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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