the topic pretty much describes what we would like to accomplish.
a) start with a possible range of integers, for example, 1 to 10000.
b) take any md5 hash, run it thru this algo.
c) result that pops out will be an integer between 1 to 10000.
we are open to using another hashing method too.
the flow would ideally look like this:
string -> md5(string) -> algo(md5(string),range) -> resulting integer within range
is something like this possible?
final note: the range will always start with 1.
if you have an answer, feel free to post just the general idea, or if you so desire, php snippet works too :)
thanks!
Since MD5 (and SHA-1, etc.) will give you 128 bits of data (in PHP, you'll get it in hexadecimal string notation, so you need to convert it to an integer first). That number modulo 10000 will give you your integer.
Note however that many different hashes will convert to the same integer; this is unavoidable with any sort of conversion to your integer range, as the modulo operation essentially maps a larger set of numbers (in this case, 128 bits, that is numbers from 0 to 340,282,366,920,938,463,463,374,607,431,768,211,456) to a smaller set of numbers (less than 17 bits, numbers from 1 to 100,000).
since the range that we want will always start at 1, the following works great. all credit goes to Piskvor, as he was the one who provided the basic idea of how to go at this.
the code below seams to accomplish what we want. please chime in if this can be (not the code, its just for reference, but if the idea) improved at all. running the code below will result in 6305 / 10000 unique results. that in our case is good enough.
<?
$final=array();
$range=10000;
for($i=1;$i<=$range;$i++){
$string='this is my test string - attempt #'.$i;
echo 'initial string: '.$string.PHP_EOL;
$crc32=crc32($string);
echo 'crc32 of string: '.$crc32.PHP_EOL;
$postalgo=$crc32%$range;
echo 'post algo: '.$postalgo.PHP_EOL;
if(!in_array($postalgo,$final)){
$final[]=$postalgo;
}
}
echo 'unique results for '.($i-1).' attempts: '.count($final).PHP_EOL;
?>
enjoy!