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file_id.diz

APLPresent

Author:   pdw
Category: Christmas Challenge
System:   IBM 5110
Language: APL
Len source code: 40 characters
Len exe file:    N/A
Len code only:   N/A
Instructions:
  Run on the IBM 5110 emulator at 
  https://norbertkehrer.github.io/ibm_5110/emu5110.html
  (Or on any other APL implementation.)

  The 5110's screen only has 16 lines, so the screenshot uses a modified
  program that draws a smaller image.
Description:
  
  The full program is:

      ⍉(11⌽19↑'\O/'),' !-+'[1+A∘.+2×A←1=9|⍳19]

  Step by step:

        ⍝ The integers 1 to 19.
        ⍳19
  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
  
        ⍝ The remainder after division by 9.
        9|⍳19
  1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 8 0 1
  
        ⍝ Find the ones.
        1=9|⍳19
  1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1
  
        ⍝ And call this vector A.
        A←1=9|⍳19
  
        ⍝ Make a table of the sum of A and A doubled.
        A∘.+2×A
  3 1 1 1 1 1 1 1 1 3 1 1 1 1 1 1 1 1 3
  2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 2
  2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 2
  2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 2
  2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 2
  2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 2
  2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 2
  2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 2
  2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 2
  3 1 1 1 1 1 1 1 1 3 1 1 1 1 1 1 1 1 3
  2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 2
  2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 2
  2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 2
  2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 2
  2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 2
  2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 2
  2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 2
  2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 2
  3 1 1 1 1 1 1 1 1 3 1 1 1 1 1 1 1 1 3
  
        ⍝ Looking good. Let's call this B in this explanation.
        B←A∘.+2×A
  
        ⍝ Map this to characters.
        ' -!+'[1+B]
  +--------+--------+
  !        !        !
  !        !        !
  !        !        !
  !        !        !
  !        !        !
  !        !        !
  !        !        !
  !        !        !
  +--------+--------+
  !        !        !
  !        !        !
  !        !        !
  !        !        !
  !        !        !
  !        !        !
  !        !        !
  !        !        !
  +--------+--------+
  
        ⍝ Take the string '\O/', enlarge it to the required length, then
        ⍝ rotate it so that the \O/ is in the right position.
        11⌽19↑'\O/'
          \O/        
  
        ⍝ Append that to our previous result
        (11⌽19↑'\O/'),' -!+'[1+B]
   +--------+--------+
   !        !        !
   !        !        !
   !        !        !
   !        !        !
   !        !        !
   !        !        !
   !        !        !
  \!        !        !
  O+--------+--------+
  /!        !        !
   !        !        !
   !        !        !
   !        !        !
   !        !        !
   !        !        !
   !        !        !
   !        !        !
   +--------+--------+
  
        ⍝ Oops, that's the wrong axis. We could fix it like this.
        (11⌽19↑'\O/'),[1]' -!+'[1+B]
          \O/        
  +--------+--------+
  !        !        !
  !        !        !
  !        !        !
  !        !        !
  !        !        !
  !        !        !
  !        !        !
  !        !        !
  +--------+--------+
  !        !        !
  !        !        !
  !        !        !
  !        !        !
  !        !        !
  !        !        !
  !        !        !
  !        !        !
  +--------+--------+
  
        ⍝ But it's actually shorter to use the transpose operator.
        ⍉(11⌽19↑'\O/'),' !-+'[1+B]
          \O/   
  +--------+--------+ 
  !        !        !
  !        !        !
  !        !        !
  !        !        !
  !        !        !
  !        !        !
  !        !        ! 
  !        !        ! 
  +--------+--------+ 
  !        !        ! 
  !        !        ! 
  !        !        ! 
  !        !        ! 
  !        !        ! 
  !        !        ! 
  !        !        ! 
  !        !        ! 
  +--------+--------+