Here we calculate y1=>y2=>y3 (#3) Y-LINE.
    Perform move from #5 to #4 in the opposite direction.
    Now calculate y4 as usual.
4) Getting z4 via x4 v y4 =1 basic chart
5) Completing Z-LINE having z4  (on line-0 160 , on line-1-456)

    System 3

    (x1^x2=>x3)^x4=>x5=1
    (y1^y2=>y3)^y4=>y5=1
    (z1^z2=>z3)^z4=>z5 =1
    (x3 =>y3)^(y4=>z4) =1
 

 
    See also
    1.  https://informatics-ege.blogspot.com/2019/11/chaining-order-reverse-pass-to-center.html
    2.  https://informatics-ege.blogspot.com/2019/11/unleash-power-of-reverse-pass-to-center.html