in reply to An informal introduction to O(N) notation
One thing that you didn't actualy give is a defintion for "order of growth", which would explain why O(1)==O(2), and why O(2N+3) == O(N).
I'd try to give a defintion, but I'd most likely get it wrong... I almost quoted Knuth here, but then I realized that I didn't understand what he meant. (For those that care, see pg 107 of vol 1, 3rd ed, of TAoCP.)
Update: Everything below this point. (I kept reading that section, and didn't want to "waste" another node.)
In purticular, note that O(n) gives an /upper/ bound on the order of growth of a purticular algo, and is by no means exact. There's also Ω(N), which gives the lowerbound.
Additionaly, saying that one algo is O(1), whereas another is O(2^N) does mean that, for a large enough dataset, the first algo will be faster then the second. It does /not/ mean that it will be for any reasonable dataset. For example, if the first algo takes 1 year to complete, but the second takes 2^N nanoseconds, it will take... well, a huge dataset before the first becomes faster.
Warning: Unless otherwise stated, code is untested. Do not use without understanding. Code is posted in the hopes it is useful, but without warranty. All copyrights are relinquished into the public domain unless otherwise stated. I am not an angel. I am capable of error, and err on a fairly regular basis. If I made a mistake, please let me know (such as by replying to this node).
Re: Re: An informal introduction to O(N) notation by dws (Chancellor) on Jan 18, 2003 at 05:15 UTC 
One thing that you didn't actualy give is a defintion for "order of growth", which would explain why O(1)==O(2), and why O(2N+3) == O(N).
From a casual perspective, "order of growth" can be thought of as the shape of the worstcase growth curve, regardless of the slope of that shape. Even it a line indicating linear (i.e., O(N)) growth has a high slope, eventuallywhen the data set is big enoughit will be overtaken by an O(N^{2}) curve. That's one of the pitfalls. People get seduced by the behavior of their code on small data sets, then get blindsided when they try a much larger data set and performance sucks.
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Re: An informal introduction to O(N) notation by AbigailII (Bishop) on Jan 18, 2003 at 12:37 UTC 
Actually, a year is less than 2^55 nanoseconds. And I wouldn't
call a set with 55 elements "huge". Beware of the power of
exponentation. Your computer needs to speed up with a factor
of 1000 to be able to increase your dataset with no more than
10 so that it will run in the same time....
Abigail  [reply] 

Sigh. That's what I get for not checking my arithmetic. You're right, of course. I stand corrected on this example, but if you change the example to be O(1) at 1 year vs. O(N^2) at 1ns*N^2, the size of the dataset for the second to become slower becomes a lot larger. (Specificly, around 178 million items.) Also, there's the consideration of O() notation being the worstcase senerio. For example, even though bubblesort is O(N^2), for nearlysorted input, it can be quit efficent.
Warning: Unless otherwise stated, code is untested. Do not use without understanding. Code is posted in the hopes it is useful, but without warranty. All copyrights are relinquished into the public domain unless otherwise stated. I am not an angel. I am capable of error, and err on a fairly regular basis. If I made a mistake, please let me know (such as by replying to this node).
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