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The optimization problems which I am familiar with (e.g. the Traveling Salesman Problem) are such that approximate solutions to these problems are still quite useful.

I'm wondering, however, if there are real-world optimization problems for which approximate solutions are useless? Or, to put it another way, are there any real-world optimization problems for which the only solution that is beter/more useful than any other is the optimal solution, and all others are equally bad/useless?

user541686
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  • Wouldn't this be more at home at cs.stackexchange.com? – tomasz Feb 24 '15 at 00:51
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    Optimization is one of the most active areas of applied math, so I think this question fits well here. – littleO Feb 24 '15 at 00:52
  • Approximately optimal, approximately feasible, or both? – Johan Löfberg Feb 24 '15 at 08:24
  • @JohanLöfberg: I meant problems for which approximately optimal solutions are useless, but I'm slightly confused why you made "both" a separate option, since approximately optimal implies approximately feasible as well. – user541686 Feb 24 '15 at 08:29
  • I am talking about optimality vs feasibility. Example: find smallest integer $x$ larger than $0$. A sub-optimal but feasible solution is $x=1$. A sub-optimal but infeasible solution is $x=0.01$. Depending on the context, they have different merits and use. – Johan Löfberg Feb 24 '15 at 08:36
  • @JohanLöfberg: I can't make sense of your example, isn't x = 1 the optimal solution? Anyway, seems like your definition of "approximately" changes with whether you're talking about feasibility or optimality. I guess I'm looking for something whose approximately optimal solution and approximately feasible solution are both useless, i.e. only the optimal solution is any more useful than any other solution. – user541686 Feb 24 '15 at 08:47
  • Sorry, should have been larger than or equal to $0$. – Johan Löfberg Feb 24 '15 at 08:55
  • Produce a proof certificate for [insert famous conjecture] containing as few errors as possible? – Chris Culter Feb 24 '15 at 09:47
  • @ChrisCulter: I've never seen people call finding a proof an "optimization problem", even though mathematically you might be able to formulate it as such. (Even that I have doubts about, since quantifying the number of "errors" is subjective.) That's just not the point of the question -- I'm hoping those reading my question will "get" the spirit of the question instead of finding loopholes in what I've asked literally. The goal here is for me to understand in which kinds of problems is the usefulness of the solution sensitive to its optimality. – user541686 Feb 24 '15 at 10:12
  • Although I am not trying to find a loophole :), one possible problem with the spirit of the question is that "usefulness" of an approximation depends on our end-goal. Think of a scenario where the NP-hard optimization problem is used to make a decision for the corresponding NP-complete decision problem. For example, consider the max-clique problem on a graph $G$. We ask what is the maximum clique size or equivalently if $G$ has a clique of size $\ge k$. Any approximation (e.g., a large clique), cannot answer the question definitively. In that sense, the approximation is useless, right? – megas Mar 10 '15 at 20:02
  • @megas: Yeah, I'm basically trying to understand what kinds of (real-world) end goals don't tolerate suboptimality. – user541686 Mar 10 '15 at 20:30

2 Answers2

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After thinking about it a bit I think that problems in PPAD may be interesting to you: http://en.wikipedia.org/wiki/PPAD_(complexity)

There are a few PPAD-complete problems: Nash equilibria, Finding Brouwer Fixed Points, and End-of-the-Line. I think that you can see the common thread among all of these: we know that the object exists (unlike in SAT or the decision version of TSP).The thing about approximating a Nash equilibrium (say a mixed equilibrium in a symmetric game) is that any solution but the exact equilibrium will result in continued play or possibly even a loss by proceeding from the approximated point of equilibrium---the approximately best strategy. So, unless one can make the approximation bound incredibly small the prospect of approximate Nash equilibria seems dubious. One might also ask whether there are problems such that the existence of approximation algorithms implies something about an entire class of problems....

Suppose we have an $N-$player fair game with symmetric utility functions. Let the space of strategies for player $i$ be $D_i$. Mixed strategies for player $i$ can consist of subsets of $D_i$. In an $N-$player game call a strategy profile $\vec{V} = \{v_i\}_{i=1}^N$ an $\epsilon-$approximate Nash Equilibrium if for all players $i \in \{1, \ldots, N\}$ and all actions $a \in v_i$ $$\mathbb{E}\left[ u_i(a | \vec{X})\right] \geq \mathbb{E}\left[ u_i(b | \vec{X})\right] - \epsilon, \; \forall b \in D_i$$ where $u_i$ is the utility function for player $i$ and the expectation is over the weights that the remaining players put on their mixed strategies. A lot of games and actually a lot of real-world scenarios turn out to have mixed equilibria, but that isn't the focus of this response.

Currently the smallest known value for which there is a partially polynomial time $\epsilon-$approximation (exponential in the approximation accuracy) is $\approx .333$ for normalized utility functions, i.e. $\operatorname{Ran}(u_i) = \left[ 0,1 \right]$. Furthermore, if there were to be a fully polynomial time algorithm, ie. $O((1/\epsilon)^d P(N))$ rather that $O(N^{1/\epsilon})$, then $PPAD \subset P$ which is a big deal. So as it stands either the approximations are bad or some hierarchy collapse happens---even then the usefulness of the approximation is very questionable.

If you are still convinced that approximate Nash equilibria don't quite fit the bill I suggest looking at End-of-the-Line (explained in the wiki article I attached).

For more on this wonderful subject I suggest Noam Nisan (and Co.'s) great book Algorithmic Game Theory.

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You can take for example the problem of sparse linear regression. Given a response variable $Y \in \mathbb{R}^n$ and a design matrix $X\in \mathcal{M}_{n,p}(\mathbb{R})$, you want to statistically estimate a sparse vector $\beta \mathbb{R}^n $ such that $Y=X\beta+\varepsilon$ where $\varepsilon$ is an isotropic Gaussian noise.

To enhance sparsity, a natural estimator is given by penalizing the least squares with a $\ell_0$ term : $$\hat{\beta} \in \operatorname{argmin}{\{ \left \| Y-X\beta \right \|_2^2+\lambda \left \| \beta \right \|_0\}} $$ where $\left \| \beta \right \|_0$ is the $\ell_0$ pseudonorm of $\beta$ (i.e. the number of nonzero coefficients in $\beta$) and $\lambda >0$ is a tuning parameter fixed beforehand. This procedures is statistically quite efficient, but is unfortunately NP-hard.

A simple relaxation of the problem is obtained by replacing the $\ell_0$ term by a $\ell_1$ penalty: $$\hat{\beta} \in \operatorname{argmin}{\{ \left \| Y-X\beta \right \|_2^2+\lambda \left \| \beta \right \|_1\}} $$ which is known as lasso, or basis pursuit. This relaxed problem is convex, solvable in polynomial time, and enhances the sparsity of the solution. A lot of work has been conducted in recent years to show the statistical efficiency of this relaxation (see for example the this article by Candès & Plan).

However, as you can see in this working paper by Lin, Foster & Ungar, the relaxation can, in certain (pretty rare) cases, give results infinitely worse that the NP-hard problem. In other words, approximating the optimization problem can lead to useless solutions.

  • Candès, E. J., & Plan, Y. (2009). Near-ideal model selection by ℓ1 minimization. The Annals of Statistics, 37(5A), 2145-2177.

  • D. Lin, D. P. Foster, and L. H. Ungar. A risk ratio comparison of ℓ0 and ℓ1 penalized regressions. Technical report, University of Pennsylvania.

PAM
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    I will kindly object to this example! The example demonstrates that a certain relaxation (approach to approximate) a hard problem can fail miserably. But I think the original question seeks examples of problems where any approximate solution (no matter how we got it) is just not good, and the only useful solution is the optimal one. – megas Mar 10 '15 at 20:16
  • What @megas said. :) – user541686 Mar 10 '15 at 20:29
  • Oh, ok, sorry for the misunderstanding ;) – PAM Mar 10 '15 at 21:33