I'm new to Gurobi in Python and I am wondering if there is way to express/code "or" in the following constraint, where $x_i$ are binary variables:

$x_i-x_i*x_{i-1} =0$


$x_i*x_{i+1} =1.$


Is there a way to express "Or" between the above two constraints?

My Approch:

I tried the following model.addConstr((x[i]-x[i]*x[i-1]) * (x[i]*x[i+1] -1 )==0) however this leads to an error:

gurobipy.GurobiError: Invalid argument to QuadExpr multiplication

I think that addConstr can take at max three multiplications.

  • $\begingroup$ 1) Take a look at these two links from gurobi: or_ function and indicator constraints. 2) There are other questions on this site about formulating "OR" constraints. Make sure to search for it $\endgroup$
    – EhsanK
    Nov 14, 2022 at 14:06
  • $\begingroup$ @EhsanK In fact, or_function doesn't work in this case, indicator constraints could work, but I do not see how. I already searched for the other questions about "OR" constraints, didn't find something suitable. $\endgroup$
    – M.Badaoui
    Nov 14, 2022 at 14:27
  • 4
    $\begingroup$ Hi @M.Badaoui May you could expand a bit on what you want to achieve. Maybe there is a better way to formulate the logic you want to enforce? The first constraint says that "if $x_i=1$ then $x_{i-1}$ must equal 1 as well" the second say "both $x_i$ and $x_{i+1}$ must be one". Note that you can formulate both of these constraints linearly: $x_i\leq x_{i-1}$ and $x_i+x_{i+1}=2$ $\endgroup$
    – Sune
    Nov 14, 2022 at 14:43
  • 1
    $\begingroup$ Boolean type constraints are more familiar to Constraint Programming type of problems. It is almost always possible to write them in MIP with a little more difficulty. OR-Tools has boolean operators for its CP-SAT module developers.google.com/optimization/reference/python/sat/python/… Btw, it seems what you are trying is nonlinear, which may be another problem depending on the solver type. $\endgroup$
    – berkorbay
    Nov 15, 2022 at 7:15
  • 1
    $\begingroup$ @M.Badaoui I might have missed something but I think I have an answer for you for your specific case. $\endgroup$
    – berkorbay
    Nov 15, 2022 at 11:37

3 Answers 3


Since $x_i$'s are binary: $$ \begin{align*} &(x_i(1 - x_{i-1}) = 0) \vee (x_ix_{i+1}=1)\\ \Leftrightarrow&(\neg x_i \vee x_{i-1})\vee(x_i \wedge x_{i+1})\\ \Leftrightarrow&\neg x_i \vee x_{i-1}\vee(x_i \wedge x_{i+1})\\ \Leftrightarrow&(\neg x_i \vee x_{i-1}\vee x_i) \wedge (\neg x_i \vee x_{i-1}\vee x_{i+1})\\ \Leftrightarrow& 1 \wedge (\neg x_i \vee x_{i-1} \vee x_{i+1})\\ \Leftrightarrow& \neg x_i \vee x_{i-1} \vee x_{i+1}\\ \Leftrightarrow& (1-x_i) + x_{i-1} + x_{i+1} \geq 1\\ \Leftrightarrow& x_{i-1} - x_i + x_{i+1} \geq 0 \end{align*} $$

  • $\begingroup$ Thank you! Using the last inequality we can formulate the constraint in gurobi. I was hopping for something similar to CPLEX in this question (to express OR constraint directly): stackoverflow.com/questions/56710025/… Seems that there no direct way in Gurobi like Cplex. $\endgroup$
    – M.Badaoui
    Nov 15, 2022 at 10:54
  • $\begingroup$ See addGenConstrOr and addGenConstrIndicator @M.Badaoui $\endgroup$
    – xd y
    Nov 16, 2022 at 4:03

Edit: This is actually the "English" version of @xd y's accepted answer.

In your specific case I think adding just $x_{i+1} \ge x_i-x_{i-1}$ works. Let's enumerate

If $x_i = 0$ then your OR constraint is satisfied since LHS will be 0 for both constraints. Other variables are free.

If $x_i = 1$ and $x_{i+1} = 1$ then the second constraint is satisfied. $x_{i-1}$ is free.

If $x_i = 1$ and $x_{i+1} = 0$ then $x_{i-1}$ should be 1.

So if you add $x_{i+1} \ge x_i-x_{i-1}$, you ensure that if $x_{i} = 1$ and $x_{i-1} = 0$ you will get $x_{i+1} = 1$. Otherwise, $x_{i+1}$ is free.


$x_i*(1-x_{i-1} = 0\ or \ x_i*x_{i+1}=1$ implies
Cons(1): $x_i*(1-x_{i-1}) >=0$;
Cons(2): $x_i*x_{i+1} <=1$;
Cons(3): $x_i*(1-x_{i-1} + x_{i+1}) =1$
So if you are looking for 0 or 1 for cons(1) & cons(2) & not either way then constraints are:
$x_i = 1$
$x_{i+1} >= x_{i-1}$


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