How to resolve the algorithm Sierpinski carpet step by step in the NetRexx programming language

Published on 12 May 2024 09:40 PM
#Netrexx

How to resolve the algorithm Sierpinski carpet step by step in the NetRexx programming language

Table of Contents

Problem Statement

Produce a graphical or ASCII-art representation of a Sierpinski carpet of order   N.

For example, the Sierpinski carpet of order   3   should look like this: The use of the   #   character is not rigidly required for ASCII art. The important requirement is the placement of whitespace and non-whitespace characters.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Sierpinski carpet step by step in the NetRexx programming language

Source code in the netrexx programming language

/* NetRexx */
options replace format comments java crossref symbols nobinary

numeric digits 1000
runSample(arg)
return

-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method runSample(arg) public static
  DARK_SHADE = '\u2593'
  parse arg ordr filr .
  if ordr = '' | ordr = '.' then ordr = 3
  if filr = '' | filr = '.' then filler = DARK_SHADE
  else                           filler = filr
  drawSierpinskiCarpet(ordr, filler)
  return

-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method drawSierpinskiCarpet(ordr, filler = Rexx '@') public static binary
  x = long
  y = long
  powr = 3 ** ordr
  loop x = 0 to powr - 1
    loop y = 0 to powr - 1
      if isSierpinskiCarpetCellFilled(x, y) then cell = filler
      else                                       cell = ' '
      say cell'\-'
      end y
      say
    end x
  return

-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method isSierpinskiCarpetCellFilled(x = long, y = long) public static binary returns boolean
  isTrue  = boolean (1 == 1)
  isFalse = \isTrue
  isFilled = isTrue
  loop label edge while x \= 0 & y \= 0
    if x // 3 = 1 & y // 3 = 1 then do
      isFilled = isFalse
      leave edge
      end
    x = x % 3
    y = y % 3
    end edge
  return isFilled

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