Find if two digits in same place are the same

[tjc19999]

I need a way to find numbers 0000-9999 where no number has more than 2 digits in the same place in common. I know PHP but this is a little beyond me, can anyone help?

[requinix]

Two digits "in the same place"?

[tjc19999]

Basically if I have the number 1234 I do not want to have anything like 1334 OR 2234, OR 1235 but if i have 1245 it is fine.

 

Here is what I currently have 0000-0099,1100-1199, 2200-2299 etc.

 

I'm not sure if I can add any other numbers otherwise there will be more than 2 numbers in the same place in common.

 

Does that make sense?

[requinix]

Are you comparing every number against every other number? Or do you want a set of numbers that don't share three or more digits with some other, specific number?

[markjoe]

Yea I'm not sure I get it either but here's my guess at a (crude) stating point for you to check the numbers.

You should be able to do it with modulus checks also.

function numPlacesMatch($a, $b){
$matches = 0;
$a = zerofill($a, 4);
$b = zerofill($b, 4);
for($i = 0; $i < 4; $i++){
	if($a{$i} == $b{$i}){
		$matches++;
	}
}
return $matches;
}

function zerofill($num, $places){
$str = (string)$num;
while(strlen($str) < $places){
	$str = '0' . $str;
}
return $str;
}

[markjoe]

my curiosity got the better of me, this one is way faster than messing with all the strings.

function numPlacesMatch2($a, $b){
$matches = 0;
if($a % 10 == $b % 10){
	$matches++;
}
if(floor($a/10) % 10 == floor($b/10) % 10){
	$matches++;
}
if(floor($a/100) % 10 == floor($b/100) % 10){
	$matches++;
}
if(floor($a/1000) % 10 == floor($b/1000) % 10){
	$matches++;
}
return $matches;
}