Solution System №129 & 116 & 156 from ege23.pdf via Mapping Method as of 11/05 2018

   (x1 => y1)^((x2 v y2) => (x1=y1)) =1
   (x2 => y2)^((x3 v y3) => (x2=y2)) =1
   (x3 => y3)^((x4 v y4) => (x3=y3)) =1
   (x4 => y4)^((x5 v y5) => (x4=y4)) =1
   (x5 => y5)^((x6 v y6) => (x5=y5)) =1
   (x6 => y6)^((x7 v y7) => (x6=y6)) =1
   x7=y7

 

  

  

(x1 v y1)^((x1^y1) v (!x2 v !y2)) =1
(x2 v y2)^((x2^y2) v (!x3 v !y3)) =1
(x3 v y3)^((x3^y3) v (!x4 v !y4)) =1
(x4 v y4)^((x4^y4) v (!x5 v !y5)) =1
(x5 v y5)^((x5^y5) v (!x6 v !y6)) =1
x6 v y6=1

  

Equations (1) && (4) set values "01" and "10" of column 1 to zeros

((x1=x2)v(x3=x4))^(¬((x1=x2) => (x3=x4))) =1

((x5=x6)v(x3=x4))^(¬((x5=x6) => (x3=x4))) =1

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

  

 

   

Solution for system from BU's Stream Queue as of 07/05/2018 via Mapping Method

    

(x1vy1)^(x2^y2) => (x1^y1) =1
(x2vy2)^(x3^y3) => (x2^y2) =1
(x3vy3)^(x4^y4) => (x3^y3) =1
(x4vy4)^(x5^y5) => (x4^y4) =1
(x5vy5)^(x6^y6) => (x5^y5) =1
(x6vy6)^(x7^y7) => (x6^y6) =1
x7vy7=1
 

  

Conversion decimal to binary in Pascal (fpc Fedora 27) or how much power provides recursion in Pascal

[boris@fedora27workstation PASCAL]$ cat ConvDecBin.pas
program ConvDecBin;

procedure Dec2Bin(var sbin : string; NUMBER : integer);
begin
    if NUMBER > 0 then begin
        Dec2Bin(sbin, NUMBER div 2);
        if NUMBER mod 2 = 0 then
            sbin := sbin + '0'
        else
            sbin := sbin + '1';
    end;
end;

var
  j : integer;
  s : string;

begin
  read(j);
  Dec2Bin(s,j);
  writeln(s);
end.
[boris@fedora27workstation PASCAL]$ fpc -oConvDecBin.exe ConvDecBin.pas
Free Pascal Compiler version 3.0.2 [2018/03/14] for x86_64
Copyright (c) 1993-2017 by Florian Klaempfl and others
Target OS: Linux for x86-64
Compiling ConvDecBin.pas
Linking ConvDecBin.exe
/usr/bin/ld: warning: link.res contains output sections; did you forget -T?
22 lines compiled, 0.1 sec
[boris@fedora27workstation PASCAL]$ ./ConvDecBin.exe
285
100011101

Switching fork diagrams and Mapping Metod for system per request 01/05/2018 (BU's Queue picked up)

System :-

(x1 => x2) v (x2 =>x1)^x3 =1
x2 v x3 v (x2^x3 => x4 ) =1
(x3 => x4) v (x4 =>x3)^x5 =1
x4 v x5 v (x4^x5 => x6) =1
(x5 => x6) v (x6 =>x5)^x7 =1
x6 v x7 v (x6^x7 => x8) =1

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
 

  

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

  

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