[Haskell-cafe] Matrices in Haskell

Vincent Kraeutler vincent at kraeutler.net
Tue Mar 20 05:02:00 EDT 2007

Matthew Brecknell wrote:
> Ivan Miljenovic:
>> As such, I'd like to know if there's any way of storing a an n-by-n
>> matrix such that the algorithm/function to get either the rows or the
>> columns is less than O(n^2) like transposition is.  I did try using an
>> Array, but my (admittedly hurried and naive) usage of them took longer
>> than a list of lists, was more difficult to manipulate, and wasn't
>> required separate inverse functions to row and cols.  Also, since I
>> need to look all throughout the list of values, it's still O(n^2), but
>> this time for both rows and columns.
> I'm not sure I see the problem, since any operation that touches all the
> elements of a n-by-n matrix will be at least O(n^2). For such an
> operation, a transposition should just add a constant factor.
> When you tried using Arrays, I presume you used an array indexed by a
> pair (i,j), and just reversed the order of the index pair to switch from
> row-wise to column-wise access? It's hard to see how that would slow you
> down. Perhaps the slowdown was caused by excessive array copying?
> Immutable arrays can be efficient for algorithms that write a whole
> array at once, but usually not for algorithms that modify one element at
> a time.
> I think you'll need to show specific functions that are performing below
> expectation before the list will be able to help.
> For problems like latin squares and sudoku, also try thinking "outside
> the matrix". Do you really need to structure the problem in terms of
> rows and columns? What about a set of mutually-intersecting sets of
> cells?
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i think you might also want to check up on


if you intend to do a significant number of incremental updates, it is my
(not particularly well-informed) understanding that you should use either
mutable arrays (Data.Array.ST  together with runST), or Data.Array.Diff
with explicit sequencing.

both approaches are discussed in the above wikipedia entry.


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