APLPresent

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/
+--------+--------+
! ! !
! ! !
! ! !
! ! !
! ! !
! ! !
! ! !
! ! !
+--------+--------+
! ! !
! ! !
! ! !
! ! !
! ! !
! ! !
! ! !
! ! !
+--------+--------+