Step by Step solution about How to resolve the algorithm Pascal's triangle/Puzzle step by step in the Haskell programming language

This code is a constraint solver, written in Haskell, that can be used to solve systems of linear equations. It is given a puzzle, that is a matrix of numbers and variables, and it returns the solution to the system of equations that is defined by the puzzle. The code uses a technique called Gaussian elimination to solve the system of equations. The first step is to convert the puzzle into a system of equations. This is done by creating a list of equations, one for each row of the puzzle. The equations are in the form of a list of tuples, where the first element is the constant term, and the remaining elements are pairs of variables and coefficients. For example, the first row of the puzzle, ["151"], is converted into the equation [(0, 1:(-1):1:replicate n 0)], where n is the number of columns in the puzzle.

Once the system of equations has been created, the code uses Gaussian elimination to solve the system. Gaussian elimination is a technique that uses row operations to transform the system of equations into an equivalent system that is easier to solve. The row operations that are used are:

• Swapping two rows
• Multiplying a row by a constant
• Adding a multiple of one row to another row

The code uses these row operations to transform the system of equations into an upper triangular matrix. An upper triangular matrix is a matrix where all the elements below the main diagonal are zero. Once the system of equations has been transformed into an upper triangular matrix, it is easy to solve the system. The solution is found by back-substitution, which starts with the last equation and solves for the variables in that equation. The values of the variables in the last equation are then used to solve for the variables in the second-to-last equation, and so on.

Once the system of equations has been solved, the code normalizes the solution. Normalization is the process of converting the solution into a form that is easier to read and understand. The code normalizes the solution by converting all the fractions in the solution into integers.

The code can be used to solve a variety of puzzles. For example, it can be used to solve Sudoku puzzles, crossword puzzles, and other types of puzzles.

puzzle = [["151"],["",""],["40","",""],["","","",""],["X","11","Y","4","Z"]]


triangle n = n * (n+1) `div` 2

coeff xys x = maybe 0 id $ lookup x xys

row n cs = [coeff cs k | k <- [1..n]]

eqXYZ n = [(0, 1:(-1):1:replicate n 0)]
 
eqPyramid n h = do
  a <- [1..h-1]
  x <- [triangle (a-1) + 1 .. triangle a]
  let y = x+a
  return $ (0, 0:0:0:row n [(x,-1),(y,1),(y+1,1)])

eqConst n fields = do
  (k,s) <- zip [1..] fields
  guard $ not $ null s
  return $ case s of
    "X" - (0, 1:0:0:row n [(k,-1)])
    "Y" - (0, 0:1:0:row n [(k,-1)])
    "Z" - (0, 0:0:1:row n [(k,-1)])
    _   - (fromInteger $ read s, 0:0:0:row n [(k,1)])

equations :: [[String]] - ([Rational], [[Rational]])
equations puzzle = unzip eqs where
  fields = concat puzzle
  eqs = eqXYZ n ++ eqPyramid n h ++ eqConst n fields 
  h = length puzzle
  n = length fields


normalize :: [Rational] - [Integer]
normalize xs = [numerator (x * v) | x <- xs] where 
  v = fromInteger $ foldr1 lcm $ map denominator $ xs

run puzzle = map (normalize . drop 3) $ answer where
  (a, m) = equations puzzle
  lr = decompose 0 m
  answer = case solve 0 lr a of
    Nothing - []
    Just x  - x : kernel lr


*Main run puzzle
[[151,81,70,40,41,29,16,24,17,12,5,11,13,4,8]]
*Main run [[""],["2",""],["X","Y","Z"]]
[[3,2,1,1,1,0],[3,0,3,-1,1,2]]