Using list elements and indices together Using list elements and indices together arrays arrays

Using list elements and indices together


Borrowing enumerate is fine and encouraged. However, it can be made a bit lazier by refusing to calculate the length of its argument:

enumerate = zip [0..]

(In fact, it's common to just use zip [0..] without naming it enumerate.) It's not clear to me why you think your second example should be costlier in either time or space. Remember: indexing is O(n), where n is the index. Your complaint about the unwieldiness of fst and snd is justified, and can be remedied with pattern-matching:

validQueens' xs = and [abs (y - x) /= j - i | (i, x) <- l, (j, y) <- l, i < j]    where l = zip [0..] xs

Now, you might be a bit concerned about the efficiency of this double loop, since the clause (j, y) <- l is going to be running down the entire spine of l, when really we just want it to start where we left off with (i, x) <- l. So, let's write a function that implements that idea:

pairs :: [a] -> [(a, a)]pairs xs = [(x, y) | x:ys <- tails xs, y <- ys]

Having made this function, your function is not too hard to adapt. Pulling out the predicate into its own function, we can use all instead of and:

validSingleQueen ((i, x), (j, y)) = abs (y - x) /= j - ivalidQueens' xs = all validSingleQueen (pairs (zip [0..] xs))

Or, if you prefer point-free notation:

validQueens' = all validSingleQueen . pairs . zip [0..]


Index-element tuples are quite a common thing to do in Haskell. Because zip stops when the first list stops, you can write them as

enumerate x = zip [0..] x

which is both more elegant and more efficient (as it doesn't compute length x up front). In fact I wouldn't even bother naming it, as zip [0..] is so short.

This is definitely more efficient than iterating by index for lists, because !! is linear in the second argument due to lists being linked lists.

Another way you can make your program more elegant is to use pattern-matching instead of fst and snd:

validQueens' :: [Int] -> BoolvalidQueens' x = and [abs (j2 - i2) /= j1 - i1 | (i1, i2) <-l, (j1, j2) <-l, j1 > i1]                   where l = zip [0..] x