Reverse and Invert a String

Question

Reverse and Invert a String

Challenge

In this challenge. You'll be writing a program which will output or return the input, reversed and inverted.

First, each character should be converted to its character code. Then, that should be converted to base-2. Following, that string should be reversed. After, the string should be inverted (1 -> 0 and 0 -> 1). Finally, that should be converted back to base 2 and then converted back to a character. If an character results to be an unprintable, you may optionally output it but they do not have to be removed.

H -> 72  -> 1001000 -> 0001001 -> 1110110 -> 118 -> v
e -> 101 -> 1100101 -> 1010011 -> 0101100 -> 44  -> ,
l -> 108 -> 1101100 -> 0011011 -> 1100100 -> 100 -> d
l -> 108 -> 1101100 -> 0011011 -> 1100100 -> 100 -> d
o -> 111 -> 1101111 -> 1111011 -> 0000100 -> 4   -> (unprintable)
, -> 44  -> 101100  -> 001101  -> 110010  -> 50  -> 2
  -> 32  -> 100000  -> 000001  -> 111110  -> 62  -> >
W -> 87  -> 1010111 -> 1110101 -> 0001010 -> 10  -> (newline)
o -> 111 -> 1101111 -> 1111011 -> 0000100 -> 4   -> (unprintable)
r -> 114 -> 1110010 -> 0100111 -> 1011000 -> 88  -> X
l -> 108 -> 1101100 -> 0011011 -> 1100100 -> 100 -> d
d -> 100 -> 1100100 -> 0010011 -> 1101100 -> 108 -> l
! -> 33  -> 100001  -> 100001  -> 011110  -> 30  -> (unprintable)

Scoring

Shortest code in bytes wins.

-15% Bonus: if your program removes un-printables from the output. This must be at least all characters below 32 except newlines (char 10)

I need to get my Simplex interpreter working again XD GBktnkZs — Conor O'Brien, Commented Nov 6, 2015 at 1:13
So the characters in the string aren't reversed, but the bits in each character are? — xnor, Commented Nov 6, 2015 at 1:36
Just to be sure: for 0010000 is the bit reverse 0000100 or 00001? — Digital Trauma, Commented Nov 6, 2015 at 2:37
@DigitalTrauma If the binary code is 0010000, it should be treated as 10000 so the reverse would be 00001 — Downgoat, Commented Nov 6, 2015 at 3:06
Can we assume just ASCII (as your examples), or should this work for whatever is a character in my language? (Also, if a language uses a different character code, should I use this instead of ASCII/Unicode)? — Paŭlo Ebermann, Commented Nov 6, 2015 at 17:30

Dennis · Accepted Answer · 2015-11-06 02:41:02Z

9

Pyth, 14 bytes

smCi!M_jCd2 2z

Try it online.

How it works

 m           z  Map over the input with lambda d:
        Cd        Cast d to character.
       j  2       Convert to base 2.
      _           Reverse the resulting array.
    !M            Mapped logical NOT.
   i        2     Convert back to integer.
  C               Cast to character.
s               Concatenate the resulting characters.

edited Nov 6, 2015 at 2:41

answered Nov 6, 2015 at 1:36

Dennis

211k41 gold badges370 silver badges825 bronze badges

2

\$\begingroup\$ Alternative solutions (all 14 bytes): smCi!MvM_.Bd2z smCi.r_.Bd`T2z smCiqR\0_.Bd2z \$\endgroup\$
– Dennis
Commented Nov 6, 2015 at 2:38
\$\begingroup\$ How about a version which removes unprintables, just out of interest/for comparison? Probably as a separate answer. \$\endgroup\$
– hyde
Commented Nov 6, 2015 at 6:51

Add a comment |

manatwork · Accepted Answer · 2015-11-07 17:32:26Z

8

Perl, 57 51 characters

(50 characters code + 1 character command-line option.)

s/./$_=unpack b8,$&;s|0+$||;"chr 0b".y|10|01|r/gee

Sample run:

bash-4.3$ perl -pe 's/./$_=unpack b8,$&;s|0+$||;"chr 0b".y|10|01|r/gee' <<< 'Hello, World!'
v,dd2>
Xdl

bash-4.3$ perl -pe 's/./$_=unpack b8,$&;s|0+$||;"chr 0b".y|10|01|r/gee' <<< 'Hello, World!' | od -tad1
0000000    v    ,    d    d  eot    2    >   nl  eot    X    d    l   rs   nl
         118   44  100  100    4   50   62   10    4   88  100  108   30   10
0000016

edited Nov 7, 2015 at 17:32

answered Nov 6, 2015 at 16:58

manatwork

20.6k5 gold badges51 silver badges81 bronze badges

1

\$\begingroup\$ 51 bytes: -p s/./$_=unpack b8,$&;s|0+$||;"chr 0b".y|10|01|r/gee. unpack b8,$& is shorter than sprintf'%b',ord$&, and additionally decodes in reverse order. Unfortunately it also produces trailing 0s, which need to be removed. \$\endgroup\$
– primo
Commented Nov 7, 2015 at 11:08
\$\begingroup\$ Thank you @primo. unpack is still unexplored terrain for me. \$\endgroup\$
– manatwork
Commented Nov 7, 2015 at 17:28
\$\begingroup\$ 42 bytes: -p s/./"chr 0b".unpack(b8,~$&)=~s|1+$||r/gee. Invert the character, no need to transliterate ;) \$\endgroup\$
– primo
Commented Nov 9, 2015 at 11:48

Add a comment |

jrich · Accepted Answer · 2015-11-06 04:08:24Z

JavaScript (ES6 ES7), 119 114 108 bytes

This turned out way longer than expected :(

Thanks to @vihan for 5 bytes saved! Thanks to @ETHProductions for another 6 bytes saved!

To test: Run the snippet below, enter input like "Hello, World!", and click Test!

x=>String.fromCharCode(...[for(y of x)+('0b'+[...y.charCodeAt().toString(2)].reverse().map(z=>z^1).join``)])

<!--                               Try the test suite below!                              --><strong id="bytecount" style="display:inline; font-size:32px; font-family:Helvetica"></strong><strong id="bytediff" style="display:inline; margin-left:10px; font-size:32px; font-family:Helvetica; color:lightgray"></strong><br><br><pre style="margin:0">Code:</pre><textarea id="textbox" style="margin-top:5px; margin-bottom:5px"></textarea><br><pre style="margin:0">Input:</pre><textarea id="inputbox" style="margin-top:5px; margin-bottom:5px"></textarea><br><button id="testbtn">Test!</button><button id="resetbtn">Reset</button><br><p><strong id="origheader" style="font-family:Helvetica; display:none">Original Code Output:</strong><p><div id="origoutput" style="margin-left:15px"></div><p><strong id="newheader" style="font-family:Helvetica; display:none">New Code Output:</strong><p><div id="newoutput" style="margin-left:15px"></div><script type="text/javascript" id="golfsnippet">var bytecount=document.getElementById("bytecount");var bytediff=document.getElementById("bytediff");var textbox=document.getElementById("textbox");var inputbox=document.getElementById("inputbox");var testbtn=document.getElementById("testbtn");var resetbtn=document.getElementById("resetbtn");var origheader=document.getElementById("origheader");var newheader=document.getElementById("newheader");var origoutput=document.getElementById("origoutput");var newoutput=document.getElementById("newoutput");textbox.style.width=inputbox.style.width=window.innerWidth-50+"px";var _originalCode=null;function getOriginalCode(){if(_originalCode!=null)return _originalCode;var allScripts=document.getElementsByTagName("script");for(var i=0;i<allScripts.length;i++){var script=allScripts[i];if(script.id!="golfsnippet"){originalCode=script.textContent.trim();return originalCode}}}function getNewCode(){return textbox.value.trim()}function getInput(){try{var inputText=inputbox.value.trim();var input=eval("["+inputText+"]");return input}catch(e){return null}}function setTextbox(s){textbox.value=s;onTextboxChange()}function setOutput(output,s){output.innerHTML=s}function addOutput(output,data){output.innerHTML+='<pre style="background-color:'+(data.type=="err"?"lightcoral":"lightgray")+'">'+escape(data.content)+"</pre>"}function getByteCount(s){return(new Blob([s],{encoding:"UTF-8",type:"text/plain;charset=UTF-8"})).size}function onTextboxChange(){var newLength=getByteCount(getNewCode());var oldLength=getByteCount(getOriginalCode());bytecount.innerHTML=newLength+" bytes";var diff=newLength-oldLength;if(diff>0){bytediff.innerHTML="(+"+diff+")";bytediff.style.color="lightcoral"}else if(diff<0){bytediff.innerHTML="("+diff+")";bytediff.style.color="lightgreen"}else{bytediff.innerHTML="("+diff+")";bytediff.style.color="lightgray"}}function onTestBtn(evt){origheader.style.display="inline";newheader.style.display="inline";setOutput(newoutput,"");setOutput(origoutput,"");var input=getInput();if(input===null){addOutput(origoutput,{type:"err",content:"Input is malformed. Using no input."});addOutput(newoutput,{type:"err",content:"Input is malformed. Using no input."});input=[]}doInterpret(getNewCode(),input,function(data){addOutput(newoutput,data)});doInterpret(getOriginalCode(),input,function(data){addOutput(origoutput,data)});evt.stopPropagation();return false}function onResetBtn(evt){setTextbox(getOriginalCode());origheader.style.display="none";newheader.style.display="none";setOutput(origoutput,"");setOutput(newoutput,"")}function escape(s){return s.toString().replace(/&/g,"&amp;").replace(/</g,"&lt;").replace(/>/g,"&gt;")}window.alert=function(){};window.prompt=function(){};function doInterpret(code,input,cb){var workerCode=interpret.toString()+";function stdout(s){ self.postMessage( {'type': 'out', 'content': s} ); }"+" function stderr(s){ self.postMessage( {'type': 'err', 'content': s} ); }"+" function kill(){ self.close(); }"+" self.addEventListener('message', function(msg){ interpret(msg.data.code, msg.data.input); });";var interpreter=new Worker(URL.createObjectURL(new Blob([workerCode])));interpreter.addEventListener("message",function(msg){cb(msg.data)});interpreter.postMessage({"code":code,"input":input});setTimeout(function(){interpreter.terminate()},1E4)}setTimeout(function(){getOriginalCode();textbox.addEventListener("input",onTextboxChange);testbtn.addEventListener("click",onTestBtn);resetbtn.addEventListener("click",onResetBtn);setTextbox(getOriginalCode())},100);function interpret(code,input){window={};alert=function(s){stdout(s)};window.alert=alert;console.log=alert;prompt=function(s){if(input.length<1)stderr("not enough input");else{var nextInput=input[0];input=input.slice(1);return nextInput.toString()}};window.prompt=prompt;(function(){try{var evalResult=eval(code);if(typeof evalResult=="function"){var callResult=evalResult.apply(this,input);if(typeof callResult!="undefined")stdout(callResult)}}catch(e){stderr(e.message)}})()};</script>

I think you could save a 4 bytes replacing the parseInt with +('0b'+<code>) as described here and another one by using w^1 instead of +!+w — Downgoat, Commented Nov 6, 2015 at 2:12
I never would have thought this possible, but I just golfed off .05 bytes: x=>String.fromCharCode(...[for(y of x)if((c=+('0b'+[...y.charCodeAt().toString(2)].reverse().map(z=>z^1).join``)) >31||c==10)c]) (127 - 15% = 107.95) Maybe this isn't legal, though; it only handles 10 == \n, not 13 == \r. @Vɪʜᴀɴ what is your opinion? — ETHproductions, Commented Nov 6, 2015 at 4:14
@ETHproductions: thanks - I only have Safari and Chrome to hand and I guess neither of these is "ES6-compliant". — Paul R, Commented Nov 6, 2015 at 14:07
@PaulR ES6, or ECMAScript 6, is one of the latest sets of new features for JavaScript. See this site for more info. There's also a compatability table that shows which features are supported by which browsers (and other programs). This answer in particular requires "arrow functions", the "spread operator", and ES7's "array comprehensions". — ETHproductions, Commented Nov 6, 2015 at 15:26

Community · Accepted Answer · 2020-06-17 09:04:33Z

JavaScript (ES7), 126 bytes - 15% = 107.1

I was playing around with this answer to see if the bonus was worth it. Apparently, it is. The test suite was stolen from the same answer, but I added my own twist: full support of the 15% bonus! :)

x=>String.fromCharCode(...[for(y of x)if((c=+('0b'+[...y.charCodeAt().toString(2)].reverse().map(z=>z^1).join``))>31|c==10)c])

<!--                               Try the test suite below!                              --><strong id="bytecount" style="display:inline; font-size:32px; font-family:Helvetica"></strong><strong id="bytediff" style="display:inline; margin-left:10px; font-size:32px; font-family:Helvetica; color:lightgray"></strong><br><strong id="score" style="display:inline; font-size:32px; font-family:Helvetica">Score:</strong><strong id="scorediff" style="display:inline; margin-left:10px; font-size:32px; font-family:Helvetica; color:lightgray"></strong><br><br><pre style="margin:0">Code:</pre><textarea id="textbox" style="margin-top:5px; margin-bottom:5px"></textarea><br><pre style="margin:0">Input:</pre><textarea id="inputbox" style="margin-top:5px; margin-bottom:5px">"Hello, World!"</textarea><br><button id="testbtn">Test!</button><button id="resetbtn">Reset</button><br><p><strong id="origheader" style="font-family:Helvetica; display:none">Original Code Output:</strong><p><div id="origoutput" style="margin-left:15px"></div><p><strong id="newheader" style="font-family:Helvetica; display:none">New Code Output:</strong><p><div id="newoutput" style="margin-left:15px"></div><script type="text/javascript" id="golfsnippet">var bytecount=document.getElementById("bytecount");var bytediff=document.getElementById("bytediff");var score=document.getElementById("score");var scorediff=document.getElementById("scorediff");var textbox=document.getElementById("textbox");var inputbox=document.getElementById("inputbox");var testbtn=document.getElementById("testbtn");var resetbtn=document.getElementById("resetbtn");var origheader=document.getElementById("origheader");var newheader=document.getElementById("newheader");var origoutput=document.getElementById("origoutput");var newoutput=document.getElementById("newoutput");textbox.style.width=inputbox.style.width=window.innerWidth-50+"px";var _originalCode=null;function getOriginalCode(){if(_originalCode!=null)return _originalCode;var allScripts=document.getElementsByTagName("script");for(var i=0;i<allScripts.length;i++){var script=allScripts[i];if(script.id!="golfsnippet"){originalCode=script.textContent.trim();return originalCode}}}function getNewCode(){return textbox.value.trim()}function getInput(){try{var inputText=inputbox.value.trim();var input=eval("["+inputText+"]");return input}catch(e){return null}}function setTextbox(s){textbox.value=s;onTextboxChange()}function setOutput(output,s){output.innerHTML=s}function addOutput(output,data){output.innerHTML+='<pre style="background-color:'+(data.type=="err"?"lightcoral":"lightgray")+'">'+escape(data.content)+"</pre>"}function getByteCount(s){return(new Blob([s],{encoding:"UTF-8",type:"text/plain;charset=UTF-8"})).size}function getScore(s){var a=1;try{b=eval('('+s+')("Hello, World!")');if(b=="v,dd2>\nXdl")a=.85}catch(e){};return getByteCount(s)*a}function onTextboxChange(){var newLength=getByteCount(getNewCode());var oldLength=getByteCount(getOriginalCode());bytecount.innerHTML=newLength+" bytes";var diff=newLength-oldLength;if(diff>0){bytediff.innerHTML="(+"+diff+")";bytediff.style.color="lightcoral"}else if(diff<0){bytediff.innerHTML="("+diff+")";bytediff.style.color="lightgreen"}else{bytediff.innerHTML="("+diff+")";bytediff.style.color="lightgray"}newLength=getScore(getNewCode());var oldLength=getScore(getOriginalCode());score.innerHTML="Score: "+newLength;var diff=Math.round((newLength-oldLength)*100)/100;if(diff>0){scorediff.innerHTML="(+"+diff+")";scorediff.style.color="lightcoral"}else if(diff<0){scorediff.innerHTML="("+diff+")";scorediff.style.color="lightgreen"}else{scorediff.innerHTML="("+diff+")";scorediff.style.color="lightgray"}}function onTestBtn(evt){origheader.style.display="inline";newheader.style.display="inline";setOutput(newoutput,"");setOutput(origoutput,"");var input=getInput();if(input===null){addOutput(origoutput,{type:"err",content:"Input is malformed. Using no input."});addOutput(newoutput,{type:"err",content:"Input is malformed. Using no input."});input=[]}doInterpret(getNewCode(),input,function(data){addOutput(newoutput,data)});doInterpret(getOriginalCode(),input,function(data){addOutput(origoutput,data)});evt.stopPropagation();return false}function onResetBtn(evt){setTextbox(getOriginalCode());origheader.style.display="none";newheader.style.display="none";setOutput(origoutput,"");setOutput(newoutput,"")}function escape(s){return s.toString().replace(/&/g,"&amp;").replace(/</g,"&lt;").replace(/>/g,"&gt;")}window.alert=function(){};window.prompt=function(){};function doInterpret(code,input,cb){var workerCode=interpret.toString()+";function stdout(s){ self.postMessage( {'type': 'out', 'content': s} ); }"+" function stderr(s){ self.postMessage( {'type': 'err', 'content': s} ); }"+" function kill(){ self.close(); }"+" self.addEventListener('message', function(msg){ interpret(msg.data.code, msg.data.input); });";var interpreter=new Worker(URL.createObjectURL(new Blob([workerCode])));interpreter.addEventListener("message",function(msg){cb(msg.data)});interpreter.postMessage({"code":code,"input":input});setTimeout(function(){interpreter.terminate()},1E4)}setTimeout(function(){getOriginalCode();textbox.addEventListener("input",onTextboxChange);testbtn.addEventListener("click",onTestBtn);resetbtn.addEventListener("click",onResetBtn);setTextbox(getOriginalCode())},100);function interpret(code,input){window={};alert=function(s){stdout(s)};window.alert=alert;console.log=alert;prompt=function(s){if(input.length<1)stderr("not enough input");else{var nextInput=input[0];input=input.slice(1);return nextInput.toString()}};window.prompt=prompt;(function(){try{var evalResult=eval(code);if(typeof evalResult=="function"){var callResult=evalResult.apply(this,input);if(typeof callResult!="undefined")stdout(callResult)}}catch(e){stderr(e.message)}})()};</script>

Awesome modification of the test snippet! It seems the snippet now automatically checks for the bonus, may I ask how you did that? (p.s. if you want the original (not on one line) source, feel free to ask, might be easier to modify that way) — jrich, Commented Nov 8, 2015 at 23:09
@UndefinedFunction Oh, sorry for not replying right away! I added a getScore() function which checks the test-case Hello, World! for compliance (it conveniently contains both a newline and unprintable chars), and returns the score multiplied by .85 or 1, depending on the result. And yes, access to the un-minified snippet would be great. :) — ETHproductions, Commented Nov 14, 2015 at 23:27
I've made the original snippet code available here. Have fun! — jrich, Commented Nov 15, 2015 at 22:02

aditsu quit because SE is EVIL · Accepted Answer · 2015-11-06 22:20:19Z

4

CJam, 14

q{i2bW%:!2bc}%

Try it online

Explanation:

Pretty straightforward:

q       read input
{…}%    convert each character
  i     convert to number
  2b    convert to base 2 (digit array)
  W%    reverse
  :!    invert each digit
  2b    convert from base 2
  c     convert to character

"Printable" version, 20 - 15% = 17

q{i2bW%:!2bc' ,N--}%

edited Nov 6, 2015 at 22:20

answered Nov 6, 2015 at 1:17

aditsu quit because SE is EVIL

22.8k4 gold badges59 silver badges102 bronze badges

\$\begingroup\$ Dangit, you. I was just about to post a CJam answer D: \$\endgroup\$
– anOKsquirrel
Commented Nov 6, 2015 at 1:20
\$\begingroup\$ @anOKsquirrel sorry ^^ was it similar? \$\endgroup\$
– aditsu quit because SE is EVIL
Commented Nov 6, 2015 at 1:21
\$\begingroup\$ It would have been if I had finished it. \$\endgroup\$
– anOKsquirrel
Commented Nov 6, 2015 at 1:23
\$\begingroup\$ How does W% work? W is -1, so... \$\endgroup\$
– anOKsquirrel
Commented Nov 6, 2015 at 1:25
1

\$\begingroup\$ @anOKsquirrel see the documentation here \$\endgroup\$
– aditsu quit because SE is EVIL
Commented Nov 6, 2015 at 1:26

| Show 2 more comments

undefined · Accepted Answer · 2015-11-09 10:27:38Z

4

PHP - 187 182 163 bytes

<?php $s=$_GET["s"];$m="array_map";echo join($m("chr",$m("bindec",$m(function($v){return strtr($v,[1,0]);},$m("strrev",$m("decbin",$m("ord",str_split($s))))))));?>

Pass the value as GET["s"].

array_map returns an array with all the elements of the second parameter (an array) after applying the callback function (first parameter) to all of them.

Not sure if I should take the 15% off, since echo doesn't output unprintable characters, but I didn't remove them.

Just glad I finished, since this is the first challenge I take part.

edited Nov 9, 2015 at 10:27

answered Nov 6, 2015 at 16:00

undefined

2211 silver badge3 bronze badges

1

\$\begingroup\$ Is shorter if you not declare those functions: $m='array_map';echo join($m("chr",$m("bindec",$m(function($v){return strtr($v,[1,0]);},$m("strrev",$m("decbin",$m("ord",str_split($s))))))));. \$\endgroup\$
– manatwork
Commented Nov 6, 2015 at 16:24
\$\begingroup\$ @manatwork completely forgot about that. Thanks. \$\endgroup\$
– undefined
Commented Nov 6, 2015 at 16:27
\$\begingroup\$ You can't just put the input in a variable. You should create a function or read the input from STDIN. By the way, you don't need to use quotes around string ("chr", "bindec", …) since we don't care about warnings. That should save you 12 bytes. \$\endgroup\$
– Blackhole
Commented Nov 7, 2015 at 11:50
\$\begingroup\$ @Blackhole thanks for the infos, I'll be aware of that next time. \$\endgroup\$
– undefined
Commented Nov 7, 2015 at 13:40
\$\begingroup\$ You'd better do the modification on this answer, which is otherwise invalid :) . It will cost you almost no bytes, just replace str_split($s) with str_split(fgets(STDIN)) for instance. \$\endgroup\$
– Blackhole
Commented Nov 7, 2015 at 17:57

| Show 1 more comment

Community · Accepted Answer · 2020-06-17 09:04:33Z

4

C function, 63

o;f(o,char *s){for(;*s;*s=o,s++)for(o=0;*s;*s/=2)o+=o+!(*s%2);}

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 6, 2015 at 5:21

Digital Trauma

73k9 gold badges112 silver badges264 bronze badges

Add a comment |

JohnE · Accepted Answer · 2015-11-06 03:57:21Z

3

K5, 28 bytes

`c${b/~|{x@&|\x}@(b:8#2)\x}'

This is a bit inconvenient because K5's decode operator performs a fixed-width base conversion, so to comply with the problem statement I have to trim leading zeroes. The lambda {x@&|\x} accomplishes this step.

Smear:

  |\0 0 1 0 1 1 0 1
0 0 1 1 1 1 1 1

Gather:

  &|\0 0 1 0 1 1 0 1
2 3 4 5 6 7

Select:

  {x@&|\x}0 0 1 0 1 1 0 1
1 0 1 1 0 1

The whole program in action:

  `c${b/~|{x@&|\x}@(b:8#2)\x}'"Hello, World"
"v,dd2>\nXdl"

I believe oK's natural behavior with unprintables makes this eligible for -15%, giving this a score of 28*0.85 = 23.8.

answered Nov 6, 2015 at 3:57

JohnE

5,0821 gold badge22 silver badges33 bronze badges

\$\begingroup\$ +1 because I tried but couldn't figure out a short way to get rid of the leading zeros! \$\endgroup\$
– kirbyfan64sos
Commented Nov 8, 2015 at 3:20
\$\begingroup\$ A few related constructions can be found here. \$\endgroup\$
– JohnE
Commented Nov 8, 2015 at 4:39

Add a comment |

Alex A. · Accepted Answer · 2015-11-06 04:28:14Z

3

Julia, 77 bytes - 15% = 65.45

s->join(filter(isprint,[Char(parse(Int,join(1-digits(Int(c),2)),2))for c=s]))

This creates an unnamed functon that accepts a string and returns a string. Unprintable characters are removed, which qualifies this for the bonus.

Ungolfed:

function f(s::AbstractString)
    # digits() returns the digits in reverse order, so no explicit
    # reverse() is needed
    x = [Char(parse(Int, join(1 - digits(Int(c), 2)), 2)) for c = s]

    # Remove unprintables, join into a string
    return join(filter(isprint, x))
end

edited Nov 6, 2015 at 4:28

answered Nov 6, 2015 at 2:37

Alex A.

24.7k5 gold badges38 silver badges119 bronze badges

\$\begingroup\$ While it certainly qualifies it for the bonus, it also costs more than the bonus saves. 16 bytes to filter(isprint,) and only 11.55 bytes saved through the bonus. \$\endgroup\$
– Glen O
Commented Nov 6, 2015 at 15:03
\$\begingroup\$ And if you give up the filter step, you can avoid the comprehension and join by using map directly on the string. s->map(c->Char(parse(Int,join(1-digits(Int(c),2)),2)),s) (for 56 bytes) \$\endgroup\$
– Glen O
Commented Nov 6, 2015 at 15:06
\$\begingroup\$ @GlenO Thanks for the suggestions, but that approach leaves in unprintables as hex codes, which the OP said is not allowed. Using filter(isprint,) both qualifies it for the bonus and makes it compliant with the rules. \$\endgroup\$
– Alex A.
Commented Nov 6, 2015 at 21:03
\$\begingroup\$ "If an character results to be an unprintable, you may optionally output it but they do not have to be removed." \$\endgroup\$
– Glen O
Commented Nov 7, 2015 at 3:50
\$\begingroup\$ And if the concern is on the other side of it (that it displays as \x04 and the like), then print() costs seven, which would bring 56 up to 63. \$\endgroup\$
– Glen O
Commented Nov 7, 2015 at 3:53

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

PowerShell, 199 175 (171 - 15%)=145.35

param([char[]]$a)($a|%{$b=[convert]::ToString(+$_,2);$c=[convert]::ToInt32("$((-join$b[$b.Length..0])-split0-replace1,0-join1)",2);if($c-gt31-or$c-eq10){[char]$c}})-join''

Uses ~~an unfortunate amount of~~ some .NET calls/built-ins, which ~~significantly~~ bloats the code.

Explained:

Takes the input param(..) and casts it as a char[] so we can work through it appropriately.

The next bit (..)-join'' collects and joins our output together.

Inside those parens, we iterate with $a|%{..} as a foreach loop.

Inside the loop:

We create a new string $b, which is our input letter cast as an int +$_ and [convert]ed to base 2
This next bit, setting $c, is tricky, so let's start inside and work our way out
We reverse the string $b with (-join$b[$b.length..0])
We leverage my previous code for inverting a binary string and recast the result as a string with "$(..)"
We feed that string into a different .NET call that [convert]s ToInt32 from base 2, which is finally stored that into $c
If $c is greater than 31, or equal to 10, we cast it as a char and that value is left on the pipeline for output (which is what gets collected and -join''ed together, above), else nothing gets left on this particular iteration

Phew.

Qualifies for the -15% bonus, as well.

Example

PS C:\Tools\Scripts\golfing> .\reverse-and-invert-a-string.ps1 "Hello, World!"
v,dd2>
Xdl

Community · Accepted Answer · 2020-06-17 09:04:33Z

2

𝔼𝕊𝕄𝕚𝕟, 39 chars / 73 bytes

ô⟦ï]ć⇝ϚĎ(+“0b`+⟦a.ü⬮ß§2]ù⬮ć⇀$^1)ø`”⸩ø⬯⦆

Try it here (Firefox only).

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 6, 2015 at 2:46

Mama Fun Roll

9,9281 gold badge36 silver badges46 bronze badges

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Razetime · Accepted Answer · 2021-05-06 08:01:51Z

2

Vyxal `K`, 5 bytes

bR†vB

Try it Online!

epic.

-2 bytes from Lyxal.

-1 byte from Lyxal.

edited May 6, 2021 at 8:01

answered May 5, 2021 at 13:26

Razetime

27.4k3 gold badges29 silver badges72 bronze badges

\$\begingroup\$ 7 bytes if it's Keg \$\endgroup\$
– lyxal
Commented May 5, 2021 at 23:10
\$\begingroup\$ Actually make that 6 bytes \$\endgroup\$
– lyxal
Commented May 6, 2021 at 0:16
\$\begingroup\$ 5 bytes. That's my best and final \$\endgroup\$
– lyxal
Commented May 6, 2021 at 7:22
\$\begingroup\$ smh inb4 0 bytes \$\endgroup\$
– Razetime
Commented May 6, 2021 at 8:01
1

\$\begingroup\$ bruh 0 bytes is way too long. -69 bytes should be the next stop \$\endgroup\$
– lyxal
Commented May 6, 2021 at 8:06

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Minkolang 0.11, 26 bytes

od?.(d2%,$r2:d)xrI1-[2*+]O

Try it here.

Explanation

od?.            Takes input as character, halting if empty
(d2%,$r2:d)x    Converts to binary, inverting digits on the way
r               Reverses stack
I1-[2*+]        Converts to decimal
O               Outputs as character (if printable)

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 6, 2015 at 2:34

El'endia Starman

16.1k3 gold badges37 silver badges71 bronze badges

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

MATLAB, 60 bytes

@(y)[arrayfun(@(x)bin2dec([97-fliplr(dec2bin(x)) '']),y) '']

Basically each character in turn is converted into a binary string (with no leading zeros). The array is flipped and is subtracted from 97 ('0'+'1') which inverts the character. This is converted back to decimal. After all characters have been processed, the whole array is then converted back to characters before being returned.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 6, 2015 at 2:55

Tom Carpenter

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Python 3, 95 91

Straightforward implementation.

print(''.join(chr(int(''.join('10'[j>'0']for j in bin(ord(i))[:1:-1]),2))for i in input()))

Ungolfed:

inp = input()
ints = (ord(i) for i in inp)
bins = (bin(i) for i in ints)
revs = (i[2:][::-1] for i in bins) #without leading '0b'
invs = (''.join('0' if j == '1' else '1' for j in i) for i in revs)
newInts = (int(i, 2) for i in invs)
newChars = (chr(i) for i in newInts)
newStr = ''.join(newChars)
print(newStr)

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 6, 2015 at 11:05

Trang Oul

7504 silver badges17 bronze badges

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manatwork · Accepted Answer · 2015-11-06 11:35:49Z

1

Ruby, 62 characters

gets.bytes{|b|$><<b.to_s(2).reverse.tr("01","10").to_i(2).chr}

Samply run:

bash-4.3$ ruby -e 'gets.bytes{|b|$><<b.to_s(2).reverse.tr("01","10").to_i(2).chr}' <<< 'Hello, World!'
v,dd2>
Xdl

bash-4.3$ ruby -e 'gets.bytes{|b|$><<b.to_s(2).reverse.tr("01","10").to_i(2).chr}' <<< 'Hello, World!' | od -tad1
0000000    v    ,    d    d  eot    2    >   nl  eot    X    d    l   rs   nl
         118   44  100  100    4   50   62   10    4   88  100  108   30   10
0000016

answered Nov 6, 2015 at 11:35

manatwork

20.6k5 gold badges51 silver badges81 bronze badges

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Trang Oul · Accepted Answer · 2016-01-12 16:01:02Z

1

C#, 156 bytes - 15% = 132.6

class P{static void Main(string[]a){try{for(int i=0,b,c;;){for(b=a[0][i++],c=0;b>0;b/=2)c=c<<1|1-b%2;if(c==10|c>31)System.Console.Write((char)c);}}catch{}}}

Indentation and new lines for clarity:

class P{
    static void Main(string[]a){
        try{
            for(int i=0,b,c;;){
                for(b=a[0][i++],c=0;b>0;b/=2)
                    c=c<<1|1-b%2;
                if(c==10|c>31)
                    System.Console.Write((char)c);
            }
        }
        catch{}
    }
}

edited Jan 12, 2016 at 16:01

Trang Oul

7504 silver badges17 bronze badges

answered Nov 8, 2015 at 22:16

Hand-E-Food

8,8205 gold badges32 silver badges51 bronze badges

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Trang Oul · Accepted Answer · 2016-01-12 16:01:07Z

1

Javascript 123 bytes

s=>[].map.call(s,s=>String.fromCharCode("0b"+s.charCodeAt().toString(2).split('').reverse().map(s=>s^1).join(''))).join('')

edited Jan 12, 2016 at 16:01

Trang Oul

7504 silver badges17 bronze badges

answered Nov 6, 2015 at 19:18

Naouak

3391 silver badge4 bronze badges

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Community · Accepted Answer · 2020-06-17 09:04:33Z

Retina, 1107 629 bytes - 15% = 534.65 (non-competing)

Uses features added after challenge date. (Implicit behavior of $*, ¶, Sorting)

Retina doesn't have a built-in to convert a character to its ASCII ordinal or back... so behold its lustrous length. This handles printable ASCII, and removes unprintables as well as newlines. Byte count assumes ISO 8859-1 encoding.

The code contains unprintable characters.


¶
±
S_`
%(S`±
{2`
$`
}T01`-`_o
)Ms`.
\d+
$*
+`(1+)\1
${1}0
01
1
%O$^`.

T`01`10
1
01
+`10
011
0

m`^1{1,31}$

M%`1
m`^0¶?

126
~
125
}
124
|
123
{
122
z
121
y
120
x
119
w
118
v
117
u
116
t
115
s
114
r
113
q
112
p
111
o
110
n
109
m
108
l
107
k
106
j
105
i
104
h
103
g
102
f
101
e
100
d
99
c
98
b
97
a
96
`
95
_
94
^
93
]
92
\
91
[
90
Z
89
Y
88
X
87
W
86
V
85
U
84
T
83
S
82
R
81
Q
80
P
79
O
78
N
77
M
76
L
75
K
74
J
73
I
72
H
71
G
70
F
69
E
68
D
67
C
66
B
65
A
64
@
63
?
62
>
61
=
60
<
59
;
58
:
57
9
56
8
55
7
54
6
32
 
33
!
34
"
35
#
36
$
37
%
38
&
39
'
40
(
41
)
42
*
43
+
44
,
45
-
46
.
47
/
48
0
49
1
50
2
51
3
52
4
53
5
¶

Try it online

If you check out the Retina tutorial for unary arithmetic, you'll recognize several different pieces of my code as coming from there.

Thanks to Martin for golfing off hundred of bytes

David Conrad · Accepted Answer · 2016-10-08 12:51:41Z

Java, 205 - 15% = 174.2

interface F{static void main(String[]a){for(int i=0,c,s;i<a[0].length();++i){c=a[0].charAt(i);s=Integer.numberOfLeadingZeros(c);c=~(Integer.reverse(c)>>s)&-1>>>s;if(c==10|c>31)System.out.print((char)c);}}}

Ungolfed:

interface F {
    static void main(String[] a) {
        for (int i = 0, c, s; i < a[0].length(); ++i) {
            c = a[0].charAt(i);
            s = Integer.numberOfLeadingZeros(c);
            c = ~(Integer.reverse(c) >> s) & -1 >>> s;
            if (c == 10 | c > 31) System.out.print((char)c);
        }
    }
}

I think this solution is a bit interesting in its use of the Integer methods Integer.reverse and Integer.numberOfLeadingZeros which do what they sound like, and the shift of -1 >>> s where s is the number of leading zeroes, to get the mask to mask off high bits that we don't want. I just regret that the name of the latter method is so damn verbose, but that's what I get for golfing in Java.

Output:

v,dd2>
Xdl

Community · Accepted Answer · 2020-06-17 09:04:33Z

Japt, 25 bytes

Want to create a golfy JavaScript program, but the shortest method involves lots of long function names? That's what Japt was made for. :)

UmX=>Xc s2 w mY=>Y^1 n2 d

Try it in the online interpreter!

How it works

         // Implicit: U = first item in input
UmX=>    // for each character X in U:
Xc s2 w  //  take the char-code of X, convert to binary, and reverse
mY=>     //  for each character Y in this:
Y^1      //   take Y XOR 1 (converts 1 to 0 and 0 to 1)
n2 d     //  convert the result back to decimal, then to a character
         // Implicit: output last expression

Using the current version of Japt (as of v1.4.4), the byte count can be cut to 14:

®c ¤w m^1 n2 d

Test it online!

Community · Accepted Answer · 2020-06-17 09:04:33Z

1

PHP, 80 bytes

while(a&$c=$argn[$i++])echo chr(bindec(strtr(strrev(decbin(ord($c))),10,"01")));

takes input from STDIN; run with -R.

bonus version, 97 110 bytes -> 93.5 score

while(a&$c=$argn[$i++])ctype_print($c=chr(bindec(strtr(strrev(decbin(ord($c))),10,"01"))))||"
"==$c?print$c:0;

prints ASCII 10 and 32 to 126 (newline and printables)

breakdown, TiO and if possible some golfing will follow; I´m tired right now.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Apr 15, 2017 at 19:53

Titus

14.8k1 gold badge24 silver badges41 bronze badges

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Shaggy · Accepted Answer · 2020-10-23 11:03:42Z

1

Japt, 9 bytes

c_¤Ô®^1ÃÍ

Try it

answered Oct 23, 2020 at 11:03

Shaggy

43.5k4 gold badges36 silver badges94 bronze badges

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Razetime · Accepted Answer · 2020-10-23 11:46:47Z

1

Husk, 9 bytes

m(cḋm¬↔ḋc

Try it online!

answered Oct 23, 2020 at 11:46

Razetime

27.4k3 gold badges29 silver badges72 bronze badges

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caird coinheringaahin g · Accepted Answer · 2021-05-05 14:12:00Z

1

Jelly, 6 bytes

OBU¬ḄỌ

Try it online!

Explanation:

OBU¬ḄỌ Main link: z
O      Convert z to a list of character codes.
 B     Convert the codes to bit lists.
  U    Reverse the bits (not the lists).
   ¬   Invert the bits.
    Ḅ  Convert back to decimal.
     Ọ Convert back to string.

edited May 5, 2021 at 14:12

caird coinheringaahin g♦

50.3k11 gold badges128 silver badges354 bronze badges

answered Oct 8, 2016 at 9:29

Erik the Outgolfer

40.6k5 gold badges45 silver badges122 bronze badges

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chunes · Accepted Answer · 2021-05-06 03:26:08Z

1

Factor, 50 46 41 bytes

[ [ >bin reverse 97 v-n vneg bin> ] map ]

Try it online!

-4 bytes thanks to @Bubbler! -5 bytes thanks to @Bubbler!

Even though removing control characters is just [ control? ] reject it's not worth the 15% discount.

[ ... ] map Map over the code points in the input string.
>bin Convert a code point to a binary string.
reverse Reverse it.
97 v-n vneg Invert the string (1s become 0s and vice-versa).
bin> Convert from a binary string to decimal.

edited May 6, 2021 at 3:26

answered May 5, 2021 at 22:03

chunes

26.8k3 gold badges24 silver badges53 bronze badges

\$\begingroup\$ Suggest 97 - neg for 48 = 49 48 ? \$\endgroup\$
– Bubbler
Commented May 5, 2021 at 23:00
\$\begingroup\$ And [ 97 - neg ] map to 97 swap n-v. Surprisingly 97 v-n vneg also works. \$\endgroup\$
– Bubbler
Commented May 6, 2021 at 1:16

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

0

Haskell, 167 bytes

import Data.Char
import Numeric
r '0'='1'
r '1'='0'
s=map (chr.fst.head.(readInt 2 (`elem` "01") digitToInt).(map r).reverse.flip (showIntAtBase 2 intToDigit . ord)"")

Unfortunately Haskell gets quite verbose when needing to read/print in another base…

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 6, 2015 at 6:23

arjanen

1615 bronze badges

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Brad Gilbert b2gills · Accepted Answer · 2015-11-08 20:07:35Z

0

Perl 6, 66 bytes

Going all-out by removing the non-printing control characters gets me to (83+1)-15% = 71.4

perl6 -ne 'print grep /<-:C+[\n]>/,.ords».base(2)».flip.map({chr :2([~] $_.comb.map(+!+*))})'

If I remove the code that strips out the control characters I save quite a bit 65+1 = 66

perl6 -ne 'print .ords».base(2)».flip.map({chr :2([~] $_.comb.map(+!+*))})'

( I used » instead of >> for clarity )

answered Nov 8, 2015 at 20:07

Brad Gilbert b2gills

13.3k1 gold badge18 silver badges36 bronze badges

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rnso · Accepted Answer · 2016-10-08 14:39:09Z

Racket 250 15% bonus = 212 bytes

(λ(s)(list->string(map integer->char(filter(λ(x)(or(> x 31)(= x 10)))(for/list((i(string->list s)))(string->number(string-append
"#b"(list->string(map(λ(i)(if(equal? #\0 i)#\1 #\0))(reverse(string->list(number->string(char->integer i)2))))))))))))

Ungolfed:

(define (f s)
    (list->string 
     (map 
      integer->char
      (filter
       (λ(x)(or(> x 31)(= x 10)))

       (for/list ((i (string->list s)))
         (string->number
          (string-append
           "#b"
           (list->string
            (map
             (λ(i)(if(equal? #\0 i) #\1 #\0))
             (reverse
              (string->list
               (number->string
                (char->integer i) 2)
               )))))))))))

Testing:

(f "Hello, World!")

Output:

"v,dd2>\nXdl"

GammaFunction · Accepted Answer · 2021-05-06 03:57:19Z

Zsh, 114 bytes \$\times\$ 0.85 \$=\$ 96.9

for c (${(s::)1})b=(${(s::)$(([##2]#c))})&&b=${(j::)${(Oa)b}}&&s+=${(#)$((2#$b^(2**$#b-1)))}
<<<${s//[^[:print:]]}

Try it online!

Without filtering for printables, this would scores 98 bytes.

for char (${(s::)1}){
    bits=(${(s::)$(([##2]#char))})     # convert to binary, split into characters
    bits=${(j::)${(Oa)b}}              # reverse order of bits, join on empty string
    string+=${(#)$((2#$b^(2**$#b-1)))} # XOR with the 2**(len(bits) - 1), convert from codepoint
}
<<<${string//[^[:print:]]}             # ${string//find/replace}

Reverse and Invert a String

Reverse and Invert a String

Challenge

Scoring

31 Answers 31

Pyth, 14 bytes

How it works

Perl, 57 51 characters

JavaScript (ES6 ES7), 119 114 108 bytes

JavaScript (ES7), 126 bytes - 15% = 107.1

CJam, 14

"Printable" version, 20 - 15% = 17

PHP - 187 182 163 bytes

C function, 63

K5, 28 bytes

Julia, 77 bytes - 15% = 65.45

PowerShell, 199 175 (171 - 15%)=145.35

Explained:

Example

𝔼𝕊𝕄𝕚𝕟, 39 chars / 73 bytes

Vyxal K, 5 bytes

Minkolang 0.11, 26 bytes

Explanation

MATLAB, 60 bytes

Python 3, 95 91

Ruby, 62 characters

C#, 156 bytes - 15% = 132.6

Javascript 123 bytes

Retina, 1107 629 bytes - 15% = 534.65 (non-competing)

Java, 205 - 15% = 174.2

Japt, 25 bytes

How it works

PHP, 80 bytes

Japt, 9 bytes

Husk, 9 bytes

Jelly, 6 bytes

Factor, 50 46 41 bytes

Haskell, 167 bytes

Perl 6, 66 bytes

Racket 250 15% bonus = 212 bytes

Zsh, 114 bytes \$\times\$ 0.85 \$=\$ 96.9

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