[m-users.] [Beginner] Optimally representing a diamond-shaped grid

[Volker Wysk]

Am Samstag, dem 15.10.2022 um 15:30 +0530 schrieb Razetime:
> I have a grid like follows:
> 
>     1
>   2 3 4
> 5 6 7 8 9
>   A B C
>     D
> 
> for a pointer that spawns on any position on this grid, the possible
> options are up, down, left and right, which move 1 step in each of
> those directions. If a direction is given and there is no spot in that
> direction, the point should remain in place.
> 
> My first idea is exhaustively defining all the possible moves like so:
> 
> :- type dir ---> u;d;l;r.
> 
> :- pred(character:in,dir::in,character::out) is det.
> move('5',u,'5').
> move('5',d,'5').
> move('5',l,'5').
> move('5',r,'6').
> % and so on.
> 
> What could be a more idiomatic method to accomplish this movement in Mercury?

I couldn't find a fitting translation of "idiomatic". But I'd try it as
follows.

I'd save the grid in a list of lists:

:- func grid_ll = list(list(char)).

grid_ll = 
[['x', 'x', '1', 'x', 'x'],
 ['x', '2', '3', '4', 'x'],
 ...
].

Then I'd use the array2d library an make an array out of it:

Grid = array2d(grid_ll)

I'd get the coordinates of the start element by searching the array. From
the coordinates and from the array element 'x', I'd get to know if a move in
the specified direction is possible. If so, looking up the array at the new
coordinates, gets the target element.

Cheers,
Volker
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