Lab 5

[BertLisser]

Ex  2
The following is Okay

constraints :: Sudoku -> [Constraint] 
constraints s = sortBy length3rd 
    [(r,c, freeAtPos' s (r,c) sudokuConstraints) | 
                       (r,c) <- openPositions s ]

but why not refactor the code

prune :: (Row,Column,Value) 
      -> [Constraint] -> [Constraint]
prune _ [] = []
prune (r,c,v) ((x,y,zs):rest)
  | r == x = (x,y,zs\\[v]) : prune (r,c,v) rest
  | c == y = (x,y,zs\\[v]) : prune (r,c,v) rest
  | sameblock (r,c) (x,y) = 
        (x,y,zs\\[v]) : prune (r,c,v) rest
  | otherwise = (x,y,zs) : prune (r,c,v) rest

sameblock :: (Row,Column) -> (Row,Column) -> Bool
sameblock (r,c) (x,y) = bl r == bl x && bl c == bl y 

I see you understand this, but you are on the half way.

Exercise 4

Nice reasoning. But still sudoko's with 4 empty blocks and an unique solution exists.

| 8 6   |     3 |       |
| 3     | 4   1 |       |
|     9 |       |       |
+-------+-------+-------+
|       |     6 |       |
|       |       |       |
|       | 5     |       |
+-------+-------+-------+
|   2 3 |       |     8 |
|   8   |       |   9 3 |
| 5     |       | 7   2 |
+-------+-------+-------+