EPTNerdy Math Problem - Page 2
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techn1cal
United States68 Posts
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kingjames01
Canada1603 Posts
If in case 1, person A knew person B was standing still, then you have to think about how person A can exhaustively cover the area of the sphere in the best possible way. As someone above has said, you can think of this like a grid covering the spherical surface. Anyway, you don't want to do it in a thoughtless way because then you'll spend time crossing your path. The answer in this case is to spiral out from your starting position. The math comes in if you consider A's initial position in spherical coordinates, being (r, theta, phi) = (R, 0, 0). Then you want to cover, ie. "integrate" over the surface, so you move by d_theta and then walk 2 * pi around phi. Repeat til you find person B. In other words, spiral out. Case 2) and 3) I haven't thought about yet. | ||
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kingjames01
Canada1603 Posts
1) staying still 2) actively wanting to be found 3) actively not wanting to be found what is the optimal search pattern that A should take. I think A doesn't know where B landed for case 1) because the answer is to walk in a straight line over the spherical surface (ie. along d_theta, see my above post) and that solution is too easy for this question. | ||
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shavingcream66
United States1219 Posts
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Malongo
Chile3479 Posts
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kingjames01
Canada1603 Posts
On April 21 2009 12:50 shavingcream66 wrote: no Why do you say no? Do you have any meaningful input that you're not sharing? Anyway, I put more thought into this and I have some things to add which actually serve to stregthen my answer for case 1). I said above that A's pattern for searching the surface should be to spiral outwards and cover the planet. (See above post.) Consider this: Person A and B both landed in random spots. For ease of illustration and in no way introduces a loss of generality, consider A's position to be the North Pole. ie. (r, theta, phi) = (R, 0, 0). Now, B could have landed anywhere. You might have objections because if B landed at the South Pole, then the spiral will be the path which takes the longest to cross the position of B, (r, theta, phi) = (R, pi, 0). However, since B's landing position is random and I would assume the probability is uniformly distributed then the probability for B to have landed near the South Pole (which leads to the longest search times) is much smaller than the probability for B to have landed near the Equator. This is due to the solid angle which is much larger near the Equator. Of course the probability that B landed very close to A is the same as the probability to have landed at the South Pole (very small). However, A can't just go search the Equator without looking near A's starting position since A would then have to backtrack if B wasn't found thus wasting time. To summarize, if we call A's starting position (R, 0, 0) then probability of B to have crashed at a position (R, theta, phi) is proportion to sin(theta). The best way to find B if B's position does not change with time is to spiral outwards. | ||
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404.Delirium
United States1190 Posts
On April 21 2009 11:30 Carnivorous Sheep wrote: This isn't math. [/QUOTE]ED FOR TRUTH This question just makes me angry. Any answer just uses assumed variables. | ||
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deL
Australia5540 Posts
Or both keep going around the circumference of the planet, passing through both poles and at different speeds and they have to meet up where their paths intersect at some point. Probably not very fast I guess xD | ||
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Deletrious
United States458 Posts
1) person B stands still 2) person B wants to find person A as well 3) person B does not want person A to find him. As said, sounds more like a straight logic problem then a math problem. 1. Move in such a way as to maximize your field of vision and never recover old ground. 2. Problem here is that they both want to find each other, but can't communicate who stands still. I think it resolves as a prisoner dilemma. If A stands still and B searches, that is best result. But if A stands still and so does B, that is worst. If A moves and B moves as well, that is middle between good and best, may keep missing each other but have a chance to find each other, unlike when both stand still. If A moves and B stands still, that is best again. Since they both face the same choice, and moving guarantees avoiding the worst scenario, A should move. It would be best if one stood still, but not being able to communicate, neither should take the risk of both standing still. 3. Prisoner dilemma again. Except I think A can consider B's analysis in the prisoner dilemma as now they face different dilemmas. In B's prisoner dilemma, not wanting to be found, he can move or stay still. If he stays still and A moves, he will be found, the worst scenario. If he stays and A stays, best scenario. If he moves and A stays, he be found, worse scenario again. If he moves and A moves, middle of the road result. Given the four possibilities, it looks best to B to stay put. Knowing this, A should definitely move as that is his best result. Even if B does not do what is best for him and moves, A still has his middle of the road result of both moving. I think A can't reasonably bet B moves. A should move. I believe the answer to all three is for A to move in his search pattern | ||
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spydernoob
Canada1066 Posts
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kingjames01
Canada1603 Posts
Just grab a round object. Let's assume you're person A. It doesn't matter where you land. Pick any place you want to start. Since the sphere is perfectly symmetric you are free to choose your coordinate system. Rotate the object so that the place you chose to start from is at the top. That makes your landing position ( r( t = 0 ), theta( t = 0 ), phi( t = 0 ) )_A = (R, 0, 0), where t represents time and t = 0 represents the time of the crash. You haven't assumed anything at all. Next, person B could have landed anywhere on the round surface with equal probability. You have no idea where but you do know that he's NOT moving. The math comes here at this step. His position in time is described by: ( r( t ), theta( t ), phi( t ) )_B = ( r( t = 0 ), theta( t = 0 ), phi( t = 0 ) )_B = ( R, theta_B0, phi_B0 ), ie constant. The problem ends when ( r( t ), theta( t ), phi( t ) )_A = ( r( t ), theta( t ), phi( t ) )_B. This is mathematics. Indeed, this is an optimization problem. Minimize: t such that B is not moving and the positions are randomly distributed on a spherical surface. OF COURSE this is mathematics! | ||
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ieatkids5
United States4628 Posts
There is a sphere. There are 2 points on the surface of the sphere, A and B. The starting locations of A and B are completely random. 1. If B does not move, what is the shortest path that should be taken by A to intersect B? Giving this just a quick thought, I'd say A should move in a spiral motion on the surface of the sphere, ending with the point on the sphere opposite of where A started. This type of movement never overlaps itself, so no distance is wasted. 2. A and B are both allowed to move. What are the shortest paths taken by both such that A intersects B? Same answer as question 1 imo. A moves in a spiral motion toward to opposite point of where A started. B stays still. This is because if B moves at all, B may move into an area that has already been covered by A's spiral movement, and then A will not find B during the first spiral run. If B doesn't move, then A will eventually intersect B. However, if B happens to move toward A during A's spiral movement, then A will intersect B with in a shorter amount of time. The chance of this happening, I think, is lower than the chance of B moving into an area already covered by A, or is lower than the chance of B moving away from A but eventually intersecting. 3. A and B are allowed to move. A should move in a way that intersects B in the shortest distance. B should move in a way that prevents intersection with B. This is harder, since it seems like A's movement depends on what B is doing, and vice versa. Hmmmmmm.... If A does the same spiral movement, if B stays still, A will intersect B some time. B should move, in a straight line, curved, spiral, I dunno. If A moves spirally, and B moves too (dont know what kind of movement here is best for avoiding A, since A and B don't know each other's location [otherwise this would be pointless]) then the chance of B avoiding A is greater than A intersecting B. Therefore, B should move to avoid A. If A knows that B will move to counter A's spiral movement, then A will have to find another way of moving that has a higher chance of finding B. If there is one. I dont know. If A moves randomly, will this have a higher chance of intersecting B than if A moved in a spiral when B is also moving? If A moves randomly, how should B move then? I think this is the kind of stuff the OP was looking for. edit - seems like i took a bit too long typing this up lol edit2 - i feel dumb cuz the highest math ive taken is ap calc bc (im a highschool senior) and people are using crazy math lingo lol | ||
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kingjames01
Canada1603 Posts
On April 21 2009 13:39 Deletrious wrote: Under the following assumptions, what is the quickest way in which person A should find person B 1) person B stands still 2) person B wants to find person A as well 3) person B does not want person A to find him. As said, sounds more like a straight logic problem then a math problem. 1. Move in such a way as to maximize your field of vision and never recover old ground. I agree with your answer for case 1) exactly. BUT if person A comes and asks you what is the path that he takes in order to achieve this suggestion, what would you have to say? Can you prove that your suggestion will result in the shortest search time? That's where the mathematics comes in! In this case you can describe the path and even prove that it will give the best search time. Also, Mathematics is a language that we use when we want to talk logic. | ||
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Deleted User 31060
3788 Posts
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Deletrious
United States458 Posts
On April 21 2009 13:48 ieatkids5 wrote: 2. A and B are both allowed to move. What are the shortest paths taken by both such that A intersects B? Same answer as question 1 imo. A moves in a spiral motion toward to opposite point of where A started. B stays still. This is because if B moves at all, B may move into an area that has already been covered by A's spiral movement, and then A will not find B during the first spiral run. If B doesn't move, then A will eventually intersect B. However, if B happens to move toward A during A's spiral movement, then A will intersect B with in a shorter amount of time. The chance of this happening, I think, is lower than the chance of B moving into an area already covered by A, or is lower than the chance of B moving away from A but eventually intersecting. You assume here that A can make B stand still, but we aren't told he has any ability to do that. The question is what should A do not knowing what B does. by my analysis A should still move, see above. | ||
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seppolevne
Canada1681 Posts
On April 21 2009 11:26 meeple wrote: 1) A quick exhaustive search of the sphere is the fastest method, If you imagine the sphere to be composed of a grid, there are a finite number of points to search and no best way to do it. Essentially its like asking whats the quickest way of looking in five holes... just look dammit. 2) The fastest way to find a person is if they stand still. So see the above answer. The reason for this is because this method is guaranteed to end, there are a finite number of places on the sphere and if they stay in one, they will eventually be found, which is why this is the method that search ad rescue crews tell people. 3) Well, you only have two options, you can either stay still or move around pretty randomly, and since staying still means you will eventually be found, then keep moving, and if you're really lucky, you can always avoid the person, but generally speaking you will avoid them for much longer than staying in the same spot. this and yes this is math, not "ooo well if B hides behind a tree then etc etc" | ||
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Deletrious
United States458 Posts
On April 21 2009 13:49 kingjames01 wrote: I agree with your answer for case 1) exactly. BUT if person A comes and asks you what is the path that he takes in order to achieve this suggestion, what would you have to say? Can you prove that your suggestion will result in the shortest search time? That's where the mathematics comes in! In this case you can describe the path and even prove that it will give the best search time. Also, Mathematics is a language that we use when we want to talk logic. Any path that covers the whole planet without overlapping the field of vision would suffice. An expanding spiral out from your current location satisfies this, since something such as walking pole to pole would result in some overlap on reaching the pole each time. I didn't really need a mathematical formula for that. Not that I am disagreeing you, and I certainly appreciate the skill to represent problems in mathematics. | ||
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Deletrious
United States458 Posts
On April 21 2009 13:59 seppolevne wrote: this and yes this is math, not "ooo well if B hides behind a tree then etc etc" Again, in part 2. we aren't told in any assumption that A can tell or otherwise make B stand still. | ||
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seppolevne
Canada1681 Posts
On April 21 2009 14:03 Deletrious wrote: Again, in part 2. we aren't told in any assumption that A can tell or otherwise make B stand still. but A never changes his pattern, so its dependent on B | ||
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kingjames01
Canada1603 Posts
On April 21 2009 13:48 ieatkids5 wrote: You guys aren't thinking about this problem right. It IS a math problem. There is a sphere. There are 2 points on the surface of the sphere, A and B. The starting locations of A and B are completely random. 1. If B does not move, what is the shortest path that should be taken by A to intersect B? Giving this just a quick thought, I'd say A should move in a spiral motion on the surface of the sphere, ending with the point on the sphere opposite of where A started. This type of movement never overlaps itself, so no distance is wasted. 2. A and B are both allowed to move. What are the shortest paths taken by both such that A intersects B? Same answer as question 1 imo. A moves in a spiral motion toward to opposite point of where A started. B stays still. This is because if B moves at all, B may move into an area that has already been covered by A's spiral movement, and then A will not find B during the first spiral run. If B doesn't move, then A will eventually intersect B. However, if B happens to move toward A during A's spiral movement, then A will intersect B with in a shorter amount of time. The chance of this happening, I think, is lower than the chance of B moving into an area already covered by A, or is lower than the chance of B moving away from A but eventually intersecting. 3. A and B are allowed to move. A should move in a way that intersects B in the shortest distance. B should move in a way that prevents intersection with B. This is harder, since it seems like A's movement depends on what B is doing, and vice versa. Hmmmmmm.... If A does the same spiral movement, if B stays still, A will intersect B some time. B should move, in a straight line, curved, spiral, I dunno. If A moves spirally, and B moves too (dont know what kind of movement here is best for avoiding A, since A and B don't know each other's location [otherwise this would be pointless]) then the chance of B avoiding A is greater than A intersecting B. Therefore, B should move to avoid A. If A knows that B will move to counter A's spiral movement, then A will have to find another way of moving that has a higher chance of finding B. If there is one. I dont know. If A moves randomly, will this have a higher chance of intersecting B than if A moved in a spiral when B is also moving? If A moves randomly, how should B move then? I think this is the kind of stuff the OP was looking for. edit - seems like i took a bit too long typing this up lol Yes! Finally, a supporter!!  Let's continue this discussion! Case 1) I think that it's important to note that since the positions are random A should try to cover the territory nearest himself and then the Equator before moving to the other hemisphere (using a spiral like you said) since that covers the greatest solid angle in the shortest amount of time. Consider the following for Case 2) A lands somewhere, we'll call that the North Pole. B lands exactly on the Equator, and for ease of discussion we'll use the spherical coordinates: theta = pi / 2 and phi = 0. Now, B has no idea where A is so if B takes a single step in any direction, B can go either towards or away from A. This direction can be any direction ex. (due North = decreasing theta, constant phi OR exactly NE = decreasing theta, increasing phi and so on). B can go in 2 * pi directions. If B only takes one step: into the NORTHERN hemisphere ie. decreasing theta, and then stands still, the search time is less as compared to not moving at all. into the SOUTHERN hemisphere ie. increasing theta, and then stands still, the search time is increased as compared to not moving at all. ALONG the Equator, the search time is not changed. With only one step the results seem balanced when weighted by percentage. Since only one step was taken, A will still always find B UNLESS that one in a million chance case occurs: B lands right next to A and takes one step towards A while A starts moving away. In this case, A won't find B in the first pass but this probability for weighting this case is vanishingly small since it is proportional to sin(theta) * d_theta and theta is approximately equal to 0 in this case. | ||
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