12
Not a definite answer but I narrowed it down to 2 possibilities.
It's either
In this case
or the second possible answer is
Because
8
An upper bound for the 3 spider case:
7
For two spiders, one of the spiders has to be at least as fast as the ant in order for the spiders to catch the ant.
There are 3 situations:
Both spiders are slower than the ant: The ant can avoid forever
At least one spider is as fast as the ant: The spiders can catch the ant
At least one spider is faster than the ant: The spiders can catch the ant
...
6
In general, with logic puzzles it's important to mark not only the parts of the solution, but also things you know can't be part of the solution. Here, that means you should mark segments that you aren't allowed to draw, because there would be no way of completing the puzzle if you did.
For example, there can't be a connection between F1 and F2, because F1 ...
6
I feel like 1/2+ε is enough for the 3 spider case. The spiders can start (or take their sweet time to get) in this configuration:
And then the two slow spiders can follow the red arrows to ensure the ant will be trapped, while the "fast" spider keeps watch on its edge.
Unless I overlooked something, this seems to work, but doesn't prove there isn'...
6
You should
because
5
Original problem statement
Consider a game played with 12 people. 9 are randomly assigned to Team Good and 3 to Team Evil. Players on Team Evil know their teammates, but players on Team Good don't. Two of the players on Team Good are secretly assigned devices with red and green buttons on them. Gameplay proceeds as follows:
All of the players talk publicly ...
4
For my analysis, I will assume that all we get to specify is the sum of the speeds of the spiders and that the sum we specify must work for any possible assignment of that total speed to the spiders where each spider must be given a non-zero speed.
Four Spiders
It wasn't asked, but first lets take a look at the case of 4 spiders. With 4, any positive sum ...
4
If you are absolutely certain that all 8 edge pieces not on the top layer are in their correct places and in their correct orientations, (this seems to be the case, since you say you have finished the first 2 layers) then there are exactly two possibilities:
either the number of correctly oriented edge pieces on the top is even (0, 2 or 4), or
you have an ...
2
Here is a strategy giving a better uppper-bound for the 3 spider case.
And there is a strategy for two spiders, one with speed exactly 1 and one with speed ε.
1
A couple of words are confusing:
By "hover" I assume you mean it's moving (generally means to stay in place)
By "retain" I assume you mean some kind of average speed needed. Clearly our speed will accelerate and not be maintained as a constant.
So, maybe:
1
If you're talking about a regulation change, then the most efficient method to place cubes will depend on the solver's strategy for memorising cubes and reviewing their memo. There won't will be an "optimal" way which can be written in the regulations that will suit everybody. The regulations currently ask for cubes to arranged as close to a square ...
1
This question has been changed in a way that invalidates my answer.
Previously, the button-pressers had their remotes from the start, and all communication was allowed as long as it was done "publicly." In this question's original state, my answer was perfectly valid, and probably the correct answer. Please feel free to check the edit history to ...
1
100% success rate with lateral-thinking (yeah, I know, I'm stupid that way, sorry):
58% success rate by an actual strategy that doesn't require prior planning:
Then
Since there's still a 2 in 9 chance that the bad guys guess right purely by chance, the probability of success goes down a bit.
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