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  1. PO
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    <p>Off the top of my head, you could look for syllables, making use of <a href="http://www.php.net/soundex" rel="nofollow noreferrer">soundex</a>. That's the direction I would explore, based on the assumption that a pronounceable word has at least one syllable.</p> <p>EDIT: Here's a function for counting syllables:</p> <pre><code>function count_syllables($word) { $subsyl = Array( 'cial' ,'tia' ,'cius' ,'cious' ,'giu' ,'ion' ,'iou' ,'sia$' ,'.ely$' ); $addsyl = Array( 'ia' ,'riet' ,'dien' ,'iu' ,'io' ,'ii' ,'[aeiouym]bl$' ,'[aeiou]{3}' ,'^mc' ,'ism$' ,'([^aeiouy])\1l$' ,'[^l]lien' ,'^coa[dglx].' ,'[^gq]ua[^auieo]' ,'dnt$' ); // Based on Greg Fast's Perl module Lingua::EN::Syllables $word = preg_replace('/[^a-z]/is', '', strtolower($word)); $word_parts = preg_split('/[^aeiouy]+/', $word); foreach ($word_parts as $key =&gt; $value) { if ($value &lt;&gt; '') { $valid_word_parts[] = $value; } } $syllables = 0; // Thanks to Joe Kovar for correcting a bug in the following lines foreach ($subsyl as $syl) { $syllables -= preg_match('~'.$syl.'~', $word); } foreach ($addsyl as $syl) { $syllables += preg_match('~'.$syl.'~', $word); } if (strlen($word) == 1) { $syllables++; } $syllables += count($valid_word_parts); $syllables = ($syllables == 0) ? 1 : $syllables; return $syllables; } </code></pre> <p>From this very interesting link:</p> <p><a href="http://www.addedbytes.com/php/flesch-kincaid-function/" rel="nofollow noreferrer">http://www.addedbytes.com/php/flesch-kincaid-function/</a></p>
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    1. POhow to check if a string looks randomized, or human generated and pronouncable?
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    1. PTAnswer
    1. USkarim79
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      1. VTUpMod
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    1. CONice, but then you need to produce a dictionary in order to be able to use it. And even after that you still can miss some cases.
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      1. USArtem Barger
    2. CO@Artem - Nothing is going to be an 100% effective solution for this problem
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      1. USkarim79
    3. CO@karim - like I've said ;)
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      1. USArtem Barger
 

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