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#1: besmirched tea (Besmirched Tea) on Oct 3, 2021 [SPOILER]

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IBT, it would depend upon whether there was air or vacuum in the hole. With no air resistance, the person would accelerate all of the way to the center, and decelerate all of the way from the center to the south pole, ending up at zero velocity whenreaching the pole (assuming that both poles are the same distance from the center of gravity of the earth and ignoring micro-gravity distortions from density fluctuations in the earth's composition. When you consider an ideal vacuum tunnel between any two points on earth, with the best "shortest-distance" parabola possible between those two points, it works out that all trips take exactly the same time - point pairs far apart get up to a higher speed because of the longer acceleration-deceleration time, closer point pairs have a shallower curve and don't accelerate as much (unless you add some sort of motor to help, of course).

With air resistance, especially with very high density air near the center of the tunnel, the falling person would overshoot the center a bit and then fall back and eventually stop at the center.

The falling person (back to going through a vacuum) is in a highly elliptical orbit (the limit of an ellipse is a straight line), with a smaller radius than any near earth orbit satellite than is not at ground level, so the satellite has a longer orbital period than the falling person. Reaching the south pole is actually a half orbit.

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No idea. Pretty image. Tough puzzle with lots of guessing at the end.

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Holy cow that was a tough solve! I think I hit deep lookahead but I'm not as good as people like Joe or Gator at determining exactly where that threshold is so I will leave that to them. I will say that it is logically solvable and that there is next-to-no chance I do a walk-through. Best of luck!

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First we would have to overcome the enormous engineering problems such as molten mantle and outer core and solid iron spinning inner core. Then we would cap both ends and pump out the air so that the falling object would not reach terminal velocity. This then becomes a calculus problem as the speed increases to maximum and the gravitational acceleration decreases to zero at the center then reverses on the way back out.

My calculus is as rusty as my memory but I did this calculation in my youth and I recall that it was about a 45 minute trip.

A satellite in low earth orbit takes about 90 minutes for a complete orbit so a half trip should be about...drumroll please...45 minutes. I think it would be pretty close to a tie.

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The puzzle is definitely logically solvable. I am not sure where the border is between moderate lookahead and deep lookahead, but it felt to me like moderate lookahead -- you just had to work really hard to find the right place to look, but then it was usually only one step deep. For example, for the bottom left-hand corner, there was a lot of "look across the bottom row for where red and black could possibly go" and then it would turn out that one of the colors could only go in one particular column, and there was only one pixel of that color in that row, so it had to go there. Repeating that on either the bottom row that had space or the right-most column that had space got me almost all the remaining pixels. So I'm going to flag it as moderate lookahead.

Found to be solvable with moderate lookahead by valerie.

Come to think of it, what I was talking about in #12 is color logic, not even lookahead.

yeah it's almost "advanced color logic" because there is color logic that you can do in a single line, but that's multi-line color logic :) i mean, there is also lookahead moves needed, but you are right about what you were talking about :)

Some tricky (but manageable) logic, for an interesting solution and interesting discussion in the comments.

I got to 83% with color logic, C2-5 still needing fixing, with the bottom of C6.

Then I got to moderate, maybe deep, lookahead but I suspect I missed a few shortcuts. There's diagonal color logic where 2 colors alternate and we can fill them in, but the left side having a bunch of red-black-red-black short-circuits that.

I went from the upper left on down. I placed the black 1 in column 2, then the 3. It was an interesting solve.

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