which one would be faster?

Mushtaque Nizamani

There are two sets one set has ten thousands elements other set has 30 element Which loop would be the faster for(i = 0; i < sizeoffirstset(10000); i++) { for (j=0; i < sizeofsecondset(20); j++) { if(set1[i] == set2[j]) //do some thing } } for(i = 0; i < sizeofsecondset(20); i++) { for (j=0; i < sizeoffirstset(10000); j++) { if(set2[i] == set1[j]) //do some thing } }

Best Regards, Mushq Mushtaque Ahmed Nizamani Software Engineer Ultimus Pakistan

Michael Dunn

Both are doing the same operations, just in a different order.

--Mike-- Visual C++ MVP :cool: LINKS~! Ericahist | PimpFish | CP SearchBar v3.0 | C++ Forum FAQ I work for Keyser Söze

Russell

Mushq wrote:

sizeofsecondset(20)...sizeoffirstset(10000)

const UINT Sizeoffirstset =sizeoffirstset (10000); //this can help the compiler
const UINT Sizeofsecondset=sizeofsecondset(20);
for(i = 0; i < Sizeofsecondset; i++)
{
for (j=0; j < Sizeoffirstset; j++)
{
if(set2[i] == set1[j])
//do some thing
}
}

I'm looking for differences, I see only that the initialization j=0 is computed Sizeofsecondset times ... so let Sizeofsecondset be the smaller number. :)

Russell

Russell

Mushq wrote:

for(i = 0; i < sizeoffirstset(10000); i++) { for (j=0; i < sizeofsecondset(20); j++)

Ahahah:laugh::laugh: faster yes .... but to go where?:laugh:

Russell

rihdus

First one is faster and because of internal smaller array. run following code in VB: Dim i, j As Integer Dim strArr1(10000) As String Dim strArr2(20) As String For i = 0 To 10000 strArr1(i) = i & "" Next For i = 0 To 20 strArr2(i) = i * i * i & "" Next Response.Write(DateTime.Now.Millisecond.ToString()) For i = 0 To 10000 For j = 0 To 20 If (strArr1(i) = strArr2(j)) Then Response.Write("
Hello" & strArr1(i) & strArr2(j)) End If Next Next Response.Write(DateTime.Now.Millisecond.ToString()) For i = 0 To 20 For j = 0 To 10000 If (strArr2(i) = strArr1(j)) Then Response.Write("
Hello " & strArr2(i) & strArr1(j)) End If Next Next Response.Write(DateTime.Now.Millisecond.ToString())

Any systematic work reflects its significance for a long time. So let's discuss the best...

Russell

Yes: finally the best way to find out who is the faster is use ... the clock:laugh: :cool:

Russell

Mushtaque Nizamani

Thanks for reply.

rihdus wrote:

First one is faster and because of internal smaller array.

can you please explain a bit more.

Best Regards, Mushq Mushtaque Ahmed Nizamani Software Engineer Ultimus Pakistan

chandu004

clock_t start,end; start=clock(); //your code snippet here end=clock(); //observe start=end here. hope you got the point.

cp9876

If time is important you could almost certainly do better than either way. For example, instead doing a simple search 10000 times on a set of 20 things, spend a bit of time and either order them or hash them, so that the search that you have to do many times is much faster.

Peter "Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."

Russell

Mushq wrote:

for(i = 0; i < sizeoffirstset(10000); i++) { for (j=0; i < sizeofsecondset(20); j++) { if(set1[i] == set2[j]) //do some thing } }

another optimization:

for(i = 0; i < sizeoffirstset(10000); i++)
{
const double set1_i_ = set1[i];
for (j=0; j < sizeofsecondset(20); j++)
{
if(set1_i_ == set2[j])
//do some thing
}
}

In this way the program not needs to reload that value every time loooking into the array. Hoping that the compiler is enough smart to better solution itself. You can only check the time spend in every solutions and see who is the faster... Important: do this time check in Release mode!!! also look to the project features...I remember that there is some flags like smallest exe file dimension vs fastest run exe (that somethimes can be bigger)....but I don't remember the name of this flags. Hope helps:)

Russell

Luc Pattyn

Sure, measuring is good, smart measuring is even better. if you have two alternatives that are likely to be almost equal, whichever you try first will take longer, due to different starting conditions (maybe code needs to be JITted, probably data needs to be loaded in memory and/or will be loaded in cache, etc). Remedy: put everything in a for loop, and run a few passes; ignore the first pass, take the average (or the best time!) of the remaining passes. :)

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blackjack2150

_Russell_ wrote:

Important: do this time check in Release mode!!!

Could you please explain a bit what the difference is? Thanks.

Russell

The difference is that in DEBUG mode the code is compiled exactly as you write it and it isn't optimizated. For example, inline functions are not inline and macros like ASSERT are inserted in the application. In RELEASE mode the code is optimized: inline functions will become inlined and ASSERTs are cutted from the code. In this case the code is really faster and it does exactly what you want, without do any internal test. Of course you have, first, to prepare the routines in DEBUG mode, to be sure that there isn't errors, but when you take the clock in your hand to see what algorithm is faster... then use the RELEASE mode. You can find details on the differences between DEBUG mode and RELEASE mode into the documentation. :-D

Russell

Chetan Patel

Both are same because the no. of iterations is same.

Best Regards, Chetan Patel