Make a recursive acronym

Clash Royale CLAN TAG#URR8PPP

up vote
2
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

You have 2 choices:

• 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)




• Make a program that takes a string of 1 or more words separated by space characters, and output a sentence that makes the input an acronym for that sentence, or output an empty string if it is impossible. It is never impossible to make a sentence if the input is 1 word long, but if we have 2 words, we get a similar problem to the one above. For example, programming puzzles, you would have to obviously have programming puzzles at the beginning of your outputted sentence, but if we try to convert it back to an acronym, we see that taking the p from programming is fine, but then the p from puzzles makes the acronym spell out pp.. which doesn't match pr..

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

---

Guidelines

• Include which of the 2 options you chose.


• Include your byte count because this is code golf.

---

Test Cases

(Finding acronym) hi igloo -> hi
(Finding acronym) ppcg paints -> (impossible)
(Finding acronym) ppcg paints cool giraffes -> ppcg
(Finding acronym) I -> I
(Finding sentence) ppcg paints -> ppcg paints cool galaxies paints air in not the sea
(Finding sentence) hi -> hi invert
(Finding sentence) I -> I

---

Scoring

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

code-golf string

share|improve this question

edited 1 hour ago

asked 2 hours ago

FireCubez

1398

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JoKing The challenge becomes trivial without that
  – FireCubez
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JoKing alright then, I can change that
  – FireCubez
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  In your test cases you have ppcg paints -> pp, but it should match the first word or words. Must the acronym merely be contained in the string, or does it need to start with it, or be one of the words?
  – Jo King
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JoKing Oh, my mistake, it should spell out the word completely
  – FireCubez
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Partially related.
  – AdmBorkBork
  1 hour ago
  

  
  
  
  

 | 
show 1 more comment

up vote
2
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

You have 2 choices:

• 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)




• Make a program that takes a string of 1 or more words separated by space characters, and output a sentence that makes the input an acronym for that sentence, or output an empty string if it is impossible. It is never impossible to make a sentence if the input is 1 word long, but if we have 2 words, we get a similar problem to the one above. For example, programming puzzles, you would have to obviously have programming puzzles at the beginning of your outputted sentence, but if we try to convert it back to an acronym, we see that taking the p from programming is fine, but then the p from puzzles makes the acronym spell out pp.. which doesn't match pr..

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

---

Guidelines

• Include which of the 2 options you chose.


• Include your byte count because this is code golf.

---

Test Cases

(Finding acronym) hi igloo -> hi
(Finding acronym) ppcg paints -> (impossible)
(Finding acronym) ppcg paints cool giraffes -> ppcg
(Finding acronym) I -> I
(Finding sentence) ppcg paints -> ppcg paints cool galaxies paints air in not the sea
(Finding sentence) hi -> hi invert
(Finding sentence) I -> I

---

Scoring

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

code-golf string

share|improve this question

edited 1 hour ago

asked 2 hours ago

FireCubez

1398

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JoKing The challenge becomes trivial without that
  – FireCubez
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JoKing alright then, I can change that
  – FireCubez
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  In your test cases you have ppcg paints -> pp, but it should match the first word or words. Must the acronym merely be contained in the string, or does it need to start with it, or be one of the words?
  – Jo King
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JoKing Oh, my mistake, it should spell out the word completely
  – FireCubez
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Partially related.
  – AdmBorkBork
  1 hour ago
  

  
  
  
  

 | 
show 1 more comment

up vote
2
down vote

favorite

up vote
2
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

You have 2 choices:

• 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)




• Make a program that takes a string of 1 or more words separated by space characters, and output a sentence that makes the input an acronym for that sentence, or output an empty string if it is impossible. It is never impossible to make a sentence if the input is 1 word long, but if we have 2 words, we get a similar problem to the one above. For example, programming puzzles, you would have to obviously have programming puzzles at the beginning of your outputted sentence, but if we try to convert it back to an acronym, we see that taking the p from programming is fine, but then the p from puzzles makes the acronym spell out pp.. which doesn't match pr..

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

---

Guidelines

• Include which of the 2 options you chose.


• Include your byte count because this is code golf.

---

Test Cases

(Finding acronym) hi igloo -> hi
(Finding acronym) ppcg paints -> (impossible)
(Finding acronym) ppcg paints cool giraffes -> ppcg
(Finding acronym) I -> I
(Finding sentence) ppcg paints -> ppcg paints cool galaxies paints air in not the sea
(Finding sentence) hi -> hi invert
(Finding sentence) I -> I

---

Scoring

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

code-golf string

share|improve this question

edited 1 hour ago

asked 2 hours ago

FireCubez

1398

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

You have 2 choices:

• 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)




• Make a program that takes a string of 1 or more words separated by space characters, and output a sentence that makes the input an acronym for that sentence, or output an empty string if it is impossible. It is never impossible to make a sentence if the input is 1 word long, but if we have 2 words, we get a similar problem to the one above. For example, programming puzzles, you would have to obviously have programming puzzles at the beginning of your outputted sentence, but if we try to convert it back to an acronym, we see that taking the p from programming is fine, but then the p from puzzles makes the acronym spell out pp.. which doesn't match pr..

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

---

Guidelines

• Include which of the 2 options you chose.


• Include your byte count because this is code golf.

---

Test Cases

(Finding acronym) hi igloo -> hi
(Finding acronym) ppcg paints -> (impossible)
(Finding acronym) ppcg paints cool giraffes -> ppcg
(Finding acronym) I -> I
(Finding sentence) ppcg paints -> ppcg paints cool galaxies paints air in not the sea
(Finding sentence) hi -> hi invert
(Finding sentence) 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 1 hour ago

asked 2 hours ago

FireCubez

1398

share|improve this question

edited 1 hour ago

asked 2 hours ago

FireCubez

1398

share|improve this question

share|improve this question

edited 1 hour ago

edited 1 hour ago

edited 1 hour ago

asked 2 hours ago

FireCubez

1398

asked 2 hours ago

FireCubez

1398

asked 2 hours ago

FireCubez

1398

1398

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JoKing The challenge becomes trivial without that
  – FireCubez
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JoKing alright then, I can change that
  – FireCubez
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  In your test cases you have ppcg paints -> pp, but it should match the first word or words. Must the acronym merely be contained in the string, or does it need to start with it, or be one of the words?
  – Jo King
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JoKing Oh, my mistake, it should spell out the word completely
  – FireCubez
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Partially related.
  – AdmBorkBork
  1 hour ago
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JoKing The challenge becomes trivial without that
  – FireCubez
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JoKing alright then, I can change that
  – FireCubez
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  In your test cases you have ppcg paints -> pp, but it should match the first word or words. Must the acronym merely be contained in the string, or does it need to start with it, or be one of the words?
  – Jo King
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JoKing Oh, my mistake, it should spell out the word completely
  – FireCubez
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Partially related.
  – AdmBorkBork
  1 hour ago
  

  
  
  
  

@JoKing The challenge becomes trivial without that
– FireCubez
2 hours ago

@JoKing The challenge becomes trivial without that
– FireCubez
2 hours ago

@JoKing alright then, I can change that
– FireCubez
2 hours ago

@JoKing alright then, I can change that
– FireCubez
2 hours ago

1

1

In your test cases you have ppcg paints -> pp, but it should match the first word or words. Must the acronym merely be contained in the string, or does it need to start with it, or be one of the words?
– Jo King
2 hours ago

In your test cases you have ppcg paints -> pp, but it should match the first word or words. Must the acronym merely be contained in the string, or does it need to start with it, or be one of the words?
– Jo King
2 hours ago

@JoKing Oh, my mistake, it should spell out the word completely
– FireCubez
1 hour ago

@JoKing Oh, my mistake, it should spell out the word completely
– FireCubez
1 hour ago

Partially related.
– AdmBorkBork
1 hour ago

Partially related.
– AdmBorkBork
1 hour ago

 | 
show 1 more comment

9 Answers
9

active

oldest

votes

up vote
1
down vote

05AB1E, 13 bytes

Finds acronym

ð¡©ηʒJ®€нJQ}»

Try it online!

Explanation

ð¡ # split input on spaces
 © # store result in register
 ηʒ } # filter the prefixes of the result
 # keep only those that
 J # when joined to one string
 Q # are equal to
 ®€н # the first letter of each word in the input
 J # joined to one string 
 » # join on spaces and newlines

share|improve this answer

edited 1 hour ago

answered 1 hour 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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hmm, you could be right it's related. I always saw stop loop if true and split on space as two different functions, both using #. So the compiler encountering a # would do a check: Is an infinite loop running? Yes -> stop it; No -> Split the top of the stack on spaces. If it's implemented like this pseudo-code, it would be strange if # as stop loop if true would be wrapped in a list, since it would never go into the else-clause in the compiler. But perhaps the pseudo-code isn't so easy to implement in Elixir as I think, in which case you're perhaps right it's related.
  – Kevin Cruijssen
  1 hour ago
  
  

  
  
  
  

 | 
show 4 more comments

up vote
1
down vote

Japt, 13 11 bytes

-2 bytes from @Shaggy

Finds acronym

¸
mά
ζV©V

Try it online!

share|improve this answer

edited 4 secs ago

answered 2 hours ago

Luis felipe De jesus Munoz

3,39711049

• 
  
  
  

  
  

  
  
  
  
  
  

  
  11 bytes
  – Shaggy
  28 mins ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

Perl 6, 37 bytes

~.words>>.ord.chrs.grep(*∈.words)

Try it online!

First option.

share|improve this answer

edited 2 hours ago

answered 2 hours ago

Jo King

17.9k24198

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

up vote
0
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 1 hour ago

answered 1 hour ago

Neil

76.9k744173

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @FireCubez Sorry, I was working off an older version of the question.
  – Neil
  1 hour 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 54 mins ago

answered 2 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
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @FireCubez fixed
  – Dead Possum
  54 mins ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

Haskell, 40 bytes

f x|w<-words x=[0|map(!!0)w==w!!0]>>w!!0

Finds acronym.

Try it online!

w<-words x -- let w be the input string split into words
 map(!!0)w==w!!0 -- check if the first character of each word are
 -- equal to the first word
 [0| <Test> ] >> <Out> -- return <Out> (here the first word) if <Test> is True
 -- else an empty string [1]

[1] taken from Tips for golfing in Haskell

share|improve this answer

edited 23 mins ago

answered 33 mins ago

nimi

30.4k31884

add a comment | 

up vote
0
down vote

Ruby -apl, 28 bytes

$_=$F.map(&:chr).join[$F[0]]

Try it online!

Option 1 - finds the acronym if it exists.

share|improve this answer

answered 17 mins ago

Kirill L.

2,9661117

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

answered 9 mins 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, 49 bytes

s=>(a=s.split` `).map(x=>x[0]).join``==a[0]&&a[0]

Try it online

share

answered 9 mins ago

Shaggy

17.4k21663

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Fail on ppcg paints cool giraffes pa ac ik ne td sr
  – l4m2
  8 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @lm42, ppcgpaints != ppcg.
  – Shaggy
  3 mins ago
  

  
  
  
  

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 ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "200"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: false,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodegolf.stackexchange.com%2fquestions%2f174766%2fmake-a-recursive-acronym%23new-answer', 'question_page');

);

Post as a guest

Name Email

9 Answers
9

active

oldest

votes

9 Answers
9

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

05AB1E, 13 bytes

Finds acronym

ð¡©ηʒJ®€нJQ}»

Try it online!

Explanation

ð¡ # split input on spaces
 © # store result in register
 ηʒ } # filter the prefixes of the result
 # keep only those that
 J # when joined to one string
 Q # are equal to
 ®€н # the first letter of each word in the input
 J # joined to one string 
 » # join on spaces and newlines

share|improve this answer

edited 1 hour ago

answered 1 hour 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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hmm, you could be right it's related. I always saw stop loop if true and split on space as two different functions, both using #. So the compiler encountering a # would do a check: Is an infinite loop running? Yes -> stop it; No -> Split the top of the stack on spaces. If it's implemented like this pseudo-code, it would be strange if # as stop loop if true would be wrapped in a list, since it would never go into the else-clause in the compiler. But perhaps the pseudo-code isn't so easy to implement in Elixir as I think, in which case you're perhaps right it's related.
  – Kevin Cruijssen
  1 hour ago
  
  

  
  
  
  

 | 
show 4 more comments

up vote
1
down vote

05AB1E, 13 bytes

Finds acronym

ð¡©ηʒJ®€нJQ}»

Try it online!

Explanation

ð¡ # split input on spaces
 © # store result in register
 ηʒ } # filter the prefixes of the result
 # keep only those that
 J # when joined to one string
 Q # are equal to
 ®€н # the first letter of each word in the input
 J # joined to one string 
 » # join on spaces and newlines

share|improve this answer

edited 1 hour ago

answered 1 hour 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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hmm, you could be right it's related. I always saw stop loop if true and split on space as two different functions, both using #. So the compiler encountering a # would do a check: Is an infinite loop running? Yes -> stop it; No -> Split the top of the stack on spaces. If it's implemented like this pseudo-code, it would be strange if # as stop loop if true would be wrapped in a list, since it would never go into the else-clause in the compiler. But perhaps the pseudo-code isn't so easy to implement in Elixir as I think, in which case you're perhaps right it's related.
  – Kevin Cruijssen
  1 hour ago
  
  

  
  
  
  

 | 
show 4 more comments

up vote
1
down vote

up vote
1
down vote

05AB1E, 13 bytes

Finds acronym

ð¡©ηʒJ®€нJQ}»

Try it online!

Explanation

ð¡ # split input on spaces
 © # store result in register
 ηʒ } # filter the prefixes of the result
 # keep only those that
 J # when joined to one string
 Q # are equal to
 ®€н # the first letter of each word in the input
 J # joined to one string 
 » # join on spaces and newlines

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Emigna

44k431133

05AB1E, 13 bytes

Finds acronym

ð¡©ηʒJ®€нJQ}»

Try it online!

Explanation

ð¡ # split input on spaces
 © # store result in register
 ηʒ } # filter the prefixes of the result
 # keep only those that
 J # when joined to one string
 Q # are equal to
 ®€н # the first letter of each word in the input
 J # joined to one string 
 » # join on spaces and newlines

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Emigna

44k431133

share|improve this answer

share|improve this answer

edited 1 hour ago

edited 1 hour ago

edited 1 hour ago

answered 1 hour ago

Emigna

44k431133

answered 1 hour ago

Emigna

44k431133

answered 1 hour 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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hmm, you could be right it's related. I always saw stop loop if true and split on space as two different functions, both using #. So the compiler encountering a # would do a check: Is an infinite loop running? Yes -> stop it; No -> Split the top of the stack on spaces. If it's implemented like this pseudo-code, it would be strange if # as stop loop if true would be wrapped in a list, since it would never go into the else-clause in the compiler. But perhaps the pseudo-code isn't so easy to implement in Elixir as I think, in which case you're perhaps right it's related.
  – Kevin Cruijssen
  1 hour ago
  
  

  
  
  
  

 | 
show 4 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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hmm, you could be right it's related. I always saw stop loop if true and split on space as two different functions, both using #. So the compiler encountering a # would do a check: Is an infinite loop running? Yes -> stop it; No -> Split the top of the stack on spaces. If it's implemented like this pseudo-code, it would be strange if # as stop loop if true would be wrapped in a list, since it would never go into the else-clause in the compiler. But perhaps the pseudo-code isn't so easy to implement in Elixir as I think, in which case you're perhaps right it's related.
  – Kevin Cruijssen
  1 hour 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
1 hour 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
1 hour ago

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

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

Hmm, you could be right it's related. I always saw stop loop if true and split on space as two different functions, both using #. So the compiler encountering a # would do a check: Is an infinite loop running? Yes -> stop it; No -> Split the top of the stack on spaces. If it's implemented like this pseudo-code, it would be strange if # as stop loop if true would be wrapped in a list, since it would never go into the else-clause in the compiler. But perhaps the pseudo-code isn't so easy to implement in Elixir as I think, in which case you're perhaps right it's related.
– Kevin Cruijssen
1 hour ago

Hmm, you could be right it's related. I always saw stop loop if true and split on space as two different functions, both using #. So the compiler encountering a # would do a check: Is an infinite loop running? Yes -> stop it; No -> Split the top of the stack on spaces. If it's implemented like this pseudo-code, it would be strange if # as stop loop if true would be wrapped in a list, since it would never go into the else-clause in the compiler. But perhaps the pseudo-code isn't so easy to implement in Elixir as I think, in which case you're perhaps right it's related.
– Kevin Cruijssen
1 hour ago

 | 
show 4 more comments

up vote
1
down vote

Japt, 13 11 bytes

-2 bytes from @Shaggy

Finds acronym

¸
mά
ζV©V

Try it online!

share|improve this answer

edited 4 secs ago

answered 2 hours ago

Luis felipe De jesus Munoz

3,39711049

• 
  
  
  

  
  

  
  
  
  
  
  

  
  11 bytes
  – Shaggy
  28 mins ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

Japt, 13 11 bytes

-2 bytes from @Shaggy

Finds acronym

¸
mά
ζV©V

Try it online!

share|improve this answer

edited 4 secs ago

answered 2 hours ago

Luis felipe De jesus Munoz

3,39711049

• 
  
  
  

  
  

  
  
  
  
  
  

  
  11 bytes
  – Shaggy
  28 mins ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Japt, 13 11 bytes

-2 bytes from @Shaggy

Finds acronym

¸
mά
ζV©V

Try it online!

share|improve this answer

edited 4 secs ago

answered 2 hours ago

Luis felipe De jesus Munoz

3,39711049

Japt, 13 11 bytes

-2 bytes from @Shaggy

Finds acronym

¸
mά
ζV©V

Try it online!

share|improve this answer

edited 4 secs ago

answered 2 hours ago

Luis felipe De jesus Munoz

3,39711049

share|improve this answer

share|improve this answer

edited 4 secs ago

edited 4 secs ago

edited 4 secs ago

answered 2 hours ago

Luis felipe De jesus Munoz

3,39711049

answered 2 hours ago

Luis felipe De jesus Munoz

3,39711049

answered 2 hours ago

Luis felipe De jesus Munoz

3,39711049

3,39711049

• 
  
  
  

  
  

  
  
  
  
  
  

  
  11 bytes
  – Shaggy
  28 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  11 bytes
  – Shaggy
  28 mins ago
  

  
  
  
  

11 bytes
– Shaggy
28 mins ago

11 bytes
– Shaggy
28 mins ago

add a comment | 

up vote
0
down vote

Perl 6, 37 bytes

~.words>>.ord.chrs.grep(*∈.words)

Try it online!

First option.

share|improve this answer

edited 2 hours ago

answered 2 hours ago

Jo King

17.9k24198

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

up vote
0
down vote

Perl 6, 37 bytes

~.words>>.ord.chrs.grep(*∈.words)

Try it online!

First option.

share|improve this answer

edited 2 hours ago

answered 2 hours ago

Jo King

17.9k24198

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

Perl 6, 37 bytes

~.words>>.ord.chrs.grep(*∈.words)

Try it online!

First option.

share|improve this answer

edited 2 hours ago

answered 2 hours ago

Jo King

17.9k24198

Perl 6, 37 bytes

~.words>>.ord.chrs.grep(*∈.words)

Try it online!

First option.

share|improve this answer

edited 2 hours ago

answered 2 hours ago

Jo King

17.9k24198

share|improve this answer

share|improve this answer

edited 2 hours ago

edited 2 hours ago

edited 2 hours ago

answered 2 hours ago

Jo King

17.9k24198

answered 2 hours ago

Jo King

17.9k24198

answered 2 hours ago

Jo King

17.9k24198

17.9k24198

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

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

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

add a comment | 

up vote
0
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 1 hour ago

answered 1 hour ago

Neil

76.9k744173

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

up vote
0
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 1 hour ago

answered 1 hour ago

Neil

76.9k744173

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
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 1 hour ago

answered 1 hour ago

Neil

76.9k744173

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 1 hour ago

answered 1 hour ago

Neil

76.9k744173

share|improve this answer

share|improve this answer

edited 1 hour ago

edited 1 hour ago

edited 1 hour ago

answered 1 hour ago

Neil

76.9k744173

answered 1 hour ago

Neil

76.9k744173

answered 1 hour ago

Neil

76.9k744173

76.9k744173

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

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

@FireCubez Sorry, I was working off an older version of the question.
– Neil
1 hour 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 54 mins ago

answered 2 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
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @FireCubez fixed
  – Dead Possum
  54 mins 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 54 mins ago

answered 2 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
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @FireCubez fixed
  – Dead Possum
  54 mins 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 54 mins ago

answered 2 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 54 mins ago

answered 2 hours ago

Dead Possum

2,805623

share|improve this answer

share|improve this answer

edited 54 mins ago

edited 54 mins ago

edited 54 mins ago

answered 2 hours ago

Dead Possum

2,805623

answered 2 hours ago

Dead Possum

2,805623

answered 2 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
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @FireCubez fixed
  – Dead Possum
  54 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @FireCubez fixed
  – Dead Possum
  54 mins ago
  

  
  
  
  

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

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

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

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

@FireCubez fixed
– Dead Possum
54 mins ago

@FireCubez fixed
– Dead Possum
54 mins ago

add a comment | 

up vote
0
down vote

Haskell, 40 bytes

f x|w<-words x=[0|map(!!0)w==w!!0]>>w!!0

Finds acronym.

Try it online!

w<-words x -- let w be the input string split into words
 map(!!0)w==w!!0 -- check if the first character of each word are
 -- equal to the first word
 [0| <Test> ] >> <Out> -- return <Out> (here the first word) if <Test> is True
 -- else an empty string [1]

[1] taken from Tips for golfing in Haskell

share|improve this answer

edited 23 mins ago

answered 33 mins ago

nimi

30.4k31884

add a comment | 

up vote
0
down vote

Haskell, 40 bytes

f x|w<-words x=[0|map(!!0)w==w!!0]>>w!!0

Finds acronym.

Try it online!

w<-words x -- let w be the input string split into words
 map(!!0)w==w!!0 -- check if the first character of each word are
 -- equal to the first word
 [0| <Test> ] >> <Out> -- return <Out> (here the first word) if <Test> is True
 -- else an empty string [1]

[1] taken from Tips for golfing in Haskell

share|improve this answer

edited 23 mins ago

answered 33 mins ago

nimi

30.4k31884

add a comment | 

up vote
0
down vote

up vote
0
down vote

Haskell, 40 bytes

f x|w<-words x=[0|map(!!0)w==w!!0]>>w!!0

Finds acronym.

Try it online!

w<-words x -- let w be the input string split into words
 map(!!0)w==w!!0 -- check if the first character of each word are
 -- equal to the first word
 [0| <Test> ] >> <Out> -- return <Out> (here the first word) if <Test> is True
 -- else an empty string [1]

[1] taken from Tips for golfing in Haskell

share|improve this answer

edited 23 mins ago

answered 33 mins ago

nimi

30.4k31884

Haskell, 40 bytes

f x|w<-words x=[0|map(!!0)w==w!!0]>>w!!0

Finds acronym.

Try it online!

w<-words x -- let w be the input string split into words
 map(!!0)w==w!!0 -- check if the first character of each word are
 -- equal to the first word
 [0| <Test> ] >> <Out> -- return <Out> (here the first word) if <Test> is True
 -- else an empty string [1]

[1] taken from Tips for golfing in Haskell

share|improve this answer

edited 23 mins ago

answered 33 mins ago

nimi

30.4k31884

share|improve this answer

share|improve this answer

edited 23 mins ago

edited 23 mins ago

edited 23 mins ago

answered 33 mins ago

nimi

30.4k31884

answered 33 mins ago

nimi

30.4k31884

answered 33 mins ago

nimi

30.4k31884

30.4k31884

add a comment | 

add a comment | 

up vote
0
down vote

Ruby -apl, 28 bytes

$_=$F.map(&:chr).join[$F[0]]

Try it online!

Option 1 - finds the acronym if it exists.

share|improve this answer

answered 17 mins ago

Kirill L.

2,9661117

add a comment | 

up vote
0
down vote

Ruby -apl, 28 bytes

$_=$F.map(&:chr).join[$F[0]]

Try it online!

Option 1 - finds the acronym if it exists.

share|improve this answer

answered 17 mins ago

Kirill L.

2,9661117

add a comment | 

up vote
0
down vote

up vote
0
down vote

Ruby -apl, 28 bytes

$_=$F.map(&:chr).join[$F[0]]

Try it online!

Option 1 - finds the acronym if it exists.

share|improve this answer

answered 17 mins ago

Kirill L.

2,9661117

Ruby -apl, 28 bytes

$_=$F.map(&:chr).join[$F[0]]

Try it online!

Option 1 - finds the acronym if it exists.

share|improve this answer

answered 17 mins ago

Kirill L.

2,9661117

share|improve this answer

share|improve this answer

answered 17 mins ago

Kirill L.

2,9661117

answered 17 mins ago

Kirill L.

2,9661117

answered 17 mins ago

Kirill L.

2,9661117

2,9661117

add a comment | 

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

answered 9 mins 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

answered 9 mins 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

answered 9 mins 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

answered 9 mins 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

share

answered 9 mins 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 9 mins ago

alpheus

214

answered 9 mins 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

JavaScript, 49 bytes

s=>(a=s.split` `).map(x=>x[0]).join``==a[0]&&a[0]

Try it online

share

answered 9 mins ago

Shaggy

17.4k21663

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Fail on ppcg paints cool giraffes pa ac ik ne td sr
  – l4m2
  8 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @lm42, ppcgpaints != ppcg.
  – Shaggy
  3 mins ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

JavaScript, 49 bytes

s=>(a=s.split` `).map(x=>x[0]).join``==a[0]&&a[0]

Try it online

share

answered 9 mins ago

Shaggy

17.4k21663

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Fail on ppcg paints cool giraffes pa ac ik ne td sr
  – l4m2
  8 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @lm42, ppcgpaints != ppcg.
  – Shaggy
  3 mins ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

JavaScript, 49 bytes

s=>(a=s.split` `).map(x=>x[0]).join``==a[0]&&a[0]

Try it online

share

answered 9 mins ago

Shaggy

17.4k21663

JavaScript, 49 bytes

s=>(a=s.split` `).map(x=>x[0]).join``==a[0]&&a[0]

Try it online

share

answered 9 mins ago

Shaggy

17.4k21663

share

share

answered 9 mins ago

Shaggy

17.4k21663

answered 9 mins ago

Shaggy

17.4k21663

answered 9 mins ago

Shaggy

17.4k21663

17.4k21663

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Fail on ppcg paints cool giraffes pa ac ik ne td sr
  – l4m2
  8 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @lm42, ppcgpaints != ppcg.
  – Shaggy
  3 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Fail on ppcg paints cool giraffes pa ac ik ne td sr
  – l4m2
  8 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @lm42, ppcgpaints != ppcg.
  – Shaggy
  3 mins ago
  

  
  
  
  

Fail on ppcg paints cool giraffes pa ac ik ne td sr
– l4m2
8 mins ago

Fail on ppcg paints cool giraffes pa ac ik ne td sr
– l4m2
8 mins ago

@lm42, ppcgpaints != ppcg.
– Shaggy
3 mins ago

@lm42, ppcgpaints != ppcg.
– Shaggy
3 mins ago

add a comment | 

 

draft saved

draft discarded

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodegolf.stackexchange.com%2fquestions%2f174766%2fmake-a-recursive-acronym%23new-answer', 'question_page');

);

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Name Email

Name

Email

Name Email

Name

Email