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Given a string n, create a pyramid of the string split into pieces relative to the current row.

The first row contains the string unmodified.

The second row contains the string separated into halves by a pipe.

The third row separates it by thirds...

And so on. The length of each substring, where l is the length of string n is equal to

floor(l/n)

Characters left over are put in their own substring. The last row used is the first one where substrings are 2 in length.

Test Cases:

Input: Hello, world.

Output:

Hello, world.

Hello,| world|.

Hell|o, w|orld|.

Hel|lo,| wo|rld|.

He|ll|o,| w|or|ld|.

Input: abcdefghij

Output:

abcdefghij

abcde|fghij

abc|def|ghi|j

ab|cd|ef|gh|ij

Input: 01234567890abcdef

Output:

01234567890abcdef

01234567|890abcde|f

01234|56789|0abcd|ef

0123|4567|890a|bcde|f

012|345|678|90a|bcd|ef

01|23|45|67|89|0a|bc|de|f

Extra rules:

  • You can write a full program or a function, whichever uses less code.

  • Input will always be at least 4 characters in length.

  • You MUST use line breaks if your language supports them. If not possible, replace line breaks with :

  • Input will always be printable ASCII.

  • Minus 100% if your program solves P vs. NP.


Leaderboard:

var QUESTION_ID=104297,OVERRIDE_USER=62384;function answersUrl(e){return"http://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function commentUrl(e,s){return"http://api.stackexchange.com/2.2/answers/"+s.join(";")+"/comments?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+COMMENT_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){answers.push.apply(answers,e.items),answers_hash=[],answer_ids=[],e.items.forEach(function(e){e.comments=[];var s=+e.share_link.match(/\d+/);answer_ids.push(s),answers_hash[s]=e}),e.has_more||(more_answers=!1),comment_page=1,getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){e.items.forEach(function(e){e.owner.user_id===OVERRIDE_USER&&answers_hash[e.post_id].comments.push(e)}),e.has_more?getComments():more_answers?getAnswers():process()}})}function getAuthorName(e){return e.owner.display_name}function process(){var e=[];answers.forEach(function(s){var r=s.body;s.comments.forEach(function(e){OVERRIDE_REG.test(e.body)&&(r="<h1>"+e.body.replace(OVERRIDE_REG,"")+"</h1>")});var a=r.match(SCORE_REG);a&&e.push({user:getAuthorName(s),size:+a[2],language:a[1],link:s.share_link})}),e.sort(function(e,s){var r=e.size,a=s.size;return r-a});var s={},r=1,a=null,n=1;e.forEach(function(e){e.size!=a&&(n=r),a=e.size,++r;var t=jQuery("#answer-template").html();t=t.replace("{{PLACE}}",n+".").replace("{{NAME}}",e.user).replace("{{LANGUAGE}}",e.language).replace("{{SIZE}}",e.size).replace("{{LINK}}",e.link),t=jQuery(t),jQuery("#answers").append(t);var o=e.language;/<a/.test(o)&&(o=jQuery(o).text()),s[o]=s[o]||{lang:e.language,user:e.user,size:e.size,link:e.link}});var t=[];for(var o in s)s.hasOwnProperty(o)&&t.push(s[o]);t.sort(function(e,s){return e.lang>s.lang?1:e.lang<s.lang?-1:0});for(var c=0;c<t.length;++c){var i=jQuery("#language-template").html(),o=t[c];i=i.replace("{{LANGUAGE}}",o.lang).replace("{{NAME}}",o.user).replace("{{SIZE}}",o.size).replace("{{LINK}}",o.link),i=jQuery(i),jQuery("#languages").append(i)}}var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe",COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk",answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page;getAnswers();var SCORE_REG=/<h\d>\s*([^\n,]*[^\s,]),.*?([\d.]+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/,OVERRIDE_REG=/^Override\s*header:\s*/i;
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<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><td>Language</td><td>Size</td></tr></thead> <tbody id="answers"> </tbody> </table> </div><div id="language-list"> <h2>Winners by Language</h2> <table class="language-list"> <thead> <tr><td>Language</td><td>User</td><td>Score</td></tr></thead> <tbody id="languages"> </tbody> </table> </div><table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table>

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  • \$\begingroup\$ 0 bytes: return: false \$\endgroup\$ – Gabriel Benamy Dec 23 '16 at 14:29
  • 3
    \$\begingroup\$ Nice first challenge! A few clarification questions -- is the input only printable ASCII (I strongly suggest "yes")? What does "line breaks are necessary when possible" mean? \$\endgroup\$ – AdmBorkBork Dec 23 '16 at 14:39
  • 3
    \$\begingroup\$ It's a joke. P vs NP is an unsolved problem in computing. The joke is that if you can solve it, I will stop caring about the fact the your program doesn't solve the challenge. \$\endgroup\$ – Julian Lachniet Dec 23 '16 at 19:24
  • 3
    \$\begingroup\$ The real unsolved problem in computing is "tabs or spaces"... \$\endgroup\$ – FlipTack Dec 23 '16 at 22:49
  • 3
    \$\begingroup\$ No, the real problem is Internet Explorer. \$\endgroup\$ – Julian Lachniet Dec 23 '16 at 22:50
0
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JavaScript (ES6), 103 101 91 84 bytes

Fixed to respect challenge requirements

f=(s,n=0,p=s.length/++n|0)=>p>1?s.match(eval('/.{1,'+p+'}/g')).join`|`+'\n'+f(s,n):''

Lambda f that takes input string as first parameter s and recursively prints to console the split string. Pretty straightforward: as long as the substring length, p, is above 1, print the string split by a '|' every p characters, then proceed with appending the following level. This then calls the function again with p being t / n floored, where t is the original string length and n being an incremented divider.

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  • \$\begingroup\$ I don't think dividing n by 2 each time is correct. \$\endgroup\$ – Neil Dec 24 '16 at 0:12
  • \$\begingroup\$ @Neil you are correct, mistake on my part. I fixed the problem and saved 2 bytes in the process. \$\endgroup\$ – XavCo7 Dec 24 '16 at 13:37
  • \$\begingroup\$ @ETHproductions I thought of that, but I don't know if that would count as STDOUT... I guess I would need to do alert(f(s)) just after right? \$\endgroup\$ – XavCo7 Dec 24 '16 at 18:33
4
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Perl, 46 + 1 = 47 bytes

Run with the -n flag

say s/.{$=}(?=.)/$&|/gr while($==y///c/++$,)-2

Try it online!

Code breakdown

-n                                              #Reads input into the $_ variable
say s/.{$=}(?=.)/$&|/gr while($==y///c/++$,)-2
                                 y///c          #Transliteration.  Implicitly operates on $_, replacing every character with itself and counting replacements
                                                #y///c effectively returns the length of $_
                                      /++$,     #Increments $, (which starts off at 0) and divides the length of $_ by $,
                              $==               #Stores the result of this division into $=
                                                #$= forces its contents to be an integer, so it truncates any decimal
                             (             )-2  #Returns 0 if $= is equal to 2
                        while                   #Evaluates its RHS as the condition.  If truthy, evaluates its LHS.
    s/          /   /gr                         #Substitution.  Implicitly operates on $_.
                                                #Searches for its first argument and replaces it with its second argument, repeating until it's done, and returns the new string.  $_ is not modified.
      .{$=}                                     #Looks for a string of $= characters...
           (?=.)                                #...that is followed by at least one non-newline character, but does not include this character in the match...
                 $&|                            #...and replaces it with itself followed by a pipe character.
say                                             #Output the result of the substitution.
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  • \$\begingroup\$ This doesn't seem to work for longer inputs. \$\endgroup\$ – Neil Dec 24 '16 at 0:16
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Pyth, 16 bytes

Vh/lQ3j\|cQ/lQhN

V                # For N in range(1, \/ )
 h/lQ3           # 1+lenght(input)/3
      j\|        # join with '|'
         cQ      # chop input in
           /lQhN # lenght(input)/(N+1) pieces

try here

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  • 1
    \$\begingroup\$ This might work for the test cases but I don't think it works for longer inputs. \$\endgroup\$ – Neil Dec 24 '16 at 0:15
2
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C, 145 131 128 125 bytes

l,n,i=1,j;f(char*s){l=strlen(s);puts(s);do{n=l/++i;for(j=0;j<l;)j&&(j%n||putchar('|')),putchar(s[j++]);puts("");}while(n>2);}

This is a function that takes a string as its argument and prints the output to STDOUT.

l,n,i=1,j;       // declare some variables
f(char*s){       // declare the function
l=strlen(s);     // get the length of the string
puts(s);         // output the initial version, with trailing newline
do{n=l/++i;      // n is the number of characters per "section",
                 //  and we'll do-while n>2 to stop at the right time
for(j=0;j<l;)    // loop through the characters of the string
j&&(             // if j != 0,
j%n||            // and j % n == 0,
putchar('|')),   // insert a | before this character
putchar(s[j++]); // print the character
puts("");        // print a newline after the loop
}while(n>2);}
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  • \$\begingroup\$ How does this work once i*i>l? It looks as if it will start to repeat sections. \$\endgroup\$ – Neil Dec 24 '16 at 0:13
  • \$\begingroup\$ @Neil I'm not sure what you mean. Could you give an example? \$\endgroup\$ – Doorknob Dec 24 '16 at 0:16
  • \$\begingroup\$ @Neil Ah, never mind, I see what you're saying. That appears to be a hole in the specification, which explicitly states that the length of each substring is floor(l/n); I'm not sure what the intended behavior is for longer inputs or if the OP anticipated that. \$\endgroup\$ – Doorknob Dec 24 '16 at 0:19
1
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Pyth, 17 bytes

jmj\|cQ/lQdSh/lQ3

Explanation

     cQ/lQ         Divide into equal pieces (with the last shorter)
  j\|              Join with pipes
 m        d        Map to each row index...
           Sh/lQ3  ... up to the first row with substrings of length 2
j                  Join with newlines
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1
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Javascript, 98 Bytes

a=>{for(b=1;2<=a.length/b;)eval("console.log(a.match(/.{1,"+(a.length/b|0)+"}/g).join('|'))"),b++}

Function x(a). Call using

console.log(x("ABCDEF"))

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0
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Ruby 60 + 1 = 61 bytes

+1 byte for -n flag.

z= ~/$/
(z/3+1).times{|n|puts $_.scan(/.{1,#{z/(n+1)}}/)*?|}

See it on Ideone: http://ideone.com/RtoReG

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0
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Python 3, 123 bytes

f=lambda s:print(*['|'.join(s[i:i+n]for i in range(0,len(s),n))for n in[len(s)//i for i in range(1,len(s)//2+1)]],sep='\n')

At longer strings some parts will be repeated, as the formula for the length of the substring is floor(l/n). For example with a string 13 chars long, the string split into 5's would be the same as the string split into 6's as floor(13/5)==floor(13/6). I'm not sure if the OP expected this or if it was an oversight.

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