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Thread: Go Dog!

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    Re: Go Dog!

    I guess I am a little confused or you have a label wrong.
    Is the dog starting at "D" 20 m West of B or is the dog at "B" as your puzzle indicates ("The dog will wait at B")?

    If the dog is waiting at B, why would the squirrel even approach B?

    Steve

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    Go Dog!

    <P ID="edit" class=small>(Edited by Timbo on 11-Feb-05 13:59. Fixed confusion with letters))</P>There are two trees, a Birch ([img]/forums/images/smilies/cool.gif[/img] and a Cedar . These two trees are 25m apart with C directly south of B. A squirrel runs between B and C, always in a straight line and always at a constant speed of 3 m/s (or zero if the squirrel is feeding). A dog (D) is standing at D', a point 20m due west of A. The dog always runs at 4m/s but can go in any direction (so true). The dog will wait at D until the squirrel stops at point T - a theoretical point that minimizes the time between then the dog starts the chase (S) and when the squirrel starts fleeing (F) to safety in either tree.

    Find T.
    (Location Australia, then UK, but now USA. Heart, outlook, attitude, etc always Australian)
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    Re: Go Dog!

    Sorry Steve - guess my proof reading is not that hot lately. Thanks for pointing this out. The dog starts at D (fixed in original post now).
    (Location Australia, then UK, but now USA. Heart, outlook, attitude, etc always Australian)
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    Re: Go Dog!

    I am still confused (and this might be part of the "puzzle")

    The fastest the dog can get to the line BC of trees is <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">20/4 = 5 sec</span hide>

    If the squirrel is halfway between the trees, he is only <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">12.5 m = 12.5/3 = 4.17s</span hide> from safety

    <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">If the squirrel leaves midpoint at the sametime the dog does the squirrel will win.If the squirrel is just on the midpoint side closer to B (at mid point he has a 50/50 chance of going to B or C) if he is sitting just a wee bit closer to B, if the dog heads straight for B, he might beat the squirrel there due to the squirrel's "response time" after seeing the dog take off</span hide>

    Though this does not seem like a "satisfying answer".

    Steve

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    Re: Go Dog!

    I think you understand the puzzle. If the dog and squirrel start running at the same time then the squirrel is ALWAYS safe. However, squirrels usually wait (lets call this 'wait time') until the last possible minute (either taunting the dog or not seeing the dog start) until it flees. The dog in question is waiting until the squirrel is at point T, a point that means that the squirrel has the minimum amount of 'wait time' before fleeing to safety. For example, if the squirrel is 1cm from B, then it can wait a touch under 5 seconds before fleeing. If the squirrel is 2m from B then it can wait about 4.33 seconds before fleeing.
    (Location Australia, then UK, but now USA. Heart, outlook, attitude, etc always Australian)
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    Re: Go Dog!

    <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">If the squirrel is running from C to B and when the squirrel is 15 m from B (10 m from C) it will take the squirrel 5 secs to finish. If the dog runs straight to B at this time, he will reach the tree at the same time as the squirrel.
    If he leaves later (<15 m) he will not catch him. If he leaves earlier, he can catch him at some distance "x" from the tree (which he needs to head straight for) but it will take longer than 5 secs to get there.(for example if he leaves right when the squirrel leaves C, he can catch him in 5.45 s, 8.66 meters from B</span hide>

    Steve

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    Re: Go Dog!

    But in that case, the squirrel will just turn around and run back to C.
    (Location Australia, then UK, but now USA. Heart, outlook, attitude, etc always Australian)
    Quote: "All Happiness is the release of internal pressure"

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    Re: Go Dog!

    Maybe the following will help clarify some of the issues.

    <UL><LI>the squirrel has no preference for either B or C except when it involves fleeing D
    <LI>the squirrel doesn't care where between B and C he goes (consider it random) as he is just eating a infinite supply of nuts along that path
    <LI>the squirrel is not on a journey to either B or C, he is just eating nuts
    <LI>the squirrel will happily eat nuts all day if the dog will let him
    <LI>expect that the squirrel will flee to the tree that gives it the best change of avoiding D[/list]
    (Location Australia, then UK, but now USA. Heart, outlook, attitude, etc always Australian)
    Quote: "All Happiness is the release of internal pressure"

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    Re: Go Dog!

    At what point will the squirrel decide to "change course" and go the opposite direction or just stop and wait?

    Is it a certain distance from the dog (stright line), a certain "vertical distance" or a certain "horizontal distance".

    As I said, if the squirrel is "smart" he never gets caught, so there is the "randomness" which my answer gives (he is already going from B to C), even the "risk factor" which my answer gave since it was a "Tie", or what? I seem to be missing some part of the "workings of the minds/instincts" of the dog and squirrel.

    Steve

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    Re: Go Dog!

    To be honest, I thought of this puzzle the other day watching my dog try to catch squirrels so some of the finer points of the 'puzzle' might not be obvious unless you spend time looking at dogs v squirrels. I'll try and answer your questions ...

    At what point will the squirrel decide to "change course" and go the opposite direction or just stop and wait?
    The squirrel will flee to the tree that gives him the best chance of escaping the dog. It might not be the closest to him but it will be the one where his time to reach the tree is less than the dogs.

    Is it a certain distance from the dog (straight line), a certain "vertical distance" or a certain "horizontal distance".
    I'm not sure what you mean by this so I cannot try to comment.

    As I said, if the squirrel is "smart" he never gets caught, so there is the "randomness" which my answer gives (he is already going from B to C), even the "risk factor" which my answer gave since it was a "Tie", or what?
    I agree, if the squirrel is watching the dog, he will never get caught. Also, he can go from B to C and from C to B as long as he watches the dog and retreats to safety (either B or C) if the dog makes a run at the squirrel. However, the squirrel doesn't always watch the dog, often he has his face in the ground rooting out a nut or just plain stuffing around. This is why the dog has a chance of catching the squirrel - the dog is hoping that the squirrel will be distracted when he starts the chase and is waiting until the squirrel reaches T, the point where the safety margin is the smallest and hence the distraction time for the squirrel is most critical.

    As I quoted earlier, you can calculate the squirrels wait time (or safety margin) for each point between B and C. At 2m from B, the squirrel can be distracted for about 4.33 seconds and still make it to safety. This means that the squirrel can do 4.33 seconds of snuffing, digging, general squirrel stuff before checking on the dog and remain save.

    Does this help?
    (Location Australia, then UK, but now USA. Heart, outlook, attitude, etc always Australian)
    Quote: "All Happiness is the release of internal pressure"

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    Re: Go Dog!

    I still seem to be missing "something" in your question (perhaps others have also, since I do not see too many other repsonses).

    1)If the squirrel always flees to the tree with best safety, he should be able to wait for the dog to start running to a tree and then go to the opposite one, giving him more "lead time"

    2) The question is how does the squirrel perceive the distance of the dog to him. if the squirrel is 15 m south of B, the squirrel is 20 m East of the dog (horizontal), 15 m South of the dog (vertical) and also 25 m (straight line) from the dog. How close will the squirrel allow the dog before he changes direction. Essentially what does the squirrel "key on" for changing direction?

    3) <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">I am assuming the dog never makes a "run at the squirrel", since running at the squirrel will guarantee he will miss him, he must be "heading off" the squirrel, trying to be where the squirrel will be. Thus he will (most likely) run towards B or towards C. Or do you imagine that the dog runs straight towards the squirrel start and when the squirrel commits to B or C the dog changes directions to also go there? But the squirrel then may goto B until and when the dog starts that direction, head for C. This seems to get overly complicated, and we would need more info on what the squirrel will "key on" for changing direction</span hide>

    I know that:
    <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">If the squirrel is 15 m from B, and the dog heads for B, he would tie the squirrel to B, But the squirrel could get to C long before the dog.
    If the squirrel is about 0.988 m from B, and the dog heads for C, he would tie the squirrel to C, But the squirrel could get to B long before the dog.

    If the squirrel is about 7.994 m from B, he has about 2.33 s safety margin when going to B or C. At this point the safety margin for either way is the same. But in this instance, even if the squirrel waited longer than the safety margin, he just had to look to see if the dog were heading for B or for C and go to the opposite. Since the squirrel can always wait for the dog always to commit first, I don't see how he will ever get caught, or what "point" you think the dog is waiting for.

    And, as you mentioned, even if the squirrel heads for one tree, at some point he can reverse and go the opposite direction.</span hide>

    Steve

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    Re: Go Dog!

    Steve,

    My thinking is that the dog will <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold"> not run directly at either tree (for the reasons you mentioned). I am currently not sure if the dog will run directly at the squirrel and then switch to either B or C depending on where the squirrel heads. I think that if the dog was 20m west of the mid point between B and C (ie the same distance from B and C), then it will run directly at R when R reaches the mid point of BC. However, placing the dog closer to B means that the dog will run directly towards a point somewhere between B and R.</span hide>
    (Location Australia, then UK, but now USA. Heart, outlook, attitude, etc always Australian)
    Quote: "All Happiness is the release of internal pressure"

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    Re: Go Dog!

    Well if the dog doesn't plan to intercept and the squirrel is daring, we still need some info on how they plan on moving. My assumption is that they will do their best, but you seem to indicate that is not what will happen...

    Steve

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    Re: Go Dog!

    My current thinking re an answer is ...

    <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">Take tree B, ignore C for the moment. When the squirrel R is T distance from the tree, the squirrel is save unless the dog gets within 4/3T of the tree.
    <IMG SRC=http://www.geocities.com/ruff_hi/woody/squirreldog2.jpg>
    So, if the dog gets within the circle then it can catch the squirrel. You can also draw a circle around C with a radius that can be expressed in terms of T.
    <IMG SRC=http://www.geocities.com/ruff_hi/woody/squirreldog3.jpg>
    So, if the dog gets within the shaded area it will catch the squirrel. So, my thinking is that the dog needs to wait until the squirrel is such that the shaded area is closest to the dog.</span hide>

    Hmmm - it seems that I cannot hide the images.
    (Location Australia, then UK, but now USA. Heart, outlook, attitude, etc always Australian)
    Quote: "All Happiness is the release of internal pressure"

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    Re: Go Dog!

    I don't follow what you are trying to show with the diagrams... Sorry.

    Steve

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