Find a recursive acronym

Question

Introduction

A recursive acronym is an acronym that contains or refers to itself, for example: Fish could be a recursive acronym for Fish is shiny hero, notice how that also contains the acronym itself. Another example is Hi -> Hi igloo. Or even ppcg paints -> ppcg paints cool galaxies pouring acid into night time stars

So basically, a sentence is a recursive acronym if the first letters of each of the words spell out the first word or words.

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Challenge

Make a program that takes a string of 1 or more words separated by a space character, and outputs a recursive acronym, or an empty string if it's impossible. It is impossible to make a recursive acronym for a string like, for example, ppcg elephant because you would start by taking the p from ppcg then adding that to the acronym, then taking the e from elephant. But now we have a contradiction, since the acronym currently spells out "pe..", which conflicts with "pp..". That's also the case with, for example, hi. You would take the h from hi, but the sentence is now over and there are no more letters to spell out hi and we are just left with h which doesn't match hi. (The string needs an amount of words more than or equal to the amount of letters in the acronym)

Input and output are not case sensitive

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Restrictions

• Anything inputted into your program will be valid English words. But you must make sure to output valid English words too (you can use a database or just store a word for each of the 26 letters)
• Standard loopholes and default IO rules apply

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Test Cases

hi igloo -> hi
ppcg paints -> (impossible)
ppcg paints cool giraffes -> ppcg
ppcg paints cool galaxies pouring acid into night time stars -> ppcgpaints
ppcg paints cool galaxies pouring acid into night time -> ppcg
ppcg questions professional pool challengers greatly -> (impossible)
I -> I

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Scoring

This is code-golf, so the smallest source code in bytes wins

Answer 1

Japt, 13 bytes

¸
mά
VøUÎ ©V

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Answer 2

05AB1E, 16 bytes

ð¡©ηʒJ®€нJηså}θJ

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Answer 3

Haskell, 51 48 bytes

Edit: -3 bytes thanks to @xnor.

(\w->[r|p<-scanl1(++)w,map(!!0)w==p,r<-p]).words

Finds acronym.

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\w->            .words -- let 'w' be the input list split into words
   p<-scanl1(++)w      -- loop 'p' through the list starting with the first word
                       --  and appending the next words one by one, e.g.
                       --  "Fish","is","shiny","hero" -> "Fish","Fishis","Fishisshiny","Fishisshinyhero"
     ,map(!!0)w==p     -- if the word made out of the first characters of the
                       --  words of 'w' equal 'p'
  [r|   r<-p]          -- return the letters of 'p' - if the check before
                       --  never evaluates to True then no letters, i.e. the
                       --  the empty string is returned

Answer 4

Perl 6, 50 42 58 49 bytes

-9 bytes thanks to nwellnhof

{~first {m:g/<<./.join~~/^$^a/},[R,] [\~] .words}

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First option. I'm exploiting the fact the ord only returns the ordinal value of first letter of a string, while chrs takes a list of ords and returns a string. Or the regex from moonheart's answer is shorter :(. For reference, the previous answer was .words>>.ord.chrs instead of [~] m:g/<<./

Explanation:

{~first {m:g/<<./.join~~/^$^a/},[R,] [\~] .words}
{                                               } # Anonymous code block
  first  # Find the first 
                                [R,] [\~] .words  # Of the reverse of the triangular joined words
         {                    }  # That matches:
          m:g/   /   # Match all from the original string
              <<.    # Single letters after a word boundary
                  .join    # Joined
                       ~~/^$^a/   # And starts with the given word
 ~  # And stringify Nil to an empty string

Answer 5

Retina 0.8.2, 60 bytes

^
$'¶
\G(\w)\w* ?
$1
+`^(.+)(\w.*¶\1 )
$1 $2
!`^(.+)(?=¶\1 )

Try it online! Finds the recursive acronym, if any. Explanation:

^
$'¶

Duplicate the input.

\G(\w)\w* ?
$1

Reduce the words on the first line to their initial letters.

+`^(.+)(\w.*¶\1 )
$1 $2

Insert spaces to match the original words, if possible.

!`^(.+)(?=¶\1 )

Output the first line if it is a prefix of the second line.

Answer 6

Perl 6, 56 bytes

$!=[~] m:g{<<.};say $! if m:g{<<\w+}.map({$_ eq $!}).any

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Previously regexes were confusing and unusable to me. Suddenly I understand them perfectly. What happened to me :P

Fulfills choice 1.

Answer 7

K (ngn/k), 40 bytes

First option:

{$[1=#:x;x;$[(*:t)~,/*:'t:" "\x;*:t;`]]}

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Answer 8

Rust, 155, try it online!

Selected: Problem 1: Finding acronym

type S=String;fn f(t:&str)->S{let l=t.to_lowercase();let w=l.split(' ').fold(S::new(),|a,b|a+&b[..1])+" ";if (l+" ").contains(w.as_str()){w}else{S::new()}}

Ungolfed, just a bit:

fn f(t: &str) -> String {
    let l = t.to_lowercase();
    let w = l.split(' ').fold(String::new(), |a, b| a + &b[0..1]) + " ";
    if (l + " ").contains(w.as_str()) {
        w
    } else {
        String::new()
    }
}

Or if we can assume that the input is all lowercase, just 130:

type S=String;fn f(l:S)->S{let w=l.split(' ').fold(S::new(),|a,b|a+&b[..1])+" ";if (l+" ").contains(&w.as_str()){w}else{S::new()}}

Answer 9

Jelly, 9 bytes

Ḳµ;\fZḢWƊ

A full-program printing the recursive abbreviation if it is possible.

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How?

Ḳµ;\fZḢWƊ - Main Link: list of characters
Ḳ         - split at space (let's call this v)
 µ        - start a new monadic chain (i.e. f(v)):
   \      - cumulative reduce v with:
  ;       -   concatenation -> [v(1), v(1);v(2), v(1);v(2);v(3); ...]
        Ɗ - last three links as a monad (i.e. f(v)):
     Z    -   transpose -> [[v(1)[1], v(2)[1], ...],[v(1)[1],v(2)[2],...],...]
      Ḣ   -   head -> [v(1)[1], v(2)[1], ...] ... i.e. 'the potential abbreviation'
       W  -   wrap in a list -> ['the potential abbreviation']
    f     - filter discard those from the left list that are not in the right list
          - implicit print -- a list of length 0 prints nothing
          -                   while a list of a single item prints that item

Answer 10

JavaScript [ES6], 74 bytes

s=>s.split` `.map(w=>(b+='('+w,e+=')?',t+=w[0]),b=e=t='')&&t.match(b+e)[0]

Creates a regular expression to match on. See examples in code.

All test cases:

let f=

s=>s.split` `.map(w=>(b+='('+w,e+=')?',t+=w[0]),b=e=t='')&&t.match(b+e)[0]

console.log(f('hi igloo'))
// 'hi'.match('(hi(igloo)?)?')[0] == 'hi'

console.log(f('ppcg paints'))
// 'pp'.match('(ppcg(paints)?)?')[0] == ''

console.log(f('ppcg paints cool giraffes'))
// 'ppcg'.match('(ppcg(paints(cool(giraffes)?)?)?)?')[0] == 'ppcg'

console.log(f('ppcg paints cool galaxies pouring acid into night time stars'))
// 'ppcgpaints'.match('(ppcg(paints(cool(galaxies(pouring(acid(into(night(time(stars)?)?)?)?)?)?)?)?)?)?')[0] == 'ppcgpaints'

console.log(f('ppcg paints cool galaxies pouring acid into night time'))
// 'ppcgpaint'.match('(ppcg(paints(cool(galaxies(pouring(acid(into(night(time)?)?)?)?)?)?)?)?)?')[0] == 'ppcg'

console.log(f('ppcg questions professional pool challengers greatly'))
// 'pqppcg'.match('(ppcg(questions(professional(pool(challengers(greatly)?)?)?)?)?)?')[0] == ''

console.log(f('I'))
// 'I'.match('(I)?')[0] == 'I'

console.log(f('increase i'))
// 'ii'.match('(increase(i)?)?')[0] == ''

console.log(f('i increase'))
// 'ii'.match('(i(increase)?)?')[0] == 'i'

Answer 11

Vyxal, 8 bytes

⌈:¦$∩h¦↔

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Answer 12

Python 2, 106 bytes

First option - finding recursive acronym.
Returns result in list.

I=input().split()
print[' '.join(I[:i])for i in range(1,-~len(I))if[j[0]for j in I]==list(''.join(I[:i]))]

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Python 2, 120 bytes

First option - finding recursive acronym.

def F(I,a=[],r=''):
 for j in I.split():
  a+=j,
  if list(''.join(a))==[i[0]for i in I.split()]:r=' '.join(a)
 return r

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Answer 13

Javascript, 71 bytes

Approach 1

l=s=>{p=s.split(' ');k=p.reduce((r,x)=>r+x[0],'');return k==p[0]?k:''}

Ungolfed:

l=s=>{
    p = s.split(' ');
    k = p.reduce((r,x)=>r+x[0],'');
    return k==p[0] ? k : '';
}

• Split the string by space.
• Create new string by taking first character from each word.
• Compare it with the first word.

Answer 14

Ruby -apl, 57 bytes

$_=y="";$F.all?{|w|$F.map(&:chr).join[/^#{y+=w}/]?$_=y:p}

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Answer 15

Python 2, 109 bytes

def f(s,J=''.join):s=s.split();return[J(s[:i])for i in range(len(s)+1)if J(zip(*s)[0]).find(J(s[:i]))==0][-1]

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Answer 16

Scala, 76 bytes

Solution for simple case (acronyms without whitespaces)

def^(s:String)={val l=s.split(" ");if(l(0)==l.map(_(0)).mkString)l(0)else""} 

Scala, 144 bytes 100 bytes (see solution by ASCII-only in the comments)

def^(s:String)={val l=s.split(" ");l.scanLeft(List[String]())(_:::List(_)).find(_.mkString==l.map(_(0)).mkString).map(_.mkString).getOrElse("")}

Test in REPL

scala> def^(s:String)={val l=s.split(" ");if(l(0)==l.map(_(0)).mkString)l(0)else""}
$up: (s: String)String

scala> ^("hi igloo")
res12: String = hi

scala> ^("ppcg paints cool giraffes")
res13: String = ppcg

scala> ^("ppcg paints Xcool giraffes")
res14: String = ""

scala> ^("ppcg paints cool galaxies pouring acid into night time stars")
res15: String = ""

scala>

scala> def^(s:String)={val l=s.split(" ");l.scanLeft(List[String]())(_:::List(_)).find(_.mkString==l.map(_(0)).mkString).map(_.mkString).getOrElse("")}
$up: (s: String)String

scala> ^("hi igloo")
res16: String = hi

scala> ^("ppcg paints cool giraffes")
res17: String = ppcg

scala> ^("ppcg paints Xcool giraffes")
res18: String = ""

scala> ^("ppcg paints cool galaxies pouring acid into night time stars")
res19: String = ppcgpaints