Mathematical (Arithmetic) representation of XOR

问题

I have spent the last 5 hours searching for an answer. Even though I have found many answers they have not helped in any way.

What I am basically looking for is a mathematical, arithmetic only representation of the bitwise XOR operator for any 32bit unsigned integers.

Even though this sounds really simple, nobody (at least it seems so) has managed to find an answer to this question.

I hope we can brainstorm, and find a solution together.

Thanks.

回答1:

XOR any numerical input

a + b - ab(1 + a + b - ab)

XOR binary input

a + b - 2ab or (a-b)²

Derivation

Basic Logical Operators

NOT = (1-x)

AND = x*y

From those operators we can get...

OR = (1-(1-a)(1-b)) = a + b - ab

Note: If a and b are mutually exclusive then their and condition will always be zero - from a Venn diagram perspective, this means there is no overlap. In that case, we could write OR = a + b, since a*b = 0 for all values of a & b.

2-Factor XOR

Defining XOR as (a OR B) AND (NOT (a AND b)):

(a OR B) --> (a + b - ab)

(NOT (a AND b)) --> (1 - ab)

AND these conditions together to get...

(a + b - ab)(1 - ab) = a + b - ab(1 + a + b - ab)

Computational Alternatives

If the input values are binary, then powers terms can be ignored to arrive at simplified computationally equivalent forms.

a + b - ab(1 + a + b - ab) = a + b - ab - a²b - ab² + a²b²

If x is binary (either 1 or 0), then we can disregard powers since 1² = 1 and 0² = 0...

a + b - ab - a²b - ab² + a²b² -- remove powers --> a + b - 2ab

XOR (binary) = a + b - 2ab

Binary also allows other equations to be computationally equivalent to the one above. For instance...

Given (a-b)² = a² + b² - 2ab

If input is binary we can ignore powers, so...

a² + b² - 2ab -- remove powers --> a + b - 2ab

Allowing us to write...

XOR (binary) = (a-b)²

Multi-Factor XOR

XOR = (1 - A*B*C...)(1 - (1-A)(1-B)(1-C)...)

Excel VBA example...

Function ArithmeticXOR(R As Range, Optional EvaluateEquation = True)

Dim AndOfNots As String
Dim AndGate As String
For Each c In R
    AndOfNots = AndOfNots & "*(1-" & c.Address & ")"
    AndGate = AndGate & "*" & c.Address
Next
AndOfNots = Mid(AndOfNots, 2)
AndGate = Mid(AndGate, 2)

'Now all we want is (Not(AndGate) AND Not(AndOfNots))
ArithmeticXOR = "(1 - " & AndOfNots & ")*(1 - " & AndGate & ")"
If EvaluateEquation Then
    ArithmeticXOR = Application.Evaluate(xor2)
End If

End Function

Any n of k

These same methods can be extended to allow for any n number out of k conditions to qualify as true.

For instance, out of three variables a, b, and c, if you're willing to accept any two conditions, then you want a&b or a&c or b&c. This can be arithmetically modeled from the composite logic...

(a && b) || (a && c) || (b && c) ...

and applying our translations...

1 - (1-ab)(1-ac)(1-bc)...

This can be extended to any n number out of k conditions. There is a pattern of variable and exponent combinations, but this gets very long; however, you can simplify by ignoring powers for a binary context. The exact pattern is dependent on how n relates to k. For n = k-1, where k is the total number of conditions being tested, the result is as follows:

c1 + c2 + c3 ... ck - n*∏

Where c1 through ck are all n-variable combinations.

For instance, true if 3 of 4 conditions met would be

abc + abe + ace + bce - 3abce

This makes perfect logical sense since what we have is the additive OR of AND conditions minus the overlapping AND condition.

If you begin looking at n = k-2, k-3, etc. The pattern becomes more complicated because we have more overlaps to subtract out. If this is fully extended to the smallest value of n = 1, then we arrive at nothing more than a regular OR condition.

Thinking about Non-Binary Values and Fuzzy Region

The actual algebraic XOR equation a + b - ab(1 + a + b - ab) is much more complicated than the computationally equivalent binary equations like x + y - 2xy and (x-y)². Does this mean anything, and is there any value to this added complexity?

Obviously, for this to matter, you'd have to care about the decimal values outside of the discrete points (0,0), (0,1), (1,0), and (1,1). Why would this ever matter? Sometimes you want to relax the integer constraint for a discrete problem. In that case, you have to look at the premises used to convert logical operators to equations.

When it comes to translating Boolean logic into arithmetic, your basic building blocks are the AND and NOT operators, with which you can build both OR and XOR.

OR = (1-(1-a)(1-b)(1-c)...)

XOR = (1 - a*b*c...)(1 - (1-a)(1-b)(1-c)...)

So if you're thinking about the decimal region, then it's worth thinking about how we defined these operators and how they behave in that region.

Non-Binary Meaning of NOT

We expressed NOT as 1-x. Obviously, this simple equation works for binary values of 0 and 1, but the thing that's really cool about it is that it also provides the fractional or percent-wise compliment for values between 0 to 1. This is useful since NOT is also known as the Compliment in Boolean logic, and when it comes to sets, NOT refers to everything outside of the current set.

Non-Binary Meaning of AND

We expressed AND as x*y. Once again, obviously it works for 0 and 1, but its effect is a little more arbitrary for values between 0 to 1 where multiplication results in partial truths (decimal values) diminishing each other. It's possible to imagine that you would want to model truth as being averaged or accumulative in this region. For instance, if two conditions are hypothetically half true, is the AND condition only a quarter true (0.5 * 0.5), or is it entirely true (0.5 + 0.5 = 1), or does it remain half true ((0.5 + 0.5) / 2)? As it turns out, the quarter truth is actually true for conditions that are entirely discrete and the partial truth represents probability. For instance, will you flip tails (binary condition, 50% probability) both now AND again a second time? Answer is 0.5 * 0.5 = 0.25, or 25% true. Accumulation doesn't really make sense because it's basically modeling an OR condition (remember OR can be modeled by + when the AND condition is not present, so summation is characteristically OR). The average makes sense if you're looking at agreement and measurements, but it's really modeling a hybrid of AND and OR. For instance, ask 2 people to say on a scale of 1 to 10 how much do they agree with the statement "It is cold outside"? If they both say 5, then the truth of the statement "It is cold outside" is 50%.

Non-Binary Values in Summary

The take away from this look at non-binary values is that we can capture actual logic in our choice of operators and construct equations from the ground up, but we have to keep in mind numerical behavior. We are used to thinking about logic as discrete (binary) and computer processing as discrete, but non-binary logic is becoming more and more common and can help make problems that are difficult with discrete logic easier/possible to solve. You'll need to give thought to how values interact in this region and how to translate them into something meaningful.

回答2:

"mathematical, arithmetic only representation" are not correct terms anyway. What you are looking for is a function which goes from IxI to I (domain of integer numbers).
Which restrictions would you like to have on this function? Only linear algebra? (+ , - , * , /) then it's impossible to emulate the XOR operator.
If instead you accept some non-linear operators like Max() Sgn() etc, you can emulate the XOR operator with some "simpler" operators.

回答3:

Given that (a-b)(a-b) quite obviously computes xor for a single bit, you could construct a function with the floor or mod arithmetic operators to split the bits out, then xor them, then sum to recombine. (a-b)(a-b) = a² -2·a·b + b² so one bit of xor gives a polynomial with 3 terms.

Without floor or mod, the different bits interfere with each other, so you're stuck with looking at a solution which is a polynomial interpolation treating the input a,b as a single value: a xor b = g(a · 2³² + b)

The polynomial has 2⁶⁴-1 terms, though will be symmetric in a and b as xor is commutative so you only have to calculate half of the coefficients. I don't have the space to write it out for you.

回答4:

I wasn't able to find any solution for 32-bit unsigned integers but I've found some solutions for 2-bit integers which I was trying to use in my Prolog program. One of my solutions (which uses exponentiation and modulo) is described in this StackOverflow question and the others (some without exponentiation, pure algebra) can be found in this code repository on Github: see different xor0 and o_xor0 implementations.

The nicest xor represention for 2-bit uints seems to be: xor(A,B) = (A + B*((-1)^A)) mod 4.

Solution with +,-,*,/ expressed as Excel formula (where cells from A2 to A5 and cells from B1 to E1 contain numbers 0-4) to be inserted in cells from A2 to E5:

(1-$A2)*(2-$A2)*(3-$A2)*($A2+B$1)/6 - $A2*(1-$A2)*(3-$A2)*($A2+B$1)/2 + $A2*(1-$A2)*(2-$A2)*($A2-B$1)/6 + $A2*(2-$A2)*(3-$A2)*($A2-B$1)/2 - B$1*(1-B$1)*(3-B$1)*$A2*(3-$A2)*(6-4*$A2)/2 + B$1*(1-B$1)*(2-B$1)*$A2*($A2-3)*(6-4*$A2)/6

回答5:

It may be possible to adapt and optimize this solution for 32-bit unsigned integers. It's complicated and it uses logarithms but seems to be the most universal one as it can be used on any integer number. Additionaly, you'll have to check if it really works for all number combinations.

回答6:

I do realize that this is sort of an old topic, but the question is worth answering and yes, this is possible using an algorithm. And rather than go into great detail about how it works, I'll just demonstrate with a simple example (written in C):

    #include <stdio.h>
    #include <stdlib.h>
    #include <math.h>   
    #include <time.h>

    typedef unsigned long
            number;

    number XOR(number a, number b)
    {
            number
                    result = 0,
            /*
                    The following calculation just gives us the highest power of 
                    two (and thus the most significant bit) for this data type.
            */          
                    power = pow(2, (sizeof(number) * 8) - 1);
    /*
            Loop until no more bits are left to test...
    */                
            while(power != 0)
            {
                    result *= 2;
            /*
                    The != comparison works just like the XOR operation.
            */                
                    if((power > a) != (power > b))
                            result += 1;
                    a %= power;
                    b %= power;
                    power /= 2;
            }
            return result;
    }

    int main()
    {
            srand(time(0));
            for(;;)
            {
                    number
                            a = rand(), 
                            b = rand();
                    printf("a = %lu\n", a);
                    printf("b = %lu\n", b);
                    printf("a ^ b     = %lu\n", a ^ b);
                    printf("XOR(a, b) = %lu\n", XOR(a, b));
                    getchar();
            }
    }