Step by Step solution about How to resolve the algorithm Reduced row echelon form step by step in the Haskell programming language

The provided Haskell code implements the Gauss-Jordan elimination algorithm to reduce a matrix to reduced row echelon form (RREF). A matrix in RREF has the following properties:

1. Each row contains exactly one leading 1 (pivot).
2. The pivots are on the diagonal, from the top-left corner.
3. All rows below a pivot row are zero.
4. All columns containing a pivot are zero except for the pivot row.

The rref function takes a matrix m as input and returns its RREF. It does this by calling the recursive function f with the matrix m, the current lead column lead, and the remaining rows rs.

The function f first checks if there are any nonzero elements in the current lead column below the current row. If not, the matrix is already in RREF and the function returns the matrix.

If there is a nonzero element, the function finds the row index of the first nonzero element in the current lead column and swaps the current row with that row. It then multiplies the new current row by the reciprocal of the pivot element (the element in the current lead column and row) to make the pivot element equal to 1.

The function then subtracts multiples of the new current row from the other rows to zero out the elements in the current lead column. It updates the matrix and calls itself recursively with the next lead column and the remaining rows.

The replace function is a helper function that replaces the element at a given index in a list. It is used to swap rows and to zero out elements in the matrix.

The time complexity of the Gauss-Jordan elimination algorithm is O(n^3), where n is the number of rows and columns in the matrix.

import Data.List (find)

rref :: Fractional a => [[a]] -> [[a]]
rref m = f m 0 [0 .. rows - 1]
  where rows = length m
        cols = length $ head m

        f m _    []              = m
        f m lead (r : rs)
            | indices == Nothing = m
            | otherwise          = f m' (lead' + 1) rs
          where indices = find p l
                p (col, row) = m !! row !! col /= 0
                l = [(col, row) |
                    col <- [lead .. cols - 1],
                    row <- [r .. rows - 1]]

                Just (lead', i) = indices
                newRow = map (/ m !! i !! lead') $ m !! i

                m' = zipWith g [0..] $
                    replace r newRow $
                    replace i (m !! r) m
                g n row
                    | n == r    = row
                    | otherwise = zipWith h newRow row
                  where h = subtract . (* row !! lead')

replace :: Int -> a -> [a] -> [a]
{- Replaces the element at the given index. -}
replace n e l = a ++ e : b
  where (a, _ : b) = splitAt n l