Sunday, February 21, 2016

xor as linear inequalities

I came across a question about using an xor condition in a MIP model. Here is a summary of the answer.

Expressing \(z=x\cdot y\) (or \(z=x\text{ and }y\) ) where \(x,y,z \in \{0,1\}\) as a set of linear constraints is well-known:

and:     \[\begin{align} &z \le x \\ &z \le y \\ &z \ge x+y-1 \end{align}\] \(\>\>\>\Longleftrightarrow\>\>\> \)
\(z\)\(x\)\(y\)

\(0\)\(0\)\(0\)
\(0\)\(0\)\(1\)
\(0\)\(1\)\(0\)
\(1\)\(1\)\(1\)
The relation \(z=x\text{ xor }y\) is slightly more complicated. The xor (exclusive-or) condition can also be written as \(z = x<>y\), i.e. \(z\) is 1 if \(x\) and \(y\) are different (and \(z\) is zero if they are the same). This we can write as the following set of linear inequalities:
xor:     \[\begin{align} & z \le x+y\\ & z \ge x-y \\ & z \ge y-x \\ & z \le 2 -x - y \end{align}\] \(\>\>\>\Longleftrightarrow\>\>\> \)
\(z\)\(x\)\(y\)

\(0\)\(0\)\(0\)
\(1\)\(0\)\(1\)
\(1\)\(1\)\(0\)
\(0\)\(1\)\(1\)
To be complete \(z=x\text{ or }y\) can be written as:
or:     \[\begin{align} & z \le x+y\\ & z \ge x \\ & z \ge y \end{align}\] \(\>\>\>\Longleftrightarrow\>\>\> \)
\(z\)\(x\)\(y\)

\(0\)\(0\)\(0\)
\(1\)\(0\)\(1\)
\(1\)\(1\)\(0\)
\(1\)\(1\)\(1\)
In all of this we assumed \(x,y\) and \(z\) are binary variables.