How to resolve the algorithm 99 bottles of beer step by step in the Processing programming language

Published on 12 May 2024 09:40 PM

How to resolve the algorithm 99 bottles of beer step by step in the Processing programming language

Table of Contents

Problem Statement

Display the complete lyrics for the song:     99 Bottles of Beer on the Wall.

The lyrics follow this form: ... and so on, until reaching   0     (zero). Grammatical support for   1 bottle of beer   is optional. As with any puzzle, try to do it in as creative/concise/comical a way as possible (simple, obvious solutions allowed, too).

Let's start with the solution:

Step by Step solution about How to resolve the algorithm 99 bottles of beer step by step in the Processing programming language

Source code in the processing programming language

for (int i = 99; i > 0; i--) {
  print(i + " bottles of beer on the wall\n"
    + i + " bottles of beer\nTake one down, pass it around\n"
    + (i - 1) + " bottles of beer on the wall\n\n");
}

int i = 99;
void setup() {
  size(200, 140);
}
void draw() {
  background(0);
  text(i + " bottles of beer on the wall\n"
     + i + " bottles of beer\nTake one down, pass it around\n"
 + (i-1) + " bottles of beer on the wall\n\n", 
    10, 20);
  if (frameCount%240==239) next();  // auto-advance every 4 secs
}
void mouseReleased() {
  next();  // manual advance
}
void next() {
  i = max(i-1, 1);  // stop decreasing at 1-0 bottles
}

i = 99
def setup():
    size(200, 140)

def draw():
    background(0)
    text("{} bottles of beer on the wall\n".format(i) +
         "{} bottles of beer\n".format(i) + 
         "Take one down, pass it around\n" +
         "{} bottles of beer on the wall\n\n".format(i - 1),
         10, 20)
    if frameCount % 240 == 239:  # auto-advance every 4 secs
         next()
         
def mouseReleased():
    next()  # manual advance

def next():
    global i
    i = max(i - 1, 1)  # stop decreasing at 1-0 bottles

setup <- function() {
  stdout$print(bottlesong(99))
}

bottlesong <- function(num) {
  verses = ""
  for(i in num:1){
    verses = paste0(verses,
        num," bottles of beer on the wall \n",
        num," bottles of beer \n",
        "Take one down, pass it around \n",
        num-1, " bottles of beer on the wall \n\n", sep="");
    num <- num - 1
  }
  return(verses)
}

  

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