Haskell- creating 2d array

Question

I am fairly new to Haskell and trying to improve myself as far as functional languages are concerned. I am attempting to create a function that will take a number of rows and columns and produce a 2D array as follows, for example:

 arrayMaker :: (Int, Int) -> [[Int]]
  -- use replicate function somehow to create each row?

The resulting matrix would appear as follows (for 5 rows, 3 columns):

 [[14, 15, 16],
 [13, 12, 11],
 [8, 9, 10],
 [7, 6, 5],
 [2, 3, 4]]

Notice the values begin in the last row at 2. Also the array should alternate between ascending and descending rows. Help would be greatly appreciated

Write a function row that given i generates the i-th row. Then map row [1..numRows]. To do that, can you write a simple arithmetical formula for the number in position (i,j) when i is even? and when i is odd? — chi
– chi, Commented Nov 14, 2014 at 18:00
Note that [Int] in haskell is not an array but a list. In practice you typically would not want to use lists to represent a matrix. — DanielM
– DanielM, Commented Nov 14, 2014 at 18:16

ErikR · Accepted Answer · 2023-12-27 06:49:30Z

3

Let's start with this example:

rowMaker1 n k = [ n .. n+k-1 ] : rowMaker1 (n+k) k

rowMaker1 creates groups of consecutive numbers. It creates an infinite list, so we use take to limit the number of elements computed:

ghci> take 5 (rowMaker1 1 3)
[ [1,2,3], [4,5,6], [7,8,9], [10,11,12], [13,14,15] ]

We can also create a descending version:

rowMaker2 n k = [ n,n-1..n-k+1 ] : rowMaker2 (n-k) k

ghci> take 5 (rowMaker2 16 3)
[[16,15,14],[13,12,11],[10,9,8],[7,6,5],[4,3,2]]

The goal will be to define arrayMaker like this:

arrayMaker (nrows,ncols) = take nrows (rowMaker start ncols)
  where start = ???
        rowMaker n k = ???

We just need a rowMaker which will alternate between ascending and descending groups.

Does this help?

If you get stuck, here's one possibility for the rowMaker function:

rowMaker n k = [ n .. n+k-1 ] : rowMaker2 (n-k) k rowMaker2 n k = [ n, n-1 .. n-k+1 ] : rowMaker (n-k-k+1) k

and you can run this code here: http://tpcg.io/_YJENGK

edited Dec 27, 2023 at 6:49

answered Nov 14, 2014 at 18:09

ErikR

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6 Comments

GregH Over a year ago

sort of. Not sure how to tie rowMaker in to rowMaker1 and rowMaker2

ErikR Over a year ago

How about having rowMaker create two groups - an ascending group and then a descending group - and then repeat by calling itself.

GregH Over a year ago

How would you do this? I am terrible with recursion and that is a large reason of starting Haskell. However, I've been having trouble finding examples pertaining to what I am working with

ErikR Over a year ago

It will have the form: rowMaker n k = (formula for asc group) : (formula for desc group) : rowMaker new-n k - the colon operator is the basic way to create lists. a:b:c prepends the elements a and b to the front of list c.

ErikR Over a year ago

@JimWitschey Ok - replaced the lpaste link with a spoiler block

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