# Tag Info

95

Stay put for about 45 days, after which the pirates would have circumnavigated the globe and returned to your current position.

59

Here is the strategy:   It works because:

56

If we assume the ocean is flat and extends indefinitely in all directions, there is a strategy that guarantees we can catch the pirates in at most 800,000 years. Put our current location as the origin of a coordinate system. We will describe our position in polar coordinates, as a function of time: $(r(t),\theta(t))$ (where we have arbitrarily chosen a ...

45

Yes, the policeman can catch the thief, although it may take a very long time. In particular, the following strategy works: The policeman starts in the center. First, he makes a counterclockwise loop around the top right quadrant. Then, he makes a counterclockwise loop around the top left quadrant. Then, he makes a counterclockwise loop around the bottom ...

39

The attack drone should fly to a virtual target that is always at 29/30 of the way from the center of the room to the target drone. That point moves 3.3% slower than the target drone. Since the attack drone flies only 3% slower than the target drone, it can reach that virtual target in a finite time regardless how the target drone moves. Since the maximum ...

30

It is possible to escape a monster that is a little more than $4.6$ times faster. Given a monster running at speed $X$ times rowing speed and a lake with radius $R$, you must first row to the circle that is $R/X$ from the centre of the lake by spiralling out while keeping the centre of the lake between you and the monster. Thereafter you should take a route ...

23

First of all, row out to a radius $R/4$ (where the lake has radius $R$) keeping you, the centre of the lake and the monster in a straight line - with you on the far side to the monster. This is always possible; radius $R/4$ is the first point where the angular speed you can achieve just matches that of the monster as he runs round to get you. You are now a ...

22

Drive 20 hours in a direction we will denote with as having $\theta=0$. Drive in a spiral pattern such that you are $20t+20$ nautical miles away from you starting position. $t$ is time in hours from right this second. After 20 hours, your distance from the starting position will be $r=20t+20$ while your angle will be... more difficult to find. If we ...

21

I have no idea who can win, but I do want to clear up one point that some other people seem to be overlooking. Namely, the lions can always surround some plurality of zebras so any strategy to keep the zebras on the outside is doomed by construction. Let $n \ge 100$ be the number of zebras and the number of lions. Let $\text{net}$ be the convex hull of the ...

16

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Number the nodes as follows: A valid solution is to check: To see that this works, first note that the fox can only go from a hole in an even-numbered level of the tree to a hole in an odd-numbered level of the tree and vice-versa. If the fox starts in an odd level, the sequence 4,2,5,2,1,3,6,3,7 guarantees the hunter will catch it. This works because the ...

14

A more simplified/intuitive way to think about it is if you have the following: No matter where the ant and spider start, the spiders go into this formation: Where the red dots are the spiders, and the green dot is the ant. Notice how the two red spiders on the edges of the cube will only move up and down. This prevents the ant from getting across to the ...

14

Here's my answer: Once this is settled, we can easily say the problem can be solved The strategy being: Bonus: Second Bonus FTW:

13

This is not an answer, but rather a counter to Micael Rize's answer. I don't intend for this to be picking on him, but it is too difficult to explain in a comment, and there are many arguments about his answer, so hopefully this will be persuasive. In my example, I am going to use only 16 zebras, but it should be obvious that this is can extend to 100 ...

13

First, run 1/3 of a unit toward the center of the arena. The radius of the arena is 1 unit and the lion runs at the same speed as you, so the lion will not catch you. Now, we run along a series of straight line paths. The $n$-th path will be orthogonal to the the line segment connecting your position and the lion's (at the start of the path), and have a ...

13

I just came back home and was about to draw a diagram illustrating this, but I can't do any better than @noedne already did.

12

I will place the Zebras as follows : Then whenever the knight attacks any piece, I will move it to its adjacent diagonal so that it is contained within these boxes : Now, the knight cannot fork two pieces inside same box(as they are of opposite colour). Also knight cannot fork two pieces from two boxes as they are quite apart...(4 places between them). So ...

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

11

Zebras can always win if the lion claw length $r=0$, but the lion wins with claws $r>0$. When the lion has claws, the safety-buffer distance $b$ would have to grow with the number of chasing turns $n$, which would mean the zebras would need to define infinitely large buffer zones to never be caught. Alternatively, if the zebras know the lion will get-...

10

Assuming there is no wind, your boat was completely still on the water before it was boarded by pirates, the pirates stepped off your boat at its center of gravity, and the pirates used motorized boats, the escape vector of the pirates will be the exact opposite of the direction your boat will be drifting in once they "step off" your boat after robbing you. ...

9

Note: The following full answer expands on the previous partial answer, which has been retained below. Full answer To analyze all the possible states, the algorithm Ken Thompson described in his 1986 paper Retrograde Analysis of Certain Endgames, which was used to develop some of the earliest chess endgame tablebases, was adapted for this question. His ...

8

For the square island, you need Not sure if I can provide an escape strategy for one less ship patrolling. Should be something like "approach one corner, lure the ships towards it, and then diagonally cross the island".

8

The monster wants to eat me, but he is also particuarly fond of my boat since "he will always run towards the closest bit of shore to your boat." So I row the boat in one direction and then jump off and swim in the opposite direction. Even the fastest monsters would never catch me!

8

Zebras win with an up (or down) and across strategy. Start with Lopsy's idea of vertical strips, with each strip 400m wide. We call a strip empty if it contains no zebras (it may contain the lion). For convenience, referring to a strip without qualification refers to the middle of the strip. The lion is near a zebra if it is within 100m of the zebra ...

8

Surprisingly, Suppose you have a very very high number of smugglers. The strategy to escape is approaching the coast as much as possible (the more the smugglers, the closest you have to be), then follow a straight line (a chord) which is always shorter than the smugglers' path to reach you. Check the picture: when you come very very close to the coast, the ...

8

An upper bound for the 3 spider case:

7

You can succeed when n = Here's how: Thoughts on proof of optimality:

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

My final assumption is following this strategy : The zebras choose to be uniformly distributed on a circle, distanced sufficiently from lions' center of gravity - which the circle is the centered on. They will not be doomed to lose the game given: Any zebra is instructed to escape by following the vector in the direction parallel to \$\sum \vec{l_i b}...

6

Edited because for some reason everything below gets flagged as spam in a de novo post, but not in edits made to an existing post! The answer is that A discussion (below) of the generality and assumptions of the solution might reveal the answer. To motivate the solution, note the following (lemmas 1 and 2 can be read as hints; 3 and 4 might reveal the ...

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