21–28 of 28 posts
It does make me chuckle the way this is reported. We answered this problem! Ok what's the answer "well about (0.143n)^n". I've figured out the value of Pi! It's about 3. Firstly, that's not an answer that's's an approximate answer, and secondly, it's wrong for a practical values of n. I really don't have the mentality of a mathematician.
It does make me chuckle the way this is reported. We answered this problem! Ok what's the answer "well about (0.143n)^n". I've figured out the value of Pi! It's about 3. Firstly, that's not an answer that's's an approximate answer, and secondly, it's wrong for a practical values of n. I really don't have the mentality of a mathematician.
I was very surprised to see a headline involving "Harvard mathematician" and "chess problem" and find that the article was not about Noam Elkies.
Related (N-Queens is NP-hard, 4 years ago): https://news.ycombinator.com/item?id=15168867
It does make me chuckle the way this is reported. We answered this problem! Ok what's the answer "well about (0.143n)^n". I've figured out the value of Pi! It's about 3. Firstly, that's not an answer that's's an approximate answer, and secondly, it's wrong for a practical values of n. I really don't have the mentality of a mathematician.
The problem of determining an efficient algorithm for every n, on the other hand, is just hopeless, which means that it is not interesting for the purpose I described above. Studying "small" n (up to 40, say) is not useless, though, if it leads to algorithmic breakthroughs (in SAT solving, for example). It is, however, an "orthogonal" problem, in the sense that it requires a completely different set of techniques.
To illustrate my point: If the answer for n = 50 appeared out of nowhere, would people care about the number itself? I bet people would be much more interested in how it was done.
It does make me chuckle the way this is reported. We answered this problem! Ok what's the answer "well about (0.143n)^n". I've figured out the value of Pi! It's about 3. Firstly, that's not an answer that's's an approximate answer, and secondly, it's wrong for a practical values of n. I really don't have the mentality of a mathematician.
I agree, you don't have the mentality of a mathematician.
Earlier quoted context omitted.
It's little o notation. It does not depend on n. For example x^2 is in O(x^2) but is not in o(x^2).
When used in that equation it's meant to implicitly depend on n as edflsafoiewq points out. o(1) stands for o(f(n)) where f(n) = 1. Hence any function g(x) in the family of functions represented by o(1) must be less than c * f(x) for every positive c and all x greater than some m. This is exactly the statement that any function in the family of functions o(1) must tend to zero. Whenever you see big-O/little-O/theta n…