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| Started by | Paul <pepstein5@gmail.com> |
|---|---|
| First post | 2016-07-08 03:53 -0700 |
| Last post | 2016-07-08 08:43 -0700 |
| Articles | 2 — 2 participants |
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A faster maximal partial sum in a rectangle. Paul <pepstein5@gmail.com> - 2016-07-08 03:53 -0700
Re: A faster maximal partial sum in a rectangle. Daniel Pitts <usenet@coloraura.com> - 2016-07-08 08:43 -0700
| From | Paul <pepstein5@gmail.com> |
|---|---|
| Date | 2016-07-08 03:53 -0700 |
| Subject | A faster maximal partial sum in a rectangle. |
| Message-ID | <f70420a7-6d08-4735-9c20-97c66e90b732@googlegroups.com> |
A rectangular matrix is filled with positive and negative numbers. Please consider the problem of finding the maximal sum over any subrectangle. (And if you don't want to consider this problem, I don't see why you're reading this post in the first place.) Let n be the maximum length of any edge. O(n ^ 3) solutions are well-known but is there an O(n ^ 2) solution? If not, does anyone know of a proof that no such solution exists. Many thanks. Paul Epstein
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| From | Daniel Pitts <usenet@coloraura.com> |
|---|---|
| Date | 2016-07-08 08:43 -0700 |
| Message-ID | <nlohmg$d8a$1@dont-email.me> |
| In reply to | #6700 |
On 7/8/16 3:53 AM, Paul wrote: > A rectangular matrix is filled with positive and negative numbers. Please consider the problem of finding the maximal sum over any subrectangle. (And if you don't want to consider this problem, I don't see why you're reading this post in the first place.) > > Let n be the maximum length of any edge. O(n ^ 3) solutions are well-known but is there an O(n ^ 2) solution? If not, does anyone know of a proof that no such solution exists. > > Many thanks. > > Paul Epstein > This sounds like a Dynamic Programming solution would reduce the complexity, but I was never very good at determining a DP solution to any given problem. You could probably get some reduction in cost by precomputing runs of various lengths, or using memoization to reduce duplicated computations, though I don't think that reduces the Big O.
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