Combine Multiple Regex into One - php

Question

I am trying to write a code to hyphenate a string into latin verses. There are a few constraints to it which I have taken care of, however I do not get the desired output. My code is given below :
<?php
$string = "impulerittantaenanimis caelestibusirae";
$precedingC = precedingConsonant($string);
$xrule = xRule($precedingC);
$consonantc = consonantCT($xrule);
$consonantp = consonantPT($consonantc);
$cbv = CbetweenVowels($consonantp);
$tv = twoVowels($cbv);
echo $tv;
function twoVowels($string)
{
return preg_replace('/([aeiou])([aeiou])/', '$1-$2', $string);
}
function CbetweenVowels($string)
{
return preg_replace('/([aeiou])([^aeiou])([aeiou])/', '$1-$2$3', $string);
}
function consonantPT($string)
{
return preg_replace('/([^aeiou]p)(t[aeiou])/', '$1-$2', $string);
}
function consonantCT($string)
{
return preg_replace('/([^aeiou]c)(t[aeiou])/', '$1-$2', $string);
}
function precedingConsonant($string)
{
$arr1 = str_split($string);
$length = count($arr1);
for($j=0;$j<$length;$j++)
{
if(isVowel($arr1[$j]) && !isVowel($arr1[$j+1]) && !isVowel($arr1[$j+2]) && isVowel($arr1[$j+3]))
{
$pc++;
}
}
function strAppend2($string)
{
$arr1 = str_split($string);
$length = count($arr1);
for($i=0;$i<$length;$i++)
{
$check = $arr1[$i+1].$arr1[$i+2];
$check2 = $arr1[$i+1].$arr1[$i+2].$arr1[$i+3];
if($check=='br' || $check=='cr' || $check=='dr' || $check=='fr' || $check=='gr' || $check=='pr' || $check=='tr' || $check=='bl' || $check=='cl' || $check=='fl' || $check=='gl' || $check=='pl' || $check=='ch' || $check=='ph' || $check=='th' || $check=='qu' || $check2=='phl' || $check2=='phr')
{
if(isVowel($arr1[$i]) && !isVowel($arr1[$i+1]) && !isVowel($arr1[$i+2]) && isVowel($arr1[$i+3]))
{
$updatedString = substr_replace($string, "-", $i+1, 0);
return $updatedString;
}
}
else
{
if(isVowel($arr1[$i]) && !isVowel($arr1[$i+1]) && !isVowel($arr1[$i+2]) && isVowel($arr1[$i+3]))
{
$updatedString = substr_replace($string, "-", $i+2, 0);
return $updatedString;
}
}
}
}
$st1 = $string;
for($k=0;$k<$pc;$k++)
{
$st1 = strAppend2($st1);
}
return $st1;
}
function xRule($string)
{
return preg_replace('/([aeiou]x)([aeiou])/', '$1-$2', $string);
}
function isVowel($ch)
{
if($ch=='a' || $ch=='e' || $ch=='i' || $ch=='o' || $ch=='u')
{
return true;
}
else
{
return false;
}
}
function isConsonant($ch)
{
if($ch=='a' || $ch=='e' || $ch=='i' || $ch=='o' || $ch=='u')
{
return false;
}
else
{
return true;
}
}
?>
I believe if I combine all these functions it will result in the desired output. However I will specify my constraints below :
Rule 1 : When two or more consonants are between vowels, the first consonant is joined to the preceding vowel; for example - rec-tor, trac-tor, ac-tor, delec-tus, dic-tator, defec-tus, vic-tima, Oc-tober, fac-tum, pac-tus,
Rule 2 : 'x' is joined to the preceding vowel; as, rex-i.
However we give a special exception to the following consonants - br, cr, dr, fr, gr, pr, tr; bl, cl, fl, gl, pl, phl, phr, ch, ph, th, qu. These consonants are taken care by adding them to the later vowel for example - con- sola-trix
n- sola-trix.
Rule 3 : When 'ct' follows a consonant, that consonant and 'c' are both joined to the first vowel for example - sanc-tus and junc-tum
Similarly for 'pt' we apply the same rule for example - scalp-tum, serp-tum, Redemp-tor.
Rule 4 : A single consonant between two vowels is joined to the following vowel for example - ma-ter, pa-ter AND Z is joined to the following vowel.
Rule 5 : When two vowels come together they are divided, if they be not a diphthong; as au-re-us. Diaphthongs are - "ae","oe","au"

Answer 1

If you look carefully at each rule, you can see that all involve a vowel at the beginning or a preceding vowel. Once you realize that, you can try to build a single pattern putting [aeiou] in factor at the beginning:
$pattern = '~
(?<=[aeiou]) # each rule involves a vowel at the beginning (also called a
# "preceding vowel")
(?:
# Rule 2: capture particular cases
( (?:[bcdfgpt]r | [bcfgp] l | ph [lr] | [cpt] h | qu ) [aeiou] x )
|
[bcdfghlmnp-tx]
(?:
# Rule 3: When "ct" follows a consonant, that consonant and "c" are both
# joined to the first vowel
[cp] \K (?=t)
|
# Rule 1: When two or more consonants are between vowels, the first
# consonant is joined to the preceding vowel
\K (?= [bcdfghlmnp-tx]+ [aeiou] )
)
|
# Rule 4: a single consonant between two vowels is joined to the following
# vowel
(?:
\K (?= [bcdfghlmnp-t] [aeiou] )
|
# Rule 2: "x" is joined to the preceding vowel
x \K (?= [a-z] | (*SKIP)(*F) )
)
|
# Rule 5: When two vowels come together they are divided, if they not be a
# diphthong ("ae", "oe", "au")
\K (?= [aeiou] (?<! a[eu] | oe ) )
)
~xi';
This pattern is designed to only match the position where to put the hyphen (except for particular cases of Rule 2), that's why it uses a lot of \K to start the match result at this position and lookaheads to test what follows without matching characters.
$string = <<<EOD
Aeneadum genetrix, hominum diuomque uoluptas,
alma Uenus, caeli subter labentia signa
quae mare nauigerum, quae terras frugiferentis
concelebras, per te quoniam genus omne animantum
EOD;
$result = preg_replace($pattern, '-$1', $string);
Ae-ne-a-dum ge-ne-trix, ho-mi-num di-u-om-qu-e u-o-lup-tas,
al-ma U-e-nus, cae-li sub-ter la-ben-ti-a sig-na
qu-ae ma-re nau-i-ge-rum, qu-ae ter-ras fru-gi-fe-ren-tis
con-ce-leb-ras, per te qu-o-ni-am ge-nus om-ne a-ni-man-tum
Note that I didn't include several letters like k, y and z that don't exist in the latin alphabet, feel free to include them if you need to handle translated greek words or other.