I have an integer linear program (ILP) with some variables $x_i$ that are intended to represent boolean values. The $x_i$'s are constrained to be integers and to hold either 0 or 1 ($0 \le x_i \le 1$).

I want to express boolean operations on these 0/1-valued variables, using linear constraints. How can I do this?

More specifically, I want to set $y_1 = x_1 \land x_2$ (boolean AND), $y_2 = x_1 \lor x_2$ (boolean OR), and $y_3 = \neg x_1$ (boolean NOT). I am using the obvious interpretation of 0/1 as Boolean values: 0 = false, 1 = true. How do I write ILP constraints to ensure that the $y_i$'s are related to the $x_i$'s as desired?

(This could be viewed as asking for a reduction from CircuitSAT to ILP, or asking for a way to express SAT as an ILP, but here I want to see an explicit way to encode the logical operations shown above.)


3 Answers 3


Logical AND: Use the linear constraints $y_1 \ge x_1 + x_2 - 1$, $y_1 \le x_1$, $y_1 \le x_2$, $0 \le y_1 \le 1$, where $y_1$ is constrained to be an integer. This enforces the desired relationship. See also https://or.stackexchange.com/q/37/2415.

Logical OR: Use the linear constraints $y_2 \le x_1 + x_2$, $y_2 \ge x_1$, $y_2 \ge x_2$, $0 \le y_2 \le 1$, where $y_2$ is constrained to be an integer.

Logical NOT: Use $y_3 = 1-x_1$.

Logical implication: To express $y_4 = (x_1 \Rightarrow x_2)$ (i.e., $y_4 = \neg x_1 \lor x_2$), we can adapt the construction for logical OR. In particular, use the linear constraints $y_4 \le 1-x_1 + x_2$, $y_4 \ge 1-x_1$, $y_4 \ge x_2$, $0 \le y_4 \le 1$, where $y_4$ is constrained to be an integer.

Forced logical implication: To express that $x_1 \Rightarrow x_2$ must hold, simply use the linear constraint $x_1 \le x_2$ (assuming that $x_1$ and $x_2$ are already constrained to boolean values).

XOR: To express $y_5 = x_1 \oplus x_2$ (the exclusive-or of $x_1$ and $x_2$), use linear inequalities $y_5 \le x_1 + x_2$, $y_5 \ge x_1-x_2$, $y_5 \ge x_2-x_1$, $y_5 \le 2-x_1-x_2$, $0 \le y_5 \le 1$, where $y_5$ is constrained to be an integer.

Another helpful technique for handling complex boolean formulas is to convert them to CNF, then apply the rules above for converting AND, OR, and NOT.

And, as a bonus, one more technique that often helps when formulating problems that contain a mixture of zero-one (boolean) variables and integer variables:

Cast to boolean (version 1): Suppose you have an integer variable $x$, and you want to define $y$ so that $y=1$ if $x \ne 0$ and $y=0$ if $x=0$. If you additionally know that $0 \le x \le U$, then you can use the linear inequalities $0 \le y \le 1$, $y \le x$, $x \le Uy$; however, this only works if you know an upper and lower bound on $x$.

Alternatively, if you know that $|x| \le U$ (that is, $-U \le x \le U$) for some constant $U$, then you can use the method described here. This is only applicable if you know an upper bound on $|x|$.

Cast to boolean (version 2): Let's consider the same goal, but now we don't know an upper bound on $x$. However, assume we do know that $x \ge 0$. Here's how you might be able to express that constraint in a linear system. First, introduce a new integer variable $t$. Add inequalities $0 \le y \le 1$, $y \le x$, $t=x-y$. Then, choose the objective function so that you minimize $t$. This only works if you didn't already have an objective function. If you have $n$ non-negative integer variables $x_1,\dots,x_n$ and you want to cast all of them to booleans, so that $y_i=1$ if $x_i\ge 1$ and $y_i=0$ if $x_i=0$, then you can introduce $n$ variables $t_1,\dots,t_n$ with inequalities $0 \le y_i \le 1$, $y_i \le x_i$, $t_i=x_i-y_i$ and define the objective function to minimize $t_1+\dots + t_n$. Again, this only works nothing else needs to define an objective function (if, apart from the casts to boolean, you were planning to just check the feasibility of the resulting ILP, not try to minimize/maximize some function of the variables).

For some excellent practice problems and worked examples, I recommend Formulating Integer Linear Programs: A Rogues' Gallery.

  • $\begingroup$ which linear programming solver can solve this? becouse in *.lp or *.mps-format one side of the constraint has to be an fixed integer and not an variable. $\endgroup$
    – boxi
    Commented Mar 25, 2015 at 14:31
  • 4
    $\begingroup$ @boxi, I don't know anything about *.lp or *.mps format, but every integer linear programming solver should be able to solve this. Note that if you have something like $x \le y$, this is equivalent to $y -x \ge 0$, which may be in the format you wanted. $\endgroup$
    – D.W.
    Commented Mar 25, 2015 at 19:44
  • $\begingroup$ -i checked it again. lp_solve can solve it, but for example qsopt can't. i don't know why. but thanks <3 $\endgroup$
    – boxi
    Commented Mar 25, 2015 at 19:46
  • 1
    $\begingroup$ @Pramod, good catch! Thank you for spotting that error. You're absolutely right. I've asked a new question about how to model that case and I'll update this answer when we get an answer to that one. $\endgroup$
    – D.W.
    Commented Dec 22, 2015 at 4:04
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    $\begingroup$ @InuyashaYagami, yes, absolutely, that is sufficient for boolean logic, and that would be fine. There might be some advantage to other translations if they introduce fewer temporary variables (for instance, in the case of logical implication), and we still need to know how to express conditional expressions or "cast to boolean", but you have an excellent point. $\endgroup$
    – D.W.
    Commented Dec 28, 2021 at 19:25

The logical AND relation can be modeled in one range constraint instead of three constraints (as in the other solution). So instead of the three constraints $$y_1\geq x_1+x_2−1,\qquad y_1\leq x_1,\qquad y_1\leq x_2\,,$$ it can be written using the single range constraint $$0 \leq x_1 + x_2-2y_1 \leq 1\,.$$ Similarly, for logical OR: $$0 \leq 2y_1 - x_1 - x_2 ≤ 1\,.$$

For NOT, no such improvement is available, because NOT is restricted to one argument.

In general for $y=x_1 \land x_2 \land \dots \land x_n$ ($n$-way AND) the constraint will be: $$0 \leq \sum x_i -ny \leq n-1\,.$$ Similarly for OR: $$0 \leq ny -\sum x_i \leq n-1\,.$$

  • $\begingroup$ A very simillar approach is in this paper: ncbi.nlm.nih.gov/pmc/articles/PMC1865583 $\endgroup$ Commented Jun 24, 2015 at 10:57
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    $\begingroup$ Keep in mind that size isn't everything when looking at integer programming! The downside of the shorter constraints you suggest is that the LP relaxation that is solved as a subproblem for the MILP problem is not strong. You can see that by noting that, e.g., the upper bounding inequality for 'AND' becomes active (i.e., equality holds) for the non integral points (x1, x2, y) = (0, 1, 0.5) or (1, 0, 0.5) or (1, 1, 0.5). $\endgroup$
    – mrclng
    Commented Dec 26, 2020 at 18:38

I found shorter solution for XOR y=x1⊕x2 (x and y are binary 0, 1)

just one line: x1 + x2 - 2*z = y (z is any integer)

  • 1
    $\begingroup$ To express the equality in ILP, you need two inequalities. Further, to avoid solution $x_1=1,x_2=0,z=200,y=-199$, you need two more inequalities, $0\leq y \leq 1$. So this answer has four inequalities and an extra variable compared to six inequalities in D.W.'s answer. $\endgroup$
    – JiK
    Commented Mar 9, 2017 at 15:22
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    $\begingroup$ To express an equality in ILP only needs one equation, this is true in both LP theory and in software such Gurobi or CPLEX . @jIk, I guess you mean "expressing "a $\neq b$" needs two inequalities. " $\endgroup$
    – whitegreen
    Commented Jan 18, 2018 at 3:11

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