What is the usage of a decision problem for an optimization problem like the longest path problem?

Question

I just read this definition for the longest path problem:

LONGEST PATH
Input: A graph $G=(V,E)$, an integer $k$.
Question: Is there a path with at least $k$ vertices in $G$

This seems a decision problem and a certificate can be verified easily. But, what is the usage of such a problem? Isn't the main concern finding the longest path in the graph? I guess that would be an optimization problem, and verifying a given certificate can't be done in polynomial time, so that is not $NP$. right?

If yes, then what is the usage of the decision problem (which is $NP$) and what is its relation to the main problem?

My question is different from "NP-complete" optimization problems because I also asked about the relationship between a decision problem and a relevant optimization problem, and how a decision problem may help us in solving the optimization problem, and basically what is their usage in such cases.

Answer 1

In complexity theory, we tend to focus on decision problems. They're conceptually simpler and the corresponding function and optimization problems can be reduced to them with only a polynomial loss of efficiency. From a theoretical point of view, polynomial factors are considered "efficient".

Let's take your example of Longest Path. We know that the longest path has length between 1 and $n$ (where $n$ is the number of vertices), so we can find the actual length of the longest path by doing binary search using a series of queries of the form "Is there a path of length at least $n/2$?", "... $3n/4$?", ... That tells us that the longest path has some length $\ell$. Now, we can loop through the edges one by one. Delete the $i$th edge and ask, "Is there still a path of length at least $\ell$?" If there is, go on to the next edge; if there isn't, undelete edge $i$ and go on to the next one. When this algorithm terminates, we'll be left with a graph that is exactly a path of length $\ell$. We used $\Theta(\log n$) calls to Longest Path to find the length of the longest path, and $O(n^2)$ calls to find the actual path.

The decision problem isn't necessarily "useful" per se: as you observe, the function problem ("Give me a solution", rather than "Is there a solution?") is the one we usually want to solve, practically speaking.