问题
How can I implement XOR using basic mathematical operators like +,-,*,/
Update: Actually, I need to track change in two matrix having Boolean values. This can be done using XORing each value with corresponding value in other matrix. But, Lp_Solve library doesn't support XOR operation. Also, it accepts only linear equations.
回答1:
(a − b)²
This works because:
(a − b)² = a * (a − b) + b * (b − a)
Since multiplication in ℤ₂ is conjuction (&), and 1 - a is negation (!), the above formula is equivalent to XOR for a, b ∈ {0, 1}:
(a & !b) | (b & !a)
See the comment below by Pascal Cuoq explaining why this cannot be a linear equation.
回答2:
The simplest expression I can come up with is: a != b.
(Previous best effort was (a + b) == 1)
回答3:
Weellllllllllll........
It's not quite as simple as that.
To model a XOR (let's call it X), we start with the logic.
X = (A & !B) | (!A & B)
In math, the above can be written as:
X = A*(1-B) + B*(1-A)
But the above expression is nonlinear (due to the bilinear terms -- to preserve linearity, we are not allowed to multiply variables with each other).
But! Because we are allowed to use constraints, we can rewrite the above expression in linear form.
First, we expand the terms:
X = A*(1-B) + B*(1-A) = A + B - 2*A*B
Now we need to take care of the A*B term (which essentially means A & B). Let a variable H represent the logical condition A & B. We can now write the AND condition as follows: (see cited reference PDF below)
H <= A
H <= B
H >= A + B - 1
H >= 0
Linear XOR Formulation
Finally, let's put everything together. This is your XOR formulation, using only linear constraints.
X = A + B - 2*H
H <= A
H <= B
H >= A + B - 1
H >= 0
I know it looks complicated (for a simple operation like a XOR). There may be a more compact formulation.
But in general, writing logical conditions in a linear programming context is complicated because one is usually severely restricted in the operations one can perform -- in order to avoid destroying the theoretical properties of the problem.
Reference
See here for a list of standard integer formulations for representing logic linearly. http://brblog.typepad.com/files/mipformref-1.pdf
Edit:
Explanation on how the H constraints model the "AND" logical condition. Essentially, in an LP, we pose inequality constraints that have to be satisfied at the solution point -- what we're doing here is playing a trick to "squeeze" H to the right value. For instance, given the tuple (A,B) = (0,0), the constraints for H would be:
H <= 0
H <= 0
H >= -1
H >= 0
In the above case, the only value H can take is 0, because H belongs in the interval [0,0]. Hence we get (A,B) = (0,0) => H = 0.
Let's try another example, (A,B) = (1,1).
H <= 1
H <= 1
H >= 1
H >= 0
From the above, you will immediately see that 1 <= H <= 1 implies that H = 1. We get (A,B) = (1,1) => H = 1.
And so on. You'll see that the H constraints model the "AND" condition exactly.
回答4:
In Brown, G. and Dell, R., Formulating linear and integer linear programs: A rogues’ gallery the following linear programming formulation for the XOR can be found:
Z3 = Z1 XOR Z2
resolves to
Z3 <= Z1 + Z2
Z3 >= Z1 - Z2
Z3 >= -Z1 + Z2
Z3 <= 2 - Z1 - Z2
回答5:
Can you do something like:
(a + b) % 2
回答6:
2 Factor XOR
While (x-y)² is a great compact equation for 2-factor XOR, this form is misleading in a few ways.
Although the evaluation of these equations is the same for values of 1 and 0, they are not algebraically equal...
(a − b)² ≠ a * (1 − b) + b * (1 − a)
Also, the logical OR operator does not translate arithmetically as + without constraints. That would give you a value of 2 for the AND condition of two 1's. If you first consider translations of NOT and AND..
NOT = (1-x)
AND = x*y
What you really need is something like this...
OR = (1-(1-a)(1-b)) = a + b - ab
Note that characteristically, OR is purely additive whereby you've joined two sets, BUT you don't want to duplicate any overlap of the sets, so you subtract out the AND condition which is found by multiplying. Thus you have your additive term a+b minus your overlap or AND condition a*b. If you are certain that your sets will not overlap, then you can use
OR = a + b, if we know that a*b = 0 for all values of a & b
Likewise, we can derive an equation for XOR. Using the composite logic (a && !b) || (!a && b) you would get...
XOR = 1 - (1 - a(1-b))(1 - b(1-a)) ≠ (a-b)²
So (a-b)² is somewhat misleading in its translation of logic and algebra. As it turns out, these errors get masked by the constraint of binary input and because the conditions a(1-b) and b(1-a) are mutually exclusive, which relaxes the constraint of the OR operator handling the AND condition and allows it to be modeled as +.
Gilead's answer helps explain why (x-y)² actually works. When you expand (x-y)² = x² + y² - 2xy you can see how it satisfies this basic model...
X = A + B - 2*H
H <= A
H <= B
H >= A + B - 1
H >= 0
Using that bit of knowledge, you can see that there are a number of equations that'll work. For instance, the actual most basic equation that satisfies these conditions is...
x + y - 2xy
This is the exact same as the equation we arrived at for OR except now we're not only removing the duplicate of the AND condition (-xy), but rejecting the AND condition all together (-2xy). As it turns out, this is also the actual algebraic equivalent for (a-b)²...
a * (1 − b) + b * (1 − a) = a + b - 2ab ≠ (a-b)².
(a-b)² can be used in place of this because,
(a-b)² = a² + b² - 2ab
and for values of 1 and 0,
a² + b² - 2ab = a + b - 2ab (to me, this is the best model to use in code for two reasons: 1.) It is computationally simpler than using powers 2.) It is easier to see the logic behind it, since a gets added to b followed by full negation (x2) of their overlap a and b)
Beyond 2 Factors
What about when you want to XOR(A,B,C...)? The problem here is that if we try to discern all truth conditions like we did in the composite logic for 2-factor XOR, it doesn't scale very nicely, as you have to add every permutation of truth. However, logic being what it is, we can arrive at XOR the complimentary way...
XOR = !(A & B & C...) & !(!A & !B & !C...)
From which you can construct an arithmetic XOR for any number of factors in the form of...
(1 - A*B*C...)(1 - (1-A)(1-B)(1-C)...)
fun fun... want to test it out? Here's some Excel VBA to XOR an entire range of cells...
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
Getting Fuzzy
It's interesting to stop and ponder for a second that if we use the multi-factor equation above for a 2-factor equation we get the following...
a + b - ab(1 + a + b - ab)
The first thing to note is that this is similar but not equal to the 2-factor equation we derived from truth conditions...
1 - (1 - a(1-b))(1 - b(1-a)) = a + b - ab(3 - a - b + ab)
in fact, the difference is in the following terms...
1 + a + b - ab ≠ 3 - a - b + ab
So what gives? I think this is an arithmetic artifact of using compliments. If you notice, these two terms compliment each other, they're doing the same thing from different directions: one rises from 1 to 2, and the other drops from 3 to 2. Both arrive at 2, but their directions of arrival differ because they're approaching as compliments.
The second thing to note is that both equations are much more complicated than the minimal 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.
Translating 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.
Translating 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%.
The take away here is that, if you relax integer constraints, then there is some meaning to the decimal region. You may want to do this to make discrete problems easier/possible to solve. You'll need to give thought to how values interact in this region and how they'll be converted back.
Any n of k
One last tidbit here. Sometimes you want a condition to be true if any n number of inputs are true. This can be viewed as a relaxed AND condition, whereby you're willing to accept a&b or a&c or b&c for instance. 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)...
That's useful on it's own, but there's also an interesting pattern when you expand the terms. 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.
回答7:
Exclusive-OR is a linear function, but the definition of 'linear' with regards to a boolean function is not the same as with a polynomial function. You will have to look through the documentation for your lp_solve library to see if it is capable of handling linear boolean functions. From what I have read, I don't suspect that it can.
Edit: After looking further into the simplex algorithm that lp_solve uses, I'm fairly certain that you can't do what you are trying to do.
回答8:
abs(A+B-1). if it doesn't do abs, then (A+B-1)*(A+B-1) should do it.
来源：https://stackoverflow.com/questions/2385389/xor-using-mathematical-operators