Step by Step solution about How to resolve the algorithm Catalan numbers/Pascal's triangle step by step in the Julia programming language

The provided Julia code calculates the Catalan numbers using Pascal's triangle. Catalan numbers are frequently encountered in combinatorial problems, such as counting the number of ways to triangulate a polygon.

Here's a detailed breakdown of the code:

1. pascal(n::Int) function:

   • This function generates Pascal's triangle up to row n, resulting in an n x n matrix r.
   • Pascal's triangle is initialized with ones(Int, n, n) matrix, meaning all values are initially set to 1.
   • It iterates through rows i and columns j from 2 to n (skipping the first row and column).
   • For each element r[i, j], it is calculated as the sum of the element above it (r[i-1, j]) and the element to its left (r[i, j-1]).
   • This process builds Pascal's triangle, with each element representing the binomial coefficient binomial(i+j-2, j-1).

2. catalan_num(n::Int) function:

   • This function calculates the nth Catalan number.
   • It first generates Pascal's triangle up to row n+2 using the pascal function.
   • It then extracts specific elements from the Pascal's triangle matrix p:
     • p[n+4:n+3:end-1]: This extracts every third element starting from row n+4 (i.e., p[n+4], p[n+7], p[n+10], ...).
     • diag(p, 2): This extracts the diagonal elements of p shifted by 2 positions to the right (i.e., p[2, 4], p[3, 5], p[4, 6], ...).
   • The difference between these two extracted elements (p[n+4:n+3:end-1] - diag(p, 2)) gives the n-th Catalan number.

3. The last line of the code:

   • @show catalan_num(15): This displays the 15th Catalan number.

# v0.6

function pascal(n::Int)
    r = ones(Int, n, n)
    for i in 2:n, j in 2:n
        r[i, j] = r[i-1, j] + r[i, j-1]
    end
    return r
end

function catalan_num(n::Int)
    p = pascal(n + 2)
    p[n+4:n+3:end-1] - diag(p, 2)
end

@show catalan_num(15)