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Hi, I really need your help. My Maths teacher wants me to do a presentation on the follwoing problem to improve my Maths mark. I just don't know how to do that task. Please help me.
There are 4 points on a plane coordinate system.
These points have the following coordinates:
1.(0;0)
2.(1;0)
3.(0;1)
4.(1;1)
These 4 points are to be repositioned.
You may only reposition points in the following way:
You may only change the position of a point, if there is another
point exactly in the middle between the original and the new
position of the point.
Now, I have to prove whether or whether not it is possible to
reposition the points, in order to get the following coordinates:
1.(0;0)
2.(1;1)
3.(2;-1)
4.(3;0)
I've tried this little game quite intensively and I'm sure it is impossible but I still can't prove it.
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I cannot upload any images...
Last edited by Thomas11 (2008-02-02 08:52:29)
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Do you understand the task?
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One way to think about this is to try and play the game backwards. Now suppose it were possible to arrive at the given final configuration. What would have been the final move that was made in order to arrive at that configuration? If it is impossible for such a move to exist, then the configuration is impossible to reach.
By the way, are diagonal moves allowed in this game?
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Yes, they are, as long as they don't break the rule: To reposition a point, there must be another point exactly in the middle between the new and the old position. Therefore diaganoal moves are allowed.
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Hmm, Jane, the method seems to be interesting, however, there are infinite possible final moves...
Moreover, how can I prove a certain move is impossible?
Last edited by Thomas11 (2008-02-02 10:14:48)
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The rules are actually the same backwards as forwards, so if you can start with them positioned at (0,0), (1,1), (2,-1), (3,0) and use the same method of moving to get them to the start position, then you can just reverse that set of moves to get your answer.
Why did the vector cross the road?
It wanted to be normal.
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Yeah, you are right.
However, I have tried it quite intensively, I tend to say it is not possible. I am just lacking a proof as I cannot just say I didn't manage to move them...
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Minor breakthrough:
Because whenever you move a point, you're moving it by twice the distance between it and another point, you're always adding (2x,2y) to its co-ordinates, where x and y depend on the distance between the two points. Hence the parity of each point is constant.
Unfortunately, both the starting and finishing positions have one each of the possible forms: (odd,odd); (odd,even); (even,odd); (even,even), so that alone doesn't give a disproof.
However, it does mean that if there is a solution, then it would have to be that (0,0) and (1,1) stay where they are, (0,1) goes to (2,-1) and (1,0) goes to (3,0).
Why did the vector cross the road?
It wanted to be normal.
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Hi.
What you wrote seems to be logic. However, I don't understand what you mean by parity. Could you enlarge on that?
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Parity is just a fancy way of saying whether something's odd or even.
Moving these points around doesn't change their parity, so (for example) a point that is at (odd,odd) before moving will be at (odd,odd) afterwards.
Why did the vector cross the road?
It wanted to be normal.
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Okay, I understand what you are saying. I think it's helpful, so one only has to prove by using this finding such moves are impossible to get to the set.
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Did someone found out s.th. helpful or anything else at least?
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Can nobody help me...-.-?
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...-.-
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Please help me...-.-
I'm desperate...
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