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?

@Oded nope, I just want to know. I'm programming in a scripting environment which only has AND and OR, but neither NOT or XOR, and I need XOR. You guys at StackOverflow sure are quick to judge something as homework. Also, I don't understand that link, I had already looked at it before posting the question. — John Zane, Jan 17 '11 at 16:13
I'm pretty sure you need a NOT to create an XOR = (a && !b) || (!a && b) is the simplest way to put it — Jesus Ramos, Jan 17 '11 at 16:17
You need NOT otherwise it's not possible. If it doesn't have NOT (what scripting environment is this that *doesn't* have NOT -- that's a fairly fundamental omission) -- then you should just be able to use conditional logic to invert the bit... `if (x == 0) x = 1`, then it should be straightforward. — Chris J, Jan 17 '11 at 16:18
No NOT? Seriously? That's not funny. (Oh ana @nixon: NOR works as well; but it's no fun with NAND and NOR) — , Jan 17 '11 at 16:19
Is it permissible to use arithmetic operators, such as `+` and `-` ? If so then see my answer below... — Paul R, Mar 21 '11 at 11:49

score 16 · Answer 1 · answered Jan 17 '11 at 16:18

16

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

answered Jan 17 '11 at 16:18

Lou Franco

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I think we all know what those operations do an how their truth tables look like. The question is, can you combine the first two to form the third, and if so, how? – Jan 17 '11 at 16:20
2

I was starting my answer in the event that it was homework. To be a direction rather than an answer. I don't think it can be done. – Lou Franco Jan 17 '11 at 16:24
Since the latter half which is (NOT (A AND B)) is the same as ((NOT A) OR (NOT B)), the entire can also be written as (A OR B) AND ((NOT A) OR (NOT B)), or (A+B)(!A+!B). The (A+B) handles the part where either A or B are true, the (!A+!B) handles when both A and B are false, the AND in between makes sure that both parts are true for the entire expression to be true. – hsbatra Mar 26 '23 at 22:14

score 9 · Answer 2 · answered Dec 01 '17 at 06:48

9

(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

answered Dec 01 '17 at 06:48

Zack

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MGoksu · Answer 3 · 2018-06-22T07:57:41.263

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"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

edited Jun 22 '18 at 07:57

answered Oct 19 '15 at 13:49

MGoksu

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score 4 · Answer 4 · answered Aug 29 '14 at 16:53

4

In C: x ^ y = (x & ~y) | (~x & y)

answered Aug 29 '14 at 16:53

user1527227

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score 4 · Accepted Answer · edited Mar 17 '16 at 13:21

4

Creating my own scripting language - ChrisScript - you just need something like:

#!/bin/chrish

bit XOR (bit A, bit B)
{
   bit notA;
   bit notB;

   IF (A == 0) notA = 1 ELSE notA = 0;
   IF (B == 0) notB = 1 ELSE notB = 0;

   F = ((A && notB) || (notA && B));

   RETURN F;
}

Even without NOT, it can be emulated like this. But this is the best solution you're going to get without having some form of inverter. I find it hard to believe you don't have some form of inverter availble -- what scripting environment are you using?

edited Mar 17 '16 at 13:21

Lou Franco

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answered Jan 17 '11 at 16:24

Chris J

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A very limited Javascript environment. – John Zane Jan 17 '11 at 16:28
Also, I need to do it int by int, not bit by bit, since I can't get individual bits, for lack of bit shifting operators. – John Zane Jan 17 '11 at 16:33
2

Define "very limited" -- do you have the actual javascript engine you're using (e.g., Microsoft JScript)? Is this part of some proprietary thing, if so, what product and version? It would be nice to have as much information about your environment as possible. – Chris J Jan 17 '11 at 16:56

Paul R · Answer 6 · 2018-10-19T23:04:42.630

4

If you have arithmetic operators such as + and - in addition to bitwise AND (&) and OR (|) then you can do bitwise XOR like this:

int bitwise_XOR(int a, int b)
{
    return (a + b) - (a & b) - (a & b);
}

The reason this works is that we are doing a full add, which is equivalent to XOR when the sum for amy given bit position is <= 1, and then we're correcting for the case where a carry is generated (1 + 1) by subtracting 2 * (a & b).

Note that this works even when the intermediate terms overflow, assuming we have "normally behaved" integers (2's complement, modulo 2 wraparound for overflow, etc).

edited Oct 19 '18 at 23:04

answered Mar 21 '11 at 11:47

Paul R

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[gcc 6.x onward is smart enough to transform this to `a ^ b`](https://godbolt.org/z/WTl4Uk) – phuclv Feb 11 '19 at 16:02

score 2 · Answer 7 · answered Jan 17 '11 at 16:19

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.

score 0 · Answer 8 · answered Jan 19 '20 at 23:44

0

in python

...:def xor(a,b):
...:     c = (not a) and b
...:     d = (not b) and a
...:     e = not c
...:     f = not d
...:     g = e and f
...:     h = not g
...:     return h

answered Jan 19 '20 at 23:44

jkobject

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score 0 · Answer 9 · answered Mar 01 '21 at 15:31

0

  PrintF('%-10s XOR',[inttobinbyte((10 OR 12)-(12 AND 10))])

00000110 XOR

answered Mar 01 '21 at 15:31

Max Kleiner

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score -1 · Answer 10 · answered Jan 17 '11 at 16:18

-1

i am pretty sure that the formula below is correct:

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

answered Jan 17 '11 at 16:18

Noray

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Read the question again. NOT is not available. – Jan 17 '11 at 16:20
+ is also not a boolean operator last I checked let alone an AND or OR operation – Bernd Wechner Jan 13 '22 at 07:38

score -3 · Answer 11 · answered Jan 17 '11 at 16:17

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.

XOR from only OR and AND

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