| Re: x*x versus x^2 [message #61287 is a reply to message #61199] |
Sat, 12 July 2008 09:25  |
dzhang49
Messages: 1
Registered: July 2008
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Junior Member |
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On Jul 9, 11:10 am, Conor <cmanc...@gmail.com> wrote:
> On Jul 9, 12:57 pm, Bruce Bowler <bbow...@bigelow.org> wrote:
>
>
>
>> On Wed, 09 Jul 2008 09:43:27 -0700, Conor wrote:
>>> On Jul 9, 12:32 pm, Conor <cmanc...@gmail.com> wrote:
>>>> So I've been looking at execution time for various algorithms, and I
>>>> found this interesting result:
>
>>>> bigarr = fltarr(1000,1000)
>
>>>> t1 = systime(/seconds)
>>>> t = bigarr^2.0
>>>> t2 = systime(/seconds)
>>>> t = bigarr*bigarr
>>>> t3 = systime(/seconds)
>
>>>> print,t2-t1
>>>> print,t3-t2
>
>>>> IDL prints:
>
>>>> 0.024163008
>>>> 0.010262012
>
>>>> Apparently multiplying an array by itself is twice as fast as using the
>>>> carat operator! Anyone know why this is? Is it a memory issue or
>>>> something?
>
>>> This also holds true for array's smaller than the multi-threading
>>> minimum size, so it isn't because multi-threading is being used in one
>>> case but not the other...
>
>> Digging into the deep dark recesses of my brain...
>
>> exponentiation with a real exponent generally uses the log function to do
>> it's thing. *some* language implementations are smart enough that if the
>> exponent is an integer, they decompose the exponentiation into
>> multiplication.
>
>> It might be worth trying your experiment with t=bigarr^2 and see how the
>> results change.
>
>> Bruce
>
> Interesting... I tried your suggestion and got this result:
>
> 0.018048048
> 0.010533094
>
> So it is still slower, but the difference is smaller. A calculation
> like this is rarely the bottleneck for speed in a program, so I
> probably won't worry about it too much, but it is an interesting fact
> to be aware of...
Actually, if you increase the dimension of the array, the result will
be reverse, here it is:
pro test_speed
bigarr = fltarr(10000,10000)
t1 = systime(/seconds)
t = bigarr^2.0
t2 = systime(/seconds)
t = bigarr*bigarr
t3 = systime(/seconds)
print,t2-t1
print,t3-t2
end
and the results are :
0.68420601
0.83076620
So if you run a larger number, the ^2 will be faster
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