@quophyie
1. It iterates 12 times so the exactly growth rate is stable, you "could" say O(12) or O(1)
2. yes, log(n) means you need to raise that number to a certain power to get that number. The smallest way to get log(n) is to multiply the increment by 2 (larger numbers also work and probably have a more exact answer, but I lump them into log(n)).
3. That is an infinite loop.
4. No idea. It's a typo, I'll fix it.
5. It's only O(n) if it iterates 'n' number of times, where 'n' is the # of elements in the data structure. Since we're multiplying n by itself we now iterate O(n^2) times.
Determining Big O Notation An easier way
MultiQuote
#17
xTorvos
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Posted 30 March 2010 - 09:42 PM
KYA, on 12 September 2009 - 04:03 PM, said:
for(int i = 0; i < n; i *= 2) {
cout << i << endl;
}
for(int i = 0; i < n; i++) { //linear
for(int j = 0; j < n; j *= 2){ // log (n)
//do constant time stuff
}
}
Last time I checked (about 1.5 minutes ago), taking 0 * 2 gives me 0. I'm thinking your Big O is a little more than log(n).
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#18 Guest_quophyie*
Posted 31 March 2010 - 02:55 AM
Cheers KYA!! It all makes sense now especially the log(n) bit. Thanks you very much for making Big-O clear and easy to understand. In all my years at university, non of my lecturers were able adequately to explain big-O and how its applied and its taken your simple blog to explain it. Thanks again man!!!
#19
KYA
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Posted 02 April 2010 - 08:59 PM
xTorvos, on 30 March 2010 - 09:42 PM, said:
Last time I checked (about 1.5 minutes ago), taking 0 * 2 gives me 0. I'm thinking your Big O is a little more than log(n).
Fixed.
@quophyie glad you found it helpful.
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#20 Guest_kpb15*
Posted 06 August 2010 - 02:58 AM
{
for(j=1; j<=n; j++)
for(k=1; k<=n; k++)
{
c[j][k] = 0;
for(l=1; l<=n; l++)
c[j][k] = c[j][k] * b[l][k];
}
}
My head hurts. Is this O(n^k)? A simple yes or no would be enough, but I'd really appreciate it if you can point me to the right direction.
#21
macosxnerd101
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Posted 06 August 2010 - 07:06 AM
Technically, it's O(n^3) as you have three nested for loops making n iterations. But it is bounded by O(n^k).
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#22 Guest_amit*
Posted 06 August 2010 - 09:57 PM
Really appreciable and doubt clearing concept u gave.
#23
ThePheonix21
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Posted 04 September 2010 - 09:22 PM
Just a question, would an if statement nested inside a while loop be considered a nested statement or a sequential statement?
For example:
That's the one thing I noticed wasn't really noted.
For example:
while(x!=0){
if(y>z){
i++;
}
}
That's the one thing I noticed wasn't really noted.
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#24
macosxnerd101
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Posted 05 September 2010 - 08:00 AM
It's additive in regards to efficiency, so it is a constant, and constants are dropped when calculating Big-O.
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#25
KYA
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Posted 05 September 2010 - 09:37 AM
Alternatively, you could view conditionals constant as well. (Unless said conditional calls a function or such that is known to be a certain runtime, i.e. strlen(), sizeof(), etc...
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#26
ThePheonix21
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Posted 05 September 2010 - 11:02 AM
So, a series of If Else statements would be additive as well.
Like:
Like:
if(x==0){
y++;
}
else if(x>0){
y--;
}
else{
return 0;
}
macosxnerd101, on 05 September 2010 - 07:00 AM, said:
It's additive in regards to efficiency, so it is a constant, and constants are dropped when calculating Big-O.
This post has been edited by ThePheonix21: 05 September 2010 - 11:07 AM
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#27
macosxnerd101
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Posted 06 September 2010 - 06:15 PM
Yes, but for Big-O they are irrelevant as they are constants, and they will get dropped at the end.
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#28 Guest_Wilko*
Posted 09 September 2010 - 08:20 PM
To write O(log(n^3)) would you do say a series of for loops the first being O(logn) then 3 O(n) loops for example
And does the form O(k^n) exist? for example O(4^n). because if you multiply nested loops it will give it the other way round like n^4 so not sure how you'd do this.
for(int a=1; a<n; a*=2)
{
for(int b=1; b<n; b++)
{
for(int c=1; c<n; c++)
{
for(int d=1; d<n; d++)
{
}
}
}
}
And does the form O(k^n) exist? for example O(4^n). because if you multiply nested loops it will give it the other way round like n^4 so not sure how you'd do this.
#29 Guest_Guest*
Posted 25 October 2010 - 12:19 AM
Wilko, on 09 September 2010 - 07:20 PM, said:
To write O(log(n^3)) would you do say a series of for loops the first being O(logn) then 3 O(n) loops for example
And does the form O(k^n) exist? for example O(4^n). because if you multiply nested loops it will give it the other way round like n^4 so not sure how you'd do this.
for(int a=1; a<n; a*=2)
{
for(int b=1; b<n; b++)
{
for(int c=1; c<n; c++)
{
for(int d=1; d<n; d++)
{
}
}
}
}
And does the form O(k^n) exist? for example O(4^n). because if you multiply nested loops it will give it the other way round like n^4 so not sure how you'd do this.
I am interested in the above mentioned problem as well, anyone knows?
#30 Guest_Asma Zeb*
Posted 05 November 2010 - 01:47 AM
sir kindly explain me running time complexity for the following piece of code:
y=0;
x=0;
for (i=n; i>0;i=i-1)
{ y=y+1;}
for (i=1;i<=n;i=i*3)
{ for (j=1;j<=3n;++j)
{
for(k=0;k<n; k=k+5)
{
x=x+5;
}
}
}
Waiting for your positive feed back...i'm getting confused how to find out Big O for nested loops...Thanks
y=0;
x=0;
for (i=n; i>0;i=i-1)
{ y=y+1;}
for (i=1;i<=n;i=i*3)
{ for (j=1;j<=3n;++j)
{
for(k=0;k<n; k=k+5)
{
x=x+5;
}
}
}
Waiting for your positive feed back...i'm getting confused how to find out Big O for nested loops...Thanks
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