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Re: Non-linear full-2d writing (again)

From:tomhchappell <tomhchappell@...>
Date:Tuesday, January 31, 2006, 18:25
--- In, Sai Emrys <sai@S...> wrote:
> Actually, no - simpler: > 123 > 4O5 > 678 > > There, you have 8 connections, all directly touching the central > character, all of the same size. > - Sai >[snip]
If you consider all of those nodes to be squares, then, the pairs {1,O}, {2,4}, {2,5}, {3,O}, {4,7}, {O,6}, {O,8}, and {5,7}, share only a single point -- a corner. In tiling, tiles aren't considered to be neighbors unless they share a non-zero-length portion of border. So at least some of the above pairs can't be considered "connected" in the "semantic domain" in which Jeff is talking about the "<=6" limitation. ---- Also, if the connections 1-2, 2-3, 4-O, O-5, 6-7, 7-8, 1-4, 4-6, 2-O, O-7, 3-5, and 5-8, are all the same length as each other, then the connections 1-O, 3-O, O-6, and O-8 cannot all also be that same length. If they were, then you would have eight non-overlapping equilateral triangles all having O as a vertex; triangles 12O, 14O, 23O, 35O, 46O, 58O, 67O, 78O. You can't fit more than six (non-overlapping) equilateral triangles all around a single shared vertex, without warping the "paper" (the writing-surface) so that it is no longer flat. You could make just 1-O and O-8 be the same length as the other eight, leaving 3-O and O-6 to be different; or, instead, make 3-O and O-6 to be the same length as the other eight, and 1-O and O-8 different. Or, you could make 1-2, 1-4, 2-3, 3-5, 4-6, 5-8, 6-7, and 7-8 be one length, shorter than the lengths of 1-O, 2-O, 3-O, 4-O, O-5, O-6, O- 7, O-8, which could all be equal. But you can't make all sixteen lengths equal. Tom H.C. in MI


Sai Emrys <sai@...>