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Topic: TOPOLOGICAL RINGS!! (Read 1266 times) |
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vaconas
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TOPOLOGICAL RINGS!!
« on: Mar 29th, 2005, 6:26am » |
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Hi guys!!! i just wanted to know this: has anyone ever solved topological rings problem?? i am doing it for days with 3 elastics atached but i really cant do it....im starting to think it is impossible!! i dont want the solution, i just want to be sure its possible...ok?
greetings to all
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vaconas
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Re: TOPOLOGICAL RINGS!!
« Reply #2 on: Mar 29th, 2005, 7:02am » |
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hey ty for your answer but my english is not very good so here it goes: u mean that with rubber bands is impossible to solve or just harder?? silly putty or clay?? u mean like plasticine?
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towr
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Re: TOPOLOGICAL RINGS!!
« Reply #3 on: Mar 29th, 2005, 7:34am » |
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Yes, like plasticine. (If I can trust the dictionary on what iot is  )
And it is indeed impossible with rubber bands.
(At least the topological solution. If you somehow are able to move objects through a fourth dimension it might work  )
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vaconas
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Re: TOPOLOGICAL RINGS!!
« Reply #4 on: Mar 29th, 2005, 7:46am » |
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DUDE!!! no offense but...i have just shreded around 15 rubber bands in last days...chesus!!  and thats very disapointing that this has no solution..i mean with plasticine its ridiculously easy..even a 4 years boy can solve it....
just one more thing: r u sure the puzzle is clear? i mean in my country language elastic=rubber band= elástico....maybe its not clear for people bad in english  my opinion is that you should give a objective example of the material to be used ( clay, plasticine, etc).
greetings once more and ty for all your help
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towr
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Re: TOPOLOGICAL RINGS!!
« Reply #5 on: Mar 29th, 2005, 8:00am » |
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Well, the solution isn't as trivial as cutting/breaking one of the rings than pushing it back together when they're unlinked, if that's what you're thinking.
And 'elastic' in this case refers to the property of being elastic, i.e. you can stretch and squish it.
What it comes down to is that topologically any object with two holes is equivalent to any other object with two holes.
If you want to make the puzzle really hard, formalize it mathematically, and prove it.
Just proving a square is topologically equivalent to a disk is hard enough for me.. (too hard even, probably )
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iyerkri
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Re: TOPOLOGICAL RINGS!!
« Reply #6 on: Mar 29th, 2005, 8:26am » |
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f(x,y) = f(rcosu, rsinu) = (r,u).
(x,y) \in unit disk centred at origin.
(r,u) \in a rectangle (0,0) X (1,2pi)
f is one to one & onto.
isn't this enough to prove disk & square are topologically equivalent?...please correct me if i am wrong.
~kris
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towr
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Re: TOPOLOGICAL RINGS!!
« Reply #7 on: Mar 29th, 2005, 8:58am » |
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on Mar 29th, 2005, 8:26am, kris wrote:| isn't this enough to prove disk & square are topologically equivalent?...please correct me if i am wrong. |
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Well, all points (0, u) map to just one point. So it's not a one to one function.
Aside from that, I think it's possible to make a bijective (one to one & onto) function from a planar object with two holes, to one with one hole, which aren't topologically equivalent.
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towr
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Re: TOPOLOGICAL RINGS!!
« Reply #8 on: Mar 29th, 2005, 9:38am » |
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I think the following transformation might work to stretch a unit circle to a 2x2 square centered on the origin.
(x,y) -> 1/cos(u) (x,y) if -pi/8 <= u < pi/8
(x,y) -> 1/sin(u) (x,y) if pi/8 <= u < 3pi/8
(x,y) -> -1/cos(u) (x,y) if 3pi/8 <= u < 5pi/8
(x,y) -> -1/sin(u) (x,y) if 5pi/8 <= u < 7pi/8
It's bijective. But more important is that it's also a continuous deformation.
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iyerkri
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Re: TOPOLOGICAL RINGS!!
« Reply #9 on: Mar 29th, 2005, 11:28am » |
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yeah..missed that point...its not bijective.
another example i found is an annulus and a square (need not be closed) , for which there exist bijective mapping, but topological equivalence is not present.
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vaconas
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Re: TOPOLOGICAL RINGS!!
« Reply #10 on: Mar 29th, 2005, 11:54am » |
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GUYS!!! WOW THAT IS COMPLETELY OVER MY HEAD!!! STOP IT OR U WILL MAKE ME HATE MATHS!! lol
towr that solution of breaking it was not my solution, however my solution is not the "popular" one..it envolves mixing the plasticine (or should i say make a fusion..) but i guess the main idea is the same
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Icarus
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Re: TOPOLOGICAL RINGS!!
« Reply #11 on: Mar 29th, 2005, 5:33pm » |
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vaconas - first of all, the word "elastic" has to be understood here in a topological sense. The key here is that you can do anything that does not suddenly separate points that were close together.
In particular, if at any stage in your process, you draw any curve on your surface, at all later stages, it should still be a curve. So for instance, tearing or cutting is not allowed. For if you draw a curve across where the cut goes, then cutting will turn your single curve into two separate curves. Even cutting, then restoring, is not allowed, as the curve needs to remain at all times, not just at the beginning and end.
Your solution is only half-acceptable, because while it does satisfy the condition that curves remain curves, it cannot be reversed while satisfying the same conditions. There are some topological problems of this type allow this, and even require it to solve. But it is far preferable to have a method that is reversable, as the solution given in the other thread is.
kris, beside the bijectivity failure, your map is not continuous along the x-axis. One side of the axis is mapped to the left side of the rectangle, while the other side is mapped to the right side of the rectangle.
To show topological equivalence, you have to come up with a homeomorphism: a bijection f such that both f and f-1 are continuous at every point.
The standard way to show equivalence between any bounded star-like region A and the circle, let x be a central point of A. (I.e. if y is any other point of A, then the line segment from x to y lies entirely inside A. A is "star-like" if such a point x exists.)
For each point y in A other than x, the line L through x and y must intersect the boundary of A in exactly two points r, s, with r <= x < y <= s, by one of the two orderings of the line. Define g(y) = s. g is continuous. Define f(y) to be the unique point on L such that f(y) lies on the same side of x as y does, and the distance d(x, f(y)) between x and f(y) is given by d(x, f(y)) = d(x, y)/d(x, g(y)). Also define f(x) = x.
Then f is a homeomorphism between A and the unit circle centered at x.
If you apply this to the square of side length 2 centered at the origin, then you will find that f is exactly the map that towr defined above.
More generally, you can show that any two simply connected domains in the plane are topologically equivalent, regardless of their shape. (Simply connected means that any loop lying in the region can be shrunk to a point without ever leaving the region - i.e., there are no holes in it.)
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