Simplify (a + b) XOR (c + b)

Question

Is it possible to simplify (a+b)xor(c+b)? What is the contribution of b to the final result? Note that I'm mixing boolean algebra with arithmetic, xor is a bitwise exclusive or on corresponding bits and + is a standard addition on 8 bits, that wraps around when overflown. a, b, c are unsigned char;

Unlikely, since XOR is purely bitwise, whereas addition affects adjacent bits. — Paul R, Mar 05 '14 at 15:51

score 5 · Answer 1 · answered Mar 05 '14 at 16:15

5

We can use an SMT solver to test our hypothesis that your formula can be simplified. You can head over to http://rise4fun.com:

x = BitVec('x', 8)
y = BitVec('y', 8)
z = BitVec('z', 8)

print simplify((x + z) ^ (y + z))

and the result, anticlimactically, is:

x + z ^ y + z

Which means your formula cannot be further simplified.

answered Mar 05 '14 at 16:15

Michael Foukarakis

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I don't see any hypothesis in the question! – Lightness Races in Orbit Mar 05 '14 at 16:17
1

This result does not make sense `x + z ^ y + z` . Original sum is symmetrical around `x` and `y`, whereas a result is not. – Salvador Dali May 31 '16 at 07:32
@SalvadorDali: It does make sense if you consider Python operator precedence, which Z3 uses: https://docs.python.org/3/reference/expressions.html#index-77 – Michael Foukarakis May 31 '16 at 08:17

score 3 · Answer 2 · edited May 08 '20 at 11:27

3

(a+b)xor(c+b) 
--------------
=((not(a+b))*(c+b))+((a+b)*(not(c+b))) 
-----------------------
=((not a)*(not b)*(c+b))+((a+b)*(not c)*(not b)) 
----
=((not a)(not b)*c) + (a*(not c)(not b)) 
----
=(not b)((not a)c + a(not c)) 
----
=(not b)(a xor c)
----

edited May 08 '20 at 11:27

Mat

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answered May 08 '20 at 11:15

anonymous

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Simplify (a + b) XOR (c + b)

2 Answers2