Step by Step solution about How to resolve the algorithm Price fraction step by step in the Haskell programming language

The provided code snippet demonstrates two functions, price_fraction, each taking a single argument n. Both functions calculate a price fraction based on the value of n.

• The first version of the function uses pattern matching to handle different ranges of n values. It assigns specific price fractions for ranges like n < 0.06, n < 0.11, n < 0.16, and so on, up to n < 1.00.

• The second implementation uses a different approach with a table list of tuples. Each tuple in the table represents a range of n values, with the first element being the upper bound of the range, and the second element being the corresponding price fraction. The function uses the head function to find the first tuple where the n value is less than or equal to the upper bound, and returns the price fraction from that tuple.

Both versions of the function handle cases where n is outside the valid range (less than 0 or greater than 1) by raising an error.

price_fraction n
  | n < 0 || n > 1 = error "Values must be between 0 and 1."
  | n < 0.06 = 0.10
  | n < 0.11 = 0.18
  | n < 0.16 = 0.26
  | n < 0.21 = 0.32
  | n < 0.26 = 0.38
  | n < 0.31 = 0.44
  | n < 0.36 = 0.50
  | n < 0.41 = 0.54
  | n < 0.46 = 0.58
  | n < 0.51 = 0.62
  | n < 0.56 = 0.66
  | n < 0.61 = 0.70
  | n < 0.66 = 0.74
  | n < 0.71 = 0.78
  | n < 0.76 = 0.82
  | n < 0.81 = 0.86
  | n < 0.86 = 0.90
  | n < 0.91 = 0.94
  | n < 0.96 = 0.98
  | otherwise = 1.00


table = [
    (0.06, 0.10),   (0.11, 0.18),   (0.16, 0.26),   (0.21, 0.32),   (0.26, 0.38),
    (0.31, 0.44),   (0.36, 0.50),   (0.41, 0.54),   (0.46, 0.58),   (0.51, 0.62),
    (0.56, 0.66),   (0.61, 0.70),   (0.66, 0.74),   (0.71, 0.78),   (0.76, 0.82),
    (0.81, 0.86),   (0.86, 0.90),   (0.91, 0.94),   (0.96, 0.98),   (1.01, 1.00),
  ]

price_fraction n
  | n < 0 || n > 1 = error "Values must be between 0 and 1."
  | otherwise = snd $ head $ dropWhile ((<= n) . fst) table