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As far as closed loops go, it makes most sense (or maybe it only makes sense) when the position of the letters are viewed as vectors. In other words, if you add the directions you have to move from the start to reach each letter in the word, you end up with 0. Or for the mathematically inclined (preferably with m>1, if y=mx+c), the resultant vector is [0,0,0].
For example, the A is one left, one up, and one towards or [-1,1,1], the N is one right [1,0,0] and the Y is one down and one away [0,-1,-1]. If you add the vectors, the result is [0,0,0] you're back where you started and the loop is closed.
As for tying knots, that's a bit more complex. I'll let you know if I come up with anything.
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logological
dodyskin 09/30/03 12:04 PM ![]()
Re: logological
sjmaxq 09/30/03 12:07 PM ![]()
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Jackie 09/30/03 02:29 PM ![]()
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Re: logological
TheFallibleFiend 10/01/03 05:14 PM ![]()
Re: logological
Buffalo Shrdlu 10/01/03 08:38 PM ![]()
Re: logological
TheFallibleFiend 10/02/03 12:59 PM ![]()
Re: logological
Jackie 10/02/03 01:26 PM
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