Big O notation algorithms — OZONE Asylum, home of the Mad Scientists

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Your Text: Insert Slimies » Insert UBB Code » Close Last Tag \| All Tags UBB Help	1. You should use a graphic calculator to graph these two functions. You will find that there are 2 intersection points: n = 0.8 and n = 3.25. n < 0.8 nlog(2n) is faster n = 0.8 they are the same n > 0.8 and n < 3.25 (x^2)/4 is faster n = 3.25 they are the same n > 3.25 nlog(2n) is faster 2. This is a false statement. When n = 4, (n/4)log(n/2) = log(2) ~= 0.3 and (n/4)log(n) = log(4) ~= 0.6 This 0.3 <> 0.6 However, if you can make use of a rule which allows you eliminate constants. You will have: (n/4)log(n/2) = (n/4)(log(n) + log(2)) = (n/4)(log(n) - 0.3) Dan @ [url=http://www.codetown.org]Code Town[/url] Loading...
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