What does bitwise XOR (exclusive OR) mean?

I'm trying to understand the binary operators in C# or in general, in particular ^ - exclusive or.

For example:

Given an array of positive integers. All numbers occur even number of times except one number which occurs odd number of times. Find the number in O(n) time and constant space.

This can be done with ^ as follows: Do bitwise XOR of all the elements. Fin开发者_StackOverflow社区ally we get the number which has odd occurrences.

How does it work?

When I do:

int res = 2 ^ 3;  
res = 1;  
int res = 2 ^ 5;  
res = 7;  
int res = 2 ^ 10;  
res = 8;  

What's actually happening? What are the other bit magics? Any reference I can look up and learn more about them?

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I know this is a rather old post but I wanted simplify the answer since I stumbled upon it while looking for something else.
XOR (eXclusive OR/either or), can be translated simply as toggle on/off.
Which will either exclude (if exists) or include (if nonexistent) the specified bits.

Using 4 bits (1111) we get 16 possible results from 0-15:

 decimal | binary | bits (expanded)
       0 | 0000   | 0
       1 | 0001   | 1
       2 | 0010   | 2
       3 | 0011   | (1+2)
       4 | 0100   | 4
       5 | 0101   | (1+4)
       6 | 0110   | (2+4) 
       7 | 0111   | (1+2+4)
       8 | 1000   | 8
       9 | 1001   | (1+8)
      10 | 1010   | (2+8)
      11 | 1011   | (1+2+8)
      12 | 1100   | (4+8)
      13 | 1101   | (1+4+8)
      14 | 1110   | (2+4+8)
      15 | 1111   | (1+2+4+8)

The decimal value to the left of the binary value, is the numeric value used in XOR and other bitwise operations, that represents the total value of associated bits. See Computer Number Format and Binary Number - Decimal for more details.

For example: 0011 are bits 1 and 2 as on, leaving bits 4 and 8 as off. Which is represented as the decimal value of 3 to signify the bits that are on, and displayed in an expanded form as 1+2.

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As for what's going on with the logic behind XOR here are some examples
From the original post

2^3 = 1

• 2 is a member of 1+2 (3) remove 2 = 1

2^5 = 7

• 2 is not a member of 1+4 (5) add 2 = 1+2+4 (7)

2^10 = 8

• 2 is a member of 2+8 (10) remove 2 = 8

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Further examples

1^3 = 2

• 1 is a member of 1+2 (3) remove 1 = 2

4^5 = 1

• 4 is a member of 1+4 (5) remove 4 = 1

4^4 = 0

• 4 is a member of itself remove 4 = 0

1^2^3 = 0
Logic: ((1^2)^(1+2))

• (1^2) 1 is not a member of 2 add 2 = 1+2 (3)
• (3^3) 1 and 2 are members of 1+2 (3) remove 1+2 (3) = 0

1^1^0^1 = 1
Logic: (((1^1)^0)^1)

• (1^1) 1 is a member of 1 remove 1 = 0
• (0^0) 0 is a member of 0 remove 0 = 0
• (0^1) 0 is not a member of 1 add 1 = 1

1^8^4 = 13
Logic: ((1^8)^4)

• (1^8) 1 is not a member of 8 add 1 = 1+8 (9)
• (9^4) 1 and 8 are not members of 4 add 1+8 = 1+4+8 (13)

4^13^10 = 3
Logic: ((4^(1+4+8))^(2+8))

• (4^13) 4 is a member of 1+4+8 (13) remove 4 = 1+8 (9)
• (9^10) 8 is a member of 2+8 (10) remove 8 = 2

• 1 is not a member of 2+8 (10) add 1 = 1+2 (3)

4^10^13 = 3
Logic: ((4^(2+8))^(1+4+8))

• (4^10) 4 is not a member of 2+8 (10) add 4 = 2+4+8 (14)
• (14^13) 4 and 8 are members of 1+4+8 (13) remove 4+8 = 1

• 2 is not a member of 1+4+8 (13) add 2 = 1+2 (3)

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To see how it works, first you need to write both operands in binary, because bitwise operations work on individual bits.

Then you can apply the truth table for your particular operator. It acts on each pair of bits having the same position in the two operands (the same place value). So the leftmost bit (MSB) of A is combined with the MSB of B to produce the MSB of the result.

Example: 2^10:

    0010 2
XOR 1010 8 + 2
    ----
    1    xor(0, 1)
     0   xor(0, 0)
      0  xor(1, 1)
       0 xor(0, 0)
    ----
 =  1000 8

And the result is 8.

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The other way to show this is to use the algebra of XOR; you do not need to know anything about individual bits.

For any numbers x, y, z:

XOR is commutative: x ^ y == y ^ x

XOR is associative: x ^ (y ^ z) == (x ^ y) ^ z

The identity is 0: x ^ 0 == x

Every element is its own inverse: x ^ x == 0

Given this, it is easy to prove the result stated. Consider a sequence:

a ^ b ^ c ^ d ...

Since XOR is commutative and associative, the order does not matter. So sort the elements.

Now any adjacent identical elements x ^ x can be replaced with 0 (self-inverse property). And any 0 can be removed (because it is the identity).

Repeat as long as possible. Any number that appears an even number of times has an integral number of pairs, so they all become 0 and disappear.

Eventually you are left with just one element, which is the one appearing an odd number of times. Every time it appears twice, those two disappear. Eventually you are left with one occurrence.

[update]

Note that this proof only requires certain assumptions about the operation. Specifically, suppose a set S with an operator . has the following properties:

Assocativity: x . (y . z) = (x . y) . z for any x, y, and z in S.

Identity: There exists a single element e such that e . x = x . e = x for all x in S.

Closure: For any x and y in S, x . y is also in S.

Self-inverse: For any x in S, x . x = e

As it turns out, we need not assume commutativity; we can prove it:

(x . y) . (x . y) = e  (by self-inverse)
x . (y . x) . y = e (by associativity)
x . x . (y . x) . y . y = x . e . y  (multiply both sides by x on the left and y on the right)
y . x = x . y  (because x . x = y . y = e and the e's go away)

Now, I said that "you do not need to know anything about individual bits". I was thinking that any group satisfying these properties would be enough, and that such a group need not necessarily be isomorphic to the integers under XOR.

But @Steve Jessup proved me wrong in the comments. If you define scalar multiplication by {0,1} as:

0 * x = 0
1 * x = x

...then this structure satisfies all of the axioms of a vector space over the integers mod 2.

Thus any such structure is isomorphic to a set of vectors of bits under component-wise XOR.

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This is based on the simple fact that XOR of a number with itself results Zero.

and XOR of a number with 0 results the number itself.

So, if we have an array = {5,8,12,5,12}.

5 is occurring 2 times. 8 is occurring 1 times. 12 is occurring 2 times.

We have to find the number occurring odd number of times. Clearly, 8 is the number.

We start with res=0 and XOR with all the elements of the array.

int res=0; for(int i:array) res = res ^ i;

    1st Iteration: res = 0^5 = 5
    2nd Iteration: res = 5^8 
    3rd Iteration: res = 5^8^12
    4th Iteration: res = 5^8^12^5 = 0^8^12 = 8^12
    5th Iteration: res = 8^12^12 = 8^0 = 8

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The bitwise operators treat the bits inside an integer value as a tiny array of bits. Each of those bits is like a tiny bool value. When you use the bitwise exclusive or operator, one interpretation of what the operator does is:

• for each bit in the first value, toggle the bit if the corresponding bit in the second value is set

The net effect is that a single bit starts out false and if the total number of "toggles" is even, it will still be false at the end. If the total number of "toggles" is odd, it will be true at the end.

Just think "tiny array of boolean values" and it will start to make sense.

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The definition of the XOR (exclusive OR) operator, over bits, is that:

0 XOR 0 = 0
0 XOR 1 = 1
1 XOR 0 = 1
1 XOR 1 = 0

One of the ways to imagine it, is to say that the "1" on the right side changes the bit from the left side, and 0 on the right side doesn't change the bit on the left side. However, XOR is commutative, so the same is true if the sides are reversed. As any number can be represented in binary form, any two numbers can be XOR-ed together.

To prove it being commutative, you can simply look at its definition, and see that for every combination of bits on either side, the result is the same if the sides are changed. To prove it being associative, you can simply run through all possible combinations of having 3 bits being XOR-ed to each other, and the result will stay the same no matter what the order is.

Now, as we proved the above, let's see what happens if we XOR the same number at itself. Since the operation works on individual bits, we can test it on just two numbers: 0 and 1.

0 XOR 0 = 0
1 XOR 1 = 0

So, if you XOR a number onto itself, you always get 0 (believe it or not, but that property of XOR has been used by compilers, when a 0 needs to be loaded into a CPU register. It's faster to perform a bit operation than to explicitly push 0 into a register. The compiler will just produce assembly code to XOR a register onto itself).

Now, if X XOR X is 0, and XOR is associative, and you need to find out what number hasn't repeated in a sequence of numbers where all other numbers have been repeated two (or any other odd number of times). If we had the repeating numbers together, they will XOR to 0. Anything that is XOR-ed with 0 will remain itself. So, out of XOR-ing such a sequence, you will end up being left with a number that doesn't repeat (or repeats an even number of times).

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This has a lot of samples of various functionalities done by bit fiddling. Some of can be quite complex so beware.

What you need to do to understand the bit operations is, at least, this:

• the input data, in binary form
• a truth table that tells you how to "mix" the inputs to form the result

For XOR, the truth table is simple:

1^1 = 0
1^0 = 1
0^1 = 1
0^0 = 0

To obtain bit n in the result you apply the rule to bits n in the first and second inputs.

If you try to calculate 1^1^0^1 or any other combination, you will discover that the result is 1 if there is an odd number of 1's and 0 otherwise. You will also discover that any number XOR'ed with itself is 0 and that is doesn't matter in what order you do the calculations, e.g. 1^1^(0^1) = 1^(1^0)^1.

This means that when you XOR all the numbers in your list, the ones which are duplicates (or present an even number of times) will XOR to 0 and you will be left with just the one which is present an odd number of times.

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As it is obvious from the name(bitwise), it operates between bits. Let's see how it works, for example, we have two numbers a=3 and b=4, the binary representation of 3 is 011 and of 4 is 100, so basically xor of the same bits is 0 and for opposite bits, it is 1. In the given example 3^4, where "^" is a xor symbol, will give us 111 whose decimal value will be 7. for another example, if you've given an array in which every element occurs twice except one element & you've to find that element. How can you do that? simple xor of the same numbers will always be 0 and the number which occur exactly once will be your output. because the output of any one number with 0 will be the same name number because the number will have set bits which zero don't have.