Haskell: Call a function with each element in a list

Question

I am making a sudoku solving program and I have a potentialNbrsAt function that gets the numbers that could be at position x y.

Now, I am trying to get the intersect of each lists of potential numbers in a column. Something like the onlyOnePlaceForNbrInCol function bellow.

Code:

potentialNbrsAt :: Int -> Int -> Sudoku -> [Int]
potentialNbrsAt x y sudoku = intersect rowMissingNbrs $ intersect colMissingNbrs sqrMissingNbrs
                where rowMissingNbrs = getMissingNbrs $ getRow y sudoku
                      colMissingNbrs = getMissingNbrs $ getCol x sudoku
                      sqrMissingNbrs = getMissingNbrs $ getSquare squareIndex sudoku
                      squareIndex    = 3 * (y `div` 3) + (x `div` 3)

onlyOnePlaceForNbrInCol :: Int -> Int -> Sudoku -> Bool
onlyOnePlaceForNbrInCol colIndex nbr sudoku = -- What goes here? Some pointers please???

I think onlyOnePlaceForNbrInCol should, at some point, call potentialNbrsAt with each numbers from 0 to 8 as an argument for y. Telling me how to do this would greatly help.

PS: whenenver the number 9, or the div operator, occur in a Sudoku solver, it shows insufficient modelling: the grid does not have two dimensions (9x9), but four (3x3x3x3). In German, "die Anschauung ist auch oft der Feind des Verständnisses." — d8d0d65b3f7cf42
– d8d0d65b3f7cf42, Commented Jun 19, 2014 at 19:08

chi · Accepted Answer · 2014-06-19 18:34:24Z

1

What about [ potentialNbrsAt x y sudoku | y <- [0..8] ] ? This gives you a list of all the results for such values of y.

answered Jun 19, 2014 at 18:34

chi

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1 Comment

Mickael Bergeron Néron Over a year ago

I did not expect it could be so easy. Thx!

leftaroundabout · Accepted Answer · 2014-06-19 18:33:17Z

1

So you're trying to determine whether all of the numbers [0..8] fulfill a given predicate.

answered Jun 19, 2014 at 18:33

leftaroundabout

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