How to represent XOR of two decimal Numbers with Arithmetic Operators

This is a summary of a similar longer answer I gave on StackOverflow...

Basic Logical Operators

NOT = (1-x)

AND = x*y

From those operators we can get...

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

OR = a + b, if we know that a*b = 0 for all values of a & b

2-Factor XOR

Deriving from set of all truth conditions...

XOR = 1 - (1 - a(1-b))(1 - b(1-a)) = a + b - ab(3 - a - b + ab)

Deriving from compliment of truth conditions...

XOR = (1 - abc)(1 - (1-a)(1-b)(1-c)) = a + b - ab(1 + a + b - ab)

Because we can write (a & !b) || (!a & b) with mutually exclusive terms we can simplify the OR translation as simply + and get...

XOR = a + b - 2ab

For binary values we can condense this expression to

XOR = (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.

I think what Sanisetty Pavan means is that he has two non-negative integers $a$ and $b$ which we assume to be in the range $0 \leq a, b < 2^{n+1}$ and thus representable as $(n+1)$-bit vectors $(a_n, \cdots, a_0)$ and $(b_n, \cdots, b_0)$ where $$ a = \sum_{i=0}^n a_i 2^i, ~~ b = \sum_{i=0}^n b_i 2^i. $$ He wants an an arithmetic expression for the integer $c$ where $$c = \sum_{i=0}^n (a_i \oplus b_i) 2^i = \sum_{i=0}^n (a_i + b_i -2 a_ib_i) 2^i = a + b - 2 \sum_{i=0}^n a_ib_i 2^i$$ in terms of $a$ and $b$ and the arithmetic operators $+, -, *, /, \%$. Presumably integer constants are allowed in the expression. The expression for $c$ above shows a little progress but I don't think it is much easier to express $\sum_{i=0}^n a_ib_i 2^i$ than it is to express $\sum_{i=0}^n (a_i \oplus b_i) 2^i$ in terms of $a$ and $b$, but perhaps Listing's gigantic formula might be a tad easier to write out, though Henning Makholm's objections will still apply.

Added note: For fixed $n$, we can express $c$ as $c = a + b - 2f(a,b)$ where $f(a, b)$ is specified recursively as $$f(a, b) = (a\%2)*(b\%2) + 2f(a/2, b/2)$$ with $a\%2$ meaning the remainder when integer $a$ is divided by $2$ (that is, $a \bmod 2$) and $a/2$ meaning "integer division" which gives the integer quotient (that is, $a/2 = (a - (a\%2))/2$). Working out the recursion gives a formula with $n+1$ terms for $f(a, b)$.