Timeline for Maximum average value within a rectangular bounding box
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 10, 2010 at 10:03 | answer | added | TonyK | timeline score: 1 | |
| Oct 29, 2010 at 6:44 | answer | added | sleepless in beantown | timeline score: 0 | |
| Oct 15, 2010 at 5:54 | answer | added | Suresh Venkat | timeline score: 0 | |
| Oct 15, 2010 at 3:35 | comment | added | Bernard | i updated the question, so any extra comments would be appreciated. Thanks! | |
| Oct 15, 2010 at 3:34 | history | edited | Bernard | CC BY-SA 2.5 |
added 495 characters in body
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| Oct 14, 2010 at 21:23 | answer | added | Aaron Meyerowitz | timeline score: 0 | |
| Oct 14, 2010 at 21:23 | comment | added | sleepless in beantown | What algorithms have you tried? What's your motivation behind this problem? Is it homework? Please take a look at the FAQ's and consider that a different forum might be more appropriate for this question, and that even on a different forum you might need to clarify and explain the problem more explicitly. | |
| Oct 14, 2010 at 21:15 | comment | added | sleepless in beantown | -1: vote to close; seems like homework to me that the student needs to think about and solve on their own. Or of course the simple way to do it if $m$ and $n$ are given is to perform a $2$-dimensional convolution of the given $M$ by $N$ matrix with the $m \times n$ sized matrix consisting of all ones. The resulting convolution, called it $X$, has a maximal entry or entries identifying the positionning of the $m$ by $n$ matrix. This looks like homework for an image processing type of class. I'd vote to close it if I had closing vote power number of magic points. | |
| Oct 14, 2010 at 21:09 | comment | added | Noah Stein | This question seems more appropriate for stackoverflow. Even if resubmitted there, you should probably give more details about the problem. For example: are m and n given? If not, you could just take the spot in the grid with the largest intensity and call that a 1x1 box with the largest average intensity. If they are given, what else have you tried? Dynamic programming somehow seems natural. | |
| Oct 14, 2010 at 20:57 | comment | added | Greg Muller | I might not understand the question. Why isn't a 1x1 box around the maximum value always the answer? | |
| Oct 14, 2010 at 20:48 | history | asked | Bernard | CC BY-SA 2.5 |