Speed up a slow loop

[Alex Tweedly]

Hmmm - I have to correct myself :-)

It's not 110; there are 144 digraphs on a 5x5 board

  - nine centre tiles each have 8 neighbours

  - four corner tiles each have 3 neighbours

  - twelve remaining edge tiles each have 5 neighbours

giving 72 + 12 + 60 - so 144 pairs (already including both directions 
for each adjacency).

Now if you allow edge-to-edge wrapping (i.e. from the top right tile you 
can move to the right - and find yourself back on the top left, etc.) 
then there are simply 25 * 8 pairs - but then you've moved a long way 
from genuine Boggle rules :-)

Alex.

On 09/03/2022 23:09, Alex Tweedly via use-livecode wrote:
> Yes, Quentin's allowing for diagonals (that's how the number of 
> digraphs on a 5x5 board gets up to 110).
>
> And it's probably a good idea, allowing an even finer filter - if you 
> aren't doing the boardwalk method..
>
> If you do use the boardwalk to generate the exact list of words, you 
> get no benefit from single-letter or digraph filtering, because the 
> tree-walk is constrained to only those valid "next char"s, and so 
> implicitly avoids using those non-present digraphs.
>
> Oh - so many different ways to do things, all interesting, and all 
> good for some variation of the problem.
>
>
> btw - that reminds me - back when I used to play real, physical Boggle 
> with friends, we often played variants of the word rules; either
>
>  - you can reuse the same tile later in a word (e.g.
> Y  L  A
> X  E  T
>
> would allow "lately" as a word.
>
> OR
>
> - you can double-up on a tile (e.g. M I L would allow 'mill')
> (more important if you're British than if you're American :-)
>
> Alex.
>
>
>
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