C Sebestyen's Comments

As my wife can tell anyone , I am not a 'grammar/syntax/spelling' guy. Unless the semi-colon is missing at the end of a line of code , I wouldn't even know it was gone.
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idahoqie , I think that you assume the rabbit can always swim out to 1/4 of the distance to the shore while being able to maintain an opposite position to the agent.

This distance will vary on the speed of the agent , getting smaller as the agents speed increases. By casual observation we can see that that rabbit can swim out to the fraction 1/max to the shore based on a max speed for the agent. This would leave the rabbit a distance of (max-1)/max to swim at rate (x) while the agent was moving at rate (max)(x).

So the time take should be equal to find the max rate of the agent.

(((max-1)/max)R)/x = (R(pi))/((max)(x))

If multiply both sides by x we get
((max-1)/max)R = (R(pi))/(max)

We can divide both side by R and get
(max-1)/max = (pi)/(max)

We can multiple both sides by (max) and get
max-1 = (pi)

add one to both side and your done
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It seems like MadMolecule answer should work. I tried a quick calculation only to realize you probably need a bit more then straigh GEO to get the job done.

But on second thought all the rabbit has to do is be able to get out to a radius away from the center that the rabbit can cover in the same time it takes the agent to run around the pond. This will take some time for the rabbit since it need to ensure the agent is always directly behind it. Using MadMolecule's idea. Once it reaches that radius (distance) from the center of the pond it can make a mad dash at the shore line.

Pond has Radius = R
Agents Speed = 4x
Rabbit Speed = x

If the rabbit moves outward to a distance of (1/4)R from the center always keeping the agent behind the rabbit can swim around in circles and the agent will always remain at the opposite shore assuming the agent never stop pursuit.

This is because the rabbit here will be covering a distance of 2(1/4)(pi)R or (1/2)(pi)R at speed x taking ((1/2)(pi)R)/x or (1/2x)(pi)R and the agent will traveling 2(pi)R at 4x which takes (2(pi)R)/4x = (1/2x)(pi)R

Now the rabbit is R - (1/4)R away from the edge or (3/4)R and can cover that in ((3/4)R)/x or (3R)/(4x). The agent is half way around the pond and will need to cover (1/2)(2(pi)R) or (pi)R at a rate of 4x. That will take him ((pi)R)/(4x). Since (pi) is about 3.14 the agent will take longer to reach the other side of the pond then the rabbit in this scenario.

The agent will have to run slower than((pi)+1) times faster than the rabbit. Hopefully I will write back the process later, its time to go home.
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Profile for C Sebestyen

  • Member Since 2012/08/08



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