See: http://kks.cabal.fi/GoodEnoughSearch
I have gone through quite many papers and sites. I have not found where this algorithm has been presented before, or that someone has made something similar, but better or more general. The algorithm is pretty simple and thus should be found quite easily by anyone facing the same kind of problem...
Hello,
Please, help me to compare complexity of two algorithms.
O(N+1000) + O(M*log(M))
O(N*5) + O(2000)
N = 100000
M = 100
I can't understand, what should I do with O(...)? Can I leave it? And just do...
(N+1000) + (M*log(M)) = 101200
(N*5) + 2000 = 502000
Is it right?
Thank you
UPDATED
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Hi, i try to find the complexity of this algorithm:
m=0;
i=1;
while (i<=n)
{
i=i*2;
for (j=1;j<=(long int)(log10(i)/log10(2));j++)
for (k=1;k<=j;k++)
m++;
}
I think it is O(log(n)*log(log(n))*log(log(n))):
The 'i' loop runs until i=log(n)
the 'j' loop runs until log(i) means log(log(n))
the 'k' loop runs until k...
for(i=0;i< m; i++)
{
for(j=i+1; j < m; j++)
{
for(k=0; k < n;k++)
{
for(l=0;l< n;l++)
{if(condition) do something}
}
}
}
...
for i := 1 to n do
j := 2;
while j < i do
j := j^4;
I'm really confused when it comes to Big-O notation, so I'd like to know if it's O(n log n). That's my gut, but I can't prove it. I know the while loop is probably faster than log n, but I don't know by how much!
Edit: the caret denotes exponent.
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Possible Duplicate:
Regarding in-place merge in an array
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Regards,
Mithun
...