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One of my professors mentioned such algorithms exist but could not think of any offhand. Obviously any algorithm will be at least $\mathcal{O(1)}$, but are there algorithms not yet proven to have a lower bound greater than this, which do have a proven upper bound?

I'm interested in both problems and algorithms, but in either case practical examples (if they exist) as opposed to artificial ones.

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  • $\begingroup$ It’s easy to construct artificial examples, say an algorithm that looks for a proof of statement X of length at most $n$. If X is provable, then this runs in (essentially) constant time, otherwise in exponential time. $\endgroup$ – Yuval Filmus Mar 20 at 15:47
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    $\begingroup$ Are you interested in algorithms or problems? $\endgroup$ – Yuval Filmus Mar 20 at 15:47
  • $\begingroup$ @YuvalFilmus Both, I suppose, but in either case practical examples (if they insist) as opposed to artificial ones $\endgroup$ – ubadub Mar 20 at 16:18
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    $\begingroup$ Can you include all this information in your post? $\endgroup$ – Yuval Filmus Mar 20 at 16:53
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    $\begingroup$ Note that saying "at least O(_)" is a nonsensical statement, just like "at least something smaller than x". Use $\Omega$, $\omega$, and $\Theta$ to properly express lower buonds. $\endgroup$ – Raphael♦ Mar 21 at 6:51
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To try and answer your question:

Yes. generally lower bound is much harder to prove than upper bound. That is, to prove an upper bound, you must only show one algorithm that solves problem $A$ in $t_1(n)$ time.

To prove a lower bound $t_2(n)$ for the same problem $A$, you must show that ANY algorithm that solves $A$ MUST work in at least $t_2(n)$ time.

Take the median for example (under the comparison model). There is a recursive algorithm that solves the median problem in $O(n)$ time. That means any greater upper bound is moot, and any greater lower bound is wrong.

However a precise lower bound for the median is an open question. The algorithm mentioned above performs $c*n$ comparisons, where $c$ ~ $17$. It has been proven that for selection of median, a total of $\Omega(\frac{3n}{2})$ comparisons is required (which means a better algorithm does not exist).

However, we don't know if a greater lower bound exist. Proving a $\frac{5n}{2}$ comparisons lower bound is an open question!

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    $\begingroup$ $\Omega$ ignores constant factors. $\endgroup$ – Yuval Filmus Mar 20 at 20:39
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    $\begingroup$ $2n$ comparison and a bit more is required to find the median as shown by Median Selection Requires (2+epsilon)n Comparisons by Dorit Dor , Uri Zwick. So, this answer does not fully answer the question. $\endgroup$ – Apass.Jack Mar 20 at 20:55

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