ASCII-Art Venn Diagram

Question

Given two lists that contain no duplicate elements a and b, find the crossover between the two lists and output an ASCII-Art Venn Diagram. The Venn Diagram will use a squarified version of the traditional circles for simplicity.

Example

Given:

a = [1, 11, 'Fox', 'Bear', 333, 'Bee']
b = ['1', 333, 'Bee', 'SchwiftyFive', 4]

Output (Order is 100% arbitrary, as long as the Venn Diagram is correct):

+-----+----+-------------+
|11   |333 |SchwiftyFive |
|Fox  |Bee |4            |
|Bear |1   |             |
+-----+----+-------------+

The program may either consider '1' == 1 or '1' != 1, up to your implementation. You may also choose to just handle everything as strings, and only accept string input.

---

Given:

a=[]
b=[1,2,3]

Output (Notice how the two empty parts still have the right-pad space):

+-+-+--+
| | |1 |
| | |2 |
| | |3 |
+-+-+--+

---

Given:

a=[1]
b=[1]

Output:

+-+--+-+
| |1 | |
+-+--+-+

Rules

• Elements of the Venn Diagram are left-aligned and padded to the max length entry plus 1.
• The ordering of the elements within sub-sections of the Venn-Diagram are arbitrary.
• Corners of the Venn Diagram (where | meets -) must be represented by a +.
• You are garuanteed that a.join(b).length() > 0, if both are empty, you may do whatever.
  • You may even print a picture of Abe Lincoln, don't care.
• This is code-golf, ascii-art and set-theory.

Bonus

Charcoal renders boxes like this naturally, but the whole set theory part... Don't know how well it does that. +100 bounty for the shortest charcoal submission before I am able to add a bounty to the question (2 days from being asked).

Answer 1

Husk, ⁷² 58 bytes

mṙ1₁Fz+m(₁TT' Ṡ:→S:öR'-→▲mL)TTømėF`-FnF-
Ṡz:(`:'+:'+R'|-2L

Try it online!

Husk does not support arrays of multiple types, so the arguments must be string arrays.

I am not sure whether I want to explain this monstrosity.

Due to it's sheer size, it takes about 28 secs(due to type inference) for the first test case mentioned. Without some of the type inference issues, this could be a lot shorter.

-14 bytes with a lot of adjustments(I got better at Husk).

Answer 2

Charcoal, 106 89 87 bytes:

Ａ⟦⟧ςＡ⟦⟧λＡ⟦⟧ρＡ⟦⟧τＷＳ⊞ςιＷＳ⊞⎇№ςιτριＦς⊞⎇№τι⟦⟧λιＦ⟦λτρ⟧«Ｆι«↓Ｐκ»ＭＬι↑←Ａ⁺⌈ＥιＬκ³ζＵＲζ⁺⌈⟦ＬλＬτＬρ⟧²Ｍζ→

Try it online! Note that link is to verbose code for explanatory purposes, with the -sl option that shows the equivalent native Charcoal code. Takes input as newline-separated strings with a blank line after each set.

Edit: Saved 11 bytes thanks to @ASCII-only. Previous version actually had a bug when the last word in the first set was not in the second set and also the first column was the tallest, which manifested as an apparently inability to optimise away a temporary. Saved 2 bytes by optimising two Move commands (the deverbosifier now does this automatically but the resulting code was always valid so the answer is still competing).

Edit: I don't think Multiprint used to work with multiline output but it does currently and making use of that would save 6 bytes, plus a further 4 bytes because current Charcoal preinitialises the u variable to the empty list: Try it online!

Answer 3

Python 2, 221 210 212 bytes

m=map
A,B=m(set,input())
d=A-B,B&A,B-A
e=[max(m(len,s))+1for s in d]
p,i,n='+|\n'
o=b=p+p.join(m('-'.__mul__,e))+p+n
while sum(m(len,d)):o+=i+i.join(m(str.ljust,[len(s)and s.pop()or''for s in d],e))+i+n
print o+b

Try it online!

Answer 4

PHP>=7.1, 287 Bytes

<?for([$a,$b]=$_GET,$x=max(($m=array_map)(count,$r=[($d=array_diff)($a,$b),array_intersect($a,$b),$d($b,$a)]));$n<3;$n++)for(sort($r[+$n]),$i=-1;$i<=$x;$i++){$o[$i].="|+"[$b=$i<0||$i==$x].str_pad($b?"":$r[+$n][$i],max($m(strlen,$r[+$n]))+1," -"[$b]).("|+"[$b][$n<2]);}echo join("
",$o);

Online Version

Expanded

for([$a,$b]=$_GET, # store input arrays in shorter variables
$x=max(($m=array_map)(count,   # get maximum of 
$r=[($d=array_diff)($a,$b),array_intersect($a,$b),$d($b,$a)])); #the set array
$n<3;$n++)
  for(sort($r[+$n]),$i=-1;$i<=$x;$i++){ # sort array to remove keys
    $o[$i].="|+"[$b=$i<0||$i==$x].   # concat line $b boolean for first and last line beginning char 
    str_pad($b?"":$r[+$n][$i]   # string of item in array if not first or last line
    ,max($m(strlen,$r[+$n]))+1  # fill till maximum length of items in array
    ," -"[$b]) # with char depends on first/last line or item line
    .("|+"[$b][$n<2]); # make end of string if last array is reach
}
echo join("   
",$o); #Output

Answer 5

05AB1E, 47 bytes

ʒå}©K®¹®Kr)õζ'|ìøε€SζøJ}ø'|δªðý¬„|+`:D¬мS'-:.ø»

Try it online or verify all test cases.

Explanation:

ʒ               # Filter the first (implicit) input-list by:
 å              #  Check if the current item is in the second (implicit) input-list
}©              # After the filter: store this overlap in variable `®` (without popping)
  K             # Pop and remove those items from the second (implicit) input-list
®               # Push overlap `®`
¹               # Push the first input-list again
 ®K             # Remove overlap `®`
r               # Reverse the three lists on the stack
 )              # And wrap them into a list
 ζ              # Zip/transpose this list, swapping rows/columns,
õ               # with "" as filler if the lists are of unequal length
  '|ì          '# Prepend an "|" in front of each inner string
ø               # Zip/transpose back, swapping rows/columns
 ε              # Map over each row:
  €             #  Map over each string in this row:
   S            #   Convert it to a list of characters
    ζ           #  Zip/transpose, swapping rows/columns, with " " as default filler
     ø          #  And then zip/transpose back
      J         #  And join the strings back together
 }ø             # After the map: zip/transpose back
  δ             # Map over each row:
'| ª           '#  And append an "|" to it
    ðý          # Then join each inner-most list by spaces
¬               # Push the first row (without popping the list)
 „|+`:          # Replace all "|" with "+"
      D         # Duplicate this string
       ¬        # Push its first characters (without popping), which is a "+"
        м       # Remove all "+" from the duplicated string
         S      # Convert what remains to a list of characters
          '-:  '# Replace each character (except for "+") with "-"
             .ø # Surround the list of rows with this string as leading/trailing items
»               # And finally join the list by newlines
                # (after which the result is output implicitly)