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

Published on 12 May 2024 09:40 PM
#Picat

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

Source code in the picat programming language

beer1(N) => 
   Beer = N,
   while (Beer > 0)
       printf("%d bottles of beer on the wall,\n", Beer),
       printf("%d bottles of beer.\n", Beer),
       printf("Take one down, pass it around.\n"),
       printf("%d bottles of beer.\n", Beer-1),
       Beer := Beer -1
   end,
   print("0 more bottles of beer on the wall.\n"),
   nl.

% With plurals.
beer2(B) = S =>
   BS = B.to_string(),
   BB = " bottle",
   BT = BB,
   if B > 1 then BB := BB ++ "s" end,
   OB = " of beer",
   NL = "\n",
   BW = OB ++ " on the wall." ++ NL,
   T = "Take one down, pass it around." ++ NL,
   S1 =  BS ++ BT ++ BW ++ BS ++ BT ++ OB ++ T ++
        cond(B > 0, (B-1).to_string() ++ BT ++ BW ++ NL, ""),
   S = S1.

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