9-hole mini-golf: Description

  • 9 (mostly fairly easy) code golfing challenges of varying difficulty
  • Penalties for using the same language more than once
  • All challenges about a specific theme (this theme: Text Manipulation)
  • Input and output can be anywhere reasonable (i.e. STDIN and STDOUT, reading from/writing to a file, function argument and return value, etc.) but must NOT be hardcoded into the program
  • Heavily inspired by 9 Hole Challenge and Text Mechanic


  1. Code-golf bag

    Take two strings as input.
    Output the first string's character count, while ignoring any occurence of any character in the second string.
    Example: f("foobarbaz", "ao") => 5
  2. A pre-text for golfing

    Take two strings as input.
    Output the first string, with every line prefixed with the second.
    Example: f("foo\nbar\nbaz", "a") => "a foo\na bar\na baz"
  3. War of tabs vs spaces

    Take a string s, a number n, and a boolean b (specified however you want) as input.
    If b is true, output s with every tab converted to n spaces.
    Else, output the s with every n spaces converted to tabs.
    Example: f("if (x) {\n\tdoStuff();\n}", 4, true) => "if (x) {\n[sp][sp][sp][sp]doStuff();\n}" ([sp] means space)
  4. Pillars of golf

    Take a string s, a number n, and another number m as input.
    Output s in columns of n lines each and m characters per column.
    Also have padding of one space between the columns.
    Example: f("this is some placeholder text, foo bar baz...", 3, 5) =>

    this  aceho  foo 
    is so lder  bar b
    me pl text, az...
  5. Friendly letters

    Take a string s and a number n as input.
    Output the most common group of n letters in s.
    If there is a tie, output any or all of them.
    Example: f("abcdeabcfghiabc", 3) => "abc"
  6. Scrambled eggs letters for breakfast

    Take a string as input.
    Output the string with all of its words scrambled (letter order randomized) except their first and last letters.
    For simplicity, assume that the input will be a list of "word"s, space separated (i.e. in @$&_():;" foo bar, @$&_():;" is considered a "word.")
    Example: f("this is a sentence that will be scrambled") => "tihs is a stcneene that wlil be sclamrbed"
  7. ASCIIfy

    Take a string as input.
    If the string only contains numbers and spaces, then replace the numbers with their respective ASCII characters (removing the spaces).
    Else, do the reverse (characters to numbers).
    Example: f("ASCIIfy challenge") => "65 83 67 73 73 102 121 32 99 104 97 108 108 101 110 103 101"
    Example 2: f("65 83 67 73 73 102 121 32 99 104 97 108 108 101 110 103 101") => "ASCIIfy challenge"
  8. Mini-mini-markdown transformation

    Take a string as input.
    Output the string converted with mini-markdown, as used in comments on Stack Exchange.
    This is an even mini-er version: you only need to handle **bold**, *italics*, and `code`.
    You need not handle invalid nesting, like **foo *bar** baz*. Also assume that when you see a delimiter (* or `), it will always mean to format (i.e. te**st**ing => te<b>st</b>ing, and foo* bar *baz => foo<i> bar </i>baz).
    Example: f("**foo** *bar **baz*** `qux`") => "<b>foo</b> <i>bar <b>baz</b></i> <code>qux</code>"
  9. Only the best characters

    Take a string s, number n, and string r as input.
    Output the nth character of each word in s. (0-indexed, words are space-separated).
    If the length of the word is less than n, use r for that word instead.
    Example: f("this is a test sentence foo bar baz", 2, "-") => "i--snorz"


Your score is the sum of the character counts of your programs. For every repeated language, multiply by 110%. For example, if you have three Ruby solutions, and the total character count of all of your solutions is 1000, your score is 1000 * 1.1 * 1.1 = 1210. Round down if you have a non-integer score.

Good luck!

  • 1
    \$\begingroup\$ Challenge 8 touches on one of the least well specified aspects of Markdown, and the one which is hardest to do really well. It needs a clear explanation of how to handle ambiguities and a good test suite. See Emphasis.text from the mdtest suite. \$\endgroup\$ Commented Jan 22, 2014 at 22:59
  • \$\begingroup\$ @PeterTaylor Well, _ doesn't matter since I specified not to include it. I've edited to clarify some of the others. \$\endgroup\$
    – Doorknob
    Commented Jan 22, 2014 at 23:04
  • \$\begingroup\$ What about **foo***bar**baz*? \$\endgroup\$ Commented Jan 22, 2014 at 23:12
  • 1
    \$\begingroup\$ Challenge 6 is identical to this one. \$\endgroup\$
    – daniero
    Commented Jan 23, 2014 at 2:34
  • 4
    \$\begingroup\$ I'm voting to close this question as off-topic because it's a multi-part challenge with insufficient interaction between the parts \$\endgroup\$ Commented Nov 9, 2019 at 1:54

4 Answers 4


Score: 382 * 1.12 = 462

Languages prone to change.

1. APL, 8 4

Thanks @marinus for shaving 4 chars off.


Called with the strings as the left and right arguments, eg.

      'foobarbaz' f 'ao'

2. Ruby, 35 31

Thanks @DoorknobofSnow for shaving 4 chars off.

f=->s,r{s.gsub(/(?<=^)/,r+' ')}

3. Python, 48

f=lambda s,n,b:s.replace(*['\t',' '*n][::2*b-1])

4. GolfScript, 20

{@//zip{' '*}%n*}:f;

Assumes that the arguments are on the stack. Test online

5. J, 50


Called with the string as the left argument and the number as the right, eg.

   'abcdeabcfghiabc' f 3

6. Ruby, 61

f=->s{s.gsub(/(?<!^| )[^ ]+(?!$| )/){[*$&.chars].shuffle*''}}

7. GolfScript, 39 35 34

{[.10,' '*-{{}/]' '*}{~]''+}if}:f;

Again, assumes the argument is on the stack. Test online

8. Perl, 98

sub f{$_=@_[0];s!\*\*(.+?)\*\*!<b>$1</b>!g;s!\*(.+?)\*!<i>$1</i>!g;s!`(.+?)`!<code>$1</code>!g;$_}

9. Haskell, 36

f s n r=[(x++cycle r)!!n|x<-words s]
  • \$\begingroup\$ Great! You can save some chars on the Ruby one by using a proc (f=->s,r{...}) instead of a function. I refuse to upvote until it is complete, though :P \$\endgroup\$
    – Doorknob
    Commented Jan 23, 2014 at 0:34
  • \$\begingroup\$ @DoorknobofSnow ah, thanks. Not that proficient with Ruby :) \$\endgroup\$
    – Volatility
    Commented Jan 23, 2014 at 0:41
  • \$\begingroup\$ Hooray, you're the first one to finish :-D +1 \$\endgroup\$
    – Doorknob
    Commented Jan 23, 2014 at 12:48
  • \$\begingroup\$ I get the impression that all of the functions are supposed to return a string, so for number 4 I think you're going to have to use one character more. \$\endgroup\$ Commented Jan 23, 2014 at 23:17
  • \$\begingroup\$ @PeterTaylor fixed :) \$\endgroup\$
    – Volatility
    Commented Jan 23, 2014 at 23:26

Python - 697 × 1.19 ≈ 1644

Gee, I sure love lambdas.

Note: 3 and 5 were shamelessly copied from Volatility's answer, as I couldn't find a better alternative. Also, this was done just for fun.

f=lambda a,b:sum([x not in b for x in a])        # 1, 41 chars
f=lambda a,b:b+' '+a.replace('\n','\n'+b+' ')    # 2, 43 chars
f=lambda s,n,b:s.replace(*['\t',' '*n][::b*2-1]) # 3, 47 chars
f=lambda s,n,m:'\n'.join([' '.join([s[x:x+m]for x in range(y*m,len(s),m*n)])for y in range(n)])
                                                 # 4, 94 chars
f=lambda s,n:max([s[x:x+n]for x in range(len(s)+1-n)],key=s.count)
                                                 # 5, 66 chars
import random;f=lambda s:' '.join([''.join(sorted(y,key=lambda*x:random.random()))for y in s.split()])
                                                 # 6, 102 chars
f=lambda s:s.replace(' ','').isdigit()and ''.join(map(chr,map(int,s.split())))or ' '.join(map(str,map(ord,s)))
                                                 # 7, 110 chars
import re;f=lambda s:re.sub('`(.*?)`','<code>\\1</code>',re.sub(r'\*(.*?)\*','<i>\\1</i>',re.sub(r'\*\*(.*?)\*\*','<b>\\1</b>',s)))
                                                 # 8, 128 chars
f=lambda s,n,r:''.join([len(x)>n and x[n]or r for x in s.split()])
                                                 # 9, 66 chars

EDIT: Thanks to Volatility for the tips.

  • \$\begingroup\$ You can use generator expressions for most of them, which will save you a bunch of characters. Also, for 1. there's no need to use int, since booleans are a subclass of them, and for 7. all([...]) can be shortened to x.replace(' ','').isdigit() \$\endgroup\$
    – Volatility
    Commented Jan 23, 2014 at 22:15
  • \$\begingroup\$ The second answer won't prefix the first line of a. \$\endgroup\$
    – daniero
    Commented Jan 24, 2014 at 2:06

Score 513 * 1.15 = 826 

Took quite a beating by the same-language penalty. Solved most of these in Ruby just to finish them as fast as I could. Might change some languages later. Added a small recap/explanation on each answer.

1: Python (46)

f=lambda a,b:len([x for x in a if not x in b])

First, shorter answer in Ruby 2.0 (30) that gives more penalty and higher overall score:

p (gets.chars-gets.chars).size

2: Ruby 1.9+ (37)

Returns each line of s prefixed with t:


3: Ruby 1.9+ (48)

Returns s with tabs replaced by n spaces or vice versa, depending on b:

f=->s,n,b{r=[" "*n,?\t];b||r.reverse!;s.gsub *r}

4: Ruby 1.9+ (95)

Somebody shoot me.

f=->s,n,m{[*s.chars.each_slice(m).map{|w|w*''}.each_slice(s.size/m/n)].transpose.map{|w|w*' '}}

5: Ruby 1.9+ (58)

Returns most common run of n characters in s:

f=->s,n{(a=s.chars.each_slice(n)).max_by{|v|a.count v}*''}

6: J (47)

Scrambles the text somehow; Shamelessly stolen verbatim from marinus:

''[1!:2&4('\w(\w+)\w';,1)({~?~@#)rxapply 1!:1[3

7: Ruby (57+1)

Prints input ASCIIfied or de-ASCIIfied. Run with the -p switch.

~/\d/?gsub(/\d+\s*/){$&.to_i.chr}:gsub(/./){"#{$&.ord} "}

8: Sed (87)

Prints input converted from (mini)markdown to HTML:


9 Ruby 1.9+ (37)

Returns a string of the nth characters of each first word in s, or r:

  • \$\begingroup\$ Wouldn't your 8 output <b>test** **test2</b> for **test** **test2**? \$\endgroup\$
    – Doorknob
    Commented Jan 23, 2014 at 14:28
  • \$\begingroup\$ @DoorknobofSnow yes it would ;) Fixed (sed doesn't have non-greedy repetition). \$\endgroup\$
    – daniero
    Commented Jan 23, 2014 at 16:30
  • \$\begingroup\$ What would it give for **foo *bar* baz**? \$\endgroup\$
    – Volatility
    Commented Jan 23, 2014 at 22:19
  • \$\begingroup\$ @Volatility Oops. I didn't read the question properly, thought there wouldn't be any nesting, but it said no invalid nesting. I'm not sure I'll bother fixing it right now. \$\endgroup\$
    – daniero
    Commented Jan 24, 2014 at 2:00
  • \$\begingroup\$ For 1st ruby code - operator works with enumerator ? \$\endgroup\$
    – Siva
    Commented Jan 24, 2014 at 7:33

Work in progress

1. Java - 66

int f(String s,String b){for(char c:b)s=s.replace(b,"");return s;}

2. Java - 64

String f(String i,String r){return i.replaceAll("(?m)^",r+" ");}

3. Python - 58

def f(s,n,b):t=" "*n;a=t,'\t';print s.replace(a[b],a[b^1])

4. Python - 84

def f(s,n,m):
 while s:a[c%n]+=s[:m]+" ";s=s[m:];c+=1
 for i in a:print i



7. Befunge 98 - 9




  • \$\begingroup\$ At the first hole it should be for(char c:b.toCharArray()) and replace(c+"",""); and return s.length(); (or something like that) to get it work. \$\endgroup\$
    – bobbel
    Commented Mar 3, 2014 at 13:03

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