Nerdy Math Problem

On April 21 2009 11:09 Byo wrote:
they are variables.......... ie. under 1) the answer is for A to move in a spiral motion so it takes the least amount of time to cover the surface of the sphere.

Under the following assumptions, what is the quickest way in which person A should find person B

3) person B does not want person A to find him.

On April 21 2009 11:13 Divinek wrote:


so shouldnt it be impossible if B is trying to hide from A?

I mean yeah I guess he's standing still, but what is point of the third option then if he cant hide?

On April 21 2009 11:13 travis wrote:



ah so these problems are to be looked at mathematically then ?


It would seem to me that 1, 2, and 3 all have the same answer.

On April 21 2009 11:13 Divinek wrote:


so shouldnt it be impossible if B is trying to hide from A?

I mean yeah I guess he's standing still, but what is point of the third option then if he cant hide?

On April 21 2009 11:20 sixghost wrote:

I think the difference in the 3rd problem is that you are supposed to assume that person B will move in a pattern that is not optimal for the two people to meet.

On April 21 2009 12:50 shavingcream66 wrote:
no

On April 21 2009 11:30 Carnivorous Sheep wrote:
This isn't math.

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.

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.

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.

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.

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"

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.

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

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.

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?

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

On April 21 2009 14:55 deathgod6 wrote:
Get on your radio and say, "Can you hear me now?"

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.

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

On April 21 2009 11:30 Carnivorous Sheep wrote:
This isn't math.

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...

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.

On April 21 2009 22:27 stenole wrote:
Spent some time figuring out how the house number thing could help me out...+ Show Spoiler +

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.

On April 22 2009 04:54 kingjames01 wrote:
Ah, I just got home and saw these. I got the same answer as above too.
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.

On April 22 2009 08:18 Monoxide wrote:


Here is my answer
+ Show Spoiler +

On April 22 2009 04:02 meeple wrote:


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.

On April 21 2009 15:28 kingjames01 wrote: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.

On April 22 2009 12:09 ninjafetus wrote:


I know we're past this, and I understand the point you're making, but I'd like to make a small correction. Yes, 'polar' tends to refer to 2d more often and 'spherical' is specifically for 3d, but for we mathematicians out there (who don't need to have pesky physical situations to justify our work) there is a generalization to n-dimensional polar coordinates. Spherical is just the 3 dimensional case.