Find a recursive acronym

Clash Royale CLAN TAG#URR8PPP

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7
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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

---

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

---

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

---

Scoring

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

code-golf string

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edited 11 mins ago

asked 22 hours ago

FireCubez

1649

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  Partially related.
  – AdmBorkBork
  21 hours ago
  

  
  
  
  


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  What's valid English? Is gat one?
  – l4m2
  19 hours ago
  

  
  
  
  


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  @l4m2 Anything that's found in a dictionary, I wont be too strict about that rule
  – FireCubez
  18 hours ago
  

  
  
  
  


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  Q, q [kyoo] noun, plural Q's or Qs, q's or qs. the 17th letter of the English alphabet, a consonant. any spoken sound represented by the letter Q or q, as in quick, acquit, or Iraq. something having the shape of a Q.
  – l4m2
  18 hours ago
  

  
  
  
  


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  Also I don't think ppcg is a word in dictionary
  – l4m2
  17 hours ago
  

  
  
  
  

 | 
show 13 more comments

up vote
7
down vote

favorite

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.

---

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

---

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

---

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

---

Scoring

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

code-golf string

share|improve this question

edited 11 mins ago

asked 22 hours ago

FireCubez

1649

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  Partially related.
  – AdmBorkBork
  21 hours ago
  

  
  
  
  


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  What's valid English? Is gat one?
  – l4m2
  19 hours ago
  

  
  
  
  


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  @l4m2 Anything that's found in a dictionary, I wont be too strict about that rule
  – FireCubez
  18 hours ago
  

  
  
  
  


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  Q, q [kyoo] noun, plural Q's or Qs, q's or qs. the 17th letter of the English alphabet, a consonant. any spoken sound represented by the letter Q or q, as in quick, acquit, or Iraq. something having the shape of a Q.
  – l4m2
  18 hours ago
  

  
  
  
  


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  Also I don't think ppcg is a word in dictionary
  – l4m2
  17 hours ago
  

  
  
  
  

 | 
show 13 more comments

up vote
7
down vote

favorite

up vote
7
down vote

favorite

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.

---

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

---

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

---

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

---

Scoring

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

code-golf string

share|improve this question

edited 11 mins ago

asked 22 hours ago

FireCubez

1649

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.

---

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

---

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

---

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

---

Scoring

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

code-golf string

code-golf string

share|improve this question

edited 11 mins ago

asked 22 hours ago

FireCubez

1649

share|improve this question

edited 11 mins ago

asked 22 hours ago

FireCubez

1649

share|improve this question

share|improve this question

edited 11 mins ago

edited 11 mins ago

edited 11 mins ago

asked 22 hours ago

FireCubez

1649

asked 22 hours ago

FireCubez

1649

asked 22 hours ago

FireCubez

1649

1649

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Partially related.
  – AdmBorkBork
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What's valid English? Is gat one?
  – l4m2
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @l4m2 Anything that's found in a dictionary, I wont be too strict about that rule
  – FireCubez
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Q, q [kyoo] noun, plural Q's or Qs, q's or qs. the 17th letter of the English alphabet, a consonant. any spoken sound represented by the letter Q or q, as in quick, acquit, or Iraq. something having the shape of a Q.
  – l4m2
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Also I don't think ppcg is a word in dictionary
  – l4m2
  17 hours ago
  

  
  
  
  

 | 
show 13 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Partially related.
  – AdmBorkBork
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What's valid English? Is gat one?
  – l4m2
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @l4m2 Anything that's found in a dictionary, I wont be too strict about that rule
  – FireCubez
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Q, q [kyoo] noun, plural Q's or Qs, q's or qs. the 17th letter of the English alphabet, a consonant. any spoken sound represented by the letter Q or q, as in quick, acquit, or Iraq. something having the shape of a Q.
  – l4m2
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Also I don't think ppcg is a word in dictionary
  – l4m2
  17 hours ago
  

  
  
  
  

Partially related.
– AdmBorkBork
21 hours ago

Partially related.
– AdmBorkBork
21 hours ago

What's valid English? Is gat one?
– l4m2
19 hours ago

What's valid English? Is gat one?
– l4m2
19 hours ago

@l4m2 Anything that's found in a dictionary, I wont be too strict about that rule
– FireCubez
18 hours ago

@l4m2 Anything that's found in a dictionary, I wont be too strict about that rule
– FireCubez
18 hours ago

1

1

Q, q [kyoo] noun, plural Q's or Qs, q's or qs. the 17th letter of the English alphabet, a consonant. any spoken sound represented by the letter Q or q, as in quick, acquit, or Iraq. something having the shape of a Q.
– l4m2
18 hours ago

Q, q [kyoo] noun, plural Q's or Qs, q's or qs. the 17th letter of the English alphabet, a consonant. any spoken sound represented by the letter Q or q, as in quick, acquit, or Iraq. something having the shape of a Q.
– l4m2
18 hours ago

2

2

Also I don't think ppcg is a word in dictionary
– l4m2
17 hours ago

Also I don't think ppcg is a word in dictionary
– l4m2
17 hours ago

 | 
show 13 more comments

13 Answers
13

active

oldest

votes

up vote
3
down vote

Japt, 13 bytes

¸
mά
VøUÎ ©V

Try it online!

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edited 16 hours ago

answered 21 hours ago

Luis felipe De jesus Munoz

3,41511049

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  11 bytes
  – Shaggy
  20 hours ago
  

  
  
  
  


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  This fails on the ppcg paints cool galaxies pouring acid into night time stars test case
  – Kamil Drakari
  18 hours ago
  

  
  
  
  


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  here's a version that works for that test case, but it's not golfed
  – Kamil Drakari
  18 hours ago
  

  
  
  
  


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  My previous 13 bytes solution was correct Dx
  – Luis felipe De jesus Munoz
  16 hours ago
  

  
  
  
  


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  The current version just checks that the acronym contains the first word, which results in some new issues
  – Kamil Drakari
  14 hours ago
  

  
  
  
  

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show 1 more comment

up vote
3
down vote

05AB1E, 16 bytes

ð¡©ηʒJ®€нJηså}θJ

Try it online!

share|improve this answer

edited 11 hours ago

answered 21 hours ago

Emigna

44k431133

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  Why did it change to ð¡ instead of # in your last edit? Some special test cases I'm not taking into account?
  – Kevin Cruijssen
  21 hours ago
  

  
  
  
  


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  @KevinCruijssen: Because # would fail for single word input outputting the input instead of an empty string.
  – Emigna
  21 hours ago
  

  
  
  
  


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  Ah yeah, that was it. I remember asking something similar before. I still think # should act the same as ð¡.. Is there a use-case you can think of where you want to split a string on spaces, but if it doesn't contain a space, it should remain the string (instead of the string wrapped in a list)? Other people reading this; FYI: Using # (split on space) on a string without spaces results in the string as is (i.e. "test" -> "test"). Using ð¡ (split on space) on a string without spaces results in the string wrapped in a list (i.e. "test" -> ["test"]).
  – Kevin Cruijssen
  21 hours ago
  
  

  
  
  
  


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  @KevinCruijssen: I think it's mainly due to # also being used as quit if true (which is its main function). If # returned false, you probably wouldn't want the value checked to be wrapped in a list, left on the stack.
  – Emigna
  21 hours ago
  

  
  
  
  


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  @KamilDrakari: Works now though.
  – Emigna
  11 hours ago
  

  
  
  
  

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up vote
2
down vote

Perl 6, 50 42 58 bytes

~first m:g/<<./.join.starts-with($^a),[R,] [~] .words

Try it online!

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

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edited 12 mins ago

answered 21 hours ago

Jo King

17.9k24198

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  You aren't required to output "IMPOSSIBLE" now
  – FireCubez
  21 hours ago
  

  
  
  
  


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  @Jo King I can do regexes, but for the life of me I can't seem to think with all the operators. I keep forgetting the x operator exists, for example :P
  – moonheart08
  16 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

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.

share|improve this answer

edited 21 hours ago

answered 21 hours ago

Neil

77k744173

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  For ppcg paints, output is invalid, it should output nothing since pp only spells a portion of the first word, instead of all of it
  – FireCubez
  21 hours ago
  

  
  
  
  


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  @FireCubez Sorry, I was working off an older version of the question.
  – Neil
  21 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

Perl 6, 56 bytes

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

Try it online!

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

Fulfills choice 1.

share|improve this answer

answered 18 hours ago

moonheart08

59210

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  Sadly, I'm still at the stage where regexes are simply madness. Unfortunately, this fails the ppcgpaints test, otherwise I would have suggested something like $!∈.words for the if condition
  – Jo King
  16 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

Rust, 155, try it online!

Selected: Problem 1: Finding acronym

type S=String;fn f(t:&str)->Sa+&b[..1])+" ";if (l+" ").contains(w.as_str())welseS::new()

Ungolfed, just a bit:

fn f(t: &str) -> String a, b

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

type S=String;fn f(l:S)->Sa+&b[..1])+" ";if (l+" ").contains(&w.as_str())welseS::new()

share|improve this answer

edited 14 hours ago

answered 19 hours ago

Hannes Karppila

2,00421136

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  Which of the 2 choices does this program do?
  – FireCubez
  18 hours ago
  

  
  
  
  


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  @FireCubez Updated.
  – Hannes Karppila
  14 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

Jelly, 9 bytes

Ḳµ;fZḢWƊ

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

Try it online!

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

share|improve this answer

edited 13 hours ago

answered 15 hours ago

Jonathan Allan

49.5k534163

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  What do you mean by "print the first word"? It needs to find the acronym if one exists, does it do that?
  – FireCubez
  14 hours ago
  

  
  
  
  


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  Fails for "ppcg paints cool galaxies pouring acid into not the sky", should print "ppcg paints" or "ppcgpaints"
  – FireCubez
  13 hours ago
  

  
  
  
  


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  Oh, I missed the adjoining words requirement :(
  – Jonathan Allan
  13 hours ago
  

  
  
  
  


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  Fixed it up to meet this requirement.
  – Jonathan Allan
  13 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

JavaScript [ES6], 64 bytes

Using the first option:

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

Matches the initial letters to each word wrapped in parentheses, eg:

ppcgpaint
(ppcg)?(paints)?(cool)?(galaxies)?(pouring)?(acid)?(into)?(night)?(time)?

All test cases:

let f=
s=>s.split` `.map(w=>t+=w[d+=`($w)?`,0],t=d='')&&t.match(d)[0]


console.log(f('hi igloo')) // hi
console.log(f('ppcg paints')) // empty string
console.log(f('ppcg paints cool giraffes')) // ppcg
console.log(f('ppcg paints cool galaxies pouring acid into night time stars')) // ppcgpaints
console.log(f('ppcg paints cool galaxies pouring acid into night time')) // ppcg
console.log(f('ppcg questions professional pool challengers greatly')) // empty string
console.log(f('I')) // I

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edited 10 hours ago

answered 10 hours ago

Rick Hitchcock

2,421515

add a comment | 

up vote
1
down vote

Haskell, 51 48 bytes

Edit: -3 bytes thanks to @xnor.

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

Finds acronym.

Try it online!

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

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edited 2 hours ago

answered 20 hours ago

nimi

30.4k31884

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  Since you're not using x, composing (w-> ...).words would be shorter.
  – xnor
  10 hours ago
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

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]))]

Try it online!

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

Try it online!

share|improve this answer

edited 20 hours ago

answered 21 hours ago

Dead Possum

2,805623

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You aren't required to output "IMPOSSIBLE" as per @JoKing 's request, that might decrease your byte count
  – FireCubez
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Single letters like 'I' don't work, it should output that single letter
  – FireCubez
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @FireCubez fixed
  – Dead Possum
  20 hours ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

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.

share|improve this answer

answered 19 hours ago

alpheus

214

New contributor

alpheus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

up vote
0
down vote

K (ngn/k), 40 bytes

First option:

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

Try it online!

share|improve this answer

edited 18 hours ago

answered 18 hours ago

Thaufeki

15117

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Which of the 2 options does this work on?
  – FireCubez
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The first, outputs acronym from string input. I'll edit my post to clarify
  – Thaufeki
  18 hours ago
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

JavaScript (Node.js), 70 bytes

s=>(s.replace(/w+/g,x=>w+=x[0]+' ?',w='^'),s.match(w+'\b')||[''])[0]

Try it online!

share|improve this answer

edited 17 hours ago

answered 19 hours ago

l4m2

3,8981431

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
);
);
, "mathjax-editing");

StackExchange.ifUsing("editor", function () improve this answer

edited 16 hours ago

edited 16 hours ago

edited 16 hours ago

answered 21 hours ago

Luis felipe De jesus Munoz

3,41511049

answered 21 hours ago

Luis felipe De jesus Munoz

3,41511049

answered 21 hours ago

Luis felipe De jesus Munoz

3,41511049

3,41511049

• 
  
  
  

  
  

  
  
  
  
  
  

  
  11 bytes
  – Shaggy
  20 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This fails on the ppcg paints cool galaxies pouring acid into night time stars test case
  – Kamil Drakari
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  here's a version that works for that test case, but it's not golfed
  – Kamil Drakari
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  My previous 13 bytes solution was correct Dx
  – Luis felipe De jesus Munoz
  16 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The current version just checks that the acronym contains the first word, which results in some new issues
  – Kamil Drakari
  14 hours ago
  

  
  
  
  

 Î¸J

Try it online!

share|improve this answer

edited 11 hours ago

answered 21 hours ago

Emigna

44k431133

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Why did it change to ð¡ instead of # in your last edit? Some special test cases I'm not taking into account?
  – Kevin Cruijssen
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen: Because # would fail for single word input outputting the input instead of an empty string.
  – Emigna
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah yeah, that was it. I remember asking something similar before. I still think # should act the same as ð¡.. Is there a use-case you can think of where you want to split a string on spaces, but if it doesn't contain a space, it should remain the string (instead of the string wrapped in a list)? Other people reading this; FYI: Using # (split on space) on a string without spaces results in the string as is (i.e. "test" -> "test"). Using ð¡ (split on space) on a string without spaces results in the string wrapped in a list (i.e. "test" -> ["test"]).
  – Kevin Cruijssen
  21 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen: I think it's mainly due to # also being used as quit if true (which is its main function). If # returned false, you probably wouldn't want the value checked to be wrapped in a list, left on the stack.
  – Emigna
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @KamilDrakari: Works now though.
  – Emigna
  11 hours ago
  

  
  
  
  

 | 
show 6 more comments

up vote
3
down vote

05AB1E, 16 bytes

ð¡©ηʒJ®€нJηså}θJ

Try it online!

share|improve this answer

edited 11 hours ago

answered 21 hours ago

Emigna

44k431133

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Why did it change to ð¡ instead of # in your last edit? Some special test cases I'm not taking into account?
  – Kevin Cruijssen
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen: Because # would fail for single word input outputting the input instead of an empty string.
  – Emigna
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah yeah, that was it. I remember asking something similar before. I still think # should act the same as ð¡.. Is there a use-case you can think of where you want to split a string on spaces, but if it doesn't contain a space, it should remain the string (instead of the string wrapped in a list)? Other people reading this; FYI: Using # (split on space) on a string without spaces results in the string as is (i.e. "test" -> "test"). Using ð¡ (split on space) on a string without spaces results in the string wrapped in a list (i.e. "test" -> ["test"]).
  – Kevin Cruijssen
  21 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen: I think it's mainly due to # also being used as quit if true (which is its main function). If # returned false, you probably wouldn't want the value checked to be wrapped in a list, left on the stack.
  – Emigna
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @KamilDrakari: Works now though.
  – Emigna
  11 hours ago
  

  
  
  
  

 | 
show 6 more comments

up vote
3
down vote

up vote
3
down vote

05AB1E, 16 bytes

ð¡©ηʒJ®€нJηså}θJ

Try it online!

share|improve this answer

edited 11 hours ago

answered 21 hours ago

Emigna

44k431133

05AB1E, 16 bytes

ð¡©ηʒJ®€нJηså}θJ

Try it online!

share|improve this answer

edited 11 hours ago

answered 21 hours ago

Emigna

44k431133

share|improve this answer

share|improve this answer

edited 11 hours ago

edited 11 hours ago

edited 11 hours ago

answered 21 hours ago

Emigna

44k431133

answered 21 hours ago

Emigna

44k431133

answered 21 hours ago

Emigna

44k431133

44k431133

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Why did it change to ð¡ instead of # in your last edit? Some special test cases I'm not taking into account?
  – Kevin Cruijssen
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen: Because # would fail for single word input outputting the input instead of an empty string.
  – Emigna
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah yeah, that was it. I remember asking something similar before. I still think # should act the same as ð¡.. Is there a use-case you can think of where you want to split a string on spaces, but if it doesn't contain a space, it should remain the string (instead of the string wrapped in a list)? Other people reading this; FYI: Using # (split on space) on a string without spaces results in the string as is (i.e. "test" -> "test"). Using ð¡ (split on space) on a string without spaces results in the string wrapped in a list (i.e. "test" -> ["test"]).
  – Kevin Cruijssen
  21 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen: I think it's mainly due to # also being used as quit if true (which is its main function). If # returned false, you probably wouldn't want the value checked to be wrapped in a list, left on the stack.
  – Emigna
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @KamilDrakari: Works now though.
  – Emigna
  11 hours ago
  

  
  
  
  

 | 
show 6 more comments

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Why did it change to ð¡ instead of # in your last edit? Some special test cases I'm not taking into account?
  – Kevin Cruijssen
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen: Because # would fail for single word input outputting the input instead of an empty string.
  – Emigna
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah yeah, that was it. I remember asking something similar before. I still think # should act the same as ð¡.. Is there a use-case you can think of where you want to split a string on spaces, but if it doesn't contain a space, it should remain the string (instead of the string wrapped in a list)? Other people reading this; FYI: Using # (split on space) on a string without spaces results in the string as is (i.e. "test" -> "test"). Using ð¡ (split on space) on a string without spaces results in the string wrapped in a list (i.e. "test" -> ["test"]).
  – Kevin Cruijssen
  21 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen: I think it's mainly due to # also being used as quit if true (which is its main function). If # returned false, you probably wouldn't want the value checked to be wrapped in a list, left on the stack.
  – Emigna
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @KamilDrakari: Works now though.
  – Emigna
  11 hours ago
  

  
  
  
  

1

1

Why did it change to ð¡ instead of # in your last edit? Some special test cases I'm not taking into account?
– Kevin Cruijssen
21 hours ago

Why did it change to ð¡ instead of # in your last edit? Some special test cases I'm not taking into account?
– Kevin Cruijssen
21 hours ago

@KevinCruijssen: Because # would fail for single word input outputting the input instead of an empty string.
– Emigna
21 hours ago

@KevinCruijssen: Because # would fail for single word input outputting the input instead of an empty string.
– Emigna
21 hours ago

Ah yeah, that was it. I remember asking something similar before. I still think # should act the same as ð¡.. Is there a use-case you can think of where you want to split a string on spaces, but if it doesn't contain a space, it should remain the string (instead of the string wrapped in a list)? Other people reading this; FYI: Using # (split on space) on a string without spaces results in the string as is (i.e. "test" -> "test"). Using ð¡ (split on space) on a string without spaces results in the string wrapped in a list (i.e. "test" -> ["test"]).
– Kevin Cruijssen
21 hours ago

Ah yeah, that was it. I remember asking something similar before. I still think # should act the same as ð¡.. Is there a use-case you can think of where you want to split a string on spaces, but if it doesn't contain a space, it should remain the string (instead of the string wrapped in a list)? Other people reading this; FYI: Using # (split on space) on a string without spaces results in the string as is (i.e. "test" -> "test"). Using ð¡ (split on space) on a string without spaces results in the string wrapped in a list (i.e. "test" -> ["test"]).
– Kevin Cruijssen
21 hours ago

@KevinCruijssen: I think it's mainly due to # also being used as quit if true (which is its main function). If # returned false, you probably wouldn't want the value checked to be wrapped in a list, left on the stack.
– Emigna
21 hours ago

@KevinCruijssen: I think it's mainly due to # also being used as quit if true (which is its main function). If # returned false, you probably wouldn't want the value checked to be wrapped in a list, left on the stack.
– Emigna
21 hours ago

1

1

@KamilDrakari: Works now though.
– Emigna
11 hours ago

@KamilDrakari: Works now though.
– Emigna
11 hours ago

 | 
show 6 more comments

up vote
2
down vote

Perl 6, 50 42 58 bytes

~first m:g/<<./.join.starts-with($^a),[R,] [~] .words

Try it online!

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

share|improve this answer

edited 12 mins ago

answered 21 hours ago

Jo King

17.9k24198

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You aren't required to output "IMPOSSIBLE" now
  – FireCubez
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Jo King I can do regexes, but for the life of me I can't seem to think with all the operators. I keep forgetting the x operator exists, for example :P
  – moonheart08
  16 hours ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

Perl 6, 50 42 58 bytes

~first m:g/<<./.join.starts-with($^a),[R,] [~] .words

Try it online!

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

share|improve this answer

edited 12 mins ago

answered 21 hours ago

Jo King

17.9k24198

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You aren't required to output "IMPOSSIBLE" now
  – FireCubez
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Jo King I can do regexes, but for the life of me I can't seem to think with all the operators. I keep forgetting the x operator exists, for example :P
  – moonheart08
  16 hours ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

Perl 6, 50 42 58 bytes

~first m:g/<<./.join.starts-with($^a),[R,] [~] .words

Try it online!

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

share|improve this answer

edited 12 mins ago

answered 21 hours ago

Jo King

17.9k24198

Perl 6, 50 42 58 bytes

~first m:g/<<./.join.starts-with($^a),[R,] [~] .words

Try it online!

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

share|improve this answer

edited 12 mins ago

answered 21 hours ago

Jo King

17.9k24198

share|improve this answer

share|improve this answer

edited 12 mins ago

edited 12 mins ago

edited 12 mins ago

answered 21 hours ago

Jo King

17.9k24198

answered 21 hours ago

Jo King

17.9k24198

answered 21 hours ago

Jo King

17.9k24198

17.9k24198

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You aren't required to output "IMPOSSIBLE" now
  – FireCubez
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Jo King I can do regexes, but for the life of me I can't seem to think with all the operators. I keep forgetting the x operator exists, for example :P
  – moonheart08
  16 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You aren't required to output "IMPOSSIBLE" now
  – FireCubez
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Jo King I can do regexes, but for the life of me I can't seem to think with all the operators. I keep forgetting the x operator exists, for example :P
  – moonheart08
  16 hours ago
  

  
  
  
  

You aren't required to output "IMPOSSIBLE" now
– FireCubez
21 hours ago

You aren't required to output "IMPOSSIBLE" now
– FireCubez
21 hours ago

@Jo King I can do regexes, but for the life of me I can't seem to think with all the operators. I keep forgetting the x operator exists, for example :P
– moonheart08
16 hours ago

@Jo King I can do regexes, but for the life of me I can't seem to think with all the operators. I keep forgetting the x operator exists, for example :P
– moonheart08
16 hours ago

add a comment | 

up vote
1
down vote

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.

share|improve this answer

edited 21 hours ago

answered 21 hours ago

Neil

77k744173

• 
  
  
  

  
  

  
  
  
  
  
  

  
  For ppcg paints, output is invalid, it should output nothing since pp only spells a portion of the first word, instead of all of it
  – FireCubez
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @FireCubez Sorry, I was working off an older version of the question.
  – Neil
  21 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

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.

share|improve this answer

edited 21 hours ago

answered 21 hours ago

Neil

77k744173

• 
  
  
  

  
  

  
  
  
  
  
  

  
  For ppcg paints, output is invalid, it should output nothing since pp only spells a portion of the first word, instead of all of it
  – FireCubez
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @FireCubez Sorry, I was working off an older version of the question.
  – Neil
  21 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

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.

share|improve this answer

edited 21 hours ago

answered 21 hours ago

Neil

77k744173

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.

share|improve this answer

edited 21 hours ago

answered 21 hours ago

Neil

77k744173

share|improve this answer

share|improve this answer

edited 21 hours ago

edited 21 hours ago

edited 21 hours ago

answered 21 hours ago

Neil

77k744173

answered 21 hours ago

Neil

77k744173

answered 21 hours ago

Neil

77k744173

77k744173

• 
  
  
  

  
  

  
  
  
  
  
  

  
  For ppcg paints, output is invalid, it should output nothing since pp only spells a portion of the first word, instead of all of it
  – FireCubez
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @FireCubez Sorry, I was working off an older version of the question.
  – Neil
  21 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  For ppcg paints, output is invalid, it should output nothing since pp only spells a portion of the first word, instead of all of it
  – FireCubez
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @FireCubez Sorry, I was working off an older version of the question.
  – Neil
  21 hours ago
  

  
  
  
  

For ppcg paints, output is invalid, it should output nothing since pp only spells a portion of the first word, instead of all of it
– FireCubez
21 hours ago

For ppcg paints, output is invalid, it should output nothing since pp only spells a portion of the first word, instead of all of it
– FireCubez
21 hours ago

@FireCubez Sorry, I was working off an older version of the question.
– Neil
21 hours ago

@FireCubez Sorry, I was working off an older version of the question.
– Neil
21 hours ago

add a comment | 

up vote
1
down vote

Perl 6, 56 bytes

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

Try it online!

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

Fulfills choice 1.

share|improve this answer

answered 18 hours ago

moonheart08

59210

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sadly, I'm still at the stage where regexes are simply madness. Unfortunately, this fails the ppcgpaints test, otherwise I would have suggested something like $!∈.words for the if condition
  – Jo King
  16 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

Perl 6, 56 bytes

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

Try it online!

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

Fulfills choice 1.

share|improve this answer

answered 18 hours ago

moonheart08

59210

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sadly, I'm still at the stage where regexes are simply madness. Unfortunately, this fails the ppcgpaints test, otherwise I would have suggested something like $!∈.words for the if condition
  – Jo King
  16 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Perl 6, 56 bytes

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

Try it online!

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

Fulfills choice 1.

share|improve this answer

answered 18 hours ago

moonheart08

59210

Perl 6, 56 bytes

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

Try it online!

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

Fulfills choice 1.

share|improve this answer

answered 18 hours ago

moonheart08

59210

share|improve this answer

share|improve this answer

answered 18 hours ago

moonheart08

59210

answered 18 hours ago

moonheart08

59210

answered 18 hours ago

moonheart08

59210

59210

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sadly, I'm still at the stage where regexes are simply madness. Unfortunately, this fails the ppcgpaints test, otherwise I would have suggested something like $!∈.words for the if condition
  – Jo King
  16 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sadly, I'm still at the stage where regexes are simply madness. Unfortunately, this fails the ppcgpaints test, otherwise I would have suggested something like $!∈.words for the if condition
  – Jo King
  16 hours ago
  

  
  
  
  

Sadly, I'm still at the stage where regexes are simply madness. Unfortunately, this fails the ppcgpaints test, otherwise I would have suggested something like $!∈.words for the if condition
– Jo King
16 hours ago

Sadly, I'm still at the stage where regexes are simply madness. Unfortunately, this fails the ppcgpaints test, otherwise I would have suggested something like $!∈.words for the if condition
– Jo King
16 hours ago

add a comment | 

up vote
1
down vote

Rust, 155, try it online!

Selected: Problem 1: Finding acronym

type S=String;fn f(t:&str)->Sa+&b[..1])+" ";if (l+" ").contains(w.as_str())welseS::new()

Ungolfed, just a bit:

fn f(t: &str) -> String a, b

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

type S=String;fn f(l:S)->Sa+&b[..1])+" ";if (l+" ").contains(&w.as_str())welseS::new()

share|improve this answer

edited 14 hours ago

answered 19 hours ago

Hannes Karppila

2,00421136

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Which of the 2 choices does this program do?
  – FireCubez
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @FireCubez Updated.
  – Hannes Karppila
  14 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

Rust, 155, try it online!

Selected: Problem 1: Finding acronym

type S=String;fn f(t:&str)->Sa+&b[..1])+" ";if (l+" ").contains(w.as_str())welseS::new()

Ungolfed, just a bit:

fn f(t: &str) -> String a, b

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

type S=String;fn f(l:S)->Sa+&b[..1])+" ";if (l+" ").contains(&w.as_str())welseS::new()

share|improve this answer

edited 14 hours ago

answered 19 hours ago

Hannes Karppila

2,00421136

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Which of the 2 choices does this program do?
  – FireCubez
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @FireCubez Updated.
  – Hannes Karppila
  14 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Rust, 155, try it online!

Selected: Problem 1: Finding acronym

type S=String;fn f(t:&str)->Sa+&b[..1])+" ";if (l+" ").contains(w.as_str())welseS::new()

Ungolfed, just a bit:

fn f(t: &str) -> String a, b

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

type S=String;fn f(l:S)->Sa+&b[..1])+" ";if (l+" ").contains(&w.as_str())welseS::new()

share|improve this answer

edited 14 hours ago

answered 19 hours ago

Hannes Karppila

2,00421136

Rust, 155, try it online!

Selected: Problem 1: Finding acronym

type S=String;fn f(t:&str)->Sa+&b[..1])+" ";if (l+" ").contains(w.as_str())welseS::new()

Ungolfed, just a bit:

fn f(t: &str) -> String a, b

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

type S=String;fn f(l:S)->Sa+&b[..1])+" ";if (l+" ").contains(&w.as_str())welseS::new()

share|improve this answer

edited 14 hours ago

answered 19 hours ago

Hannes Karppila

2,00421136

share|improve this answer

share|improve this answer

edited 14 hours ago

edited 14 hours ago

edited 14 hours ago

answered 19 hours ago

Hannes Karppila

2,00421136

answered 19 hours ago

Hannes Karppila

2,00421136

answered 19 hours ago

Hannes Karppila

2,00421136

2,00421136

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Which of the 2 choices does this program do?
  – FireCubez
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @FireCubez Updated.
  – Hannes Karppila
  14 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Which of the 2 choices does this program do?
  – FireCubez
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @FireCubez Updated.
  – Hannes Karppila
  14 hours ago
  

  
  
  
  

Which of the 2 choices does this program do?
– FireCubez
18 hours ago

Which of the 2 choices does this program do?
– FireCubez
18 hours ago

@FireCubez Updated.
– Hannes Karppila
14 hours ago

@FireCubez Updated.
– Hannes Karppila
14 hours ago

add a comment | 

up vote
1
down vote

Jelly, 9 bytes

Ḳµ;fZḢWƊ

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

Try it online!

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

share|improve this answer

edited 13 hours ago

answered 15 hours ago

Jonathan Allan

49.5k534163

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What do you mean by "print the first word"? It needs to find the acronym if one exists, does it do that?
  – FireCubez
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Fails for "ppcg paints cool galaxies pouring acid into not the sky", should print "ppcg paints" or "ppcgpaints"
  – FireCubez
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh, I missed the adjoining words requirement :(
  – Jonathan Allan
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Fixed it up to meet this requirement.
  – Jonathan Allan
  13 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

Jelly, 9 bytes

Ḳµ;fZḢWƊ

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

Try it online!

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

share|improve this answer

edited 13 hours ago

answered 15 hours ago

Jonathan Allan

49.5k534163

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What do you mean by "print the first word"? It needs to find the acronym if one exists, does it do that?
  – FireCubez
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Fails for "ppcg paints cool galaxies pouring acid into not the sky", should print "ppcg paints" or "ppcgpaints"
  – FireCubez
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh, I missed the adjoining words requirement :(
  – Jonathan Allan
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Fixed it up to meet this requirement.
  – Jonathan Allan
  13 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Jelly, 9 bytes

Ḳµ;fZḢWƊ

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

Try it online!

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

share|improve this answer

edited 13 hours ago

answered 15 hours ago

Jonathan Allan

49.5k534163

Jelly, 9 bytes

Ḳµ;fZḢWƊ

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

Try it online!

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

share|improve this answer

edited 13 hours ago

answered 15 hours ago

Jonathan Allan

49.5k534163

share|improve this answer

share|improve this answer

edited 13 hours ago

edited 13 hours ago

edited 13 hours ago

answered 15 hours ago

Jonathan Allan

49.5k534163

answered 15 hours ago

Jonathan Allan

49.5k534163

answered 15 hours ago

Jonathan Allan

49.5k534163

49.5k534163

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What do you mean by "print the first word"? It needs to find the acronym if one exists, does it do that?
  – FireCubez
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Fails for "ppcg paints cool galaxies pouring acid into not the sky", should print "ppcg paints" or "ppcgpaints"
  – FireCubez
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh, I missed the adjoining words requirement :(
  – Jonathan Allan
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Fixed it up to meet this requirement.
  – Jonathan Allan
  13 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What do you mean by "print the first word"? It needs to find the acronym if one exists, does it do that?
  – FireCubez
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Fails for "ppcg paints cool galaxies pouring acid into not the sky", should print "ppcg paints" or "ppcgpaints"
  – FireCubez
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh, I missed the adjoining words requirement :(
  – Jonathan Allan
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Fixed it up to meet this requirement.
  – Jonathan Allan
  13 hours ago
  

  
  
  
  

What do you mean by "print the first word"? It needs to find the acronym if one exists, does it do that?
– FireCubez
14 hours ago

What do you mean by "print the first word"? It needs to find the acronym if one exists, does it do that?
– FireCubez
14 hours ago

Fails for "ppcg paints cool galaxies pouring acid into not the sky", should print "ppcg paints" or "ppcgpaints"
– FireCubez
13 hours ago

Fails for "ppcg paints cool galaxies pouring acid into not the sky", should print "ppcg paints" or "ppcgpaints"
– FireCubez
13 hours ago

Oh, I missed the adjoining words requirement :(
– Jonathan Allan
13 hours ago

Oh, I missed the adjoining words requirement :(
– Jonathan Allan
13 hours ago

Fixed it up to meet this requirement.
– Jonathan Allan
13 hours ago

Fixed it up to meet this requirement.
– Jonathan Allan
13 hours ago

add a comment | 

up vote
1
down vote

JavaScript [ES6], 64 bytes

Using the first option:

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

Matches the initial letters to each word wrapped in parentheses, eg:

ppcgpaint
(ppcg)?(paints)?(cool)?(galaxies)?(pouring)?(acid)?(into)?(night)?(time)?

All test cases:

let f=
s=>s.split` `.map(w=>t+=w[d+=`($w)?`,0],t=d='')&&t.match(d)[0]


console.log(f('hi igloo')) // hi
console.log(f('ppcg paints')) // empty string
console.log(f('ppcg paints cool giraffes')) // ppcg
console.log(f('ppcg paints cool galaxies pouring acid into night time stars')) // ppcgpaints
console.log(f('ppcg paints cool galaxies pouring acid into night time')) // ppcg
console.log(f('ppcg questions professional pool challengers greatly')) // empty string
console.log(f('I')) // I

share|improve this answer

edited 10 hours ago

answered 10 hours ago

Rick Hitchcock

2,421515

add a comment | 

up vote
1
down vote

JavaScript [ES6], 64 bytes

Using the first option:

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

Matches the initial letters to each word wrapped in parentheses, eg:

ppcgpaint
(ppcg)?(paints)?(cool)?(galaxies)?(pouring)?(acid)?(into)?(night)?(time)?

All test cases:

let f=
s=>s.split` `.map(w=>t+=w[d+=`($w)?`,0],t=d='')&&t.match(d)[0]


console.log(f('hi igloo')) // hi
console.log(f('ppcg paints')) // empty string
console.log(f('ppcg paints cool giraffes')) // ppcg
console.log(f('ppcg paints cool galaxies pouring acid into night time stars')) // ppcgpaints
console.log(f('ppcg paints cool galaxies pouring acid into night time')) // ppcg
console.log(f('ppcg questions professional pool challengers greatly')) // empty string
console.log(f('I')) // I

share|improve this answer

edited 10 hours ago

answered 10 hours ago

Rick Hitchcock

2,421515

add a comment | 

up vote
1
down vote

up vote
1
down vote

JavaScript [ES6], 64 bytes

Using the first option:

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

Matches the initial letters to each word wrapped in parentheses, eg:

ppcgpaint
(ppcg)?(paints)?(cool)?(galaxies)?(pouring)?(acid)?(into)?(night)?(time)?

All test cases:

let f=
s=>s.split` `.map(w=>t+=w[d+=`($w)?`,0],t=d='')&&t.match(d)[0]


console.log(f('hi igloo')) // hi
console.log(f('ppcg paints')) // empty string
console.log(f('ppcg paints cool giraffes')) // ppcg
console.log(f('ppcg paints cool galaxies pouring acid into night time stars')) // ppcgpaints
console.log(f('ppcg paints cool galaxies pouring acid into night time')) // ppcg
console.log(f('ppcg questions professional pool challengers greatly')) // empty string
console.log(f('I')) // I

share|improve this answer

edited 10 hours ago

answered 10 hours ago

Rick Hitchcock

2,421515

JavaScript [ES6], 64 bytes

Using the first option:

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

Matches the initial letters to each word wrapped in parentheses, eg:

ppcgpaint
(ppcg)?(paints)?(cool)?(galaxies)?(pouring)?(acid)?(into)?(night)?(time)?

All test cases:

let f=
s=>s.split` `.map(w=>t+=w[d+=`($w)?`,0],t=d='')&&t.match(d)[0]


console.log(f('hi igloo')) // hi
console.log(f('ppcg paints')) // empty string
console.log(f('ppcg paints cool giraffes')) // ppcg
console.log(f('ppcg paints cool galaxies pouring acid into night time stars')) // ppcgpaints
console.log(f('ppcg paints cool galaxies pouring acid into night time')) // ppcg
console.log(f('ppcg questions professional pool challengers greatly')) // empty string
console.log(f('I')) // I

let f=
s=>s.split` `.map(w=>t+=w[d+=`($w)?`,0],t=d='')&&t.match(d)[0]


console.log(f('hi igloo')) // hi
console.log(f('ppcg paints')) // empty string
console.log(f('ppcg paints cool giraffes')) // ppcg
console.log(f('ppcg paints cool galaxies pouring acid into night time stars')) // ppcgpaints
console.log(f('ppcg paints cool galaxies pouring acid into night time')) // ppcg
console.log(f('ppcg questions professional pool challengers greatly')) // empty string
console.log(f('I')) // I

let f=
s=>s.split` `.map(w=>t+=w[d+=`($w)?`,0],t=d='')&&t.match(d)[0]


console.log(f('hi igloo')) // hi
console.log(f('ppcg paints')) // empty string
console.log(f('ppcg paints cool giraffes')) // ppcg
console.log(f('ppcg paints cool galaxies pouring acid into night time stars')) // ppcgpaints
console.log(f('ppcg paints cool galaxies pouring acid into night time')) // ppcg
console.log(f('ppcg questions professional pool challengers greatly')) // empty string
console.log(f('I')) // I

share|improve this answer

edited 10 hours ago

answered 10 hours ago

Rick Hitchcock

2,421515

share|improve this answer

share|improve this answer

edited 10 hours ago

edited 10 hours ago

edited 10 hours ago

answered 10 hours ago

Rick Hitchcock

2,421515

answered 10 hours ago

Rick Hitchcock

2,421515

answered 10 hours ago

Rick Hitchcock

2,421515

2,421515

add a comment | 

add a comment | 

up vote
1
down vote

Haskell, 51 48 bytes

Edit: -3 bytes thanks to @xnor.

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

Finds acronym.

Try it online!

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

share|improve this answer

edited 2 hours ago

answered 20 hours ago

nimi

30.4k31884

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since you're not using x, composing (w-> ...).words would be shorter.
  – xnor
  10 hours ago
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

Haskell, 51 48 bytes

Edit: -3 bytes thanks to @xnor.

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

Finds acronym.

Try it online!

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

share|improve this answer

edited 2 hours ago

answered 20 hours ago

nimi

30.4k31884

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since you're not using x, composing (w-> ...).words would be shorter.
  – xnor
  10 hours ago
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Haskell, 51 48 bytes

Edit: -3 bytes thanks to @xnor.

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

Finds acronym.

Try it online!

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

share|improve this answer

edited 2 hours ago

answered 20 hours ago

nimi

30.4k31884

Haskell, 51 48 bytes

Edit: -3 bytes thanks to @xnor.

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

Finds acronym.

Try it online!

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

share|improve this answer

edited 2 hours ago

answered 20 hours ago

nimi

30.4k31884

share|improve this answer

share|improve this answer

edited 2 hours ago

edited 2 hours ago

edited 2 hours ago

answered 20 hours ago

nimi

30.4k31884

answered 20 hours ago

nimi

30.4k31884

answered 20 hours ago

nimi

30.4k31884

30.4k31884

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since you're not using x, composing (w-> ...).words would be shorter.
  – xnor
  10 hours ago
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since you're not using x, composing (w-> ...).words would be shorter.
  – xnor
  10 hours ago
  
  

  
  
  
  

Since you're not using x, composing (w-> ...).words would be shorter.
– xnor
10 hours ago

Since you're not using x, composing (w-> ...).words would be shorter.
– xnor
10 hours ago

add a comment | 

up vote
0
down vote

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]))]

Try it online!

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

Try it online!

share|improve this answer

edited 20 hours ago

answered 21 hours ago

Dead Possum

2,805623

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You aren't required to output "IMPOSSIBLE" as per @JoKing 's request, that might decrease your byte count
  – FireCubez
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Single letters like 'I' don't work, it should output that single letter
  – FireCubez
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @FireCubez fixed
  – Dead Possum
  20 hours ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

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]))]

Try it online!

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

Try it online!

share|improve this answer

edited 20 hours ago

answered 21 hours ago

Dead Possum

2,805623

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You aren't required to output "IMPOSSIBLE" as per @JoKing 's request, that might decrease your byte count
  – FireCubez
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Single letters like 'I' don't work, it should output that single letter
  – FireCubez
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @FireCubez fixed
  – Dead Possum
  20 hours ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

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]))]

Try it online!

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

Try it online!

share|improve this answer

edited 20 hours ago

answered 21 hours ago

Dead Possum

2,805623

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]))]

Try it online!

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

Try it online!

share|improve this answer

edited 20 hours ago

answered 21 hours ago

Dead Possum

2,805623

share|improve this answer

share|improve this answer

edited 20 hours ago

edited 20 hours ago

edited 20 hours ago

answered 21 hours ago

Dead Possum

2,805623

answered 21 hours ago

Dead Possum

2,805623

answered 21 hours ago

Dead Possum

2,805623

2,805623

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You aren't required to output "IMPOSSIBLE" as per @JoKing 's request, that might decrease your byte count
  – FireCubez
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Single letters like 'I' don't work, it should output that single letter
  – FireCubez
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @FireCubez fixed
  – Dead Possum
  20 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You aren't required to output "IMPOSSIBLE" as per @JoKing 's request, that might decrease your byte count
  – FireCubez
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Single letters like 'I' don't work, it should output that single letter
  – FireCubez
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @FireCubez fixed
  – Dead Possum
  20 hours ago
  

  
  
  
  

You aren't required to output "IMPOSSIBLE" as per @JoKing 's request, that might decrease your byte count
– FireCubez
21 hours ago

You aren't required to output "IMPOSSIBLE" as per @JoKing 's request, that might decrease your byte count
– FireCubez
21 hours ago

Single letters like 'I' don't work, it should output that single letter
– FireCubez
21 hours ago

Single letters like 'I' don't work, it should output that single letter
– FireCubez
21 hours ago

@FireCubez fixed
– Dead Possum
20 hours ago

@FireCubez fixed
– Dead Possum
20 hours ago

add a comment | 

up vote
0
down vote

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.

share|improve this answer

answered 19 hours ago

alpheus

214

New contributor

alpheus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

up vote
0
down vote

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.

share|improve this answer

answered 19 hours ago

alpheus

214

New contributor

alpheus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

up vote
0
down vote

up vote
0
down vote

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.

share|improve this answer

answered 19 hours ago

alpheus

214

New contributor

alpheus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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.

share|improve this answer

answered 19 hours ago

alpheus

214

New contributor

alpheus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this answer

share|improve this answer

answered 19 hours ago

alpheus

214

New contributor

alpheus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 19 hours ago

alpheus

214

answered 19 hours ago

alpheus

214

214

New contributor

alpheus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

New contributor

alpheus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

alpheus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

add a comment | 

up vote
0
down vote

K (ngn/k), 40 bytes

First option:

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

Try it online!

share|improve this answer

edited 18 hours ago

answered 18 hours ago

Thaufeki

15117

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Which of the 2 options does this work on?
  – FireCubez
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The first, outputs acronym from string input. I'll edit my post to clarify
  – Thaufeki
  18 hours ago
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

K (ngn/k), 40 bytes

First option:

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

Try it online!

share|improve this answer

edited 18 hours ago

answered 18 hours ago

Thaufeki

15117

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Which of the 2 options does this work on?
  – FireCubez
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The first, outputs acronym from string input. I'll edit my post to clarify
  – Thaufeki
  18 hours ago
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

K (ngn/k), 40 bytes

First option:

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

Try it online!

share|improve this answer

edited 18 hours ago

answered 18 hours ago

Thaufeki

15117

K (ngn/k), 40 bytes

First option:

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

Try it online!

share|improve this answer

edited 18 hours ago

answered 18 hours ago

Thaufeki

15117

share|improve this answer

share|improve this answer

edited 18 hours ago

edited 18 hours ago

edited 18 hours ago

answered 18 hours ago

Thaufeki

15117

answered 18 hours ago

Thaufeki

15117

answered 18 hours ago

Thaufeki

15117

15117

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Which of the 2 options does this work on?
  – FireCubez
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The first, outputs acronym from string input. I'll edit my post to clarify
  – Thaufeki
  18 hours ago
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Which of the 2 options does this work on?
  – FireCubez
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The first, outputs acronym from string input. I'll edit my post to clarify
  – Thaufeki
  18 hours ago
  
  

  
  
  
  

Which of the 2 options does this work on?
– FireCubez
18 hours ago

Which of the 2 options does this work on?
– FireCubez
18 hours ago

The first, outputs acronym from string input. I'll edit my post to clarify
– Thaufeki
18 hours ago

The first, outputs acronym from string input. I'll edit my post to clarify
– Thaufeki
18 hours ago

add a comment | 

up vote
0
down vote

JavaScript (Node.js), 70 bytes

s=>(s.replace(/w+/g,x=>w+=x[0]+' ?',w='^'),s.match(w+'\b')||[''])[0]

Try it online!

share|improve this answer

edited 17 hours ago

answered 19 hours ago

l4m2

3,8981431

add a comment | 

up vote
0
down vote

JavaScript (Node.js), 70 bytes

s=>(s.replace(/w+/g,x=>w+=x[0]+' ?',w='^'),s.match(w+'\b')||[''])[0]

Try it online!

share|improve this answer

edited 17 hours ago

answered 19 hours ago

l4m2

3,8981431

add a comment | 

up vote
0
down vote

up vote
0
down vote

JavaScript (Node.js), 70 bytes

s=>(s.replace(/w+/g,x=>w+=x[0]+' ?',w='^'),s.match(w+'\b')||[''])[0]

Try it online!

share|improve this answer

edited 17 hours ago

answered 19 hours ago

l4m2

3,8981431

JavaScript (Node.js), 70 bytes

s=>(s.replace(/w+/g,x=>w+=x[0]+' ?',w='^'),s.match(w+'\b')||[''])[0]

Try it online!

share|improve this answer

edited 17 hours ago

answered 19 hours ago

l4m2

3,8981431

share|improve this answer

share|improve this answer

edited 17 hours ago

edited 17 hours ago

edited 17 hours ago

answered 19 hours ago

l4m2

3,8981431

answered 19 hours ago

l4m2

3,8981431

answered 19 hours ago

l4m2

3,8981431

3,8981431

add a comment | 

add a comment | 

 

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