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linestein
90 Posts
A guard walks around a pool three times faster than you swim. You are in the pool swimming. Can you escape? The pool is a square, and you are swimming in the pool. I thought this was a great puzzle when I finally figured it out. I hope my answer is right!    | ||
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Alejandrisha
United States6148 Posts
edit: upon further thought i was thinking the guard could counter this by hugging a corner while you are in the middle of the pool. but the guard gets to the opposite corner just a tad slower than you do if you both go to the opposite corner at the same time. so i stand by my answer. curious what you came up with | ||
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Harris1st
Germany2828 Posts
On September 06 2019 10:19 Alejandrisha wrote: i believe so. if you start an infinitesimally small distance from the middle and then go the opposite direction, you will travel ~.5000001 the length of one side of the pool and the guard has to travel 4 times that distance to stop you I believe you are wrong. I'm to tired to do higher maths right now so I used number ^^ + Show Spoiler + Worst case: You are in the middle, guard is in a corner Pool is 4x4 Your distance (going to the opposite corner) is 4/2 ² + 4/2 ² = Root of 8 Guards distance is 4+4 =8 Root of 8 *3 > 8 | ||
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Alejandrisha
United States6148 Posts
On September 06 2019 15:08 Harris1st wrote: I believe you are wrong. I'm to tired to do higher maths right now so I used number ^^ + Show Spoiler + Worst case: You are in the middle, guard is in a corner Pool is 4x4 Your distance (going to the opposite corner) is 4/2 ² + 4/2 ² = Root of 8 Guards distance is 4+4 =8 Root of 8 *3 > 8 ah true. in the corner guard case i was using root 2 when i shouldn't have been. should have used side length of 2 rather than 1. thanks! | ||
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calh
381 Posts
On September 06 2019 10:19 Alejandrisha wrote: i believe so. if you start an infinitesimally small distance from the middle and then go the opposite direction, you will travel ~.5000001 the length of one side of the pool and the guard has to travel 4 times that distance to stop you edit: upon further thought i was thinking the guard could counter this by hugging a corner while you are in the middle of the pool. but the guard gets to the opposite corner just a tad slower than you do if you both go to the opposite corner at the same time. so i stand by my answer. curious what you came up with That is essentially correct. Assuming you start in the center: - If the guard is not at any corner, you swim to the opposite side to him and escape. - If he is at one corner, you swim to the opposite corner slowly while watching him. If he never moves, you escape. If he commits to one side, you swim to the opposite side and escape (basically reduces to the previous case with extra advantage for you). | ||
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Harris1st
Germany2828 Posts
On September 06 2019 15:33 calh wrote: That is essentially correct. Assuming you start in the center: - If the guard is not at any corner, you swim to the opposite side to him and escape. - If he is at one corner, you swim to the opposite corner slowly while watching him. If he never moves, you escape. If he commits to one side, you swim to the opposite side and escape (basically reduces to the previous case with extra advantage for you). You are absolutely right. With changing directions mid-swim you can easily evade the guard | ||
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Ej_
45998 Posts
There are 2 "boundary" scenarios: -the guard's in the corner (1) -the guard's in the exact middle of a pool side (2) Let t_s be the time it takes the swimmer to reach the side and t_g the time it takes the guard to reach the same spot. Let v be the speed of the swimmer and a the lenght of the pool side. For (2), the solution is trivial, as the desired spot of the swimmer is directly across the guard and t_s=a/2v; t_g=2/3v, then t_s<t_g For (1), the swimmer has to change their trajectory in relation to the guard (if a solution exists at all) Let us examine then: Let (0,0) denote the middle of a perpendicular coordinate system OXOY, let us set (0,0) in the center of the pool. Let's assume that the guard is initially in (a/2, - a/2) (bottom right). Then, the swimmer's initial trajectory is y=-x (towards top left). The guard will follow, traversing x=a/2. The swimmer should then change their trajectory to such that the guard has to travel the longest possible path, let's denote that as A. Let's denote swimmer's path as B. A solution exists only if: t_s<t_g => B/v<A/3v => B<A/3 (*). For B>2a, the guard would have to walk past a point of where the optimal strategy would be to walk around and approach the corner from the other side. Therefore, he would need to be in (a/2, -a/2+x) such that x<z and the swimmer's final trajectory point is (-a/2, a/2-z). Therefore, the swimmer's optimal trajectory is the initial y=-x, but after certain time, assuming y=a/2-z. That time is equal the time that the guard walks the distance of x (boundary case for x=z), therefore it's equal to z/3v. In z/3v, the swimmer would have swam a distance of z/3 along y=-x, so his position would be (-z/3/sqrt2, z/3/srtq2). So, the above solution exists if (B=a/2-z/3/sqrt2 and A=a-z+a+a/2-z/3/sqrt2) (1) and (B=a/2-z/3/sqrt2 and A=z+a+a/2+z/3/sqrt2) (2) Both satisfy (*). Let's check then: For (1) and (*) z<3sqrt2/14 * a Satisfies the solution For (2) and (*) We arrive at a/z<(3+4/3/sqrt2)/4<1 Which is bollocks and even if the proof above had been without errors (and I'm sure some are there), the swimmer wouldn't ne safe using a non-continously changing trajectiry :D Let me look for a simpler solution | ||
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Ej_
45998 Posts
The actual solution is also much much eaaier than what I initially did and boiled down to basic trig in my case :D | ||
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Harris1st
Germany2828 Posts
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raynpelikoneet
Finland36610 Posts
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Harris1st
Germany2828 Posts
On September 06 2019 21:09 raynpelikoneet wrote: Depends on how fast can the guard swim. He is obviously afraid of water  | ||
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brickrd
United States4803 Posts
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Nakajin
Canada5473 Posts
On September 06 2019 21:09 raynpelikoneet wrote: Depends on how fast can the guard swim. Also depend how fast you can run once your outside of the water. | ||
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Clazziquai10
Singapore1925 Posts
Assuming the swimmer starts at the centre and the guard starts at one of the corners and you try to escape by swimming towards the opposite corner, you can escape by making a juke towards the opposite edge that the guard is running along. ![[image loading]](https://i.imgur.com/hVq22pz.png) As long as you make the juke anytime between the start of your 45-degree diagonal swim and when you are (1 - 2 / (3 Sqrt(2) - 2) ) X from the edge of the pool, you're safe | ||
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linestein
90 Posts
here were what i thought of as the two bounding cases. you swim to the middle of the pool and then either A) ![[image loading]](https://i.imgur.com/jD8eNoW.png) or B) ![[image loading]](https://i.imgur.com/sZ7vVhG.png) now obviously in case A you escape but in case B you don't. e.g. 1.41421356237*3 = 4.2426 which is greater than 4. so finally i would refer to clazziquai's solution where you make a little turn to gain ground | ||
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Chairman Ray
United States11469 Posts
For the problem "What is the fastest speed the guard can walk such that the swimmer is still able to escape", here is the optimal solution: ![[image loading]](https://i.imgur.com/QH2kimb.png) The swimmer should start in middle and guard should start in corner. The swimmer should always swim towards to point on the edge opposite of the guard such that the guard is indifferent whether to go forward or turn around, with a slight adjustment towards the point on the edge closest to the swimmer. Using this solution, the swimmer should swim in a curve like in the picture. Calculating that curve involves some calculus that I have forgotten since college. For the problem "What is the shortest time that the swimmer can escape in", here is the optimal solution: ![[image loading]](https://i.imgur.com/xbeSRKZ.png) You also need some tricky calculus to get the exact curve and end point, but the optimal path for the swimmer is to travel below the indifference curve from the previous solution, such that the guard will walk one direction, and then turn around so that the swimmer reaches the edge just before the guard reaches it. For the original problem, using guess and check, all I did was have the swimmer swim towards the opposite corner, and then head straight left after traveling 1/4 of the way. The swimmer makes it. ![[image loading]](https://i.imgur.com/kXh5T9n.png) | ||
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Alejandrisha
United States6148 Posts
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Alejandrisha
United States6148 Posts
On September 06 2019 21:09 raynpelikoneet wrote: Depends on how fast can the guard swim. he's a cat person. and i don't mean he prefers cats to dogs i mean he can't go in water | ||
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fluidrone
France1475 Posts
+ Show Spoiler + | ||
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Chef
10738 Posts
I think the best strategy in this case is to try to catch the guard running to get to you, splash water in his path, and hope he slips and splits his skull or otherwise really hurts himself. Then you've got a shot to find some clothes and skedaddle. Alternatively maybe you can quickly pull his leg and get him to fall into the pool and buy yourself some time that way, that is probably easier to pull off but less likely to incapacitate him. Also risks getting hit in the head with a baton when you go to do it, depends how vicious the guard is. The mathematics are kind of irrelevant to the problem, you don't give yourself much of a head start even in the best case. | ||
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