Visualize a Recursive Acronym

Question

Background

Famously, the acronym GNU stands for GNU's Not Unix. ¹

It's recursive because, after expanding it once, it still contains the acronym GNU, and so must be exanded again:

(GNU's Not Unix)'s Not Unix

And so on, ad infinitum. Visualizing this, we get a kind of Droste effect:

┌────────────────────────────────────────────┬───────────┐
│┌──────────────────────────────┬───────────┐│'s Not Unix│
││┌────────────────┬───────────┐│'s Not Unix││           │
│││┌──────────────┐│'s Not Unix││           ││           │
││││GNU's Not Unix││           ││           ││           │
│││└──────────────┘│           ││           ││           │
││└────────────────┴───────────┘│           ││           │
│└──────────────────────────────┴───────────┘│           │
└────────────────────────────────────────────┴───────────┘ 

Recursive acronyms need not recurse on the first word, or only once. For example:

• YOPY: Your Own Personal YOPY
• PIPER: PIPER Is PIPER Expanded Recursively

Visualized:

Challenge

Input

You will be given two inputs:

1. A string whose space-delimited words form a recursive acronym. That is, if you form a string from the first letter of each word, that string is guaranteed to be either:
   • One of the words of the input string (it may occur more than once).
   • A prefix of one or more of those words (e.g. GNU is a prefix of GNU's)
   • The casing will match exactly
2. A non-negative integer -- the number of times to recursively expand. Given 0, you'll return the input unaltered (or "framed" once, in its entirety). Given 1, you'll expand once. Etc.

Output

The output is the input string with all instances of the acronym visually expanded, recursively, the specified number of times.

You must use some visual effect to "frame" the nesting -- at minimum, distinct start and end delimiters like parentheses. Ascii boxing of some sort, as in the examples above, is also fine. As would be outputting an actual image that showed the nesting.

I'm flexible as long as the nesting is in fact visualized.

For clarity, parenthesized output would like this:

(((GNU's Not Unix)'s Not Unix)'s Not Unix)'s Not Unix

You are guaranteed that parentheses will never be part of acronym. Other than alphanumeric characters, the acronym will only contain apostrophes, commas, quotes, question marks and exclamation points, and those will only occur as valid punctuation (e.g., a question mark will not appear at the beginning of a word).

This is code golf, fewest bytes wins, no loopholes.

Test Cases

This assumes you're using a parentheses visualization.

Format for test cases:

1. Input string (the acronym)
2. Input integer (recursion)
3. Expected Output

---

1. GNU's Not Unix
2. 0
3. GNU's Not Unix

---

1. GNU's Not Unix
2. 2
3. ((GNU's Not Unix)'s Not Unix)'s Not Unix

---

1. YOPY Own Personal YOPY
2. 1
3. (YOPY Own Personal YOPY) Own Personal (YOPY Own Personal YOPY)

---

1. YOPY Own Personal YOPY
2. 2

3. ((YOPY Own Personal YOPY) Own Personal (YOPY Own Personal YOPY)) Own Personal ((YOPY Own Personal YOPY) Own Personal (YOPY Own Personal YOPY))
   

---

1. YourYOPY Own Personal YOPY
2. 2

3. YourYOPY Own Personal (YourYOPY Own Personal (YourYOPY Own Personal YOPY))
   

Answer 1

JavaScript (ES9),  88 85  82 bytes

Takes input as (string)(n).

s=>g=n=>n?s.replace(eval(`/\\b${s.match(/(?<=^| )./g).join``}/g`),`(${g(n-1)})`):s

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Commented

s =>                       // s = string
  g = n =>                 // g is a recursive function taking the counter n
    n ?                    // if n is not equal to 0:
      s.replace(           //   replace in s:
        eval(              //     each word matching:
          `/\\b${          //       a word boundary followed by
            s.match(       //       the acronym which consists of
              /(?<=^| )./g //         the letters that are preceded by
                           //         either a space or nothing at all
            ).join``       //       joined together
          }/g`             //
        ),                 //     with:
        `(${               //     the result of
          g(n - 1)         //       a recursive call
        })`                //     surrounded with parentheses
      )                    //   end of replace()
    :                      // else:
      s                    //   end of recursion: just return s

Answer 2

APL (Dyalog Extended), 33 32 bytes

-1 byte thanks to Bubbler.

Anonymous infix lambda. Takes count as left argument and string as right argument.

{('\b',⊃¨⍵⊆⍨≠⍵)⎕R(1⌽')(',⍵)⍣⍺⊢⍵}

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{…} "dfn"; number is ⍺ and text is ⍵:

 ⊢⍵ on the text:

 ⍣⍺ repeat given number of times

 (…)⎕R(…) PCRE Replace:

  ')(',⍵ the text prefixed by ")("

  1⌽ cyclically rotated one step left (puts ")" at end)

  … with:

  ≠⍵ The mask of non-spaces in the text

  ⍵⊆⍨ characters runs in the text corresponding to runs of 1s in that

  ⊃¨ first character of each

 '\b', prepend PCRE for word boundary

Answer 3

05AB1E, 27 25 22 bytes

ðìÐ#€нJðìs" (ÿ )"Iи.:¦

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To make sure we only match the acronym at the beginning of a word, we match an extra leading space. This results in extra spaces in the output, which is allowed.

ðì                      # prepend a space to the input
  Ð                     # triplicate
   #€нJ                 # acronymize (split on spaces, head of each, join)
       ðì               # prepend a space to the acronym
         s              # swap
          " (ÿ )"       # surround with parentheses
                 Iи     # repeat the following second-input times:
                   .:   #  replace the acronym with the parenthesized string
                     ¦  # remove the leading space

Answer 4

Perl 6, 48 bytes

{($^s,{S:g/$([~] $s~~m:g/<<./)/($s)/}...*)[$^x]}

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Explanation

{                                              }  # Anonymous block
 (   ,                               ...*)  # Construct infinite sequence
  $^s                                       # starting with input string.
      {                             }  # Compute next item by
       S:g/                   /    /   # globally replacing
           $(                )         # interpolated
                 $s~~m:g/<<./          # first letters of each word
             [~]                       # joined (the acronym)
                               ($s)    # with input string in parens
                                          [$^x]  # Take nth item

Answer 5

Python 3, 104 96 bytes

-7 bytes thanks to @DeathIncarnate, and another -1 byte thanks to @wastl

import re
f=lambda s,n:n and re.sub('\\b'+''.join(c[0]for c in s.split()),f'({s})',f(s,n-1))or s

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Explanation:

re.sub(
  '\\b'+''.join(c[0]for c in s.split()), # match all acronyms that starts on a word boundary
  '('+s+')',                             # replace with the original string in parenthesis
  f(s,n-1)                               # on the result of the (n-1)th recursion
)

This is the acronym expanded n times.

return n and recursive_case or base_case

This returns base_case if n is 0, and recursive_case otherwise. This works because in Python, 0 is Falsy and non-empty string is Truthy.

• If n is 0, n and recursive_case short circuits to False, then False or base_case is evaluated which returns base_case.
• If n is not 0, n and recursive_case evaluates to recursive_case which is Truthy, so recursive_case or base_case is short-circuited to recursive_case.

Answer 6

Perl 5 -apl, 49 43 bytes

/./,$p.=$&for@F;for$a(1..<>){s/\b$p/(@F)/g}

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Answer 7

Retina, 48 bytes

~(`$
 ¶($%`)
(.).*? (?=.*¶)
$1
^(.+)¶
1A`¶$1+`\b

Try it online! Takes the expansion count and acronym string on separate lines. Explanation:

~(`

After performing the script, interpret the result as a new script to be evaluated on the original input.

$
 ¶($%`)

Append a space to the string, and append a copy of the string wrapped in ()s on a new line.

(.).*? (?=.*¶)
$1

Turn the string into the acronym.

^(.+)¶
1A`¶$1+`\b

Replace the count with a Retina repeat instruction, and also prefix an instruction to delete the count from the original input.

For example, the result of evaluating the inner script on the input 2 GNU's Not Unix is

1A`
2+`\bGNU
(GNU's Not Unix)

This deletes the count from the original input, and then performs the acronym expansion twice.

Answer 8

Python 2, 93 bytes

f=lambda s,n:n and re.sub('\\b'+''.join(zip(*s.split())[0]),'('+s+')',f(s,n-1))or s
import re

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Similar to a few other answers.

If simple substitutions were allowed, I could do...

Python 2, 78 bytes

f=lambda s,n:n and f(s,n-1).replace(''.join(zip(*s.split())[0]),'('+s+')')or s

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Except I didn't see the YourYOPY example when I put this together...

Answer 9

sed 4.2.2, 180 bytes

h
s/(.)[^ ]* */\1/g
G
s/$/\n/
G
h
N
s/.*\n//
tx
:l
s/^(.*)\n(.*)\b\1(.*)\n(.*)\n(.*)/\1\n\2\n(\5)\3\4\n\5/
tl
s/\n(.*)\n(.*)\n/\n\1\2\n\n/
x
tx
:x
s/.//
x
tl
s/.*\n(.*)\n.*\n.*/\1/

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Well, finding one part of input in another part of input is not that simple in sed. Input is on two lines: first line is the acronym string; second line is number of expansions in unary.

Answer 10

Python 3, ^{123 106} 118 bytes

Rolled back to my initial post.
Chas Brown kindly pointed out I was going down a rabbit hole.
Added 17 bytes to fix word boundary bug.

import re
def f(a,n):
 o=a
 for i in[1]*n:a=re.sub(r'\b'+''.join([i[0]for i in o.split()])+r'\b',f"({o})",a)
 return a

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