note
vladb
Hi ;-).
<br><br>
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:<br>
10^2 = 10x10 = 10x(2x2x2+2).
<br><br>
Notice all the 2's there? Here's where the left shift
operator '<<' comes in handy (and it's pretty fast by the way).
<br><br>
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:
<br>
<code>
10^2 = 10<<3 + 10<<1; (by the way, this is may not be
written as 10<<4! :)
</code>
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:
<br><br>
10x10x10.
<br><br>
From this you'll also note that 10x10 is really <code>(10<<3 + 10<<1)</code> therefore, each 'x10' could be replaced accordingly.
<br><br>
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... ;-))
<br><br>
<br><br>
<table border=1 align=center>
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<i>"There is no system but GNU, and Linux is one of its kernels." -- Confession of Faith</i>
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</table>
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