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Add LLMs note, source of problem, to Dead Pixels post
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Anthony Wang 2023-12-20 00:27:54 +00:00
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@ -36,7 +36,7 @@ OK, if you insist on reading the solution, here it is:
{{< rawhtml >}}
<details>
<summary>Click to show solution</summary>
The subrectangle with the smallest average brightness is just the pixel with the lowest brightness! This solution runs in O(NM) time.
The subrectangle with the smallest average brightness is just the pixel with the lowest brightness! We can find that in O(NM) time by scanning through the entire array of pixels. I also asked ChatGPT to solve this problem, and its initial attempt was an O(N^3 M^3) brute force. After asking it to improve its solution, it wrote an O(N^2 M^2) program using prefix sums. Yay, I guess that means I have a new favorite problem for stumping LLMs, in addition to my old favorite, "What's the average aspect ratio of a human?"
</details>
{{< /rawhtml >}}
@ -55,3 +55,5 @@ At this point, you might think we can just search through all 1 by 2 subrectangl
The best 1 by 2 subrectangle would include a pixels with brightnesses 0 and 1, which gives an average of 1/2. However, we can do better by taking the whole 1 by 3 rectangle, which gives an average of 1/3. While any rectangle with an even dimension can be divided up into 1 by 2 rectangles, odd by odd rectangles cannot. With both 1 by 2 and 1 by 3 rectangles, you can build any 1 by i rectangle for i greater than 2. With that, you can then make any j by i rectangle. Thus, we only have to search through all 1 by 2 and 1 by 3 subrectangles, which takes O(NM) time.
You can also consider the variant of requiring the answer subrectangle to by at least 2 by 2, but it's just more casework.
*Note: This problem was previously asked in the [Ladue Computing Contest](https://codeberg.org/LadueCS/L.cc/) that sadly never happened. I first heard this problem from [Eric Zhang](https://www.ekzhang.com/) a few years ago who had written it for a programming contest at his high school. So yes, this problem has quite a storied history!*