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On some terminals, pressing backspace generates the control code ^H to delete the previous character. This gave rise to a snarky idiom where edits are feigned for comedic effect:

Be nice to this fool^H^H^H^Hgentleman, he's visiting from corporate HQ.

Given a string with one or more ^H's, output the result of backspacing on each ^H. The input will use only printable characters (ASCII 32-126), and ^ will only appear as ^H. Backspaces will never happen on empty text.

You may not assume that the output environment supports control codes, in particular the backspace code \x08.

>> Horse^H^H^H^H^HCow
Cow

>> Be nice to this fool^H^H^H^Hgentleman, he's visiting from corporate HQ.
Be nice to this gentleman, he's visiting from corporate HQ.

>> 123^H45^H^H^H78^H
17

>> Digital Trauma^H^H^H^H^H^H^H^H^H^H^H^H^H^HMaria Tidal Tug^H^H^H^H^H^H^H^H^H^H^H^H^H^H^HDigital Trauma
Digital Trauma

Leaderboard

Here's a by-language leaderboard, courtesy of Martin Büttner.

To make sure that your answer shows up, please start your answer with a headline, using the following Markdown template:

# Language Name, N bytes

where N is the size of your submission. If you improve your score, you can keep old scores in the headline, by striking them through. For instance:

# Ruby, <s>104</s> <s>101</s> 96 bytes

function answersUrl(e){return"https://api.stackexchange.com/2.2/questions/52946/answers?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function getAnswers(){$.ajax({url:answersUrl(page++),method:"get",dataType:"jsonp",crossDomain:true,success:function(e){answers.push.apply(answers,e.items);if(e.has_more)getAnswers();else process()}})}function shouldHaveHeading(e){var t=false;var n=e.body_markdown.split("\n");try{t|=/^#/.test(e.body_markdown);t|=["-","="].indexOf(n[1][0])>-1;t&=LANGUAGE_REG.test(e.body_markdown)}catch(r){}return t}function shouldHaveScore(e){var t=false;try{t|=SIZE_REG.test(e.body_markdown.split("\n")[0])}catch(n){}return t}function getAuthorName(e){return e.owner.display_name}function process(){answers=answers.filter(shouldHaveScore).filter(shouldHaveHeading);answers.sort(function(e,t){var n=+(e.body_markdown.split("\n")[0].match(SIZE_REG)||[Infinity])[0],r=+(t.body_markdown.split("\n")[0].match(SIZE_REG)||[Infinity])[0];return n-r});var e={};var t=1;answers.forEach(function(n){var r=n.body_markdown.split("\n")[0];var i=$("#answer-template").html();var s=r.match(NUMBER_REG)[0];var o=(r.match(SIZE_REG)||[0])[0];var u=r.match(LANGUAGE_REG)[1];var a=getAuthorName(n);i=i.replace("{{PLACE}}",t++ +".").replace("{{NAME}}",a).replace("{{LANGUAGE}}",u).replace("{{SIZE}}",o).replace("{{LINK}}",n.share_link);i=$(i);$("#answers").append(i);e[u]=e[u]||{lang:u,user:a,size:o,link:n.share_link}});var n=[];for(var r in e)if(e.hasOwnProperty(r))n.push(e[r]);n.sort(function(e,t){if(e.lang>t.lang)return 1;if(e.lang<t.lang)return-1;return 0});for(var i=0;i<n.length;++i){var s=$("#language-template").html();var r=n[i];s=s.replace("{{LANGUAGE}}",r.lang).replace("{{NAME}}",r.user).replace("{{SIZE}}",r.size).replace("{{LINK}}",r.link);s=$(s);$("#languages").append(s)}}var QUESTION_ID=45497;var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe";var answers=[],page=1;getAnswers();var SIZE_REG=/\d+(?=[^\d&]*(?:&lt;(?:s&gt;[^&]*&lt;\/s&gt;|[^&]+&gt;)[^\d&]*)*$)/;var NUMBER_REG=/\d+/;var LANGUAGE_REG=/^#*\s*((?:[^,\s]|\s+[^-,\s])*)/
body{text-align:left!important}#answer-list,#language-list{padding:10px;width:290px;float:left}table thead{font-weight:700}table td{padding:5px}
<script src=https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js></script><link rel=stylesheet type=text/css href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b"><div id=answer-list><h2>Leaderboard</h2><table class=answer-list><thead><tr><td></td><td>Author<td>Language<td>Size<tbody id=answers></table></div><div id=language-list><h2>Winners by Language</h2><table class=language-list><thead><tr><td>Language<td>User<td>Score<tbody id=languages></table></div><table style=display:none><tbody id=answer-template><tr><td>{{PLACE}}</td><td>{{NAME}}<td>{{LANGUAGE}}<td>{{SIZE}}<td><a href={{LINK}}>Link</a></table><table style=display:none><tbody id=language-template><tr><td>{{LANGUAGE}}<td>{{NAME}}<td>{{SIZE}}<td><a href={{LINK}}>Link</a></table>

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13
  • 4
    \$\begingroup\$ Is AAA^HB^H^H valid? \$\endgroup\$ Jul 10, 2015 at 21:47
  • \$\begingroup\$ @NathanMerrill Yes, and it results in A. \$\endgroup\$
    – xnor
    Jul 10, 2015 at 21:54
  • 3
    \$\begingroup\$ I suspect retina would do well here. \$\endgroup\$
    – Claudiu
    Jul 10, 2015 at 21:55
  • 1
    \$\begingroup\$ @Fatalize: "Backspaces will never happen on empty text." \$\endgroup\$
    – Maltysen
    Jul 10, 2015 at 22:06
  • 16
    \$\begingroup\$ @Maria Tidal Tug comes back to haunt me \$\endgroup\$ Jul 10, 2015 at 23:18

49 Answers 49

1
2
2
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Jelly, 9 bytes

œṣ⁾^HṖ;¥/

Try it online!

Not a new approach, but the question was missing a Jelly answer

How it works

œṣ⁾^HṖ;¥/ - Main link. Takes a string S on the left
  ⁾^H     - Yield "^H"
œṣ        - Split on "^H"
       ¥/ - Reduce by the following:
     Ṗ    -   Remove the last character
      ;   -   And append the right argument
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1
  • \$\begingroup\$ ping me in an hour :P \$\endgroup\$
    – Wezl
    Apr 8, 2021 at 22:14
2
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Vim, 38 35 bytes

Saved 3 bytes thanks to Leo!

qa:%s/\([^H]\|\^\@<!.\)\^H//g
@aq@a

Try it online!

See Razetime's answer for a shorter and smarter version of this.

qa:%s/\([^H]\|\^\@<!.\)\^H//g
@aq@a

qa                                     Start recording a macro a
  :%s/                                 Substitute in entire file
      \([^H]\|\^\@<!H\)                Regex for the character to be deleted
        [^H]                           A character that isn't H
            \|                         or
                    H                  an H
              \^\@<!                   that doesn't have a ^ before it
                       \^H             All of that followed by ^H
                          //           Replace with empty string
                            g          Global flag so it replaces multiple times
                                       Enter the command
@a                                     Recursively call macro a
  q                                    Stop recording
   @a                                  Run the macro
```
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1
  • 1
    \$\begingroup\$ You need to pass V the -v flag in order to use verbose mode that turns <cr> into a carriage return Try it online! You could also directly insert a newline in the code instead Try it online! \$\endgroup\$
    – Leo
    Apr 9, 2021 at 0:45
1
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Python 2, 74 + 2 = 76 Bytes

I've tried a few approaches so far, this is the best I've been able to come up with so far.

n=input();o='';c=0
for l in n:d=l=='^';o=[o+l*(1-c),o[:-1]][d];c=d
print o
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  • 4
    \$\begingroup\$ Where are those 2 extra bytes coming from? \$\endgroup\$
    – xnor
    Jul 11, 2015 at 9:17
  • \$\begingroup\$ @xnor input has to be surrounded by quotes for this to work. I forgot to put that in the post. \$\endgroup\$
    – Kade
    Jul 11, 2015 at 13:42
  • 1
    \$\begingroup\$ I think the usual convention has been to allow string arguments to be taken in quotes for free, but I'm not totally sure. \$\endgroup\$
    – xnor
    Jul 13, 2015 at 7:19
1
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Mumps, 84 Bytes

R Z S T="",Y=$L(Z,"^H") F I=1:1:Y{S T=T_$P(Z,"^H",I) S:I<Y T=$E(T,1,$L(T)-1)} W !,T

This could probably be made shorter as a function (1 byte I was able to save in quick testing) but I kinda like the one-liner aspect... :-)

The braces come from the Intersystems Cache flavour of Mumps which is what I'm most versed in.

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Java - 123 bytes

I personally like the g---1 part the best.

String f(char[] a){String b="";for(int g=-1;++g<a.length;b=(a[g++]=='^'?b.substring(0,b.length()-1):b+a[g---1]));return b;}

expanded (slightly):

  String f(char[] a) {
      String b = "";
      for (int g = -1;
           ++g < a.length;
           b = (a[g++]=='^' 
                ? b.substring(0, b.length() - 1) 
                : b + a[g---1])
      );
      return b;
  }
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1
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Batch - 138 bytes

@!! 2>nul||cmd/q/v/c%0 %1&&exit/b
set s=%1&for /F %%a in ('"prompt $H&echo on&for %%b in (1)do rem"')do set D=%%a
echo %s:^H=!D! !D!%

The first line is a way of saving a few bytes over the lengthy @echo off&setLocal enableDelayedExpansion (which turns echo off and enables the delayed expansion of variables, in case you were wondering). I explained it in Tips for Golfing in Batch.

The second line is a neat little trick to save the a backspace control character into a variable. It's pretty hacky, and I can't pretend to take credit for it. It's sort of explained here. Basically uses the prompt command to generate a backspace character and captures it in a variable - in this case !D!.

The final line then performs the simple string manipulation of - replace ^H with !D!<SPACE>!D!.

C:\>bsp.bat "testing^H^H^H test"
"test test"

Unfortunately it breaks with cases like "AAA^HB^H^H" - where it should produce "A", it instead produces "A"B. Which is somewhat confusing. I'll have to look into how Batch string manipulation works in some more depth.

C:\>bsp.bat "AAA^HB^H^H"
"A"B

Thanks to to some helpful people over here - I now realize that I was only saving the backspace character (0x08), and so was only overwriting the characters. It now works with examples like the following:

C:\>bsp.bat "AAA^HB^H^H"
"A"
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1
  • \$\begingroup\$ there's a shorter version: set s=%1&for /F %%a in ('prompt $H^&cmd /k^<nul')do set D=%%a \$\endgroup\$
    – ScriptKidd
    Apr 3, 2020 at 5:31
1
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JavaScript, 60 bytes

b=n=>(v=n.search('^H'))>0?b(n.replace(n.slice(v-1,v+2),'')):n

Not as good as edc65's answer, but I wanted to try something that didn't use Regex.

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bash, 50 bytes

while [ "$a" != "${a/?^H/}" ];do a=${a/?^H/};done

Sample:

a=$'Horse^H^H^H^H^HCow'
while [ "$a" != "${a/?^H/}" ];do a=${a/?^H/};done
echo $a
Cow

a="Be nice to this fool^H^H^H^Hgentleman, he's visiting from corporate HQ."
while [ "$a" != "${a/?^H/}" ];do a=${a/?^H/};done
echo $a
Be nice to this gentleman, he's visiting from corporate HQ.

This could work with real (binaries) backspaces as well:

a=$'Be nice to this fool\b\b\b\bgentleman, he'\'$'s visiting from corporate HQ.'
echo $a.. but:
Be nice to this gentleman, he's visiting from corporate HQ... but:
printf %q\\n "$a"
$'Be nice to this fool\b\b\b\bgentleman, he\'s visiting from corporate HQ.'

while [ "$a" != "${a/?$'\b'/}" ];do a=${a/?$'\b'/};done
printf %q\\n "$a"
Be\ nice\ to\ this\ gentleman\,\ he\'s\ visiting\ from\ corporate\ HQ.
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1
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><>, 39 38 36 bytes

There was no ><> answer so I thought I'd add one.

i:0(?v:'^'=?\
     ~   ~~i/
;!?lr<ro
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1
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JavaScript, 60 Bytes

I know there is already a JS answer here, but wanted to do it without regexes, because regex is really almost a language by itself. I'm sorry if you feel I shouldn't post this, this is my first post.

X is the string to be operated upon.

while((z=x.indexOf('^D'))>0){x=x.slice(0,z-1)+x.slice(z+2);}

Test

Put this into your browser's address bar.

javascript:x=prompt('Enter the sentence to be erased');while((z=x.indexOf('^D'))>0){x=x.slice(0,z-1)+x.slice(z+2);}alert(x);

I love JS because it will let you do things like saying (z=x.indexOf('^D))>0, and it will both assign z.indexOf('^D'); to x and evaluate z.indexOf('^D') in the condition, which saves me precious bytes in this problem. It's not very useful otherwise, but it's fun!

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1
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Emacs Lisp, 45 73 bytes

(lambda(s)(while(string-match".^H"s)(set's(replace-match"" nil nil s)))s)

Searches for the first occurence of anything else and ^H as long as it exists and replaces it with an empty string.

Old, incorrect version

(lambda(s)(replace-regexp-in-string".^H"""s))
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1
  • \$\begingroup\$ Sorry, but I'm afraid this is not enough – you have to do the replacing in multiple steps so “one^Hff^H^H^Huch” gets transformed into “ouch”. \$\endgroup\$
    – manatwork
    Dec 25, 2015 at 17:03
1
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jq 1.5, 41 37 bytes

(34 characters code + 3 characters command line option.)

reduce(./"^H")[]as$t("";.[:-1]+$t)

Sample run:

bash-4.3$ bin/jq -R -r 'reduce(./"^H")[]as$t("";.[:-1]+$t)' <<< "Be nice to this fool^H^H^H^Hgentleman, he's visiting from corporate HQ."
Be nice to this gentleman, he's visiting from corporate HQ.

On-line test (Passing -R through URL is not supported – so input passed as JSON string literal. Passing -r through URL is not supported – check Raw Output yourself.)

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1
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Oracle SQL 11.2, 172 bytes

WITH v(s,i)AS(SELECT:1,INSTR(:1,'^')FROM DUAL UNION ALL SELECT LPAD(s,i-2)||SUBSTR(s,i+2),INSTR(s,'^',1,2)-3 FROM v WHERE i>0AND'^'<>s)SELECT s FROM v WHERE INSTR(s,'^')=0;

Un-golfed

WITH v(s,i) AS                        -- Recursive view, s-> string, i->pos of first ^
(
  SELECT :1,INSTR(:1,'^')             -- Initialisation view 
  FROM   DUAL
  UNION ALL
  SELECT LPAD(s,i-2)||SUBSTR(s,i+2),  -- Remove the ^ at pos i and the characters before and after 
         INSTR(s,'^',1,2)-3           -- Compute the pos of the next ^ (the 2nd of s as before the remove just above)
  FROM   v 
  WHERE  i>0                          -- Exit clause : no more ^
    AND  s<>'^'                       -- Needed to circumvent oracle's cycle detection, without it 123^H45^H^H^H78^H will fail
)
SELECT s FROM v WHERE INSTR(s,'^')=0; -- Keep only the row without any ^
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1
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K (ngn/k), 23 16 bytes

-7 bytes by rereading question (specifically, "^ will only appear as ^H")

{x_/3#-1+x?"^"}/

Try it online!

Removes one instance of ?^H on each invocation, and is run until convergence (i.e. no more ^'s are present).

  • {...}/ run function on (implicit) input until it converges
  • -1+x?"^" get the index of the character immediately prior to the first ^ in x
  • x_/3# drop the value occurring at that index three times
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1
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Vyxal, 9 bytes

‛^H/ƒλ$Ṫp

Try it Online! Port of caird coinheringaahing's Jelly answer.

   /      # Split on
‛^H       # "^H"
    ƒλ--- # Reduce by
        p # Prepend to the first item
      $Ṫ  # The second with its last item removed
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0
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Javascript ES6, 41 39 bytes

f=s=>s==(s=s.replace(/.\^H/,""))?s:f(s)
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0
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𝔼𝕊𝕄𝕚𝕟, 8 chars / 10 bytes (noncompetitive)

ïė/.\^H/

Try it here (Firefox only).

Works like all of the other JSGL's.

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0
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APL (Dyalog Unicode), 18 bytes

'.\^H'⎕R''⍠'ML'1⍣≡

Try it online!

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0
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Swift, 81 bytes

let f={(s:inout _)in while(s+"").contains("^"){s={s.replacing}()(#/.\^H/#,"",1)}}

Call it like this:

var s = "Horse^H^H^H^H^HCow"
f(&s)
print(s) // "Cow"
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