Agreed, under those circumstances it doesn't.

By the way, by tetrahedron do you mean a true 4 sided object (traigle base with three triangular sides) or do you mean that classic pyramid everyone calls a tetrahedron but isn't actually one (four triangular sides and a square base).

My guess is that he means a tetrahedron. :-P

Some of the solutiuons I've seen proposed online say "tetrahedron" but then do the classic pyramid, so I just wanted to clarify.

I know a solution, and it is on a tretrahedron. I'm pretty sure you can prove that it is impossible on a pyramid, actually. I might think about this.

Spiders/fly tentative solution:
Let the tetrahedron vertices be called 1, 2, 3 and 4.
Position the spiders A, B, C respectively on 1, 2 and 3.
Move spider A continuously from 1 to 2 to 3 to 4 to 1 and so on.
Move spider B from 2 to 4 and back.
Move spider C from 1 to 3 and back.
Eventually, the slower fly won't have anywhere to run.

I should add that spiders B and C would also move continuously, and at half the speed of spider A, such that A and B would always meet at 2 and 4, and A and C would always meet at 1 and 3.

@uclabb, ghug, did you guys REALLY figure it out? I mean I straight up READ the answer and I would still have to write it out to understand it.

@PE no, prisoners have no idea what goes on outside their cell unless they are being escorted to the switch room

I must have misinterpreted. I thought the spiders moved at the same speed, all three of them.

@YJ- Yeah, I'm a math major focusing in algebra so this kind of stuff is the sort of thing I spend all day thinking about, if in a slightly more formal and abstract form.

@Draug- JP's solution doesn't actually require that B and C move at half speed.

@JP- Cool! That seems right to me, which is cool because I know of a different way to do it!

Um, yes it does. I can visually see how the fly could start just behind one of the single vertice spiders and wait until the 4 vertex spider passes, then slip back down and rest for a time. If they don't move at different speeds then two corners never has two spiders on them at once and he can just sit on the vertex where the spider is moving away as the other approaches.

Cool man, that's really impressive. Will you tell us the answer uclabb?

Even if the spiders have only one speed, B and C can kill some time by making back and forth moves along their respective edges so they'll always meet A at the vertices.

@Draug- All that matters is that whenever the "chasing spider" hits a vertex, the other "enforcer spider" is there at the same time. That would be the case even without the spiders being half speed. So yeah, what JP said.

The way that I had thought of was having one "chasing spider" going around one of the triangles, then having the other two spiders flush the fly out with the following movement (iterated):

Chaser: A -> B -C -> A
Enforcer1: A ->D -> C-> D -> B -> D -> A
Enforcer2: D-> B -> D -> A -> D -> C -> D

@YJ- Do you mean to the light switch question?

Uclabb, I'll have to check, but I'm pretty sure that doesn't work. You solved it JP's way last time.

I didn't remember what I came up with last time so I tried to think of another way. It seems to me like this way works, but I could be wrong.

And if that way does work, it is actually faster than the original solution!

The fly can survive by staying very close to point D and being on the following lines at each step of your strategy:

AD - BD - CD - AD...

Do you guys want the answer to YJ's prisoner one?

@uclab - It was stated that the fly could stop, slow down, or reverse direction at any time.There fore it stops just before hitting the enforcer corner and waits for the enforcer to change edges.

uclab, yeah I mean tthe lightswitch

@Draug- I don't know what you mean by that, and it doesn't seem true to me. Can you be a little more specific?

@YJ- My idea is that you basically just designate one of the (I think there were 100?) 100 people as the "counter", and then what you can do is the following:
1. If someone who is not the counter goes into the room and the light switch is off and they have never turned the light switch on before, then they turn it on. Otherwise, they do nothing.
2. If the counter goes into the room and the light switch is on, he turns it off and increments his count by 1.
3. Once the counter gets to 99, then he reports that everyone has been to the room, and everyone is released.

That is obviously incredibly slow, but it would work. I can't think of a faster way, though (although you can be much, much faster with a very high probability of being right).

Oh, ghug, you are right, if the fly starts between the two enforcers my strategy doesn't work. I think if we make a slight modification to the beginning where the two enforcers both start at D then Enforcer 1 goes to A and then the whole process starts, that is addressed, no?

@uclabb The fly can start on CD and slip in between them once the repeating pattern starts,

I gave this problem to somebody a lot smarter than me, and I think he proved that you need a chaser spider to have a loop of length four while in the process of solving it, but I unfortunately don't remember it.

RE: The Switch Problem:

uclabb's solution works for the single switch, once per day version, but because we don't know the starting position of the switch the count could go off by one. If you care, a more efficient solution to that problem is having tiers of counters:
Group 1: 90 people
Group 2: 9 people
Group 3 (high counter): 1 person
First 100 days:
Group 1 people switch light on if it is off and they haven't flipped it before.
Group 2 people increment their count if the switch is on, and revert it to off.
Group 3 person does nothing.
Next 100 days:
Group 1 people do nothing.
Group 2 people flip the switch to on if their count is at 10, and return their count to 0.
Group 3 person increments his count by 10 each time he finds the switch on and turns it off.
Repeat.

I'm not sure what the optimal tier grouping, time period, or increment numbers are, but that's the general idea.

As for the problem YJ posed:
There might be a complex solution, but I haven't found it yet, so I'm giving my simple solution, with 1 counter and 99 normal people.

Basically the idea here is that switch B is a buffer switch, and then we just have to do a little fixing to account for the uncertainty of the initial position.

Normal people will flip switch 2 unless switch 1 is off, they have previously seen it on, and they have never flipped it. The counter will always flip switch 1, and will increment his count if the switch is on when he comes in but was off when he last left the room.

@uclab - ghug said the fly could do that. Look back up in the thread.

@Draug, I don't really see what you're saying either. The fly survives by staying between the two enforcers, not by staying on another edge.

There are always 3 edges at any vertex. The fly is allowed to change direction or stop mid edge. You said so. The fly is psychic. You said so. There fore he knows the pattern and knows which vertex the spiders will meet at.

But if you want to do a drawing, I will show you a path the fly can use to continuosly avoid your spiders, both chaser and enforcer.

Which method are you talking about, JesusPetry's or uclabb's?

JP's works fine, and I already showed how uclabb's doesn't, what are you arguing about?

I was talking (as I have been since it was posted) about the one where uclab (or somebody) argued that the back and forth spiders didn't have to move at twice the speed of the chaser spider. But then I am back home for a day as I am heading back to Canada tomorrow evenign and I spent all day yesterday at a client or in the air, so I'm a bit behind.

Oh, well, what I think uclabb is saying there is that they can move at full speed if they do some direction reversal and/or stoppage of their own, but you're right that the simplest way is to merely move the enforcers at half speed.