Not friends on the earth
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Not friends on the earth
For the purpose of these puzzles, the Earth is a uniform, smooth, sphere of circumference 40,000km.
Easy to start:
a. Two people who dislike each other very much stand at the North Pole. One moves to be as far away from the other person as they can. How far is that?
Now, the real problems:
b. Three people who each dislike the other two stand at the North Pole. Two move away so that all three of them are as far away from the other two as they can be (i.e. the three distances are equal and maximal). How far is that?
c. Four people who each dislike the other three stand at the North Pole. Three move away so that all four of them are as far away from the other three as they can be (i.e. the six distances are equal and maximal). How far is that?
Easy to start:
a. Two people who dislike each other very much stand at the North Pole. One moves to be as far away from the other person as they can. How far is that?
Now, the real problems:
b. Three people who each dislike the other two stand at the North Pole. Two move away so that all three of them are as far away from the other two as they can be (i.e. the three distances are equal and maximal). How far is that?
c. Four people who each dislike the other three stand at the North Pole. Three move away so that all four of them are as far away from the other three as they can be (i.e. the six distances are equal and maximal). How far is that?
Re: Not friends on the earth
To clarify a possible misreading: everyone stays on the surface of the earth at all times.

 ThinkGravyTrain
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Re: Not friends on the earth
Out of curiosity, does this make sense if of the group with three people who dislike the original first two people, one of them intentionally stays at the north pole when there's someone from the first group there? Because that's the scenario painted by what you said. It doesn't create a situation of being maximally away from the first two people. Just saying.
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 phlip
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Re: Not friends on the earth
I think they meant that each one of the group of three hates the other two people from that same group of three... the group being independent of the group of two in the first part.
Code: Select all
enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};
void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}

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Re: Not friends on the earth
As far as I know, it's the same idea as molecular shapes.
http://www.sparknotes.com/testprep/book ... ion8.rhtml
http://www.sparknotes.com/testprep/book ... ion8.rhtml
Re: Not friends on the earth
I agree with this and I don't find this puzzle very interesting. But I am a little curious to know the answer to the next extrapolated case, the configuration of 5 enemies on the sphere.Qaanol wrote:As I read itSpoiler:
Spoiler:
 phlip
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Re: Not friends on the earth
Ralp wrote:Spoiler:
Spoiler:
Code: Select all
enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};
void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}
Re: Not friends on the earth
Another variant would be, “Given N equal classical pointcharges, how can they be arranged on a conducting sphere (or other shape—right circular cylinder anyone?) in a (stable?) equilibrium?”
wee free kings
 jestingrabbit
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Re: Not friends on the earth
Qaanol wrote:Another variant would be, “Given N equal classical pointcharges, how can they be arranged on a conducting sphere (or other shape—right circular cylinder anyone?) in a (stable?) equilibrium?”
This is very nearly the thomson problem (there the condition is for least energy, whereas you state it as an equilibrium, and there could easily be local equilibria, especially for large numbers of charges).
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Re: Not friends on the earth
I find the 8 node case the most interesting, since the answer is not what most people's first instinct says it is.
Spoiler:
Spoiler:
Re: Not friends on the earth
jaap wrote:I find the 8 node case the most interesting, since the answer is not what most people's first instinct says it is.Spoiler:Spoiler:
Excellent observation.
wee free kings
Re: Not friends on the earth
Ralp.....The 2, 3, 4 solutions may not be very interesting, but I do not think there are any solutions for 5 and above.
The original puzzle asks for the distances to be maximal and equal. Your suggested 5 solution has four people around the equator. They are not all the same distance from each other, even though they are all the same distance from the North Pole person.
I am not averse to people relaxing the conditions for a more interesting puzzle; just want to ensure that the original puzzle is not forgotten in the switch to a wider case. Thanks everyone for playing, whichever puzzle you are solving
The original puzzle asks for the distances to be maximal and equal. Your suggested 5 solution has four people around the equator. They are not all the same distance from each other, even though they are all the same distance from the North Pole person.
I am not averse to people relaxing the conditions for a more interesting puzzle; just want to ensure that the original puzzle is not forgotten in the switch to a wider case. Thanks everyone for playing, whichever puzzle you are solving
Re: Not friends on the earth
Like ARandomDude said, this simply corresponds to the arrangements of atoms in a molecule. They are distributed exactly according to this principle, so that the total distance between each pair is maximized.
3 Corresponds to sp2 hybridization, linear so they are 120 degrees apart on the equator.
5 corresponds to sp3 , so they form a tetrahedron, with one of the vertices at one of the poles.
That also explains why 8 corresponds to the square antiprism.
Equal distances are possible if there is a hybridization scheme which is symmetric.
3 Corresponds to sp2 hybridization, linear so they are 120 degrees apart on the equator.
5 corresponds to sp3 , so they form a tetrahedron, with one of the vertices at one of the poles.
That also explains why 8 corresponds to the square antiprism.
Equal distances are possible if there is a hybridization scheme which is symmetric.
Re: Not friends on the earth
can some of the distances decrease temporarily at some moment during the moving? In other words, only the ending positions are evaluated?
Re: Not friends on the earth
Thanks for asking for clarification.
My unstated assumptions in the original question were for
My unstated assumptions in the original question were for
 final distance
 as measured on the surface (ie no tunnelling with tape measures)
Re: Not friends on the earth
So then I guess a better way would be saying "Distribute N points on a sphere surface..." rather than having them all start at the north pole
Re: Not friends on the earth
charonme wrote:So then I guess a better way would be saying "Distribute N points on a sphere surface..." rather than having them all start at the north pole
Puzzles are often characterized by being expressed in a narrative form, and wrapped in a story. Some can be recast into a formal mathematical form as a first stage in solving.
What is perhaps interesting about this puzzle is that it has now had nearly two weeks of being reformulated in various ways, but no solutions. So, just possibly, recasting some puzzles into a formal mathematical form is a dead end that is not conducive to solutions.
 jestingrabbit
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Re: Not friends on the earth
bair, I think people were just holding off from solving because it was pretty straightforward.
Just to be clear,
Just to be clear,
Spoiler:
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
Re: Not friends on the earth
balr wrote:ie no tunnelling with tape measures.
The great arch distance and the tunneling distance depends monotonely on each other, so this does not change the puzzle.
charonme wrote:can some of the distances decrease temporarily at some moment during the moving?
For the simple cases asked in the OP it does not make a difference. I am unsure if it ever does. If so what is the lowest number for which we can get a stable equilibrium that is not optimal?
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