Bravo (or brava, depending...). My solution was different, but could (and should) have been optimized into yours. Mine uses the notion that any matching string looks like a series of strings like "11..1100..00" concatenated, with a string of 1's at the end. (Or vice-versa, with 0's...) But yours is far simpler -- in fact, it states the very fact: if there's a 1 (or 0) at both ends, the string will succeed. Oh, and no backreferences, but that's ok.

Here's mine. It seems like even more of a behemoth now...

/^((1+0+)*1+|(0+1+)*0+)$/
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"Ewww," Tom said sheepishly.

$_="goto+F.print+chop;\n=yhpaj";F1:eval