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

The provided Julia code defines two functions named pascal to generate and print Pascal's triangle. Here's a detailed explanation of both functions:

1. First pascal function:

• iround(x): This function converts a floating-point number x to its nearest rounded integer value of type Int64.

• triangle(n): This function generates the coefficients of the rows of Pascal's triangle up to row n using a matrix exponential and returns them as an array of integers.

• pascal(n): This function prints the first n rows of Pascal's triangle using the triangle function and the println function. It iterates through the rows from 1 to n+1 and prints the non-zero elements of each row, separated by spaces.

2. Second pascal function:

• pascal(n): This is an alternative implementation of the Pascal's triangle generation function. It takes a positive integer n as input. It checks if n is less than or equal to 0, and if so, it throws an error because Pascal's triangle cannot have zero or negative rows.

• The function initializes two vectors r and pr with n elements, both initially filled with undefined values. It then sets the first element of both vectors to 1.

• It enters a loop that iterates from 2 to n. In each iteration, it updates the first element of r and pr to 1.

• For each element in r from index 2 to i-1, it updates the value by adding the corresponding elements from pr.

• It prints the current row of Pascal's triangle by joining the elements of r from index 1 to i using the join function and separating them with spaces.

• Finally, it updates r and pr by swapping their roles (r becomes pr, and pr becomes r) and continues to the next iteration.

iround(x) = round(Int64, x)

triangle(n) = iround.(exp(diagm(-1=> 1:n)))

function pascal(n)
   t=triangle(n)
   println.(join.([filter(!iszero, t[i,:]) for i in 1:(n+1)], " "))
end


function pascal(n)
   (n<=0) && error("Pascal trinalge can not have zero or negative rows")
   r=Vector{Int}(undef,n)
   pr=Vector{Int}(undef,n)
   pr[1]=r[1]=1
   println(@view pr[1])
   for i=2:n
      r[1]=r[i]=1
      for j=2:i-1
         r[j]=pr[j-1]+pr[j]
      end
      println(join(view(r,1:i), " "))
      r,pr=pr,r
   end
end