"be consistent"  
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Re: Finding the max()/min()by etcshadow (Priest) 
on Nov 11, 2004 at 04:55 UTC ( #406924=note: print w/replies, xml )  Need Help?? 
Here's an interesting one that doesn involve a comparison operator (even an obfuscated one):
The math behind that is: take the average of the two numbers, and then add half their difference (start halfway between, and then go up by half). So: (x+y)/2 + abs(xy)/2, and then factor out the division by two.
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