D**u
发帖数: 204 | 1 Question:
Can you partition a 3-dim Euclidean space R^3 into a union of a set of
circles. Namely, for every point A in R^3, A is on 1 and only 1 circle from
the set. | a*******h
发帖数: 123 | 2 R^3 = R * R^2。 再把 R^2 分成无数个同心圆。
from
【在 D**u 的大作中提到】
: Question:
: Can you partition a 3-dim Euclidean space R^3 into a union of a set of
: circles. Namely, for every point A in R^3, A is on 1 and only 1 circle from
: the set.
| D**u
发帖数: 204 | 3 Then how to deal with the points that are centers of those circles?
【在 a*******h 的大作中提到】
: R^3 = R * R^2。 再把 R^2 分成无数个同心圆。
:
: from
| J****g
发帖数: 103 | 4 aftermath的意思是不是就是： 把一个球切一刀， 切出来的面是一个disk， 然后那个
disk呢， 就是由无数个一圈一圈的同心的circle组成的.
中心轴我不知道怎么处理。。。
【在 D**u 的大作中提到】
: Then how to deal with the points that are centers of those circles?
| p*****k
发帖数: 318 | 5 not sure what exactly DuGu had in mind, but im guessing that
he wants the radii of these circles nonzero and bounded.
otherwise, e.g., the infinite str8 line R could be considered
as a circle with infinite radius:
(x-R)^2+y^2=R^2 with R->infty | D**u
发帖数: 204 | 6 yes, the "circle" here follows the traditional definition which does not
include straight lines.
【在 p*****k 的大作中提到】
: not sure what exactly DuGu had in mind, but im guessing that
: he wants the radii of these circles nonzero and bounded.
: otherwise, e.g., the infinite str8 line R could be considered
: as a circle with infinite radius:
: (x-R)^2+y^2=R^2 with R->infty
| A****s
发帖数: 129 | 7 Consider any two points A and B on a sphere, which are symmetric w.r.t.
some great circle of this sphere. So there are two hemispheres. And I think
it could be shown that there always exists a family of cicrles parallelling
the great circle which covers one of the hemispheres without the given point
by some analytical geometry.So it's concluded that any sphere without two
points could be covered by a family of disjoint circles.
Now we just take the family of spheres centered at origin. Construct
【在 D**u 的大作中提到】
: Question:
: Can you partition a 3-dim Euclidean space R^3 into a union of a set of
: circles. Namely, for every point A in R^3, A is on 1 and only 1 circle from
: the set.
| J**********g
发帖数: 213 | 8 I am not sure of what kind of problem I should regard it at, I mean, a
brainteaser, a math one, or whatever.
Problem reduction.
1. R^3=R*R^2 can be dealed in Dugu's way. but you don't know how to
deal with the axis/center line, ie, how would map a line into a set of
separate circles.
or
2. R^3=R^2*R, ie, If a line could be mapped into a set of circles,
then so is R^3: R^3 is the set of lines indexed by (a,b), ie,
R^3={(a,b,x) with x varies in R, ie, the real line | (a, b) belong to R^2}
It's obv | p*****k
发帖数: 318 | 9
Allens, this is not apparent to me. could you clarify a bit more? thx.
【在 A****s 的大作中提到】
: Consider any two points A and B on a sphere, which are symmetric w.r.t.
: some great circle of this sphere. So there are two hemispheres. And I think
: it could be shown that there always exists a family of cicrles parallelling
: the great circle which covers one of the hemispheres without the given point
: by some analytical geometry.So it's concluded that any sphere without two
: points could be covered by a family of disjoint circles.
: Now we just take the family of spheres centered at origin. Construct
| J**********g
发帖数: 213 | 10 There is one point you don't know how to deal with: the origin. Actually
that's why it's hard to get an relatively explicit way to map R^3 into
circles......
think
parallelling
point
another
【在 A****s 的大作中提到】
: Consider any two points A and B on a sphere, which are symmetric w.r.t.
: some great circle of this sphere. So there are two hemispheres. And I think
: it could be shown that there always exists a family of cicrles parallelling
: the great circle which covers one of the hemispheres without the given point
: by some analytical geometry.So it's concluded that any sphere without two
: points could be covered by a family of disjoint circles.
: Now we just take the family of spheres centered at origin. Construct
| | | p*****k
发帖数: 318 | 11
in Allens's construction, the origin is only on the circle
with center (-r,0) and radius r.
【在 J**********g 的大作中提到】
: There is one point you don't know how to deal with: the origin. Actually
: that's why it's hard to get an relatively explicit way to map R^3 into
: circles......
:
: think
: parallelling
: point
: another
| J**********g
发帖数: 213 | 12 Acutally I got those two links when I google "a line into a circle", and
unfortunately neither of them works: I do not see the map from the first
link at all; the second link meets same problem I meet: there is always an "
extra point" that we don't know how to map with, same as Allen's, same as
the one I meet.
Also, #1, I do not see how he chose the radius r, if we put it in a
continuous way, I don't see this one works. I could be wrong, but I ever
thought in a similar way but it fails me due t
【在 p*****k 的大作中提到】
:
: in Allens's construction, the origin is only on the circle
: with center (-r,0) and radius r.
| J**********g
发帖数: 213 | 13 The question is that the origin in R^3 corresponds to which point in which
circle?
【在 p*****k 的大作中提到】
:
: in Allens's construction, the origin is only on the circle
: with center (-r,0) and radius r.
| p*****k
发帖数: 318 | 14 hmm... JonathanWang, i would be happy to discuss with you, but
would also like to stick to the original question.
in Allens's construction, r is a fixed positive number, and
whatever value you choose works. the origin would only be on
the circle i mentioned. | D**u
发帖数: 204 | 15 The answer is nice.
A picture is given in the link below.
http://www.cut-the-knot.org/proofs/tessellation.shtml
think
parallelling
point
another
【在 A****s 的大作中提到】
: Consider any two points A and B on a sphere, which are symmetric w.r.t.
: some great circle of this sphere. So there are two hemispheres. And I think
: it could be shown that there always exists a family of cicrles parallelling
: the great circle which covers one of the hemispheres without the given point
: by some analytical geometry.So it's concluded that any sphere without two
: points could be covered by a family of disjoint circles.
: Now we just take the family of spheres centered at origin. Construct
| J**********g
发帖数: 213 | 16 First, I might need to say thanks to you since I indeed did not read his
post carefully for first time, and then I actually found another big ***,
whatever you call it: there are no two circles that touch each other, but
in his construction it's obvious that any two adjacent circles touch each
other, which obviously does not work......
Go to shopping now and nice discusion with you pcasnik....
PS. I modified my post too since my construction does not work for the same
reason......
【在 p*****k 的大作中提到】
: hmm... JonathanWang, i would be happy to discuss with you, but
: would also like to stick to the original question.
: in Allens's construction, r is a fixed positive number, and
: whatever value you choose works. the origin would only be on
: the circle i mentioned.
| D**u
发帖数: 204 | 17
I regard it as a problem harder than a brainteaser, use some math, but not
too much.
【在 J**********g 的大作中提到】
: I am not sure of what kind of problem I should regard it at, I mean, a
: brainteaser, a math one, or whatever.
: Problem reduction.
: 1. R^3=R*R^2 can be dealed in Dugu's way. but you don't know how to
: deal with the axis/center line, ie, how would map a line into a set of
: separate circles.
: or
: 2. R^3=R^2*R, ie, If a line could be mapped into a set of circles,
: then so is R^3: R^3 is the set of lines indexed by (a,b), ie,
: R^3={(a,b,x) with x varies in R, ie, the real line | (a, b) belong to R^2}
| J**********g
发帖数: 213 | 18 It's trivial if we can regard a line (we can simply regard R^3 as set of
lines for sure) or a point as a circle(we can of course regard R^3 as set of
points), and thanks for the link. On the other hand, I read the proof, and
he might need to put more words there, ie, I do not see the support for the
disjointness of any two circles there, though I would not say his proof does
not work.
Also, my proof does not work either. I am math major PhD (candidate), and I
thought in a purely mathematical way
【在 D**u 的大作中提到】
: The answer is nice.
: A picture is given in the link below.
: http://www.cut-the-knot.org/proofs/tessellation.shtml
:
: think
: parallelling
: point
: another
| J**********g
发帖数: 213 | 19 and intuitively, the answer is kinda NO for me. Mathematically we can change
the wording a little bit: we are trying to find a map from R^3 into set of
closed connected set, ie the circles. If we think a reduced (and therefere
assumed to be simpler) problem is: can we map a line into a set of disjoint
circles? here is a map, and we don't need to PARTITION it into disjoint
circles. Though I love math and love geomtry so much, I cannot give a direct
proof here. Maybe I am stupid to study math, but
【在 J**********g 的大作中提到】
: It's trivial if we can regard a line (we can simply regard R^3 as set of
: lines for sure) or a point as a circle(we can of course regard R^3 as set of
: points), and thanks for the link. On the other hand, I read the proof, and
: he might need to put more words there, ie, I do not see the support for the
: disjointness of any two circles there, though I would not say his proof does
: not work.
: Also, my proof does not work either. I am math major PhD (candidate), and I
: thought in a purely mathematical way
| D**u
发帖数: 204 | 20 Let me fill the gap to prove "A sphere with 2 points removed can be
partitioned with disjoint circles", and hope it will help.
One a sphere, let the 2 removed points be A and B.
(1) If A and B are opposite to each other (like north pole and south pole),
then the remaining sphere can be partitioned with circles that are
perpendicular to line AB.
(2) If A and B are not opposite to each other, let P_1 be the tangent plane
(to the sphere) at point A, and P_2 be the tangent plane at point B. Assume
P
【在 J**********g 的大作中提到】
: It's trivial if we can regard a line (we can simply regard R^3 as set of
: lines for sure) or a point as a circle(we can of course regard R^3 as set of
: points), and thanks for the link. On the other hand, I read the proof, and
: he might need to put more words there, ie, I do not see the support for the
: disjointness of any two circles there, though I would not say his proof does
: not work.
: Also, my proof does not work either. I am math major PhD (candidate), and I
: thought in a purely mathematical way
| | | J**********g
发帖数: 213 | 21 Thank you but this part is kinda obvious for me, I mean, geometrically. as I
mentioned, what I did not see is the part that all circles are all disjoint
from each other...
,
plane
Assume
【在 D**u 的大作中提到】
: Let me fill the gap to prove "A sphere with 2 points removed can be
: partitioned with disjoint circles", and hope it will help.
: One a sphere, let the 2 removed points be A and B.
: (1) If A and B are opposite to each other (like north pole and south pole),
: then the remaining sphere can be partitioned with circles that are
: perpendicular to line AB.
: (2) If A and B are not opposite to each other, let P_1 be the tangent plane
: (to the sphere) at point A, and P_2 be the tangent plane at point B. Assume
: P
| D**u
发帖数: 204 | 22 All these spheres are centered at the origin, so all these SPHERES are
disjoint from each other. This implies that circles on different spheres are
also disjoint.
I
disjoint
【在 J**********g 的大作中提到】
: Thank you but this part is kinda obvious for me, I mean, geometrically. as I
: mentioned, what I did not see is the part that all circles are all disjoint
: from each other...
:
: ,
: plane
: Assume
| J**********g
发帖数: 213 | 23 This part is straightforward too, but the black circles in the following
link actually intersect at more than two points with orange cirles (or
sphere) with certain radii. This is true because the radius of sphere has to
be taken in a continous way to cover "most part" of R^3, so actually they
are not disjoint at all....
http://www.cut-the-knot.org/proofs/tessellation.shtml#solution
are
【在 D**u 的大作中提到】
: All these spheres are centered at the origin, so all these SPHERES are
: disjoint from each other. This implies that circles on different spheres are
: also disjoint.
:
: I
: disjoint
| A****s
发帖数: 129 | 24 first it should be clear that only one black circle and one orange circle
are considered when checking intersection. and it is constructed that each
orange circle either is tangent with two black circles at x-axis, or only
one
black circle.
to
【在 J**********g 的大作中提到】
: This part is straightforward too, but the black circles in the following
: link actually intersect at more than two points with orange cirles (or
: sphere) with certain radii. This is true because the radius of sphere has to
: be taken in a continous way to cover "most part" of R^3, so actually they
: are not disjoint at all....
: http://www.cut-the-knot.org/proofs/tessellation.shtml#solution
:
: are
| D**u
发帖数: 204 | 25 Each orange sphere (centered at the origin) intersects with exactly 2 points
from all the black circles. So I don't know why you claim that "but the
black circles in the following link actually intersect at more than two
points with orange sphere with certain radii."
Also, "disjoint" simply means "no intersection", and does not mean "not
continuous". Is this part of the ambiguity that caused the confusion?
to
【在 J**********g 的大作中提到】
: This part is straightforward too, but the black circles in the following
: link actually intersect at more than two points with orange cirles (or
: sphere) with certain radii. This is true because the radius of sphere has to
: be taken in a continous way to cover "most part" of R^3, so actually they
: are not disjoint at all....
: http://www.cut-the-knot.org/proofs/tessellation.shtml#solution
:
: are
| J**********g
发帖数: 213 | 26 Let's take r=1 in your original post to consist with the link provided by DuGu. Using that graph, if it's what you meant, suppose those two points that you
choose are all on x-axis as shown in the graph, ie, those points should be 0
,2,4,etc, then do you have a sphere with radius .5 as one of your sphere? If
yes, this sphere would cut the first black cicle (the one closest to origin
from right in the graph) in the first and fouth quadrants. Similarly, if
you contain sphere with radius 0
【在 A****s 的大作中提到】
: first it should be clear that only one black circle and one orange circle
: are considered when checking intersection. and it is constructed that each
: orange circle either is tangent with two black circles at x-axis, or only
: one
: black circle.
:
: to
| J**********g
发帖数: 213 | 27 First the black circle has to touch/be tangent with orange circle to complete proof there, so I assume by "intersects" you mean "touches". Second, by "Each orange sphere (centered at the origin) intersects with exactly 2
points from all the black circles" you mean this even include orange circle
with radius r=1 there? It's obvious that they are intersecting instead of touching...
The idea in that link, is to get spheres first, then, determine two points
on each sphere, then get the black circle
【在 J**********g 的大作中提到】
: Let's take r=1 in your original post to consist with the link provided by DuGu. Using that graph, if it's what you meant, suppose those two points that you
: choose are all on x-axis as shown in the graph, ie, those points should be 0
: ,2,4,etc, then do you have a sphere with radius .5 as one of your sphere? If
: yes, this sphere would cut the first black cicle (the one closest to origin
: from right in the graph) in the first and fouth quadrants. Similarly, if
: you contain sphere with radius 0
| J**********g
发帖数: 213 | 28 I wanna wrap it up anyway, since it's really not interesting question and
kinda wasting my time now.
If this conclusion is true for R^3, it'd be for R^n, n>=3.
If we think it's a simpler problem for the case of R^2, since R^2 has less
dimension than R^3, then I do not it's true for R^2 either, though I could
be wrong again. | A****s
发帖数: 129 | 29 I can't get you at all.
At very least,
How could a sphere intersect another circle at more than 3 points?
If so, this circle belongs to this sphere, I think? Since both the circle
and the plane which cut the sphere are fixed.
Can you give an example where a sphere intersects a circle at more than 3
points but this circle is not part of this sphere?
DuGu. Using that graph, if it's what you meant, suppose those two points
that you
0
If
origin
black
【在 J**********g 的大作中提到】
: Let's take r=1 in your original post to consist with the link provided by DuGu. Using that graph, if it's what you meant, suppose those two points that you
: choose are all on x-axis as shown in the graph, ie, those points should be 0
: ,2,4,etc, then do you have a sphere with radius .5 as one of your sphere? If
: yes, this sphere would cut the first black cicle (the one closest to origin
: from right in the graph) in the first and fouth quadrants. Similarly, if
: you contain sphere with radius 0
| A****s
发帖数: 129 | 30 surely not for R^2, draw one circle and you have draw another one inside,
and so on...
【在 J**********g 的大作中提到】
: I wanna wrap it up anyway, since it's really not interesting question and
: kinda wasting my time now.
: If this conclusion is true for R^3, it'd be for R^n, n>=3.
: If we think it's a simpler problem for the case of R^2, since R^2 has less
: dimension than R^3, then I do not it's true for R^2 either, though I could
: be wrong again.
| | | J**********g
发帖数: 213 | 31 Let's forget this problem since I am really not interested in it now, though
I would say "No, it cannot be partitioned in that way" if I were asked of
this question.
I ever wanted to ask my advisor for this problem since his research field is
just Algebraic Geometry and he focus more on their geometrical properties,
but I gave up since I think that, if I spend the time I spend on this one
studying stochastic calculus or differential equations, it would be much
more valuable.
Hope people here get
【在 A****s 的大作中提到】
: I can't get you at all.
: At very least,
: How could a sphere intersect another circle at more than 3 points?
: If so, this circle belongs to this sphere, I think? Since both the circle
: and the plane which cut the sphere are fixed.
: Can you give an example where a sphere intersects a circle at more than 3
: points but this circle is not part of this sphere?
:
: DuGu. Using that graph, if it's what you meant, suppose those two points
: that you
| p*****k
发帖数: 318 | 32
DuGu, that is a very nice proof. thx.
inspired by the official solution, here is another choice of the
"black circles" (i.e., the family of circles on R^2 in Allens's post):
(x-r/2)^2 + y^2 = (n*r-r/2)^2, where n is a natural number.
【在 D**u 的大作中提到】
: Let me fill the gap to prove "A sphere with 2 points removed can be
: partitioned with disjoint circles", and hope it will help.
: One a sphere, let the 2 removed points be A and B.
: (1) If A and B are opposite to each other (like north pole and south pole),
: then the remaining sphere can be partitioned with circles that are
: perpendicular to line AB.
: (2) If A and B are not opposite to each other, let P_1 be the tangent plane
: (to the sphere) at point A, and P_2 be the tangent plane at point B. Assume
: P
| t*******e
发帖数: 172 | 33 There is a hopf fibration of S^3, with each fiber is exactly S^1.
So take R^3 be some open neighbourhood in S^3. Everything done | D**u
发帖数: 204 | 34 The idea of using hopf fibration is very interesting, can you give more
detail?
There are 2 gaps that I could not fill now.
(1) An open neighbourhood in S^3 might only contains "partials" of some S^1,
how to make sure the neighbourhood contains "full" circles only?
(2) Even if (1) is fixed, we still need to map the open neighbourhood
continuously to a R^3. Then how to ensure that we can keep the circle still
as a circle (not just only topologically)?
【在 t*******e 的大作中提到】
: There is a hopf fibration of S^3, with each fiber is exactly S^1.
: So take R^3 be some open neighbourhood in S^3. Everything done
| D**u
发帖数: 204 | 35 Sorry that the link I gave might bring more confusions than it clears up for
you. The picture might be confusing in the sense that they draw it in black
and orange circles, but indeed they are
black circles (not sphere) and orange spheres (not circles, with all radii 0
Also the black circles are NOT drawn after picking some points on the orange
spheres. The black ones and oranges are drawn independently.
though
is
,
something
【在 J**********g 的大作中提到】
: Let's forget this problem since I am really not interested in it now, though
: I would say "No, it cannot be partitioned in that way" if I were asked of
: this question.
: I ever wanted to ask my advisor for this problem since his research field is
: just Algebraic Geometry and he focus more on their geometrical properties,
: but I gave up since I think that, if I spend the time I spend on this one
: studying stochastic calculus or differential equations, it would be much
: more valuable.
: Hope people here get
| D**u
发帖数: 204 | 36 This Tanghulu solution is very nice, it is very clear.
【在 p*****k 的大作中提到】
:
: DuGu, that is a very nice proof. thx.
: inspired by the official solution, here is another choice of the
: "black circles" (i.e., the family of circles on R^2 in Allens's post):
: (x-r/2)^2 + y^2 = (n*r-r/2)^2, where n is a natural number.
| p*****k
发帖数: 318 | 37 (i think the discussion now goes way beyond this board's scope,
so sorry about the "spam")
about the Hopf fi bration idea, stereographic projection of
this partition into R^3 gives very interesting tiling with
Villarceau circles, but unfortunately with one additional line,
which makes it less attractive. not sure how to circumvent this.
however, i was pointed to this paper by a friend:
Conway & Croft, 'Covering a Sphere with Congruent Great-Circle Arcs',
Proc. Camb. Phil. Soc. 60 (1964), 787-80 | D**u
发帖数: 204 | 38 The R^3 analogue of Hopf fi bration is indeed surprising.
Here is my thoughts on the new question "how to partition R^3 into pairwise
non-parallel lines?".
We know that the hyperboloid of one sheet
(x^2/a^2 + y^2/b^2 - z^2/c^2 = 1)
can be partitioned with pairwise non-parallel lines.
If we partition R^3 into hyperboliods
x^2/a^2 + y^2/a^2 - z^2 = 1 (0
PLUS the z-axis,
and then partition each hyperboliod with line, then these lines should be
pairwise non-parallel.
【在 p*****k 的大作中提到】
: (i think the discussion now goes way beyond this board's scope,
: so sorry about the "spam")
: about the Hopf fi bration idea, stereographic projection of
: this partition into R^3 gives very interesting tiling with
: Villarceau circles, but unfortunately with one additional line,
: which makes it less attractive. not sure how to circumvent this.
: however, i was pointed to this paper by a friend:
: Conway & Croft, 'Covering a Sphere with Congruent Great-Circle Arcs',
: Proc. Camb. Phil. Soc. 60 (1964), 787-80
| J**********g
发帖数: 213 | 39 read the paper on Amer. math. mothly and seen what it means now. I did not
understand idea correctly before. It's a smart construction, and it's a
pretty nice end now. Thanks to "pcasnik" for the references since it tells
the idea in a much clearer way. | p*****k
发帖数: 318 | |
|
