Bitwise2 vs E.A. Mironchik "Solving problems EGE-18 with per bit operations " (2017)

  
Treat problem 5 from original manuscript
"Решение задания ЕГЭ - 18 с битовыми операциями"    with Bitwise2

Follow Bitwise2 rules

(x&26 ¬=10 ) = ¬Z(16) + Z(8) + Z(2)
(x&27 = 11 )   =   Z(16)*¬Z(8)*¬Z(2)*¬Z(1)

26 = 11010 
10 = 01010
27 = 11011
11 = 01011

Due to
  ¬A + A*B = (¬A + A)*(¬A + B) = ¬A + B
    A + ¬A*B = (A + ¬A)*(A + B) = A + B   
After applying conversions mentioned above
We are going to obtain :-

¬Z(16) + Z(8) + Z(2) +  Z(16)*¬Z(8)*¬Z(2)*¬Z(1) + A = 1
¬Z(16) + Z(8) +  Z(2) + ¬Z(1) + A = 1
¬(Z(16)*Z(1)) + Z(8) + Z(2) + A = 1
Z(16 OR 1) => Z(8) + Z(2) + A = 1
(Z(17)=> A) + (Z(17)=>Z(8)) + (Z(17)=>Z(2)) = 1
Z(17) => A = 1

Thus A(max) = 17
So issue when "b1-c1=b2-c2" might be resolved
via Bitwise2 technique and Problem 5 gets resolved without any new development efforts. Also notice that in general (b-c) doesn't have to
be a basic predicate. Sample bellow has (b-c) = 36. But the multitude
of "c" bits "1"  must be included in the multitude of "b" bits "1"

First notice

¬A + A*B = (¬A + A)*(¬A + B) = ¬A + B (1)
A + ¬A*B = (A + ¬A)*(A + B) = A + B     (2)

and  follow "Решение задания ЕГЭ - 18 с битовыми операциями"


***********************************
For instance consider equation
***********************************

(x&46 ¬= 10) v (x&47 = 11) v (x&A =0) = 1
(x&47 = 11) = ¬E(36)*E(8)*E(2)*E(1)
(x&46 ¬= 10) = E(36) + ¬E(8) + ¬E(2)


Hacking "mysterious" rules of Boolean algebra ( fragment of original
manuscript posted bellow )

 
According Boolean algebra rules (1),(2) :-

Per (1)

E(36) + ¬E(8) + ¬E(2) + ¬E(36)*E(8)*E(2)*E(1)+¬E(A) = 1
E(36) + ¬E(8) + ¬E(2) + ¬E(36)*E(2)*E(1) + ¬E(A) = 1
E(36) + ¬E(8) + ¬E(2) + ¬E(36)*E(1) + ¬E(A) = 1

Per (2)

E(36) + ¬E(8) + ¬E(2) + E(1) + ¬E(A) = 1

 46= 101110 (binary)
 10 =001010
 47 =101111
 11 =001011 

Now we see that A(max) = 37

**************************************
Another sample done both ways
**************************************

(x&62 ¬= 26)v(x&63 = 27)v(x&A = 0) = 1

62 = 111110
26 = 011010

63 = 111111
27 = 011011

(x&63 = 27) = ¬E(36)*E(16)*E(8)*E(2)*E(1)
(x&62 ¬= 26) = E(36) + ¬E(16) + ¬E(8) + ¬E(2)
E(36) + ¬E(16) + ¬E(8) + ¬E(2) + ¬E(36)*E(16)*E(8)*E(2)*E(1) + ¬E(A) = 1
E(36) + ¬E(16) + ¬E(8) + ¬E(2) + E(1) + ¬E(A) = 1
E(37) + ¬E(16) + ¬E(8) + ¬E(2) + E(1) + ¬E(A) = 1

Thus A(max)  = 37
*********************
Bitwise solution
*********************
(x&62 ¬= 26)v(x&63 = 27)v(x&A = 0) = 1

62 = 111110
26 = 011010

63 = 111111
27 = 011011

(x&63 = 27) = Z(36)*¬Z(16)*¬Z(8)*¬Z(2)*¬Z(1)
(x&62 ¬= 26) = ¬Z(36) + Z(16) + Z(8) + Z(2)
¬Z(36) + Z(16) + Z(8) + Z(2) + Z(36)*¬Z(16)*¬Z(8)*¬Z(2)*¬Z(1) + A = 1
¬Z(36) + Z(16) + Z(8) + Z(2) + ¬Z(1) + A = 1
¬(Z(36)*Z(1)) + Z(16) + Z(8) + Z(2) + A = 1
Z(37) => Z(16) + Z(8) + Z(2) + A = 1

(Z(37)=>A) + (Z(37)=>16) + (Z(37)=>Z(8)) + (Z(37)=>Z(2)) = 1

Thus A(max)  = 37

********************************
Keep staying with Bitwise
********************************
(x&62 = 30)v(x&60 ¬= 28)v(x&A = 0) = 1

62 = 111110
30 = 011110
60 = 111100
28 = 011100

(x&62 = 30) = Z(32)*¬Z(16)*¬Z(8)*¬Z(4)*¬Z(2)
(x&60 ¬= 28) = ¬Z(32) + Z(16) + Z(8) + Z(4)
¬Z(32) + Z(16) + Z(8) + Z(4) + Z(32)*¬Z(16)*¬Z(8)*¬Z(4)*¬Z(2) +A = 1
¬Z(32) + Z(16) + Z(8) + Z(4) + ¬Z(2) + A = 1
¬(Z(32)*Z(2)) + Z(16) + Z(8) + Z(4) + A = 1
Z(34) => Z(16) + Z(8) + Z(4) + A = 1
(Z(34)=>Z(16)) + (Z(34)=>Z(8)) + (Z(34)=>Z(4)) + (Z(34)=>A) = 1
Z(34)=>A = 1

Thus A(max) = 34

*******************************************************
(x&120 ¬= 56) v (x&124 = 60) v (x&A=0) = 1
*******************************************************
120 = 1111000
56  =  0111000
124 = 1111100
60  =  0111100

(x&120 ¬= 56) = ¬Z(64) + Z(32) + Z(16) + Z(8) 
(x&124 = 60)   =   Z(64)*¬Z(32)*Z(16)*¬Z(8)*¬Z(4) 

(x&120 ¬= 56) v (x&124 = 60) v A =
¬Z(64) + Z(32) + Z(16) + Z(8) + Z(64)*¬Z(32)*Z(16)*¬Z(8)*¬Z(4) + A = 1
¬Z(64) + Z(32) + Z(16) + Z(8) + ¬Z(4) + A = 1
¬(Z(64)*Z(4)) + Z(32) + Z(16) + Z(8)   + A = 1
¬(Z(64 OR 4)) + Z(32) + Z(16) + Z(8) + A = 1
¬Z(68) + Z(32) + Z(16) + Z(8)  + A = 1
Z(68) => Z(32) + Z(16) + Z(8)  + A = 1
(Z(68) => Z(32)) + (Z(68)=>Z(16)) + (Z(68)=>Z(8)) + (Z(68)=> A) = 1
Z(68) => A = 1

Thus A(max)=68

************************************************************
(x&248 ¬= 152) v (x&252 = 156) v (x&A = 0) = 1
************************************************************
252 = 11111100
156 = 10011100
248 = 11111000
152 = 10011000

(x&248 ¬= 152) = ¬Z(96) + Z(128) + Z(16) + Z(8)
(x&252 =  156) =  Z(96)*¬Z(128)*¬Z(16)*¬Z(8)*¬Z(4)

¬Z(96) + Z(128) + Z(16) + Z(8) + Z(96)*¬Z(128)*¬Z(16)*¬Z(8)*¬Z(4) + A = 1
¬Z(96) + Z(128) + Z(16) + Z(8) + ¬Z(4) + A =  1
¬(Z(96)*Z(4))   + Z(128) + Z(16) + Z(8) + A = 1
Z(100) => Z(128) + Z(16) + Z(8) + A = 1
(Z(100)=>Z(128)) + (Z(100)=>Z(16)) + (Z(100)=>Z(8)) + (Z(100)=>A) = 1
Z(100) => A = 1

Thus A(max) = 100

***************************************************************************
(x&248 ¬= 152)v(x&252 = 156)v(x&250 = 154)v(x&A = 0) = 1
***************************************************************************
252 = 11111100
156 = 10011100
248 = 11111000
152 = 10011000
250 = 11111010
154 = 10011010

(x&248 ¬= 152) = ¬Z(96) + Z(128) + Z(16) + Z(8)
(x&252 =  156) =  Z(96)*¬Z(128)*¬Z(16)*¬Z(8)*¬Z(4)
(x&250 =  154) =  Z(96)*¬Z(128)*¬Z(16)*¬Z(8)*¬Z(2)

¬Z(96) + Z(128) + Z(16) + Z(8) + Z(64)*¬Z(128)*¬Z(16)*¬Z(8)*¬Z(4)
       + Z(96)*¬Z(128)*¬Z(16)*¬Z(8)*¬Z(2) + A = 1

¬Z(96) + Z(128) + Z(16) + Z(8) + ¬Z(4)
       + Z(96)*¬Z(128)*¬Z(16)*¬Z(8)*¬Z(2)  + A =  1

¬Z(96) + Z(128) + Z(16) + Z(8) + ¬Z(4) + ¬Z(2)  + A =  1      
¬(Z(96)*Z(4)*Z(2)) + Z(128) + Z(16) + Z(8) + A = 1

Z(102) => Z(128) + Z(16) + Z(8) + A = 1

(Z(102)=>Z(128)) + (Z(102)=>Z(16)) + (Z(102)=>Z(8)) + (Z(102)=>A) = 1
Z(102) => A = 1

Thus A(max) = 102