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My question is about how usual/strange defining constraints compactly can be (in a scientific document). Formally, I define constraints $$f_{i,j,k}(x) \leq 0, \quad \forall i \in [I], \ \forall j \in [I], \ \forall k \in K(i,j)​$$ where $K(i,j) := \{ k \ : \ A_k \cap (A_i \cap A_j) \neq \emptyset \}$. However, for some combinations of $(i,j)$, we have $A_{i} \cap A_j = \emptyset$ so we have $K(i,j) = \emptyset$. I have two options:

  1. Define the constraints for all $i \in [I]$ and for all $j,k$ in some (slightly more) complicated set parametrized by $i$, or,
  2. Just note that if we have some constraints where $k$ is not defined, we ignore these constraints (i.e., these constraints are not defined).

I was wondering, can I "get away" with the second option? I do not want to introduce a set for $(j,k)$, rather keep it as it is, and just ignore the constraints for $i,j$ where $K(i,j) = \emptyset$.

End-notes: $f$ is an arbitrary function, $x$ is a decision variable, $I$ is an abitrary natural number, e.g., $I = 5$, and $|I|$ denotes $1,\ldots, I$. The sets $A_i$ for $i \in [I]$ are not important in this context.

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  • $\begingroup$ Why not define a sparse set of triples $(i,j,k)$ and use that instead? $\endgroup$
    – RobPratt
    Nov 15 '21 at 21:25
  • $\begingroup$ @RobPratt thanks for your answer. What do you mean by sparse set? It sounds like it is point 1. above which I am trying to avoid. $\endgroup$ Nov 15 '21 at 21:37
  • $\begingroup$ The answer also depends on the modeling tool being used (not all tools allow empty constraints, some have great support for sparse sets). In general, I prefer to make things as explicit as possible (not in the least to document that this is a sparse structure) and to generate small models instead of relying only on the presolver. $\endgroup$ Nov 16 '21 at 0:50
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The constraint is defined only for triples $(i,j,k)$ satisfying the domain conditions, including the requirement that $k\in K(i,j)$. So I think it is fine to just point out that $K(i,j)$ can be empty and remind the reader that $K(i,j)=\emptyset$ implies there are no instances of this constraint for that combination of $i$ and $j$. In other words, I think your option 2 is perfectly acceptable.

Be warned, though, that not all referees are reasonable people. :-)

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  • $\begingroup$ Haha, exactly the answer that I was looking for :) Thank you, and your experience! $\endgroup$ Nov 15 '21 at 22:58
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One approach is to define a sparse set $T$ of triples: $$T=\{i\in [I], j \in [I], k\in [I]: A_i \cap A_j \cap A_k \not= \emptyset\}$$ or $$T=\{(i,j,k)\in [I]^3: A_i \cap A_j \cap A_k \not= \emptyset\}.$$ Then write the constraint as $$f_{i,j,k}(x) \le 0 \quad \text{for $(i,j,k)\in T$}$$

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