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Topic: Infinite B&W plane (Read 1137 times) |
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BNC
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Infinite B&W plane
« on: Apr 29th, 2004, 12:42pm » |
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A plane is painted in two colors, such that each point is painted either black or white.
Prove it always possible to find:
(1) Three points of the same colors, one of which is at the exact middle of the segment connecting the other two;
(2) Three points of the same color, forming an equilateral triangle .
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rmsgrey
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Re: Infinite B&W plane
« Reply #1 on: Apr 29th, 2004, 1:24pm » |
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I've managed to solve both of them - the line by considering 5 points, and the triangle 7. I'll hold off on actually posting the constructions for a while so others can have a go...
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Noke Lieu
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Re: Infinite B&W plane
« Reply #2 on: Apr 29th, 2004, 7:32pm » |
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Only a fledgling at infinities, so excuse me.
I think I thought the same thought as rmsgrey. But now I don't know.
I assume the plane isn't definately uniformly distributed bwbwbw like a chess board?
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BNC
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Re: Infinite B&W plane
« Reply #3 on: Apr 29th, 2004, 10:43pm » |
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on Apr 29th, 2004, 7:32pm, Noke Lieu wrote:Only a fledgling at infinities, so excuse me.
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Don't worry about it. You don't have to work with infinities here. all the calculations are finite; infinite plane is there to indicate unbounded area.
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I assume the plane isn't definately uniformly distributed bwbwbw like a chess board?
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It wounldn't be interesting then.. 
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Noke Lieu
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Re: Infinite B&W plane
« Reply #4 on: Apr 29th, 2004, 11:00pm » |
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That means the points are uniformly distributed- other wise it would be very easy. Actually, no it doesn't- a unit-cell of
.w.b
w.b.
would do it. And that's pretty obvious to see. Okay- am with you now.
So any random distribution of points, labelled W or B will make it work. Ahh. Much more interesting, you're right.
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| « Last Edit: Apr 29th, 2004, 11:02pm by Noke Lieu » |
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TenaliRaman
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Re: Infinite B&W plane
« Reply #5 on: May 1st, 2004, 4:34am » |
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how does one get to the solution?
i can think of cases to consider
1>entire plane of one color ... conditions proved
2>distributed coloring
now here comes the problem i cannot think of breaking it down further??
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Icarus
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Re: Infinite B&W plane
« Reply #6 on: May 1st, 2004, 7:08am » |
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The trick is to find arrangements of a finite number of points such that no matter how these points are colored, there must be a subset of them that satisfies the condition. Note that rmsgrey says he solved (1) by considering an arrangement of 5 points, and (2) by an arrangement of 7 points.
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BNC
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Re: Infinite B&W plane
« Reply #7 on: May 1st, 2004, 7:42am » |
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hint: (2) can be solved using the result from (1), and considering only 6 points.::
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towr
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Re: Infinite B&W plane
« Reply #8 on: May 1st, 2004, 9:33am » |
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on May 1st, 2004, 4:34am, TenaliRaman wrote:how does one get to the solution?
i can think of cases to consider
1>entire plane of one color ... conditions proved
2>distributed coloring
now here comes the problem i cannot think of breaking it down further??
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I can think of other cases. Like,
A finite number of black points.
Or all black points on a hyperbolic curve.
The results will differ for different distributions..
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rmsgrey
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Re: Infinite B&W plane
« Reply #9 on: May 2nd, 2004, 6:52am » |
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BNC: While using the result of the first part reduces the number of points needed for the second, proving the first part as part of the proof of the second requires more than 7 points. Anyway, I think the trick in the 7 point construction is much more interesting...
As an extension, if, instead of having the entire plane (or any finite region of it) entirely painted, instead you can specify in advance a finite number of points to be painted either black or white, what's the minimum number of points (and in what arrangement) required so that you can guarantee:
1) At least one point lies precisely midway between two others of its colour.
2) Three points of the same colour form an equilateral triangle.
I have answers of ::9 and 15:: respectively. The arrangements are again left unstated.
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rmsgrey
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Re: Infinite B&W plane
« Reply #10 on: May 2nd, 2004, 7:04am » |
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Some hints/preliminary results for the original problems:
::
If there is any finite convex area all one single colour, then it's always possible to pick a finite line segment and an equilateral triangle that lie entirely within that region and so both parts of the original problem become trivial.
::
::
Rather than proving that all possible arrangements of points must satisfy each condition, it's probably easier to show that there is no possible arrangement that falsifies either condition. This can be done by trying to construct an arrangement that doesn't satisfy one or other condition, and showing that you can't.
::
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BNC
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Re: Infinite B&W plane
« Reply #11 on: May 2nd, 2004, 8:11am » |
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rmsgrey,
A clarification regarding your extension:
Are you allowed to state a specific arrangement of N points, and just the colors are random?
[e] Got 9 for (1), so I guess it's the right question[/e]
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| « Last Edit: May 2nd, 2004, 8:19am by BNC » |
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TenaliRaman
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Re: Infinite B&W plane
« Reply #12 on: May 2nd, 2004, 11:06pm » |
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on May 1st, 2004, 7:08am, Icarus wrote:| The trick is to find arrangements of a finite number of points such that no matter how these points are colored, there must be a subset of them that satisfies the condition. Note that rmsgrey says he solved (1) by considering an arrangement of 5 points, and (2) by an arrangement of 7 points. |
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Thanks Icarus!
on May 1st, 2004, 9:33am, towr wrote:I can think of other cases. Like,
A finite number of black points.
Or all black points on a hyperbolic curve.
The results will differ for different distributions.. |
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That's exactly the dilemma i had. But now cleared.
----------------------------------------------------------
I have managed to solve #2 by using 5 points. But i have not yet managed to solve #1. 
I am looking deeply into rmsgrey's hints but i am not sure if i understand that. Any pointers would be helpful.
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| « Last Edit: May 2nd, 2004, 11:10pm by TenaliRaman » |
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Sameer
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Re: Infinite B&W plane
« Reply #13 on: May 3rd, 2004, 7:10am » |
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This looks harder to me as I may lack some basic knowledge about something.
Just one question. Are we looking for a gernal method that would work for any distribution? From rmsgrey's hint it definitely feels like it, but just want to confirm.
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BNC
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Re: Infinite B&W plane
« Reply #14 on: May 3rd, 2004, 10:13am » |
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on May 3rd, 2004, 7:10am, Sameer wrote:This looks harder to me as I may lack some basic knowledge about something.
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No special knowledge is required, only logic, and the "a-ha" moment.
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Just one question. Are we looking for a gernal method that would work for any distribution? From rmsgrey's hint it definitely feels like it, but just want to confirm. |
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Yes, indeed.
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yadayada
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Here is one way to solve the midpoint problem
First assume that there is no line whose midpoint is same colour as the vertices. Now,
Consider points (-1,0), (0,0) and (1,0). We can assume that (-1,0) and (1,0) are both Black. Then (0,0) is white.
Consider (-3,0). Must be White. similarly (3,0). This implies (0,0) must be black. Contradiction.
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TenaliRaman
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Re: Infinite B&W plane
« Reply #16 on: May 4th, 2004, 2:23am » |
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yadayada,
i did hit that particular contradiction earlier but i think the assumptions seem faulty.
I think its equivalent to saying,
consider a rhombus with two opposite vertices colored black and other two opposite vertices colored white.The intersection of the diagonals will be black or white.QED.
Don't you think the assumptions are unfair??
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Barukh
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Re: Infinite B&W plane
« Reply #17 on: May 4th, 2004, 9:04am » |
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on May 4th, 2004, 2:23am, TenaliRaman wrote:yadayada,
i did hit that particular contradiction earlier but i think the assumptions seem faulty. |
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Yadayada's assumptions are perfectly correct: you can always pick 2 points of the same color and build a coordinate system on them (i.e. say their coordinates will be (-1, 0) and (1, 0).
Quote:I think its equivalent to saying,
consider a rhombus with two opposite vertices colored black and other two opposite vertices colored white.The intersection of the diagonals will be black or white.QED.
Don't you think the assumptions are unfair?? |
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That's something different: it’s not immediately obvious you can choose that many specific points to define a coordinate system on them. If you prove it's possible, than this argument is as sound as yadayada's.
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| « Last Edit: May 4th, 2004, 9:04am by Barukh » |
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rmsgrey
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Re: Infinite B&W plane
« Reply #18 on: May 4th, 2004, 9:17am » |
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yadayada's construction is easily fixed - instead of saying "... we can assume that ..." say "... we can define our co-ordinate system so that ..."
::
If you don't want to assume that there are (at least) two black points in the plane, pick two points of the same colour, call that colour "Fred" and rewrite the proof replacing "black" and "white" with "Fred" and "Barney" respectively...
::
And for the requested clarification: yes, you are allowed to specify the specific points to be painted (though not the colour of each!  ) not just the number of them. In fact, you can find an arrangement of any number of points so that no three of them even lie on the same straight line (eg the edge of a circle), or so that no three form an equilateral triangle (eg a straight line), so you need the arrangement of points to be specified for ti even to be possible.
[e]Barukh posted faster... [/e]
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| « Last Edit: May 4th, 2004, 9:18am by rmsgrey » |
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TenaliRaman
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Re: Infinite B&W plane
« Reply #19 on: May 4th, 2004, 11:22am » |
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Thx for the replies!
however my point was not that yadayada's construction was wrong but the assumptions he made in between.
i have no problems in assuming two points which are of the same color but my argument is that "this" is the limit of assumptions and that any more assumptions made after this would yield nothing.
To clarify what i am trying to say , lets go along yadayada's construction,
1>take two points which are of same color (perfectly valid system)
2>midpoint not of same color (perfectly valid assumption)
3>suppose there is -3,0 and 3,0 of the white color then contradiction...(here is where i have my problem ... i can distribute the whites such that no pair of white point has 0,0 as the midpoint) obviously such a distribution doesn't help much to contradiction proof.
sorry if the above is unclear and feels absurd.any comments or criticism on the above post is warmly welcomed.
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towr
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Re: Infinite B&W plane
« Reply #20 on: May 4th, 2004, 12:22pm » |
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If those points weren't white, you'd allready have a contradiction earlier.. because
(-1,0) , (1,0) and (3,0) would be black
or (-3,0) , (-1,0) and (1,0) would be black
Thus (3,0) , (0,0) and (3,0) must be white, and you get a contradiction anyway
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| « Last Edit: May 4th, 2004, 12:27pm by towr » |
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Icarus
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Re: Infinite B&W plane
« Reply #21 on: May 4th, 2004, 4:45pm » |
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For clarity, let me recast yadayada's argument:
Out of any three points, 2 must be of the same color, so let B and D be two such points and let C be the midpoint between them. Let A and E be the points for which B is the mid point of AD and D is the midpoint of BE. It follows that C is also the mid point of AE.
By our choice, B and D are both the same color (call it 1 and the other color 2).
If A is of color 1, then A,B,D form the desired triple.
Assume A is of color 2.
If E is of color 1, then B, D, E form the desired triple.
Assume E is of color 2.
If C is of color 1, then B, C, D form the desired triple.
If C is of color 2, then A, C, E form the desired triple.
Therefore in all cases, three of A, B, C, D, E must be equally spaced points of the same color.
Does that make the argument a little clearer?
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TenaliRaman
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Re: Infinite B&W plane
« Reply #22 on: May 5th, 2004, 3:35am » |
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Thanks towr and Icarus!
(i feel so dumb right now  )
========================================
I have also solved rmsgrey's extension problems (which are quite easy once u solve the original problems nonetheless they were still pretty nice extensions),
the first one i did with 9 points
the second i finished with 12 points
(if i have my drawings correct hopefully)
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rmsgrey
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Re: Infinite B&W plane
« Reply #23 on: May 5th, 2004, 8:17am » |
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I'm curious about your 12 point solution to the second part of my extension. I got 15 points easily enough, and assumed that was optimal. If there's a 12 point construction, I'd like to know what I missed...
[e]OK, just had a brainwave and got it down to 10...[/e]
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| « Last Edit: May 5th, 2004, 8:22am by rmsgrey » |
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TenaliRaman
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Re: Infinite B&W plane
« Reply #24 on: May 5th, 2004, 9:02am » |
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yeehaw! me too! i removed 2 redundant points from my diagram.But 10 pts is the limit and i think it is the optimal.
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