XOR from only OR and AND

Question

How do you do the XOR bitwise operation if you only have available the AND and the OR operations?

Answer 1

i am pretty sure that the formula below is correct:

a xor b = not((a and b) or not(a+b))

Answer 2

Truth table for AND

  A  B  AND
  T  T  T
  T  F  F
  F  T  F
  F  F  F
  

Truth table for OR

  A  B  OR
  T  T  T
  T  F  T
  F  T  T
  F  F  F
  

Truth table for XOR

  A  B  XOR
  T  T  F
  T  F  T
  F  T  T
  F  F  F
  

So, XOR is just like OR, except it's false if A and B are true.

So, (A OR B) AND (NOT (A AND B)), which is (A OR B) AND (A NAND B)

  A  B  OR  AND NAND [(A OR B) AND (A NAND B)]
  T  T  T    T    F        F
  T  F  T    F    T        T
  F  T  T    F    T        T
  F  F  F    F    T        F
  

Not sure if it can be done without NOT or NAND

Answer 3

Best advice is to look up XOR in reference manuals and encyclopedia sites on the net and then write code or script which does the same as the description of what a XOR built-in function does and use your own return or status values. We can not tell you how to do that type of bit compares from within the software community.

Answer 4

(a XOR b) = ((a OR b) - (a AND b)), or in other words, the union set minus the intersection set.

Code example (in javascript):

var a = 5;
var b = 12;
var xor = (a | b) - (a & b); // result: 9

Answer 5

Wikipedia's entry on XOR goes over this in detail. Probably a good first place to check before aksing a SO question.

If you already have bits you don't care about masked off, it seems to me the easiest way to do it (as far as writing the code goes anyway) is to just use your not equal operator.

Answer 6

"The systems ({T, F}, and) and ({T, F}, or) are monoids."

"The system ({T, F}, xor) is an abelian group" which has the property of invertibility unlike monoids.

Therefore, 'and' and 'or' fail to construct 'xor' operation.

Source: https://en.wikipedia.org/wiki/Exclusive_or#Relation_to_modern_algebra