# Discrete problems versus Combinatorial problems

I have seen one explanation about the difference between discrete and combinatorial optimization problems in such a way that "all combinatorial optimization problems are also discrete optimization problems but the converse is not true" [1].

I am not aware of the existence of such a problem. Also, looking at the literature, from vehicle routing to the scheduling problems, all of them are discrete (of course, some of them are mixed-integer). Could you enlighten me about how it is possible for a problem that is discrete, but not combinatorial?

• From the paper you mention, the difference is that a combinatorial problem has a finite set of solutions, while a discrete problem might have a infinite number of solutions Oct 25, 2021 at 7:57
• Could you extend your explanation "infinite number of solutions"? If a problem has constraints and boundaries, the problem will have a finite set of solutions. Otherwise, I don't know whether there is a method in the literature to solve this type of problem.
– YcK
Oct 25, 2021 at 8:07
• Yes, if all variables are integer and bounded, then the number of solutions is finite. In practice, it is often the case. It is still possible to apply a branch-and-bound algorithm with unbounded variables, but there's no guarantee it will terminate Oct 25, 2021 at 11:35
• Thank you for your answer. So do you know any problem in the literature that refers to "unbounded" types?
– YcK
Oct 25, 2021 at 11:38
• @A.Omidi I'm not sure that I understand what you mean. Why wouldn't they be COP if it is possible to enumerate all their solutions? Then, I'm not sure it's so important to determine if a problem has the tag COP or not. And I think that it is easy to find border cases where the problem has continuous decisions, but easily deduced from the discrete decisions, so which could or couldn't be considered as part of the decisions Oct 26, 2021 at 7:53