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From: Jeffrey Carter <spam.jrcarter.not@spam.not.acm.org>
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Subject: Re: My Invention of "Bug Sort".
Date: Thu, 21 Jun 2012 11:45:21 -0700
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Date: 2012-06-21T11:45:21-07:00
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On 06/21/2012 04:50 AM, Austin Obyrne wrote:
>
> In general,
>
> A change of base in logarithms:
>
> Log(base(a)M = Log (base(b))M / Log (base(b))'a'.
>
> Applying that,In particular here,
>
> Ln of N = Log2N / Ln 2

No. log2 N = ln N / ln 2, so ln N = log2 N � ln 2. Not that it matters to Big-O 
notation.

> Unless it is true that there is some constant incorporated in O that verifies this, then
>
> N(Ln N) /=  N(Log2 N)
>
> Which is it?
>
> On face value it appears totally incorrect arithmetically, I'm surprised that such an anomaly would be acceptable.

Big-O notation is about the order of the complexity, not its exact value. If you 
have a super fast algorithm that only does 1 pass over the data, it takes time 
proportional to N. A slower algorithm that makes 2 passes over the data takes 
time proportional to 2N. Both have time complexity of O(N); a constant factor 
doesn't affect the Big-O value.

Another way to think about it is that Big-O indicates how the time required 
increases as N increases, not what the time is for a specific value of N.

-- 
Jeff Carter
"Death awaits you all, with nasty, big, pointy teeth!"
Monty Python & the Holy Grail
20

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