Question

Suppose I have the following nested list:

list =   
  [[0, 1, 0],  
   [1, 9, 1],  
   [1, 1, 0]]

Assuming you are only given the x and y coordinate of 9. How do I use Haskell code to find out how many 1’s surrounds the number 9?

Let me clarify a bit more, assume the number 9 is positioned at (0, 0).
What I am trying to do is this:

int sum = 0;
for(int i = -1; i <= 1; i++){
  for(int j = -1; j <= 1; j++){
    if(i == 0 || j == 0) continue;
    sum += list[i][j];
  }
}

The positions surrounding (0,0) are the following coordinates:

 (-1, -1) (0, -1) (1, -1)
 (-1,  0)         (1,  0)
 (-1,  1) (0,  1) (1,  1)  

Answer 1

list = [[0,1,0],[1,9,1],[1,1,0]]
s x y = sum [list !! j !! i | i <- [x-1..x+1], j <- [y-1..y+1], i /= x || j /= y]
--s 1 1 --> 5

Note that I there is no error correction if the coordinates are at the edge. You could implement this by adding more conditions to the comprehension.

A list of lists isn’t the most efficient data structure if things get bigger. You could consider vectors, or a Map (Int,Int) Int (especially if you have many zeros that could be left out).

[Edit]

Here is a slightly faster version:

s x y xss = let snip i zs = take 3 $ drop (i-1) zs 
                sqr = map (snip x) $ snip y xss
            in sum (concat sqr) - sqr !! 1 !! 1     

First we “snip out” the 3 x 3 square, then we do all calculations on it. Again, coordinates on the edges would lead to wrong results.