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Topic: A topological problem
Replies: 6   Last Post: Mar 2, 2013 9:39 AM

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dan.ms.chaos@gmail.com

Posts: 409
Registered: 3/1/08
Re: A topological problem
Posted: Feb 28, 2013 1:03 AM
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On Feb 26, 7:17 pm, <fc3a...@uni-hamburg.de> wrote:
> Warning, incoming lousy ASCII, change to fixed font :-)
>
> o  o  o
> |  |  |
> |  O  |
> \ /|\ /
>  X | X
> / \|/ \
> |  O  |
> |  |  |
> o  o  o
>
> The lines |/\ are tied to the unmovable nodes Oo.
> (As you see, four lines come out of O and one out of o.)
> X denotes a crossing, which is like a virtual crossing
> from knot theory, i.e. you can move it ad lib and any
> line over any other. Should they cross in the process,
> well duh, then you have more crossings.)
>
> Can you move the lines around such that no horizontal
> line going through this graph cuts more than three
> of these lines? I think no, but can you lend me a
> formal proof?
> --
> Hauke Reddmann <:-EX8    fc3a...@uni-hamburg.de
> Die Weltformel: IQ+dB=const.


Proofs in knot theory aren't my thing , but here's a shot :
Label the knots with numbers :

1 1 1
| | |
| 2 |
\ /|\ /
X | X
/ \|/ \
| 3 |
| | |
4 4 4

denoting four 'levels of height' .
1 path directly connects 4 to 3
2 paths directly connect 4 to 2

If you choose a horizontal coordinate for your line situated between k
and (k+1) levels of height, your line will cross at least all the
paths taken from m to n , for all m <= k and all n>=(k+1) .

So the minimum number of paths cut between 4 and 3 is 2+1 = 3
Analogous reasoning shows the minimum number of paths between 2 and 3
is :
2 paths from 4 - 2 +
2 paths from 1 - 3 +
1 path from 2- 3 = five paths .

As shown in a previous post , it is indeed possible to rearrange the
lines so that paths 'appear to coincide' ,
o o o
\|/
O
|
O
/|\
o o o

but whether this is considered 'cheating' or not is another manner .



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