Is O(1/2 X log N) = O(logN)

Question

Suppose I have an algorithm with memory requirement of logN+1 where N is the size of the problem (number of bits to process). I propose a 2nd version that reduces this memory requirement to (logN)/2+1. I have learnt that constants are ignored in Big-O analysis, so both the algorithm versions have complexity of O(logN).

Now, if I calculate the memory that I saved using the 2nd version of the algorithm, I get

Memory saved at N = M(N) = 1 - [(logN)/2+1]/[logN+1]
lim N→∞ M(N) = 1/2

which shows that asymptotically I will always save 50% of the memory. I am confused why I am unable to see this gain in Big-O analysis?

My second question is: If my understanding about Big-O notation is wrong, what is a proper way of highlighting the memory saved in 2nd version of the algorithm?

Answer 1

Remember that big-O notation does not include constant factors. The functions f(n) = n and g(n) = 10¹⁰⁰n are both O(n), though f(n) is a much, much smaller function than g(n).

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 absolute amount of space used, constants don't control the long-term growth rate of the space usage.

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 Cholesky factorization algorithm and Gaussian elimination both give O(n³) 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(n³), the former is preferable to the latter when it's possible to use it.

Hope this helps!