|Do you know where your variables are?|
Re: Help w/ Code Optimizationby vladb (Vicar)
|on Dec 26, 2001 at 21:41 UTC ( #134427=note: print w/replies, xml )||Need Help??|
Let me throw around my math skills... Recalling some binary math I figured that 10^2 (10 to the power of 2) may be simplified into this:
10^2 = 10x10 = 10x(2x2x2+2).
Notice all the 2's there? Here's where the left shift operator '<<' comes in handy (and it's pretty fast by the way).
So, every multiplication by 2 could be replaced by a left shift by one (in binary it's equivalent to multiplying by 2 ;) like this:
So, I've replaced 10x10 by a few left shift operators. The key here is to determine how many left shifts will have to be performed for given power. Say, if you were to raise 10 to the power of 3, you'd look down at this:
From this you'll also note that 10x10 is really (10<<3 + 10<<1) therefore, each 'x10' could be replaced accordingly.
Hopefully I've given you some food for thought ;-). I"ll try to look for a few ways to guess the left shift number for you. Evidently, it depends on the number being multiplied. My thinking is that left shift should work faster than multiplication, althought, folks who've developed Perl might have taken that into consideration... ;-))
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