Step by Step solution about How to resolve the algorithm Numbers with equal rises and falls step by step in the Haskell programming language

This Haskell code defines a function riseEqFall that checks if the sum of the digits that increase is equal to the sum of the digits that decrease in a given number.

1. pairs function: It takes a list and returns a list of pairs of consecutive elements.

2. riseEqFall function:

   • It takes an integer n and converts it to a string using show n.
   • It then maps each character in the string to an integer using map digitToInt.
   • It computes the pairs of digits using pairs.
   • It defines rel as a function that takes a comparator (e.g., > or <) and applies it to each pair of digits.
   • It calculates the sum of the results of applying rel to each pair using sum.
   • Finally, it checks if the sum of increasing digits is equal to the sum of decreasing digits.

3. a296712 list:

   • It is a list comprehension that generates a list of all integers n from 1 to infinity that satisfy the riseEqFall condition.

4. main function:

   • It prints the first 200 numbers in the a296712 list.
   • It prints the 10,000,000th number in the a296712 list.

import Data.Char

pairs :: [a] -> [(a,a)]
pairs (a:b:as) = (a,b):pairs (b:as)
pairs _        = [] 

riseEqFall :: Int -> Bool
riseEqFall n = rel (>) digitPairs == rel (<) digitPairs
    where rel r = sum . map (fromEnum . uncurry r)
          digitPairs = pairs $ map digitToInt $ show n

a296712 :: [Int]
a296712 = [n | n <- [1..], riseEqFall n]

main :: IO ()
main = do
	putStrLn "The first 200 numbers are: "
	putStrLn $ unwords $ map show $ take 200 a296712
	putStrLn ""
	putStr "The 10,000,000th number is: "
	putStrLn $ show $ a296712 !! 9999999