5 ms·
For some large n, integers in the algorithm may be so large that operations on them cease to be constant time.
by crashocaster 6y ago
For some large n, integers in the algorithm may be so large that operations on them cease to be constant time.
- d0mine 6y agothat is obvious. But how does it change big O? (why any manual implementation would have a better big O compared to existing arithmetics implementation in CPython?)
- kaba0 6y agoI think he meant that naively implementing an algorithm may not be bounded by the O notation he/she originally wanted due to code calling other functions “hidden” from the programmer.