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Normally to show that an optimization problem is hard, we show the corresponding decision version of the problem is hard. However, is this sufficient to support the conclusion? Does there exist any optimization problem which is easy but its decision version is hard?

user2477759
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No. The optimization problem is "How big is the biggest $X$?" and the decision problem is "Is there an $X$ that is bigger than $y$?" Solving the decision problem simply involves comparing $y$ with the size of the biggest $X$. You can certainly compare two numbers in polynomial time so, if you can solve the optimization problem (compute the size of the biggest $X$) in polynomial time, you can solve the decision problem.

David Richerby
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    I'm thinking about the case, where we are asked to solve $max_{x \in X} f(x)$, such that $f(x)$ is a constant function but its value is hard to compute (for example, $f(x) = max_{y \in Y} g(y)$). Then any $x \in X$ is optimal, but we don't know if any $x$ makes $f(x) > k$ for a given $k$. But this is perhaps more about the definition of what it means to "solve an optimization problem"... – user2477759 May 28 '18 at 16:07
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Maybe it depends on what it means by solving an optimization problem. If it is to find "how big is the biggest $f(x)$", then the answer is no (see the answer of @David Richerby). If it is to find "the $x$ that maximizes $f(x)$", then consider the function $f(x) = \max_{y\in Y}g(y)$ such that $\max_{y\in Y}g(y)$ is hard to compute. The optimization problem is easy to solve since $f(x)$ is constant w.r.t. $x$, but the decision problem — decide whether there exists an $x$ such that $f(x)>k$ — is hard to solve.

xskxzr
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user2477759
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No, because you are supposed to choose the decision version so that any solution to the optimization problem can be used to solve the decision problem (with at most a polynomial increase in running time).

You can't choose just any unrelated decision problem; for this methodology to work, there has to be a correspondence between the optimization problem and the decision problem. The correspondence is that you need to make sure that any algorithm A for the optimization problem can be used to solve the decision problem, with at most polynomially many invocations of A.

D.W.
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No... if the optimal solution is $1101$ in binary, then that inherently already represents the solution to the four corresponding decision problems.

So if finding $1101$ is easy, that means computing the respective $1$, $1$, $0$, and $1$ was easy too.

user541686
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  • The decision version of an optimisation problem is “Is the optimum at least/most $x$?” not “Is the $k$th bit of the optimum $1$?” The latter can be a much easier problem, especially given that “Is the last bit of the optimum $1$?” Is equivalent to “Is the optimum an odd number?” – David Richerby May 07 '18 at 07:38
  • @DavidRicherby: Where did I claim the corresponding decision problem is "Is the k'th bit of the optimum 1?"? – user541686 May 07 '18 at 07:57
  • Your second paragraph talks about computing individual bits of the optimum. That doesn’t seem relevant unless “the decision problem” is to compute one of those bits. – David Richerby May 07 '18 at 08:01
  • @DavidRicherby: Sure it does? First question could be "is the optimum at least 1000 [yes]", second one "is the optimum at least 1100 [yes]", third one "is the optimum at least 1110 [no]", fourth one "is the optimum at least 1101 [yes]". – user541686 May 07 '18 at 08:04
  • @DavidRicherby: Would appreciate it if you could take off your downvote now that your confusion is cleared up... – user541686 May 07 '18 at 08:15
  • I don't think the confusion has been cleared up. Unless you're interested in the bits themselves, there is literally no difference between computing a number and computing the bits of that number. For example, the value of the least significant bit "almost never" affects whether the optimum is achieved or not, so it may not even be necessary to compute every bit to solve the optimization problem. – David Richerby May 08 '18 at 09:49
  • And, by the way, the downvote that my answer to this question received yesterday is a total coincidence, right? – David Richerby May 08 '18 at 09:53
  • @DavidRicherby: "Unless you're interested in the bits themselves, there is literally no difference between computing a number and computing the bits of that number. For example, the value of the least significant bit 'almost never' affects whether the optimum is achieved or not, so it may not even be necessary to compute every bit to solve the optimization problem." Uh...huh? Literally in your own answer you say "The optimization problem is* 'How big is the biggest X?'"* and now you're objecting "just kidding!!! my answer is wrong, you usually don't really want the biggest X"? – user541686 May 08 '18 at 14:39
  • Sorry, typo -- I meant that it may not even be necessary to compute every bit of the optimal value to solve the decision problem. Is it really necessary to yell at me in this way for making a simple, easily fixed mistake? – David Richerby May 08 '18 at 14:45
  • @DavidRicherby: I'm not yelling at you (?) I'm just bolding to point out the specific words. Oh, so you meant the decision problem? Exactly where in my answer did I suggest it is "necessary to compute every bit of the optimal value to solve the decision problem"? – user541686 May 08 '18 at 14:52
  • OK. Posting a comment that's almost entirely italic and boldface looks a lot like yelling. But, anyway, let's get back to the point. Your answer says that afour-bit answer to an optimization problem "inherently already represents the solution to the four corresponding decision problems." What do you mean by these four problems, if not "compute the first bit", "compute the second bit", "compute the third bit" and "compute the fourth bit"? None of those is "the decision problem associated with the optimization problem." The decision problem is "Is the optimum at least $x$?" – David Richerby May 08 '18 at 15:18
  • @DavidRicherby: I already explained what the 4 problems are very explicitly here. – user541686 May 08 '18 at 15:20
  • OK. None of those is the decision problem associated with the optimization problem. I'm done here. My downvote stands. – David Richerby May 08 '18 at 15:31