I'm guessing you all know of the knot puzzle
**plagarise and paste from wordways**
Word worms and knots

A word can be diagrammed in three-dimensional space by identifying its letters with vectors, each pointing in a different direction; each word thus becomes a segmented worm twisting through space. If one considers the three-dimensional lattice of points with integer coordinates, then, by a stroke of luck, there are 26 directions by which one can step from one point to another adjacent one (whose coordinates differ by at most one from the previous point). We can identify these 26 directions with the letters of the alphabet as in the diagram below:

A B C J K L R S T

D E F M * N U V W

G H I O P Q X Y Z

A small number of word worms such as ANY, ROIL and DISINGENUOUSNESS form closed loops, returning to their starting points. Some closed loops (such as the non-word TYDBNYRDI or the two-word "phrase" YO ELUVIAL) actually manage to tie a knot in space. It has been proven that nine letters is the minimum needed to form a knot, but so far no one has found a single word that by itself ties a knot. Does one exist? Large word lists have been searched by computer without finding one, so it won't be easy! Perhaps in another language?
**/plagarise and paste**

is it saying that you should search through words and then plot the course? can you assign vectors yourself? d'you reckon you can plot knot courses to generate strings of letters and then make words out of the letters anagram stylee? AAAAAAARRRRRRGGGGGGHHHHHH, my brain is melting. I reckon it should be possible to write a simple program that can generate nine point knots. Assigning letter values to the points has got to be possible. What do you all think?

> What do you all think?

I'm thinking that I shouldn't be trying to cope with this at 01:06.

Never heard of this. Very interesting.

k

It took me a bit, but I finally realized that the

ABC
DEF
GHI

is the front face of a cube. The

RST
UVW
XYZ

is the back face and the

JKL
M*N
OPQ

is in the middle. Thanks, tsuwm

A little hard to diagram in the two space universe we have on our computer monitors but.

Man, I stared at this so long, my computer went to screen saver! First, Faldage, if I may--I would like to add my description to yours, since my first response to yours was to wonder what was on the other 3 sides of the cube. It makes more sense (ha--I first typed that as snese) to me to visualize a hollow box with 3 vertical slots: one at the front, one halfway back, and the last at the back. Then I slide "hanging file folders" into the slots with the letters as you gave them; that is, the front folder shows
ABC
DEF
GHI.

However, looking at the examples given, I don't think we have any choice but to think of them as numbers (coordinates, actually, but I don't care to expend my brain power first assigning them and then figuring out the differentials) rather than letters. This is what I was staring at for so long: in this kind of grid, I can't make it work that i, say, is "next to" s; nor can I make it work that ANY is a closed loop--that is, the fact that they say ANY is a "closed loop" tells me that I don't know what they mean by closed loop! How can something at the back (y) be "next to" something at the front (a)? However, if the coordinate for a is only one away from the coordinate for y, then that would work. (Obviously.) Ohhh, my brain is starting to hurt! Am I off the track completely now, or what?

Dody (how are your toes, Sweet Thing? Not too bad, I hope?)--I will venture to say, though, that from reading what I assume were the "rules" you posted, that you can either decide on a word and then vector it, OR plot a vector and then see if it makes a word. I would guess, though, that the letters would have to be in order, and not anagrammed; mainly because anagramming (sp?) would allow too many words.

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.

I think the letters are irrelevant, unless one is going to restrict the space to those paths which form, say, words found in the AHD or OED or what have you.
Otherwise, it's a distraction.

It seems obvious that 9 is the number - say 4 for the base of the loop, 4 to loop around and 1 to pass through. The knot starts in the drawing plane. The ampersand is the "hole" through which the "string" must loop to form the knot.

1
*---*
\
\ 2
\
* & *
\ /
4 \ / 3
\ /
*


Dotted line 5 comes out of the plane to a point in front of point A.
Dotted line 6 returns to the plane.

1 A 6
*---*...*
. \
5. \ 2
. \
* *
\ /
4 \ / 3
\ /
*

The segment from B to a point BEHIND & is number 7.

1 A 6
*---*...* B
. \.
5. .\ 2
. \
* & *
\ /
4 \ / 3
\ /
*


Eight is from the point behind & to & itself.
Nine is from & to any point in the plane in front
of the drawing plane.

Intuitively, one has to have a loop with a hole in it.
The catch here is whether points 3 and 4 really need to
be in the plane. Seems like with suitable refinements,
the space of possible solutions is not that great -
perhaps one could have a computer check into it.

k

hehe. yeah, right. what he said.

Crap. I just noticed that that didn't come out the way I typed it in. I know I should check before posting, but I never do. Usually a waste of time, except this time I wasted all that time typing in the diagrams and they don't show up. Crap.

k

I've had the same thing happen, Keith. Argh. Just on the chance, you might try it using the pre in brackets then the /pre in brackets. But this hasn't always worked, for me.