ANNNOUNNNCCCEEERRR VOICCCEEE

Question

Write a function that takes a string of letters and spaces (no need to handle non-letters) and performs the ANNOUNCER VOICE translation algorithm as follows:

• First, uppercase everything.
• For each word,
  • Elongate each consonant cluster by tripling each letter; except, if the word begins with a consonant cluster, do not elongate that cluster. For example, other should become OTTTHHHEEERRR but mother should become MOTTTHHHEEERRR.
  • Elongate the final vowel by tripling it.
• In both cases of elongation, if you're tripling a letter, first coalesce it with duplicate letters on either side. For example, hill should become HIIILLL and bookkeeper should become BOOKKKEEPPPEEERRR.
• For the purposes of this challenge, y counts as a consonant.
• Clarification/simplification: You may assume that each pair of words is separated by a single space, and that the input contains no consecutive spaces, and that the input will not be the empty string.
• Shortest code wins!

Test vectors:

> sunday sunday
SUNNNDDDAAAYYY SUNNNDDDAAAYYY
> mia hamm
MIAAA HAAAMMM
> chester alan arthur
CHESSSTTTEEERRR ALLLAAANNN ARRRTTTHHHUUURRR
> attention please
ATTTENNNTTTIOOONNN PLEASSSEEE
> supercalifragilisticexpialidocious
SUPPPERRRCCCALLLIFFFRRRAGGGILLLISSSTTTICCCEXXXPPPIALLLIDDDOCCCIOUUUSSS
> moo
MOOO
> Aachen
AACCCHHHEEENNN
> Oooh
OOOHHH
> grifffest
GRIFFFEEESSSTTT
> k
K
> aaaabbbbc
AAAABBBBCCC

---

Here's a reference implementation which I would move to an answer except that as of this morning the question's been closed. :P

import itertools,re
def j(s):return re.match('^[AEIOU]+$',s)
def c(s):return ''.join(sum(([h,h,h]for h in[k for k,g in itertools.groupby(s)]),[]))
def v(s):
 while len(s)>=2 and s[-2]==s[-1]:s=s[:-1]
 return s+s[-1]+s[-1]
def a(n):
 r=''
 for w in n.split():
  if r:r+=' '
  ss=re.split('([AEIOU]+)', w.upper())
  for i,s in enumerate(ss):
   r += [v(s),s][any(j(t) for t in ss[i+1:])]if j(s)else[s,c(s)][i>0]
 return r
while 1:print a(raw_input('> '))

Answer 1

APL (Dyalog), 175 bytes

1↓' +'⎕R' '⊢'[AEIOU][^AEIOU]+ '⎕R{m/⍨1,3×2≠/m←⍵.Match}'([AEIOU])\1*([^AEIOU]*? )' ' [AEIOU]' ' [^ AEIOU]+' '([^AEIOU ])\1*'⎕R'\1\1\1\2' '&' '&' '\1\1\1'⊢'$| |^'⎕R'  '⊢1(819⌶)⍞

Try it online!

⍞ prompt for character input

1(819⌶) convert to uppercase (819 ≈ Big)

⊢ pass the result further (serves to separate the strings and the 1)

'$| |^'⎕R' ' Replace:
 the end, any space, and the beginning
 → two spaces

⊢ pass the result further (serves to separate two groups of strings)

'([AEIOU])\1*([^AEIOU]*? )' ' [AEIOU]' ' [^ AEIOU]+' '([^AEIOU ])\1*'⎕R'\1\1\1\2' '&' '&' '\1\1\1' Replace:
 any number of identical vowels and any number of non-vowels and a space
 → the vowel thrice and the unmodified consonants
 a space and a vowel
 → themselves
 a space and a consonant cluster
 → themselves
 a run of identical consonants
 → three of those vowels

'[AEIOU][^AEIOU]+ '⎕R{…} Replace:
 a run of non-vowels and a space
 → the result of the following anonymous function with the namespace ⍵ as argument:
  ⍵.Match the text that was found
  m← assign that to m
  2≠/ pair-wise different-from
  3× multiply by three
  1, prepend one
  m/⍨ use that to replicate m

⊢ pass the result further (serves to separate two strings)

' +'⎕R' ' Replace:
 one or more spaces
 → with a single space

1↓ drop the initial letter (a space)

Answer 2

APL, 90 bytes

{1↓∊{s←{⍵⊂⍨1,2≠/⍵}⋄x↑⍨¨(3⌈≢¨s⍵)⌊≢¨x←s⍵/⍨(1+2×{⌽<\⌽⍵}∨~∧∨\)⍵∊'AEIOU'}¨w⊂⍨w=⊃w←' ',1(819⌶)⍵}

Explanation:

• 1(819⌶)⍵: convert to uppercase
• w⊂⍨w=⊃w←' ',: split on spaces
• {...}¨: for each word...
  • s←{⍵⊂⍨1,2≠/⍵}: s is a function that splits a string into groups of contiguous matching characters
  • ⍵∊'AEIOU': mark the vowels
  • (...): see which characters to triplicate
    • ~∧∨\: all consonants past the first vowel,
    • {⌽<\⌽⍵}: the last vowel.
    • 2×: multiply the bit vector by two,
    • 1+: and add one. Now all selected characters have 3 and the rest have 1.
  • ⍵/⍨: replicate each character in ⍵ by the given amount
  • x←s: split it up into strings of matching characters, and store this in x.
  • (3⌈≢¨s⍵): the length of each group of matching characters in the input word, with a maximum of 3.
  • ⌊≢¨: the minimum of that and the lengths of the groups in x.
  • x↑⍨¨: make each group be that length
• 1↓∊: flatten the result and drop the first character (the space that was added at the beginning to help with splitting)

Answer 3

JS (ES6), 138 134 129 bytes

s=>s.toUpperCase()[r="replace"](/(\w)\1/g,"$1")[r](/[AEIOU](?=[^AEIOU]*( |$))/g,s=>s+s+s)[r](/\B./g,s=>/[AEIOU]/.test(s)?s:s+s+s)

WAAAYYY TOOO MAAANNNYYY BYYYTTTEEESSS. Contains AEIOU 3 times, but I can't golf those into one.

-4 bytes thanks to HyperNeutrino

Ungolfed

function v(str){
    return str.toUpperCase().replace(/(\w)\1/g,"$1").replace(/[AEIOU](?=[^AEIOU]*( |$))/g,s=>s+s+s).replace(/\B./g,s=>[..."AEIOU"].includes(s)?s:s+s+s);
}

I like to write, not read code.

Answer 4

Python, 417 bytes

Here's a reference implementation in Python. Not terribly golfed.

import itertools,re
def j(s):return re.match('^[AEIOU]+$',s)
def c(s):return ''.join(sum(([h,h,h]for h in[k for k,g in itertools.groupby(s)]),[]))
def v(s):
 while len(s)>=2 and s[-2]==s[-1]:s=s[:-1]
 return s+s[-1]+s[-1]
def a(n):
 r=''
 for w in n.split():
  if r:r+=' '
  ss=re.split('([AEIOU]+)',w.upper())
  for i,s in enumerate(ss):
   r+=[v(s),s][any(j(t) for t in ss[i+1:])]if j(s)else[s,c(s)][i>0]
 return r

Test with:

while True:
 print a(raw_input('> '))

Answer 5

Python 3, 238 bytes

def f(s):
 s=s.upper();k=[s[0]];s=''.join(k+[s[i]for i in range(1,len(s))if s[i]!=s[i-1]])
 for i in range(1,len(s)):k+=[s[i]]*(3-2*(s[i]in'AEIOU'and i!=max(map(s.rfind,'AEIOU'))))
 return''.join(k)
print(' '.join(map(f,input().split())))

Try it online!

Answer 6

Perl 5, 139 + 1 (-p) = 140 bytes

$_=uc,s/^([^AEIOU]*)//,$s=$1,s/([^AEIOU])\1*/$1x(($q=length$&)>3?$q:3)/ge,s/.*?\K([AEIOU])\1*/$1x(($q=length$&)>3?$q:3)/e,print"$s$_ "for@F

Try it online!

Even handles the "aaaabbbbc" test case in accordance with the example.