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Slithe
United States985 Posts
Answer: Option #2 is the fastest. Assume that there is a predetermined cycle that A will traverse to visit all possible locations on the planet. If B simply goes in the reverse direction along that cycle, then A and B will always meet. It takes half the time it would normally take if A just went along the path and B stood still. | ||
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kingjames01
Canada1603 Posts
On April 21 2009 14:17 Slithe wrote: My assumption: They have a pre-determined strategy that they will follow in the case that they have to find each other on the planet. Answer: Option #2 is the fastest. Assume that there is a predetermined cycle that A will traverse to visit all possible locations on the planet. If B simply goes in the reverse direction along that cycle, then A and B will always meet. It takes half the time it would normally take if A just went along the path and B stood still. BUT if A and B land in random locations, how does B know which direction A is going to be traversing the surface? | ||
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Wonders
Australia753 Posts
It seems that there's no answer to 3) though. Suppose that A has an optimal path to take. They both have the same amount of information, so B knows this path too. Then B could just take exactly the same path as A, and would never be found because the two are on the same path, and always at different points on that path. But I guess this could be resolved with lots of self-intersection, or an element of randomness in there somewhere, such as starting off in a random direction. | ||
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Deletrious
United States458 Posts
On April 21 2009 14:23 kingjames01 wrote: BUT if A and B land in random locations, how does B know which direction A is going to be traversing the surface? Lol, more importantly we don't just get to add our own assumptions to the problem. I would like A to have a Harley, and A and B just meet at tgi friday's for a blooming onion. | ||
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deathgod6
United States5064 Posts
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meeple
Canada10211 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 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 Although I agree that it can be thought of as a math problem... there is no fastest way... I've said it before, but seriously, there are a finite number of places on this planet... so just think of it like a grid... any method that covers all the ground will take the same time as any other method. Why? because you can check only check say 15 locations per minute, and there are 30 locations... so you will finish in 2 minutes. It doesn't matter what way you check it. Saying checking it like a spiral is faster is like saying if I look into 5 holes starting from the right side first it will finish faster than starting from the left. Lastly, just because you can assign locations with high school math, using polar coordinates, doesn't make it an optimization problem. There is no optimization in the pattern to search if both people are moving, to say there might be is ridiculous... Its a probabilistic problem... one is an exhaustive search, the other, potentially endless... never go with the potentially endless search | ||
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Underwhelmed
United States207 Posts
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kingjames01
Canada1603 Posts
On April 21 2009 14:55 deathgod6 wrote: Get on your radio and say, "Can you hear me now?" lol | ||
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mjh
United States133 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. intuitively, this seems right. also, seems that without knowing which of (1) or (2) is right, option (3) is equivalent to option (1) if (2) is the right answer, or (2) if (1) is the right answer. either way, it's the wrong answer. | ||
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kingjames01
Canada1603 Posts
On April 21 2009 15:06 meeple wrote: Although I agree that it can be thought of as a math problem... there is no fastest way... I've said it before, but seriously, there are a finite number of places on this planet... so just think of it like a grid... any method that covers all the ground will take the same time as any other method. Why? because you can check only check say 15 locations per minute, and there are 30 locations... so you will finish in 2 minutes. It doesn't matter what way you check it. Saying checking it like a spiral is faster is like saying if I look into 5 holes starting from the right side first it will finish faster than starting from the left. Lastly, just because you can assign locations with high school math, using polar coordinates, doesn't make it an optimization problem. There is no optimization in the pattern to search if both people are moving, to say there might be is ridiculous... Its a probabilistic problem... one is an exhaustive search, the other, potentially endless... never go with the potentially endless search Someone should just Monte Carlo this and post the answer. Yes, I'm not disagreeing with you that this is not a probabilistic problem. But just because it is one does not make it one that you can't model with formulae. If you can model this problem which we've shown you can, and you have some constraints, at least one degree of freedom and this is a max/min problem (which it is) with a upper/lower bound then this is indeed an optimization problem for which there is at least one solution. Also, I'm not using high school math or even university level math... If you're familiar with the term POLAR coordinates then you realize that means you're talking about a 2-D circle. I've been describing a 3-D sphere, thus I need spherical coordinates. Note as well that the convention that mathematicians use is to call the polar direction phi and the azimuthal direction theta, whereas I've been calling the polar angle theta and the azimuthal angle phi... | ||
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stenole
Norway868 Posts
Option 1 is an action (or a non-action). Option 2 and 3 are desires. Also there is no clear way that the situation can be modelled mathematically to the degree where it seems much more intuitive to solve this question in a non-mathematical manner. | ||
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jgad
Canada899 Posts
On April 21 2009 11:30 Carnivorous Sheep wrote: This isn't math. Aye. It's not logic either. It's just a ridiculous question with no complete answer. You can calculate the first situation (B stationary) for the full general case. For the others, B's motion is not defined - all you know is that he wants or does not want to find you. If both people don't move, nobody finds each other. If only one moves, they will definitely find each other. If both move, they may or may not find each other. If they can't communicate, they don't know if they should be the one to move or the one to stay still. Regardless of what B wants, he can't know what to do if he doesn't know what A is doing and vice-versa. | ||
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stenole
Norway868 Posts
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meeple
Canada10211 Posts
On April 21 2009 15:28 kingjames01 wrote: Someone should just Monte Carlo this and post the answer. Yes, I'm not disagreeing with you that this is not a probabilistic problem. But just because it is one does not make it one that you can't model with formulae. If you can model this problem which we've shown you can, and you have some constraints, at least one degree of freedom and this is a max/min problem (which it is) with a upper/lower bound then this is indeed an optimization problem for which there is at least one solution. Also, I'm not using high school math or even university level math... If you're familiar with the term POLAR coordinates then you realize that means you're talking about a 2-D circle. I've been describing a 3-D sphere, thus I need spherical coordinates. Note as well that the convention that mathematicians use is to call the polar direction phi and the azimuthal direction theta, whereas I've been calling the polar angle theta and the azimuthal angle phi... Hurray for the difference between 2D and 3D. Can I assume a thorough understanding of non-linear dynamics and chaos? Two people are searching for something that (1) The location is unknown (2) The direction is unknown Just because you can model it with mathematics, doesn't mean there is a solution. However, here's a real problem, one that can be solved... and without any spherical coordinates. At a party, a guest asks the age of the host’s three children. The host tells the guest the following information: Assume each child is a whole number age; The product of their ages is 72; The sum of their ages is the same as my house number. The guest goes outside and looks at the house number. Upon returning, the guest indicates that he still needs additional information. The host then says "Oops, I forgot to tell you that the oldest is the only one that likes vanilla ice cream." You now have all the information to determine the ages of the children. How old is each child? I'll post the answer in a day or so | ||
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Klockan3
Sweden2866 Posts
On April 21 2009 15:06 meeple wrote: Although I agree that it can be thought of as a math problem... there is no fastest way... I've said it before, but seriously, there are a finite number of places on this planet... so just think of it like a grid... any method that covers all the ground will take the same time as any other method. Why? because you can check only check say 15 locations per minute, and there are 30 locations... so you will finish in 2 minutes. It doesn't matter what way you check it. Saying checking it like a spiral is faster is like saying if I look into 5 holes starting from the right side first it will finish faster than starting from the left. You can't take a continuous system and make a discrete model and say that they are the same. The thing is that when you move in a straight line you will cover more ground per time than if you are turning, which means that the optimal solution is the solution where you turn as little as possible. which gives us the smooth spiral, it wastes a bit in the first turn but after that it barely wastes anything at all while any other form wastes a ton at every turn. For question B the answer is that A should still search, even though if both moves it have a potential to go on forever it is on average roughly as fast as if either stands still but you will not have to take the risk that both stays still(Can't really be arsed to calculate how much the chance per time gets increased if both moves, but if it is twice it goes faster and if it is none it goes slower but it is neighter, it goes faster but not twice). For question C therefore is also for A to move. | ||
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stenole
Norway868 Posts
On April 21 2009 21:06 meeple wrote: Spent some time figuring out how the house number thing could help me out...+ Show Spoiler +At a party, a guest asks the age of the host’s three children. The host tells the guest the following information: Assume each child is a whole number age; The product of their ages is 72; The sum of their ages is the same as my house number. The guest goes outside and looks at the house number. Upon returning, the guest indicates that he still needs additional information. The host then says "Oops, I forgot to tell you that the oldest is the only one that likes vanilla ice cream." You now have all the information to determine the ages of the children. How old is each child? 72 = 2*2*2*3*3 Possible combinations: 2,2,18 = 22 2,3,12 = 17 2,6,6 = 14 2,4,9 = 15 3,4,6 = 13 3,3,8 = 14 We can assume the guest would have been able to figure it out after seeing the house number - unless more of the age combinations add up to the same house number. Because there is an oldest child, the two others have to be younger. Ergo, the answer is 3, 3 and 8. | ||
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Chromyne
Canada561 Posts
On April 21 2009 22:27 stenole wrote: Spent some time figuring out how the house number thing could help me out...+ Show Spoiler + 72 = 2*2*2*3*3 Possible combinations: 2,2,18 = 22 2,3,12 = 17 2,6,6 = 14 2,4,9 = 15 3,4,6 = 13 3,3,8 = 14 We can assume the guest would have been able to figure it out after seeing the house number - unless more of the age combinations add up to the same house number. Because there is an oldest child, the two others have to be younger. Ergo, the answer is 3, 3 and 8. stenole beat me to it. I Got the same answer. | ||
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stenole
Norway868 Posts
On April 21 2009 22:37 Chromyne wrote: stenole beat me to it. I Got the same answer. What is my prize? There has to be a win the internet prize. The difference between this riddle/puzzle and the one in the OP is that this one is carefully worded, it tells you that you have enough information to solve it, and it practically strings you along to a single answer by stating to use whole numbers, and giving you three different clues, all of which you know need to give you some sort of info. Here is another unsolvable problem: Assume you have 2 herds of cows that have accidentally joined into one herd, randomly dispersed over a circular area. There are two herder teams, A and B. A tries to herd as many of his cows west of the oval. What is the fastest way for herding team B to get all their cows to the east: A) B sits and watches A herd their cows westward. B) B wants to herd cows towards the east. C) B likes A's cows. | ||
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meeple
Canada10211 Posts
On April 21 2009 23:15 stenole wrote: What is my prize? There has to be a win the internet prize. The difference between this riddle/puzzle and the one in the OP is that this one is carefully worded, it tells you that you have enough information to solve it, and it practically strings you along to a single answer by stating to use whole numbers, and giving you three different clues, all of which you know need to give you some sort of info. Here is another unsolvable problem: Assume you have 2 herds of cows that have accidentally joined into one herd, randomly dispersed over a circular area. There are two herder teams, A and B. A tries to herd as many of his cows west of the oval. What is the fastest way for herding team B to get all their cows to the east: A) B sits and watches A herd their cows westward. B) B wants to herd cows towards the east. C) B likes A's cows. zizi yo... here's another in the same vein, Find positive integers a1, a2, ... , an whose sum is 100 and whose product a1a2 ... an is maximal over every possible choice of n and every possible choice of a1, a2, ... , an. Now this one is a optimization problem. | ||
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kingjames01
Canada1603 Posts
I'll work on the integers one. Here's one I just remembered when I was on my way home. A shipwreck strands 10 people on a desert island whose surrounding waters are filled with the world's most deadliest man-eating sharks. These 10 people think using perfect logic. On the island they find a treasure chest filled with a 1000 gold pieces. They decide to divide the gold. Here is the process that they come up with. They each drew straws to figure out the order in which they will attempt to divide the gold. Whoever goes first can share the gold however that person chooses. The catch is that after the person has finished sharing the gold there is a vote. If the majority (greater than but not equal to 50%) of the people are not happy they can vote to have that person tossed off the island. It would then fall onto the person who was selected to be number 2 to attempt to share the gold, and so on. Now the conditions are: 1) These people will choose life over death. 2) These people will choose more gold than equal or lesser amounts. 3) These people are perfect logicians. Question: Can person one divide the gold in such a way that he survives the vote? If so, what is the maximum amount of gold that he can get? If not, show why. | ||
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