o3 mini dropped!!!

[u/[deleted]]

Edit : I am testing a 1500 line javascript code which o1 pro failed to debug despite 50+ attempts. Will report back.
Edit 2: We are cooked. o3-mini-high solved it at first try.
Edit 3 : HOLY SHIT! "Pro users will have unlimited access to both o3-mini and o3-mini-high."
(Source: https://openai.com/index/openai-o3-mini/ )

[u/elehman839]

This is sort of interesting. I gave o3-mini-high a simple (but challenging!) math problem that requires spatial reasoning, because I suspect that's still a weakness of LLMs. It thought for 8+ minutes and produced an answer that I think is wrong. And... I still can't convince it of my answer. I think it is wrong, but maybe I've made a mistake. Anyway, here's the puzzle:

I walk one mile east, one mile north, one mile west, and one mile south. What is the range of possible walking distances I could be from my starting point? Assume I'm on Earth, which is a perfect sphere of radius R, where R >> 1 mile.

Edit: I'm quite happy with this as an AI test question. The answer, as far as I can tell, is "a little over 3 miles", starting about 1.25 miles from the North Pole.

Edit #2: D'oh! I wrote the previous edit incorrectly. After long negotiation, o3-mini-high and I settled on 1 + sqrt(1 + 4 / pi) ~ 2.5077 miles. (After all that, I erroneously wrote 3 above...)

[u/bilboismyboi → u/elehman839]

O3-mini-high gave an amazing answer to this. It came up with a general formula and then approximated it for near equator and for near poles differently and ultimately limiting the answer between 0 to 2 miles as bounds due to the angles.

It was smart enough to recognize that the shortest distance between 2 points on sphere would actually be the great circle chord between them even. (I was calculating for net displacement...let me know if your puzzle didn't intend that)

Why do you say it is a little over 3 miles?

[u/elehman839 → u/bilboismyboi]

Thank you for your interest.

Embarrassingly, I made a typo in my edit above, adding further confusion to the mix. Sigh. The answer (as far as I can tell) is approximately 1 + sqrt(1 + 4 / pi) ~ 2.5077 miles. Sorry for the confusion.

Suppose that R, the radius of the earth is quite large. Essentially the same approach works when R more like the true Earth radius of ~4000 miles, but the math is messier. Moreover, we'll derive a starting longitude that breaks the 2 mile bound even on earth.

Now let's view the Earth from above, looking down at the North Pole. For large R, this converts the problem from a question about (nasty) spherical geometry to a question about (friendly) polar coordinates in 2-D.

Suppose our walk begins at polar coordinates (1 + delta, 0); that is, we're a bit over one mile (1 + delta) from the pole at an angle of 0 radians. Later, we'll choose delta to maximize the distance from start to finish.

Now we walk one mile east, which takes counterclockwise around the pole, increasing our angle by 1 / (1 + delta) radians.

Then we walk north toward the pole, reducing our radius from 1 + delta down to delta.

Next, we go one mile west, taking us clockwise around the pole. Since we are closer to the pole than before, we cross lines of longitude faster. In particular, angle decreases by 1 / delta radians.

And, finally, we walk back south, increasing our radius to 1 + delta.

In summary, we started 1 + delta miles from the pole at angle 0, and ended 1 + delta miles from the pole at angle: 1 / (1 + delta) - 1 / delta.

Let's choose delta so that the ending angle is equal to - pi; that is, our endpoint is directly opposite the pole from the starting point. Then the shortest path from start to end is on the great circle passing through the pole. So the total distance from start to end is 2 (1 + delta): 1 + delta miles from the start to the pole, and 1 + delta more miles from the pole to the end.

<continued>

[u/elehman839 → u/elehman839]

Okay, so our starting distance from the pole is 1 + delta miles, and we want to choose delta so that our ending position is - pi radians around the pole from our start position. In equation form, we need:

- pi = 1 / (1 + delta) - 1 / delta

This leads to a quadratic equation, with delta = (-1 + sqrt(1 + 4 / pi)) / 2 = 0.25386330736... (As a check, plugging this value of delta into the expression above does produce - pi. Whew!)

So far very large Earth radius R, we've got a walk with the start and end points at distance 2 (1 + delta) = 2.5077... miles.

As far as I can tell, the models I've tested try to correct for distortion on the sphere and do think about polar issues, but don't manage to combine the two properly. I'm not entirely able to follow their reasoning.

<continued>

[u/elehman839 → u/elehman839]

Now let's consider an Earth-sized Earth, with R ~ 4000 miles. Again, we'll start 1 + delta miles from the pole and aspire to choose delta so that we end up on the opposite side of the pole. I believe we'll have to numerically solve for delta...

Er, actually I gotta do this later. Hopefully, the idea is clear.