Step by Step solution about How to resolve the algorithm Sum and product puzzle step by step in the Haskell programming language

Puzzle:

The code solves a mathematical puzzle: find the set of pairs of integers (x, y) such that:

• 1 < x < y < 100
• x + y < 100
• There exists only one other pair within the set of solutions (s1) that has the same sum as the pair (x, y)
• There exists only one other pair within the set of solutions that has the same product as the pair (x, y)

Solution:

The code follows these steps:

1. Generate Pairs (s1):

   • Generate all pairs of integers (x, y) that satisfy the initial constraints (line 14).

2. Filter Pairs by Sum (s2):

   • For each pair p, find all other pairs q that have the same sum as p (line 16).
   • Filter out pairs p where the number of pairs q with the same sum is not equal to 1 (line 17).

3. Filter Pairs by Product (s3):

   • For each pair p from s2, find all other pairs q that have the same product as p (line 19).
   • Filter out pairs p where the number of pairs q with the same sum is not equal to 1 (line 20).

4. Filter Pairs by Intersection (s4):

   • For each pair p from s3, find all other pairs q that have the same sum as p (line 22).
   • Filter out pairs p where the number of pairs q with the same sum is not equal to 1 (line 23).

5. Print Result:

   • Print the list of pairs s4 that satisfy all the conditions (line 26).

Additional Notes:

• The intersect function from Data.List is used to find the intersection of two lists.
• The >>= operator is the "bind" operator from Haskell's monadic syntax, which allows for chaining together computations.
• The filter function is used to filter a list based on a predicate function.
• The all function checks if all elements in a list satisfy a predicate function.
• The uncurry function converts a function that takes a tuple as an argument to a function that takes individual arguments.

import Data.List (intersect)

s1, s2, s3, s4 :: [(Int, Int)]
s1 = [(x, y) | x <- [1 .. 100], y <- [1 .. 100], 1 < x && x < y && x + y < 100]

add, mul :: (Int, Int) -> Int
add (x, y) = x + y
mul (x, y) = x * y

sumEq, mulEq :: (Int, Int) -> [(Int, Int)]
sumEq p = filter (\q -> add q == add p) s1
mulEq p = filter (\q -> mul q == mul p) s1

s2 = filter (\p -> all (\q -> (length $ mulEq q) /= 1) (sumEq p)) s1
s3 = filter (\p -> length (mulEq p `intersect` s2) == 1) s2
s4 = filter (\p -> length (sumEq p `intersect` s3) == 1) s3

main = print s4


import Data.List (intersect)

------------------ SUM AND PRODUCT PUZZLE ----------------

s1, s2, s3, s4 :: [(Int, Int)]
s1 =
  [2 .. 100]
    >>= \x ->
      [succ x .. 100]
        >>= \y ->
          [ (x, y)
            | x + y < 100
          ]

s2 = filter (all ((1 /=) . length . mulEq) . sumEq) s1

s3 = filter ((1 ==) . length . (`intersect` s2) . mulEq) s2

s4 = filter ((1 ==) . length . (`intersect` s3) . sumEq) s3

sumEq, mulEq :: (Int, Int) -> [(Int, Int)]
sumEq p = filter ((add p ==) . add) s1
mulEq p = filter ((mul p ==) . mul) s1

add, mul :: (Int, Int) -> Int
add = uncurry (+)
mul = uncurry (*)

--------------------------- TEST -------------------------
main :: IO ()
main = print $ take 1 s4