Why is XOR the default way to combine hashes?

Say you have two hashes H(A) and H(B) and you want to combine them. I've read that a good way to combine two hashes is to XOR them, e.g. XOR( H(A), H(B) ).

The best explanation I've found is touched briefly here on these hash function guidelines:

XORing two numbers with roughly random distribution results in another number still with roughly random distribution*, but which now depends on the two values.

...

* At each bit of the two numbers to combine, a 0 is output if the two bits are equal, else a 1. In other words, in 50% of the combinations, a 1 will be output. So if the two input bits each have a开发者_如何学C roughly 50-50 chance of being 0 or 1, then so too will the output bit.

Can you explain the intuition and/or mathematics behind why XOR should be the default operation for combining hash functions (rather than OR or AND etc.)?

xor is a dangerous default function to use when hashing. It is better than and and or, but that doesn't say much.

xor is symmetric, so the order of the elements is lost. So "bad" will hash combine the same as "dab".

xor maps pairwise identical values to zero, and you should avoid mapping "common" values to zero:

So (a,a) gets mapped to 0, and (b,b) also gets mapped to 0. As such pairs are almost always more common than randomness might imply, you end up with far to many collisions at zero than you should.

With these two problems, xor ends up being a hash combiner that looks half decent on the surface, but not after further inspection.

On modern hardware, adding usually about as fast as xor (it probably uses more power to pull this off, admittedly). Adding's truth table is similar to xor on the bit in question, but it also sends a bit to the next bit over when both values are 1. This means it erases less information.

So hash(a) + hash(b) is better than hash(a) xor hash(b) in that if a==b, the result is hash(a)<<1 instead of 0.

This remains symmetric; so the "bad" and "dab" getting the same result remains a problem. We can break this symmetry for a modest cost:

hash(a)<<1 + hash(a) + hash(b)

aka hash(a)*3 + hash(b). (calculating hash(a) once and storing is advised if you use the shift solution). Any odd constant instead of 3 will bijectively map a "k-bit" unsigned integer to itself, as map on unsigned integers is math modulo 2^k for some k, and any odd constant is relatively prime to 2^k.

For an even fancier version, we can examine boost::hash_combine, which is effectively:

size_t hash_combine( size_t lhs, size_t rhs ) {
  lhs ^= rhs + 0x9e3779b9 + (lhs << 6) + (lhs >> 2);
  return lhs;
}

here we add together some shifted versions of lhs with a constant (which is basically random 0s and 1s – in particular it is the inverse of the golden ratio as a 32 bit fixed point fraction) with some addition and an xor. This breaks symmetry, and introduces some "noise" if the incoming hashed values are poor (ie, imagine every component hashes to 0 – the above handles it well, generating a smear of 1 and 0s after each combine. My naive 3*hash(a)+hash(b) simply outputs a 0 in that case).

Extending this to 64 bits (using the expansion of pi as our constant for 64 bits, as it is odd at 64 bits):

size_t hash_combine( size_t lhs, size_t rhs ) {
  if constexpr (sizeof(size_t) >= 8) {
    lhs ^= rhs + 0x517cc1b727220a95 + (lhs << 6) + (lhs >> 2);
  } else {
    lhs ^= rhs + 0x9e3779b9 + (lhs << 6) + (lhs >> 2);
  }
  return lhs;
}

(For those not familiar with C/C++, a size_t is an unsigned integer value which is big enough to describe the size of any object in memory. On a 64 bit system, it is usually a 64 bit unsigned integer. On a 32 bit system, a 32 bit unsigned integer.)

Assuming uniformly random (1-bit) inputs, the AND function output probability distribution is 75% 0 and 25% 1. Conversely, OR is 25% 0 and 75% 1.

The XOR function is 50% 0 and 50% 1, therefore it is good for combining uniform probability distributions.

This can be seen by writing out truth tables:

 a | b | a AND b
---+---+--------
 0 | 0 |    0
 0 | 1 |    0
 1 | 0 |    0
 1 | 1 |    1

 a | b | a OR b
---+---+--------
 0 | 0 |    0
 0 | 1 |    1
 1 | 0 |    1
 1 | 1 |    1

 a | b | a XOR b
---+---+--------
 0 | 0 |    0
 0 | 1 |    1
 1 | 0 |    1
 1 | 1 |    0

Exercise: How many logical functions of two 1-bit inputs a and b have this uniform output distribution? Why is XOR the most suitable for the purpose stated in your question?

In spite of its handy bit-mixing properties, XOR is not a good way to combine hashes due to its commutativity. Consider what would happen if you stored the permutations of {1, 2, …, 10} in a hash table of 10-tuples.

A much better choice is m * H(A) + H(B), where m is a large odd number.

Credit: The above combiner was a tip from Bob Jenkins.

Xor may be the "default" way to combine hashes but Greg Hewgill's answer also shows why it has its pitfalls: The xor of two identical hash values is zero. In real life, there are identical hashes are more common than one might have expected. You might then find that in these (not so infrequent) corner cases, the resulting combined hashes are always the same (zero). Hash collisions would be much, much more frequent than you expect.

In a contrived example, you might be combining hashed passwords of users from different websites you manage. Unfortunately, a large number of users reuse their passwords, and a surprising proportion of the resulting hashes are zero!

There's something I want to explicitly point out for others who find this page. AND and OR restrict output like BlueRaja - Danny Pflughoe is trying to point out, but can be better defined:

First I want to define two simple functions I'll use to explain this: Min() and Max().

Min(A, B) will return the value that is smaller between A and B, for example: Min(1, 5) returns 1.

Max(A, B) will return the value that is larger between A and B, for example: Max(1, 5) returns 5.

If you are given: C = A AND B

Then you can find that C <= Min(A, B) We know this because there is nothing you can AND with the 0 bits of A or B to make them 1s. So every zero bit stays a zero bit and every one bit has a chance to become a zero bit (and thus a smaller value).

With: C = A OR B

The opposite is true: C >= Max(A, B) With this, we see the corollary to the AND function. Any bit that is already a one cannot be ORed into being a zero, so it stays a one, but every zero bit has a chance to become a one, and thus a larger number.

This implies that the state of the input applies restrictions on the output. If you AND anything with 90, you know the output will be equal to or less than 90 regardless what the other value is.

For XOR, there is no implied restriction based on the inputs. There are special cases where you can find that if you XOR a byte with 255 than you get the inverse but any possible byte can be output from that. Every bit has a chance to change state depending on the same bit in the other operand.

If you XOR a random input with a biased input, the output is random. The same is not true for AND or OR. Example:

00101001 XOR 00000000 = 00101001
00101001 AND 00000000 = 00000000
00101001 OR  11111111 = 11111111

As @Greg Hewgill mentions, even if both inputs are random, using AND or OR will result in biased output.

The reason we use XOR over something more complex is that, well, there's no need: XOR works perfectly, and it's blazingly stupid-fast.

Cover the left 2 columns and try to work out what the inputs are using just the output.

 a | b | a AND b
---+---+--------
 0 | 0 |    0
 0 | 1 |    0
 1 | 0 |    0
 1 | 1 |    1

When you saw a 1-bit you should have worked out that both inputs were 1.

Now do the same for XOR

 a | b | a XOR b
---+---+--------
 0 | 0 |    0
 0 | 1 |    1
 1 | 0 |    1
 1 | 1 |    0

XOR gives away nothing about it inputs.

XOR does not ignore some of the inputs sometimes like OR and AND.

If you take AND(X, Y) for example, and feed input X with false, then the input Y does not matter...and one probably would want the input to matter when combining hashes.

If you take XOR(X, Y) then BOTH inputs ALWAYS matter. There would be no value of X where Y does not matter. If either X or Y is changed then the output will reflect that.

The source code for various versions of hashCode() in java.util.Arrays is a great reference for solid, general use hashing algorithms. They are easily understood and translated into other programming languages.

Roughly speaking, most multi-attribute hashCode() implementations follow this pattern:

public static int hashCode(Object a[]) {
    if (a == null)
        return 0;

    int result = 1;

    for (Object element : a)
        result = 31 * result + (element == null ? 0 : element.hashCode());

    return result;
}

You can search other StackOverflow Q&As for more information about the magic behind 31, and why Java code uses it so frequently. It is imperfect, but has very good general performance characteristics.