in reply to Re^8: Modified Binary Search
in thread Modified Binary Search

In big O notation O(logN - 1) and O(log N) are equivalent. They denote the same complexity order.

Though, that does not mean that the two algorithms are equally efficient. Actually they are not: Re^3: Modified Binary Search.

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Re^10: Modified Binary Search
by BrowserUk (Patriarch) on Jan 16, 2010 at 09:56 UTC

    I'm aware that the theorists will categorise them as having the same order of complexity, but when additional conditional checks are required, the complexity has increased.

    And at some point it is necessary to decide whether you need to find the lowest value greater or equal to the search term or the highest value less than or equal to the search term. And that adds to the (actual, real-world), complexity of the code.

    I know you know this--as your many Sort::* packages assert--in Perl, it is the number of source-level operations that is most relevant to efficiency: 

    @a = 1 .. 1e6; 
cmpthese -1, {
   a=>q[ my $total = sum @a; ],
   b=>q[ my $total = 0; $total += $_ for @a ],
};;
    Rate    b    a
b 10.8/s   -- -73%
a 40.9/s 277%   --

    [download]

    Identical algorithms but a significant difference in efficiency.

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