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    <p>Remember that big-O notation does not include constant factors. The functions f(n) = n and g(n) = 10<sup>100</sup>n are both O(n), though f(n) is a much, much smaller function than g(n).</p> <p>Your analysis is correct - if you can make the space usage (log n) / 2 - 1, then you will (in the limit) be halving the amount of memory required. However, this will not show up in a big-O analysis, since big-O ignores the constant factors. As mentioned in some of the other answers, big-O notation captures long-term growth rates, and although constants might tell you more about the <em>absolute</em> amount of space used, constants don't control the long-term growth rate of the space usage.</p> <p>If you want to do a more precise analysis, you could give the exact memory usages before and after and then say that you have reduced the memory usage by 50%. Many papers on algorithms and data structures do actually include the constant factors and mention that they're getting a constant speedup. For example, the <a href="http://en.wikipedia.org/wiki/Cholesky_decomposition" rel="nofollow">Cholesky factorization</a> algorithm and Gaussian elimination both give O(n<sup>3</sup>) algorithms for solving linear systems, but when the Cholesky factorization can be used it does so with about 50% fewer operations. Most textbooks covering these topics will mention that although both algorithms are O(n<sup>3</sup>), the former is preferable to the latter when it's possible to use it.</p> <p>Hope this helps!</p>
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    1. POIs O(1/2 X log N) = O(logN)
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    1. POIs O(1/2 X log N) = O(logN)
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    1. COThanks for the answer. I just realized even at the modest N = 32k, M(N) = .45 which is good enough to brag about.
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