Recursive acronyms

Regex, .NET flavour, 62 bytes

(?i)(?<=^\w(?<c>\w)*)( \k<c>(?<-c>)\w+| (?<w>\w+))*$(?(c)(?!))

You can test it here. If the input is a recursive acronym, this will yield a match, and capturing group w will contain all function words. If it isn't, then there will be no match.

This does preserve capitalisation of the function words (but matches case-insensitively).

Unfortunately, the tester doesn't display the entire stack of a named capturing group, but if you used it anywhere in .NET, the w group would contain all function words in order.

Here is a C# snippet to prove that:

var pattern = @"(?i)(?<=^\w(?<c>\w)*)( \k<c>(?<-c>)\w+| (?<w>\w+))*$(?(c)(?!))";
var input = new string[] {
    "RPM Package Manager",
    "Wine is not an emulator",
    "GNU is not Unix",
    "Golf is not an acronym",
    "X is a valid acronym"

var r = new Regex(pattern);
foreach (var str in input)
    var m = r.Match(str);
    for (int i = 0; i < m.Groups["w"].Captures.Count; ++i)

Here is a quick explanation. I'm using .NET's balancing groups to build a stack of the acronym letters in the named group c, with this snippet


The trick is that I need the second letter on top of the stack and the last one at the bottom. So I put all of this in a lookbehind that matches the position after the acronym. This helps, because .NET matches lookbehinds from right to left, so it encounters the last letter first.

Once I got that stack, I match the rest of the string word for word. Either the word begins with the letter on top of the acronym stack. In that case I pop that letter from the stack:


Otherwise, I match the word anyway and push onto the w stack which will then contain all function words:


At the end I make sure I reached the end of the string with $ and also make sure that I've used up all letters from the acronym, by checking that the stack is empty:


Test it on ideone.

Python (158, without regex)

It's not that I don't like regexes. It's that I don't know them.

def f(x):
 if not w:return 1,s
 [w.pop(0)if i[0]==w[0]else o.append(i)for i in s]
 return(0,)if w else(1,o)

Oh, I also had an ungolfed version:

def acronym(string):
    scentence = string.lower().split()
    word = scentence[0][1:]
    scentence = scentence[1:]
    over = []
    if not word: return 1, scentence
    for item in scentence:
        if item[0] == word[0]:
            word = word[1:]
    if word:
        return 0,
    return 1,over

GolfScript, 51 50 chars

{32|}%" "/(1>\{.1<2$1<={;1>}{\}if}/{]!}{]`1" "@}if

It probably can be golfed further. Takes input on STDIN. The boolean is 0/1.

Test online


{32|}%      # change everything to lower-case
" "/        # splits the string by spaces
(1>         # takes the first word out and removes the first letter
\           # moves the list of remaining words in front of the acronym word
{           # for every word:
  .1<2$1<=    # compares the first letter of the word with
              # the next unmatched letter of the acronym
  {;1>}       # if they are the same, discard the word and the now-matched letter
  {\}         # otherwise store the word in the stack
  if          # NB. if all letters have been matched, the comparison comes out as false
{]!}        # if there are still unmatched letters, return 0 (`!` non-empty list)
{]`1" "@}   # otherwise, return 1, and display the list of function words


Code Golf