In this case they have c = 1-0.5(\log \log n)/(\log n) which is approaching 1 for large n.
58 karma · joined April 20, 2022
In this case they have c = 1-0.5(\log \log n)/(\log n) which is approaching 1 for large n.
If subexponential just means 2^{o(n)}, (as it is here, O(2^{n/\sqrt{\log n}})), then no. E.g. if the cost of the reduction took you from n to \sqrt{\log n}n, then you'd be back to exponential time. The Subexponential Reduction Family is the class of problems that preserve subexponential time in the second sense.
I am also a little worried about what it means for your training as a junior PhD. Often you would try and solve a problem your advisor thinks is doable that they assign to you as a learning exercise. It may be more and more difficult to find problems that a junior PhD can solve but that AI can not. Tim Gowers has written about that here: https://gowers.wordpress.com/2026/05/08/a-recent-experience-...
Do you really think that is the purpose of those departments?